Class 9

Haryana State Board HBSE 9th Class Maths Solutions Chapter 1 संख्या पद्धति Ex 1.5 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 1 संख्या पद्धति Exercise 1.5

प्रश्न 1.
बताइए नीचे दी गई संख्याओं में कौन-कौन परिमेय हैं और कौन-कौन अपरिमेय हैं

हल :
(i) 2 – \(\sqrt{5}\) = 2 – 2.236…….

क्योंकि यह एक असांत तथा अनावर्ती संख्या है।
इसलिए 2 – \(\sqrt{5}\) एक अपरिमेय संख्या है। उत्तर

(ii) \((3+\sqrt{23})-\sqrt{23}=3+\sqrt{23}-\sqrt{23}\) = 3
जोकि एक परिमेय संख्या है। उत्तर

(iii)

क्योंकि यह एक असांत तथा आवर्ती संख्या है।
इसलिए यह एक परिमेय संख्या है। उत्तर ।

(iv) \(\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}=\frac{1.4142 \ldots \ldots}{2}\) = 0.7071………

क्योंकि यह एक असांत तथा अनावर्ती संख्या है। इसलिए यह एक अपरिमेय संख्या है। उत्तर

(v) 2π = 2 × 3.14……… = 6.28………
क्योंकि यह एक असांत व अनावर्ती संख्या है।
इसलिए यह एक अपरिमेय संख्या है। उत्तर

प्रश्न 2.
निम्नलिखित व्यंजकों में से प्रत्येक व्यंजक को सरल कीजिए

हल :

प्रश्न 3.
आपको याद होगा कि श को एक वृत्त की परिधि (मान लीजिए c) और उसके व्यास (मान लीजिए d) के अनुपात से परिभाषित किया जाता है, अर्थात π = \(\frac {c}{d}\) है । यह इस तथ्य का अंतर्विरोध करता हुआ प्रतीत होता है कि π अपरिमेय है। इस अंतर्विरोध का निराकरण आप किस प्रकार करेंगे ?
हल :
इसका कोई अंतर्विरोध नहीं है। स्मरण रहे कि जब कभी-भी एक स्केल से या किसी अन्य युक्ति से लंबाई मापते हैं, तब आपको केवल एक सन्निकट परिमेय मान प्राप्त होता है। अतः आप यह अनुभव नहीं कर पाते कि c या d अपरिमेय है।

प्रश्न 4.
संख्या रेखा पर \(\sqrt{9.3}\) को निरूपित कीजिए।
हल :
सर्वप्रथम संख्या रेखा l पर धनात्मक एवं ऋणात्मक संख्याएँ लीजिए। फिर बिंदु O पर 0 प्रदर्शित करें। इसी प्रकार – 9.3 बिंदु A पर तथा 1 बिंदु B पर है।

अब एक अर्द्धवृत्त AB व्यास पर खींचते हैं। बिंदु O पर एक लंब खींचो जो अर्द्धवृत्त को बिंदु C पर काटे।
अब केंद्र O से OC त्रिज्या पर एक चाप प्रदर्शित करते हैं, जो बिंदु P = \(\sqrt{9.3}\) को प्रदर्शित करती है। उत्तर

प्रश्न 5.
निम्नलिखित के हरों का परिमेयकरण कीजिए

हल :

Haryana State Board HBSE 9th Class Maths Solutions Chapter 1 संख्या पद्धति Ex 1.4 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 1 संख्या पद्धति Exercise 1.4

प्रश्न 1.
उत्तरोत्तर आवर्धन करके संख्या रेखा पर 3.765 को देखिए।
हल :
3.765 को संख्या रेखा पर निम्न प्रकार आवर्धित किया जा सकता है-

प्रश्न 2.
4 दशमलव स्थानों तक संख्या रेखा पर \(4 . \overline{26}\) को देखिए।
हल :
\(4 . \overline{26}\) को संख्या रेखा पर निम्न प्रकार से प्रदर्शित किया जा सकता है-

Haryana State Board HBSE 9th Class Maths Solutions Chapter 1 संख्या पद्धति Ex 1.3 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 1 संख्या पद्धति Exercise 1.3

प्रश्न 1.
निम्नलिखित भिन्नों को दशमलव रूप में लिखिए और बताइए कि प्रत्येक का दशमलव प्रसार किस प्रकार का है –
(i) \(\frac {36}{100}\)
(ii) \(\frac {1}{11}\)
(iii) 4\(\frac {1}{8}\)
(iv) \(\frac {3}{13}\)
(v) \(\frac {2}{11}\)
(vi) \(\frac {329}{400}\)
हल:
\(\frac {36}{100}\) = 0.36; यह दशमलव प्रसार सांत है।

(ii) \(\frac {1}{11}\)

अतः \(\frac {1}{11}\) = 0.0909…… = \(0 . \overline{09}\);
यह दशमलव प्रसार अनवसानी पुनरावृत्ति है।

(iii) 4\(\frac {1}{8}\)

अतः 4\(\frac {1}{8}\) = 4.125
यह दशमलव प्रसार सांत है।

(iv) \(\frac {3}{13}\)

अतः \(\frac {3}{13}\) = \(0 . \overline{230769}\) ; यह दशमलव प्रसार अनवसानी पुनरावृत्ति है। (पुनरावृत्ति)

(v) \(\frac {2}{11}\)

अतः \(\frac {2}{11}\) = \(0 . \overline{18}\) ; यह दशमलव प्रसार अनवसानी पुनरावृत्ति है।
(पुनरावृत्ति)

(vi) \(\frac {329}{400}\)

अतः \(\frac {329}{400}\) = 0.8225, यह दशमलव प्रसार सांत है।

प्रश्न 2.
आप जानते हैं कि \(\frac {1}{7}\) = \(0 . \overline{142857}\) है। वास्तव में, लंबा भाग दिए बिना क्या आप यह बता सकते हैं कि \(\frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}\) के दशमलव प्रसार क्या हैं? यदि हाँ, तो कैसे?
हल :
हाँ, हम \(\frac {1}{7}\) के मान की सहायता से लंबा भाग किए बिना व \(\frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}\) का मान क्रमशः 2, 3, 4, 5 व 6 से \(\frac {1}{7}\) के मान को गुणा करके ज्ञात कर सकते हैं अर्थात

प्रश्न 3.
निम्नलिखित को \(\frac {p}{q}\) के रूप में व्यक्त कीजिए, जहाँ p और q पूर्णाक हैं तथा q ≠ 0 है-
(i) \(0 . \overline{6}\)
(ii) \(0 . 4\overline{7}\)
(iii) \(0 . \overline{001}\)
हल :
(i) माना x = \(0 . \overline{6}\)
⇒ x = 0.6666………
दोनों ओर 10 से गुणा करने पर
या 10x = 6.6666……….
या 10r = 6 + 0.6666………
या 10x = 6 + x
या 10x – x = 6
या 9x = 6
या x = \(\frac{6}{9}=\frac{2}{3}\)
अतः \(0 . \overline{6}\) = \(\frac {2}{3}\) उत्तर

(ii) माना
x = \(0 . 4\overline{7}\)
x = 0.47777……..
दोनों ओर 10 से गुणा करने पर
10x = 4.7777………
पुनः दोनों ओर 10 से गुणा करने पर
या 100x = 47.777……..
या 100x = 43 + 4.7777……..
या 100x = 43 + 10x
या 100x – 10x = 43
या 90x = 43
या x = \(\frac{43}{90}\)
अतः \(0 . 4\overline{7}\) = \(\frac{43}{90}\) उत्तर

(iii) माना x = \(0 . \overline{001}\)
x = 0.001001001……….
दोनों ओर 1000 से गुणा करने पर
या 1000x = 1.001001………
या 1000x = 1 + 0.001001001………
या 1000x = 1 + x
या 1000x – x = 1
या 999x = 1
या x = \(\frac{1}{999}\)
अतः \(0 . \overline{001}\) = \(\frac{1}{999}\) उत्तर

प्रश्न 4.
0.99999… को \(\frac{p}{q}\) के रूप में व्यक्त कीजिए। क्या आप अपने उत्तर से आश्चर्यचकित हैं ? अपने अध्यापक और कक्षा के सहयोगियों के साथ उत्तर की सार्थकता पर चर्चा कीजिए।
हल :
माना x = 0.99999……..
दोनों ओर 10 से गुणा करने पर 10x = 9.99999…………
या 10x = 9 + 0.99999……….
या 10x = 9 + x
या 10x – x = 9
या 9x = 9
या x = \(\frac{9}{9}\) = 1
अतः 0.99999……. = \(\frac{1}{1}\) उत्तर

प्रश्न 5.
\(\frac{1}{17}\) के दशमलव प्रसार में अंकों के पुनरावृत्ति खंड में अंकों की अधिकतम संख्या क्या हो सकती है ? अपने उत्तर की जाँच करने के लिए विभाजन-क्रिया कीजिए।
हल :
\(\frac{1}{17}\) के दशमलव प्रसार में अंकों के पुनरावृत्ति खंड में अंकों की अधिकतम संख्या हो सकती है = 17 – 1 = 16

अतः \(\frac{1}{17}\) = \(0 . \overline{05882352941176471}\)
इस प्रकार \(\frac{1}{17}\) में 16 अंकों की पुनरावृत्ति होती है। उत्तर

प्रश्न 6.
\(\frac{p}{q}\)(q ≠ 0) के रूप की परिमेय संख्याओं के अनेक उदाहरण लीजिए, जहाँ Pऔर 4 पूर्णाक हैं, जिनका 1 के अतिरिक्त अन्य कोई उभयनिष्ठ गुणनखंड नहीं है और जिसका सांत दशमलव निरूपण (प्रसार) है। क्या आप यह अनुमान लगा सकते हैं कि q को कौन-सा गुण अवश्य संतुष्ट करना चाहिए ?
हल :
\(\frac{p}{q}\) के रूप में परिमेय संख्याएँ = \(\frac{3}{4}, \frac{3}{5}, \frac{1}{2}, \frac{1}{4}, \frac{1}{10}\) हैं तथा इनका दशमलव निरूपण = 0.75, 0.6, 0.5, 0.25, 0.1 सांत है।
इससे पता चलता है कि जिन संख्याओं के हर में 2 या 5 अथवा दोनों के गुणनखंड हों तो उन्हें सात दशमलव के रूप में निरूपित कर सकते हैं।

प्रश्न 7.
ऐसी तीन संख्याएँ लिखिए जिनके दशमलव प्रसार अनवसानी अनावर्ती हों।
हल:
पहली अनवसानी अनावर्ती संख्या = 0.01001000100001………
दूसरी अनवसानी अनावर्ती संख्या = 0.202002000200002………
तीसरी अनवसानी अनावर्ती संख्या = 0.003000300003………..

