Author name: Prasanna

Haryana State Board HBSE 10th Class Maths Important Questions Chapter 5 समांतर श्रेढ़ियाँ Important Questions and Answers.

Haryana Board 10th Class Maths Important Questions Chapter 5 समांतर श्रेढ़ियाँ

परीक्षोपयोगी अन्य महत्त्वपूर्ण प्रश्न :

प्रश्न 1.
निम्नलिखित AP के लिए प्रथम पद a और सार्व अंतर d लिखिए-
(i) -225, -425, -625, -825,
(ii) \(\frac{3}{2}, \frac{1}{2},-\frac{1}{2},-\frac{3}{2}\), ………………..
हल :
(i) यहाँ पर AP = – 225, – 425, – 625, – 825, ………..
अतः प्रथम पद (a) = – 225
और सार्व अंतर (d) = – 425 – (225)
= – 425 + 225 = – 200

(ii) यहाँ पर AP = \(\frac{3}{2}, \frac{1}{2},-\frac{1}{2},-\frac{3}{2}\), …………….
अतः प्रथम पद (a) = \(\frac{3}{2}\)
और सार्व अंतर (d) = \(\frac{1}{2}-\frac{3}{2}\)
= \(\frac{1-3}{2}\)
= \(\frac{-2}{2}\) = – 1

प्रश्न 2.
AP का सार्व अंतर ज्ञात कीजिए और उसके अगले दो पद लिखिए-
(i) 51, 59, 67, 75, …………
(ii) 75, 67, 59, 51, ………..
हल :
(i) यहाँ पर
AP = 51, 59, 67, 75, ……….
⇒ प्रथम पद (a₁) = 51
सार्व अंतर (d) = 59 – 51 = 8
∴ a₅ = a₄ + d = 75 + 8 = 83
a₆ = a₅+ d = 83 + 8 = 91

(ii) यहाँ पर
⇒ प्रथम पद (a₁) = 75
सार्व अंतर (d) = 67 – 75 = – 8
∴ a₅ = a₄ + d = 51 + (-8) = 51 – 8 = 43
a₆ = a₅ + d = 43 + (-8) = 43 – 8 = 35

प्रश्न 3.
समांतर श्रेढ़ी 2, 4, 6, 8, 10, …………… का 15वाँ पद ज्ञात कीजिए ।
हल :
यहाँ पर दिया गया अनुक्रम है-
2, 4, 6, 8, 10, …….
प्रथम पद (a) = 2
और सार्व अंतर (d) = 4 – 2 = 2
∵ a_n = a + (n – 1)d
a₁₅ = a + (15 – 1)d
= 2 + 14 × 2
= 2 + 28 = 30

प्रश्न 4.
उस AP का 12 वाँ पद ज्ञात कीजिए जिसका –
(i) प्रथम पद 9 और सार्व अंतर 10 है ।
(ii) प्रथम पद – 20 और सार्व अंतर 4 है ।
हल :
(i) यहाँ पर
प्रथम पद (a) = 9
सार्व अंतर (d) = 10
∴ a₁₂ = a + 11d
= 9 + 11 × 10
= 9 + 110
= 119

(ii) यहाँ पर
प्रथम पद (a) = – 20
सार्व अंतर (d) = 4
∴ a₁₂ = a + 11d
= – 20 + 11 × 4
= – 20 + 44
= 24

प्रश्न 5.
समांतर श्रेढ़ी 7, 3, -1, -5, -9, …………. का n वाँ पद ज्ञात कीजिए ।
हल :
यहाँ पर दिया गया अनुक्रम है-
7, 3, -1, -5, -9, ……….
प्रथम पद (a) = 7
सर्व अंतर (d) = 3 – 7 = – 4
a_n = a + (n – 1)d
= 7 + (n – 1) (-4)
= 7 – 4n + 4 = 11 – 4n

प्रश्न 6.
प्रथम 1000 धन – पूर्णांकों का योग ज्ञात कीजिए।
हल :
यहाँ पर प्रथम 1000 धन पूर्णांक = 1, 2, 3, 4, …………, 1000
∴ a = 1
l = 1000
n = 1000
A.P. के प्रथम n पदों का योगफल (S_n) = \(\frac {n}{2}\)(a + l)
⇒ S¹⁰⁰⁰ = \(\frac {1000}{2}\)(1 + 1000)
= 500 × 1001
= 500500

प्रश्न 7.
उस समांतर श्रेढ़ी का पता करें जिसका तीसरा पद 16 और सातवाँ पद 40 है ।
हल :
माना समांतर श्रेढ़ी = a, a + d, a + 2d, …………….
तो a_n = a + 2d
तथा a₇ = a + 6d
प्रश्नानुसार,
a₃ = 16
⇒ a + 2d = 16 ………(i)
तथा a₇ = 40
⇒ a + 6d = 40 …….(ii)
समीकरण (i) को समीकरण (ii) में से घटाने पर,
4d = 24
या d = \(\frac {24}{4}\) = 6
d का मान समीकरण (i) में रखने पर,
a + 2(6) = 16
या a = 16 – 12 = 4
∵ समांतर श्रेढ़ी का प्रथम पद 4 और इसका सार्व अंतर 6 है ।
∴ समांतर श्रेढ़ी = 4, 10, 16, 22, 28, 34, …………….

प्रश्न 8.
क्या संख्याओं की सूची 5, 11, 17, 23, …………. का कोई पद 301 है, क्यों?
हल :
यहाँ पर सूची = 5, 11, 17, 23, ……………
a₂ – a₁ = 11 – 5 = 6
a₃ – a₂ = 17 – 11 = 6
a₄ – a₃ = 23 – 17 = 6
क्योंकि दो क्रमागत पदों के बीच अंतर समान है, इसलिए दी गई सूची A . P . है ।
⇒ प्रथम पद (a) = 5
सार्व अंतर (d) = 11 – 5 = 6
माना AP का n वाँ पद 301 है तो
a_n = a + (n – 1)d
301 = 5 + (n – 1) × 6
(n – 1) = \(\frac{301-5}{6}=\frac{296}{6}=\frac{148}{3}\)
n = \(\frac {148}{3}\) + 1 = \(\frac{148+3}{3}=\frac{151}{3}\)
क्योंकि एक धनात्मक पूर्णांक नहीं है, इसलिए दी गई A. P. कोई पद 301 नहीं हो सकता ।

प्रश्न 9.
एक A. P. का 8वाँ पद – 23 है तथा 12वाँ पद – 39 है। A. P. ज्ञात कीजिए ।
हल :
माना A.P. = a, a + d, a + 2d, ……………..
तो a₈ = a + 7d
तथा a₁₂ = a + 11d
प्रश्नानुसार,
a₈ = – 23
⇒ a + 7d = – 23 (i)
तथा a₁₂ = – 39
⇒ a + 11d = – 39 (ii)
समीकरण (i) को समीकरण (ii) में से घटाने परं
4d = – 16
या d = \(\frac {-16}{4}\) = – 4
d का मान समीकरण (i) में रखने पर
a + 7 (- 4) = – 23
या a = – 23 + 28 = 5
∵ A.P. a = 5 तथा d = – 4
∴ A.P. = 5, 1, -3, -7.

प्रश्न 10.
किसी AP का प्रथम पद 5 और 100वाँ पद – 292 है । इस A. P. का 50वाँ पद ज्ञात कीजिए ।
हल :
यहाँ पर दी गई समांतर श्रेढ़ी के लिए-
प्रथम पद (a) = 5
100 वाँ पद (a₁₀₀) = – 292
⇒ a + 99d = – 292
या 5 + 99d = – 292
या 99d = – 292 – 5
या 99d = – 297
या d = \(\frac {- 297}{99}\) = – 3
अतः समांतर श्रेढ़ी का 50 वाँ पद (a₅₀) = a + 49d
= 5 + 49 (-3)
= 5 – 147 = – 142

प्रश्न 11.
उस AP के पहले 50 पदों का योगफल ज्ञात करें जिसका दूसरा पद 14 और 5वाँ पद 26 है।
हल :
माना
AP का प्रथम पद = a
और सार्व अंतर = d
प्रश्नानुसार
a₂ = 14
⇒ a + d = 14 …………(i)
तथा a₅ = 26
⇒ a + 4d = 26 …………..(ii)
समीकरण (i) को समीकरण (ii) में से घटाने पर,
3d = 12
या d = \(\frac {12}{3}\) = 4
d का मान समीकरण (i) में रखने पर,
a + 4 = 14
या a = 14 – 4 = 10
हम जानते हैं कि
S_n = \(\frac {n}{2}\)[2a + (n – 1)d]
⇒ S₅₀ = \(\frac {50}{2}\)[2(10) + (50 – 1) (4)]
= 25[20 + 196]
= 25 × 216 = 5400

प्रश्न 12.
यदि एक A. P. के पहले 6 पदों का योग 12 और पहले 10 पदों का योग 60 है, तो उस A. P. के n पदों का योग ज्ञात कीजिए ।
हल :
यहाँ पर, A. P. के लिए
S₆ = 12
S₁₀ = 60
S_n = ?
हम जानते हैं कि S_n = \(\frac {n}{2}\)[2a + (n – 1)d]
S₁₀ = \(\frac {10}{2}\) [2a + (10 – 1) d]
60 = 5 [2a + 9d] ………..(i)
2a + 9b = 12
तथा S₆ = \(\frac {6}{2}\)[2a + (6 – 1)]d
⇒ 12 = 3[2a + 5d]
⇒ 2a + 5d = 4 ……..(ii)

समीकरण (ii) को समीकरण (i) में से घटाने पर,
4d = 8
या d = \(\frac {8}{4}\) = 2
d का मान समीकरण (i) में रखने पर,
2a + 9(2) = 12
या 2a = 12 – 18
या a = \(\frac {-6}{2}\) = – 3
अब S_n = \(\frac {n}{2}\)[2a + (n – 1)d]
= \(\frac {n}{2}\)[2(-3) + (n – 1)(2)]
= \(\frac {n}{2}\)[-6 + 2n – 2]
या S_n = \(\frac {n}{2}\)(2n – 8)
= n² – 4n

प्रश्न 13.
समांतर श्रेढ़ी – 4, – 1, 2, 5, 8,………………. के पहले 8 पदों और पदों का योग ज्ञात कीजिए ।
हल :
यहाँ पर दी हुई समांतर श्रेढ़ी है-
-4, -1, 2, 5, 8, ……….
∴ प्रथम पद (a) = – 4
और सार्व अंतर (d) = – 1 – (-4) = – 1 + 4 = 3
हम जानते हैं कि S_n = \(\frac {n}{2}\)[2a + (n – 1)d]
⇒ S₈ = \(\frac {8}{2}\)[2(-4) + (8 – 1)(3)]
= 4[-8 + 21]
= 4 × 13 = 52
और S₁₂ = \(\frac {12}{2}\)[2(-4) + (12 – 1)(3)]
= 6[-8 + 33] = 6 × 25 = 150

प्रश्न 14.
योगफल ज्ञात कीजिए प्रथम –
(i) 100 प्राकृत संख्याओं का ।
(ii) n प्राकृत संख्याओं का ।
हल :
(i) यहाँ पर दी हुई समांतर श्रेढ़ी है-
1, 2, 3, 4, …… 100
∴ प्रथम पद (a) = 1
और सार्व अंतर (d) = 2 – 1 = 1
पदों की संख्या (n) = 100
हम जानते हैं कि
S_n = \(\frac {n}{2}\)[2a + (n – 1) d]
⇒ S₁₀₀ = \(\frac {100}{2}\)[2 × 1 + (100 – 1) × 1]
= 50[2 + 99]
= 50 × 101 = 5050

(ii) यहाँ पर दी हुई समांतर श्रेढ़ी है–
1, 2, 3, 4, ……. n
∴ प्रथम पद (a) = 1
सार्व अंतर (d) = 2 – 1 = 1
पदों की संख्या = n
हम जानते हैं कि
S_n = \(\frac {n}{2}\)[2a + (n – 1) d]
⇒ S_n = \(\frac {n}{2}\)[2 × 1 + (n – 1) (1)]
= \(\frac {n}{2}\)[2 + n – 1]
= \(\frac {n}{2}\)[n + 1]

प्रश्न 15.
दो अंकों वाली कितनी संख्याएँ 3 से विभाज्य हैं?
हल :
हम जानते हैं कि
3 से विभाज्य दो अंकों की सबसे छोटी संख्या = 12
3 से विभाज्य दो अंकों की सबसे बड़ी संख्या = 99
अतः AP = 12, 15, 18, ………….. 99
प्रथम पद (a) = 12
सार्व अंतर (d) = 3
हम जानते हैं कि a_n = a + (n – 1)d
⇒ 99 = 12 + (n – 1) × 3
⇒ n – 1 = \(\frac{99-12}{3}\)
⇒ n – 1 = \(\frac {87}{3}\)
⇒ n = 29 + 1
⇒ n = 30
अतः दो अंकों वाली 3 से विभाज्य संख्याओं की संख्या = 30

प्रश्न 16.
योगफल ज्ञात कीजिए – (-5) + (-8) + (-11) + ………………. + (-230)
हल :
यहाँ पर दी हुई समांतर श्रेढ़ी है-
(-5) + (-8) + (-11) + ……… + (-230)
∴ प्रथम पद (a) = – 5
और सार्व अंतर (d) = (-8) – (-5) = – 8 + 5 = – 3
अंतिम पद (l) = – 230
हम जानते हैं कि
l = a + (n – 1) d
⇒ – 230 = 5 + (n – 1) (-3)
या – 230 = – 5 – 3n + 3
या – 230 + 5 – 3 = – 3n
या – 3n = – 228
या n = \(\frac {-228}{-3}\) = 76
अतः S_n = \(\frac {n}{2}\)(a + l)
⇒ S₇₆ = \(\frac {76}{2}\)[- 5 – 230]
= 38 × (- 235) = – 8930

प्रश्न 17.
यदि A. P. के पहले 6 पदों का योग 96 है और पहले 10 पदों का योग 240 है, तो उस A. P. के n पदों का योग ज्ञात कीजिए ।
हल :
यहाँ पर, A. P. के लिए S₆ = 96
S₁₀ = 240
S_n = ?
हम जानते हैं कि S_n = \(\frac {n}{2}\)[2a + (n – 1) d]
⇒ S₁₀ = \(\frac {10}{2}\)[2a + (10 – 1)d]
240 = 5[2a + 9d]
2a + 9d = 48 ……………..(i)
तथा S₆ = \(\frac {6}{2}\) = 2[2a + (6 – 1)d]
96 = 3[2a +5d]
2a + 5d = 32 ……………..(ii)
समीकरण (ii) को समीकरण (i) में से घटाने पर प्राप्त होता है,
4d = 16
d = \(\frac {16}{4}\) = 4
d का मान समीकरण (i) में प्रतिस्थापित करने पर,
2a + 9(4) = 48
या 2a = 48 – 36
या a = \(\frac {12}{2}\) = 6
अब S_n = \(\frac {n}{2}\)[2a + (n – 1)d]
= \(\frac {n}{2}\)[2(6) + (n – 1)4]
= \(\frac {n}{2}\)[12 + 4n – 4]
= \(\frac {n}{2}\)[4n + 8] = 2n² + 4n

