Class 9

Haryana State Board HBSE 9th Class Maths Solutions Chapter 11 रचनाएँ Ex 11.1 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 11 रचनाएँ Exercise 11.1

प्रश्न 1.
एक दी हुई किरण के प्रारंभिक बिंदु पर 90° के कोण की रचना कीजिए और कारण सहित रचना की पुष्टि कीजिए।
हल :

रचना के चरण-

1. एक किरण AB खींचिए।
2. A को केंद्र मानकर तथा किसी त्रिज्या की परकार खोलकर एक चाप लगाओ जो किरण AB को D पर प्रतिच्छेद करे।
3. परकार को इतना ही खुला रखकर D से चाप DE लगाओ। इसी प्रकार E से चाप EF लगाओ।
4. अब परकार कुछ अधिक खोलकर E तथा F से चापें लगाओ जो परस्पर K पर काटें।
5. अब K से गुजरती हुई एक किरण AC खींचिए।
6. ∠CAB, अभीष्ट कोण है जो 90° का है।

सत्यापन- रचना से EA = AD = ED
∴ AED एक समबाहु त्रिभुज है।
अतः ∠EAD = 60° …….(i)
इसी प्रकार, AE = EF = FA
⇒ AEF एक समबाहु त्रिभुज है।
∠EAF = 60°
क्योंकि AC, ∠EAF का समद्विभाजक है।
∴ ∠EAC = \(\frac{1}{2}\) × 60° = 30° …….(ii)
समीकरण (i) व (ii) से,
∠CAB = ∠EAC + ∠EAD
= 30° + 60° = 90°

प्रश्न 2.
एक दी हुई किरण के प्रारंभिक बिंदु पर 45° के कोण की रचना कीजिए और कारण सहित रचना की पुष्टि कीजिए।
हल :

रचना के चरण-

1. एक किरण AB खींचिए।
2. A को केंद्र मानकर तथा किसी त्रिज्या की परकार खोलकर एक चाप लगाओ जो किरण AB को D पर प्रतिच्छेद करे।
3. परकार को इतना ही खुला रखकर D से चाप DE लगाओ। इसी प्रकार E से चाप EF लगाओ। अब परकार कुछ अधिक खोलकर E तथा F से चापें लगाओ जो परस्पर K पर काटें।
4. अब K से गुजरती हुई एक किरण AL खींचो। ∠LAB का माप 90° होगा।
5. अब M तथा D से चापें लगाओ जो परस्पर N पर काटें।
6. अब N से गुजरती हुई एक किरण AC खींचिए।
7. ∠CAB, अभीष्ट कोण है जो 45° का है।

सत्यापन- रचना में ∠LAB = 90° है और AC, ∠LAB का समद्विभाजक है।
∠CAB = \(\frac{1}{2}\)∠LAB = \(\frac{1}{2}\) × 90° = 45°

प्रश्न 3.
निम्न मापों के कोणों की रचना कीजिए- [B.S:E.H. March, 2018]
(i) 30°
(ii) \(22 \frac{1^{\circ}}{2}\)
(iii) 15°
हल :
(i) रचना के चरण-

1. एक किरण AB खींचिए।
2. A को केंद्र मानकर तथा किसी त्रिज्या की परकार खोलकर एक चाप लगाओ जो किरण AB को D पर प्रतिच्छेद करे।
3. परकार को इतना ही खुला रखकर D से DE चाप लगाओ।
4. अब परकार कुछ अधिक खोलकर D व E से चापें लगाओ जो परस्पर F पर काटें।
5. F में से गुजरती हुई एक किरण AC खींचिए।
6. ∠CAB, अभीष्ट कोण है जो 30° का है।

(ii) रचना के चरण- [B.S.E.H. March, 2019]

1. एक किरण AB खींचिए।
2. A को केंद्र मानकर तथा किसी त्रिज्या की परकार खोलकर एक चाप लगाओ जो किरण AB को D पर प्रतिच्छेद करे।
3. परकार को इतना ही खुला रखकर D से DE चाप लगाओ।
4. अब परकार कुछ अधिक खोलकर D व E से चापें लगाओ जो परस्पर K पर काटें।
5. K से गुजरती हुई एक किरण AH खींचिए। इस प्रकार ∠HAB = 30° का होगा।
6. अब D व L से चापें लगाओ जो परस्पर M पर काटें।
7. M से गुजरती हुई किरण AF खींचिए। इस प्रकार ∠FAB = 15° का होगा।
8. अब N तथा L से चापें लगाओ जो परस्पर P पर काटें।
9. P में से गुजरती हुई किरण AC खींचिए।
10. ∠CAB, अभीष्ट कोण है जो \(22 \frac{1^{\circ}}{2}\) का है।

(iii) रचना के चरण-

1. एक किरण AB खींचिए।
2. A को केंद्र मानकर तथा किसी त्रिज्या की परकार खोलकर एक चाप लगाओ जो किरण AB को D पर प्रतिच्छेद करे।
3. परकार को इतना ही खुला रखकर D से DE चाप लगाओ।
4. अब परकार कुछ अधिक खोलकर D व E से चापें लगाओ जो परस्पर K पर काटें।
5. K से गुजरती हुई एक किरण AH खींचिए। इस प्रकार ∠HAB = 30° का होगा।
6. अब F व D से चापें लगाओ जो परस्पर L पर काटें।
7. L में से गुजरती हुई एक किरण AC खींचिए।
8. ∠CAB, अभीष्ट कोण है जो 15° का है।

प्रश्न 4.
निम्न कोणों की रचना कीजिए और चाँदे द्वारा मापकर पुष्टि कीजिए-
(i) 750
(ii) 105° [March, 2020]
(iii) 135°
हल :
(i) रचना के चरण-

1. एक किरण AB खींचिए।
2. A को केंद्र मानकर तथा किसी त्रिज्या की परकार खोलकर एक चाप लगाओ जो किरण AB को D पर प्रतिच्छेद करे।
3. परकार को इतना ही खुला रखकर D से DE चाप लगाओ। इसी प्रकार E से चाप EF लगाओ।
4. अब परकार कुछ अधिक खोलकर E व F से चापें लगाओ जो परस्पर K पर काटें।
5. K से गुजरती हुई एक किरण AH खींचिए।
6. बिंदु L व E से चापें लगाओ जो परस्पर N पर प्रतिच्छेद करें।
7. N से गुजरती हुई एक किरण AC खींचिए।
8. ∠CAB, अभीष्ट कोण है जो 75° का है।
9. ∠CAB को चाँदे से नापने पर पता चलता है कि यह 75° का है।

(ii) रचना के चरण-

1. एक किरण AB खींचिए।
2. A को केंद्र मानकर तथा किसी त्रिज्या की परकार खोलकर एक चाप लगाओ जो किरण AB को D पर प्रतिच्छेद करे।
3. परकार को इतना ही खुला रखकर D से DE चाप लगाओ। इसी प्रकार E से चाप EF लगाओ।
4. अब परकार कुछ अधिक खोलकर E व F से चापें लगाओ जो परस्पर G पर काटें।
5. G से गुजरती हुई एक किरण AH खींचिए।
6. बिंदु F तथा L से चापें लगाओ जो परस्पर M पर काटें।
7. M से गुजरती हुई किरण AC खींचिए।
8. ∠CAB, अभीष्ट कोण है जो 105° का है।
9. इस कोण को चाँदे से नापने पर पता चलता है कि यह 105° का है।

(iii) रचना के चरण-

1. एक किरण AB खींचिए।
2. A को केंद्र मानकर तथा किसी त्रिज्या की परकार खोलकर एक चाप लगाओ जो किरण AB को D पर प्रतिच्छेद करे।
3. परकार को इतना ही खुला रखकर D से DE चाप लगाओ। इसी प्रकार E से EF तथा F से FG चाप लगाओ।
4. अब परकार को पहले से अधिक खोलकर E व F से चापें लगाओ जो परस्पर K पर काटें।
5. K में से गुजरती हुई एक किरण AH खींचिए।
6. बिंदु L तथा G से चापें लगाओ जो परस्पर M पर काटें।
7. अब M से गुजरती हुई एक किरण AC खींचिए।
8. ∠CAB, अभीष्ट कोण है जो 135° का है।
9. ∠CAB को चाँदे से नापने पर पता चलता है कि यह 135° का है।

प्रश्न 5.
एक समबाहु त्रिभुज की रचना कीजिए, जब इसकी भुजा दी हो तथा कारण सहित रचना कीजिए।
हल :
रचना के चरण-

1. एक रेखाखंड BC = 5.5 cm खींचिए।
2. B और C को केंद्र मानकर BC = 5.5 cm त्रिज्या की परकार खोलकर दो चापें लगाएँ जो परस्पर A पर प्रतिच्छेद करें।
3. A को B व C से मिलाओ।
4. इस प्रकार ABC अभीष्ट समबाहु त्रिभुज है।

सत्यापन-क्योंकि रचना द्वारा AB = BC = CA = 5.5 cm
⇒ ∠C = ∠B = ∠A = 60°
∴ ΔABC समबाहु त्रिभुज है।

Haryana State Board HBSE 9th Class Maths Solutions Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.8 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Exercise 13.8

[नोट-जब तक अन्यथा न कहा जाए, π = \(\frac{22}{7}\) लीजिए।]

प्रश्न 1.
उस गोले का आयतन ज्ञात कीजिए जिसकी त्रिज्या निम्न है :
(i) 7 सें०मी०
(ii) 0.63 मी०
हल :
(i) यहां पर,
गोले की त्रिज्या (r) = 7 सें०मी०
गोले का आयतन (V) = \(\frac{4}{3}\)πr³
= \(\frac{4}{3} \times \frac{22}{7}\) 7 × 7 × 7 सें०मी०³
= \(\frac{4312}{3}\) सें०मी०³ = 1437\(\frac{1}{3}\)सें०मी०³ उत्तर

(ii) यहां पर,
गोले की त्रिज्या (r) = 0.63 मी० = \(\frac{63}{100}\) मी०
गोले का आयतन (V) = \(\frac{4}{3}\)πr³
= \(\frac{4}{3} \times \frac{22}{7} \times \frac{63}{100} \times \frac{63}{100} \times \frac{63}{100}\) मी०³
= \(\frac{1047816}{1000000}\) मी०³
= 1.05 मी०³ उत्तर

प्रश्न 2.
उस ठोस गोलाकार गेंद द्वारा हटाए गए (विस्थापित) पानी का आयतन ज्ञात कीजिए, जिसका व्यास निम्न है:
(i) 28 सें०मी०
(ii) 0.21 मी०
हल :
(i) यहां पर,
गोले का व्यास (d) = 28 सें०मी०
गोले की त्रिज्या (r) = \(\frac{28}{2}\) सें०मी० = 14 सें०मी०
गोले का आयतन (V) = \(\frac{4}{3}\)πr³
= \(\frac{4}{3} \times \frac{22}{7}\) × 14 × 14 × 14 सें०मी०³
\(\frac{34496}{3}\) सें०मी०³ = 11498\(\frac{2}{3}\) सें०मी०³ उत्तर
अतः गोले द्वारा हटाए गए पानी का आयतन = 11498\(\frac{2}{3}\) सें०मी०³ उत्तर

(ii) यहां पर,
गोले का व्यास (d) = 0.21 मी० = \(\frac{21}{100}\) मी०
गोले की त्रिज्या (r) = \(\frac{21}{2 \times 100}\) = \(\frac{21}{100}\)
∴ गोले का आयतन (V) = \(\frac{4}{3}\)πr³
= \(\frac{4}{3} \times \frac{22}{7} \times \frac{21}{200} \times \frac{21}{200} \times \frac{21}{200}\) सें०मी०³
= \(\frac{4851}{1000000}\) सें०मी०³ = 0.004851 मी०³
अतः गोले द्वारा हटाए गए पानी का आयतन = 0.004851 मी०³ उत्तर

प्रश्न 3.
धातु की एक गेंद का व्यास 4.2 सें०मी० है। यदि इस धातु का घनत्व 8.9 ग्राम प्रति सें०मी०³ है, तो इस गेंद का द्रव्यमान ज्ञात कीजिए।
हल :
यहां पर,
धातु की गेंद का व्यास (d) = 4.2 सें०मी० = \(\frac{21}{100}\) सें०मी०
धातु की गेंद की त्रिज्या (r) = \(\frac{21}{100}\) सेंमी० = \(\frac{21}{100}\) सें०मी०
धातु की गेंद का आयतन (V) = \(\frac{4}{3}\)πr³
= \(\frac{4}{3} \times \frac{22}{7} \times \frac{21}{10} \times \frac{21}{10} \times \frac{21}{10}\) सें०मी०³
= \(\frac{38808}{1000}\) सें०मी०³
1 सें०मी० धातु का द्रव्यमान = 8.9 ग्राम
दी गई धातु की गेंद का द्रव्यमान = \(\frac{38808}{1000} \times \frac{89}{10}\) ग्राम
\(\frac{3453912}{10000}\) ग्राम = 345.3912 ग्राम
= 345.39 ग्राम उत्तर

प्रश्न 4.
चंद्रमा का व्यास पृथ्वी के व्यास का लगभग एक-चौथाई है। चंद्रमा का आयतन पृथ्वी के आयतन की कौन-सी भिन्न है ?
हल :
माना पृथ्वी का व्यास (d) = 2x मी०
पृथ्वी की त्रिज्या (r₁) = \(\frac{2 x}{2}\)
पृथ्वी का आयतन (V) = \(\frac{4}{3}\)πr₁³
= \(\frac{4}{3}\)π × x × x × x मी०³
= \(\frac{4}{3}\)πx³
चंद्रमा का व्यास (d) = \(\frac{2 x}{4}\) मी०
चंद्रमा की त्रिच्या (r₂) = \(\frac{2 x}{2 \times 4}=\frac{x}{4}\) मी०
चंद्रमा का आयतन (V) = \(\frac{4}{3}\)πr₂³
= \(=\frac{4}{3} \pi \times \frac{x}{4} \times \frac{x}{4} \times \frac{x}{4}\) मी०³
= \(\frac{1}{48}\)πx³ मी०³
चंद्रमा का आयतन पृथ्वी के आयतन का जितना भाग है

प्रश्न 5.
व्यास 10.5 सें०मी० वाले एक अर्धगोलाकार कटोरे में कितने लीटर दूध आ सकता है ?
हल :
यहां पर,
अर्धगोलाकार कटोरे का व्यास (d) = 10.5 सें०मी० = \(\frac{105}{10}\) = \(\frac{21}{2}\) सें०मी०
अर्धगोलाकार कटोरे की त्रिज्या (r) = \(\frac{21}{2 \times 2}\) सें०मी० = \(\frac{21}{4}\) सें०मी०
∴ अर्धगोलाकार कटोरे का आयतन (V) = \(\frac{2}{3}\)πr³
= \(\frac{2}{3} \times \frac{22}{7} \times \frac{21}{4} \times \frac{21}{4} \times \frac{21}{4}\) सें०मी०³
= \(\frac{4851}{16}\) सें०मी०³
अतः अर्धगोलाकार कटोरे की धारिता = \(\frac{4851}{16 \times 1000} \ell\)
= \(\frac{4851}{16000} \ell\) = 0.303 l (लगभग) उत्तर

प्रश्न 6.
एक अर्धगोलाकार टंकी 1 सें०मी० मोटी एक लोहे की चादर (sheet) से बनी है। यदि इसकी आंतरिक त्रिज्या 1 मी० है, तो इस टंकी के बनाने में लगे लोहे का आयतन ज्ञात कीजिए।
हल :
यहां पर,
अर्धगोलाकार टंकी की अंतः त्रिज्या (r) = 1 मी० = 100 सें०मी०
अर्धगोलाकार टंकी की बाह्य त्रिज्या (R) = 100 + 1 = 101 सें०मी०
∴ टंकी में लगी लोहे की चादर का आयतन = बाह्य आयतन – आंतरिक आयतन

प्रश्न 7.
उस गोले का आयतन ज्ञात कीजिए जिसका पृष्ठीय क्षेत्रफल 154 सें०मी०² है।
हल :
यहां पर,
गोले का पृष्ठीय क्षेत्रफल = 154 सें०मी०²
⇒ 4πr² = 154
या 4 × \(\frac{22}{7}\) × r² = 154
या \(\frac{88}{7}\)r² = 154

प्रश्न 8.
किसी भवन का गुंबद एक अर्धगोले के आकार का है। अंदर से, इसमें सफेदी कराने में ₹ 4989.60 व्यय हुए। यदि सफेदी कराने की दर ₹ 20 प्रति वर्ग मीटर है, तो ज्ञात कीजिए :
(i) गुंबद का आंतरिक वक्र पृष्ठीय क्षेत्रफल,
(ii) गुंबद के अंदर की हवा का आयतन।
हल :
(i) गुंबद के अंतः तल पर सफेदी कराने का खर्च = ₹ 4989.60
सफेदी कराने की दर = ₹ 20 प्रति वर्ग मीटर
4989.60 इस प्रकार गुंबद के अंतः तल का क्षेत्रफल = \(\frac{4989.60}{2}\)
= 249.48 वर्ग मीटर उत्तर
(ii) 2πr² = 249.48

प्रश्न 9.
लोहे के 27 ठोस गोलों को पिघलाकर, जिनमें से प्रत्येक की त्रिज्या r है और पृष्ठीय क्षेत्रफल S है, एक बड़ा गोला बनाया जाता है जिसका पृष्ठीय क्षेत्रफल S’ है। ज्ञात कीजिए :
(1) नए गोले की त्रिज्या r’
(ii) S और S’ का अनुपात
हल :
(i) यहां पर,
गोले की त्रिज्या (r) = r
गोले का पृष्ठीय क्षेत्रफल (S) = 4πr²
गोले का आयतन (V) = \(\frac{4}{3}\)πr³
इस प्रकार 27 ठोस गोलों को पिघलाकर बने
गोले का आयतन (V) = 27 × \(\frac{4}{3}\)πr³ = 36πr³
नए गोले की त्रिज्या = r’
नए गोले का आयतन = \(\frac{4}{3}\)π(r’)³
प्रश्नानुसार \(\frac{4}{3}\)π(r’)³ = 36πr³
या (r’)³ = 36r³ × \(\frac{3}{4}\)
या (r’)³ = 27(r)³
या (r’)³ = (3r)³
⇒ r’ = 3r
नए गोले की त्रिज्या r’ = 3r उत्तर

(ii) यहां पर,
नए गोले का पृष्ठीय क्षेत्रफल (S’) = 4π(r’)²
= 4π(3r)²
= 4π(9r²)
S’ = 36πr²
S तथा S’ का अनुपात = 4πr² : 36πr²
= 1 : 9 उत्तर