प्रश्न 8.
परिमेय संख्याओं \(\frac {5}{7}\) और \(\frac {9}{11}\) के बीच की तीन अलग-अलग अपरिमेय संख्याएँ ज्ञात कीजिए।
हल :
यहाँ पर,

अतः \(\frac {5}{7}\) = \(0 . \overline{714285}\) तथा \(\frac {9}{11}\) = \(0 . \overline{81}\)
इस प्रकार \(\frac {5}{7}\) और \(\frac {9}{11}\) के बीच तीन अलग अपरिमेय संख्याएँ होंगी –
I = 0.750750075000750000…….
II = 0.760760076000760000……..
III = 0.770770077000770000………

प्रश्न 9.
बताइए कि निम्नलिखित संख्याओं में कौन-कौन-सी संख्याएँ परिमेय और कौन-कौन-सी संख्याएँ अपरिमेय हैं
(i) \(\sqrt{23}\)
(ii) \(\sqrt{225}\)
(iii) 0.3796
(iv) 7.478478……
(v) 1.101001000100001…….
हल :

अतः \(\sqrt{23}\) = 4.795831523…….
क्योंकि यह दशमलव प्रसार अनवसानी अनावर्ती है।
∴ \(\sqrt{23}\) एक अपरिमेय संख्या है। उत्तर

(ii)

अतः \(\sqrt{225}\) = 15 जोकि एक परिमेय संख्या है।
इस प्रकार \(\sqrt{225}\) एक परिमेय संख्या है। उत्तर

(iii) 0.3796
क्योंकि यह दशमलव प्रसार सांत है। इसलिए 0.3796 एक परिमेय संख्या है। उत्तर

(iv) 7.478478………
क्योंकि यह दशमलव प्रसार अनवसानी आवर्ती है। इसलिए 7.478478…….. एक परिमेय संख्या है। उत्तर

(v) 1.101001000100001……
क्योंकि यह दशमलव प्रसार अनवसानी अनावर्ती है। इसलिए यह एक अपरिमेय संख्या है। उत्तर

Haryana State Board HBSE 9th Class Maths Solutions Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.2 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Exercise 13.2

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

प्रश्न 1.
ऊंचाई 14 सें०मी० वाले एक लंब वृत्तीय बेलन का वक्र पृष्ठीय क्षेत्रफल 88 सें०मी०’ है। बेलन के आधार का व्यास ज्ञात कीजिए। [B.S.E.H. March, 2018]
हल :
यहां पर,
माना बेलन के आधार का व्यास (d) = 2r सें०मी०
बेलन के आधार की त्रिज्या (r) = r सें०मी०
बेलन की ऊंचाई (h) = 14 सें०मी०
बेलन का वक्र पृष्ठीय क्षेत्रफल = 88 सें०मी०²
⇒ 2πrh = 88
या 2 × \(\frac{22}{7}\) × r × 14 = 88
या 88r = 88
या r = \(\frac{88}{88}\) = 1 सें०मी०
अतः बेलन के आधार का व्यास (d) = 2 × 1 = 2 सें०मी० उत्तर

प्रश्न 2.
धातु की एक चादर से 1 मी० ऊंची और 140 सें०मी० व्यास के आधार वाली एक बंद बेलनाकार टंकी बनाई जानी है। इस कार्य के लिए कितने वर्ग मीटर चादर की आवश्यकता होगी ?
हल :
यहां पर,
बेलनाकार टंकी का व्यास (d) = 140 सें०मी०
बेलनाकार टंकी की त्रिज्या (r) = \(\frac{140}{2}\) = 70 सें०मी० = 0.7 मी०
बेलनाकार टंकी की ऊंचाई (h) = 1 मी०
∴ बेलनाकार टंकी का संपूर्ण पृष्ठीय क्षेत्रफल = 2 πr (r + h)
= 2 × \(\frac{22}{7}\) × 0.7 [0.7 + 1] मी०²
= 2 × 22 × 0.1 × 1.7 मी०²
= 7.48 मी०²
अतः टंकी बनाने के लिए आवश्यक चादर = 7.48 मी०² उत्तर

प्रश्न 3.
धातु का एक पाइप 77 सें०मी० लंबा है। इसके एक अनुप्रस्थकाट का आंतरिक व्यास 4 सें०मी० है और बाहरी व्यास 4.4 सें०मी० है (आकृति अनुसार) ज्ञात कीजिए :

(i) आंतरिक वक्र पृष्ठीय क्षेत्रफल
(ii) बाहरी वक्र पृष्ठीय क्षेत्रफल
(iii) कुल पृष्ठीय क्षेत्रफल
हल :
(i) यहां पर,
पाइप का आंतरिक व्यास (dl) = 4 सें०मी०
पाइप की आंतरिक त्रिज्या (r) = \(\frac{4}{2}\) = 2 सें०मी०
पाइप की ऊंचाई (h) = 77 सें०मी०
∴ पाइप का आंतरिक वक्र पृष्ठीय क्षेत्रफल = 2πrh
= 2 × \(\frac{22}{7}\) × 2 × 77 सें०मी०²
= 968 सें०मी०² उत्तर

(ii) पाइप का बाह्य व्यास (d2) = 4.4 सें०मी०
पाइप की बाह्य त्रिज्या (R) = \(\frac{4.4}{2}\) = 2.2 सें०मी०
पाइप की ऊंचाई (h) = 77 सें०मी०
∴ पाइप का बाह्य वक्र पृष्ठीय क्षेत्रफल = 2πRh
= 2 × \(\frac{22}{7}\) × 2.2 × 77 सें०मी०²
= 1064.8 सें०मी०² उत्तर

(iii) पाइप के दोनों सिरों का क्षेत्रफल = 2π(R² – r²)
= 2 × \(\frac{22}{7}\) × [(2.2 )² – (2)²] सें०मी०²
= 2 × \(\frac{22}{7}\) × [4.84 – 4] सें०मी०²
= 2 × \(\frac{22}{7}\) × 0.84 सें०मी०² = 5.28 सें०मी०²
पाइप का संपूर्ण पृष्ठीय क्षेत्रफल = [968 + 1064.8 + 5.28 ] सें०मी०²
= 2038.08 सें०मी०² उत्तर

प्रश्न 4.
एक रोलर (roller) का व्यास 84 सें०मी० है और लंबाई 120 सें०मी० है। एक खेल के मैदान को एक बार · समतल करने के लिए 500 चक्कर लगाने पड़ते हैं। खेल के मैदान का मी० में क्षेत्रफल ज्ञात कीजिए।
हल :
यहां पर,
रोलर का व्यास (d) = 84 सें०मी०
रोलर की त्रिज्या (r) = \(\frac{84}{2}\) = 42 सें०मी०
रोलर की लंबाई (h) = 120 सें०मी०
∴ रोलर का वक्र पृष्ठीय क्षेत्रफल = 2πrh
= 2 × \(\frac{22}{7}\) × 42 × 120 सें०मी०²
= 31680 सें०मी०²
रोलर द्वारा 1 चक्कर लगाने में जितना क्षेत्रफल तय होता है = 31680 सें०मी०²
रोलर द्वारा 500 चक्कर लगाने में जितना क्षेत्रफल तय होता है = 31680 × 500 सें०मी०²
= \(\frac{31680 \times 500}{100 \times 100}\)मी०²
= 1584 मी०²
अतः खेल के मैदान का क्षेत्रफल = 1584 मी०² उत्तर

प्रश्न 5.
किसी बेलनाकार स्तंभ का व्यास 50 सें०मी० है और ऊंचाई 3.5 मी० है। ₹ 12.50 प्रति m2 की दर से इस स्तंभ के वक्र पृष्ठ पर पेंट कराने का व्यय ज्ञात कीजिए।
हल :
यहां पर,
बेलनाकार स्तंभ का व्यास (d) = 50 सें०मी०
बेलनाकार स्तंभ की त्रिज्या (r) = \(\frac{50}{2}\) सें०मी० = 25 सें०मी० = \(\frac{25}{100}\) m = \(\frac{1}{4}\)m
बेलनाकार स्तंभ की ऊंचाई (h) = 3.5 मी० = \(\frac{35}{10}\)m
बेलनाकार स्तंभ का वक्र पृष्ठीय क्षेत्रफल = 2πrh
= 2 × \(\frac{22}{7}\) × \(\frac{1}{4}\) × \(\frac{35}{10}\) × 275 मी०²
= \(\frac{55}{10}\) = 5.5 मी०²
1 मी०2स्तंभ पर पेंट करने का व्यय = ₹ 12.50
5.5 मी०2 स्तंभ पर पेंट करने का व्यय = ₹ (5.5 × 12.50)
= ₹ 68.75 उत्तर

प्रश्न 6.
एक लंब वृत्तीय बेलन का वक्र पृष्ठीय क्षेत्रफल 4.4 मी०² है। यदि बेलन के आधार की त्रिज्या 0.7 मी० है, तो उसकी ऊंचाई ज्ञात कीजिए। [B.S.E.H. March, 2020]
हल :
यहां पर,
लंब वृत्तीय बेलन की त्रिज्या (r) = 0.7 मी०
लंब वृत्तीय बेलन की ऊंचाई (h) = ?
लंब वृत्तीय बेलन का वक्र पृष्ठीय क्षेत्रफल = 4.4 मी०²
⇒ 2πrh = 4.4
या 2 × \(\frac{22}{7}\) × 0.7 × h = 4.4
या 4.4h = 4.4
या h = \(\frac{4.4}{4.4}\) = 1 मी०
अतः लंब वृत्तीय बेलन की ऊंचाई = 1 मी० उत्तर

प्रश्न 7.
किसी वृत्ताकार कुएं का आंतरिक व्यास 3.5 मी० है और यह 10 मी० गहरा है। ज्ञात कीजिए :
(i) आंतरिक वक्र पृष्ठीय क्षेत्रफल।
(ii) ₹ 40 प्रति m² की दर से इसके वक्र पृष्ठ पर प्लास्टर कराने का व्यय।
हल :
यहां पर,
(i) कुएं का आंतरिक व्यास (d) = \(\frac{35}{10}\) मी० = \(\frac{7}{2}\) मी०
कुएं की आंतरिक त्रिज्या (r) = \(\frac{7}{2}\) × \(\frac{1}{2}\) = \(\frac{7}{4}\) मी०
कुएं की ऊंचाई (h) = 10 मी०
∴ कुएं का आंतरिक वक्र पृष्ठीय क्षेत्रफल = 2πrh
= 2 × \(\frac{22}{7}\) × \(\frac{7}{4}\) × 10 मी०²
= 110 मी०² उत्तर

(ii) 1 मी०² वक्र पृष्ठ पर प्लास्टर कराने का व्यय = ₹ 40
110 मी०² वक्र पृष्ठ पर प्लास्टर कराने का व्यय = 110 × 40
= ₹ 4400 उत्तर

प्रश्न 8.
गरम पानी द्वारा गरम रखने वाले एक संयंत्र में 28 मी० लंबाई और 5 सें०मी० व्यास वाला एक बेलनाकार पाइप है। इस संयंत्र में गर्मी देने वाला कुल कितना पृष्ठ है ?
हल :
यहां पर, बेलनाकार पाइप का व्यास (d) = 5 सें०मी० = \(\frac{5}{100}\)मी० = \(\frac{1}{20}\)मी०
बेलनाकार पाइप की त्रिज्या (r) = \(\frac{1}{2}\) × \(\frac{1}{20}\) मी० = \(\frac{1}{40}\) मी०
बेलनाकार पाइप की ऊंचाई (h) = 28 मी०
बेलनाकार पाइप का वक्र पृष्ठीय क्षेत्रफल = 2πrh
= 2 × \(\frac{22}{7}\) × \(\frac{1}{40}\) × 28 मी०²
= 4.4 मी०²
अतः संयन्त्र में गर्मी देने वाला कुल पृष्ठ = 4.4 मी०² उत्तर

प्रश्न 9.
ज्ञात कीजिए:
(i) एक बेलनाकार पेट्रोल की बंद टंकी का पार्श्व या वक्र पृष्ठीय क्षेत्रफल, जिसका व्यास 4.2 मी० है और ऊंचाई 4.5 मी० है।
(ii) इस टंकी को बनाने में कुल कितना इस्पात (steel) लगा होगा, यदि कुल इस्पात का \(\frac{1}{12}\) भाग बनाने में नष्ट हो गया है?
हल :
यहां पर,
(i) बेलनाकार टंकी का व्यास (d) = 4.2 मी०
बेलनाकार टंकी की त्रिज्या (r) = \(\frac{4.2}{12}\) = 2.1 m = \(\frac{21}{10}\) m
बेलनाकार टंकी की ऊंचाई (h) = 4.5 मी० = \(\frac{45}{10}\) मी०
बेलनाकार टंकी का वक्र पृष्ठीय क्षेत्रफल = 2πrh
= 2 × \(\frac{22}{7} \times \frac{21}{10} \times \frac{45}{10}\) मी०²
= 59.4 मी०² उत्तर

(ii) बेलनाकार टंकी का संपूर्ण पृष्ठीय क्षेत्रफल = 2π(r + h)

= 87.12 मी०² उत्तर
माना आवश्यक कुल इस्पात = x मी०²
जितना इस्पात बनाने में नष्ट हुआ = x × \(\frac{1}{12}\) = \(\frac{x}{12}\) मी०²
= x – \(\frac{x}{12}\) = 87.12
या 12x – x = 87.12 × 12
या 11x = 87.12 × 12
या x = \(\frac{87.12 \times 12}{11}\) = 95.04 मी०²
अतः टंकी को बनाने के लिए आवश्यक इस्पात = 95.04 मी०² उत्तर