प्रश्न 18.
यदि A. P. के पहले 10 पदों का योग -60 और पहले 15 पदों का योग -165 है, तो उसके पहले n पदों का योग ज्ञात कीजिए ।
हल :
यहाँ पर, A.P. के लिए
S₁₀ = -60
S₁₅ = -165
S_n = ?
हम जानते हैं कि S_n = \(\frac {n}{2}\)[2a + (n – 1)d]
⇒ S₁₀ = \(\frac {10}{2}\)[2a + (10 – 1)d]
⇒ – 60 = 5[2a + 9d]
⇒ 2a + 9d = – 12 ……………(i)
तथा S₁₅ = \(\frac {15}{2}\)[2a + (15 – 1)d]
⇒ – 165 × \(\frac {2}{15}\) = 2a + 14d
⇒ 2a + 14d = – 22 …………… (ii)
समीकरण (i) को समीकरण (ii) में से घटाने पर प्राप्त होता है,
5d = – 10
या d = – 2
d का मान समीकरण (i) में प्रतिस्थापित करने पर,
2a + 9(-2) = -12
या 2a = – 12 + 18
या 2a = 6
या a = \(\frac {6}{2}\) = 3
अब S_n = \(\frac {n}{2}\)[2a + (n – 1)d]
= \(\frac {n}{2}\)[2(3) + (n – 1)(-2)]
= \(\frac {n}{2}\)[6 – 2n + 2]
= \(\frac {n}{2}\)[8 – 2n]
= 4n – n²

बहुविकल्पीय प्रश्न

प्रश्न 1.
रीना ने एक पद के लिए आवेदन किया और उसका चयन हो गया। उसे यह पद 8000 रु० के मासिक वेतन और 500 रु० वार्षिक की वेतन वृद्धि के साथ दिया गया। इसके लिए (रुपयों में) उचित स मांतर श्रेढ़ी (AP) होगी –
(A) 8000, 8500, 9000, ……..
(B) 8000, 7500, 7000, ……….
(C) 8500, 9000, 9500, ……….
(D) 8500, 8000, 7500, ……….
हल :
(A) 8000, 8500, 9000, ………..

प्रश्न 2.
एक सीढ़ी के डंडों की लंबाइयाँ नीचे से ऊपर की ओर एक समान रूप से 2cm घटती जाती हैं। सबसे नीचे वाला डंडा लंबाई में 45cm है। नीचे से, पहले, दूसरे, तीसरे, ……. डंडों की लंबाइयाँ (cm में) क्रमशः होगी-
(A) 45, 47, 49, 51, …………..
(B) 45, 43, 41, 39, …………
(C) 45, 41, 37, 33, …………..
(D) 45, 49, 53, 57, ………….
हल :
(B) 45, 43, 41, 39, …………..

प्रश्न 3.
किसी बचत योजना में, कोई धनराशि प्रत्येक 3 वर्षों के बाद स्वयं की \(\frac {5}{4}\) गुनी हो जाती है। 8000 रु० के निवेश की 3, 6, 9 और 12 वर्षों के बाद परिपक्वता राशियाँ (रुपयों में) क्रमशः होंगी-
(A) 10000, 30000, 90000 और 2,70000
(B) 10000, 7500, 5000 और 4000
(C) 10000, 12500, 15625 और 19531.25
(D) उपरोक्त में से कोई नहीं
हल :
(C) 10000, 12500, 15625 और 19531.25

प्रश्न 4.
शकीला अपनी पुत्री की गुल्लक में 100 रु० तब डालती है, जब वह एक वर्ष की हो जाती है तथा प्रत्येक वर्ष इसमें 50 रु० की वृद्धि करती जाती है। उसके पहले, दूसरे, तीसरे, चौथे, ……………. जन्म दिवसों पर उसकी गुल्लक में डाली गई राशियाँ (रुपयों में) क्रमशः होगी-
(A) 200, 250, 300, 350, ………..
(B) 50, 100, 150, 200, ………….
(C) 100, 200, 300, 400, …………
(D) 100, 150, 200, 250, ………..
हल :
(D) 100, 150, 200, 250, …………

प्रश्न 5.
इनमें से कौन-सी A. P. श्रेणी है?
(A) 2, 4, 8, 12, ……..
(B) 0.2, 0.22, 0.222, ……..
(C) – 10, – 6, – 2, 2 ………….
(D) 1, 3, 9, 27, ……..
हल :
(C) – 10, – 6, – 2, 2 ………

प्रश्न 6.
समांतर श्रेढ़ी 7, 3, – 1, – 5, – 9 का सार्व अंतर है-
(A) 4
(B) – 4
(C) \(\frac {3}{7}\)
(D) \(\frac {7}{3}\)
हल :
(B) – 4

प्रश्न 7.
इनमें से कौन-सी A. P. श्रेणी है?
(A) a, a², a³, ……………
(B) 1², 3², 5², 7², ………
(C) a, 2a, 3a, 4a, ………….
(D) 1, 3, 9, 27, …………
हल :
(C) a, 2a, 3a, 4a, ……..

प्रश्न 8.
समांतर श्रेढ़ी 12, 2, – 8, – 18 ………. का सार्व अंतर है-
(A) -10
(B) 10
(C) \(\frac {3}{7}\)
(D) – 4
हल :
(A) – 10

प्रश्न 9.
निम्नलिखित में से कौन-सा क्रम समांतर श्रेढ़ी में नहीं है ?
(A) 3, 3, 3, 3, ……..
(B) 2, 2, 2, 2, ………..
(C) 3, 5, 8, 12, 20, ………
(D) 5, 5, 5, 5, ………..
हल :
(C) 3, 5, 8, 12, 20 ……..

प्रश्न 10.
एक समांतर श्रेढ़ी संख्याओं की एक ऐसी सूची होती है जिसमें प्रत्येक पद (पहले पद के अतिरिक्त) अपने पद में एक …………. संख्या जोड़ने पर प्राप्त होता है ।
(A) निश्चित
(B) विषम
(C) सम
(D) अभाज्य
हल :
(A) निश्चित

प्रश्न 11.
एक समांतर श्रेढ़ी संख्याओं की एक ऐसी सूची होती है जिसमें प्रत्येक पद (पहले पद के अतिरिक्त) अपने पद में एक निश्चित संख्या जोड़ने पर प्राप्त होता है । यह निश्चित संख्या AP का क्या कहलाती है?
(A) प्रथम पद
(B) अंतिम पद
(C) सार्व अंतर
(D) n वाँ पद
हल :
(C) सार्व अंतर

प्रश्न 12.
प्रथम पद a तथा सार्व अंतर d वाली समांतर श्रेढ़ी का n वाँ पद होगा-
(A) \(\frac {n}{2}\)(a + l)
(B) a + (n – 1)d
(C) d + (n – 1)a
(D) \(\frac {n}{2}\)(a + d)
हल :
(B) a + (n – 1) d

प्रश्न 13.
समांतर श्रेढ़ी – 5 – 1, 3, 7 का सार्व अंतर है-
(A) – 4
(B) 4
(C) 6
(D) इनमें से कोई नहीं
हल :
(B) 4

प्रश्न 14.
समांतर श्रेढ़ी p, p + 90, p + 180, p + 270, ……… [ जहाँ p = (999)⁹⁹⁹] का सार्व अंतर है –
(A) 90
(B) – 90
(C) p
(D) शून्य
हल :
(A) 90

प्रश्न 15.
समांतर श्रेढ़ी – 125, – 325, – 525, – 725 ………… का सार्व अंतर है-
(A) – 125
(B) 125
(C) 200
(D) – 200
हल :
(D) – 200

प्रश्न 16.
प्रथम पद f तथा सार्व अंतर d वाली समांतर श्रेढ़ी का p वाँ पद होगा-
(A) f + (n – 1) d
(B) \(\frac {n}{2}\) (f + d)
(C) d + (p – 1) f
(D) f + (p – 1) d
हल :
(D) f + (p – 1 ) d

प्रश्न 17.
समांतर श्रेढ़ी 2.8, 3.0, 3.2, 3.4, …………. के अगले दो पद होंगे-
(A) 3.8 व 4.0
(B) 2.8 व 3.0
(C) 3.0 व 3.2
(D) 3.6 व 3.8
हल :
(D) 3.6 व 3.8

प्रश्न 18.
उस समांतर श्रेढ़ी (AP) के प्रथम 51 पदों का योगफल क्या होगा जिसके दूसरे और तीसरे पद क्रमशः 14 और 18 हैं?
(A) 5410
(B) 5510
(C) 5610
(D) 5710
हल :
(C) 5610

प्रश्न 19.
A.P. 13, 15\(\frac {1}{2}\), 18, 20\(\frac {1}{2}\), ……………. का 11वाँ पद है :
(A) 38
(B) 40\(\frac {1}{2}\)
(C) 43
(D) 45\(\frac {1}{2}\)
हल :
(A) 38

प्रश्न 20.
समांतर श्रेढ़ी 10.0, 10.5, 11.0, 11.5, …………… का 10वाँ पद होगा-
(A) 15.5
(B) 14.0
(C) 14.5
(D) 15.0
हल :
(C) 14.5

प्रश्न 21.
यदि एक समांतर श्रेढ़ी में S_n = 256, a = 1 और d = 2 तो n का मान होगा-
(A) 18
(B) 14
(C) 15
(D) 16
हल :
(D) 16

प्रश्न 22.
यदि A.P. का तीसरा पद 5 और 7 वाँ पद 13 है, तो उसका सार्व अंतर (common difference) है :
(A) 1
(B) 2
(C) 3
(D) 4
हल :
(D) 16

प्रश्न 23.
समांतर श्रेढ़ी (A.P.) – 11, – 7, – 3, ……………….. का सार्व अन्तर (common difference) है-
(A) 4
(B) – 4
(C) – 11
(D) – 18
हल :
(A) 4

प्रश्न 24.
A.P. 5, 6\(\frac {1}{2}\), 8, 9\(\frac {1}{2}\), ……………. का 15 वाँ पद है :
(A) 15\(\frac {1}{2}\)
(B) 14\(\frac {1}{2}\)
(C) 26
(D) 27\(\frac {1}{2}\)
हल :
(C) 26

प्रश्न 25.
समांतर श्रेढ़ी 4, 10, 16, 22, ……………. के अगले दो पद होंगे-
(A) 26 व 32
(B) 28 व 34
(C) 34 व 40
(D) 28 व 32
हल :
(B) 28 व 34

प्रश्न 26.
समांतर श्रेढ़ी 1, – 1, – 3, – 5, ………………. के अगले दो पद होंगे-
(A) – 7 व – 9
(B) – 7 व – 8
(C) – 9 व – 11
(D) – 7 व – 11
हल :
(A) – 7 व – 9

प्रश्न 27.
संख्याओं की निम्नलिखित सूची में से कौन-सी सूची AP नहीं है ?
(A) 2, 8, 14, 20, …………….
(B) 1, 2, 3, 4, ……………..
(C) 2, 2, 2, 2, ……………..
(D) 1, 1, 2, 2, 3, 3, …………..
हल :
(D) 1, 1, 2, 2, 3, 3, …………..

प्रश्न 28.
यदि एक A.P. का तीसरा पद 12 और 10 वाँ पद 26 है, तो उसका 20वाँ पद है :
(A) 46
(B) 52
(C) 50
(D) 44
हल :
(A) 46

प्रश्न 29.
प्रथम पद 4 तथा सार्व अंतर – 3 वाली AP के प्रथम दो पद होंगे-
(A) 4 व 1
(B) 4 व 7
(C) 4 व – 2
(D) 4 व 10
हल :
(A) 4 व 1

प्रश्न 30.
प्रथम पद – 2 तथा सार्व अंतर 0 वाली AP के प्रथम चार पद होंगे-
(A) -2, -4, -6, -8
(B) -2, -2, -2, -2
(C) -2, 0, 2, 4
(D) 0, 0, 0, 0
हल :
(B) -2, -2, -2, -2

प्रश्न 31.
AP : 3, 1, – 1, – 3, ………. के प्रथम पद व सार्व अंतर क्रमशः होंगे-
(A) 3 व 2
(B) 3 व 4
(C) 3 व – 2
(D) – 2 व 3
हल :
(C) 3 व – 2

प्रश्न 32.
A.P. 0.6, 1.7, 2.8, 3.9, …………… का 14 वाँ पद है :
(A) 14.9
(B) 16.0
(C) 17.1
(D) 18.2
हल :
(A) 14.9

प्रश्न 33.
दी हुई A. P. का a = 10 तथा d = 10 है, तो इस A. P. का द्वितीय पद ……………. है ।
(A) 10
(B) 20
(C) 30
(D) 40
हल :
(B) 20

प्रश्न 34.
AP : \(\sqrt{m}\), \(\sqrt{m}\), \(\sqrt{m}\), \(\sqrt{m}\), ………… का सार्व अंतर होगा-
(A) \(\sqrt{6}\)
(B) 2
(C) \(\sqrt{2}\)
(D) \(\sqrt{4}\)
हल :
(C) \(\sqrt{2}\)

प्रश्न 35.
A. P. 13, 15\(\frac {1}{2}\), 18, 20\(\frac {1}{2}\), …………… का 11 वाँ पद है :
(A) 38
(B) 40\(\frac {1}{2}\)
(C) 43
(D) 45\(\frac {1}{2}\)
हल :
(A) 38

प्रश्न 36.
यदि किसी A. P. का तीसरा पद 4 और 9वाँ पद – 8 है, तो उनका सार्व अंतर (common difference) है :
(A) – 2
(B) 2
(C) 4
(D) – 8
हल :
(A) – 2

प्रश्न 37.
यदि किसी समांतर श्रेढ़ी (AP) का प्रथम पद \(\frac {1}{2}\) और सार्व अंतर \(\frac {1}{12}\) हो तो उसका 12 वाँ पद होगा-
(A) \(\frac {17}{12}\)
(B) \(\frac {15}{12}\)
(C) \(\frac {16}{12}\)
(D) \(\frac {13}{12}\)
हल :
(A) \(\frac {17}{12}\)

प्रश्न 38.
\(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}\) …………….. A. P. के लिए, निम्न से सार्वअन्तर ज्ञात कीजिए ।
(A) \(\frac {-4}{3}\)
(B) \(\frac {2}{3}\)
(C) \(\frac {4}{3}\)
(D) इनमें से कोई नहीं
हल :
(C) \(\frac {4}{3}\)