प्रश्न 10.
दवाई का एक कैपसूल (capsule) 3.5 मि०मी० व्यास का एक गोला (गोली) है। इस कैपसूल को भरने के लिए कितनी दवाई (mm³ में) की आवश्यकता होगी ?
हल :
यहां पर,
कैपसूल का व्यास (d) = 3.5 मि०मी० = \(\frac{35}{10}=\frac{7}{2}\) मि०मी०
कैपसूल की त्रिज्या (r) = \(\frac{7}{2} \times \frac{1}{2}\) मि०मी० = \(\frac{7}{4}\) मि०मी०
कैपसूल का आयतन (V) = \(\frac{4}{3}\)πr³
= \(\frac{4}{3} \times \frac{22}{7} \times \frac{7}{4} \times \frac{7}{4} \times \frac{7}{4}\) मि०मी०
= \(\frac{539}{24}\) मि०मी०³ = 22.46 मि०मी०³ (लगभग)
इस प्रकार कैपसूल को भरने के लिए जितनी दवाई की आवश्यकता होगी।
= 22.46 मि०मी०³ उत्तर

Haryana State Board HBSE 9th Class Maths Solutions Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.5 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Exercise 13.5

प्रश्न 1.
माचिस की डिब्बी के माप 4 सें०मी० × 2.5 सेंमी० × 1.5 सें०मी० हैं। ऐसी 12 डिब्बियों के एक पैकेट का आयतन क्या होगा ?
हल :
यहां पर,
माचिस की डिब्बी की लंबाई (l) = 4 सें०मी०
माचिस की डिब्बी की चौड़ाई (b) = 2.5 सें०मी०
माचिस की डिब्बी की ऊंचाई (h) = 1.5 सें०मी०
माचिस की 1 डिब्बी का आयतन (V) = l × b × h = 4 × 2.5 × 1.5 सें०मी०³
= 15 सें०मी०³
अतः माचिस की 12 डिब्बियों का आयतन = 15 × 12 = 180 सें०मी०³ उत्तर

प्रश्न 2.
एक घनाभाकार पानी की टंकी 6 मी० लंबी, 5 मी० चौड़ी और 4.5 मी० गहरी है। इसमें कितने लीटर पानी आ सकता है ? (1 मी०³ = 1000 l)
हल :
यहां पर,
घनाभाकार टंकी की लंबाई (l) = 6 मी०
घनाभाकार टंकी की चौड़ाई (b) = 5 मी०
घनाभाकार टंकी की गहराई (h) = 4.5 मी०
घनाभाकार टंकी का आयतन (V) = l × b × h= 6 × 5 × 4.5 मी०³
= 135 मी०³
1 मी०³ टंकी में पानी आ सकता है = 1000 l
135 मी०³ टंकी में पानी आ सकता है = 135 × 1000
= 135000 लीटर उत्तर

प्रश्न 3.
एक घनाभाकार बर्तन 10 मी० लंबा और 8 मी० चौड़ा है। इसको कितना ऊंचा बनाया जाए कि इसमें 380 घन मीटर द्रव आ सके ?
हल :
यहां पर,
घनाभाकार बर्तन की लंबाई (l) = 10 मी०
घनाभाकार बर्तन की चौड़ाई (b) = 8 मी०
घनाभाकार बर्तन की ऊंचाई (h) = ?
घनाभाकार बर्तन का आयतन (V) = 380 मी०³
⇒ l × b × h = 380
या 10 × 8 × h = 380
या 80h = 380
या h = \(\frac{380}{80}\) = 4.75 मी०
अतः घनाभाकार बर्तन की ऊंचाई (h) = 4.75 मी० उत्तर

प्रश्न 4.
8 मी० लंबा, 6 मी० चौड़ा और 3 मी० गहरा एक घनाभाकार गड्ढा खुदवाने में ₹ 30 प्रति मी०³ की दर से होने वाला व्यय ज्ञात कीजिए।
हल :
यहां पर,
घनाभाकार गड्ढे की लंबाई (l) = 8 मी०
घनाभाकार गड्ढे की चौड़ाई (b) = 6 मी०
घनाभाकार गड्ढे की गहराई (h) = 3 मी०
घनाभाकार गड्ढे का आयतन (V) = l × b × h = 8 × 6 × 3 मी०³
= 144 मी०³
अतः घनाभाकार गड्ढे को खुदवाने का व्यय = 144 × 30 = ₹ 4320 उत्तर

प्रश्न 5.
एक घनाभाकार टंकी की धारिता 50000 लीटर पानी की है। यदि इस टंकी की लंबाई और गहराई क्रमशः 2.5 मी० और 10 मी० हैं, तो इसकी चौड़ाई ज्ञात कीजिए।
हल :
यहां पर,
घनाभाकार टंकी का आयतन (V) = 50000 लीटर
= \(\frac{50000}{1000}\) मी०³
= 50 मी०³
घनाभाकार टंकी की लंबाई (l) = 2.5 मी०
घनाभाकार टंकी की चौड़ाई (b) = ?
घनाभाकार टंकी की गहराई (h) = 10 मी०
घनाभाकार टंकी का आयतन (V) = 50 मी०³
⇒ l × b × h = 50
या 2.5 × b × 10 = 50
या 25 b = 50
या b = \(\frac{50}{25}\) = 2 मी०
अतः घनाभाकार टंकी की चौड़ाई = 2 मी० उत्तर

प्रश्न 6.
एक गांव जिसकी जनसंख्या 4000 है, को प्रतिदिन प्रति व्यक्ति 150 लीटर पानी की आवश्यकता है। इस गांव में 20 मी० × 15 मी० × 6 मी० मापों वाली एक टंकी बनी हुई है। इस टंकी का पानी वहां कितने दिन के लिए पर्याप्त होगा ?
हल :
यहां पर, एक व्यक्ति को प्रतिदिन जितने पानी की आवश्यकता है = 150 लीटर
4000 व्यक्तियों को प्रतिदिन जितने पानी की आवश्यकता है = 4000 × 50 = 6,00,000 लीटर
दी गई टंकी की लंबाई (l) = 20 मी०
दी गई टंकी की चौड़ाई (b) = 15 मी०
दी गई टंकी की गहराई (h) = 6 मी०
दी गई टंकी में उपस्थित पानी का आयतन (V) = l × b × h = 20 × 15 × 6 = 1800 मी०³
= 1800 × 1000 लीटर
= 18,00,000 लीटर
अतः टंकी का पानी जितने दिन के लिए पर्याप्त है = \(\frac{1800000}{600000}\) = 3 दिन उत्तर

प्रश्न 7.
किसी गोदाम की माप 40 मी० × 25 मी० × 15 मी० हैं। इस गोदाम में 1.5 मी० × 1.25 मी० × 0.5 मी० की माप वाले लकड़ी के कितने अधिकतम क्रेट (crate) रखे जा सकते हैं ? [B.S.E.H. March, 2018]
हल :
यहां पर,
गोदाम का आयतन (V₁) = 40 × 25 × 15 = 15000 मी०³
1 क्रेट का आयतन (V₂) = 1.5 × 1.25 × 0.5 = 0.9375 मी०³

प्रश्न 8.
12 सें०मी० भुजा वाले एक ठोस धन को बराबर आयतन वाले 8 घनों में काटा जाता है। नए घन की क्या भुजा होगी? साथ ही, इन दोनों घनों के पृष्ठीय क्षेत्रफलों का अनुपात भी ज्ञात कीजिए।
हल :
यहां पर,
ठोस धन की भुजा = 12 सें०मी०
ठोस घन का आयतन = 12 × 12 × 12 = 1728 सें०मी०³
नए घन का आयतन = \(\frac{1728}{8}\) = 216 सें०मी०³
अतः नए घन की भुजा = \(\sqrt[3]{\text { आयतन }}=\sqrt[3]{216}=\sqrt[3]{6 \times 6 \times 6}\)
= 6 सें०मी० उत्तर
पहले घन का पृष्ठीय क्षेत्रफल = 6 × भुजा
= 6 × (12)² सें०मी०²
= 6 × 144 = 864 सें०मी०²
नए घन का पृष्ठीय क्षेत्रफल = 6 × भुजा²
= 6 × (6)2 सें०मी०²
= 6 × 36 = 216 सें०मी०²
दोनों घनों के पृष्ठीय क्षेत्रफलों का अनुपात = 864 : 216
= 4 : 1 उत्तर

प्रश्न 9.
3 मी० गहरी और 40 मी० चौड़ी एक नदी 2 km प्रति घंटा की चाल से बह कर समुद्र में गिरती है। एक मिनट में समुद्र में कितना पानी गिरेगा ?
हल :
यहां पर,
नदी की गहराई (h) = 3 मी०
नदी की चौड़ाई (b) = 40 मी०
1 घंटे में बहने वाले पानी की लंबाई (l) = 2 km = 2000 मी०
अतः 1 घंटे में बहने वाले पानी का आयतन (V) = l × b × h = 2000 × 40 × 3 मी०³
= 24,0000 मी०³
1 मिनट में बहने वाले पानी का आयतन = \(\frac{240000}{60}\) मी०³
= 4000 मी०³
अतः 1 मिनट में जितना पानी समुद्र में गिरेगा = 4000 मी०³ उत्तर

Haryana State Board HBSE 9th Class Maths Solutions Chapter 2 बहुपद Ex 2.1 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 2 बहुपद Exercise 2.1

प्रश्न 1.
निम्नलिखित व्यंजकों में कौन-कौन एक चर में बहुपद हैं और कौन-कौन नहीं हैं ? कारण के साथ अपने उत्तर दीजिए
(i) 4x² – 3x + 7
(ii) y² + \(\sqrt{2}\)
(iii) 3\(\sqrt{t}\) + t\(\sqrt{2}\)
(iv) y + \(\frac {2}{y}\)
(v) x¹⁰ + y³ + t⁵⁰
हल :
(i) व्यंजक 4x² – 3x + 7 एक चर में बहुपद है क्योंकि चर का प्रत्येक घातांक पूर्ण संख्या है।
(ii) व्यंजक y² + \(\sqrt{2}\) एक चर में बहुपद हैं क्योंकि चर का प्रत्येक घातांक पूर्ण संख्या है।
(iii) व्यंजक 3\(\sqrt{t}\) + t\(\sqrt{2}\) बहुपद नहीं है क्योंकि चर का प्रत्येक घातांक पूर्ण संख्या नहीं है।
(iv) व्यंजक y +  \(\frac {2}{y}\) = y + 2y^-1 बहुपद नहीं है क्योंकि चर का घातांक ऋणात्मक संख्या है अर्थात पूर्ण संख्या नहीं है।
(v) व्यंजक x¹⁰ + y³ + t⁵⁰ तीन चरों में एक बहुपद है क्योंकि चरों का प्रत्येक घातांक पूर्ण संख्या है।

प्रश्न 2.
निम्नलिखित में से प्रत्येक में का गुणांक लिखिए
(i) 2 + x² + x
(ii) 2 – x² + x³
(iii) \(\frac {π}{2}\)x² + x
(iv) \(\sqrt{m}\) – 1
हल :
2 + x² + x में x² का गुणांक = 1 उत्तर
(ii) 2 – x² + x³ में x² का गुणांक = – 1 उत्तर
(iii) \(\frac {π}{2}\)x² + x में x² का गुणांक = \(\frac {π}{2}\) उत्तर
(iv) \(\sqrt{2}\)x – 1 में x² का गुणांक = 0 उत्तर

प्रश्न 3.
35 घात के द्विपद का और 100 घात के एकपदी का एक-एक उदाहरण दीजिए।
हल :
35 घात का द्विपद = 3x³⁵ – 4 उत्तर
100 घात का एकपद = \(\sqrt{2}\)y¹⁰⁰ उत्तर

प्रश्न 4.
निम्नलिखित बहुपदों में से प्रत्येक बहुपद की घात लिखिए
(i) 5x³ + 4x² + 7x
(ii) 4 – y²
(iii) 5t – \(\sqrt{7}\)
(iv) 3
हल :
(i) 5x³ + 4x² + 7x में बहुपद की घात = 3 उत्तर
(ii) 4 – y² = – y² + 4 में बहुपद की घात = 2 उत्तर
(iii) 5t – \(\sqrt{7}\) में बहुपद की घात = 1 उत्तर
(iv) 3 में बहुपद की घात = शून्य उत्तर

प्रश्न 5.
बताइए कि निम्नलिखित बहुपदों में कौन-कौन बहुपद रैखिक हैं, कौन-कौन द्विघाती हैं और कौन-कौन त्रिघाती
(i) x² + x
(ii) x – x³
(iii) y + y² + 4
(iv) 1 + x
(v) 3t
(vi) r²
(vii) 7x³
हल :
(i) बहुपद x² + x एक द्विघाती बहुपद है। उत्तर
(ii) बहुपद x – x³ एक त्रिघाती बहुपद है। उत्तर
(iii) बहुपद y + y² + 4 एक द्विघाती बहुपद है। उत्तर
(iv) बहुपद 1 + x एक रैखिक बहुपद है। उत्तर
(v) बहुपद 3t एक रैखिक बहुपद है। उत्तर
(vi) बहुपद r² एक द्विघाती बहुपद है। उत्तर
(vii) बहुपद 7x³ एक त्रिघाती बहुपद है। उत्तर

Haryana State Board HBSE 9th Class Maths Solutions Chapter 1 संख्या पद्धति Ex 1.2 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 1 संख्या पद्धति Exercise 1.2

प्रश्न 1.
नीचे दिए गए कथन सत्य हैं या असत्य हैं? कारण के साथ अपने उत्तर दीजिए।
(i) प्रत्येक अपरिमेय संख्या एक वास्तविक संख्या होती है।
(ii) संख्या रेखा का प्रत्येक बिंदु \(\sqrt{m}\) के रूप का होता है, जहाँ m एक प्राकृत संख्या है।
(iii) प्रत्येक बास्तविक संख्या एक अपरिमेय संख्या होती है।
हल :
(i) सत्य है, क्योंकि वास्तविक संख्याओं का संग्रह परिमेय और अपरिमेय संख्याओं से बना होता है। अतः प्रत्येक अपरिमेय संख्या एक वास्तविक संख्या होती है।
(ii) असत्य है, क्योंकि कोई भी ऋण संख्या किसी प्राकृत संख्या का वर्गमूल नहीं हो सकती।
(iii) असत्य है, क्योंकि 2 वास्तविक संख्या है परंतु अपरिमेय नहीं है।

प्रश्न 2.
क्या सभी धनात्मक पूर्णाकों के वर्गमूल अपरिमेय होते हैं? यदि नहीं, तो एक ऐसी संख्या के वर्गमूल का उदाहरण दीजिए जो एक परिमेय संख्या है।
हल :
नहीं, सभी धनात्मक पूर्णांकों के वर्गमूल अपरिमेय नहीं होते हैं। उदाहरण के लिए, \(\sqrt{4}\) = 2 एक परिमेय संख्या है।

प्रश्न 3.
दिखाइए कि संख्या रेखा पर \(\sqrt{5}\) को किस प्रकार निरूपित किया जा सकता है?
हल :

\(\sqrt{5}=\sqrt{4+1}=\sqrt{(2)^2+(1)^2}\)
एक संख्या रेखा l खींचो तथा 0 को बिंदु O एवं 2 को बिंदु A मानों। अब OA पर लंब AB = 1 इकाई खींचो तथा OB को मिलाओ। परकार की सहायता से केंद्र O तथा त्रिज्या OB से एक चाप खींचो, जो रेखा को बिंदु P पर काटती है। इस प्रकार बिंदु P ही अभीष्ट बिंदु है अर्थात OP = \(\sqrt{5}\) को प्रदर्शित करता है।

प्रश्न 4.
कक्षा के लिए क्रियाकलाप (वर्गमूल सर्पिल की रचना) : कागज की एक बड़ी शीट लीजिए और नीचे दी गई विधि से “वर्गमूल सर्पिल” (square root spiral) की रचना कीजिए। सबसे पहले एक बिंदु O लीजिए और एकक लंबाई का रेखाखंड (line segment) OP₁ खींचिए। एकक लंबाई वाले OP₁ पर लंब रेखाखंड P₁P₂ खींचिए (देखिए आकृति)। अब OP₂ पर लंब रेखाखंड P₂P₃ खींचिए। तब OP₃ पर लंब रेखाखंड P₃P₄ खींचिए।

इस प्रक्रिया को जारी रखते हुए OP_{n – 1} पर एकक लंबाई वाला लंब रेखाखंड खींचकर आप रेखाखंड P_{n – 1}P_n प्राप्त कर सकते हैं। इस प्रकार आप बिंदु O, P₁, P₂, P₃, ……………, P_n…………………..प्राप्त कर लेंगे और उन्हें मिलाकर \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{4}\),….. को दर्शाने वाला एक सुंदर सर्पिल प्राप्त कर लेंगे।

हल :
इस क्रियाकलाप को विद्यार्थी स्वयं करें।

Haryana State Board HBSE 9th Class Maths Solutions Chapter 14 सांख्यिकी Ex 14.2 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 14 सांख्यिकी Exercise 14.2

प्रश्न 1.
आठवीं कक्षा के 30 विद्यार्थियों के रक्त समूह ये हैं :
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O
इन आंकड़ों को एक बारंबारता बंटन सारणी के रूप में प्रस्तुत कीजिए। बताइए कि इन विद्यार्थियों में कौन-सा रक्त समूह अधिक सामान्य है और कौन-सा रक्त समूह विरलतम रक्त समूह है।
हल :
आठवीं कक्षा के 30 विद्यार्थियों के रक्त समूह की बारंबारता सारणी निम्न होगी-

विद्यार्थियों में सामान्य रक्त समूह = O
विद्यार्थियों में विरलतम रक्त समूह = AB उत्तर

प्रश्न 2.
40 इंजीनियरों की उनके आवास से कार्य-स्थल की (किलोमीटर में) दूरियां ये हैं :

0-5 को (जिसमें 5 सम्मिलित नहीं है) पहला अंतराल लेकर ऊपर दिए हुए आंकड़ों से वर्ग-माप 5 वाली एक वर्गीकृत बारंबारता बंटन सारणी बनाइए। इस सारणी बद्ध निरूपण में आपको कौन-से मुख्य लक्षण देखने को मिलते हैं?
हल :
आंकड़ों की वर्गीकृत बारंबारता सारणी अग्रांकित है-

मुख्य अभिलक्षण :
(1) 20 कि०मी० से कम दूरी तय करने वाले अभियन्ता = 5 + 11 + 11 + 9 = 36
(2) 20 कि०मी० से अधिक दूरी तय करने वाले अभियन्ता = 1 + 1 + 2 = 4

प्रश्न 3.
30 दिन वाले महीने में एक नगर की सापेक्ष आर्द्रता (% में) यह रही है :