प्रश्न 10.
आकृति में, आप एक लैंपशेड का फ्रेम देख रहे हैं। इसे एक सजावटी कपड़े से ढका जाना है। इस फ्रेम के आधार का व्यास 20 सें०मी० है और ऊंचाई 30 सें०मी० है। फ्रेम के ऊपर और नीचे मोड़ने के लिए दोनों ओर 2.5 सें०मी० अतिरिक्त कपड़ा भी छोड़ा जाना है। ज्ञात कीजिए कि लैंपशेड को ढकने के लिए कुल कितने कपड़े की आवश्यकता होगी।

हल :
यहां पर,
फ्रेम के आधार का व्यास (d) = 20 सें०मी०
फ्रेम के आधार की त्रिज्या (r) = \(\frac{20}{2}\) = 10 सें०मी०
फ्रेम की ऊंचाई = 30 सें०मी०
फ्रेम को लपेटने के लिए आवश्यक कपड़े की लंबाई (h) = (30 + 2.5 + 2.5) सें०मी०
= 35 सें०मी०
∴ आवश्यक कपड़े का वक्र पृष्ठीय क्षेत्रफल = 2πrh
= 2 × \(\frac{22}{7}\) × 10 × 35 सें०मी०²
= 2200 सें०मी०² उत्तर

प्रश्न 11.
किसी विद्यालय के विद्यार्थियों से एक आधार वाले बेलनाकार कलमदानों को गत्ते से बनाने और सजाने की प्रतियोगिता में भाग लेने के लिए कहा गया। प्रत्येक कलमदान को 3 सें०मी० त्रिज्या और 10.5 सें०मी० ऊंचाई का होना था। विद्यालय को इसके लिए प्रतिभागियों को गत्ता देना था। यदि इसमें 35 प्रतिभागी थे, तो विद्यालय को कितना गत्ता खरीदना पड़ा होगा?
हल :
यहां पर,
कलमदान के आधार की त्रिज्या (r) = 3 सें०मी०
कलमदान की ऊंचाई (h) = 10.5 सेंमी० = \(\frac{105}{10}\) सें०मी०
1 कलमदान बनाने के लिए आवश्यक गत्ता = वक्र पृष्ठीय क्षेत्रफल + आधार का क्षेत्रफल
= 2πrh + πr²
= [2 × \(\frac{22}{7}\) × 3 × \(\frac{105}{10}\) + \(\frac{22}{7}\) × 3 × 3] सें०मी०²
= \(\left[\frac{198}{1}+\frac{198}{7}\right]\) सें०मी०²
= \(\left[\frac{1386+198}{7}\right]\) सें०मी०²
= \(\frac{1584}{7}\) सें०मी०²
35 कलमदान बनाने के लिए आवश्यक गत्ता = 35 × \(\frac{1584}{7}\) सें०मी०²
= 7920 सें०मी०²
अतः विद्यालय को जितना गत्ता खरीदना पड़ेगा = 7920 सें०मी०² उत्तर

Haryana State Board HBSE 9th Class Science Solutions Chapter 15 Improvement in Food Resources Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 15 Improvement in Food Resources

HBSE 9th Class Science Improvement in Food Resources Intext Questions and Answers

Question from Sub-section 15.1

Question 1.
What do we get from cereals, pulses, fruits and vegetables ?
Answer:
From these, we get proteins, carbohydrates, fats, vitamins and minerals, which are the main and required constituents of our food. These all constituents are helpful for our body development, growth and health.

Questions from Sub-section 15.1.1

Question 1.
How do biotic and abiotic factors affect crop production ?
Answer:
Biotic factors: Diseases, insects and nematodes.
Abiotic factors: Drought, salinity, water logging, heat, cold, frost, etc., decrease the production of crops.

Question 2.
What are the desirable agronomic characteristics for crop improvements ?
Answer:
Tallness and profuse branching are desirable characters for fodder crops. Dwarfness is required in cereals and less nutrients are consumed by these crops. Thus, developing varieties of desired agronomic characters help to give higher productivity.

Questions from Sub-section 15.1.2

Question 1.
What are macro-nutrients and why are they called macro-nutrients?
Answer:
Macro-nutrients :
Such nutrients which are required by the plants in large quantities are called macro-nutrients. Due to their requirement in large quantities, these are called macro-nutrients, e.g., sulfur, nitrogen, phosphorus potassium, calcium, magnesium, etc;

Question 2.
How do plants get nutrients?
Answer:
Plants get the nutrients in the form of liquid from earth. These nutrients are also available in manures and fertilizers, which plants get easily and in large quantities.

Questions from Sub-section 15.1.2

Question 1.
Compare the use of manures and fertilizers in maintaining soil fertility.
Answer:
The comparison of manures and fertilizers to maintain the fertility of soil is as following ;
Manures
1. Manures are required in large quantities.
2. Continuous use of manures is not harmful to the soil.
3. By the use of manures, production always increases.
4. Using process of manure and crop production is delayed.
5. Use of manures is not costly.
6. Transportation and storage is unconvenient.

Fertilizers:
1. Fertilizers are required in small quantities.
2. Continuous use of fertilizers is harmful to the soil.
3. Unnecessary use of fertilizers can decreased the production.
4. By using fertilizers, crop production can be get early.
5. Use of fertilizers is costly.
6. Transportation and storage is convenient.

Questions from Sub-section 15.1.3

Question 1.
Which of the following conditions will give the most benefits? Why?
(a) Farmers use high-quality seeds, do not adopt irrigation or use fertilizers.
(b) Farmers use ordinary seeds, adopt irrigation and use fertilizers.
(c) Farmers use good quality seeds, adopt irrigation, use fertilizers and use crop protection measures.
Answer:
In the above three conditions, the third condition (c) is best because in this condition, seeds are of
high quality, i.e., high-level production. Irrigation is helpful in the nutrition of plants, fertilizers supply the nutrients to plants that are required for the development, growth and production of the plants. Crop protection prevents the crops from various harms and in fact, production increases. Hence, condition (c) is best for farmers.

Questions from Sub-section 15.1.3

Question 1.
Why should preventive measures and biological control methods be preferred for protecting crops?
Answer:
The preventive measures for the protection of crops are: timely sowing of crops, proper seedbed preparation, intercropping and crop rotation also help in weed control. Any type of poisonous chemical such as pesticides, weedicides, etc. is not used. Some other preventive measures against pests are the use of resistant varieties and summer plowing, in which fields are plowed deep in summers to destroy weeds and pests. In this way, soil pollution is not taking place and no extra amount is to be consumed on these measures. Hence, these are also useful economically.

Question 2.
What factors may be responsible for the losses of grains during storage?
Answer:
During storage, two factors are responsible for the loss of grains:
(1) Biotic factors: Insects, rodents, fungi, mites, bacteria, etc.
(2) Abiotic factors: Lack of appropriate moisture and temperature.

Questions from Sub-section 15.2.1

Question 1.
Which method is commonly used for improving cattle breeds and why?
Answer:
The cross-breed method is used for improving cattle breeds so that the lactation and resisting power against diseases increase in such cattle.

Questions from Sub-section 15.2.2

Question 1.
Discuss the implications of the following statement:
“It is interesting to note that poultry is India’s most efficient converter of low fiber foodstuff (which is unfit for human consumption) into highly nutritious animal protein food.”
Answer:
Poultry has this quality that low fiber foodstuffs which are unfit for human consumption that means on the basis of quality these are totally unfit for human consumption, to those it has the ability to convert into highly nutritious animal protein food. Poultry eggs and meat are enriched with proteins and this diet is considered to be a protein-rich diet for humans

Questions from Sub-section 15.2.2

Question 1.
What management practices are common in dairy and poultry farming?
Answer:
There is a big similarity between animal husbandry and poultry farming :
(1) Both are required clean, well-ventilated, and moisture-less homes.
(2) Both are supplied well balanced and enough diet.
(3) Both suffer from a number of diseases that are prevented by appropriate vaccination and both are also treated by the physicians
(4) Both have economic importance, i.e., both are domesticated for economic profit.
(5) Both the high-quality breeds are domesticated to earn a big profit.
Hence, both are similar under the management system.

Question 2.
What are the differences between broilers and layers in their management?
Answer:
Broiler farming is done for meat purposes, whereas layer farming is undertaken for egg production. This is the main difference between them. Broilers and chickens are fed with protein, fats, and vitamin-rich supplementary food for good growth rate and better feed efficiency. Care is taken to avoid mortality and to maintain feathering and carcass quality, whereas the layer requires this type of feeding in less quantity. The housing, nutritional and environmental requirements of broilers are somewhat different from those of egg-producing layers.

Questions from Sub-section 15.2.3

Question 1.
How are fish obtained?
Answer:
Fish are obtained in two ways:
(1) From natural resources,
(2) From fish farming or culture fishery.
Fish are available both in freshwater (non-saline water) and seawater (saline water). Freshwater fish varieties are: catla, rohu, mullets, millions, etc. whereas marine fish varieties include: pomfret, mackerel, tuna, sardines, Bombay duck, etc. Mullets, bhetki, pearl spots, prawns, mussels, and oysters are cultured in seawater. Fish farming is done in freshwater or non-saline water.

Question 2.
What are the advantages of composite fish culture?
Answer:
More intensive fish farming can be done in composite fish culture systems. In such a system, a combination of five or six fish species is used in a single fishpond. These species are selected so that they do not compete for food among them having different types of food habits:
(1) The food available in all the parts of the pond is used.
(2) Weeds can be controlled by bio-practices in composite fish culture.
(3) Without competing with each other, fish yield increases in the pond.

Questions from Sub-section 15.2.4

Question 1.
What are the desirable characteristics of bee varieties suitable for honey production?
Answer:
It is the natural desirable character in variety that it is suitable for increasing honey production. Honey collection and the defense of the beehive are both qualities in Italian bees. They stay in a given beehive for long periods and are freed very well, it increases honey production.

Question 2.
What is pasturage and how is it related to honey production?
Answer:
Pasturage is the natural grassy field where grass for animals and other flowering plants are available. By these flowers, the bees make honey from nectar and pollen collection. The taste of honey depends upon the pasturage, or the flowers available to the bees.
For Examples; honey of bruce flowers, honey of neem flowers, and honey of eucalyptus flowers. The quality of honey is determined by the taste of honey.

HBSE 9th Class Science Improvement in Food Resources Textbook Questions and Answers

Question 1.
Explain any one method of crop production which ensures high yield.
Answer:
One way of incorporating desirable characters into crop varieties is by hybridization. Hybridization is the mating or crossing of genetically different plants. The progeny resulting from hybridization is known as a hybrid. Hybridization refers to crossing between genetically dissimilar plants. This crossing may be intervarietal (between different varieties), interspecific (between two different species of the same genus), or intergeneric (between different genera). Another way of improving the crop is by introducing a gene that would provide the desired characteristics. This results in genetically modified crops.

Question 2.
Why are manures and fertilizers used in fields?
Answer:
Manures and fertilizers have macro-nutrients. There is a lack of nutrients in the soil as crops absorb nutrients from the soil continuously. Manures and fertilizers are mixed in the fields to fulfill the loss of nutrients. The land proves to be barren and the fertility of the land decreases if we do not fulfill the loss. The loss of these nutrients can also be completed by crop rotation.

Question 3.
What are the advantages of intercropping and crop rotation?
Answer:
The following are the advantages of intercropping and crop rotation:
(1) It maintains the fertility of the land. It does not become barren.
(2) Two or more crops can be grown in a year with good harvests.
(3.) Weeds and insects are destroyed.
(4) With the increase of production, farmers’ income also increases.
(5) Threat of unsuccessful crop production also decreases.

Question 4.
What is genetic manipulation? How is it useful in agricultural practices?
Answer:
The high yield of the crops is dependent on the characteristics of their varieties. If we want to improve the crop varieties, then genes with desirable characteristics are used because only genes establish what characteristics of the plants will be, i.e., crop production will increase or decrease. Hence, a change in genes is called genetic manipulation. Today, by using this technique we have got high-quality seeds and crop production has increased the number of times. Especially farmers and gardeners have got an advantage by this technique.

Question 5.
How do storage grain losses occur?
Answer:
(A) Biotic Factors: Following biotic factors are responsible for storage losses of grain :
1. Rodents, birds, and other insects: Rats, squirrels, etc. make a loss to the cereals kept in fields, houses, and in warehouses. Loss is caused to foodgrains due to their hair, feathers, and excreta.
2. Micro-organisms: Various insects and molds change the structure of the storage even after the chemical changes and in this way they cause loss of foodgrains.
3. Insects: Various types of insects damage the raw food storage, e.g., cereals and pulses.
4. Enzymes: These are biocatalysts that are found in nerve cells. These destroy the fruits, and vegetables which are stored a long time.