प्रश्न 39.
-0.1, -0.2, -0.3, …………… A. P. का 10वा पद हैं-
(A) – 0.9
(B) – 0.8
(C) – 1.0
(D) – 1.1
हल :
(C) – 1.0

प्रश्न 40.
A. P. 2, 7, 12, ………. का 10वाँ पद है-
(A) -47
(B) 47
(C) 57
(D) इनमें से कोई नहीं
हल :
(B) 47

प्रश्न 41.
दी हुई A. P. का a = – 2 तथा d = 0 है, तो इस A. P. का द्वितीय पद ……………. है-
(A) – 2
(B) 0
(C) 2
(D) 4
हल :
(A) – 2

प्रश्न 42.
A.P. – 10, – 6, – 2, 2, …………… का 20वाँ पद है :
(A) 66
(B) – 66
(C) 77
(D) इनमें से कोई नहीं
हल :
(A) 66

प्रश्न 43.
यदि 11, x, 5 किसी AP के पद हों तो x का मान होगा-
(A) 6
(B) 7
(C) 8
(D) 9
हल :
(C) 8

प्रश्न 44.
A.P. \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}\), ……………. का 15 वाँ पद है :
(A) \(\frac {61}{3}\)
(B) 6
(C) 5
(D) 19
हल :
(D) 19

प्रश्न 45.
समांतर श्रेढ़ी 18, 15 \(\frac {1}{2}\), 13, ………. – 47 में पदों की संख्या होगी-
(A) 27
(B) 28
(C) 29
(D) 30
हल :
(A) 27

प्रश्न 46.
दी हुई A. P. का a = 4 तथा d = – 3 है, तो इस A. P. का द्वितीय पद ……………… है :
(A) 4
(B) 1
(C) – 2
(D) – 5
हल :
(B) 1

प्रश्न 47.
समांतर श्रेढ़ी 25, 50, 75, 100, …………. का कौन-सा पद 1000 होगा ?
(A) 37वाँ
(B) 38वाँ
(C) 39वाँ
(D) 40वाँ
हल :
(D) 40वाँ

प्रश्न 48.
34 + 32 + 30 +…………………+ 10 का मान है :
(A) 294
(B) 289
(C) 286
(D) 386
हल :
(C) 286

प्रश्न 49.
समांतर श्रेढ़ी (AP) – 6, 0, 6, ……………. के प्रथम 13 पदों का योगफल होगा-
(A) 390
(B) 396
(C) 402
(D) 408
हल :
(A) 390

प्रश्न 50.
2 + 4 + 6 + …….. + 200 का योगफल होगा-
(A) 1010
(B) 10100
(C) 101000
(D) 101
हल :
(B) 10100

प्रश्न 51.
(-5) + (-8) + (-11) + …… + (-230) का योगफल होगा-
(A) -8730
(B) -8830
(C) -8930
(D) -9030
हल :
(C) – 8930

प्रश्न 52.
प्रथम 100 प्राकृत संख्याओं का योगफल होगा-
(A) 5050
(B) 4950
(C) 5150
(D) 5250
हल :
(A) 5050

प्रश्न 53.
यदि किसी AP का प्रथम पद a तथा सार्व अंतर d हो तो इसके n पदों का योग होगा-
(A) \(\frac {n}{2}\)[a + (n – 1)d]
(B) \(\frac {n}{2}\)[2a + (n – 1)d]
(C) \(\frac {n}{2}\)[4a+ (n – 1)d]
(D) \(\frac {n}{2}\)[3a + (n – 1)d]
हल :
(B) \(\frac {n}{2}\)[2a + (n – 1)d]

प्रश्न 54.
AP : 8, 3, -2, …………. के प्रथम 22 पदों का योग होगा-
(A) – 969
(B) – 974
(C) – 979
(D) – 984
हल :
(C) – 979

प्रश्न 55.
प्रथम n धन पूर्णांकों का योग होगा-
(A) \(\frac{(n-1)(n+1)}{2}\)
(B) \(\frac{n(n-1)}{2}\)
(C) \(\frac{n(n+2)}{2}\)
(D) \(\frac{n(n+1)}{2}\)
हल :
(D) \(\frac{n(n+1)}{2}\)

प्रश्न 56.
(A) – 24
(B) – 30
(C) – 36
(D) – 42
हल :
(B) – 30

प्रश्न 57.
किसी अनुक्रम का n वाँ पद 9 – 5n है,
(A) 4, – 1, – 6, – 11, ……….
(B) 5, 4, 3, 2, 1, …………..
(C) 7, 9, 11, 13, ………….
(D) 7, 11, 15, ……………
हल :
(A) 4, – 1, – 6, – 11, ……….

प्रश्न 58.
यदि किसी समांतर श्रेढ़ी का 9वाँ तथा 5वाँ पद क्रमशः 28 और
16 हो तो इस श्रेढ़ी के 15 पदों का योग होगा-
(A) 325
(B) 350
(C) 375
(D) 400
हल :
(C) 375

प्रश्न 59.
किसी AP का सार्व अंतर वही जो एक अन्य AP का, इनमें से एक का प्रथम पद 3 और दूसरी का 8 है । इनके चौथे पदों में अंतर होगा-
(A) 3
(B) 11
(C) 5
(D) 4
हल :
(C) 5

प्रश्न 60.
प्रथम पद f और अंतिम पद l वाली समांतर श्रेढ़ी के m पदों का योगफल होगा-
(A) \(\frac {l}{2}\)(m + f)
(B) \(\frac {m}{2}\)(f + l)
(C) \(\frac {f}{2}\) (m + l)
(D) f + (m – 1)l
हल :
(B) \(\frac {m}{2}\)(f + l)

प्रश्न 61.
यदि किसी AP में a = 5, d = 3, a_n = 50 ह्ये तो n का मान होगा-
(A) 16
(B) 15
(C) 17
(D) 18
हल :
(A) 16

प्रश्न 62.
यदि किसी AP में a₁₂ = 37 व d = 3 हो तो a का मान होगा-
(A) 6
(B) 5
(C) 4
(D) 3
हल :
(C) 4

प्रश्न 63.
यदि किसी AP में a = 8, a_n = 62, n = 6 हो तो d का मान होगा-
(A) \(\frac {51}{5}\)
(B) \(\frac {52}{5}\)
(C) \(\frac {53}{5}\)
(D) \(\frac {54}{5}\)
हल :
(D) \(\frac {54}{5}\)

प्रश्न 64.
अनुक्रम a_n = 3n + 2 का तीसरा पद होगा-
(A) 8
(B) 11
(C) 14
(D) 17
हल :
(B) 11

प्रश्न 65.
अनुक्रम a_n = (-1)^n-1 . 2ⁿ का चौथा पद होगा-
(A) 8
(B) – 8
(C) – 16
(D) 16
हल :
(C) – 16

Haryana State Board HBSE 10th Class Maths Important Questions Chapter 4 द्विघात समीकरण Important Questions and Answers.

Haryana Board 10th Class Maths Important Questions Chapter 4 द्विघात समीकरण

परीक्षोपयोगी अन्य महत्त्वपूर्ण प्रश्न :

प्रश्न 1.
जाँच कीजिए कि समीकरण (2x + 1) (3x + 2) = 6(x – 1) (x – 2) द्विघात है अथवा नहीं?
हल :
यहाँ पर
(2x + 1) (3x + 2) = 6 (x – 1) (x – 2)
या 6x² + 4x + 3x + 2 = 6(x2 – 2x – x + 2)
या 6x² + 7x + 2 = 6x² – 18x + 12
या 6x² + 7x + 2 – 6x² + 18x – 12 = 0
या 25x – 10 = 0
∵ इस समीकरण की घात एक है
∵ दिया गया समीकरण द्विघात नहीं है ।

प्रश्न 2.
जाँच कीजिए कि x = – 1 व x = – 5 दिए गए समीकरण x² + 6x + 5 = 0 के हल हैं अथवा नहीं?
हल :
यहाँ पर दिया गया समीकरण है-
x² + 6x + 5 = 0
x = – 1 समीकरण में रखने पर,
बायां पक्ष
= x2 + 6x + 5
= (-1)² + 6 (-1) + 5
= 1 – 6 + 5 = 6 – 6 = 0 = दायां पक्ष
∴ x = – 1 दिए गए समीकरण का हल है ।
अब x = – 5 समीकरण में रखने पर,
बायां पक्ष = x² + 6x + 5
= (-5)² + 6(-5) + 5
= 25 – 30 + 5 = 30 – 30 = 0 = दायां पक्ष
∴ x = – 5 भी दिए गए समीकरण का हल है
अतः x = – 1 व x = – 5 दिए गए समीकरण के हल हैं।

प्रश्न 3.
निम्नलिखित स्थितियों को द्विघात समीकरणों के रूप में निरूपित कीजिए –
(i) एक आयत की एक भुजा उसकी दूसरी भुजा से 2cm बड़ी है । यदि आयत का क्षेत्रफल 195cm² हो तो आयत की भुजाएँ ज्ञात करनी हैं ।
(ii) तीन क्रमागत धनात्मक पूर्णांक ऐसे हैं कि प्रथम के वर्ग तथा अन्य दो के गुणनफल का योग 154 है । वे पूर्णांक ज्ञात करने हैं।
हल :
(i) माना आयत की पहली भुजा = x cm
तो आयत की दूसरी भुजा = (x + 2) cm
आयत का क्षेत्रफल = 195 cm²
प्रश्नानुसार,
x(x + 2) = 195
⇒ x² + 2x – 195 = 0
अतः आयत की भुजाएँ ज्ञात करने के उचित समीकरण है – x² + 2x – 195 = 0

(ii) माना तीन क्रमागत धनात्मक पूर्णांक = x,x + 1 व x + 2
प्रश्नानुसार,
(x)² + (x + 1)(x + 2) = 154
⇒ x² + x² + 3x + 2 – 154 = 0
⇒ 2x² + 3x – 152 = 0
अतः तीन पूर्णांक ज्ञात करने के लिए उचित समीकरण है – 2x² + 3x – 152 = 0

प्रश्न 4.
गुणनखंडन विधि से निम्न द्विघात समीकरणों के मूल ज्ञात कीजिए-
(i) ax² – 2abx = 0
(ii) \(\frac {1}{9}\)x² – \(\frac {2}{3}\)x + 1 = 0
हल :
(i) यहाँ पर ax² – 2abx = 0
या ax(x – 2b) = 0
⇒ ax = 0
⇒ x = \(\frac {0}{a}\)
⇒ x = 0
अतः अभीष्ट हल x = 0 x = 2b

(ii) यहाँ पर
\(\frac {1}{9}\)x² – \(\frac {2}{3}\)x + 1 = 0
या x² – 6x + 9 = 0
या x² – 3x – 3x + 9 = 0
या x (x – 3) – 3(x – 3) = 0
या (x – 3) (x – 3) = 0
⇒ (x – 3)² = 0
⇒ x – 3 = 0
⇒ x = 3
अतः अभीष्ट हल x = 3

प्रश्न 5.
गुणनखंडन विधि से निम्नलिखित द्विघात समीकरणों के मूल ज्ञात कीजिए-
(i) 8x² – 22x – 21 = 0
(ii) abx² + (b² – ac) x – bc = 0
हल :
(i) यहाँ पर 8x² – 22x – 21 = 0
या 8x² – 28x + 6x – 21 = 0
या 4x (2x – 7) + 3 (2x – 7) = 0
या (2x – 7) (4x + 3) = 0
⇒ 2x – 7 = 0 या 4x + 3 = 0
⇒ 2x = 7 या 4x = – 3
⇒ x = \(\frac {7}{2}\) या x = \(\frac {-3}{4}\)
अतः अभीष्ट हल x = \(\frac {7}{2}\), \(\frac {-3}{4}\)

(ii) यहाँ पर abx² + (b² – ac) x – bc = 0
या abx² + xb² – acx – bc = 0
या bx (ax + b) – c (ax + b) = 0
या (bx – c) (ax + b) = 0
⇒ bx – c = 0 या ax + b = 0
⇒ bx = c या ax = – b
⇒ x = \(\frac {c}{b}\) या x = \(\frac {-b}{a}\)
अतः अभीष्ट हल x = – \(\frac {-b}{a}\), \(\frac {c}{b}\)

प्रश्न 6.
दो क्रमागत सम धनात्मक पूर्णांक ज्ञात कीजिए जिनके वर्गों का योग 100 है।
हल :
माना पहला सम धन पूर्णांक = 2x
तो दूसरा सम धन पूर्णांक = 2x + 2
प्रश्नानुसार, (2x)² + (2x + 2)² = 100
या 4x² + 4x² + 4 + 8x = 100
या 8x² + 8x + 4 – 100 = 0
या 8x² + 8x – 96 = 0
या x² + x – 12 = 0
या x² + 4x – 3x – 12 = 0
या x (x + 4) – 3 (x + 4) = 0
या (x + 4) (x – 3) = 0
⇒ x + 4 = 0 या x – 3 = 0
⇒ x = – 4 या x = 3
परंतु x = – 4 संभव नहीं है ।
∴ x = 3
∴ पहला समधन पूर्णांक = 2 × 3 = 6
∴ दूसरा समधन पूर्णांक = 2 × 3 + 2 = 8

प्रश्न 7.
विक्रम तीन लकड़ी की छड़ों से एक समकोण त्रिभुज बनाना चाहता है । समकोण त्रिभुज का कर्ण उसके आधार से 2 सें०मी० तथा शीर्षलंब से 4 सें०मी० बड़ा होना चाहिए। उसे लकड़ी की छड़ें कितनी लंबी लेनी चाहिएँ ?
हल :
माना समकोण त्रिभुज का कर्ण = x सें०मी०

तो आधार = (x – 2) सें०मी०
शीर्षलंब = (x – 4) सें०मी०
∴ पाइथागोरस प्रमेय द्वारा
(कर्ण)² = (आधार)² + (लंब)²
⇒ (x)² = (x – 2)² + (x – 4)²
या (x)² = (x)² – 4x + 4 + (x)² – 8x + 16
या (x)² = 2x² – 12x + 20
या 2x² – 12x + 20 – x² = 0
या (x)² – 12x + 20 = 0
या (x)² – 2x – 10x + 20 = 0
या x (x – 2 ) – 10 (x – 2) = 0
या (x – 10) (x – 2) = 0
⇒ x – 10 = 0 या x – 2 = 0
⇒ x = + 10 या x = 2
परंतु x = 2 नहीं हो सकता क्योंकि इस अवस्था में x – 2 = 2 – 2 = 0 होगा ।
∴ x = 10
तो कर्ण = 10 सें०मी०
आधार = 10 – 2 = 8 सें०मी०
शीर्षलंब = 10 – 4 = 6 सें०मी०
अतः लकड़ी की छड़ों की लंबाई = 6 सें०मी०; 8 सें०मी०; 10 सें०मी०