(i) वर्ग 84-86, 86-88 आदि लेकर एक वर्गीकृत बारंबारता बंटन बनाइए।
(ii) क्या आप बता सकते हैं कि ये आंकड़े किस महीने या ऋतु से संबंधित हैं?
(iii) इन आंकड़ों का परिसर क्या है?
हल :
दिए गए आंकड़ों को दी गई वर्ग बारंबारता के अनुसार वर्गीकृत करने पर प्राप्त बारंबारता बंटन निम्न होगा-
(i)

(ii) ये आंकड़े वर्षा ऋतु में लिए हुए प्रतीत होते हैं क्योंकि सापेक्ष आर्द्रता बहुत अधिक है।
(iii) आंकड़ों का परिसर = 99.2 – 84.9 = 14.3 उत्तर

प्रश्न 4.
निकटतम सेंटीमीटरों में मापी गई 50 विद्यार्थियों की लंबाइयां ये हैं :

(i) 160-165, 165-170 आदि का वर्ग अंतराल लेकर ऊपर दिए गए आंकड़ों को एक वर्गीकृत बारंबारता बंटन सारणी के रूप में निरूपित कीजिए।
(ii) इस सारणी की सहायता से आप विद्यार्थियों की लंबाइयों के संबंध में क्या निष्कर्ष निकाल सकते हैं?
हल :
दिए गए आंकड़ों को दिए गए वर्ग अंतरालों के अनुसार बांटने पर निम्न बारंबारता बंटन सारणी प्राप्त होगी-
(i)

(ii) इन आंकड़ों से यह निष्कर्ष निकलता है कि 35 बच्चे ऐसे हैं जिनकी लंबाई 165 सें०मी० से कम है। अर्थात 70% विद्यार्थियों की लंबाई 165 सें०मी० से कम है।

प्रश्न 5.
एक नगर में वायु में सल्फर डाइ-ऑक्साइड का सांदण भाग प्रति मिलियन [parts per million (ppm)] में ज्ञात करने के लिए एक अध्ययन किया गया। 30 दिनों के प्राप्त किए गए आंकड़े ये हैं :

(i) 0.00-0.04, 0.04-0.08 आदि का वर्ग अंतराल लेकर इन आंकड़ों की एक वर्गीकृत बारंबारता बंटन सारणी बनाइए।
(ii) सल्फर डाइ-ऑक्साइड की सांद्रता कितने दिन 0.11 भाग प्रति मिलियन से अधिक रही?
हल :
दिए गए वर्ग अंतरालों के अनुसार दिए गए आंकड़ों की बारंबारता बंटन सारणी निम्न होगी-
(i)

(ii) जितने दिन सल्फर डाइ-ऑक्साइड की सांद्रता 0.11 भाग प्रति मिलियन से अधिक रही = 2 + 4 + 2 = 8 दिन

प्रश्न 6.
तीन सिक्कों को एक साथ 30 बार उछाला गया। प्रत्येक बार चित (Head) आने की संख्या निम्न है :

ऊपर दिए गए आंकड़ों के लिए एक बारंबारता बंटन सारणी बनाइए।
हल :
दिए गए आंकड़ों की बारंबारता बंटन सारणी निम्न है-

प्रश्न 7.
50 दशमलव स्थान तक शुद्ध π का मान नीचे दिया गया है : [B.S.E.H. March, 2019]
3.14159265358979323846264338327950288419716939937510
(i) दशमलव बिंद्र के बाद आने वाले 0 से 9 तक के अंकों का एक बारंबारता बंटन बनाइए।
(ii) सबसे अधिक बार और सबसे कम बार आने वाले अंक कौन-कौन से हैं?
हल :
दिए गए π के मान में 0 से 9 तक के अंकों की बारंबारता बंटन सारणी निम्न होगी-
(i)

(ii) सबसे अधिक बार आने वाले अंक = 3 व 9
सबसे कम बार आने वाले अंक = 0.

प्रश्न 8.
तीस बच्चों से यह पूछा गया कि पिछले सप्ताह उन्होंने कितने घंटों तक टी.वी. के प्रोग्राम देखे। प्राप्त परिणाम ये रहे हैं-

(i) वर्ग-चौड़ाई 5 लेकर और एक वर्ग अंतराल को 5-10 लेकर इन आंकड़ों की एक वर्गीकृत बारंबारता बंटन सारणी बनाइए।
(ii) कितने बच्चों ने सप्ताह में 15 या अधिक घंटों तक टेलीविजन देखा?
हल :
(i) दिए गए वर्ग-अंतरालों के अनुसार दिए गए आंकड़ों की बारंबारता सारणी अग्रांकित होगी-

(ii) जितने बच्चों ने सप्ताह में 15 घंटे या 15 घंटे से अधिक टेलीविजन देखा = 2

प्रश्न 9.
एक कंपनी एक विशेष प्रकार की कार-बैट्री बनाती है। इस प्रकार की 40 बैट्रियों के जीवन-काल (वर्षों में) ये रहे हैं :

0.5 माप के वर्ग-अंतराल लेकर तथा अंतराल 2-2.5 से प्रारंभ करके इन आंकड़ों की एक वर्गीकृत बारंबारता बंटन सारणी बनाइए।
हल :
दिए गए वर्ग-अंतरालों के अनुसार दिए गए आंकड़ों की बारंबारता बंटन सारणी निम्न होगी-

Haryana State Board HBSE 9th Class Maths Solutions Chapter 14 सांख्यिकी Ex 14.1 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 14 सांख्यिकी Exercise 14.1

प्रश्न 1.
उन आंकड़ों के पांच उदाहरण दीजिए जिन्हें आप अपने दैनिक जीवन से एकत्रित कर सकते हैं।
हल :
हमारे दैनिक जीवन में एकत्रित किए जाने वाले आंकड़ों के पांच उदाहरण निम्नलिखित हैं-

• अपनी कक्षा में छात्रों की संख्या।
• अपने विद्यालय में पंखों की संख्या।
• पिछले दो वर्षों के घर की बिजली का बिल।
• टेलीविजन या समाचार पत्रों से प्राप्त चुनाव के परिणाम।
• शैक्षिक सर्वेक्षण से प्राप्त साक्षरता दर के आंकड़े।

प्रश्न 2.
ऊपर दिए गए प्रश्न 1 के आंकड़ों को प्राथमिक आंकड़ों या गौण आंकड़ों में वर्गीकृत कीजिए।
हल :
1. प्राथमिक आंकड़े : अन्वेषक द्वारा स्वयं एकत्रित किए गए आंकड़े प्राथमिक आंकड़े (Primary Data) कहलाते हैं। इसलिए प्रश्न (1) में (i), (ii) व (iii) प्राथमिक आंकड़ें हैं।
2. गौण आंकड़े : किसी अन्य स्रोत से प्राप्त किए गए आंकड़े गौण आंकड़े (Secondary data) कहलाते हैं। इसलिए प्रश्न (1) में (iv) व (v) गौण आंकड़े हैं।

Haryana State Board HBSE 9th Class Maths Solutions Chapter 15 प्रायिकता Ex 15.1 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 15 प्रायिकता Exercise 15.1

प्रश्न 1.
एक क्रिकेट मैच में, एक महिला बल्लेबाज खेली गई 30 गेदों में 6 बार चौका मारती है। चौका न मारे जाने की प्रायिकता ज्ञात कीजिए।
हल :
यहां पर,
कल संभाव्य परिणामों की संख्या = 30
कुल अनुकूल परिणामों की संख्या = 6
इसलिए चौका मारने की प्रायिकता = P (E) = \(\frac{6}{30}\) = \(\frac{1}{5}\)
P(चौका न मारा जाना) = 1 – \(\frac{1}{5}\)
= \(\frac{5-1}{5}=\frac{4}{5}\) उत्तर

प्रश्न 2.
2 बच्चों वाले 1500 परिवारों का यदृच्छया चयन किया गया है और निम्नलिखित आंकड़े लिख लिए गए हैं :

यदृच्छया चुने गए उस परिवार की प्रायिकता ज्ञात कीजिए, जिसमें
(i) दो लड़कियां हों
(ii) एक लड़की हो
(iii) कोई लड़की न हो
साथ ही, यह भी जांच कीजिए कि इन प्रायिकताओं का योगफल 1 है या नहीं।
हल :
(i) यहां पर,
कुल संभाव्य परिणामों (कुल परिवारों) की संख्या = 475 + 814 + 211 = 1500
दो लड़कियों वाले परिवारों (अनुकूल परिणामों) की संख्या = 475
P (दो लड़कियों वाले परिवार) = \(\frac{475}{1500}=\frac{19}{60}\) उत्तर

(ii) यहां पर,
कुल संभाव्य परिणामों (कुल परिवारों) की संख्या = 475 + 814 + 211 = 1500
1 लड़की वाले परिवारों (अनुकूल परिणामों) की संख्या = 814
P (एक लड़की वाले परिवार) = \(\frac{814}{1500}=\frac{407}{750}\) उत्तर

(iii) यहां पर,
कुल संभाव्य परिणामों (कुल परिवारों) की संख्या = 475 + 814 + 211 = 1500
0 लड़की वाले परिवारों (अनुकूल परिणामों) की संख्या = 211
P (कोई लड़की न हो) = \(\frac{211}{1500}\) उत्तर

प्रश्न 3.
अध्याय 14 के अनुच्छेद 14.4 का उदाहरण 5 लीजिए। कक्षा के किसी एक विद्यार्थी का जन्म अगस्त में होने की प्रायिकता ज्ञात कीजिए।
हल :
प्रश्नानुसार,
वर्ष में जन्में कुल विद्यार्थी (संभाव्य परिणामों की संख्या) = 40
अगस्त मास में जन्में विद्यार्थी (अनुकूल परिणामों की संख्या) = 6
∴ P (अगस्त में जन्म लेने वाला विद्यार्थी) = \(\frac{6}{40}=\frac{3}{20}\) उत्तर

प्रश्न 4.
तीन सिक्कों को एक साथ 200 बार उछाला गया है तथा इनमें विभिन्न परिणामों की बारंबारताएं ये हैं :

यदि तीनों सिक्कों को पुनः एक साथ उछाला जाए, तो दो चित के आने की प्रायिकता ज्ञात कीजिए। [B.S.E.H. March, 2017, 2020]
हल :
यहां पर,
कुल संभाव्य परिणामों की संख्या = 200 (23 + 72 + 77 + 28)
दो चित आने के अनुकूल परिणामों की संख्या = 72
∴ P (दो चित आएं) = \(\frac{72}{200}=\frac{9}{25}\) उत्तर

प्रश्न 5.
एक कंपनी ने यदृच्छया 2400 परिवार चुनकर एक घर की आय स्तर और वाहनों की संख्या के बीच संबंध स्थापित करने के लिए उनका सर्वेक्षण किया। एकत्रित किए गए आंकड़े नीचे सारणी में दिए गए हैं :

मान लीजिए एक परिवार चुना गया है। प्रायिकता ज्ञात कीजिए कि चुने गए परिवार
(i) की आय ₹ 10000 – 13000 प्रति माह है और उसके पास ठीक-ठीक दो वाहन हैं।
(ii) की आय प्रति माह ₹ 16000 या इससे अधिक है और उसके पास ठीक 1 वाहन है।
(iii) की आय ₹ 7000 प्रति माह से कम है और उसके पास कोई वाहन नहीं है।
(iv) की आय ₹ 13000 – 16000 प्रति माह है और उसके पास 2 से अधिक वाहन हैं।
(v) जिसके पास 1 से अधिक वाहन नहीं है।
हल :
कुल संभाव्य परिणामों की संख्या = 2400
(i) ₹ 10000 – 13000 प्रति माह कमाने वाले परिवारों में 2 वाहनों वाले परिवारों (अनुकूल परिणामों) की संख्या = 29
∴ P (₹ 10000 – 13000 प्रति माह कमाने वाले 2 वाहनों वाले परिवार) = \(\frac{29}{2400}\) उत्तर

(ii) ₹ 16000 या इससे अधिक प्रति माह कमाने वाले परिवारों में 1 वाहन वाले परिवारों (अनुकूल परिणामों) की संख्या = 579
∴ P (₹ 16000 या इससे अधिक प्रतिमाह कमाने वाले 1 वाहन वाले परिवार) = \(\frac{579}{2400}\) उत्तर

(iii) ₹ 7000 प्रति माह से कम कमाने वाले परिवारों में कोई वाहन न होने वाले परिवारों (अनुकूल परिणाम) की संख्या = 10
∴ P (₹ 7000 प्रति माह से कम कमाने वाले शून्य वाहन वाले परिवार) = \(\frac{10}{2400}=\frac{1}{240}\)

(iv) ₹ 13000 – 16000 प्रति माह कमाने वाले परिवारों में 2 से अधिक वाहनों वाले परिवारों (अनुकूल परिणाम) की संख्या = 25
∴ P(₹ 13000 – 16000 प्रति माह कमाने वाले 2 से अधिक वाहन वाले परिवार) = \(\frac{25}{2400}=\frac{1}{96}\)

(v) 1 से अधिक वाहन न रखने वाले परिवारों की संख्या
(अनुकूल परिणाम) = 10 + 0 + 1 + 2 + 1 + 160 + 305 + 535 + 469 + 579 = 2062
∴ P (1 से अधिक वाहन न रखने वाले परिवार) = \(\frac{2062}{2400}=\frac{1031}{1200}\) उत्तर

प्रश्न 6.
अध्याय 14 की सारणी 14.7 लीजिए।
(i) गणित की परीक्षा में एक विद्यार्थी द्वारा 20 से कम अंक प्राप्त करने की प्रायिकता ज्ञात कीजिए।
(ii) एक विद्यार्थी द्वारा 60 या इससे अधिक अंक प्राप्त करने की प्रायिकता ज्ञात कीजिए।

हल :
(i) यहां पर,
कुल विद्यार्थियों की संख्या (संभाव्य परिणाम) = 90
20 से कम अंक प्राप्त करने वाले विद्यार्थी (अनुकूल परिणाम) = 7
∴ P (20 से कम अंक प्राप्त करने वाले विद्यार्थी) = \(\frac{7}{90}\) उत्तर

(ii) 60 या उससे अधिक अंक प्राप्त करने वाले विद्यार्थी
(अनुकूल परिणाम) = 15 + 8 = 23
∴ P (60 या उससे अधिक अंक प्राप्त करने वाले विद्यार्थी) = \(\frac{23}{90}\) उत्तर

प्रश्न 7.
सांख्यिकी के बारे में विद्यार्थियों का मत जानने के लिए 200 विद्यार्थियों का सर्वेक्षण किया गया। प्राप्त आंकड़ों को नीचे दी गई सारणी में लिख लिया गया है : [B.S.E.H. March, 2019]

प्रायिकता ज्ञात कीजिए कि यदृच्छया चुना गया विद्यार्थी
(i) सांख्यिकी पसंद करता है
(ii) सांख्यिकी पसंद नहीं करता है।
हल :
यहां पर,
कुल विद्यार्थियों की संख्या (संभाव्य परिणाम) = 135 + 65 = 200
(i) सांख्यिकी पसंद करने वाले विद्यार्थी (अनुकूल परिणाम) = 135
∴ P (सांख्यिकी पसंद करने वाले विद्यार्थी) = \(\frac{135}{200}=\frac{27}{40}\) उत्तर

(ii) सांख्यिकी पसंद न करने वाले विद्यार्थी (अनुकूल परिणाम) = 65
∴ P (सांख्यिकी पसंद न करने वाले विद्यार्थी) = \(\frac{65}{200}=\frac{13}{40}\)

प्रश्न 8.
प्रश्नावली 14.2 का प्रश्न 2 देखिए। इसकी आनुभविक प्रायिकता क्या होगी कि इंजीनियर :
(i) अपने कार्यस्थल से 7 कि०मी० से कम दूरी पर रहते हैं?
(ii) अपने कार्यस्थल से 7 कि०मी० या इससे अधिक दूरी पर रहते हैं?
(iii) अपने कार्यस्थल से \(\frac{1}{2}\) कि०मी० या इससे कम दूरी पर रहते हैं?
हल :
कल इंजीनियरों की संख्या (संभाव्य परिणाम) = 40
(i) अपने कार्यस्थल से 7 कि०मी० से कम दूरी पर रहने वाले इंजीनियरों की संख्या = 9
∴ P (एक इंजीनियर अपने कार्यस्थल से 7 कि०मी० से कम दूरी पर रहता है) = \(\frac{9}{40}\) उत्तर

(ii)
अपने कार्यस्थल से 7 कि०मी० या इससे अधिक दूरी पर रहने वाले इंजीनियरों की संख्या = 31
∴ P (एक इंजीनियर 7 कि०मी० या इससे अधिक दूरी पर रहता है) = \(\frac{31}{40}\) उत्तर

(iii) अपने कार्यस्थल से \(\frac{1}{2}\) कि०मी० या इससे कम दूरी पर रहने वाले इंजीनियरों की संख्या = 0
∴ P (एक इंजीनियर अपने कार्यस्थल से \(\frac{1}{2}\) कि०मी० या इससे कम दूरी पर रहता है) = \(\frac{0}{40}\) = 0 उत्तर

प्रश्न 9.
क्रियाकलाप : अपने विद्यालय के गेट के सामने से एक समय-अंतराल में गुजरने वाले दो पहिया, तीन पहिया और चार पहिया वाहनों की बारंबारता लिख लीजिए। आप के द्वारा देखे गए वाहनों में से किसी एक वाहन का दो पहिया वाहन होने की प्रायिकता ज्ञात कीजिए।
हल :
विद्यार्थी स्वयं हल करें।

प्रश्न 10.
क्रियाकलाप : आप अपनी कक्षा के विद्यार्थियों से एक 3 अंक वाली संख्या लिखने को कहिए। आप कक्षा से एक विद्यार्थी को यदृच्छया चुन लीजिए। इस बात की प्रायिकता क्या होगी कि उसके द्वारा लिखी गई संख्या 3 से भाज्य है? याद रखिए कि कोई संख्या 3 से भाज्य होती है, यदि उसके अंकों का योग 4 से भाज्य हो।
हल :
विद्यार्थी स्वयं हल करें।

प्रश्न 11.
आटे की उन ग्यारह थैलियों में, जिन पर 5 कि०ग्रा० अंकित है, वास्तव में आटे के निम्नलिखित भार (कि०ग्रा० में हैं:
4.97 5.05 5.08 5.03 5.00 5.06 5.08 4.98 5.04 5.07 5.00
यदृच्छया चुनी गई एक थैली में 5 कि०ग्रा० से अधिक आटा होने की प्रायिकता क्या होगी?
हल :
यहां पर,
आटे की कुल थैलियों की संख्या (संभाव्य परिणाम).= 11
5 कि०ग्रा० से अधिक भार वाली थैलियों की संख्या (अनुकूल परिणाम) = 7
P(5 कि०ग्रा० से अधिक भार वाली थैली) = \(\frac{7}{11}\) उत्तर

प्रश्न 12.
प्रश्नावली 14.2 के प्रश्न 5 में आपसे 30 दिनों तक एक नगर की प्रति वायु में सल्फर डाई-ऑक्साइड की भाग प्रति मिलियन में सांद्रता से संबंधित एक बारंबारता बंटन सारणी बनाने के लिए कहा गया था। इस सारणी की सहायता से इनमें से किसी एक दिन अंतराल (0.12-0.16) में सल्फर डाई-ऑक्साइड के सांदण होने की प्रायिकता ज्ञात कीजिए।

हल :
सारणी के अनुसार दिनों की कुल संख्या (संभाव्य परिणाम) = 30
जितने दिन (0.12 – 0.16) सांद्रता रही (अनुकूल परिणाम) = 2
∴ P (जितने दिन 0.12 – 0.16 सांद्रता रही) = \(\frac{2}{30}=\frac{1}{15}\) उत्तर

प्रश्न 13.
प्रश्नावली 14.2 के प्रश्न 1 में आपसे एक कक्षा के 30 विद्यार्थियों के रक्त-समूह से संबंधित बारंबारता बंटन सारणी बनाने के लिए कहा गया था। इस सारणी की सहायता से इस कक्षा से यदृच्छया चुने गए विद्यार्थी का रक्त-समूह AB होने की प्रायिकता ज्ञात कीजिए।

हल :
सारणी के अनुसार विद्यार्थियों की कुल संख्या (संभाव्य परिणाम) = 30
AB रक्त-समूह वाले विद्यार्थियों की संख्या (अनुकूल परिणाम) = 3
∴ P (AB रक्त-संमूह वाला विद्यार्थी) = \(\frac{3}{30}=\frac{1}{10}\) उत्तर

Haryana State Board HBSE 9th Class Science Solutions Chapter 9 Force and Laws of Motion Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 9 Force and Laws of Motion

HBSE 9th Class Science Force and Laws of Motion Intext Questions and Answers

Questions from Sub-section 9.1, 9.2, 9.3

Question 1.
Which of the following has more inertia:
(a) a rubber ball and a stone of same size ?
(b) a bicycle and a train ?
(c) a five-rupee coin and a one-rupee coin ?
Answer:
(a) The inertia of stone will be more, in a rubber ball and a stone of same size.
(b) In a bicycle and a train, the inertia of the train will be more.
(c) In five-rupee coin and a one-rupee coin, the inertia of five-rupee coin will be more.