(B) Abiotic Factors: Following abiotic factors are responsible in storage losses of grain:
1. Moisture: There should not be high moisture at the time of storage. Moisture in the foodgrains should not exceed 14% in relation to its weight.

At high moisture in the foodgrains:
(1) Size of the grains develop.
(2) Microorganisms and enzymes cause more activity.
(3) Possibility of insect-infection increases.
(4) Possibility of growing molds in the moist air increases.

2. Temperature: At low temperatures, enzymes, insects and other micro-organisms do not set more active. For this reason, profligate food materials are kept in cold storage.

3. Pots for storage: Pots used in storage should not be made of lead, metal or nickel because these are toxic and the possibility of toxicity in foodgrains increases.

Question 6.
How do good animal husbandry practices benefit farmers?
Answer:
Animal husbandry practices benefit farmers as follows:
(1) Farmers get milk from dairy animals (milch animals).
(2) In animal husbandry, draught animals are useful for agricultural work and carting.
(3) In animal husbandry, due to economical advantages, farmers get occupational work.
(4) Agriculture by animals provides an extra and round year income.
(5) Biogas and organic matter are prepared from the excreta of animals domesticated by the farmers.
(6) Animal husbandry provides wool, meat, skins, and bones.

Question 7.
What are the benefits of cattle farming?
Answer:
Two main benefits of cattle farming are as under:
1. To get foodstuffs:
The main objective of cattle farming is to produce food products in which the main production is of milk and meat. Furthermore, we get wool, skins and bones from cattle farming. In this way, the objective of cattle farming is related to economic reasons.

2. To get assistance in agricultural activities:
To get assistance in agricultural activities is the main second objective of cattle farming. Oxen, mule, camel, etc. are used in agricultural works such as tilling and carting. Even today low-level farmers are also dependent on animals for agricultural work.

Question 8.
For increasing production, what is common in poultry, fisheries, and bee-keeping?
Answer:
This is only the main common factor to increase the production that the varieties of organisms should be of having disease resistance efficiency, be favorable to the environment, and have high yielding capacity. The major objective for the domestication of these three organisms is to earn the money for the highest yielding. All three organisms are used in agricultural activities as well as in supplementary occupations where the farmer earns income and there is no need to devote their full time.

Question 9.
How do you differentiate between capture fishing, mariculture and aquaculture?
Answer:
Some marine fish of high economic value are farmed in seawater. It is called capture fishing mariculture. The provision of natural environment in the ponds for fish farming is called ‘aquaculture’. In ponds, fish farming is done in non-saline water or freshwater whereas in marine fisheries the fishing is done in saline water.

Haryana State Board HBSE 9th Class Science Solutions Chapter 14 Natural Resources Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 14 Natural Resources

HBSE 9th Class Science Natural Resources Intext Questions and Answers

Questions from Sub-section 14.1

Question 1.
How is our atmosphere different from the atmospheres on Venus and Mars?
Answer:
The components of life air (breathe of life – oxygen) are present in our atmosphere while this lacks on Venus and Mars.

Question 2.
How does the atmosphere act as a blanket?
Answer:
Air present in the atmosphere is a bad conductor of heat which keeps the average temperature of the atmosphere fairly steady during day-night and even during the course of the whole year, such that a blanket prevents the heat to go out, just like that atmosphere prevents the heat of earth surface to go out and also does not let the temperature to be increased, therefore, the atmosphere acts as a blanket.

Question 3.
What causes winds?
Answer:
With uneven movements, the warming of the atmosphere of the earth creates somewhere low-pressure area and somewhere high-pressure area. The difference of this pressure creates airflow which is called breeze or wind.

Question 4.
How are clouds formed?
Answer:
Water vapours in the atmosphere when a pass from the cold areas then water vapours are condensed in the form of droplets. These droplets grow bigger and form the shape of clouds.

Question 5.
List any three human activities that you think would lead to air pollution.
Answer:
Following are the human activities which cause air pollution :
(1) Burning of fossil fuels produces pollutants, which pollute the air after mixing with it.
(2) Smoke from vehicles also pollutes the air.
(3) Gases from the industries/factories established by man also pollute the air after mixing with it.

Questions from Sub-section 14.2

Question 1.
Why do organisms need water?
Answer:
Water is the fundamental need of all living organisms. Every organism’s body contains 75% amount of water. In organisms, all biogeochemical reactions take place with water. Water medium is compulsory for all cellular reactions. Transportation of the substances in the bodies of the organisms takes place in a water medium and excretion activities also take place by water. Terrestrial organisms require freshwater to survive. Water is a living place at a lot of organisms.

Question 2.
What is the major source of freshwater in the city/town/village where your live ?
Answer:
In villages, towns and cities sources of freshwater are underground water, ponds, lakes and rivers. After melting the snow, water is available in the rivers coming from the mountains. In ponds, lakes, etc., mostly we find rainwater.

Question 3.
Do you know of any activity which may be polluting this water source?
Answer:
Sewage waste is mostly dumped into water resources (rivers, canals, lakes) which causes water pollution. Sewage excreta of metropolitan cities and towns situated on the banks of rivers is being dumped into rivers directly. Likewise, half burnt or unburnt dead bodies are thrown in the water of the Ganga by the habitats of the banks of Ganga due to which the water of Ganga is being polluted.

Questions from Sub-section 14.3

Question 1.
How is soil formed?
Answer:
Soil is formed by the breaking and grinding-rubbing of rocks. Soil is formed by the mixing of small pieces of rocks, fresh and decomposition of leaves, dead decayed remains of organisms and sand particles etc. The formation of topsoil of the earth takes a minimum 300 to 800 years. Soil formation is a continuous process.

Question 2.
What is soil erosion?
Answer:
Cutting or flowing away of the soil by any of the reasons is called soil erosion. Soil erosion may take place by various reasons, for example:
1. Flood: Flood is the main reason of soil erosion by which soil flows away from one place to another.
2. Fast Winds: The second main reason of soil erosion is the fast-blowing winds.
3. Deforestation: Deforestation is on high scale due to growing population and industrialisation by which soil erosion increases.
4. Uncontrolled and Overgrazing: Uncontrolled and overgrazing increases soil erosion.
5. Unscientific Techniques of Agriculture: Mostly farmers use to grow one variety of crops continuously in the fields by which fertility of soil decreases and soil erosion increases.
6. Deeper Cultivation of the Land: Soil erosion increases due to deeper cultivation of the land for farming.

Question 3.
What are the methods of preventing or reducing soil erosion?
Answer:
Following are the methods to prevent soil erosion:
1. By Growing Grass: Leaves and roots of grass retard the flow of water by which soil particles may not be flown by water.
2. By Levelling the Fields: By distributing the tapered land into small levelled pieces for agriculture, the flow of water is checked by which cutting of land is also prevented.
3. By Growing Trees: Roots of the trees bind the soil by which water cannot flow away the soil.
4. Better Techniques of Farming: In spite of ploughing in the sloping direction, if ploughing is being done in left-right sides of the field, then flow of water decreases, and erosion of soil is checked.
5. By Making Small Dams: Small dams constructed on slopes decrease the flow of fast-moving water
and soil erosion is stopped.
6. By Making Ladder-type Field: In hilly areas, farming should be done by making steps opposite to
slope.
7. By Using Manure: Use of manure in field increases the water retention efficiency of the soil by which soil erosion is checked.
8. By Fixing Air-protectors: Where the land is sandy and winds flow fast, there plants should be grown profusely around the field, i.e., fast winds may not flow away the soil particles.
9. By Making High Fences: Fences of the fields should be high because the water of field will not flow out and cutting of land is not possible.
10. By Growing Particular Crops : Soil erosion can be decreased by growing particular crops such as groundnut.

Questions from Sub-section 14.4,14.5

Question 1.
What are the different states in which water is found during the water cycle ?
Answer:
In water cycle, three states of water are found :
(i) Water in liquid state,
(ii) Water in vapour state and
(iii) Ice as solid state.

Vapours form clouds and these clouds bring rain on the earth and sometimes water of clouds freezes in ice, i.e., converts into hailstone. Ice on the peaks of the mountains is also water in solid state, which after melting comes in rivers and there by converting into vapours, mix with air.

Question 2.
Name two biologically important compounds that contain both oxygen and nitrogen.
Answer:
Nitrates and nitrites.

Question 3.
List any three human activities which would lead to an increase in the carbon dioxide content in air.
Answer:
Following are the reasons of human activities which cause an increase in the carbon dioxide content in air.
(i) By combustion of fuel.
(ii) By using petroleum products in vehicles.
(iii) CO₂is also released by respiration.

Question 4.
What is the greenhouse effect?
Answer:
The earth receives different types of radiation from the sun. The ozone layer in the atmosphere absorbs ultraviolet radiation and let go to other radiations, but radiations which come on the earth’s surface, some part of it in the form of infrared radiation reflected back. Molecules of CO₂ have the efficiency to absorb infrared radiations, hence carbon dioxide present in the atmosphere can absorb radiation, that reason atmosphere becomes warmer. The warming of the atmosphere in this way is called the greenhouse effect or glasshouse effect.

Question 5.
What are the two forms of oxygen found in the atmosphere?
Answer:
CO₂ (Oxygen) and O₃ (Ozone).

HBSE 9th Class Science Natural Resources Textbook Questions and Answers

Question 1.
Why is the atmosphere essential for life?
Answer:
Life is dependent only on the atmosphere because
(i) The atmosphere keeps the average temperature of the earth fairly steady.
(ii) Oxygen is available in the atmosphere in the form of life air.
(iii) Ozone layer of the atmosphere protects from the harmful ultraviolet radiation coming from the sun.
(iv) Atmosphere protects the earth from meteors, by which life is also protected.
(v) CO₂ gas which is used by green plants in photosynthesis is also found in the atmosphere.
(vi) Water in the atmosphere is found in the form of water vapours which is the foundation of life.
Hence, the atmosphere is essential for life.

Question 2.
Why is water essential for life?
Answer:
Water is the fundamental need of all living organisms. Every organism’s body contains 75% amount of water. In organisms, all biogeochemical reactions take place with water. Water medium is compulsory for all cellular reactions. Transportation of the substances in the bodies of the organisms takes place in a water medium and excretion activities also take place by the water. Terrestrial organisms require freshwater to survive. Water is a living place at a lot of organisms.

Question 3.
How are living organisms dependent on the soil ? Are organisms that live in water totally independent of soil as a resource?
Answer:
Soil is the base of life. All living organisms are dependent on soil directly or indirectly. Plants get nutrients from the soil in which main are water and minerals and all animals are dependent on plants. Animals get food and oxygen to respire from plants. Life of living organisms is not possible without plants and plants cannot survive without soil. Aquatic animals are also dependent on soil because they fulfil their requirements from plants.

Question 4.
You have seen weather reports on television and in newspapers. How do you think we are able to predict the weather?
Answer:
Yes, we are able to predict the weather. By collecting the data through artificial satellites established in earth orbit, it is very easy to predict the forthcoming weather. After each half an hour reports and photographs are collected by INSAT-IB. Through these photographs, prediction is relayed which prove to be very useful.

Question 5.
We know that many human activities lead to increasing levels of pollution of the air, water bodies and soil. Do you think that isolating these activities to specific and limited areas would help in reducing pollution?
Answer:
Yes, isolating these activities to specific and limited areas would help in reducing pollution, e.g.,
(i) Air pollution will decrease by running the vehicles based on solar system batteries rather then fossil fue/ petroleum.
(ii) These will be a check on pollution if cooking of food, heating of water, etc. are done by the use of solar energy in place of coal, L.P.G., etc. fuels in houses.
(iii) These will be a check on soil pollution by using compost manure in the agricultural field rather than chemical fertilizers.
(iv) The water pollution will be decreased by using useful substances in the treatment plants inspite of throwing the industrial waste directly in the water bodies.