प्रश्न 8.
एक समकोण त्रिभुज की समकोण बनाने वाली एक भुजा दूसरी से 17 सें०मी० कम है । यदि कर्ण की लम्बाई 25 सें०मी० है, तो दोनों भुजाओं की लम्बाई ज्ञात कीजिए ।
हल :
माना समकोण त्रिभुज की एक भुजा तो समकोण त्रिभुज की दूसरी भुजा = x सें०मी०
तो समकोण त्रिभुज =(x – 17) सें०मी०
कर्ण की लम्बाई = 25 सें०मी०.
प्रश्नानुसार,
या (x)² + (x – 17)² = (25)²
या x² + x² + 289 – 34x = 625
या 2x² – 34x + 289 – 625 = 0
या 2x² – 34x – 336 = 0
या x² – 17x – 168 = 0 (दोनों ओर 2 से भाग करने पर)
या x² – 24x + 7x – 168 = 0
या (x – 24) + 7 (x – 24) = 0
(x – 24 ) (x + 7) = 0
⇒ x – 24 = 0 या x + 7 = 0
⇒ x = 24 या x = – 7 ( सम्भव नहीं है )
अतः समकोण त्रिभुज की भुजाएँ = 24 सें०मी० व (24 – 17) सें०मी०
= 24 सें०मी० व 7 सें०मी०

प्रश्न 9.
दो ऐसे क्रमागत विषम धनात्मक पूर्णांक ज्ञात कीजिए, जिनके वर्गों का योग 290 हो ।
हल :
माना दो क्रमागत विषम धनात्मक पूर्णांक = x व x + 2
प्रश्नानुसार
(x)² + (x + 2)² = 290
⇒ x² + x² + 4x + 4 – 290 = 0
⇒ 2x² + 4x – 286 = 0
⇒ x² + 2x – 143 = 0 (दोनों ओर 2 से भाग करने पर)
⇒ x² + 13x – 11x – 143 = 0
⇒ x(x + 13) – 11 (x + 13) = 0
⇒ (x + 13) (x – 11) = 0
⇒ x + 13 = 0 या x – 11 = 0
⇒ x = – 13 या x = 11
परन्तु x एक धनात्मक विषम पूर्णांक है अतः x ≠ – 13
∴ x = 11
अतः क्रमागत विषम धनात्मक पूर्णांक = 11 व 13

प्रश्न 10.
एक किसान 100 मी०² क्षेत्रफल वाला आयताकार सब्जी का बगीचा बनाना चाहता है। क्योंकि उसके पास घेराबंदी के लिए 30 मी० लंबाई का काँटेदार तार है, इसलिए वह आयताकार बगीचे की तीन भुजाओं की घेराबंदी इस तार से करता है तथा चौथी भुजा की घेराबंदी के लिए अपने सहन की दीवार का उपयोग करता है । बगीचे की विमाएँ (dimensions) ज्ञात कीजिए ।
हल :
माना पहली भुजा की लंबाई = x मी०
तो दूसरी भुजा की लंबाई = (30 – 2x) मी०
प्रश्नानुसार,
या x(30 – 2x) = 100
या 30x – 2x² = 100
या – 2x² + 30x – 100 = 0
या x² – 15x + 50 = 0
या x² – 10x – 5x + 50 = 0
या x (x – 10) – 5 (x – 10) = 0
या (x – 10) (x – 5) = 0
⇒ x – 10 = 0 या x – 5 = 0
⇒ x = 10 या x = 5
परंतु x = 10 संभव नहीं है क्योंकि इससे बाग वर्गाकार हो जाएगा।
∴ x = 5
∴ पहली भुजा की लंबाई = 5 मी०
तथा दूसरी भुजा की लंबाई = 30 – 2x = 30 – 2 × 5 = 20 मी०

प्रश्न 11.
ज्ञात कीजिए कि क्या द्विघात समीकरण 3x² – 5x + 2 = 0 के मूल वास्तविक हैं? यदि हैं तो उन्हें ज्ञात कीजिए ।
हल :
यहाँ पर
3x² – 5x + 2 = 0
a = 3, b = – 5, c = 2
विविक्तकर b² – 4ac
= (-5)² – 4(3)(2)
= 25 – 24 = 1 > 0
अतः समीकरण के वास्तविक मूल हैं
अब द्विघाती सूत्र के उपयोग से,

प्रश्न 12.
पूर्ण वर्ग बनाकर समीकरण 5x² – 6x – 2 = 0 के मूल ज्ञात कीजिए।
हल :
यहाँ पर, 5x² – 6x – 2 = 0
दोनों ओर 5 से गुणा करने पर
⇒ 25x² – 30x – 10 = 0
⇒ (5x)² – 2 × 5x × 3 + (3)² – (3)² – 10 = 0
⇒ (5x – 3)² = 19
⇒ 5x – 3 = ± \(\sqrt{19}\)
⇒ x = \(\frac{3 \pm \sqrt{19}}{5}\)
अतः दी गई समीकरण के अभीष्ट मूल = \(\frac{3+\sqrt{19}}{5}\) व \(\frac{3-\sqrt{19}}{5}\)

प्रश्न 13.
P के वे मान ज्ञात कीजिए जिनके लिए द्विघात समीकरण px² – 6x – 2 = 0 के मूल वास्तविक हों ।
हल :
दिया गया समीकरण px² – 6x – 2 = 0
यहाँ पर a = p, b = – 6, c = – 2
∴ विविक्तकर = b² – 4ac
= (-6)² – 4(p)(-2) = 36 + 8p
वास्तविक मूल के लिए D ≥ 0 होना चाहिए ।
⇒ 36 + 8p ≥ 0
या 8p ≥ – 36
या p ≥ \(\frac {-36}{8}\)
या p ≥ \(\frac {-9}{2}\)

प्रश्न 14.
समीकरण 3x² – 2x + \(\frac {1}{3}\) = 0 का विविक्तकर ज्ञात कीजिए और फिर मूलों की प्रकृति ज्ञात कीजिए। यदि वे वास्तविक हैं, तो उन्हें ज्ञात कीजिए ।
हल :
यहाँ पर 3x² – 2x + \(\frac {1}{3}\) = 0
⇒ a = 3, b = – 2, c = \(\frac {1}{3}\)
∴ विविक्तकर = b² – 4ac
= (-2)² – 4(3)(\(\frac {1}{3}\))
= 4 – 4 = 0
∴ द्विघात समीकरण के दो बराबर वास्तविक मूल हैं ।
अब द्विघाती सूत्र के उपयोग से,

प्रश्न 15.
एक नाव को, जिसकी शांत जल में चाल 15 कि०मी० / घंटा है, धारा की दिशा में 30 कि०मी० जाने तथा फिर धारा की दिशा के विपरीत लौटने में कुल 4 घंटे 30 मिनट का समय लगता है । धारा की चाल ज्ञात कीजिए ।
हल :
शांत जल में नाव की चाल = 15 कि०मी० / घंटा
माना जल की धारा की चाल = x कि०मी० / घंटा
पहली अवस्था में जब नाव जल की धारा की दिशा में जाएगी
कुल तय की गई दूरी = 30 कि०मी० / घंटा
जल की धारा की दिशा में नाव की चाल = (15 + x) कि०मी० / घंटा
जल की धारा की दिशा में 30 कि०मी० दूरी तय तरने में
लिया गया समय = \(\frac{30}{(15+x)}\) घंटा
दूसरी अवस्था में जब नाव जल की धारा के विपरीत दिशा में जाएगी
कुल तय की गई दूरी = 30 कि०मी० / घंटा
विपरीत दिशा में नाव की चाल = (15 – x) कि०मी० / घंटा
जल की धारा की विपरीत दिशा में 30 कि०मी० दूरी तय तरने में
लिया गया समय = \(\frac{30}{(15- x)}\) घंटा
प्रश्नानुसार,

प्रश्न 16.
दो स्टेशनों के बीच 168 किमी यात्रा करने में एक एक्सप्रेस रेलगाड़ी, सवारी गाड़ी से 1 घंटा कम समय लेती है ( स्टेशनों पर ठहरने का समय ध्यान में न लिया जाए) यदि एक्सप्रेस गाड़ी की चाल सवारी गाड़ी से 14 किमी / घण्टा अधिक है, तो दोनों रेलगाड़ियों की औसत चाल ज्ञात कीजिए ।
हल :
माना सवारी गाड़ी की औसत चाल = x किमी / घंटा
तो एक्सप्रेस गाड़ी की औसत चाल = (x + 14) किमी / घंटा
सवारी गाड़ी द्वारा 168 किमी दूरी तय करने में लिया गया समय = \(\frac {168}{x}\) घंटे
एक्सप्रेस गाड़ी द्वारा 168 किमी दूरी तय करने में लिया गया समय = \(\frac{168}{(x+14)}\) घंटे
प्रश्नानुसार,
\(\frac{168}{x}-\frac{168}{x+14}\) = 1
⇒ 168 (x + 14) – 168x = x(x + 14) (दोनों ओर x (x + 14) से गुणा करने पर)
⇒ 168x + 2352 – 168x = x² + 14x
या x² + 14x – 2352 = 0
या x² + 56x – 42x – 2352 = 0
या x(x + 56) – 42(x + 56) = 0
⇒ (x + 56)(x – 42) = 0
⇒ x + 56 = 0 या x – 42 = 0
⇒ x = – 56 या x = 42
परंतु x = – 56 असंभव है, क्योंकि चाल ऋणात्मक नहीं हो सकती,
अतः सवारी गाड़ी की औसत चाल = 42 किमी / घंटा
तथा एक्सप्रेस गाड़ी की औसत चाल = (42 + 14) = 56 किमी / घंटा

प्रश्न 17.
एक रेलगाड़ी एकसमान चाल से 180 किमी की दूरी चलती है । यदि उसकी चाल 6 किमी / घण्टा अधिक हो, तो उसे उतनी ही दूर जाने में 1 घंटा कम समय लगता है। गाड़ी की चाल ज्ञात कीजिए ।
हल :
माना रेलगाड़ी की सामान्य चाल रेलगाड़ी की बढ़ी हुई चाल = x किमी / घंटा
रेलगाड़ी द्वारा चली गई कुल दूरी = x + 6 किमी / घंटा
रेलगाड़ी द्वारा चली गई कुल दूरी = 180 किमी
रेलगाड़ी द्वारा सामान्य चाल से 180 किमी दूरी तय करने में लिया गया समय = \(\frac {180}{x}\) घंटे
रेलगाड़ी द्वारा बढ़ी हुई चाल से 180 किमी दूरी तय करने में लिया गया समय = \(\frac{180}{x+6}\) घंटे
प्रश्नानुसार,
\(\frac{180}{x}-\frac{180}{x+6}\) = 1
⇒ 180(x + 6) – 180x = x(x + 6) (दोनों ओर x (x + 6) से गुणा करने पर)
⇒ 180x + 1080 – 180x = x² + 6x
⇒ x² + 6x – 1080 = 0
⇒ x² + 36x – 30x – 1080 = 0
⇒ x(x + 36) – 30(x + 36) = 0
⇒ (x + 36) (x – 30) = 0
⇒ x + 36 = 0 या x – 30 = 0
⇒ x = – 36 या x = 30
परंतु x = – 36 असंभव है, क्योंकि चाल ऋणात्मक नहीं हो सकती,
अतः रेलगाड़ी की सामान्य चाल 30 किमी / घंटा

प्रश्न 18.
बहुपद p(x) = x⁴ – 3x² + 4x + 5 को बहुपद g (x) = x² – x + 1 से भाग कीजिए। भागफल तथा शेषफल ज्ञात कीजिए ।
हल :
यहाँ पर,
p(x) = x⁴ – 3x² + 4x + 5
g(x) = x² – x + 1
क्योंकि p(x) की घात 4 तथा g (x) की घात 2 है ।
इसलिए भागफल q(x) की घात = 4 – 2 = 2 तथा शेषफल की घात 2 से कम होगी ।
माना q(x) = ax² + bx + c (भागफल)
तथा r(x) = dx + e (भागफल)
विभाजन एल्गोरिथ्म के प्रयोग से
p(x) = g (x) × q (x) + r (x)
⇒ x⁴ – 3x² + 4x + 5 = (x² – x + 1) × (ax² + bx + c) + (dx + e)
⇒ x⁴ + 0.x³ – 3x² + 4x + 5 = ax⁴ + bx³ + cx² – ax³ – bx² – cx + ax² + bx + c + dx + e
⇒ x⁴ + 0.x³ – 3x² + 4x + 5 = ax⁴ +(b – a)x³ + (c – b + a)x² + (b – c + d)x + (c + e)
दोनों ओर x की समान घातों के गुणांकों को बराबर करने पर
a = 1 (i)
b – a = 0 ⇒ b = a = 1 (ii)
c – b + a = – 3 ⇒ c – 1 + 1 = – 3 या c = – 3 (iii)
b – c + d = 4 ⇒ 1 – (-3) + d = 4 ⇒ 4 + d = 4 या d = 4 – 4= 0 (iv)
c+ e = 5 ⇒ – 3 + e = 5 या ⇒ = 5 + 3 = 8
a, b, c, d व e के मान प्रतिस्थापित करने पर
भागफल = q(x) = ax² + bx + c
= x² + x – 3
शेषफल = r(x) = dx + e = 0.x + 8 = 8

प्रश्न 19.
पूर्ण वर्ग बनाने की विधि से समीकरण 4x² + 3x + 5 = 0 के वास्तविक मूल ज्ञात कीजिए ।
हल :
यहाँ पर

\(\frac {-71}{64}\) अर्थात् R.H.S. ऋणात्मक है।
(x + \(\frac {3}{8}\))², x के किसी भी वास्तविक मान के लिए ऋणात्मक नहीं हो सकता है।
अतः दिए गए समीकरण के कोई वास्तविक मूल नहीं है।

बहुविकल्पीय प्रश्न :

प्रश्न 1.
निम्नलिखित में से कौन-सी समीकरण द्विघात है ?
(A) x³ – 6x² + 2x – 1 = 0
(B) x² + \(\frac{1}{x^2}\) = 2(x ≠ 0)
(C) (2x + 1)(3x + 2) = 6 (x – 1)(x – 2)
(D) 16x² – 3 = (2x + 5) (5x – 3)
हल :
(D) 16x² – 3 = (2x + 5) (5x – 3)

प्रश्न 2.
निम्नलिखित में से कौन-सी समीकरण द्विघात नहीं है ?
(A) (x – 2) (x + 3) + 1 = 0
(B) x + \(\frac {1}{x}\) = x² (x ≠ 0)
(C) 7x = 2x²
(D) (x + 1) (x + 3) = 0
हल :
(B) x + \(\frac {1}{x}\) = x² (x ≠ 0)