Question 2.
In the following example, try to identify the number of times the velocity of the ball changes: “A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team.” Also identify the agent supplying the force in each case.
Answer:
The velocity of the ball changes four times as explained below:
(i) When the player kicks the ball.
(ii) When the second player kicks the ball towards the goal.
(iii) When the goalkeeper collects the ball.
(iv) When the goalkeeper kicks the ball towards the player of his own team.
The agent of force in each case is the muscular effort of the respective player.

Question 3.
Explain why some of the leaves may get detached from a tree if we vigorously shake its branch ?
Answer:
When branches of a tree are shaken vigorously, the leaves get detached from the tree. This is because the branches of the tree come into motion whereas the leaves tends to remain at rest due to inertia of rest. Thus, the leaves get detached from the branches and fall down under the action of gravity.

Question 4.
Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest ?
Answer:
When a moving bus stops suddenly, the lower part of our body in contact with the bus comes to rest immediately, whereas the upper part of the body tends to move in forward direction due to inertia of motion and we fall in the forward direction.
But, when the bus accelerates from rest, the lower part of our body comes into motion alongwith the bus but the upper part of the body has tendency to remain in the state of rest due to inertia of rest and we fall in backward direction.

Questions from Sub-section 9.4, 9.5, 9.6

Question 1.
If action is always equal to the reaction, explain how a horse can pull a cart ?
Answer:
According to third law of motion, the action is always equal to reaction. When the horse pulls the cart, it acts on earth by a force F, the earth offers it reaction R, which has two components R cos A and R sin A. The horizontal part R cos A pulls the cart in forward direction, whereas the vertical part R sin A reduces the frictional force between the road and cart. In this way, the horse can pull the cart.

Question 2.
Explain, why it is difficult for a Fireman to hold a hose, which ejects large amounts of water at a high velocity ?
Answer:
According to third law of motion, the action is always equal to reaction. When the excess water comes out of the hose with high velocity, then it recoils the hose with same amount of velocity in backward direction. Therefore, it becomes difficult for a fireman to hold a hose.

Question 3.
From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 ms^-1. Calculate the initial recoil velocity of the rifle.
Solution:
Here,
Mass of the bullet (m₁) = 50g = 0.05 kg
Mass of the rifle (m₂) = 4 kg
The initial velocity of bullet (u₁) and the initial velocity of the rifle (u₂) is zero.
or u₁ = u₂ = 0

Velocity of bullet (v,) = + 35 ms^-1
The direction of bullet, from left to right is taken + ve according to figure.
Let the recoil velocity of rifle = v ms^-1
The total momentum of bullet and rifle before firing = (4 + 0.05) kg x 0 ms^-1 = 0
Total momentum of bullet and rifle after firing = 0.05 kg × 35 ms^-1 + 4 kg x v ms^-1 = (1.75 + 4 v) kg ms^-1
According to the law of conservation of momentum,
Total momentum after firing = Total momentum before firing
1.75 + 4 v = 0
v = –\(\frac{-1.75}{4}\) = -0.4375 = -0.44 ms^-1
Negative sign shows that rifle will recoil in the opposite direction or right to left.

Question 4.
Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 ms-1 and 1 ms”1, respectively. They collide and after collision, the first object moves at a velocity of 1.67 ms”1. Determine the velocity of the second object.
Solution:
Here,
Mass of the first object (m₁) = 100g = 0.1 kg
Mass of second object (m₂) = 200 g = 0.2 kg
Initial velocity of first object (u₁) = 2 ms^-1
Initial velocity of second object (u₂) = 1 ms^-1
Total momentum of two objects before collision = m₁u₁ + m₂u₂
= (0.1 x 2 + 0.2 x 1) kg ms^-1 = 0.4 kg ms^-1
The velocity of first object after collision (v₁) = 1.67 ms^-1
Let, the velocity of second object after collision (v₂) = v ms^-1
Total momentum of two objects after collision = m₁v₁ + m₂v₂
= [0.1 x 1.67 + 0.2 x v₂] kg ms^-1
= [0.167 + 0.2 v₂] kg ms^-1
According to the law of conservation of momentum
Total momentum after collision = Total momentum before collision
0.167 + 0.2 v₂ = 0.4
or 0.2v₂ = 0.4 – 0.167
0.2v₂ = 0.233
or v₂ = \(\frac {0.233}{0.2}\) = 1.165 ms^-1
The velocity of second object after collision = 1.165 ms^-1

HBSE 9th Class Science Force and Laws of Motion Textbook Questions and Answers

Question 1.
An object experiences a net zero external unbalanced force. Is it possible for the object to be traveling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason.
Answer:
Here, the external unbalanced force on an object is zero. It means,
F = 0
⇒ m a = 0
But, m ≠ 0 ∴ a = 0
It is possible to move in this state, but the motion will be in the same magnitude and in one direction or straight path.

Question 2.
When a carpet is beaten with a stick, dust comes out of it. Explain.
Answer:
When a carpet is beaten with a stick, it moves forward, whereas dust particles have a tendency to remain at rest due to the inertia of rest. Thus, dust particles come out and the carpet gets clean.

Question 3.
Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer:
It is advised to tie any luggage kept on the roof of a bus with a rope because when the moving buses or cars suddenly stop or change direction, the luggage becomes safe from falling due to inertia of motion.

Question 4.
A batsman hits a cricket ball which then rolls on level ground. After covering a short distance, the ball comes to rest The ball slows to stop because:
(а) the batsman did not hit the ball hard enough.
(b) velocity is proportional to the force exerted on the ball.
(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball, so the ball would want to come to rest
(Choose the correct option)
Answer:
(c) There is a force on the ball opposing the motion.

Question 5.
A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes. [Hint: 1 tonne = 1000 kg]
Solution:
Here, Initial velocity in first stage (u) = 0
Distance covered (s) = 400 m
Acceleration (a) = ?
Time (t) = 20s
According to formula, s = ut + \(\frac {1}{2}\) at²
ut + \(\frac {1}{2}\) at² = s
0 x (20) + \(\frac {1}{2}\) a x (20)² = 400
200 a = 400
a = \(\frac {400}{200}\) = 2 ms^-2
Mass in second stage (m) = 7 tones = 7 x 1000 = 7000kg
force (F) = m x a = 7000 kg x 2 ms^-2 = 14000 N

Question 6.
A stone of 1 kg is thrown with a velocity of 20 ms^-1 across the frozen surface of a lake and it comes to rest after traveling a distance of 50 ni. What i the force of friction between the stone and the ice?
Solution:
Here,
Mass of stone (m) = 1 kg
Initial velocity of stone (u) = 20 ms^-1
Final velocity of stone (v) = 0
Distance covered by stone (s) = 50 m
Acceleration (a) = ?

We know that,
v² – u² = 2as .
a = \(\frac{v^2-u^2}{2 s}=\frac{(0)^2-(20)^2}{2(50)}\)
= \(\frac{-400}{100}\) = -4 ms^-1
A negative (-ve) sign shows that acceleration is working against the motion.
The frictional force between the stone and the ice I (-4) = – 4N

Question 7.
An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and (be track offers a friction force of 5000 N, then calculate:
(a) the net accelerating force,
(b) the acceleration of the train; and
(c) the force of wagon I on wagon 2.
Solution:
Here,
The mass of train engine (m₁) = 8000 kg
Mass of 5 wagons(m₂) 2000kg x 5 = 10000 kg
Mass of train engine with wagons (m) = m₁ + m₂ = (8000 + 10000)kg = 18000 kg
The force applied by the engine on wagons (F₁) = 40000 N
The friction force applied by track on wagons (F₂) = 5000 N
(a) The net acceleration force (F) = F₁ – F₂
40000N – 5000N = 35000N

(b) The acceleration of the train (a) = \(\frac{F}{m_2}=\frac{35000}{10000}\) ms^-2 = 3.5 ms^-2

(c) The force of wagon I ou wagon 2 35000 N – \(\frac{35000}{5}\) = (35000 – 7000) N = 28000N Ans

Question 8.
An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is tobe stopped with a negative acceleration of 1.7 ms^-2?
Solution:
Here,
Mass of automobile vehicle (m) = 1500 kg
The acceleration on vehicle (a) = 17 ms^-2
Final velocity (v) = 0
The force between the vehicle and road (F) = m x a = 1500 × (-1.7) N = -2550 N
Negative (-ve) sign shows that the force is working against the motion of vehicle.

Question 9.
What is the momentum of an object of mass m, moving with a velocity v ?
(a) (mv)²
(b) mv²
(c) \(\frac {1}{2}\) mv²
(d) mv (choose correct option from above)
Answer:
(d) mv.

Question 10.
Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Answer:
If we push the wooden cabinet with constant velocity on the floor by applying a force of 200 N, then the frictional force of the wooden cabinet will also be 200 N.

Question 11.
Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 ms-1 before the collision during which they stick together. What will be the velocity of the combined object after collision?
Answer:
Here, the mass and velocity of both objects are the same, but both are moving in opposite directions, so they bend after the collision. Therefore, their combined velocity will be zero.

Question 12.
According to the third law of motion, when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Answer:
The truck parked along the roadside will not move on applying force by us because the force applied by us is much less than the frictional force applied by the truck between the truck and the road. If our force increases to frictional force, the truck will move.

Question 13.
A hockey ball of mass 200 g traveling at 10 ms-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms^-1. Calculate the change of momentum that occurred in the motion of the hockey ball by the force applied by the hockey stick.
Solution:
Here,
Mass of hockey ball (m) = 200 g = 0.2 kg
Initial velocity of hockey ball (v₁) = 10 ms
The initial momentum of hockey ball travels in the opposite direction with velocity (v₂) = 5 ms^-1
Final Movemntum Of Hockey Boll = mv₂
Change in momentum = mv₁ – mv₂ = 0.2 [10 – (-5)] = 0.2(15) = 3 kg ms^-1
Thus, the change in momentum of the hockey ball is 3 kg ms^-1

Question 14.
A bullet mass Logis traveling horizontally with a velocity of 150 m1 strikes a stationary wooden block and comes to at rest in 0.03 s. Calculate the distance of penetration of the bullet into the Mock. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
Solution:
Here,
Mass of bullet (m) = 10g
= \(\frac {10}{1000}\) = 0.01kg
Initial velocity of bullet (u) = 150 ms^-1
Final velocity of bullet (v) = 0 ms^-1
Time (t) = 0.03 s
Acceleration (a) = ?
Distance (s) = ?

We know that,
a = \(\frac{v-u}{t}=\frac{0-150}{0.03}\) ms^-2 = -5000 ms^-2
and s = ut + at²
= (150)(0.03) + \(\frac {π}{2}\) × (-5000) x (0.03)²
= 4,5 – 2.25 = 2.25m
Force(F) = ma
= 0.01 x (-5000)N = -50 N
The negative (-ve) sign shows that force is working in the opposite direction of motion.

Question 15.
An object mass 1 kg traveling In a straight line with a velocity of 10 ms collides with and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the Impact. Also, calculate the velocity of the combined object.
Solution:
Here,
Mass of object (m) = 1 kg
Velocity of object (v₁) = 10 ms^-1
∴Momentum before collision = m x v₁
= 1 x 1o kg ms^-1 = 1o kg ms^-1
The momentum after collision = 10 kg ms^-1
Combined mass of wooden block and object (m₁) = 5 + 1 = 6kg
Suppose. the final velocity of object and wooden block = v₂ ms^-1
According to the law of conservation of momentum,
m₁ v₂ = 10
or 6 x v₂ = 10
0r v₂ = \(\frac{10}{6}\) ms^-1 = \(\frac{10}{6}\) ms^-1

Question 16.
An object of mass loo kg Is accelerated uniformly from a velocity of S mr1 to 8 mr’ in 6s. object. Also, find the magnitude of the force exerted on the object
Solution:
Here.
Mass of object (m) = 100 kg
Initial velocity (u) = 5 ms^-1
Final velocity (y) = 8 ms^-1
Time(t) = 6s
Initial momentum = mu = 100 x 5 = 500 kg ms^-1
Final momentum = mv = 100 x 8 = 800 kg ms^-1
We know that,
a = \(\frac{v-u}{t}=\frac{8-5}{6}=\frac{3}{6}\) = 0.5 ms^-2
The magnitude of the force (F) = m x a
= l00 x 0.5N = 1.50 N

Question 17.
Akhtar, ((Iran and Rbul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorca (because the change in the velocity of the Insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change In their momentum. Comment on these suggestions.
Answer:
Akhtar was right because both the velocity and mass of the motorcar were moveso the momentum of the motorcar was more than that of flying insect. Therefore, the motorcar imposed much force on flying insect due to the insect died.

Question 18.
How much momentum will a dumbbell of mass 10 kg transfer to the floor if It falls from a height of 80cm? Take its downward acceleration to be 10 ms² ?
Solution:
Here,
Mass of dumb-bell (m) = 10 kg
Initial velocity of dumb-bell (u) = 0 ms^-1
Final velocity of dumbbell (v) = ?
Height of dumb-bell (s) = 80 cm = 0.8 m
Acceleration (a) = 10^-2
We know that.
v² – u² = 2as
v² – (0)² = 2(10) (0.8)
v² = 16
v = \(\sqrt{16}\) = 4 ms
Momentum transferred to the floor by dumb-bell (p) = mv = 10 x 4 = 40 kg ms

Additional Exercise Questions

Question 1.
The following is the distance-time table of an object in motion:

(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, orzero?
(b) What do you infer about the force acting on the object 1
Answer:
(a) It is clear from the distance-time table that acceleration is continuously increasing.
(b) The force is also increasing due to the increase in acceleration on body.

Question 2.
Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 ms^-2.With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)
Solution:
Here,
Mass of motorcar (m) = 1200 kg
The acceleration produced on pushing by three persons
(a) = 0.2 ms^-2
∴ The force imposed on car by each person (F) = m a = 1200 x 0.2N. = 240N Ans

Question 3.
A hammer of mass 500 g, moving at 50 m1, to strike a nail. The nail stops the hammer In a very short time of 0.01s. What Is the force of the nail on the hammer?
Solution:
Here,
Mass of hammer (m) = 500 g = \(\frac {500}{1000}\) kg = 0.5kg
Initial velocity of hammer (u) = 50 ms^-1
Time (t) = 0.01 s
Final velocity of hammer (v) = 0
We know that,
Acceleration (a) = \(\frac{v-u}{t}=\frac{0-50}{0.01}\) = 0.01 ms^-2
= -5000 ms^-2
(-ve sign shows that acceleration is decreasing, i.e., retarded)
The force of the nail on the hammer (F) = m x a
= O.5 x (5000)N
= 2500N

Question 4.
A motorcar of mass 1200 kg Is moving along a straight line with a uniform velocity of 90 km/h. Its velocity Is slowed down to 18 km/h In 4 s by an unbalanced external force. Calculate the acceleration and change ¡n momentum. Also calculate the magnitude of the force required.
Solution:
Here,
The mass of a motorcar (m) is 1200 kg
Initial velocity (u) = 90 km h^-1= \(\frac{90 \times 1000}{3600}\) = 25 ms^-1
Final velocity (v) = 18 km/h = \(\frac{18 \times 1000}{3600}\) = 5 ms^-1
Tíme(t) = 4s
Acceleration (a) = \(\frac{v-u}{t}=\frac{5-25}{4}=\frac{-20}{4}\)
= -5 ms^-2
Change ¡n momentum mv – mu
= m(v – u) = 1200(5 – 25) = 1200 x (-20) = -24000 kg ms^-1
The magnitude of the force required (F) = m a
1200 x – 5 = – 6000N

Question 5.
A large truck and a car, both moving with a velocity of magnitude w, have a head-on collision and both of them come to a halt after that. If the collision for is:
(a) Which vehicle experiences the greater force of impact?
(b) Which vehicle experiences the greater change ¡n momentum?
(C) Which vehicle experiences the greater acceleration?
(d) Why is the car likely to suffer more damage than the truck?
Answer:
(a) Here, the mass of the truck is more than that of car, and the force of the truck will be more which affects the car more.
(b) Truck will experience a greater change in momentum.
(c) The acceleration of the car will be greater.
(d) Car will suffer from more damage because the mass of truck is more.

Haryana State Board HBSE 9th Class Science Important Questions Chapter 2 Is Matter Around Us Pure Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 2 Is Matter Around Us Pure

Very Short-Answer Type Questions

Question 1.
What is purification?
Answer:
Separation of useful substances from unwanted and harmful substances is called as purification.