Question 6.
Write a note on how forests influence the quality of our air, soil and water resources.
Answer:
Forests are the main components in environmental balance. Forests equalize the formation (78% nitrogen, 21% oxygen) of air. This balance is disturbed by deforestation. Temperature increases with the increase of C02 and the content of condensed particles in air while the air becomes, dust free, pure and cool by the deep forestation. Forest decreases pollution. Forests help in maintaining the quality of soil by increasing its fertility and also preventing soil erosion. By deforestation land becomes barren. The temperature of the environment increases. Rains are excess by excess forestation. Water level increases in water bodies and makes the availability of useful fresh water in large quantities. Therefore, forests are helpful in improving the quality of air, soil and water resources.

Haryana State Board HBSE 9th Class Science Solutions Chapter 13 Why Do We Fall Ill Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 13 Why Do We Fall Ill

HBSE 9th Class Science Why Do We Fall Ill Intext Questions and Answers

Questions from Sub-section 13.1

Question 1.
State any two conditions essential for good health.
Answer:
Following are the two conditions essential for good health :
(i) Community cleanliness
(ii) Balanced diet.

Question 2.
State any two conditions essential for being free of diseases.
Answer:
Following are the two conditions essential for being free of diseases :
(1) Clean environment
(ii) Good habits and regular exercise.

Question 3.
Are the answers to the above questions necessarily the same or different ? Why ?
Answer:
The answers to the above questions are necessarily same because to keep good health or diseases free are almost same situation.

Questions from Sub-section 13.2

Question 1.
List any three reasons why you would think that you are sick and ought to see a doctor. If only one of these symptoms were present, would you still go to the doctor ? Why or why not ?
Answer:
There can be many
(i) cough
(ii) fever
(iii) weakness.

If the given symptoms appear for a small period of time, there is no need to go to a doctor, because the immune system of our body stops these quickly and makes the body free of the effects of any disease. But if persistent cough prevails, lungs may be infected. If fever persists, it can be due to a serious ailmant T.B. If there is nutritious deficiency, it can lead to general weakness. In conditions like these, it is proper to go to a physician, so that the real course of the symptoms may be as certained and it can be cured.

Question 2.
In which of the following case do you think the long term effects on your health are likely to be most unpleasant ?

• If you get jaundice.
• If you get lice.
• If you get acne. Why?

Answer:
If we’re suffering from jaundice, it is most likely to have ill effects on our health. Jaundice is a disease that effect the liver, which is an important organ of the body and which has an important role to play in our body. So the body of the person suffering from jaundice is predominantly affected. The effect of getting lice or acne are not so predominant.

Questions from Sub-section 13.3

Question 1.
Why are we normally advised to take bland and nourishing food when we are sick?
Answer:
Bland and nourishing food is recommended to sick people so that it is easily digested and there is reimbursement of the nutrients lost due to sickness.

Question 2.
What are the different means by which infectious diseases are spread?
Answer:
Infectious diseases spread through the following methods:
1. Through ai: Many disease particles spread in the air due to sneezing, coughing, talking, spitting or defecating. When a healthy person breathes in the air containing these disease particles, the person catches the disease.

2. Through water: Pathogens of cholera, tuberculosis etc., get mixed in drinking water and enter the body. The disease spreads by washing utensils or vegetables and fruits in polluted water. It also spreads when the clothes of an infected person are washed near a source of water or he defecates near it.

3. Through food:
Pathogens present in foodstuffs also enter the body and make the person infected. It is harmful to take stale or cold food.

4. Individual contact:
Coming in contact with the patient’s clothes, bedding, utensils, towel or directly with the patient also transmits the pathogens.

5. Through animals: Some animals like houseflies, mosquitoes, mice etc., also play the role of vectors. By polluting food the pathogens are left there which ultimately enter the human body. Malaria is spread by mosquito bite while cholera spreads through houseflies and plague through mice.

Question 3.
What precautions can you take in your school to reduce the incidence of infectious diseases?
Answer:
Following precautions should be taken:
(i) Chlorination of drinking water should be done.
(ii) All the eatables in the school canteen should be kept covered.
(iii) The sale of stale fruits and other eatables should be prohibited.
(iv) If a student catches an infectious disease he should not be allowed to mix up with his classmates in the classroom as well as in the playground till he fully recovers.
(v) Children should be properly educated about personal as well as community hygiene.
(vi) To prevent communicable diseases all children should be given proper vaccination.

Question 4.
What is immunisation?
Answer:
The strength produced in the body to resist diseases is called immunity for example; disease producer is got entered into the body. This disease producer combats with the blood cells. In reaction to that some specific chemical substances are produced. These various substances destroy the pathogens in the body and thus, produce the ability of resistivity to combat diseases. It is called as immunity of the body and this process is called as immunisation.

Question 5.
What are the immunisation programmes available at the nearest health centre in your locality? Which of these diseases are the major health problems in your area ?
Answer:
In our town health centre following vaccination programmes are available:

• National tuberculosis eradication programme.
• National, cholera eradication programme.
• National leprosy eradication programme.
• National polio eradication campaign.
• National child communicable disease eradication (vaccination of children).

Generally, in all cities and towns problems in relation to hygienic environment community hygienic and pure drinking water one quite common. In proper disposal of garbage in cities and towns is responsible for polluting the environment. Pure water is not being available. Air and water pollution have been adding to health problems.

HBSE 9th Class Science Why Do We Fall Ill Textbook Questions and Answers

Question 1.
How many times did you fall ill in the last one year? What were the illnesses?
(i) Think of one change you could make in your habits in order to avoid any of/most of the above illnesses.
(ii) Think of one change you would wish for in your surroundings in order to avoid any of/most of the above illnesses.
Answer:
Every person can suffer from a minor or major diseases at any time. Hardly a few persons can be there who do not develop any disease, but even then due to some reasons health can be affected. I also fell ill in the last year. I had suffered from malaria.
(i) Different safety measures can be adopted to get rid of malaria. It can be avoided by filling the pits and landfills around our neighbourhood with soil and sand, drain blockage should be removed, water should be drained out from the flower pots, desert cooler-tanks and empty vessels in our houses, at night net and mosquito repellent cream should be applied to the skin. The doors and windows in the houses should be equipped with wire-gauze.

(ii) The growth of mosquitoes can be checked by keeping our surroundings clean and tidy pits and drains must not be let overflow or choked with stagnant water. Unwanted bushes or weeds should be destroyed.

Question 2.
A doctor/nurse/health worker is exposed to more sick people than others in the community. Find out how she/he avoids getting sick herself/himself.
Answer:
Doctors, nurses and health workers immensely remain in contact with patients than other people. To protect themselves from harmful bacteria or microbes following safety measures are adopted by them :
(1) They are habitual wearing washed and clean clothes every day.
(2) While examining a patient, doctor use to cover their mouth and nose with a mask so that the least microbes carry ing infection should not enter their body through breathing.
(3) Nurses and other medical workers wear rubber gloves while injecting or dressing a patient, so that they may not come in direct contact with the patient.
(4) After examining a patient, a doctor or another person at help wash up their hands with an antiseptic solution.
(5) Whole of the staff in the hospital is much concerned about common hygiene. The floors of the wards and rooms are swabbed with disinfectant phenyl regularly.
(6) The garbage and waste should be destroyed with scientific methods.

Question 3.
Conduct a survey in your neighbourhood to find out what the three most common diseases are. Suggest three steps that could be taken by your local authorities to bring down the incidence of these diseases.
Answer:
On conducting a survey in the neighbourhood it was found that the three most common diseases in my locality are:
(i) Diarrhoea and expectoration (vomiting)
(ii) Malaria
(iii) Cough and cold.

Following are the three suggestions to immunise ourselves against these diseases :
(1) There is a great need to take care of the eatables and drinking water they must be protected against any kind of pollution.
(2) Ensure that the water is covered so that mosquitoes can’t breed and we can have safe drinking water.
(3) Maintaining cleanliness and proper disposal system is needed.

Question 4.
A baby is not able to tell her/his caretakers that she/he is sick. What would help us to find out
(a) that the baby is sick?
(b) what is the sickness?
Answer:
(a) If a baby is not able to tell his/her caretaker that he/she is sick then body symptoms should be observed by the caretaker. The caretaker should take the baby to the doctor, if she or he finds a clue that the baby is sick. Some symptoms that can be observed are :
1. If the baby cries continuously.
2. If the baby doesn’t want to eat anything.
3. If the baby feels irritated all the time.

(b) Kind of sickness can be known by certain symptoms e.g. yellowness of skin and eyes, redness in eye may indicate conjunctivitis.

Question 5.
Under which of the following conditions is a person most likely to fall sick?
(a) when she is recovering from malaria.
(b) when she has recovered from malaria and is taking care of someone suffering from chicken pox.
(c) when she is on a four-day fast after recovering from malaria and is taking care of someone suffering from chicken pox. Why?
Answer:
A person is most likely to fall sick when she is on a four-day fast after recovering from malaria and is taking care of someone suffering from chicken pox. This is because since she has suffered from malaria and is in the recovery phase, it means that she is still not completely disease-free. Her chances of getting chicken pox are also high as her body has lowered.

Question 6.
Under which of the following conditions are you most likely to fall sick?
(a) when you are taking examinations.
(b) when you have travelled by bus and train for two days.
(c) when your friend is suffering from measles. Why?
Answer:
(a) I am most likely to fall sick when my friend is suffering from measles. Measles is a viral infection. It is a contagious disease. It spreads through contact with infected mucus and saliva. The virus can sustain for several hours. Even coming in contact or sharing items with an infected person results in the spread of disease.

Haryana State Board HBSE 9th Class Science Solutions Chapter 12 Sound Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 12 Sound

HBSE 9th Class Science Sound Intext Questions and Answers

Questions from Sub-section 12.2

Question 1.
How does the sound produced by a vibrating object in a medium reach your ear?
Answer:
The sound produced by a vibrating objcct reaches our ear through medium.

Questions from Sub-section 12.2.1

Question 1.
Explain how sound is produced by your school bell.
Answer:
The school bell produces sound through vibration, because when the hammer strikes the bell, it produces vibration in bell.

Question 2.
Why are sound waves called mechanical waves?
Answer:
The sound waves are produced by the motion of the particles of the medium, and mechanical work is possible by them, so sound waves are called mechanical waves.

Question 3.
Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Answer:
No, because sound needs a medium to propagate There is no medium on moon and hence it is not possible to hear any sound on Che moon.

Questions from Sub-section 12.2.3

Question 1.
Which wave property determines
(a) loudness
(b) patch?
Answer:
Pitch.

Question 2.
Guess which sound has a higher pitch : guitar or car horn?
Answer:
Car horn.

Questions from Sub-section 12.2.3

Question 1.
What arc wavelength, frequency, time period and amplitude of a sound wave?
Answer:
(i) Wavelength: The distance between two consecutive compresssions or two consecutive
rarefactions is called wavelength. It is denoted by a Greek alphabet, (lambda)
(ii) Frequency: The number of complete oscillations per unit second is called frequency.
(iii) TIme Period: The time taken by the wave for one complete oscillation is called time period.
(iv) AmplItude: The magnitude of maximum disturbance in the medium on either side of the mean value is called amplitude.

Question 2.
How are the wavelength and frequency of a sound wave related to its speed?
Answer:
υ = λv
or
Velocity wavelength × frequency.

Question 3.
Calculate the wavelength ora sound wave whose frequency is 220 Hz and speed Is 440 m/s in a given medium.
Solution:
Frequency of wave (υ) = 220 Hz
Velocity of wave (v) = 440 m/s
We know that, = λ × 220
λ = \(\frac {220}{440}\) = 2m

Question 4.
A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions from the source?
Solution:
Frequency of wave (υ) = 500 Hz
Time (T) =?
We know that,
v = \(\frac {1}{T}\)
Or
T = \(\frac {1}{v}\)
T = \(\frac {1}{500}\) = 0.002 sec

Questions from Sith-section 12.2.3

Question 1.
Distinguish between loudness and intensity of sound.
Answer:
The loudness of sound: Loudness is a measure of the response of the ear to the sound. It depends upon low and high vibrations.
The intensity of sound: The amount of sound energy passing each second through unit area is called the intensity of sound.

Questions from Sub-section 12.2.4

Question 1.
In which of three media, air, water or iron, does sound travel the fastest at a particular temperature?
Answer:
The sound travels the fastest in iron. the speed of sound at 25°C in air is 346 m/s, in water 1531 m/s and in iron 5950 m/s.

Questions from Sub-section, 12.3.