प्रश्न 3.
निम्नलिखित में से कौन-सी समीकरण द्विघात है ?
(A) 3x² – 4x + 2 = 2x² – 2x + 4
(B) x + \(\frac {3}{x}\) = 5x²
(C) x³ + 5x² + x – 5 = 0
(D) (x + 4)(x – 4) = x (x + 2) + 8
हल :
(A) 3x² – 4x + 2 = 2x² – 2x + 4

प्रश्न 4.
निम्नलिखित में से कौन-सी समीकरण द्विघात नहीं है ?
(A) x² – 6x – 4 = 0
(B) 6 – x (x² + 2) = 0
(C) 3x² – 4 = 0
(D) x² + \(\sqrt{2}\)x – 4 = 0
हल :
(B) 6 – x (x² + 2) = 0

प्रश्न 5.
किन्हीं दो क्रमागत धनात्मक पूर्णांकों का गुणनफल 306 है । इसके लिए उचित द्विघात समीकरण होगी-
(A) x² + x – 306 = 0
(B) x² – x + 306 = 0
(C) x² + x + 306 = 0
(D) – x² + x – 306 = 0
हल :
(A) x² + x – 306 = 0

प्रश्न 6.
एक आयताकार भूखंड का क्षेत्रफल 528m² है । यदि क्षेत्र की लंबाई (मीटरों में) चौड़ाई x मी० के दुगुने से एक अधिक है। इसके लिए उचित द्विघात समीकरण होगी-
(A) 2x² – x – 528 = 0
(B) 2x² + x – 528 = 0
(C) 2x² + x + 528 = 0
(D) 2x² – x + 528 = 0
हल :
(B) 2x² + x – 528 = 0

प्रश्न 7.
समीकरण (x + 1)² = 2 (x – 3) कैसी समीकरण है?
(A) एकल घात
(B) त्रिघात
(C) द्विघात
(D) शून्य घात
हल :
(C) द्विघात

प्रश्न 8.
द्विघात समीकरण (x – 2)² – 25 = 0 के हल होंगे-
(A) -7, 3
(B) -7, – 3
(C) 7, – 3
(D) 7, 3
हल :
(C) 7, – 3

प्रश्न 9.
द्विघात समीकरण 5x² – 30 = 0 के …………….. हल होंगे।
(A) ± \(\sqrt{6}\)
(B) ± 6
(C) ± \(\sqrt{5}\)
(D) इनमें से कोई नहीं
हल :
(A) ± \(\sqrt{6}\)

प्रश्न 10.
यदि a, b तथा c धनात्मक वास्तविक संख्याएँ हों तो चर में द्विघात समीकरण का व्यापक रूप होगा-
(A) ay² + c = 0
(B) ay² + by + c = 0
(C) ay² = 0
(D) ay² + by = 0
हल :
(B) ay² + by + c = 0

प्रश्न 11.
द्विघात समीकरण 6x² – 5x – 21 = 0 के हल होंगे-
(A) \(\frac{3}{2}, \frac{-7}{3}\)
(B) \(\frac{-3}{2}, \frac{7}{3}\)
(C) \(\frac{-3}{2}, \frac{-7}{3}\)
(D) \(\frac{3}{2}, \frac{7}{3}\)
हल :
(B) \(\frac{-3}{2}, \frac{7}{3}\)

प्रश्न 12.
k का मान जिसके लिए, द्विघात समीकरण 2x² – kx + 5 = 0 के दोनों मूल बराबर हैं, वह है-
(A) 0
(B) ± 2\(\sqrt{10}\)
(C) 40
(D) 10
हल :
(B) ± 2\(\sqrt{10}\)

प्रश्न 13.
x² – 10x + 21 = 0 के मूल हैं-
(A) 7, 3
(B) – 7, – 3
(C) – 7, 3
(D) 7, – 3
हल :
(A) 7, 3

प्रश्न 14.
द्विघात समीकरण 2x² + 2\(\sqrt{3}\)x + 3 = 0 के मूलों की प्रकृति निम्न में से किस प्रकार की है ?
(A) दो भिन्न-भिन्न, वास्तविक मूल
(B) दो बराबर वास्तविक मूल
(C) कोई वास्तविक मूल नहीं
(D) इनमें से कोई नहीं
हल :
(C) कोई वास्तविक मूल नहीं

प्रश्न 15.
द्विघात समीकरण y² – 8y + 16 = 0 का अभीष्ट हल होगा-
(A) y = 4
(B) y = – 4
(C) y = 2
(D) y = – 2
हल :
(A) y = 4

प्रश्न 16.
द्विघात समीकरण x² – 4qx + 4q² = 0 का अभीष्ट हल होगा –
(A) – 2q
(B) 2q
(C) q
(D) – 9
हल :
(B) 2q

प्रश्न 17.
द्विघात समीकरण 25x² – 30x + 9 = 0 का अभीष्ट हल होगा-
(A) \(\frac {3}{5}\)
(B) \(\frac {-3}{5}\)
(C) \(\frac {5}{3}\)
(D) \(\frac {-5}{3}\)
हल :
(A) \(\frac {3}{5}\)

प्रश्न 18.
द्विघात समीकरण 2x² – 5x + 3 = 0 के मूल होंगे-
(A) – 1 व \(\frac {-3}{2}\)
(B) – 1 व \(\frac {3}{2}\)
(C) 1 व \(\frac {3}{2}\)
(D) 1 व \(\frac {-3}{2}\)
हल :
(C) 1 व \(\frac {3}{2}\)

प्रश्न 19.
द्विघात समीकरण 3x² – 2x – 1 = 0 के दो मूलों का गुणनफल होगा-
(A) \(\frac {-1}{3}\)
(B) \(\frac {1}{3}\)
(C) \(\frac {2}{3}\)
(D) इनमें से कोई नहीं
हल :
(A) \(\frac {-1}{3}\)

प्रश्न 20.
द्विघात समीकरण x² – 3x – 10 = 0 के मूल होंगे-
(A) – 2 व 5
(B) 2 व – 5
(C) – 2 व – 5
(D) 2 व 5
हल :
(A) – 2 व 5

प्रश्न 21.
द्विघात समीकरण x² + 6x + 5 = 0 के दो मूलों का गुणनफल निम्नलिखित में से ज्ञात कीजिए-
(A) – 5
(B) 5
(C) \(\frac {1}{5}\)
(D) इनमें से कोई नहीं
हल :
(A) – 5

प्रश्न 22.
द्विघात समीकरण 3x² + 2x – 5 = 0 के दो मूलों का गुणनफल होगा-
(A) \(\frac {-5}{3}\)
(B) \(\frac {5}{3}\)
(C) \(\frac {3}{5}\)
(D) इनमें से कोई नहीं
हल :
(A) \(\frac {-5}{3}\)

प्रश्न 23.
द्विघात समीकरण 8x² – 22x – 21 = 0 के हल होंगे-
(A) \(\frac{7}{2}, \frac{3}{4}\)
(B) \(\frac{-7}{2}, \frac{3}{4}\)
(C) \(\frac{7}{2}, \frac{-3}{4}\)
(D) इनमें से कोई नहीं
हल :
(C) \(\frac{7}{2}, \frac{-3}{4}\)

प्रश्न 24.
k का मान जिसके लिए, द्विघांत समीकरण 3x² – kx + 5 = 0 के दोनों मूल बराबर हैं, वह है-
(A) 0
(B) 60
(C) ± 2\(\sqrt{15}\)
(D) 15
हल :
(C) ± 2\(\sqrt{15}\)

प्रश्न 25.
x² – 5x + 6 = 0 के मूल हैं-
(A) 5, – 6
(B) 2, 3
(C) 6, – 1
(D) -2, – 3
हल :
(B) 2, 3

प्रश्न 26.
द्विघात समीकरण x² + 4x + 1 = 0 का विविक्तकर क्या होगा ?
(A) 12
(B) 14
(C) 16
(D) -12
हल :
(A) 12

प्रश्न 27.
द्विघात समीकरण 4x² – ax + 2 = 0 का विविक्तकर ……………………. होगा ।
(A) a² – 32
(B) a² + 32.
(C) a² = 32
(D) a² = – 32
हल :
(A) a² – 32

प्रश्न 28.
\(\sqrt{3}\)x² – 2\(\sqrt{2}\)x – 2\(\sqrt{3}\) = 0 का विविक्तकर होगा-
(A) 8
(B) 16
(C) 32
(D) 24
हल :
(C) 32

प्रश्न 29.
निम्नलिखित में से किस समीकरण के मूल वास्तविक हैं ?
(A) 2x² + x – 1 = 0
(B) 3x² + 2x – 1 = 0
(C) x² + 4x + 4 = 0
(D) 2x² + 5x + 5 = 0
हल :
(D) 2x² + 5x + 5 = 0

प्रश्न 30.
किस भारतीय गणितज्ञ ने सर्वप्रथम व्यापक द्विघात समीकरण के मूलों के लिए सूत्र प्रतिपादित किया ?
(A) आर्यभट्ट
(B) ब्रह्मगुप्त
(C) महावीर
(D) श्री धराचार्य
हल :
(D) श्री धराचार्य

प्रश्न 31.
निम्नलिखित में से किस समीकरण के मूल वास्तविक हैं ?
(A) x² + x + 1 = 0
(B) x² + 4x + 4 = 0
(C) 2x² + 5x + 5 = 0
(D) उपरोक्त सभी के
हल :
(B) x² + 4x + 4 = 0

प्रश्न 32.
k के किस मान के लिए द्विघात समीकरण kx² + 4x + 1 = 0 के दो मूल बराबर हैं-
(A) – 4
(B) 4
(C) 16
(D) इनमें से कोई नहीं
हल :
(B) 4

प्रश्न 33.
दो क्रमागत धनात्मक पूर्णांकों का गुणनफल 306 हो, तो वे पूर्णांक हैं-
(A) 18 तथा 19
(B) 12 तथा 13
(C) 16 तथा 17
(D) 17 तथा 18
हल :
(D) 17 तथा 18

प्रश्न 34.
p के किस मान के लिए द्विघात समीकरण 3x² – 5x + p = 0 के मूल बराबर होंगे ?
(A) p = \(\frac {25}{12}\)
(B) p = \(\frac {-25}{12}\)
(C) p = \(\frac {12}{25}\)
(D) p = \(\frac {-12}{25}\)
हल :
(A) p = \(\frac {25}{12}\)

प्रश्न 35.
द्विघात समीकरण 4x² + kx + 9 = 0 में k के किस मान के लिए उसके दो मूल बराबर है-
(A) ± 4
(B) ± 6
(C) ± 12
(D) इनमें से कोई नहीं
हल :
(C) ± 12

प्रश्न 36.
p के किस मान के लिए द्विघात समीकरण 2x² + px + 18 = 0 के मूल वास्तविक होंगे ?
(A) p ≥ 13
(B) p ≤ 18
(C) p ≥ 18
(D) p ≥ 12
हल :
(D) p ≥ 12

प्रश्न 37.
द्विघात समीकरण जिसके मूलों का योग 5 तथा गुणनफल 6 है होगी-
(A) x² – 5x + 6 = 0
(B) x² – 5x – 6 = 0
(C) x² + 5x + 6 = 0
(D) x² + 5x – 6 = 0
हल :
(A) x² – 5x + 6 = 0

प्रश्न 38.
एक हॉल की लंबाई उसकी चौड़ाई से 5 मी० अधिक है । यदि हॉल के फर्श का क्षेत्रफल 84 वर्ग मी० हो
तो हॉल की लंबाई होगी-
(A) 12 मी०
(B) 7 मी०
(C) 14 मी०
(D) 6 मी०
हल :
(A) 12 मी०

प्रश्न 39.
12 को दो ऐसे भागों में विभक्त कीजिए जिनके वर्गों का योग 74 है-
(A) 8, 4
(B) 7, 5
(C) 9, 3
(D) 10, 2
हल :
(B) 7, 5

प्रश्न 40.
19 को दो ऐसे भागों में विभक्त करें जिनके वर्गों का योग 193 है
(A) 14, 5
(B) 12, 7
(C) 13, 6
(D) 11, 8
हल :
(B) 12, 7

प्रश्न 41.
द्विघात समीकरण x² – kx + 9 = 0 में k के किस मान के लिए उसके दो मूल बराबर हैं-
(A) ± 5
(C) ± 4
(B) ± 6
(D) इनमें से कोई नहीं
हल :
(B) ± 6

प्रश्न 42.
निम्नलिखित में से क्रमागत सम धनात्मक पूर्णांक जिनके वर्गों का योग 100 है-
(A) 4, 6
(B) 5, 7
(C) 6, 8
(D) 8, 10
हल :
(C) 6, 8

प्रश्न 43.
दो क्रमागत विषम प्राकृत संख्याएँ जिनके वर्गों का योग 202 हो तो इन संख्याओं का योग होगा-
(A) 22
(B) 9
(C) 11
(D) 20
हल :
(D) 20

प्रश्न 44.
3 से आरंभ करके n क्रमागत विषम प्राकृत संख्याओं के योग का सूत्र S = n (n + 2) होता है । यदि S = 168 हो तो n का मान होगा-
(A) 16
(B) 14
(C) 13
(D) 12
हल :
(D) 12

प्रश्न 45.
प्रथम n क्रमागत सम प्राकृत संख्याओं का योग S निम्नलिखित संबंध द्वारा दर्शाया जाता है- S = n (n + 1) यदि योग 420 हो, तो n का मान-
(A) 18
(B) 19
(C) 20
(D) 21
हल :
(C) 20

प्रश्न 46.
यदि द्विघात समीकरण 5x² – 8x + 4k = 0 के मूल समान हों, तो k का मान होगा-
(A) \(\frac {5}{4}\)
(B) \(\frac {1}{2}\)
(C) \(\frac {4}{5}\)
(D) 2
हल :
(C) \(\frac {4}{5}\)

प्रश्न 47.
x² – 7x + 12 = 0 के मूल हैं-
(A) -3, -4
(B) 3, 4
(C) 6, 2
(D) -6, 2
हल :
(B) 3, 4

प्रश्न 48.
द्विघात समीकरण 2x² – 2\(\sqrt{3}\)x – 3 = 0 के मूलों की प्रकृति निम्न में से किस प्रकार की है ?
(A) दो भिन्न-भिन्न, वास्तविक मूल
(B) दो बराबर वास्तविक मूल
(C) कोई वास्तविक मूल नहीं
(D) इनमें से कोई नहीं
हल :
(A) दो भिन्न-भिन्न, वास्तविक मूल

प्रश्न 49.
समीकरण 3x² – 2x + \(\frac {1}{3}\) = 0 के मूल हैं-
(A) 3 तथा 2
(B) \(\frac {1}{3}\) तथा \(\frac {1}{3}\)
(C) \(\frac {1}{3}\) तथा \(\frac {1}{2}\)
(D) इनमें से कोई नहीं
हल :
(B) \(\frac {1}{3}\) तथा \(\frac {1}{3}\)

प्रश्न 50.
द्विघात समीकरण 2x² – 2\(\sqrt{2}\)x + 1 = 0 के मूल होंगे-
(A) वास्तविक तथा भिन्न
(B) वास्तविक तथा समान
(C) वास्तविक तथा शून्य
(D) इनमें से कोई नहीं
हल :
(B) वास्तविक तथा समान

Haryana State Board HBSE 10th Class Science Notes Chapter 16 Management of Natural Resources Notes.