Question 2.
Write the names of any four pure substances.
Answer:
(i) Sugar
(ii) Common salt
(iii) Gold
(iv) Mercury.

Question 3.
Write the names of any four elements.
Answer:
(i) Gold
(ii) Silver
(iii) Mercury
(iv)) Sulphur.

Question 4.
Write the names of any four compounds.
Answer:
(i) Water
(if) Camphor
(iii) Pond-water
(iv) Carbon dioxide.

Question 5.
Write the names of any four mixtures.
Answer:
(i) Soil
(ii) Air
(iii) Pond-water
(iv) Rock salt.

Question 6.
Name two solute substances, soluble in water that is used in daily life.
Answer:
(i) Common salt
(ii) Sugar.

Question 7.
Write the names of two liquid mixtures.
Answer:
(i) Pond or river water
(ii) Soda water.

Question 8.
Give an example of the gaseous mixture.
Answer:
Air.

Question 9.
What is called ‘separation’?
Answer:
The separation of a substance from the other in a mixture is called ‘separation’.

Question 10.
How many components does a mixture have?
Answer:
A mixture has two or more two components.

Question 11.
Which of the useful components do we bring into use in the form of breathing from the air?
Answer:
We use oxygen from the air in the form of breathing.

Question 12.
Which property of the magnet is brought into use in magnetic separation?
Answer:
A magnet attracts iron towards it.

Question 13.
What is the principle of sedimentation?
Answer:
A combination of tiny particles forms a big and heavy molecule.

Question 14.
What is called a ‘solution’ ?
Answer:
The homogeneous mixture of two or more two substances is called a ‘solution’.

Question 15.
What is meant by ‘alloys’?
Answer:
Those homogeneous mixtures of metals which cannot be isolated from components by physical processes are called ‘alloys’.

Question 16.
Define ‘solvent’.
Answer:
That component of the solution (which has more quantity than the other) which dissolves another component in the solution is called ‘solvent’.

Question 17.
What is meant by ‘solute’?
Answer:
That component of the solution which is remained dissolved in the solvent is called as solute.

Question 18.
In the solution of sugar and water state which of them is solute and solvent?
Answer:
Solute – sugar; Solvent – water.

Question 19.
What is meant by tincture iodine?
Answer:
The solution of iodine and alcohol is called as tincture iodine.

Question 20.
Write down the solute and solvent in the tincture iodine.
Answer:
Solute – iodine; Solvent – alcohol.

Question 21.
Write down solute and solvent in soda water and coke.
Answer:
Solute – Carbon dioxide; Solvent – water.

Question 22.
Give two examples of solutions of solid within solid.
Answer:
(i) Mixture of gold and silver.
(ii) Mixture of copper and gold.

Question 23.
Give two examples of an aqueous solution.
Answer:
(i) Solution of sugar and water.
(ii) Solution of salt and water.

Question 24.
Give two examples of non-aqueous solutions.
Answer:
(i) Solution of alcohol and iodine.
(ii) Solution of sulphur and carbon dioxide.

Question 25.
Give an example of a solution of gas within the gas.
Answer:
Air is a solution of gas within the gas.

Question 26.
Which are the main components of air?
Answer:
The main components of air are nitrogen (78%) and oxygen (21%).

Question 27.
In air, which components are supposed to be solvent?
Answer:
In air, nitrogen gas is supposed to be solvent.

Question 28.
What is the diameter of the molecules of the solution?
Answer:
The diameter of the molecules of the solution is even less than 1 nm (10^-9m).

Question 29.
What is meant by the unsaturated solution?
Answer:
When the quantity of the solute substance is less than saturation in a solution, then it is called an unsaturated solution.

Question 30.
What is meant by the supersaturated solution?
Answer:
If in a solution strength of solute substance is more than the strength of saturated strength, then it is called as the supersaturated solution.

Question 31.
Why is water called as a universal solution?
Answer:
Water is called a universal solution because maximum substances dissolve into it.

Question 32.
Why seawater cannot be used for domestic purposes?
Answer:
Because in seawater, the strength of mineral salts is more.

Question 33.
Give the formula to express the strength of a solution. ‘ Mass of solute
Answer:

Question 34.
What is meant by 10% glucose solution?
Answer:
10%glucose solution means, 100 g solution contains 10 g of glucose.

Question 35.
What is the size of molecules of suspension?
Answer:
The size of molecules of suspension is more than 100 nm (10^-7m).

Question 36.
By which method can the molecules of suspension be separated?
Answer:
The molecules of suspension can be separated by the filtration method.

Question 37.
Is the colloidal solution, homogeneous or heterogeneous?
Answer:
The colloidal solution is heterogeneous.

Question 38.
Give four examples of colloidal solutions used in our daily life.
Answer:
(i) Milk
(ii) Shaving cream
(iii) Toothpaste
(iv) Jelly.

Question 39.
What is the ‘Tyndall effect?
Answer:
When a narrow beam of light penetrates and enters the room through a small hole in the ceiling, the Tyndall effect can be noticed, because, inside the room, dust and smoke scatter light.

Question 40.
What is the size of colloid particles?
Answer:
The size of colloid particles is between 1 nm to 100 nm.

Question 41.
Give an example of the Tyndall effect
Answer:
When a fine beam of light enters in a room through a small hole in the roof, the Tyndall effect can be observed because the particles of dust and smoke scattered the light.

Question 42.
Can colloidal particles be seen by the naked eyes?
Answer:
No.

Question 43.
Give the name of a common colloid found in nature.
Answer:
Fog, is the common colloid found in nature, which is formed of water vapours present in the air.

Question 44.
What is the fundamental base of the classification of colloids?
Answer:
The fundamental base of classification of colloids is the state of dispersion medium and state of the dispersed phase.

Question 45.
Write two uses of colloids.
Answer:
In the manufacturing of medicines and in understanding the industrial processes.

Question 46.
Can water and mustard oil be separated by filtration?
Answer:
No, they can be separated either by evaporation or decantation method.

Question 47.
What is the principle of agitation?
Answer:
The method of separation of thin solid substance from a thick solid substance with a porous device is called as agitation method. No liquid is used in this method.

Question 48.
By which method contaminated water of rivers can be purified?
Answer:
By sedimentation contaminated water of rivers can be purified.

Question 49.
How is the melting point or boiling point of a substance useful?
Answer:
The purity or impurity of a substance can be known by these characteristic properties.

Question 50.
Which is the best method to separate solute substances in liquid?
Answer:
Distillation is the best method to separate solute substances in the liquid.

Question 51.
What is called as sublimation?
Answer:
The process in which a solid substance directly changes into the gaseous state is called sublimation.

Question 52.
Write the names of the two sublimated substances.
Answer:
Camphor and ammonium chloride are sublimated substances.

Question 53.
How can the mixture of water and kerosene oil be separated?
Answer:
The mixture of water and kerosene oil can be separated by a separating funnel.

Question 54.
How can cream (butter) can be skimmed from milk?
Answer:
Cream (butter) can be skimmed from milk by centrifugation method.

Question 55.
By which process distilled water is obtained?
Answer:
Distilled water can be obtained by the distillation process.

Question 56.
By which method can a mixture of ammonium chloride and sand be separated?
Answer:
A mixture of ammonium chloride and said can be separated by the sublimation method.

Question 57.
By which method water is purified in homes?
Answer:
By filtration method, water is purified in homes.

Question 58.
Which method helps to check adulteration in petrol?
Answer:
The evaporation method helps to check adulteration in petrol.

Question 59.
By which method, can a mixture of water and acetone be separated?
Answer:
By distillation method mixture of water and acetone can be separated.

Question 60.
Where is the distillation process applied?
Answer:
When there is much difference between the boiling points of the components of a mixture of two miscible liquids.

Question 61.
Which method is applied to separate different components of petroleum products?
Answer:
Different components of petroleum products can be separated by the fractional distillation process.

Question 62.
When is the fractional distillation process practised?
Answer:
When the difference of boiling points of two or more than two miscible liquids is less than 25 K.

Question 63.
By which method can different components of air be separated?
Answer:
By fractional distillation method, different components of air can be separated.

Question 64.
What is the boiling point of oxygen?
Answer:
The boiling point of oxygen is -183°C.

Question 65.
When the components of air are cooled down, which component changes into liquid at first?
Answer:
Oxygen.

Question 66.
By which method, pure copper sulphate is obtained from a sample of impure copper sulphate?
Answer:
By crystallisation method.

Question 67.
What is meant by physical changes?
Answer:
In physical changes, change takes place in the form of the substances and not in chemical form or in these changes a substance can be obtained again.

Question 68.
Give any two examples of physical changes.
Answer:
(i) Solution of sugar and water.
(ii) Solution of salt and water.

Question 69.
Can sugar be obtained from the solution of sugar?
Answer:
Yes. by evaporation of water from a solution of sugar, sugar can be obtained.

Question 70.
What is meant by chemical change?
Answer:
ln chemical change, one or more than one type of substance changes into one or many new substances or in these changes original substance cannot be obtained again.

Question 71.
Which are the two chemical changes taking place in daily life?
Answer:
(i) Formation of milk into curd.
(ii) Digestion of food.

Question 72.
What is an element?
Answer:
An element is a basic form of matter which cannot be broken into simple substances by chemical reactions.

Question 73.
Write any special property of an element.
Answer:
An element is composed of only a single type of atom.

Question 74.
How many elements are found in a gaseous state at room temperature?
Answer:
11 elements are found in a gaseous state at room temperature, like hydrogen and oxygen.

Question 75.
Write the names of a metal and a non-metal found in liquid state at room temperature.
Answer:
(i) Metal-Mercury.
(ii) Non-Metal-Bromine.

Question 76.
What is meant by alloys ?
Answer:
Elements displaying properties between metals and non-metals are called as alloys.

Question 77. Give names of any two alloys.
Answer:
(i) Boron
(ii) Silicon.

Question 78.
What is meant by compound?
Answer:
A substance formed by a chemical combination of two or more two elements in an equal ratio is called a compound.

Question 79.
Air is a homogeneous mixture of mainly which two gases?
Answer:
Air is mainly a mixture of oxygen and nitrogen gases because in air other gases are available in a little quantity.

Question 80.
What is called a heterogeneous mixture?
Answer:
A heterogeneous mixture is the one in which physically there are different parts and every part has different properties.

Question 81.
Which of the following are elements, compounds and mixtures:
(a) Na
(b) Soil
(c) Ag
(d) Sugar
(e) Urea
Answer:
(a) Element: (a) Na (c) Ag
(b) Compounds: (d) Sugar (e) Urea
(c) Mixture: (b) Soil

Question 82.
When two or more than two dements or compounds in any ratio combine together, then what is obtained ?
Answer:
Mixture is obtained when two or more than two elements or compounds in any ratio combine together.

Short-Answer Type Questions

Question 1.
What is called as a mixture? How many types of the mixture are there? Explain briefly.
Answer:
A mixture constitutes more than one substance (element/compound). A mixture can be separated by physical process into two or more two substances. Mixtures are mainly of two types:

1. Homogeneous Mixture: The mixture where composition remains constant everywhere, is called a homogeneous mixture e.g., a mixture of salt and water, a mixture of sugar and water.

2. Heterogeneous Mixture: The mixture in which physically there are different parts and each part is of different characteristics, is called a heterogeneous mixture e.g., a mixture of sodium chloride and iron peeling, a mixture of salt and sulphur and a mixture of water and oil.

Question 2.
Why is the solution of salt into water considered as a mixture but not a compound?
Answer:
Solution of salt into water is considered as a mixture but not a compound because:
(1) In salt water, salt and water can be separated by the distillation process.
(2) Saltwater displays the properties of of its both components – water and salt.
(3) Salt water has a different composition. By dissolving different quantity of salt into a definite quantity of water, salt water of different compositions can be obtained.
(4) Salt water does not have any fixed formula.

Question 3.
Give an example of such a mixture:
(i) Where both substances are compounds.
(ii) Where both element and compound are combined.
(iii) Where both are elements.
Answer:
(i) Such a mixture where both substances are compounds: On dissolving sugar in water, the syrup is formed which is a mixture in which sugar and water both are compounds.
(ii) Such a mixture where both element and compound are combined: Air is such a mixture in which oxygen and nitrogen are elements and carbon dioxide and water vapours both are compounds.
(iii) Such a mixture where both elements: Brass, copper and zinc are formed by mixing two elements.

Question 4.
What is called a solution ? Make a list of different types of solutions.
Answer:
A solution is a homogeneous mixture in which two or more two substances are there. Different types of solutions are as follows:

Question 5.
Give any four characteristics of solutions
Answer:
Different characteristics of the solution are as follows:
(i) Solution is a homogeneous mixture.
(ii) As molecules of solution are smaller in diameter than lnm (10^-9 m), so they cannot be seen with naked eyes.
(iii) Due to their smaller size, solutions do not scatter the passing over the beam of light, thus no path of light is seen in solutions.
(iv) Particles of solute cannot be separated from the solution by the filtration method. On leaving the solution, untouched even then the particles of solute do not settle down, thus the solution is stable.

Question 6.
What are aqueous and non-aqueous solutions?
Answer:
Aqueous Solutions: These are those solutions which are prepared by dissolving substances into the water; like syrup and soda water etc.
Non-aqueous Solutions: These are those solutions which are prepared by dissolving apart from water in other solvents (like alcohol, acetone etc.) e.g., solution in benzene of tincture iodine nap/itfiafene.

Question 7.
Why is it not possible to identify solute and solvent particles separately in a solution?
Answer:
Particles of solute are so closely combined with a solvent that they cannot be separately identified. Solutions are homogeneous, thus the composition of solutions is uniform. If salt-solution and sugar-solution are mixed, as the result in the obtained homogeneous solutions both solutions combine with each other well and to make difference between basic solute and solvent is not possible.

Question 8.
How is the strength of a solution expressed? Explain with an example with its meaning.
Answer:
The strength of solutions is expressed in the context of the quantity of solute present in the mass or volume of a given solution or in the context of the quantity of dissolved solute in the mass or volume of a given solvent The strength of solutions can also be expressed in the form of per cent volume of solutes which give the mass of solutes in the 100 mass per unit. Its mass unit is gram.

For example, if we take a solution of 10% glucose (C₆H₁₂O₆) by mass. It contains 10 g of glucose in 100 g of solution. It can also be called in the form as 10 g of glucose in 90 g of water. If something is not said in a special manner, then the meaning of per cent in the form of mass is per cent and water is solvent in it.

Question 9.
Define saturated solution, unsaturated solution and supersaturated solution.
Answer:
Saturated Solution: When at a given temperature in a solution the extra solute that gets dissolved in it beyond its capacity, then it is called a saturated solution. In this solution, dissolved and undissolved solutes are in equilibrium together.
Unsaturated Solution: If in a solution, the quantity of a solute already present in it is less than the saturated level, then the solution is called an unsaturated solution.
Supersaturated Solution: If in a solution, the strength of the solution is more than the saturated strength, this solution is called a supersaturated solution.

Question 10.
A certain solution, in 320 g of water keeps 40 g of salt. Give the strength of solution.
Solution:
Mass of solute substance (salt) = 40 g
Mass of solvent (water) = 320 g
Mass of solution = Mass of solute substance + Mass of solution = 40g + 320 g = 360 g

\(\frac {40}{360}\) × 100 = 11.1%

Question 11.
Define suspension and write its different characteristics.
Answer:
Suspension: Suspension is a heterogeneous mixture, in which the particles of a solute substance do not dissolve, rather they remain suspended in the equilibrium of the medium.

Its main characteristics are as follows:
(i) It is a heterogeneous mixture.
(ii) The suspended particles are bigger than 100 nm (10^-7m) as a result they can be seen with naked eyes.
(iii) These suspended particles scatter the beam of light, due to this its path is known.
(iv) Suspension is temporary. By filtration method, its particles can be separated from the mixture.

Question 12.
What is the difference between solution and colloid?
Answer:
Following is the difference between solution and colloid:
Solution:
1. Solution does not reflect light.
2. Solution is a homogeneous mixture.
3. They are totally transparent.
4. In the real state the size of particles of the solution is less than 10^-9m.
5. They do not display the Tyndall effect.

Colloid:
1. Colloid reflects light.
2. Colloid is a heterogeneous mixture.
3. Colloid is a little transparent.
4. The size of colloid particles is between 10^-9 to 10^-7 m.
5. They display the Tyndall effect.

Question 13.
What is the difference between solution and suspension?
Answer:
Following is the difference between solution and suspension:
Solution:
1. It is homogeneous.
2. It is transparent.
3. Its particles are comparatively smaller in size.
4. It passes through filter paper and leaves no residue behind.
5. Its particles are invisible.

Suspension:
1. It is heterogeneous.
2. It is opaque (hazy).
3. Its particles are comparatively bigger in size.
4. Its particles cannot pass through filter paper.
5. Its particles can be seen with eyes or a compound microscope.

Question 14.
What happens when a beam of light is passed through a colloidal solution?
Answer:
When a beam of light is passed through a colloidal solution, its path gets illumined, because the size of the colloidal particles is bigger, therefore these particles scatter the falling light on them in all directions. This is called as Tyndall effect. Colloidal solution and the actual solution are differentiated on this basis. For example, milk shows the Tyndall effect. In daily life when a beam of light passes through a small hole into a room, then we can notice Tyndall’s effect there.

Question 15.
What is meant by the Brownian movement?
Answer:
Colloidal molecules always keep travelling in zig-zag paths in all directions. The unending movement of colloidal molecules in zig-zag paths is called as Brownian movement.

Question 16.
Describe the characteristics of colloids.
Answer:
The characteristics of colloid are as follows:
(i) It is a heterogeneous mixture.
(ii) The size of colloidal molecules is between lnm to 100 nm and they cannot be seen with eyes.
(iii) They are so large that they scatter the ray of light and make Its path visible.
(iv) When they are left standstill, then they do not settle at the bottom which means they are stable.
(v) They cannot be separated by a filtration process but can be separated by the centrifugation method.

Question 17.
How are colloids classified? Explain by giving examples.
Answer:
Colloids are classified according to the state (solid, liquid or gas) of the dispersing medium and the dispersed phase, which is clarified by following examples:

Question 18.
Give any five differences between compound and mixture.
Answer:
Five differences between compound and mixture are as follows:
Compound:
1. It is composed of two or more than two elements in a definite ratio by chemical combination.
2. It is homogeneous.
3. Its physical and chemical properties are definite.
4. The components of compound cannot be seen separately.
5. Its components cannot be separated by physical methods.

Mixture:
1. It is composed of two or more two substances in any ratio combined together. It is not a chemical combination.
2. It is heterogeneous.
3. Its properties are not definite.
4. Its components can be seen separately.
5. Its components can be separated by physical methods.