Question 1.
An echo returned in 3s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 ms^-1?
Solution:
Speed nf sound(v) = 342 ms^-1
lime taken by echo (t) = 3s
Distance covered by sound = velocity × time = 342 × 3 m = 1026 m
Thus. the distance of the reflecting surface from the source
\(\frac {1026}{2}\) m = 513 m

Questions from Sub-section 12.3.3

Question 1.
are the ceilings nl concert halls curved?
Answer:
The ceilings of concert halls arc curved so that sound after reflection reaches all corners of the hail.

Questions from Sub-section 12.4

Question 1.
What is the audible range of the average human ear?
Answer:
20 Hz to 20.000 Hz.

Question 2.
What is the range of frequencies associated with
(a) Infrasound?
(h) Ultrasound?
Answer:
(a) The range of infrasound is less than 20 Hz.
(b) The range of ultrasound is more than 20 kHz

Questions from Sub-section 12.5

Question 1.
A submarine emits a sonar pulse, which returns from an underwater cliff I.02s. If the speed of sound in salt water is 1531 mIs, how far awa is the cliff?
Solution:
Here, Time between transmission and detection (t) 1.02 s
Speed of sound in seawater (v) = 1531 m/s
Distance traveled by the ultrasound = 2 x distance of cliff from submarine = 2d
Where. d = distance of cliff from submarine
Total distance (2d) = speed x time = 153 1 m/s x 1.02 s = 1561.62rn
d = \(\frac {1561.62}{2}\) = 780.81m.

HBSE 9th Class Science Sound Textbook Questions and Answers

Question 1.
What is sound and how is it produced?
Answer:
Sound:
It is a form of energy that produces a sensation of hearing. Sound is produced by any vibrating material body. For example. the sound of the human voice is produced due to vibrations in the vocal cords. The strings of trumpets and shehnais vibrate when they are played. The vibration is also produced when the bell is rung. The vibration in bell can be felt by touching it. In this way sound is produced.

Question 2.
Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of the sound.
Answer:
When a vibrating object moves forward, it pushes and compresses the air in front of it creating a region of high pressure which is called compression (C). This compression starts to move away from the vibrating object. When the vibrating object moves backward, it creates a region of low pressure called rarefaction (R). Thus, due to the vibration of an object, a series of compressions and rarefactions is created in the air. These make the sound wave that propagates through the medium.

Question 3.
Cite an experiment to show that sound needs a material medium for its propagation.
Answer:
Sound is a mechanical wave. It requires a material medium for propagation. ¡t can
be proved by the following experiment:
Experiment:
Take an electric bell and an airtight glass bell jar. The electric bell is suspended inside the airtight bell jar. Connect the bell jar to a vacuum pump. Switch on the bell. The sound of bell is heard. Now pump out the air of bell jar by vacuum pump gradually. The sound becomes fainter. When there is full vacuum in bell jar, the sound of bell will not be heard. Thus, the experiment proves that sound cannot travel in vacuum. Sound need a material medium to travel. The air was the medium in bell jar so the sound of the electric bell was being heard.

Question 4.
Why is a sound wave called a longitudinal wave?
Answer:
In sound waves, the individual particles of the medium move in a direction parallel to the direction of propagation of the disturbance. The particles do not move from one place to another but they oscillate back and forth about their position of rest. Hence, sound waves are longitudinal waves.

Question 5.
Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
Answer:
Pitch of sound. How the brain interprets the frequency of an emitted sound is called pitch. Pitch helps us to identify the sound.

Question 6.
Flash and thunder are produced simultaneously. But thunder ¡s heard a few seconds after the flash is seen, why?
Answer:
At room temperature. the speed of sound is 344 metres per second and the velocity of light is 300000 km per second. Flash and thunder are produced simultaneously but due to the difference in the speed of both, flash is seen first and thunder is heard after it.

Question 7.
A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 ms^-1.
Solution:
The velocity of sound in air (y) = 344 ms^-1
First frequency of person (v₁) = 20 Hz
Second frequency of person (v₂) = 20 kHz = 20,000 Hz
First wavelength (λ₁) = \(\frac{v}{v_1}=\frac{344}{20}\) = 17.2 m
Second wavelength (λ₂) = \(\frac{v}{v_2}=\frac{344}{20,000}\) = 0.0172 m

Question 8.
Two children are at opposite ends of an aluminum rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminum to reach the second child.
Solution:
Suppose the length of rod = 1m
Speed of sound in air (v₁) = 346 m/s
Time taken in Air = = \(\frac {1m}{346m}\)s 1/346s
The speed of sound in aluminium (v₂) = 6420 m/s
Time taken in Aluminium = = \(\frac {1m}{6420m}\) = I /6420s
Ratio of both = \(\frac{1}{346}: \frac{1}{6420}\) = 18.55:1

Question 9.
The frequency of a source of sound is loo Hz. How many times does it vibrate in a minute?
Solution:
Frequency of sound (v) = 100 Hz
Number of vibrations in 1 second = 100
Number of vibrations in 60 seconds = 100 x 60 = 6000

Question 10.
Does sound follow the same laws of reflection as light does? Explain.
Answer:
Yes, sound follows the same laws of reflection as light does. Sound is reflected from the surface of
Laws:
(1) The angle of incidence is equal to the angle of reflection.
(2) Incident wave, reflected wave and the normal lie in the same plane. An obstacle of larger size which may be polished or rough is needed for the reflection of sound waves.

Question 11.
When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remains the same. Do you hear an echo sound on a hotter day?
Answer:
Yes, because the speed of sound is more on a hotter day than on a colder day.

Question 12.
Give two practical applications of reflection of sound waves.
Answer:
(1) During night, the bats feel their way by the reflection of sound.
(2) Ultrasound is used to break small ‘stones’ formed in the kidneys into fine grains.

Question 13.
A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g = 10 m/s^-2 and speed of sound = 340 ms^-1.
Solution:
Here, u 0, s 500 m, g 10 rns2
Buts = ut + gt² = 0 + \(\frac {1}{2}\) x 1o × t² ⇒ 5t² = 500 or = 10s
Distance of the top of the tower from surface faster = 500 m
Speed of sound = 340 ms^-1

\(\frac {500}{340}\) = 1.47s
The time taken by the splash to reach at the top = 10 + 1 .47 = 11.47 s

Question 14.
A sound wave travels at a speed of 339 ms^-1. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?
Solution:
Here, the Speed of the sound wave = 339 ms^-1
Wavelength = 1.5 cm = \(\frac {15}{10}\) = \(\frac {15}{1000}\) m
Frequency Wavelength = img = \(\frac {339}{15}\) x 1000 = 22600 Hz
This sound is not audible because a person cannot hear a sound more than 20,000 Hz.

Question 15.
What is reverberation? How can it be reduced?
Answer:
Reverberation:
The repeated reflection that results in the persistence of sound is called reverberation. Methods of reducing:
To reduce reverberation, the roof and walls of the auditorium are generally covered with sound absorbent materials like compressed fibreboard, rough plaster or draperies. The seat materials are also selected on the basis of their sound-absorbing properties.

Question 16.
What is the loudness of sound? What factors does it depend on ?
Answer:
The measure of the response of the ear to the sound is called the loudness of sound. It is the sensation of hearing. This level of sensation depends upon the loud or faint vibrations of the source. When vibrations are loud, the loudness of sound is more and when vibrations are faint, the loudness of sound is faint. The loudness or softness of the sound is found by the amplitude. The sound of high energy has more loudness.

Question 17.
Explain how bats use ultrasound to catch a prey.
Answer:
The bat searches out prey in dark night by emitting and detecting reflections of ultrasonic waves. The high pitch ultrasonic squeaks of the hat are reflected from the prey and returned to the bat’s ear as shown in fig. The nature of reflections tells that bat where the prey is and it catches it. In this way the bats use ultrasound to catch prey.

Question 18.
How is ultrasound used for cleaning?
Answer:
Objects to be cleaned are placed in a cleansing solution and ultrasonic waves are sent into the solution. Due to the high frequency. the particles of the dust, grease and dirt get detected and drop out. This technique can clean those objects which are difficult to clean due to their large size.

Question 19.
Explain the working and application of a sonar.
Answer:
Procedure:
SONAR-Sound Navigation And Ranging is such a device in which ultrasonic waves are used to measure the distance, direction and speed of underwater objects. Sonar consists of a transmitter and a detector and is installed in a boat or a ship as shown in fig. The transmitter produces and transmits ultrasonic waves and travel through water and after striking the object on the seaboat, get reflected back and are sensed by the detector. The distance of the object that reflected the sound wave can he calculated by knowing the speed of sound in water and the time interval between transmission and reception of the ultrasound.

Applications: Following are its applications:

(1) To determine the depth of the sea and oil wells.
(2) To find the distance of groups of fish.
(3) To locate the enemy submarine or torpedo.
(4) To cate underwater hill valleys. icebergs and sunken slp.

Question 20.
A sonar device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of sound in water, if the distance of the object from the submarine is 3625 m.
Solution:
Distance of the object from the submarine (d) = 3625 m
The echo covers the double distance.
2d = 2 x 3625 = 7250 m
Time (t) = 5 s
Speed of sound in water = \(\frac {2d}{t}\) = imh = \(\frac {7250}{5}\) = 1450 m/s
Thus, speed of sound in water 1450 m/s

Question 21.
Explain how defects in a metal block can be detected using ultrasound.
Answer:
Ultrasound can be used to detect cracks and flaws in metal blocks. The cracks or holes inside the metal blocks, which are invisible from the outside reduce the strength of the structure of the building or bridge.

Ultrasonic waves are allowed to pass through the metal block and detectors are used to detect the transmitted waves. If there is even a small defect, the ultrasound gets reflected back indicating the presence of the flaw or defect. Ordinary sound of longer wavelengths cannot be used for such purpose as it will bend around the corners of the defective location and enter the detector and instead of detecting the flaw, they create a state of confusion.

Question 22.
E%plaìn how the human ear works.
Answer:
The outer ear of human being collects the sound from surroundings and passed through the auditory canal, and the pressure on the outside of the membrane increases. The pressure from inside is less. Due to the difference in pressure, the eardrum starts vibrating. The vibrations are amplified several times by three bones (the hammer, anvil and stirrup) in middle ear. In the inner ear, the pressure variations are turned into electric signals by the cochlea. These electrical signals are sent to the brain via the auditory nerve, and the brain interprets them as sound.

Haryana State Board HBSE 9th Class Science Solutions Chapter 11 Work and Energy Textbook Exercise Questions and Answers.

Haryana Board 9th Class Science Solutions Chapter 11 Work and Energy

HBSE 9th Class Science Work and Energy Intext Questions and Answers

Questions from Sub-Section 11.1

Question 1.
A force of 7N acts on an object. The displacement is, say 8 m in the direction of the force (Fig. 11.1). Let us take it that the force acts on the object through the displacement. What is the work done in this case?

Solution:
Here,
Force (F) = 7N
Displacement (s) = 8m
Work done (W) =?
We know that, Work done (W) = Force (F) × Displacement (s) = 7 × 8N.m = S6 Joule

Questions from Sub-Section 11.1

Question 1.
When do we say that work is done?
Answer:
When an object is displaced by applying force, it is said to be work done.
Thus, work done (W) = Force (F) × Displacement (s)

Question 2.
Write an expression for the work done when a force is acting on an object in the direction of its displacement?
Answer:
When a force is acting on an object in the direction of its displacement. The equation of work will be as follows:
Work done (W) = Force (F) X Displacement (s)
Thus, work done ¡s positive when the force is ¡n the direction of displacement.

Question 3.
Define 1 J of work.
Answer:
Work is said to be done 1 joule when a force of 1 N acting on an object and it is displaced through 1 m in the direction of force.
Thus, 1 J = 1 N x lm

Question 4.
A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in plowing the length of the field?
Solution:
Here,
Force (F) = 140 N
Displacement (s) = 15 m
Work done (W) = F × S = 140 × 15 = 21001

Questions from Sub-Section 11.2

Question 1.
What is the kinetic energy of an object?
Answer:
The energy possessed by an object due to its motion is called kinetic energy. A body of mass m moving with velocity v has kinetic energy (E) equal to \(\frac {1}{2}\)mv².