Haryana Board 10th Class Science Notes Chapter 16 Management of Natural Resources

What Happens When we Add Our Waste to the Environment?

Natural resources:

• The resources which occur naturally, and which cannot be created by man, are called natural resources. The components of the atmosphere, hydrosphere and lithosphere which are used for the maintenance of life are called natural resources.
• Water, land, air, forests, minerals, plants and animals are also natural resources.

Five R’s for saving the environment:
Refuse:
Refuse refers to the concept of saying No to things that people offer you but you do not need. In other words one must refuse unnecessary things. Example: Refuse to use single-use plastic carry bags.

Reduce:
Reduce means to use less. One can reduce the burden on environment by reducing consumption. Example: Instead of washing the car with pipes wipe it with wet cloth. Reducing wastage of food

Reuse:
The method of reuse refers to using an item again and again rather than throwing it away. Example: Rather than using plastic bags, buy proper shopping bags made out of cloth.

Repurpose:
At times when a product cannot be used further for its original purpose than use it for some other use. This is called repurpose. Example: We can make cloth bags from old trousers.

Recycle:
The action or process of converting waste into reusable material is called recycling.
Under recycling one can collect material such as plastic, paper, glass and metal items and recycle them to make required things instead of synthesizing or extracting fresh plastic, paper, glass or metal.

Sustainable development:
Economic development done without depleting the natural resources is called sustainable development.

Biodiversity:
The diversity of plant and animal life in a particular habitat or in the world as a whole is called biodiversity.

Stakeholder:
A person with an interest or concern in something is called the stake holder.

Stakeholders of forests:

• People who live in or around forests are dependent on forest produce for various aspects of their life.
• The Forest Department of the Government which owns the land and controls the resources from forests.
• The industrialists. All the industrialists covering those who use ‘tendu’ leaves to make bidis to the ones who owns paper mills. All those who use various forest produce, but are not dependent on the forests of any one area. In other words, they may source raw material from any forest.
• The wildlife and nature enthusiasts who want to conserve nature in its pristine form.

Dependency on water:

• Usually man depends upon fresh water for fulfilling his daily requirements.
• Irregularities in the rainfall may create flood or drought, thus imbalancing the quantity of fresh water on earth and causing disasters. Thus, life on earth is fully dependent on fresh water.

Construction of large dams is criticized mainly because of the following three problems:

• Social problems: Construction of large dams requires displacing large number of peasants and tribals residing nearby without paying them adequate compensation or rehabilitation. This leads to social problems.
• Economic problems: Such dams swallow huge amount of public money and do not generate proportionate benefits.
• Environmental problems: Massive construction of dams leads to large scale deforestation and huge loss to biological diversity.
• People who have been displaced by various development projects are largely poor tribals. They face dual loss – first they have to give their land and access to forests without receiving proper compensation and second they do not even get any benefit from such projects.

Watershed:
Any surface area from which draining of water resulting from rainfall is collected and drained through a common point is called a watershed. Watershed is similar to drainage basin or catchment area.

Watershed management:
The process of adopting practices of ‘land use’ and ‘water management’ in order to protect and improve the quality of the water and other natural resources within a watershed is known as watershed management.

Water-harvesting:
In general, water harvesting is the activity of collecting the water directly. The rainwater so collected can be stored for direct use or can be recharged into the groundwater.

Coal and petroleum:
Coal and petroleum are fossil fuels. Thesewere formed due to the degradation of bio mass millions of years ago. It is estimated that petroleum discovered so far will last for about 40 years and coal for about 200 years. Hence, it is utmost important to preserve them. Methods of conserving fossil fuels.

• Making maximum use of renewable energy
• Using public transport, constructing efficient building, etc.
• Developing more efficient engines for the vehicles
• Protecting fossil fuels from accidental fires

Haryana State Board HBSE 10th Class Science Notes Chapter 15 Our Environment Notes.

Haryana Board 10th Class Science Notes Chapter 15 Our Environment

What Happens When we Add Our Waste to the Environment?

Environment:

• The air, water and land in or on which people, animals and plants live is called environment.
• Biotic components such as plants and animals and abiotic components such as air, water, soil,etc. are the two components of environment.

Waste:
Unwanted or unusable items, remains or leftovers or by-products of household garbage together are termed as waste.

Biodegradable and non-biodegradable waste:
The waste materials, which are broken down to simpler substances by biological processes, are called biodegradable wastes, while those which cannot be broken down to simpler substances are called non-biodegradable wastes. For example, vegetables and fruits are biodegradable, while glass, plastics and polythene, etc. are non-biodegradable.

Ecosystem – What are its Components?

Food Chains and Webs

Food chain:
Living organisms of an ecosystem depend on each other for their food requirement and form a chain which is known as food chain.

In a food chain, the producers are at the first trophic level, the primary consumers (herbivores) at second, small carnivores i.e. secondary consumers at the third level and large carnivores or tertiary consumers at the fourth trophic level. The flow of energy in an ecosystem is always unidirectional.

Food web:

• The network of interconnected food chains functioning in an ecosystem is called food web.
• The relationship among individuals in the food chain is not represented by a straight line but as a series of branching lines which together form the food web.

Biological magnification:
When harmful chemicals and pesticides such as DOT, BHC, etc. enter in the food chain, level. The phenomenon is called biological magnification. Humans occupy the top level of any food chain. Hence, there is maximum concentration of harmful chemicals in humans. In other words blo-magnification is highest at the human level.

Ozone:

• Ozone (O₃) is a gaseous molecule formed by three atoms of oxygen. It is a deadly poisonous gas. Ozone layer is located at the higher levels of the atmosphere i.e. in stratosphere.
• The ozone layer absorbs harmful ultraviolet rays coming from the sun and protects us. Unfortunately, this layer is depleting and this is posing a serious problem for our planet.

Haryana State Board HBSE 10th Class Science Notes Chapter 14 Sources of Energy Notes.

Haryana Board 10th Class Science Notes Chapter 14 Sources of Energy

What is a Good Source of Energy

Source of energy:
A source from which useful energy can be extracted either directly or indirectly by means of a conversion or transformation is known as the source of energy. For example, sources of energy to provide us heat for cooking are LPG, kerosene, sunlight, etc.

Factors to consider while selecting an energy source:
The source of energy should be

• Availability,
• Output,
• Accessibility
• Transportation and storage and
• Economical.

Conventional (Non-renewable) source of energy:
A source of energy which we cannot regenerate or reuse once we have used it is called a conventional or non-renewable source of energy. For example, coal, petroleum and natural gas, thermal power plants, hydro power plants, bio-mass and wind energy.

Non-conventional (Renewable) source of energy:
Those sources of energy which are inexhaustible or say can be renewed are called non-conventional (renewable) sources of energy. For example, solar energy, tidal energy, wind energy, geothermal energy, etc.

Fossil fuels:
Fuels such as coal natural gas and petroleum formed in the earth crust due to decaying of plants and animal remains millions of years ago are known as fossil fuels.

Disadvantages of fossil fuels:

• Produces smoke
• Cause severe air pollution,
• Causes acid rain.
• Causes global warming.

Thermal power plant:
Thermal power plant is a set-up which converts heat energy into electrical energy on a large scale basis.

Hydropower plant:
Hydropower or hydroelectric plants use the potential energy of water stored at height (in dams) and the kinetic energy of the falling water for generating electricity. The power produced is called hydroelectricity or hydel electricity.

Disadvantages:

• Dams can be constructed only in some hilly areas,
• People living in low areas have to be relocated.
• Large ecosystems get destroyed
• Methane gas gets produced in large quantity.

Biomass:
The waste material from plants or animals such as cattle dung, dried leaves, etc. which is not used for food or feed is called biomass.

Biogas:
The gas prepared by decomposing cow dung, plant residue such as dead plants, dried leaves, residue obtained after harvesting a crop, vegetable waste and sewage in a pit is called biogas or gobar gas.

Advantages:

• An excellent fuel used for cooking and lightening
• Burns without smoke and leaves no residue
• High thermal capacity
• The slurry is an excellent manure

Biogas can also be used for lightening the villages.

Wind:
Moving air is called wind. Wind possesses kinetic energy. This kinetic energy can be used to do work or obtain electricity through a machine called wind mill.

Advantages:

• Does not cause any pollution
• It is renewable,
• Does not cause any recurring expense.

Limitations:

• Wind farm can be established only at those places where wind blows for most of the year that too with a minimum speed of 15 km/h
• Requires power back-up facility
• Required large area,
• Initial set-up cost is quite high,so is the maintenance.

Alternetive or Non-conventional Sources of Energy

Alternate (non-conventional) sources of energy:
Sources of energy which we have not yet started using on a regular and routine basis are called non-conventional sources of energy. Example: Solar energy, oceanic energy, geothermal energy and nuclear energy.

Solar energy.
The energy obtained from the sun is called solar energy. This energy is available in two forms namely, light and heat.

Solar appliances and their uses:

Energy from sea:
Tidal energy:
The level of ocean water rises and falls due to the gravitational pull of moon on the earth. The difference ¡n the tides give us energy which is called tidal energy.

Wave energy:
The winds blowing over ocean produces waves. These waves possess large amount of kinetic energy. Several devices have been developed to trap wave energy for rotating the turbine and hence producing electricity.

Ocean Thermal Energy Conversion (OTEC):
The energy available due to the difference in the temperature of water at the surface of the ocean and at deeper levels of the ocean is called Ocean Thermal Energy (OTE). The plant set-up to harnass this energy is called Ocean Thermal Energy Conversion (OTEC) Plant.

Geothermal energy:
The deep interior region of the earth where magma is found is very hot. The energy utilized from this heat is called geothermal energy. At some places, steam and hot water ooze out on their own through cracks of the rocks. Such sites serve as hot water springs or natural geysers.

Nuclear energy:

• When the nucleus of a heavy atom (such as uranium, plutonium or thorium) is bombarded with low-energy neutrons, it gets split into lighter nuclei. This process is called nuclear fission.
• During the splitting of nucleus, tremendous amount of energy is released. This energy is called nuclear energy.

Advantages:

• Produces 10 million times the energy produced by the combustion of an atom of carbon from coal,
• The nuclear fuel can itself go on chain reaction and release energy at a controlled rate.

Limitations:

• The biggest problem is the storage and disposal of used nuclear fuel.
• In case of accident if the radiations leak, they can cause very large and widespread effect.
• The installation cost is very high.

Haryana State Board HBSE 10th Class Science Notes Chapter 13 Magnetic Effects of Electric Current Notes.

Haryana Board 10th Class Science Notes Chapter 13 Magnetic Effects of Electric Current

Magnetic Field and Magnetic Field Lines

Magnet and magnetism:
Substances that have property of attracting metals such as iron, nickel, cobalt, etc. are called magnets. This property of attraction possessed by magnets is called magnetism.

Bar magnet:
A magnet in the shape of a bar having two magnetic poles, nam&y the north pole and the south pole is known as a bar magnet.

Magnetic field:
The region surrounding the magnet in which magnetic force can be experienced is called magnetic field. Field lines are drawn to describe the extent of the magnetic peki.

Magnetic compass:
Magnetic compass is a small instrument which helps us to locate the -direction. The compass has a dial on which directions namely North N, South S, East E, and West W are marked.

Magnetic Field Due to a current-carrying conductor

Magnetic field formed due to a linear (straight) conductor carrying current:

It depends upon the following two factors —

• The magnetic field produced due to the current passing in the conductor is proportional to the electric current i.e. magnetic field intensity α electric current.
• The magnetic field intensity goes on decreasing as the distance from the conductor increases i.e. magnetic field intensity α 1/distance.

Right Hand Thumb Rule:

• The direction of magnetic field associated with the electric current can be found with the help of the ‘Right Hand Thumb Rule’.
• For knowing the direction of magnetic field, imagine that you are holding the conducting wire in your right hand so that your thumb points in the direction of the current (I), then the direction in which your fingers encircle the wire will give the direction of magnetic field around the wire.

Magnetic field produced due to a current carrying loop of wire:
The magnetic field B produced at the centre of the coil is —

• Directly proportional to the current (I) flowing through it i.e. B α I
• Inversely proportional to the radius (r) of the loop i.e. B α 1
• Directly proportional to the total number of turns N of the coil i.e. B α N. This means if a ring is made up of closely spaced N turns, the magnetic field at the centre of the ring will be N times stronger. Thus, magnetic field intensity
  points in the direction of the current (I), then the direction in which your fingers encircle the wire will give the direction of magnetic field around the wire.

Magnetic field produced due to a current carrying loop of wire:

The magnetic field B produced at the centre of the coil is —

• Directly proportional to the current (I) flowing through it i.e. B α I
• Inversely proportional to the radius (r) of the loop i.e. B α 1/r
• Directly proportional to the total number of turns N of the coil i.e. B α N. This means if a the magnetic field at the centre of the ring will be N times stronger. Thus, magnetic field intensity

Solenoid:

• A long met wire turned several times to form the structure of a coiled cylinder is known as a solenoid.
• On passing electric current, a magnetic field is produced inside the solenoid. The magnetic field resulting due to N turns will be N times stronger than the magnetic Seki resulting by each circular coil.

The factors on which the strength of the magnetic field produced by a current carrying solenoid depends are —

• Number of turns (N) In the solenoid: More the number of turns (N) in the solenoid, stronger will be the magnetic field produced.
• Strength of the current (I): Greater the current (I) passing through solenoid, stronger will be the magnetic field produced.
• Nature of core material: If material such as a soft iron cylinder is used in making the solenoid core, the magnetic field will be quite strong.

Electromagnet:

A soft iron core placed inside a solenoid behaves like a powerful magnet when a current is passed through solenoid. Such a magnet (with magnetic field around) is called an electromagnet.

Force on a Current-Carn’in Conductor in a Magnetic Field

Fleming’s left hand rule:

Arrange your left hand such that the tore finger, the center finger and thumb remain at right angle to each other. Adjust your hand in such a way that the forefinger points in the direction of magnetic field and the centre finger points in the direction of current, then the direction in which the thumb points will be the direction of magnetic force.