Question 19.
What purposes are the components of mixtures separated for?
Answer:
Components of mixtures are separated for the following purposes:
(i) To separate unwanted components.
(ii) To separate any harmful component.
(iii) To obtain pure sample of any substance.
(iv) To obtain any useful component.

Question 20.
Classify the following into metals, non-metals and metalloids:
Silicon, Germanium, Iodine, Sodium, Iron, Carbon.
Answer:
Metals: Sodium, Iron
Non-Metals: Iodine, Carbon
Metalloids: Silicon, Germanium

Question 21.
Explain the threshing method.
Answer:
A peasant, when stands at a raised platform and the method that he adopts to separate chaff and wheat grains, it is called as threshing. Chaff being lighter in weight flies away with wind and the wheat grains being heavier in weight, directly fall onto the ground.

Question 22.
Explain briefly ‘weaning with hands’ a method of separation.
Answer:
Separating of unwanted materials like-fine pieces of stone pebbles from wheat, rice and pulses, by weaning with hands is called as ‘weaning or gleaning’. This method is also adopted to wean fruit and vegetables.

Question 23.
What is the evaporation process? Explain.
Answer:
If in a liquid, some other solute substance is got dissolved and if we are supposed to separate the components of that mixture, then the mixture is evaporated by heating in the sun. The liquid evaporates in the form of water vapours and the miscible solute is left behind. This method is called as evaporation. Salt is obtained from seawater by means of this method.

Question 24.
What is called a centrifugation method?
Answer:
In this method, that mixture is rotated all around in an enclosed vessel which has very small suspended particles. The heavy material settles down at the bottom of the vessel and the lighter material comes up. This is called as centrifugation method. In dairy, the cream is skimmed from milk by this method.

Question 25.
Describe the uses of the centrifugation method in day-to-day life.
Answer:
Uses of the centrifugation method in day-to-day life are as follows:
(i) In the diagnostic laboratory, this method is adopted to test blood and urine.
(ii) It is adopted in dairy and houses to extract butter from cream.
(iii) It is adopted to squeeze water from wet clothes in washing machines.

Question 26.
How will you separate kerosene oil from water ?
Answer:
Water and kerosene oil, both are immiscible liquids. Both liquids form different layers. Kerosene oil being lighter, floats on the surface of water. Both the components can be separated by separating funnel according to the given figure. Invert the mixture into the separating funnel. After sometime both the liquids will form separate layer. By opening the stopcock of the funnel the water is separated into the beaker kept just below to the stem of the funnel, whereas kerosene oil leaves behind in the funnel.

Question 27.
Which principle is used in a separating funnel to separate mixture of two immiscible liquids ? Give any two uses of it.
Answer:
The principle that is used in the separating funnel, it is that two immiscible liquids according to their densities get separated into two separate layers and by opening the stopcock are got separated.
Following are the two uses of it:
(i) In separation of two components i.e., water and oil.
(ii) During refining of a metal in separation of iron. By this method the lighter slug is collected from the top and molten iron leaves behind at the lower surface of the furnace.

Question 28.
What is sublimation ? How will you separate the mixture of salt and ammonium chloride ?
Answer:
The process of changing a solid material directly into gaseous state by heating, is called as sublimation. Generally, on heating, solid materials melt away, but, certain solid materials do not change into liquid on heating. They directly convert into gaseous state, such materials are called as volatile materials. Ammonium chloride, camphor, iodine and nephthalene are volatile materials.

Separation of Ammonium Chloride and Salt Mixture: To separate the mixture of ammonium chloride and salt, sublimation method is used.

Place the mixture of ammonium chloride and salt in China dish and put it on a tripod stand according the figure. Now, keep a glass funnel in an inverted position on the China dish as shown in the figure. Close the mouth of the stem of the funnel with a cotton plug. Start heating the China dish with a spirit lamp at it. On heating ammonium chloride will convert in vapour state. The vapours deposited on the inner wall of the funnel will again come back into solid state and salt will remain behind in the China dish. By this method the mixture of ammonium chloride and salt can be separated.

Question 29.
What do you understand bv chromatography ? What are its uses in daily life ?
Answer:
Chromatography is that process which is used to separate those solute materials which are dissolved in just one kind of solvent. It is used to separate the following:
(i) To separate the colours in dyes.
(ii) In separation of colour from natural colours.
(iii) To separate sugar from urine.
(iv) To separate medicine from blood.

Question 30.
What is called as distillation ? Where is it used ?
Answer:
Conversion of a liquid into vapour by heating and again converting it into liquid by cooling is called as distillation. This distilled water used by the doctors obtained by this method. This method is used to separate mixture of those two soluble liquids which have much difference between their boiling points (More than 25 k).

Question 31.
Where is fractional distillation practised ? How is it different from distillation process ?
Answer:
Fractional distillation is practised when the boiling point difference between two or more than two miscible liquids is less than 25 K which are mixed together. For example, separation of different gases from the air and separation of different components from petroleum products. Its apparatus resembles to that of the common distillation process-apparatus. Only a fractionating column is installed in between distillation flask and condenser. The common fractionating column is equipped with a tube which is filled up with glass powdered pieces. These glass powdered pieces provide surface to vapours to cool down and condense as shown in the figure.

Question 32.
What is called as crystallisation ? Give its two usages.
Answer:
Obtaining of pure and of definite sized pieces of a solid material from a mixture of solution is called as crystallisation. Alum, common salt, copper sulphate (blue vitrol) can be obtained in pure state through this method.To obtain crystals, firstly an impure sample is dissolved in maximum quantity in the hot liquid, then this solution is filtered and the remaining impurities are removed. This solution is let cooled down for sometime and therefore, pure solid crystals are obtained.
Minor Usages:
(i) In purifying the salt obtained from sea water.
(ii) In separation of alum from impure sample.

Question 33.
How will you prepare a big crystal from the pulverised sugar ?
Answer:
To obtain a big crystal from the pulverised sugar, prepare a solution of it in hot water. This solution is kept to cool down. Then, the solution is filtered and with the help of a thread a crystal of sugar is got suspended into this solution. The solution is left standstill in the open as usual. After a few days we notice the sugar crystal starts increasing in its size. Thus, we obtain a big crystal of sugar.

Question 34.
Why is crystallisation method considered better to common evaporation method ?
Answer:
In the following respect, crystallisation method is considered better to common evaporation method:
(i) Some solid particles breakdown or some get spoiled as sugar.
(ii) On dissolving solute material into solvent there are left behind some impurities in solution.

Question 35.
What do you understand by physical and chemical changes ? Make it clear by giving examples.
Answer:
Physical Change: These are temporary in which change takes place only in physical state. In these changes chemical changes do not occur and they can be converted into their basic state. For example, changing of water in steam, dissolving of sugar in water, dissolving of salt into water etc.
Chemical Change: These are permanent in which along with physical change chemical changes too occur. They cannot be brought back into their basic state. For example, corroding of iron, igniting of magnesium wire in presence of oxygen.

Question 36.
Define and classify an element into different groups.
Answer:
An element is that basic state of matter, which cannot be further subdivided into tiny particles through chemical process. This can be divided into the following three groups:
1. Metals: Gold, Silver, Copper, Iron, Sodium etc.
2. Non-Metals: Hydrogen, Oxygen, Iodine, Carbon, Coal etc.
3. Alloys: Elements lie in between the properties of metals and non-metals are called as alloys like- boron, silicon.

Question 37.
What do you understand by malleability and ductility ? Write names of two malleable and ductile metals each.
Answer:
Malleability: This is that characteristic of metals by virtue of which they are battered into thin sheets. Gold and silver are the most malleable metals. They can be battered with hammer yet into very thin foils than a leaf-paper.
Ductility: This is that characteristic of metals by virtue of which metals are drawn in the form of long cables. Silver and copper are the most ductile metals.

Question 38.
Write down the salient physical properties of metals.
Answer:
Following are the salient physical properties of metals:
1. Physical State: Except mercury, all other metals are solid at normal temperature.
2. Metallic lustre: All metals have a peculiar metallic lustre or shining.
3. Structure: In the outermost shell of metals there are 1, 2 or 3 electrons.
4. Conductivity: Metals are generally good conductors of heat and electricity.
5. Malleability and Ductility: Metals are generally malleable and ductile.
6. Solidity: Generally metals are solid. Sodium and potassium are soft metals, they can be cut with knife.
7. Density: Except, sodium and potassium, the density of all the metals generally high.
8. Melting point and Boiling point: The melting points and boiling points of metals are comparatively high.

Question 39.
Give a brief account of information regarding elements available so far.
Answer:
Information regarding elements available so far is as follows :
(i) The number of elements known so far is more than 112. Out of them 92 elements are natural, whereas rest of them are man-made.
(ii) Maximum elements are solids.
(iii) 11 elements are gases at room temperature.
(iv) Two elements, bromine and mercury are liquid at room temperature.
(v) Gallium and cesium can remain in liquid state at the temperature above 303K.

Question 40.
Explain the Conduction method.
Answer:
When the size of the components of mixture is different, then the sieves with different pores are used to separate components. This method is known as conduction method. In separation of bran from flour, separation of cashewnuts in cashewnut industries, separation of different pearls of different sizes by the goldsmiths and separation of different foodgrains by the peasants, in all these cases conduction method is applicable.

Question 41.
What is magnetic separation method ?
Answer:
By making use of magnetic characteristic into real practice to separate the iron fillings or iron particles from a mixture is called as magnetic separation method.
Mineral ore like: iron ore is obtained from underneath the ground by this method.

Question 42.
State briefly the method of sedimentation.
Answer:
At times, in the water from ponds or lakes the suspended particles like sand or soil particles present in the dirty water do not quickly settle down at the bottom that is, it takes times to decant them. Therefore, in order to make such particles settle down at the bottom at the earliest or to make the particles heavier in weight, alum is mixed into the water and water is decanted soon. It is called as sedimentation method. The water of pond or lake can be purified by this method.

Essay Type Questions

Question 1.
Explain the method to obtain different gases from air.
Answer:
Air is a homogeneous mixture of different gases and its components can be separated by fractional distillation. Different stages or phases of this method have been shown diagrammatically ahead:

For instance, if we want to obtain oxygen gas from the air (according to the figure), then we will have to isolate other gases present in the air. In order to obtain liquefied air, initially pressure is increased on the air and then by decreasing temperature and cooling it down and it is compressed. This liquefied gas is further heated up in fractionating distillating column, where all gases get separated in accordance with their boiling points at different altitudes, as it is shown in the fig. below:

Question 2.
Name the method applicable to separate the following mixtures respectively:
(i) Wheat grains, sugar crystals and chaff.
(ii) Rice, grams and iron powder.
(iii) Sand, phaseolies mungo (dal mash) and chaff.
(iv) Sand, camphor and iron powder.
(v) Sand, sugar and iron powder.
Answer:
(i) This mixture can be separated by more than one methods. Mixture of chaff is separated by threshing method. After that the mixture is filtered by immersing it into water. Sugar due to possessing solution quality makes solution in water and the wheat grains leaves behind on the sieve. They are got dried up. Thereafter, solution is heated up in the porcelain dish. On heating water evaporates in the form of steam and sugar leaves behind in the porcelain dish. Hence, all the three components get separated.

(ii) To separate this mixture too more than one methods are applied. Firstly, creep the magnet into the mixture, with the result the iron powder get stuck to the magnet and is got separated. Again, the remaining rice and grams can be separated by filtration method. In this method, rice will leave at the bottom and grams will come up.

(iii) To separate this mixture, at first chaff is separated by threshing method. Sand and phaseolies mungo will be left behind. These can be separated by filtration process.

(iv) Firstly, magnet is moved above the mixture, the iron powder gets stuck to the magnet and thus is got separated. After that, the mixture is separated by sublimation process. Camphor turns into vapours, which get deposited on the inner walls of the funnel when they cool down and sand remains behind in the vessel. Thus, all the three components get separated.

(v) The iron powder is got separated from the mixture by using the magnet. Then mixture is dissolved into water. Sugar gets dissolved in water. Thereafter it is filtered, sugar being solute goes down to the bottom in the form of solution, and the sand leaves behind on the filter-paper. Then this solution is heated up. On heating water evaporates and sugar is left behind in the container.

Question 3.
Explain water supply system in towns diagrammatically, also tell which separating methods are adopted ?
Answer:
In towns, water is supplied by water-works. A flow diagram of typical water works is shown in the figure. In these water works, by means of distillation decantation, sedimentation and filtration processes unwanted material is separated from water. In loading process, you will certainly be reminded of usage of alum. In water works, to kill the harmful germs, chlorine is used. This very’ purified water is supplied in houses through pipe-lines.

Question 4.
What is the difference between ‘distillation’ and ‘fractional distillation’ ?
Answer:
Distillation:
1. Mixture is heated upto the temperature of the component with least boiling pointfor sometime.
2. Only single component evaporates from the mixture at a time.
3. Mixture has to be heated up again and again at different temperatures.
4. The vapours formed by heating the mixture are condensed by passing them through the delivery tube.
5. Pure liquid is not obtained just at the first time.
6. On heating the mixture at different temperatures components are obtained.

Fractional distillation:
1. Mixture is heated upto the temperature of the liquid with maximum boiling point.
2. All the evaporating components of the mixture evaporate at a time.
3. Mixture is only once evaporated by heat process.
4. While heating the mixture, the steam that is formed is sent into the fractionating column.
5. Pure components are obtained just atthe first time.
6. On sending the steam of the mixture into fractionating columns, at different fractionating levels, different fractions are obtained.

Question 5.
Differentiate between the physical properties of metals and non-metals.
Answer:
Following are the differences between the physical properties of metals and non-metals:

Practical Work:

Experiment – 1:
Prepare two homogeneous mixtures in the laboratory.
Procedure:
(i) Take 50 ml of water in a beaker, add two spoonful of salt into it and stir the solution with a spoon. Thus, the obtained mixture will be homogeneous because in it, the particles of solute and solvent cannot be identified separately.
(ii) Take 50 ml of water in a beaker, add one spoon of sugar to it and stir it. Thus, by stirring the obtained solution will be homogeneous mixture, since, in it solute and solvent particles cannot be separately identified

Experiment – 2:
Prove with an experiment that solution is homogeneous.
Procedure:
Take a solution of salt and water. Filter it through filter paper and taste the filtered solution. It will be salty as before. Now, look at the filter paper. It does not have any residue. Thus, it proves that solutions are homogeneous which escape through the filter paper and leaves no residue.

Experiment- 3:
How can components of colour (dye) be separated from blue or black ink ?
Procedure:
Take a beaker half-filled with water, now keep a watch glass on its mouth as shown in the figure. Trickle down a few drops of ink on it. Now, start heating up the beaker. We don’t want to directly heat the ink. You will notice evaporation taking place in the watch glass. Heating is done constantly till evaporation. When we do not notice any further change in the watch glass, then we stop heating. Thus, a blue or black component remains left behind in the watch glass as a residue. In this process in the ink-water coloured mixture water gets evaporated by means of evaporation.

Experiment – 4:
How will you separate iodine and sand from its mixture ?
Procedure:
The mixture of iodine and sand is separated by sublimation process. In this process, the mixture is put into a porcelain dish and a separating funnel is kept on it in an inverted position. The opening of the stem of the funnel is blocked with a cotton plug. Now, this mixture is heated from at the bottom. On heating iodine directly converts into vapour state which on cooling down, gets deposited on to the inner walls of the funnel and the sand is left behind in the dish. Thereby, iodine is scratched down from the funnel. Thus, the components of mixture get separated.

Experiment – 5:
How is acetone and water separated from their solution in the laboratory ?
Procedure:
(i) Pour the mixture into the distillation flask. Joins a thermometer with this.
(ii) Set the apparatus according to the given diagram.
(iii) Gradually heat up the flask and carefully keep eyes on the thermometer.
(iv) Acetone evaporates and on condensing, can be collected into the vessel by condensation after taking
(v)Water is remain left in the distillation tiask. Thus, acetone and water will get separated.

Quick Review of the Chapter

1. Which of the following is a pure material?
(A) sugar
(B) milk
(C) air
(D) pond water
Answer:
(A) sugar

2. The size of the particles in the solution is:
(A) smaller than 10^-9m
(B) smaller than 10^-7m
(C) smaller than 10^-5m
(D) smaller than 10^-2m
Answer:
(A) smaller than 10^-9m

3. The size of the particles in colloid is:
(A) 10^-5m to 10^-4m
(B) 10^-6 to 10^-5m
(C) lCr9mto 10^-7m
(D) more than 10^-7m
Answer:
(C) 10^-9m to 10^-7m

4. The size of particles in suspension is:
(A) more than 10^-5m
(B) more than 10^-7m
(C) more than 10^-4m
(D) more than 10^-3
Answer:
(B) more than 10^-7m

5. An example of suspension is:
(A) salt solution
(B) ink
(C) paint
(D) milk
Answer:
(C) paint

6. Butter is extracted from curd by:
(A) centrifugation method
(B) fractional distillation process
(C) evaporation method
(D) crystallisation method
Answer:
(A) centrifugation method

7. Salt is separated from sea-water by:
(A) sifting/agitation method
(B) evaporation method
(C) centrifugation method
(D) sublimation process
Answer:
(B) evaporation method

8. In a mixture of salt and camphor, camphor and salt are separated by :
(A) evaporation method
(B) centrifugation method
(C) sublimation process
(D) filtration method
Answer:
(C) sublimation process

9. From the mixture of copper sulphate and iron powder, its components can be separated
(A) separating funnel
(B) sublimation process
(C) magnetic process
(D) centrifugation method
Answer:
(C) magnetic process

10. In a mixture of water and oil, water and oil are separated by :
(A) separating funnel
(B) sublimation
(C) evaporation
(D) filtration
Answer:
(A) separating funnel

11. Which of these is not a compound ?
(A) blood
(B) CO₂
(C) methane
(D) soap
Answer:
(A) blood

12. Which of these is not a mixture ?
(A) rock salt
(B) blood
(C) coal
(D) soap
Answer:
(D) soap

13. The dirty pond water is purified by:
(A) decantation
(B) sedimentation
(C) centrifugation
(D) filtration
Answer:
(B) sedimentation

14. In water, solute is:
(A) sand
(B) sulphur
(C) salt
(D) soot
Answer:
(C) salt

15. Solvent present in tincture iodine is:
(A) water
(B) rose elixir (gulabjal)
(C) carbon dioxide
(D) alcohol
Answer:
(D) alcohol

16. Number of known elements is:
(A) 92
(B) 108
(C) 112
(D) 118
Answer:
(C) 112

17. At room temperature number of elements found in gaseous state is:
(A) 06
(B) 09
(C) 11
(D) 23
Answer:
(C) 11

18. Physical change is:
(A) evaporation of water
(B) baking of wheat cake
(C) ripening of fruit
(D) rusting of iron
Answer:
(A) evaporation of water

19. Which of these is a metalloid ?
(A) sodium
(B) silicon
(C) iodine
(D) carbon
Answer:
(B) silicon

20. By which method can the mixture of iodine and sand be separated ?
(A) centrifugation
(B) distillation
(C) sublimation
(D) filtration
Answer:
(C) sublimation

21. In the outermost shell of metals the number of electrons not present is:
(A) 1
(B) 2
(C) 3
(D) 4
Answer:
(D) 4

22. Non-metal which is a good conductor of electricity is :
(A) iodine
(B) sulphur
(C) graphite
(D) phosphorus
Answer:
(C) graphite

23. A matter composed of particles of same kind is called :
(A) pure matter
(B) mixture
(C) solution
(D) impure matter
Answer:
(A) pure matter

24. A mixture whose composition is same, is called :
(A) homogeneous mixture
(B) heterogeneous mixture
(C) colloidal
(D) suspension
Answer:
(A) homogeneous mixture

25. Which of the following is not a homogeneous mixture ?
(A) solution of sugar in water
(B) mixture of sand and salt
(C) solution of salt in water
(D) mixture of water and alcohol
Answer:
(B) mixture of sand and salt

26. The number of naturally occurring elements is:
(A) 112
(B) 22
(C) 92
(D) 108
Answer:
(C) 92

27. Which of the following is not an element ?
(A) sodium
(B) silver
(C) tin
(D) soap
Answer:
(D) soap

28. Which of the following is not a compound ?
(A) calcium carbonate
(C) carbon dioxide
Answer:
(D) air

29. Which one is a mixture ?
(A) methane
(C) blood
Answer:
(C) blood

30. The solution obtained after dissolving a solute in any other liquid besides water is called
(A) non-aqueous solution
(B) homogeneous solution
(C) heterogeneous solution
(D) aqueous solution
Answer:
(A) non-aqueous solution

31. Main constituents of air are:
(A) C02 and Ar
(B) N, and H2
(C) C02 and vapour
(D) He and Ne
Answer:
(B) N2 and H2

32. If the quantity of a solute gets reduced less than saturation in a solution, then it is called
(A) saturated solution
(B) supersaturated solution
(C) unsaturated solution
(D) homogeneous solution
Answer:
(C) unsaturated solution

33. Which the following is property of Metals?
(A) Ductivity
(C) metallic lusture
(B) malleability
(D) all of the above
Answer:
(D) all of the above

34. Which of the following element is non-metal?
(A) Bromine
(C) Silver
(B) Gold
(D) Sodium
Answer:
(A) Bromine

35. Which of the following metal exist in liquid state at room temperature?
(A) Gold
(B) Silver
(C) Mercury
(D) Bromine
Answer:
(C) Mercury

Haryana State Board HBSE 9th Class Science Important Questions Chapter 1 Matter in Our Surroundings Important Questions and Answers.