Question 2.
Write an expression for the kinetic energy of an object.
Answer:
The kinetic energy of a body of mass m moving with velocity v is; K.E = \(\frac {1}{2}\) v²

Question 3.
The kinetic energy of an object of mass m moving with a velocity of 5 ms^-1 is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times ?
Answer:
The kinetic energy of an object of mass, m moving with a velocity of 5 ms^-1 is 25 J. If its velocity is doubled (10 ms^-1), its kinetic energy will increase four (2^-1) times i.e. 100 J. But if its velocity is tripled (15 ms^-1), its kinetic energy will increase nine (3²) times i.e. 225 J.

Questions from Sub-Section 11.3

Question 1.
What is power ?
Answer:
Power is defined as the rate of doing work or the rate of transfer of energy. If an agent does a work W in time t, then power is given by :

Unit of power is watt (W).

Question 2.
Define 1 watt of power.
Answer:
1 watt is the power of an agent, which does work at the rate of 1 joule per second.

Question 3.
A lamp consumes 1000 J of electric energy in 10 s. What is its power ?
Solution:
Here,
Work done (W) = 1000J Time(t) = 10 s Power (P) = ?
We know that,
Work done(W) = 1000 J
Time(t) = 10 s
Power(P) = ?
We know that,
P = \(\frac {W}{t}\) = \(\frac {1000}{10}\) Js^-1= 100 W(Watt)

Question 4.
Define average power.
Answer:
The ratio of total energy used to total time given is called average power. Thus,

HBSE 9th Class Science Work and Energy Textbook Questions and Answers

Question 1.
Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.
(i) Suma is swimming in a pond.
(ii) A donkey is carrying a load on its back.
(iii) A wind-mill is lifting water from a well.
(iv) A green plant is carrying out photosynthesis.
(v) A engine is pulling a train.
(vi) Food grains are getting dried in the sun.
(vii) A sailboat is moving due to wind energy.
Answer:
(i) There is work when suma is swimming in a pond, because she is displacing in the direction of force.
(it) No work is done because its displacement is zero.
(iii) Work is done in lifting the water because it displaces in the direction of force.
(iv) No work is done because displacement is zero.
(v) Work is done in pulling the train because the train is displacing in the direction of engine.
(vi) No work is done because the displacement is zero.
(vii) Work is done by the wind energy in producing motion in sailboat because sail boat is displacing in the direction of force.

Question 2.
An object was thrown at a certain angle to the ground moves in a curved path and falls back to. the ground. The initial and the final point of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?
Answer:
The work done by the force of gravity on the object is zero because the displacement of the object is zero.

Question 3.
A battery lights a bulb. Describe the energy changes involved in the process.
Answer:
When a battery lights a bulb, chemical energy is converted in electric energy.

Question 4.
The certain force acting on a 20 kg mass changes its velocity from 5 ms-1 to 2 ms^-1. Calculate the work done by the force.
Solution:
Here,
Mass of the body (m) = 20 kg
Initial velocity of the body (u) = 5 ms^-1
Final velocity of the body (v) = 2 ms^-1
Initial kinetic energy of the body (E₁) = \(\frac {1}{2}\) mu² = \(\frac {1}{2}\) × 20 × (5)² J = 250J
Final kinetic energy of the body (E₂) = \(\frac {1}{2}\)m(v)² = \(\frac {1}{2}\) × 20 × (2)² J = 40 J
Thus, Work done = Change in the Kinetic Enegry = E₁ – E₂
= 250 J – 40J = 210 J

Question 5
A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal. What is the work done on the object by the gravitational force ? Explain your answer.
Solution:
Here,
Mass of the object (m) = 10 kg
Force of gravity (g) = 9.8 ms^-2
Height (h) = 0
Work done by force of gravity (w) = mgh = 10 × 9.8 × 0 = 0
Thus, the work done by force of gravity is zero because displacement is horizontal.

Question 6.
The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy ? Why ?
Answer:
The potential energy of a freely falling object decreases progressively but this does not violate the law of conservation of energy because this energy is transferred in kinetic energy. The total potential energy nearest to earth changes in kinetic energy and on reaching the earth, it converts in potential energy.

Question 7.
What are the various energy transformations that occur when you are riding a bicycle?
Answer:
When we ride a bicycle, our muscular energy changes in kinetic energy and we get muscular energy from the food after converting in chemical energy.

Question 8.
Does the transfer of energy take place when you push a huge rock with all your might and fail to move it ? Where is the energy you spend going?
Answer:
When we push a huge rock with all our might and fail to move it, the transformation of energy in this stage wastes against friction and the work done is considered as zero.

Question 9.
A certain household has consumed 250 units of energy during a month. How much energy is , in joules?
Solution:
Here,
Energy consumed = 250 units = 250 kWh = 250 × 1000 × 3600s = 900000000 watt second = 9 × 108J

Question 10.
An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy ? If the object is allowed to fall, find its kinetic energy when it is halfway down, (g = 10 ms^-2 )
Solution:
Here,
The mass of the object (m) = 40 kg
Height of the object from the earth (h) = 5 m
Acceleration due to gravity (g) = 10 ms^-2
Potential energy of the object (w) = mgh
40 × 5 × 10 J = 200J
If the object is allowed to fall freely, the half of the potential energy of the object in the halfway will convert in kinetic energy. Therefore the kinetic energy of the object when it is half-way down = \(\frac {2000}{2}\) = 1000J

Question 11.
What is the work done by the force of gravity on a satellite moving around the earth? Justify your answer.
Answer:
The work done by the force of gravity on a satellite moving around the earth is zero because the displacement of both is zero.

Question 12.
Can there be displacement of an object in the absence of any force acting on it ? Think. Discuss this question with your friends and teacher.
Answer:
Ho, there cannot be displacement of an object in the absense of any force acting on it because the displacement is always due to unbalanced force.

Question 13.
A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not ? Justify your answer.
Answer:
The work done by the person is zero when he hold a bundle of hay over his head for 30 minutes because the displacement is zero.

Question 14.
An electric heater is rated 1500 W. How much energy does it use in 10 hours ?
Solution:
Here,
The power of the electric heater (P) = 1500W
Time (t) = 10 hours
Energy (w) = P × t = 15000 Wh = \(\frac {15000}{1000}\) kWh = 15 k Wh

Question 15.
Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest ? What happens to its energy eventually? Is it a violation of the law of conservation of energy.gy ?
Solution:
When we take the bob or pendulum from its middle position to one side (let left side) upto some height, the work done by it is conserved in the form of potential energy. When the bob is allow to swing, then the total potential energy of the bob changes into kinetic energy on reaching left-to-right and middle position. This kinetic energy takes the bob upto that height at which the kinetic energy is changed into potential energy. Due to this potential energy, the bob oscillates again from right to middle. This continuous and bob or pendulum moves from right-left.

After sometime bob comes in rest because frictional force of air works on it and the energy is wasted to work against it. This is not the violation of the law of conservation of energy.

Question 16.
An object of mass m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest ?
Answer:
An object of mass is moving with a constant velocity v, its kinetic energy is \(\frac {1}{2}\) mv². The work is equal to kinetic energy should be done on the object in order to bring the object to rest.

Question 17.
Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h ?
Solution:
Here,
Mass of the car (m) = 1500 kg
Velocity of the car (v) = 60 km/h
= \(\frac{60 \times 1000}{3600}\) ms^-1 = \(\frac{50}{3}\) ms^-1
Thus, kinetic energy of the car = \(\frac {1}{2}\) mv² = \(\frac {1}{2}\) = 1500 × \(\frac {1}{2}\) × \(\frac {1}{2}\)J = 208333.3 J
∴ The Work Required to be done to stop the car = 208333.3 J

Question 18.
In each of the following, a force F is acting on an object of mass m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.

Answer:
The work done in the first stage is zero, in the second stage is positive and in the third stage is negative.

Question 19.
Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?
Answer:
Yes, we agree with Soni because when several forces are acting on an object, the total effect of the force becomes zero, Thus, F = 0, from this, ma = 0, but m cannot be zero, Therefore, the acceleration can be zero.

Question 20.
Find the energy in kW h consumed in 10 hours by four devices of Dower 500 W each.
Solution:
Here,
The total power of four devices (P) = 500 W × 4 = 2000W
Time (t) = 10 hours
The energy consumed = Power × Time
= 2000 W × 10 h
\(\frac {20000}{1000}\) k Wh = 20kWh

Question 21.
A freely falling object eventually stops on reaching the ground. What happenes to its kinetic energy ?
Answer:
Its kinetic energy is transferred in potential energy.

Haryana State Board HBSE 9th Class Maths Solutions Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Ex 13.1 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 13 पृष्ठीय क्षेत्रफल एवं आयतन Exercise 13.1

प्रश्न 1.
1.5 मी० लंबा, 1.25 मी० चौड़ा और 65 सें०मी० गहरा प्लास्टिक का एक डिब्बा बनाया जाना है। इसे ऊपर से खुला रखना है। प्लास्टिक शीट की मोटाई को नगण्य मानते हुए, निर्धारित कीजिए :
(i) डिब्बा बनाने के लिए आवश्यक प्लास्टिक शीट का क्षेत्रफल।
(ii) इस शीट का मूल्य, यदि 1 मी०² शीट का मूल्य ₹ 20 है।
हल :
यहां पर,
l = 1.5 मी०
b = 1.25 मी०
h = 65 सें०मी० = 0.65 मी०
(i) खुला डिब्बा बनाने के लिए आवश्यक प्लास्टिक शीट का क्षेत्रफल
= 2 [l + b] × h + l × b
= 2 [1.5 + 1.25] × 0.65 + 1.5 × 1.25
= [2 × 2.75 × 0.65 + 1.875 ] मी०²
= [3.575 + 1.875] मी०²
= 5.45 मी०² उत्तर
(ii) 1 मी०² प्लास्टिक शीट का मूल्य = ₹ 20
5.45 मी०² प्लास्टिक शीट का मूल्य = ₹ (5.45 × 20)
= ₹ 109 उत्तर

प्रश्न 2.
एक कमरे की लंबाई, चौड़ाई और ऊंचाई क्रमशः 5 मी०, 4 मी० और 3 मी० हैं। ₹ 7.50 प्रति मी० की दर से इस कमरे की दीवारों और छत पर सफेदी कराने का व्यय ज्ञात कीजिए।
हल :
यहां पर,
l = 5 मी०
b = 4 मी०
h = 3 मी०
कमरे की चारदीवारी व छत का क्षेत्रफल = 2 [l + b] × h + l × b
= 2 [5 + 4] × 3 + 5 × 4
= [2 × 9 × 3 + 20 ] मी०²
= [54 + 20] मी०²
= 74 मी०²
1 मी०² क्षेत्रफल पर सफेदी कराने का व्यय = ₹ 7.50
74 मी०² क्षेत्रफल पर सफेदी कराने का व्यय = ₹ (74 × 7.50)
= ₹ 555 उत्तर

प्रश्न 3.
किसी आयताकार हॉल के फर्श का परिमाप 250 मी० है। यदि ₹ 10 प्रति मी०² की दर से चारों दीवारों पर पेंट कराने की लागत ₹ 15000 है, तो इस हॉल की ऊंचाई ज्ञात कीजिए।
हल :
यदि खर्च ₹ 10 हो तो चारों दीवारों का क्षेत्रफल = 1 मी०²
यदि खर्च ₹ 1 हो तो चारों दीवारों का क्षेत्रफल = \(\frac{1}{10}\) मी०²
यदि खर्च ₹ 15000 हो तो चारों दीवारों का क्षेत्रफल = \(\frac{1}{10}\) × 15000 = 1500 मी०² .
प्रश्नानुसार
आयताकार हॉल के फर्श का परिमाप = 250 मी०
⇒ 2 (l + b) = 250 मी०
या l + b = \(\frac{250}{2}\) मी० = 125 मी०
चारों दीवारों का क्षेत्रफल = 1500 मी०²
⇒ 2 (2 + b) × h = 1500
या 2 × 125 × h = 1500
या h = \(\frac{1500}{2 \times 125}\) = 6 मी०
अतः हॉल की ऊंचाई (h) = 6 मी० उत्तर