Electric motor:
An electric motor is a rotating device which converts electric energy into mechanical energy.

Principie of electric motor:

• When a current carrying conductor (or a coil) is placed in a magnetic field, the conductor experiences force and it rotates.
• It is used in electric fan, mixer-grinder, washing machine, DVD players, etc.

Electromagnetic induction:

• When the magnetic line of force passing through a closed circuit change, voltage and hence a current is induced in It. This phenomenon is called electromagnetic induction. The voltage produced is called induced force (e.m.f.) and the current is called induced current.
• The direction of electric current can be found out with the help of Fleming’s right hand rule.

Fleming’s right hand rule:
Arrange the forefinger, middle finger and thumb of a right hand at right angle to one another. Adjust the forefinger in the direction of magnetic field, and thumb pointing in the direction of motion of conductor. The direction of middle finger will then indicate the direction of induced electric current.

Galvanometer:
A galvanometer is an instrument that can detect the presence of current in a circuit i.e. whether the current is flowing in the circuit or not.

Electric generator:

An electric generator is a device which converts mechanical energy into electric energy. The electric generator works on the principle of electromagnetic induction.

Types of electric current:

• DC current: The current that always flows in the same direction is known as direct current (DC). Batteries used in radio/cell phone, watch, laptop, etc. works on DC.
• AC current: The current that changes direction after equal intervals of time is known as alternating current (AC). Appliances such as fan, tubelight, TV, refrigerator. etc. run on AC Current.

Domestic Electric Circuits:
The current that comes to our house from the power stations is Alternate Current (AC). From power station, this power is transmitted to our houses through thick underground cables or over head electric poles and cables.

Three main wires enter the main board of our house. They are :

• Wire with red coloured insulation. This wire is known as live wire or positive.
• Wire with black coloured insulation. This wire is known as neutral wire or ‘negative’.
• Wire with green coloured insulation. This wire is known as earth (or earthing) wire.

Short circuit:
If positive (live) and negative wires touch each other accidently, such a situation is called short-circuit

Fuse:
A fuse is a safety device made out of small metallic strip with a low melting point. It works on the heating effect of the electric current.

Overloading:
If a large number of electrical appliances are switched on simultaneously, they start drawing large amount of current from the mains. When the current drawn exceeds the safety limit it is called over loading. Overloading may cause short- circuit. Fuse can help in preventing short-circuit.

Haryana State Board HBSE 10th Class Science Notes Chapter 11 Human Eye and Colourful World Notes.

Haryana Board 10th Class Science Notes Chapter 11 Human Eye and Colourful World

Human eye:
The human eye is one of the most valuable and sensitive sense organs. It enables us to see the wonderful world and its colours.

Working of eyes:

• The light rays coming from the object enter the eye through cornea.
• The muscular diaphragm called iris controls the amount of light that enters the eye.
• There is a hole in the iris called the pupil of eye.
• After passing through the pupil, the light rays are incident on the convex eye-lens.
• Ciliary muscles hold the eye lens. They change the thickness of the eye lens while focusing which helps in proper viewing of the objects.
• The screen on which the image is formed in the eye is called retina. Retina consists of a large number of light sensitive cells.
• The retina sends these signals to the brain through optic nerve.

Power of Accommodation

Accommodation power of an eye:
The ability of the eye lens to adjust its focal length as per requirement so that objects can be seen clearly is called accommodation power of an eye.

Near point of an eye:
The minimum distance at which the objects can be seen clearly without contracting the eye lens i.e. without any strain is called the least distance of the distinct vision or near point of an eye. For a young adult having normal vision, the near point of the eye is 25 cm.

Far point of an eye :
The farthest distance up to which the eye can see objects clearly is called far point of an eye. The farthest point of a person with normal vision lies at an infinite distance. Thus, a person with normal vision can see objects clearly from 25 cm to infinite distance.

Cataract:
In old aged people when the eye lens becomes milky and cloudy, the vision becomes hazy or even opaque due to the formation of a membrane over the lens. This condition is called cataracts.

Defects of Vision and Their Correction

Types of defect of vision:

• Near sightedness or Myopia
• Far sightedness or Hypermetropia
• Presbyopia

Near-sightedness or Myopia:
When the lens is unable to become thin, the light rays converge more than they should. So, the image gets formed before the retina rather than on it. Hence, distant objects cannot be seen clearly. This defect is known as near-sightedness or myopia.

Far-sightedness (hypermetropia):
If eye lens does not become thick as per the requirement, then the rays coming from nearby objects gets less converged and hence are focused behind the retina. Hence, nearby objects cannot be seen clearly. This type of defect is known as far-sightedness or hypermetropia.

Presbyopia:
As a person grows older, the power of accommodation of an eye usually decreases.
→ The near point of aged people recedes and they find it difficult to see nearby objects clearly without spectacles. Such a defect is called presbyopia.

Refraction of Light through a Prism and Dispersion of White Light by a Glass Prism

Prism:
A prism is a portion of a transparent medium bounded by two plane faces inclined to each other at a certain angle.

Dispersion of light:
Splitting of white light into its seven constituent colours on passing through a transparent medium like a glass prism is called dispersion of light. The band of colours formed after dispersion is known as spectrum.

On the screen, we get a band of seven colours in the following order from bottom to top:
Violet, Indigo, Blue, Green, Yellow, Orange and Red (VIBGYOR).

Rainbow:
A rainbow is a natural spectrum visible in the sky after rain shower. Rainbow is formed when the water droplets present in the atmosphere disperse the sunlight falling on them.

Atmospheric Refraction:

Atmospheric refraction is the deviation of light or other electromagnetic waves from a straight line as it passes through the atmosphere due to the variation in air density. Phenomena such as twinkling of stars, early sunrise and delayed sunset occur due to this effect.

Scattering of light:
The deflection of light by minute particles and molecules in all the directions is known as scattering of light. The colour of scattered light depends upon the size of scattering particles.

Tyndall effect:
The scattering of light in the nature due to small particles present in the atmosphere is called Tyndall effect.

Haryana State Board HBSE 10th Class Science Notes Chapter 10 Light Reflection and Refraction Notes.

Haryana Board 10th Class Science Notes Chapter 10 Light Reflection and Refraction

Light:

• When light falls on an object let us say a book, or a table, etc., it gets reflected, The reflected light rays reach our eyes and enable us to see the object.
• Light waves do not require a material medium (like solid, liquid or gas) for their propagation and hence are called non-mechanical waves.
• When light is incident on a surface separating two medias such as air and water, then some part of incident light is reflected, some part is transmitted and some part of it is absorbed by that surface.

Reflection of Light

Reflection :

The phenomenon of sending back the light rays which fall on the surface of an object when light is incident on it, is called reflection of light.

Types:

• Regular reflection and
• Irregular (diffused) reflection

Image:

When a number of rays emerging from a point of an object, after reflection or refraction, ‘meet’ or ‘appear to meet’ at another point, then the point of meeting is called the image of the first point.

Types:

• Real image and
• Virtual image

Real Image: When the rays of light after reflection or refraction ‘actually meet’ at a point, the image formed is called a real image.

Virtual Image: When the rays of light after reflection or refraction ‘do not actually meet’ but ‘appear to meet’ at a point, the image formed is called a virtual image.

A polished/shining surface which reflects almost all the light incident on it is called a mirror.

Types of mirror:

(1) Plane mirror and (2) Spherical mirror.
Further, spherical mirrors are of two types. They are: (a) Concave mirror and (b) Convex mirror.

Spherical Mirrors

Plane mirror :
A plane mirror is a thin, flat and smooth sheet of glass having a shining coat of silver metal on one side. For example, mirror used in vehicles,in dressing tables, etc.

Spherical mirror :
Unlike plane mirrors, spherical (curved) mirrors converge or diverge the parallel light rays incident on them.

Concave mirror:
The spherical mirror in which light rays converge to form an image is called a concave mirror (or converging mirror).

Convex mirror:
A spherical mirror in which light rays diverge to form an image is called a convex mirror (or diverging mirror).

Image Formation by Spherical Mirrors and Representation of Images Formed by Spherical Mirrors Using Ray Diagrams

• The image of an object formed by a spherical mirror (or a lens) can be obtained and studied by constructing a ray diagram.
• Two rays are sufficient to draw a ray diagram because by intersecting two reflected rays we can obtain the position of the image formed.

Converging (Concave) mirror:

When parallel rays are incident on concave mirror,on reflection, they converge i.e. come closer to each other and meet at a point called principal focus (F). Hence, concave mirror is also called a converging mirror.

Diverging (Convex) mirror:

When parallel rays are incident on convex mirror, on reflection, they diverge i.e. spread out from each other. Hence, convex mirror is also called a diverging mirror.

Relation between radius of curvature (R) and focal length (f) of a spherical mirror:

Focal length\((f)=\frac{1}{2} R=\frac{R}{2}\)
Obtaining Images through concave., and convex mirrors

Positions at which we can place an object to obtain images from a concave mirror and summary of the images formed:

Summary of images formed in a concave mirror

Positions at which we can place an object to obtain Images from a convex mirror and summary of the images formed:

Summary of images formed in a convex mirror

Sign Convention for Reflection by Spherical Mirrors

Summary of New Cartesian Sign Convention for reflection by spherical mirror:

Mirror Formula and Magnification :

• The relationship between the object distance (u), image distance (v) and focal length (f) of a mirror is called the mirror formula.
  As per mirror formula:\(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\)
• The ratio of image height (h’) to object height (h) is called the magnification (m).
  Thus, for spherical mirrors, magnification \((m)=\frac{\text { Height of image }\left(h^{\prime}\right)}{\text { Height of object }(h)} \quad \text { i.e. } m=\frac{h^{\prime}}{h}\)
•  In terms of object distance (u) and image distance (y), magnification (m) can be expressed as, \(m=\frac{h^{\prime}}{h}=-\frac{v}{u}\)

Summary of size of image, type of image and magnification for various mirrors

Refraction of Light

Refraction:
When a ray of light enters obliquely from one transparent medium to another transparent medium, its velocity changes due to which it gets deviated from its original direction at the surface separating two medias. This is called refraction.

Medium:
A transparent substance in which light can travel is called a medium. Air, water, gas, glass, kerosene, etc. are examples of mediums.

• Optically rarer medium: A medium in which the speed of light is more is called optically rarer medium. For example, air.
• Optically denser medium: A medium in which the speed of light is less (compared to optically rarer medium) is called an optically denser medium. For example glass and water (in comparison to air).

Relative Refractive index :

• The ratio of speed of light v₁ in medium 1 to the speed of light v₂ in medium 2 is known as the relative refractive index of medium 2 with respect to medium 1 (except the medium of vacuum).
• It is denoted by η₂₁ i.e. refractive index of medium 2 with respect to medium 1.

∴ Refractive index \(\eta_{21}=\frac{\text { Speed of light in medium } 1}{\text { Speed of light in medium } 2}=\frac{v_1}{v_2}\)

Moreover, Absolute refractive index \(\eta_{\mathrm{m}}=\frac{\text { Speed of light in vacuum }}{\text { Speed of light in medium }}=\frac{\mathrm{c}}{\mathrm{v}}\)

Snell’s law:
The ratio of sine of angle of incidence to the ratio of sine of angle of refraction is constant.
∴ \(\eta_{21}=\frac{\sin \theta_1}{\sin \theta_2}\)

Refraction by Spherical Lenses and Formation of Images

Lens: A lens is a piece of transparent glass bound by two spherical surfaces.

Types: Convex (Converging) lens:
A lens which is thick at the center and tapered at the upper and lower ends is called a convex lens.

Concave (Diverging) lens:
A lens which is thin in the middle but thicker at the edges is called concave lens.

Positions at Which we can place an object in front of convex lens and the summary of images formed:

Summary of images formed by convex lens

Summary of images formed by concave lens

Summary of New Cartesian Sign Convention followed for refraction of light through spherical lenses

Power of Lens

The power of lens (P) is the “Reciprocal of the focal length (f) of lens”.
∴ Power of lens (P) = \(\frac{1}{f}\)

Haryana State Board HBSE 10th Class Science Notes Chapter 9 Heredity and Evolution Notes.

Haryana Board 10th Class Science Notes Chapter 9 Heredity and Evolution

Accumulation of Variation During Reproduction

During reproduction, the offspring inherit two things from their previous generation. They are –
1. A common basic body design and
2. Some fine (i.e. minor) variations. This way with each new generation born, the variations get accumulated.
Parental generation (P): The first set of parents crossed is called parental generation. The parental generation is denoted by ‘P’.

First generation (F₁): All the offspring born from the first set of parents crossed are said to be belonging to the First generation. The First generation is denoted by F₁

Second generation (F₂): All the offspring born from the F₁ generation are said to be belonging to the Second generation. The Second generation is denoted by F₂.

Heredity:

• Heredity means the transmission of characters from parents to offsprings. For example, eggs laid by a sparrow will hatch sparrow and not any other bird. Similarly, a dog gives birth only to pups.
• Thus, in this sense, it can be said that heredity is the continuity of features from one generation to another. This is the essence of heredity.

Rules for the Inheritance of Traits – Mendel’s Contributions

There are two versions for each trait in each child –
In a sexual reproduction, the father, as well as the mother, contributes equal amounts of genetic material to the child.

Mendel’s contribution in the field of inheritance of characteristics:
Mendel studied garden pea plants for the expression of a character. The character under study was ‘the height of plants’. Mendel took pure tall plants (TT) and pure short plants (tt).

Cross-pollination between parent (P) generation plants:
On performing cross-pollination between TT – pure tall plants and tt – pure short plants, all the plants of F₁ generations were as tall as TT of P generation.

Self-pollination of (a) Tall Parental Generation plants and (b) F₁ generation plants:
(a) The new plants produced by the self-pollination of Tall i.e. TT Parental Generation plants were all tall.
(b) The new plants produced i.e. the second (F₂) generation plants by the self-pollination of F₁ generation plants were a mix of tall and short plants. 75% of the plants were tall and 25% of the plants were short. This means the ratio of tall : short plant in the F₂ generation was 3:1.

How do these Traits get Expressed?

In plants as”well as organisms, the characters (traits) get expressed due to specific reactions. These reactions take place with the help of enzymes. When a specific gene undertakes synthesis of a specific protein it results in expression of a specific character.

Sex determination:

• In sexual reproduction, the mechanism to determine the sex of an organism is known as sex determination. In human beings, the sex is determined by genes located on the chromosomes.
• In humans, each cell contains 23 pairs of chromosomes.
• In females, the 23rd pair contains two similar ‘X’ sex chromosomes i.e. the 23rd pair is ‘XX’. In males, the 23rd pair contains one ‘X’ sex chromosome and one ‘Y’ sex chromosome i.e. the 23rd pair is ‘XY’.