Haryana Board 9th Class Science Important Questions Chapter 1 Matter in Our Surroundings

Very Short-Answer Type Questions

Question 1.
What is meant by matter?
Answer:
Matter are substances which have mass and which occupy space (volume).

Question 2.
What is every object in the world made up of?
Answer:
Every object in the world is made up of matter.

Question 3.
How many basic elements have the ancient philosophers in India classified matter into ?
Answer:
The ancient philosophers in India classified matter into five basic elements.

Question 4.
What is meant by ‘Panch Tatvas’?
Answer:
Five basic elements-air, earth, fire, water and sky are called as ‘Panch Tatvas’.

Question 5.
How many basic elements did the Greek philosophers agree that matter possesses?
Answer:
The greek philosophers agreed that matter is composed of four basic elements i.e., earth, fire, air and water.

Question 6.
What is the basis of the classification of matter in modern science?
Answer:
In modem science, the basis of classification of the matter is physical properties and chemical nature.

Question 7.
Write down four examples of solid materials.
Answer:
Stone, wood, iron and ice.

Question 8.
Write down four examples of liquid materials.
Answer:
Water, kerosene, spirit and petrol.

Question 9.
Write down four examples of gaseous materials.
Answer:
Oxygen, nitrogen, carbon dioxide and water vapour.

Question 10.
On the basis of physical state how many types of matter are there?
Answer:
On the basis of physical state, matter is available in three states
(i) solid
(ii) liquid
(iii) gas.

Question 11.
How many types of matter are there on the basis of chemical formation?
Answer:
On the basis of chemical formation matter is of three types –
(i) element
(ii) compound
(iii) mixture.

Question 12.
Write the name of the solid state of water.
Answer:
The solid-state of water is ice.

Question 13.
Which are the three different states of water?
Answer:
The three different states of water are –
(i) ice (solid)
(ii) liquid (water)
(iii) Eteam (gaseous).

Question 14.
Write down the physical properties of solids.
Answer:
Solids are completely incompressible, they have definite size, shape and volume.

Question 15.
What are the physical properties of liquids?
Answer:
Liquids are comparatively incompressible fluids. They have a definite volume, but the shape and size is indefinite.

Question 16.
What are the physical properties of gas?
Answer:
Gas is an excessively compressible fluid. The given quantity of gas will fill up in containers of any size and shape. .

Question 17.
What property of gas enables it to inflate maximum air in the type?
Answer:
Due to the property of compressibility of gas.

Question 18.
What is meant by CNG?
Answer:
CNG means Compressed Natural Gas.

Question 19.
What do you mean by intermolecular force?
Answer:
The forces of cohesion applicable in between the particles of substances (molecules or atoms) is called intermolecular force.

Question 20.
Which has the maximum intermolecular force: solid, liquid or gas?
Answer:
Solid has the maximum intermolecular force.

Question 21.
Which has the minimum intermolecular force: solid, liquid or gas?
Answer:
Gas has the minimum intermolecular force.

Question 22.
Why do solids have definite shapes?
Answer:
Due to random intermolecular force solids have definite shapes.

Question 23.
Do the molecules of matter have hollow spaces in between them?
Answer:
Yes, the molecules of matter have fair hollow space in between them.

Question 24.
What is meant by diffusion?
Answer:
Self-intermixing of molecules of two different substances into one another is called diffusion.

Question 25.
Why does the aroma of a burning incense stick spread at a far distance?
Answer:
Due to diffusion, the aroma of a burning incense stick spreads at a far distance.

Question 26.
What characteristic of gases enables us to find the leakage of LPG?
Answer:
Due to the diffusion characteristics of gases leakage of LPG is found out.

Question 27.
How does the flavour of ether and cooking reach us?
Answer:
Gases diffuse in air very quickly. Due to this very property, the flavour of ether and cooking reaches us.

Question 28.
How is the purity of honey tested?
Answer:
If on pouring down a drop of honey in a glass of water the drop of honey creeps down in the form of a coloured streak, then the honey is supposed to be pure otherwise the honey is supposed to be impure.

Question 29.
What effect does the temperature leave on the speed of molecules of matter?
Answer:
With the increase in temperature, the speed of the molecules accelerates.

Question 30.
Give an example of such a solid which changes its shape on stretching it.
Answer:
Rubber band.

Question 31.
Give an example of a compressible solid.
Answer:
The sponge is a compressible solid.

Question 32.
Which one is related to flowing, out of solid or liquid?
Answer:
The liquid is related to flow.

Question 33.
Write down names of two miscible gases.
Answer:
Oxygen and carbon dioxide are miscible gases.

Question 34.
Where do the aquatic animals receive oxygen from to breathe?
Answer:
The aquatic animals make use of oxygen dissolved in water to breathe.

Question 35.
Which gas does the balloon-seller fill into the balloon ?
Answer:
The balloon-seller fills hydrogen gas in the balloon.

Question 36.
Which gas is used in houses to cook?
Answer:
Liquefied Petroleum Gas (LPG).

Question 37.
Which gas cylinders are used in hospitals for artificial respiration?
Answer:
Oxygen gas cylinders are used in hospitals for artificial respiration.

Question 38.
In solids, liquids and gases which one has the maximum compressibility property?
Answer:
Gases have the maximum compressibility.

Question 39.
What is called the irregular speed of dust particles in air?
Answer:
The irregular speed of dust particles in air is called Brownian speed.

Question 40.
What happens when ice is heated up?
Answer:
When ice is heated up, it converts into (liquid) water.

Question 41.
What is formed when liquid (water) is heated up?
Answer:
On heating up liquid (water) vapours (steam) is formed.

Question 42.
Out of solid, liquid or gas in which case molecules are totally free to move?
Answer:
In gas the molecules are totally free to move.

Question 43.
What is the use of stirring a liquid while heating it up?
Answer:
While heating up if the liquid is stirred constantly it will get heated equally overall.

Question 44.
Write down name of the device to measure temperature.
Answer:
Thermometer.

Question 45.
What is meant by melting point?
Answer:
That fixed temperature at which a solid becomes liquid on melting is called melting point.

Question 46.
What is SI unit of temperature?
Answer:
SI unit of temperature is ‘Kelvin’ (K).

Question 47.
What is done to convert Kelvin temperature into Celsius temperature?
Answer:
In order to convert Kelvin temperature into Celsius temperature 273 is to be subtracted from the given temperature.

Question 48.
Convert 373 K into Celsius temperature.
Answer:
373 K = (373 – 273)° C = 100° C

Question 49.
How is celsius temperature converted into Kelvin temperature?
Answer:
To convert celsius temperature into Kelvin temperature 273 is added to the given temperature.

Question 50.
Convert 0° C into Kelvin temperature.
Answer:
0° C = (0 + 273) K = 273 K

Question 51.
What is the melting point of ice?
Answer:
The melting point of ice is 273.16 K.

Question 52.
What is meant by melting?
Answer:
The process of melting, i.e., conversion of a solid into a liquid state is called melting.

Question 53.
What is meant by the dormant heat energy of melting?
Answer:
At atmospheric pressure, the heat energy that is required to convert 1 kg of solid at its melting point into a liquid state, is called dormant heat energy of melting.

Question 54.
Which of the two, water or ice’s molecules will possess more energy at 0° C?
Answer:
At 0° C, the molecules of water will possess more energy than ice.

Question 55.
What is meant by boiling point ?
Answer:
At atmospheric pressure, the temperature at which the liquid starts boiling is called boiling point.

Question 56.
What is the boiling point of water?
Answer:
The boiling point of water is 373 K.

Question 57.
What is meant by in exposed evaporation or dormant heat energy of evaporation?
Answer:
At atmospheric pressure, the heat energy that is required to convert 1 kg of liquid at its boiling point into water vapours, is called inexposed evaporation or dormant heat.

Question 58.
Which of the two, steam or water will render more burning sensation at 373 K temperature?
Answer:
At 373 K temperature, steam will render more burning sensation than water, for its molecules possess yet more in exposed heat of evaporation.

Question 59.
What is meant by sublimation?
Answer:
Without changing into liquid state, the process of changing of solid, directly into gaseous and again in solid state is called sublimation.

Question 60.
Write names of two sublime substances.
Answer:
(i) Camphor
(ii) ammonium chloride.

Question 61.
What happens when pressure is increased on any gas?
Answer:
On increasing pressure on any gas it changes into liquid.

Question 62.
What is meant by dry ice?
Answer:
Dry carbon dioxide is called dry ice.

Question 63.
What is the SI unit of pressure?
Answer:
The SI unit of pressure is Pascal (Pa).

Question 64.
What is the atmospheric pressure on the sea surface?
Answer:
On the sea surface the atmospheric pressure used to be 1 atmosphere.

Question 65.
What is meant by ‘evaporation’?
Answer:
The process of changing of liquid into vapours below the temperature of boiling point is called evaporation.

Question 66.
Write down one advantage of evaporation in daily life.
Answer:
In daily life, wet clothes dry up because of evaporation.

Question 67.
What is the effect of the level region on evaporation?
Answer:
An increase in the level region, the evaporation rate increases.

Question 68.
What is the relation of evaporation with temperature?
Answer:
With the increase in temperature, the evaporation rate increases.

Question 69.
What is meant by humidity?
Answer:
The quantity of water vapours present in the air is called humidity.

Question 70.
What is the effect of humidity on evaporation?
Answer:
With the increase in humidity, the evaporation rate decreases.

Question 71.
Why do people sprinkle water on their roofs and in the open places after the scorching bright sunshine day comes to an end?
Answer:
Because the dormant heat of evaporation cools down the hot surface.

Question 72.
Which of the following substances will you expect to possess the strongest and the weakest intermolecular force: water, alcohol, sugar, sodium chloride, carbon dioxide?
Answer:
Sodium chloride will have the strongest and carbon dioxide will have the weakest intermolecular force.

Question 73.
Compression in gases is possible, whereas that of liquids is not. Why is it so?
Answer:
Because of too much intermolecular gap distance in gases, they can be compressed. But in liquids, due to less intermolecular gas the distance they cannot be compressed.

Short-Answer Type Questions

Question 1.
How has the classification of matter been done in ancient period in context with the modern science?
Answer:
Classification of matter:
(1) In ancient times, the ancient philosophers had the concept that matter is composed of five basic elements i.e.,
(i) air
(ii) earth
(iii) fire
(iv) water
(v) sky (all were said to be Panch Tatvas)
all the living or non-living things according to them were bom out of these five elements, whereas the Greek Philosophers are of the view that there have been only four elements in all i.e.,
(i) air
(ii) earth
(iii) fire and
(iv) water.
(2) The modem scientists on the other hand, on the bases of physical properties and chemical nature have classified matter into two kinds.

Question 2.
How is the classification of matter done according to the physical properties?
Answer:
According to physical properties matter are divided into three categories:
1. Solids: Solids are completely incompressible, of definite shape and volume; like stone, wood, iron, ice, salt, sugar, etc.
2. Liquids: Liquids are comparatively less incompressible. They have a definite volume but have an indefinite shape like water, kerosene, spirit, petrol, milk, etc.
3. Gases: Gas is an excessively compressible fluid. Its volume and shape is indefinite which means the gas of a given quantity can be filled in any container of any shape like oxygen, nitrogen, carbon dioxide, water vapours, methane, etc.

Question 3.
Write four special properties of solid matter.
Answer:
Special properties of solid matter are as follows:
(1) They can be collected in the form of piles.
(2) They have a definite volume.
(3) They can be scratched.
(4) Solid substances are hard to touch.
(5) They have definite shapes.

Question 4.
What are the general properties of liquid matter?
Answer:
The general properties of liquids are as follows:
(1) Liquids cannot be placed in piles.
(2) All liquids can flow.
(3) Liquids cannot be scratched.
(4) All liquids have definite volumes.
(5) All liquids form the shape of that very container in which they are kept that is they do not have a definite shape.

Question 5.
Which of the properties are common in all in solids, liquids and gases?
Answer:
The following properties are common in all in solids, liquids and gases:
(1) All are composed of molecules and atoms collectively.
(2) All occupy space.
(3) All have mass.
(4) All can be felt through our senses.

Question 6.
Give the general properties of gases.
Answer:
The general properties of gases are:
(1) Gases do not have definite shape and volume.
(2) Gases dissolve into air.
(3) Gases cannot be scratched.
(4) Gases can be collected only in enclosed containers.
(5) The gas spreads in the container in which it is kept, it spreads at that very place where it is released.
(6) Gases can flow.

Question 7.
Clarify your answer with suitable examples that the properties of all the liquid matter do not resemble one another.
Answer:
The following examples make it clear that the properties of all the liquid matter do not resemble with one another:
(1) Some liquids have low ignition point like petrol, but there are yet certain liquids which do not catch fire like water.
(2) Some liquids are comparatively heavier in weight like mercury. Some liquids are not too heavier like kerosene oil.
(3) Some liquids flow’ quickly like water. Some liquids flow slowly, for example, molten sugar and honey.
(4) Some liquids have the property to saturate salt, sugar, etc. in them. Some liquids cannot saturate these in them like mustard oil.
(5) Some liquids are colourless like kerosene oil, whereas some liquids are coloured like mustard oil.
(6) Some liquids evaporate earlier like spirit. Some liquids evaporate later like turpentine oil.

Question 8.
Compare the intermolecular force in molecules or atoms in solids, liquids and gases.
Answer:
In solids, molecules or atoms are densely and compactly arranged and they are tied up with one another by means of force of attraction whereas in liquids, atoms or molecules are thinly and loosely arranged. In solids, molecules or atoms oscillate about their respective positions, while in liquids, molecules or atoms can move about here and there within their fixed path. In gases, molecules or atoms are quite far away from one another and they move randomly. These molecules or atoms are free to move about within maximum space.

Question 9.
Clarify the reason of generating of pressure in gases.
Answer:
Since in a gaseous state, the motion of molecules is free at all. they can move about in a container, they can collide with one another and also, with the walls of the container. Due to the regular collision of them against the walls, a definite force gets produced. Consequently, pressure is generated in gases.

Question 10.
When we turn ice into water by heating process and further convert water into steam by heating it up, then what differences do we observe in the compactness of the molecules of water ?
Answer:
Ice is a solid substance and its molecules are associated with one another due to the force of cohesion. On heating ice, its molecules get active and start moving on receiving energy with the result, their force of cohesion gets loosened and molecules start flowing that means the solid ice forms the shape of liquid. When water is heated up the random speed of molecules speeds up very much and they generally leave apart from one another to a great extent and become free to move about that is, their intermolecular force weakens too much, such a state of water is said to be steam (gas).

Question 11.
Fill up the blanks choosing the suitable word or words from the list given below: too much, definite, another, very closely, freely, move, container, force, attraction, shape
(i) In gaseous state the molecules move.
(ii) In liquid state molecules can in it.
(iii) Molecules of a solid are packed and can move at a speed.
(iv) The containers in which the liquids are kept, they form the of them.
(v) The gap distance between the molecules of a gas is
(vi) In solids the molecules are joined together with strong forces of.
Answer:
(i) freely
(ii) move
(iii) very closely, definite
(iv) shape
(v) too much
(vi) attraction.

Question 12.
Match the statements given in column A by selecting appropriate words or phrases from column B:

Answer:

Question 13.
The shape of a rubber band can be changed by stretching it, but even then it is not considered as a solid, why ?
Answer:
The shape of the rubber band can be changed by stretching it, but even then it is not considered as a solid, it is because of the reason that on applying external force the rubber band changes its shape and on the removal of the external force it again retains its original shape, whereas it breaks up if the applied force is beyond a limit.

Question 14.
Sugar and salt form the shape exactly of the vessels of different shapes in which they are kept, but still they are considered as solids, why ?
Answer:
Sugar and salt form the shape of the vessels of different shapes in which they are kept, but still they are considered as solids, it is because the shape of their crystals do not change.

Question 15.
A sponge, even being a solid gets compressed, why?
Answer:
Even being a solid, a sponge gets compressed, it is because the sponge consists of small pores in it, which keep trapped air in them. When we press or compress it, the air escapes out from these pores and the sponge gets compressed.