प्रश्न 4.
किसी डिब्बे में भरा हुआ पेंट 9.375 मी०² के क्षेत्रफल पर पेंट करने के लिए पर्याप्त है। इस डिब्बे के पेंट से 22.5 सें०मी० × 10 सें०मी० × 7.5 सें०मी० विमाओं वाली कितनी ईंट पेंट की जा सकती हैं ?
हल :
यहां पर एक ईंट के लिए
l = 22.5 सें०मी०
b = 10 सें०मी०
h = 7.5 सें०मी०
∴ एक ईंट का पृष्ठीय क्षेत्रफल = 2 [l × b + b × h + h × l]
= 2 [22.5 × 10 + 10 × 7.5 + 7.5 × 22.5]
= 2 [225 + 75 + 168.75 ] सें०मी०²
= 2 × 468.75 सें०मी०²
= 937.5 सें०मी०²
हम जानते हैं कि
9.375 मी०² = 9.375 × 10000 सें०मी०² = 93750 सें०मी०²
1 डिब्बा पेंट से जितना क्षेत्रफल पेंट किया जा सकता है = 93750 सें०मी०²
∴ ईंटों की संख्या = \(\frac{93750}{937.5}=\frac{93750 \times 10}{9375}\)
= 100 उत्तर

प्रश्न 5.
एक घनाकार डिब्बे का एक किनारा 10 सें०मी० लंबाई का है तथा एक अन्य घनाभाकार डिब्बे की लंबाई, चौड़ाई और ऊंचाई क्रमशः 12.5 सें०मी०, 10 सें०मी० और 8 सें०मी० हैं।
(i) किस डिब्बे का पार्श्व पृष्टीय क्षेत्रफल अधिक है और कितना अधिक है ?
(ii) किस डिब्बे का कुल पृष्ठीय क्षेत्रफल कम है और कितना कम है ?
हल:
घनाकार डिब्बे का किनारा (a) = 10 सें०मी०
घनाभाकार डिब्बे के लिए
l = 12.5 सें०मी०
b = 10 सें०मी०
h = 8 सें०मी०
(i) घनाभाकार डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल
= 2 [l + b] × h
= 2 [12.5 + 10] × 8 सें०मी०²
= 2 × 22.5 × 8 सें०मी०²
= 360 सें०मी०²
घनाकार डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल = 2 [a + a] × a
= 2 [10 + 10] × 10 सें०मी०²
= 2 × 20 × 10 सें०मी०²
= 400 सें०मी०²
अतः घनाकार डिब्बे का पार्श्व पृष्ठीय क्षेत्रफल अधिक है।
जितना अधिक है = (400 – 360 ) सें०मी० = 40 सें०मी०² उत्तर

(ii) घनाभाकार डिब्बे का कुल पृष्ठीय क्षेत्रफल = 2 [lb + bh + hl]
= 2 [12.5 × 10 + 10 × 8 + 8 × 12.5 ] सें०मी०²
= 2 [125 + 80 + 100 ] सें०मी०²
= 2 × 305 सें०मी०²
= 610 सें०मी०²
घनाकार डिब्बे का कुल पृष्ठीय क्षेत्रफल = 6 a²
= 6 (10)2 सें०मी०²
= 6 × 100 सें०मी०²
= 600 सें०मी०²
घनाकार डिब्बे का कुल पृष्ठीय क्षेत्रफल कम है।
जितना कम है = (610 – 600 ) सें०मी०² = 10 सें०मी०² उत्तर

प्रश्न 6.
एक छोटा पौधा घर (green house) संपूर्ण रूप से शीशे की पट्टियों से (आधार भी सम्मिलित है) घर के अंदर ही बनाया गया है और शीशे की पट्टियों को टेप द्वारा चिपका कर रोका गया है। यह पौधा घर 30 सें०मी० लंबा, 25 सें०मी० चौड़ा और 25 सें०मी० ऊंचा है।
(i) इसमें प्रयुक्त शीशे की पट्टियों का क्षेत्रफल क्या है ?
(ii) सभी 12 किनारों के लिए कितने टेप की आवश्यकता है ?
हल :
यहां पर, पौधा घर की लंबाई (l) = 30 सें०मी०
पौधा घर की चौड़ाई (b) = 25 सें०मी०
पौधा घर की ऊंचाई (h) = 25 सें०मी०
(i) शीशे की प्रयुक्त पट्टियों का क्षेत्रफल = 2 [lb + bh + hl]
= 2 [30 × 25 + 25 × 25 + 25 × 30 ] सें०मी०²
= 2 [750 + 625 + 750] सें०मी०²
= 2 × 2125 सें०मी०²
= 4250 सें०मी०² उत्तर

(ii) 12 किनारों के लिए आवश्यक टेप = 4 [l + b + h]
= 4 [30 + 25 + 25] सें०मी०
= 4 × 80 सें०मी०
= 320 सें०मी० उत्तर

प्रश्न 7.
शांति स्वीट स्टाल अपनी मिठाइयों को पैक करने के लिए गत्ते के डिब्बे बनाने का ऑर्डर दे रहा था। दो मापों के डिब्बों की आवश्यकता थी। बड़े डिब्बों की माप 25 सें०मी० 20 सें०मी० × 5 सें०मी० और छोटे डिब्बों की माप 15 सें०मी० × 12 सें०मी० × 5 सें०मी० थीं। सभी प्रकार की अतिव्यापिकता (overlaps) के लिए कुल पृष्ठीय क्षेत्रफल के 5% के बराबर अतिरिक्त गत्ता लगेगा। यदि गत्ते की लागत ₹ 4 प्रति 1000 सें०मी०² है, तो प्रत्येक प्रकार के 250 डिब्बे बनवाने की कितनी लागत आएगी ?
हल :
पहली प्रकार के डिब्बे के लिए
l = 25 सें०मी०
b = 20 सें०मी०
h = 5 सें०मी०
1 डिब्बे का पृष्ठीय क्षेत्रफल = 2 [lb + bh + hl]
= 2 [25 × 20 + 20 × 5 + 5 × 25 ] सें०मी०²
= 2 [500 + 100 + 125 ] सें०मी०²
= 2 × 725 सें०मी०²
= 1450 सें०मी०²
250 डिब्बों का पृष्ठीय क्षेत्रफल = 250 × 1450 सें०मी०²
= 362500 सें०मी०²
दूसरी प्रकार के डिब्बे के लिए
l = 15 सें०मी०
b = 12 सें०मी०
h = 5 सें०मी०
1 डिब्बे का पृष्ठीय क्षेत्रफल = 2 [lb + bh + hl]
= 2 [15 × 12 + 12 × 5 + 5 × 15] सें०मी०²
= 2 [180 + 60 + 75] सें०मी०²
= 2 × 315 सें०मी०²
= 630 सें०मी०²
250 डिब्बों का पृष्ठीय क्षेत्रफल = 250 × 630 सें०मी०²
= 157500 सें०मी०²
दोनों प्रकार के (250 + 250) = 500 डिब्बों का पृष्ठीय क्षेत्रफल = [362500 + 157500] सें०मी०²
= 520000 सें०मी०²
अतिव्यापिकता (overlaps) के आवश्यक गत्ते का पृष्ठीय क्षेत्रफल = \(\frac{520000 \times 5}{100}\) =26000 सें०मी०²
500 डिब्बों के कुल आवश्यक गत्ते का पृष्ठीय क्षेत्रफल = [520000 + 26000] सें०मी०²
= 546000 सें०मी०²
1000 सें०मी०2 गत्ते का व्यय = ₹ 4
1 सें०मी०2 गत्ते का व्यय = ₹ \(\frac{4}{1000}\)
546000 सें०मी०2 गत्ते का व्यय = \(\frac{4}{1000}\) × 546000 सें०मी०²
= ₹ 2184 उत्तर

प्रश्न 8.
परवीन अपनी कार खड़ी करने के लिए, एक संदूक के प्रकार के ढांचे जैसा एक अस्थाई स्थान तिरपाल की सहायता से बनाना चाहती है, जो कार को चारों ओर से और ऊपर से ढक ले (सामने वाला फलक लटका हुआ होगा जिसे घुमाकर ऊपर किया जा सकता है)। यह मानते हुए कि सिलाई के समय लगा तिरपाल का अतिरिक्त कपड़ा नगण्य होगा, आधार विमाओं 4 मीटर × 3 मीटर और ऊंचाई 2.5 मीटर वाले इस ढांचे को बनाने के लिए कितने तिरपाल की आवश्यकता होगी?
हल :
यहां पर,
l = 4 मी०
b = 3 मी०
h = 2.5 मी०
कार ढकने के लिए बनाए जाने वाले ढांचे के लिए आवश्यक तिरपाल
= 2 [l + b] × h + l × b
= 2 [4 + 3] × 2.5 + 4 × 3
= [2 × 7 × 2.5 + 12 ] मी०²
= [35.0 + 12] मी०²
= 47 मी०² उत्तर

Haryana State Board HBSE 9th Class Maths Solutions Chapter 14 सांख्यिकी Ex 14.4 Textbook Exercise Questions and Answers.

Haryana Board 9th Class Maths Solutions Chapter 14 सांख्यिकी Exercise 14.4

प्रश्न 1.
एक टीम ने फुटबाल के 10 मैचों में निम्नलिखित गोल किए :
2, 3, 4, 5, 0, 1, 3, 3, 4, 3
इन गोलों के माध्य, माध्यक और बहुलक ज्ञात कीजिए।
हल :
माध्य (\(\bar{x}\)) = \(\frac{2+3+4+5+0+1+3+3+4+3}{10}=\frac{28}{10}\) = 2.8 उत्तर
आंकड़ों का आरोही क्रम = 0, 1, 2, 3, 3, 3, 3, 4, 4, 5

बहुलक = 3 [∵ 3 सबसे अधिक बार आता है। उत्तर

प्रश्न 2.
गणित की परीक्षा में 15 विद्यार्थियों ने (100 में से) निम्नलिखित अंक प्राप्त किए :
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
इन आंकड़ों के माध्य, माध्यक और बहुलक ज्ञात कीजिए।
हल :
माध्य (\(\bar{x}\)) = \(\frac{41+39+48+52+46+62+54+40+96+52+98+40+42+52+60}{15}\)
= \(\frac{822}{15}\) = 54.8 उत्तर
आंकड़ों का आरोही क्रम = 39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98
माध्यक = \(\frac{15+1}{2}\) वां आंकड़ा = \(\frac{16}{2}\) = 8वां आंकड़ा
= 52 उत्तर
बहुलक = 52 [∵ 52 का आंकड़ा सबसे अधिक बार आता है।] उत्तर

प्रश्न 3.
निम्नलिखित प्रेक्षणों को आरोही क्रम में व्यवस्थित किया गया है। यदि आंकड़ों का माध्यक 63 हो, तो x का मान ज्ञात कीजिए :
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
हल :
दिए गए आंकड़े = 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
क्योंकि यहां पर आंकड़ों की संख्या = 10

या 126 = 2x + 2
या 2x = 126 – 2 = 124
x = \(\frac{124}{2}\) = 62 उत्तर

प्रश्न 4.
आंकड़ों 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18 का बहुलक ज्ञात कीजिए।
हल :

क्योंकि सबसे अधिक बारंबारता आंकड़े 14 की 4 है।
∴ बहुलक = 14 उत्तर

प्रश्न 5.
निम्न सारणी से एक फैक्टरी में काम कर रहे 60 कर्मचारियों का माध्य वेतन ज्ञात कीजिए :

हल :

माध्य वेतन (\(\bar{x}\)) = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\)
= \(\frac{305000}{60}\)
= 5083.33 रुपये उत्तर

प्रश्न 6.
निम्न स्थिति पर आधारित एक उदाहरण दीजिए :
(i) माध्य ही केंद्रीय प्रवृत्ति का उपयुक्त माप है।
(ii) माध्य केंद्रीय प्रवृत्ति का उपयुक्त माप नहीं है, जबकि माध्यक एक उपयुक्त माप है।
हल :
(i) माध्य अपने अद्वितीय मान के कारण केंद्रीय प्रवृत्ति का एक उपयुक्त माप है तथा इसका उपयोग अलग-अलग आंकड़ों के समूह की तुलना करने के लिए किया जा सकता है। जैसे 3, 4, 5, 6, 7, 8, 9 [माध्य = माध्यक = 6]
(ii) माध्य का उपयोग गुण-दोषों जैसे-सुंदरता, ईमानदारी, बुद्धिमानी आदि को मापने के लिए नहीं किया जा सकता है।