Sex of the foetus:
If a sperm carrying ‘X’ chromosome fuses with the ‘X’ chromosome of female egg i.e. if ‘XX’ combination occurs, female will be produced. If a sperm carrying ‘Y’ chromosome fuses with the ‘X’ chromosome of female egg, i.e. if ‘XY’ combination occurs, male will be produced.

Evolution

Natural selection:
Natural selection is a central concept of evolution. In general terms it is also called ‘survival of the fittest’. Natural selection can be considered as the one ‘selected by nature’.

Genetic drift:
The sudden and random change in the gene frequency that occurs by chance in a small population is known as genetic drift.

Acquired and Inherited Traits

Evolution:

• Evolution is the sequence of gradual changes, which have taken place over millions of years in primitive plants and animals from which new species are formed.
• All the varieties of organisms, which we see around us, have evolved from some ancestors that lived on this earth long time ago.

Acquired traits:
Acquired trait means a trait or characteristic of an organism that it has not inherited but has developed in response to the environment. For example, if an organism starves for some time and reduces its weight, then it is called acquired trait.

Inherited trait:
A trait of an organism, which is caused by a change in its DNA, is known as inherited trait.

Variation:
Any difference that occurs between cells, individual organisms, or groups of organisms of any species either by genetic differences or by the effect of environmental factors on the expression of the genetic potential is called variation.

Acquired variation:
An acquired characteristic (variation) is a change which is acquired not due to heredity but due to response to the environmental factors.

Acquired variation may occur in the function or structure of an organism caused after birth due to reasons such as disease, injury, accident, repeated use or misuse, or other environmental factors.

Speciation:
The process of evolution of new species that occurs when members of similar population no longer interbreed to produce fertile off spring is known as speciation.

Reasons of speciation:
Geographical isolation, changes in DNA, change in number of chromosomes, gene cells of two isolated groups of populations, which cannot fuse with each other, etc.

Evolution and Classification

How are organisms classified?

• Classifying organisms by grouping them in certain way helps us to study them properly.
• One of the basic methods of classifying the organisms is on the basis of similarities they possess. A characteristic means a particular type of appearance (form) or behaviour (function).

Example:
We humans have four limbs (two hands and two legs) and it is our appearance (form characteristic. Plants perform photosynthesis is a behavioral (function) characteristic.

Tracing Evolutionary Relationships

Evidence for evolution:
Certain significant sources which provide proofs for evolution are called evidences for evolution. The main ones are:

Evidences through homologous organs: Those organs that have same internal structure but different functions are called homologous organs. For example, the basic design of internal structure of bones of forelimbs of a frog, lizard, bird, bat and man is same, even though these organs perform different functions.

Evidence through analogous organs: Those organs, which have different designs but similar appearance and carry out similar functions are called analogous organs. For example, wings of insects and birds have different structures but perform similar functions.

Evidence through fossils: The remains of dead organisms buried under the earth for millions of years are known as fossils. By studying these fossils, scientists learn how organisms evolved over time.

Evolution by Stages

Evolution of eyes:
The eyes of animals have been created in stages after many generations. First of all, eye was formed in planaria (flat-worm) which was in the form of a simple spot. Gradually, it became a complex organ.

Evolution of cabbage:
Wild cabbage is the ancestor of cabbage, broccoli, cauliflower, kohlrabi and kale varieties. These varieties have evolved from wild cabbage because farmers performed artificial selection on the wild cabbage and the varieties produced from it. Today, all the said varieties look different from their ancestor i.e. wild cabbage.

Evolution Should Not be Equated With ‘Progress’

Actually, no real ‘progress’ has taken place in the idea of evolution. The only progressive trend that is seen in evolution is that with time more and more complex body designs have emerged. This does not mean that the older designs were inefficient. Many older and simpler designs still survive.

Human Evolution: Research suggests that all the humans have evolved from a single species called Homo sapiens and initially belonged to Africa.

Haryana State Board HBSE 10th Class Science Notes Chapter 8 How do Organisms Reproduce Notes.

Haryana Board 10th Class Science Notes Chapter 8 How do Organisms Reproduce

Reproduction:

The production of new organisms from the existing organisms of same species is known as reproduction. Organisms do not need to undergo reproduction for maintaining their lives, even though they reproduce so that they may continue the existence of their species. This is the reason why reproduction is vital for the survival of the species.

Do Organisms Create Exact Copies of Themselves?

All organisms of a particular species look similar. For example, dogs look like dogs and a human looks like another. This happens because the designs of their bodies are similar. Hence, the first and foremost task of the reproduction process is to make ‘copies of blueprints of body design’. This is done by creating copies of DNA.

During cell division, a cell divides to give rise to two cells. Although, the two cells will be similar but will not be exactly identical. It is quite obvious that the process of copying the DNA will have some variations. Hence, the generated DNA copies will be similar, but may not be identical to the original.

In conclusion, the cells that survive after the cell division are

• Similar to each other as well as
• different i.e. show variation. This inbuilt tendency for variation during reproduction is the basis for evolution which is useful for the organism to adapt to ever changing environment.

The Importance of Variation

Niche:
A niche is the role and position a species plays in the environment, how the species meets its needs for food and shelter, how it survives and how it reproduces. Different species of organisms have different niches. For example, the way a lion gathers its food or shelter or reproduces are quite different from that of a crocodile i.e. the niches of two organisms vary.

Modes of Reproduction Used by Sinsle Oreanism

Asexual reproduction: The method of producing a new plant (or animal) without the act of fertilization of gametes is called asexual reproduction.

Types :

• Fission
• Fragmentation
• Regeneration
• Budding
• Vegetative propagation and
• Spore formation.

Fission: In fission process, a unicellular organism divides to form two or more new organisms.

Types :

• Binary fission and
• Multiple fission

Binary fission: In binary fission, the nucleus lengthens and then divides into two parts. After that the cytoplasm divides into two parts. This results in two daughter cells. Example: amoeba and leishmania.

Multiple fission: In multiple fission, the parent organism divides to form many new organisms at the sarrje time. Example: Malarial parasite plasmodium.

Fragmentation:
The process wherein the body of a multi-cellular organism breaks (fragments) into two or more pieces and on maturing each piece grows to form a complete new organism is known as fragmentation. Example: spirogyra.

Regeneration:
In some plants and animals, if some parts of their bodies get cut then such parts have the ability to regenerate and form a complete new organism. This method of reproduction is called regeneration. Example: Planaria.

Budding:
In budding, a small part of the body of the parent organism grows out as a ‘bud’. It then detaches from parent and becomes a new organism. Example: Hydra and planaria.

Vegetative propagation:
In vegetative propagation, new plants are reproduced from the plant parts such as roots, stem and leaves of old plants, without taking help of any reproductive organs. Example: bryophyllum.

Spore formation:
When the coat of spore bursts, the spores spread into air. These airborne spores settle on food and under favourable condition, germinate to produce new plants. Example: Production of spores in rhizopus on bread.

Sexual reproduction:

• The mode of reproduction that takes place with the help of both male sex cells and female sex cells is known as sexual reproduction.
• Flowering plants belong to the group of angiosperms. The reproductive parts of angiosperms are located in the flower. So, we can say that flower is the reproductive organ of the plant.

The four main reproductive parts of a flower are –

• Stamen
• Pistil
• Petal and
• Sepal

Unisexual flower: If the flower contains only one part out of stamen or pistil then such a flower is called unisexual. For example, papaya and watermelon.

Bisexual flower: If the flower contains both stamen as well as pistil the flower is called bisexual. For example, hibiscus and mustard.

Main processes of sexual reproduction:

• Pollination: The travel of the pollen from stamen to the stigma with the purpose of formation of zygote and hence a new plant is called pollination.
• Fertilization: The process of fusing of male gamete from the pollen grain and the female gamete in the ovary is called fertilization. Fertilization leads to formation of zygote.
• Seed germination: The seed contains the future plant or embryo which develops into a seedling under appropriate conditions. This process is known as germination.

Reproduction in Human Beings

Changes that human body experiences:

• The changes that occur in humans from infancy to childhood, to teenage and finally to fully grown matured body takes place quite slowly over several years. Moreover, neither do all these changes happen together at the same time in one person, nor do they happen at an exact age.
• In some people, the changes happen early and quickly, while in others, they can happen slowly.
• Each change takes its own time to get completed.

Puberty:
The age at which a body starts producing gametes and girls and boys become biologically capable to reproduce is known as puberty. Generally, girls attain puberty at the age of 12 years, while boys reach puberty at the age of 13 to 14 years.

Male Reproductive System

Main reproductive organs: Ureter, Seminal vesicle, Vas deferens, Scrotum, Testes, Bladder, Penis and Urethra

Female Reproductive System

Main reproductive organs:
A pair of ovaries, oviduct, fallopian tube, uterus and vagina. The female reproductive system is more complicated than male because it takes care of fertilization and the development of embryo till the birth.

Reproductive Health

Sexually transmitted diseases:
The infectious diseases which are spread from an infected person to a healthy person through sexual contact are called sexually transmitted diseases (STDs). These disease can be caused either by bacteria or by virus. Example: Syphilis, gonorrhoea and AIDS.

What is contraception? List out the methods of contraception and explain them.
Contraception:
The method to prevent pregnancy in women is called contraception. Methods adopted to prevent pregnancy are called contraceptive methods.

Methods:
1. Birth control tools: Under this method, a mechanical barrier is created which prevents the entry of sperm into the genital tract. As a result, fertilization does not occur.

• The tools include condoms, a diaphragm and Copper-T.

2. Birth control pills: There are oral pills that a women can take. These pills change the hormonal balance of the body so that the eggs are not released by the ovaries and fertilization is prevented.

3. Surgical methods: These include vasectomy to prevent the sperms from entering urethra and tubectomy to prevent It will not allow the sperm to reach the uterus.

4. Abortion: Another method is to surgically remove the foetus from the body of pregnant women.

Haryana State Board HBSE 10th Class Science Notes Chapter 7 Control and Coordination Notes.

Haryana Board 10th Class Science Notes Chapter 7 Control and Coordination

Animals – Nervous System

Importance of control and co-ordination:
Multicellular organisms are made-up of various organs and organ system. It is a basic need of the body to control and co-ordinate these organs and organ systems, so that they can function properly and accomplish the voluntary as well as involuntary functions.

Stimulus:

• An event that encourages action or creates sensation is called a stimulus. (Plural : Stimuli)
• All living organisms – humans, plants and animals respond to changes occurring in their surroundings. These changes work as stimuli.

Example: On seeing sudden bright sunlight, our eyes gets closed. Here, the sudden bright light is stimulus whereas closing of eyes is response.

Receptors:
A specialized structure in the human body that receives external stimuli is called a receptor. These receptors are located in our sense organs.

The main ones are:

• Gustatory receptors
• Olfactory receptors
• Thigmo receptors
• Photo receptors and
• Audio (phono) receptors

Nerve cell:
Nerve cells or neurons are the building as well as functional units of the nervous system. They carry information from one part of the body to another.

Nerve cell has three components:

• Cell body
• Dendrites and
• Axon

Nervous tissue:
Nervous tissue is made up of organized network of nerve cells or say neurons. It specializes in conducting messages through electrical impulses from one part of the body to another.

What Happens in Reflex Action

Reflex action:

• Reflex action is an involuntary and instant response of the muscles or glands to a stimulus.
• It takes place without involving the brain.

Example: We immediately pull-back our hands when we suddenly touch a hot vessel or when someone pricks us a pin.

Reflex arc:
The pathway or say the route taken by the nerve impulses in a reflex action is known as the reflex : arc. Reflex arcs allow rapid response.

Human Brain 

There are two systems which co-ordinate different activities in humans:

• Nervous system and
• Endocrine system (or) Hormonal system)

Nervous system:
It controls and co-ordinates all the parts of  the body. The nervous system co-ordinates  muscles, thus allowing a person to perform  activities such as dancing, reading, writing, etc.  It also co-ordinates certain involuntary functions  like heart beat and breathing.

Endocrine system (or Hormonal system):
There arexiertain glands in the endocrine system  which release chemical substances (chemical  messengers) called hormones in the body. Generally, hormones regulate the slow activities of the body such as growth, metabolism, etc.

Organization of human nervous system can be done under three classes. They are:
(A) Organs of Human Nervous System:

• Brain
• Spinal cord
• Nerves

Human Nervous System :
CNS (Central Nervous System) –

• Brain
• Spinal cord

PNS (Peripheral Nervous System) –

• 12 pairs of Cranial nerves
• 31 pairs of spinal nerves

Nerves:

• Sensory nerve
• Motor nerve
• Mixed nerve

Human brain:  It is divided into three major parts-

• Fore brain
• Mid brain and
• Hind brain.

Fore brain: Fore brain is the main thinking part of the brain. It has regions which receive sensory impulses from various receptors.
Midbrain: Mid brain connects the forebrain and hind brain. It is the centre for visual and auditory reflexes.
Hind brain: It consists of

• Cerebellum which lies on dorsal side and
• Pons and
• Medulla oblongata.

Co-ordination in Plants:

Plant movement:  Movement of individual parts of a plant such as root, leaves, etc. is possible when they are subjected to some external stimuli like sunlight, gravitational force, water, touch, etc. Such movements of plants are called plant movements. Types: (I) Tropism and (II) Nastism.

Movement due to Growth:

The movement (response) in a plant organ is due to the effect of an external and directional stimuli is called tropism or tropic movement. Stimulus and type of movement shown by the plants:

• Light – Photoperiodism
• Gravity – Geotropism
• Chemical – Chemotropism
• Water – Hydrotropism
• Touch – Thigmotropism

Plant hormone:
Plant hormones are special chemical compounds which are synthesized at one place/organ of the plants and migrate to the target organ to act. They play an important role in control and co-ordination in plants, as well as in growth, development and responses to the environment.

Types of plant hormones:

• Growth promoting hormones – (i) Auxins, (ii) Cytokinins, (iii) Gibberellins
• Growth resisting hormones – (i) Absicic acid (ABA)

Hormones in Animals:

Hormones:

• Basically, hormones are the chemical substances (chemical messengers).
• They play an important role in various metabolic processes.

Glands: There are two types of glands. They are –

• Exocrine glands and
• Endocrine glands

Glands and hormones they release:

• Adrenal – Adrenaline
• Thyroid – Thyroxine,
• Pituitary – Growth hormone and many other hormones
• Testes (in males) – Testosterone
• Ovaries (in females) – Estrogen and progesterone
• Pancreas – Insulin

Regulation of secretion of hormones:
Hormone secretion is regulated by the feedback mechanism. In general, a particular gland secretes a specific hormone which reaches to the target organ and performs its function. After the function is over, the brain receives a message to stop the gland to secrete the hormone further. This is how hormone secretion is regulated.