Question 16.
In fluids, diffusion of three states (solid, liquid and gas) is possible, why?
Answer:
In fluids, diffusion of solids, liquids and gases is possible. In liquids the diffusion rate is comparatively higher, in a liquid state, the molecules of a substance move freely and in comparison to solids, the molecules of liquids occupy more vacant space.

Question 17.
The temperature remains equal when a solid undergoes the process of melting, where does the heat energy go during the process?
Answer:
During the melting process any solid (like ice), on reaching the melting point, till the solid melts as a whole, the temperature does not change. Despite providing heat to the beaker, it happens so. By bringing under control the intermolecular force of the molecules, while in changing the state of the matter this heat energy is utilised, because without indicating any increase in the temperature, this heat energy is absorbed by the solid (ice), this thing is assumed that it remains hidden in the material of the beaker, which is called latent heat. At atmospheric pressure, the amount of heat energy that is required to convert 1 kg of solid at its melting point into liquid, is called dormant heat energy of melting, at 0° centigrade (273 K) the energy of the molecules of water is more than that of the molecules of ice at a similar temperature.

Question 18.
Convert the following Celsius measures into Kelvin measuring sequence:
-273°C; -100°C; -40°C; 30°C; 2000°C
Answer:
We know that K = °C + 273
Therefore,
(i) -273°C = -273 + 273 = 0 K
(ii) -100°C = -100 + 273 = 173K
(iii) – 40°C = – 40 + 273 = 233 K
(iv) 30°C = 30 + 273 = 303 K
(v) 2000°C = 2000 + 273 = 2273

Question 19.
Air, at 82 K temperature changes into liquid and at 61K temperature, converts into solid Convert these temperatures in Celsius measuring sequence:
Answer:
We know that
(i) While keeping K = 82
K = °C + 273
82 = °C + 273
⇒ °C = 82 – 273 -191
So, the temperature of the liquefied air = 82 K = -191°C

(ii) Assuming K = 61
61 = °C + 273
⇒ °C = 61 – 273 = -212
So, the temperature of solidified air = 61 K = – 212°C

Question 20.
Why is ice, with a temperature 0°C more effective for cooling than water at 0°C ?
Answer:
We know, the dormant heat energy of melting of ice is 335 J/g. Therefore, ice with 0°C does have comparatively more heat than water at 0°C. Hence, ice with 0PC is more effective than the water with 0°C for cooling.

Question .21.
Why does our body feel cooler in the Himalayan region during winters?
Answer:
In the Himalayan region during winters, the temperature of environment drops below 0°C, whereas the normal temperature of our body used to be 37°C or 98.6° F. Just because of this drastic difference in temperature we feel more chilled in those regions during winters. Covering the body with woollen clothes the inner heat of the body gets maintained and the low temperature outside does not affect our body.

Question 22.
What is meant by evaporation and boiling ?
Answer:
Evaporation: During this process the liquid converts into water vapours before it gets boiled. Evaporation generally occurs on the outer surface of a liquid at all temperatures.

Boiling: When the pressure of any liquids surface equalises the atmospheric pressure, then the liquid starts converting into vapours. This process is called the boiling of a liquid.

Question 23.
Why do the wet clothes dry up sooner if they are spread out ?
Answer:
The speed of evaporation depends upon the surface of the wet article. When the wet clothes are spread out to dry up, the size of the wet surface area gets increased with that evaporation takes place fastly. For that reason the wet clothes dry up earlier when they are spread out and put to dry.

Question 24.
When during winters the water in the tank thaws, then how the fishes living in the tank survive ?
Answer:
During chilly winter days the water in the tank thaws at 0°C into ice, but in this process, only the outer layer of the water surface freezes and the water below to that layer does not thaw. Due to this non-uniform expansion when water is cooled down below 4°C temperature, it instead of contracting expands up to 0°C. For that reason, water thaws into lighter ice and ice start floating on the water. The fish found in such water and the other aquatic animals move to the liquid water below to ice and they survive there.

Question 25.
Where do water vapours come from in the atmosphere?
Answer:
When we put wet clothes in the sun, they dry up sooner. The water present in the fibres of the clothes on evaporating becomes a part of the atmosphere. Similarly, water evaporates in a huge quantity from lakes, rivers and oceans Air cannot sustain water vapours at a given temperature more than a fixed amount. When there is the maximum amount of water vapours in air, the air is said to be quenched.

Question 26.
Why do people sprinkle water on floor and roofs during summers ?
Answer:
People sprinkle water on floor and roofs during summer because water evaporates for which it gets heat from the floor and the roof and thereby keeps them cool. Therefore to keep floor and roofs cool, people sprinkle water on them in summer.

Essay Type Questions

Question 1.
Write down the difference in the following :
(i) heat and temperature
(ii) boiling and evaporation
Answer:
(i) Following are the differences between heat and temperature
Heat:
1. Heat is that form of energy in which we feel the sensation of hot and cool.
2. This is a form of energy.
3. It is measured in calorie or kilocalorie. SI unit is joule (J).
4. It depends upon the form, temperature and nature of a substance.
5. It is a reason.
6. It is the amount of energy in a substance.
7. It is measured by calorimeter.

Temperature:
1. This is that property of a substance which fixes the flow of heat.
2. This is a state by which we come to know the direction of flow of heat.
3. It is measured in parts1 like centigrade and kelvin.
4. It does not depend upon these things.
5. It is the effect of heat.
6. It is a physical property of a substance.
7. It is measured with thermometer.

(ii) Following are the differences between boiling and evaporation
Boiling:
1. It is a random process.
2. It exists only at a fixed temperature.
3. It exists in the entire liquid.
4. This process produces sound.
5. Coolness does not produce in it.
6. Steam is produced in it which is visible.
7. The properties of liquids, open base and the direction of wind do not affect this process.
8. In this process, bubbles are seen moving up wards.

Evaporation:
1. It is a slow process.
2. It exists at all temperatures.
3. It exists only at the bottom level of the liquid.
4. This is a mute process.
5. Coolness produces in it.
6. Water vapours are formed which get mixed into air. Water vapours are invisible.
7. This process depends upon all these things.
8. In this process, bubbles are not produced.

Question 2.
On what factors does the process of evaporation depend ?
Answer:
The process of evaporation is affected by the following factors:
(1) On being the air become dr}’, the evaporation process goes rapidly.
(2) Also, due to fast speed of wind the process of evaporation speeds up.
(3) Even when the base of the liquid is open the process of evaporation will take place randomly.
(4) With the increase in the temperature of the liquid evaporation process will be faster.
(5) If the temperature of air increases the speed of the evaporation process will be rapid.
(6) In more volatile liquids like spirit, petrol, etc. evaporation takes place quickly.
(7) Due to low pressure evaporation process accelerates.

Question 3.
Why does coolness occur with the evaporation of water ? Give some of its uses in our daily life.
Answer:
Coolness occurs due to evaporation, because when water evaporates, it needs latent heat energy of evaporation. This heat can be obtained from that substance which is in the contact of water, thus that substance becomes somewhat cool.

Uses of Evaporation in Daily Life:
(1) Water remains cool in an earthen pot, but not in a bucket, because there are pores in the pot. Water seeps out of them and it evaporates which produces coolness. Thus, water remains cool.
(2) During summer the trees bear more leaves and from the stomata of these leaves evaporation takes place, as a result coolness produces and the trees remain cool.
(3) During perspiration, fanning brings the sensation of coolness, because by fanning the sweat evaporates which produces coolness. For evaporation of sweat, heat is obtained from the body itself.
(4) On sprinkling water on the ground there is coolness for when evaporation of water takes place, then it gains heat from the earth, so the earth becomes cool.
(5) After bathing we feel cool, it is because the body is wet. The evaporation of water takes place and for evaporation, heat is obtained from the body, hence we feel cold.

Practical Work

Experiment 1:
Prove through an activity that matter are composed of molecules and molecules have vacant space or gap in between them.

Procedure:
Take a graduated measuring cylinder. Fill % part of it with water and note down the surface level. Now add some sugar to it and note down its surface level. You will notice that the surface level of water has increased somewhat.

Now, stir well sugar in the water. Sugar will dissolve in water and the surface level of water will again reduce. This proves, that there is gap between molecules of water. While stirring the molecules of sugar-filled up that vacant space, consequently, the water level inside the cylinder was reduced. Since matter is composed of molecules, hence sugar got dissolved in the whole of the water.

Experiment 2:
Prove experimentally, the molecules of matter are so small in size, that we cannot even imagine.

Procedure:
Take a beaker, add 100 ml of water in it. Add two or three crystals of potassium permanganate and dissolve. Draw out approximately 10 ml of solution from the beaker, add 90 ml of pure water to it. Again draw out 10 ml of solution and add 90 ml of pure water to it. Repeat this process 5 to 8 times. You will notice, the water will still remain coloured. This experiment shows that with very small amount of crystals of potassium permanganate, the huge amount of water (1000 litres) too becomes coloured. Thus, we come to the conclusion that just one crystal of potassium permanganate will have a number of minute molecules. They are so minute that we cannot even imagine.

Experiment 3:
Prove experimentally that diffusion takes place more rapidly in gases than in liquids. Procedure: Make a preparation of concentrated film by adding a spoonful of potassium permanganate in 100 ml of water. Keep this solution in the beaker. Now, keep the beaker in a little bit tilted position and add 100 ml of water to it. While adding water to the beaker, be careful that the water should flow down along the wall of the beaker in such a way, so that the concentrated solution of potassium permanganate remains undisturbed. After some time you will observe two layers are formed separately which are quite clear.

Nevertheless, you will see the water gradually flowing into coloured film and the coloured film gradually flowing into water. After some time both the layers will not be seen isolated quite clearly. Note down the tune taken by both the layers to mix up with each other. Now, light a candle in one of the comers in your classroom. You stand far away at the other comer. Now, ask one of your friends to light a few incense sticks. You will very soon experience the fragrance of the incense sticks. Note down the time taken right from the time of lighting of the incense sticks, till the time taken by fragrance to reach to you. We will notice that in the latter stage total time taken was very less than that in the former stage. It proves that in gases diffusion was more rapid than in the liquids.

Experiment 4:
Illustrate experimentally, what effect does pressure have on solids, liquids and gases.

Procedure:
Take a syringe of 100 ml volume. Insert its needle (nozzle) into a rubber cork and close it up, as shown in the figure. Take away the piston and let air fill completely inside the syringe. Now, insert the piston back into the syringe carefully and make sure there is no leakage around the ends of the syringe. It would be better to apply a little bit of vaseline on the piston. Now, try to compress the air. You will notice that air will be compressed.

Now, fill the syringe with water and again repeat the same process. In the experiment done with water you will notice in water, compression is comparatively less than in the first one. Now, repeat the same experiment with piece of chalk in place of water, there will be no compressibility. On the bases of this experiment, we can say gases have maximum effect of pressure, liquids have lesser than that and on solids, it is almost negligible; that is, gases are the most compressible.

Experiment 5:
Experimentally show that with the change in temperature, the state of matter changes.

Procedure:
Take a piece of ice of 150 gm in a beaker and according to the figure hang the thermometer used in the laboratory into it such that the bulb of the thermometer should touch the ice. Start heating the beaker on mild flame. When the ice starts melting, then note down the temperature.
(1) When the whole of the ice change into water, then again note down the temperature.
(2) Note down the downfall in temperature in the change of solid into a liquid state.
(3) Now, place a glass rod into the beaker and heat it while stirring, till the water boils.
(4) Keep a constant eye on the degree of temperature in the thermometer, till most of the water converts into water vapours.
(5) Note down the drop in temperature right from the conversion of water in a liquid state into the gaseous state

Experiment 6:
Experimentally show that ammonium chloride is a sublime substance.

Procedure:
Take a small amount of powdered ammonium chloride in a dish. Place a funnel in an inverted position on this dish. Insert a cotton plug into the mouth of the stem of the funnel as shown in the figure. Now gently heat the dish carefully. We will notice, that ammonium chloride, without getting converted into a liquid state gets converted directly from a solid state to a gaseous state and gets deposited on the walls of the funnel. This proves that ammonium chloride is a sublime substance.

Quick Review of the Chapter

Question 1.
According to Creek philosophers, which fundamental element is not included in the matter?
(A) sky
(B) earth
(C) fire
(D) air and water
Answer:
(A) sky

Question 2.
The solution in the following is:
(A) ocean water
(B) soda water
(C) air
(D) pond water
Answer:
(B) soda water

Question 3.
On the basis of physical state, the types of matter are:
(A) elements
(B) compounds
(C) mixtures
(D) none of the above
Answer:
(D) none of the above

Question 4.
On the basis of chemical composition, the types of matter are:
(A) solids
(B) liquids
(C) gases
(D) noneoftheabove
Answer:
(D) none of the above

Question 5.
S.l. unit of weight is:
(A) Kelvin
(B) Newton
(C) Meter
(D) Pascal
Answer:
(B) Newton

Question 6.
A balloon-seller fills the balloons with:
(A) oxygen gas
(B) hydrogen gas
(C) nitrogen gas
(D) carbon gas
Answer:
(B) hydrogen gas

Question 7.
The property of compressibjIjt is highly present:
(A) in a wooden block
(B) in sponge
(C) in water
(D) in hydrogen gas
Answer:
(D) in hydrogen gas

Question 8.
Dry ice means:
(A) compressed oxygen gas
(B) compressed nitrogen gas
(C) solid carbon dioxide
(D) Acetone
Answer:
(C) solid carbon dioxide

Question 9.
To convert a kelvin temperature into a celsius temperature, what should we abstract from kelvin temperature?
(A) 80
(B) 212
(C) 273
(D) 373
Answer:
(C) 273

Question 10.
The melting point of Ice In kelvin is:
(A) 0K
(B) 80K
(C) 212K
(D) 273.16K
Answer:
(D) 273.16 K

Question 11.
Whose compression is possible?
(A) solid
(B) liquid
(C) gas
(D) none of the above
Answer:
(C) gas

Question 12.
Which cannot pile as a heap?
(A) ice
(B) salt
(C) water
(D) sugar
Answer:
(C) water

Question 13.
Which substance gets spread ¡n the complete/whole pot?
(A) ‘CC
(B) petrol
(C) kerosene
(D) methane
Answer:
(D) methane

Question 14.
Which liquid flows quickly?
(A) molasses
(B) mustard oil
(C) water
(D) honey
Answer:
(C) water

Question 15.
Which prepares liquid quickly?
(A) turpentine oil
(B) mustard oil
(C) spirit
(D) diesel
Answer:
(C) spirit

Question 16.
The normal temperature of our body Is:
(A) 370 F
(B) 37.6° F
(C) 80°F
(D) 98.6° F
Answer:
(D) 98.6° F

Question 17.
98.6°Fisequaito:
(A) 20°C
(B) 25°C
(C) 37°C
(D) 73°C
Answer:
(C) 37° C

Question 18.
Below which temperature at the cooling of water spreads instead of shrinking?
(A) 0°C
(B) 4°C
(C) 37° C
(D) none of the above
Answer:
(B) 4° C

Question 19.
Heat is measured:
(A) in centigrade
(B) in kelvin
(C) in calorie
(D) in Fahrenheit
Answer:
(C) in calorie

Question 20.
Diffusion is possible in:
(A) liquid
(B) solid
(C) gas
(D) all of the above
Answer:
(D) all of the above

Question 21.
Which substance will not cause a feeling of coolness when pouring on the palm?
(A) acetone
(B) petrol
(C) perfume
(D) honey
Answer:
(D) honey

Question 22.
The substance found in nature ¡n free state ¡n all the three states (solid, liquid, gas) is:
(A) petroleum
(B) sulphur
(C) water
(D) oxygen
Answer:
(C) water

Question 23.
Scientifically the materials by which all the things of the universe is formed, are called:
(A) wealthy
(B) resources
(C) substances
(d)) materials
Answer:
(C) substances

Question 24.
Intermixing of particles of two different substances by themselves is called:
(A) spreading
(B) diffusion
(C) compression
(D) smelling
Answer:
(B) diffusion

Question 25.
The gas which ¡s formed by compression of butane at high pressure for use in homes for making food, Is called:
(A) Liquefied Petroleum Gas
(B) Hydrogen Gas
(C) Compressed Natural Gas
(D) Natural Gas
Answer:
(A) Liquefied Petroleum Gas

Question 26.
The energy that is required to convert 1 kg of s solid st atmospheric pressure at its melting point into
liquid is called:
(A) latent heat of melting (B) latent heat of vaporisation
(C) latent heat of liquefication (D) none of these
Answer: (A) latent heat of meltmg

Question 27.
The energy that is required to convert 1 kg of liquid at atmospheric pressure at its boiling point is called:
(A) latent heat of fusion
(B) latent heat of vaporisation
(C) latent heat of Jiquefication
(D) none of the above
Answer:
(B) latent heat of vaporisation

Question 28.
Which energy Is present in particles of matter?
(A) magnetic energy
(B) static energy
(C) kinetic energy
(D) electrical energy
Answer:
(C) kinetic energy

Question 29.
What is called, that definite temperature at which a liquid starts changing into a solid?
(A) freezing point
(B) boiling point
(C) melting point
(D) ignition point
Answer:
(A) freezing point

Question 30.
The process ¡n a liquid changes into a vapour state below the temperature of its boiling point is called:
(A) meltinisation
(B) vaporisation
(C) fusionism
(D) sublimation
Answer:
(B) vaporisation

Question 31.
In which state the volume and size are not definite?
(A) solid
(B) liquid
(C) gas
(D) all of the above
Answer:
(C) gas

Question 32.
The property of hardness finds:
(A) in solids
(B) in liquids
(C) in gases
(D) in all of – liquid and gases
Answer:
(A) in solids

Question 33.
Whose compression is possible in the following?
(A) stone
(B) aluminium
(C) brick
(D) sponge
Answer:
(D) sponge

Question 34.
Lowest compression occurs:
(A) in solids
(B) in liquids
(C) in gases
(D) in gases and liquids
Answer:
(A) in solids

Question 35.
Highest compression occurs:
(A) in solids
(B) in liquids
(C) in gases
(D) in solids and liquids
Answer:
(C) in gases

Question 36.
After changing 25°C into kelvin we get:
(A) 248 K
(B) 298 K
(C) – 248 K
(D) -298 K
Answer:
(B) 298 K

Question 37.
Among water, sugar and oxygen which has the maximum intermolecular force?
(A) oxygen
(B) water
(C) sugar
(D) oxygen and water
Answer:
(C) sugar

Question 38.
What is the physical state of water at 100° C?
(A) Solid
(B) Liquid
(C) Gas
(D) Both (B) and (C)
Answer:
(C) Gas

Question 39.
Which of the following is sublimation matter?
(A) sodium chloride
(B) ammonium chloride
(C) calcium chloride
(D) all of the above
Answer:
(B) ammonium chloride