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NCERT Solutions for Class 8 Maths Chapter 12 - In Hindi

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NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers in Hindi PDF Download

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We, at Vedantu, offer free NCERT Solutions in English medium and Hindi medium for all the classes as well. Created by subject matter experts, these NCERT Solutions in Hindi are very helpful to the students of all classes.


Class:

NCERT Solutions For Class 8

Subject:

Class 8 Maths in Hindi

Chapter Name:

Chapter 12 - Exponents and Powers

Content Type:

Text, Videos, Images and PDF Format

Academic Year:

2024-25

Medium:

English and Hindi

Available Materials:

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Other Materials

  • Important Questions

  • Revision Notes

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Access NCERT Solutions for Class 8 maths Chapter 12 – घातांक और घात

प्रश्नावली 12.1

1. मान ज्ञात कीजिए:

(i) $3^{-2}$

उत्तर $3^{-2}=\left(\dfrac{1}{3^{2}}\right)=\left(\dfrac{1}{9}\right)$


(ii) $(-4)^{-2}$

उत्तर: $\quad(-4)^{-2}=\left(\dfrac{1}{-4^{2}}\right)=\left(\dfrac{1}{16}\right)$


(iii) $\left(\dfrac{1}{2}\right)^{-5}$

उत्तर: 

$\quad\left(\dfrac{1}{2}\right)^{-5}=\left\{\dfrac{1^{-5}}{2^{-5}}\right\}=\left\{\begin{array}{l}\left.{\dfrac{1}{\dfrac{1}{5}}}{\dfrac{1}{2^{5}}}\right\}=\left\{\dfrac{\dfrac{1}{1}}{\dfrac{1}{32}}\right\}=\dfrac{32}{1}=32\end{array}\right.$


2. सरल कीजिए एवं उत्तर को धनात्मक घातांक के रूप में व्यक्त कीजिए :

(i)$\quad(-4)^{5} \div(-4)^{8}$

Ans: $\quad(-4)^{5} \div(-4)^{8}=(-4)^{5-8}=(-4)^{-3}=\left(\dfrac{1}{-4^{3}}\right)$


(ii) $\left(\dfrac{1}{2^{3}}\right)^{2}$

Ans: $\left(\dfrac{1}{2^{3}}\right)^{2}=\left(2^{-3}\right)^{2}=2^{-3 \times 2}=2^{-6}$


(iii) $\quad(-3)^{4} \times\left(\dfrac{5}{3}\right)^{4}$

Ans: $\quad(-3)^{4} \times\left(\dfrac{5}{3}\right)^{4}=\left[-3 \times\left(\dfrac{5}{3}\right)\right]^{4}=[-5]^{4}$


(iv) $\left(3^{-7} \div 3^{-10}\right) \times 3^{-5}$

Ans: $\quad\left(3^{-7} \div 3^{-10}\right) \times 3^{-5}=\left[3^{-7+10} \times 3^{-5}\right]=\left[3^{3} \times 3^{-5}\right]=$ $3^{[3+(-5)]}=3^{-2}$


(v) $\quad 2^{-3} \times(-7)^{-3}$

Ans: $\quad 2^{-3} \times(-7)^{-3}=[2 \times(-7)]^{-3}=[-14]^{-3}=\left[\dfrac{1}{(-14)^{3}}\right]$


3. मान ज्ञात कीजिए :

(i) $\left(3^{0}+4^{-1}\right) \times 2^{2}$

उत्तर: (i) $\left(3^{0}+4^{-1}\right) \times 2^{2}=\left(1+\dfrac{1}{4^{1}}\right) \times 4=\left(1+\dfrac{1}{4}\right) \times 4=\left(\dfrac{5}{4}\right) \times 4=5$


(ii) $\quad\left(2^{-1} \times 4^{-1}\right) \div 2^{-2}$

उत्तर:(ii)$\left(2^{-1} \times 4^{-1}\right) \div 2^{-2}=(2 \times 4)^{-1} \div 4^{-1}$ $\left\{\left(\dfrac{1}{8^{1}}\right) \div\left(\dfrac{1}{2^{2}}\right)\right\}=\left\{\left(\dfrac{1}{8}\right) \div\left(\dfrac{1}{4}\right)\right\}=\dfrac{4}{8}=\dfrac{1}{2}$


(iii) $\left(\dfrac{1}{2}\right)^{-2}+\left(\dfrac{1}{3}\right)^{-2}+\left(\dfrac{1}{4}\right)^{-2}$

उत्तर:                                                                                                                                         (iii)$\left(\dfrac{1}{2}\right)^{-2}+\left(\dfrac{1}{3}\right)^{-2}+\left(\dfrac{1}{4}\right)^{-2}=\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)^{-2}=\left(\dfrac{6}{12}+\dfrac{4}{12}+\right.$ $\left.\dfrac{3}{12}\right)^{-2}=\left(\dfrac{13}{12}\right)^{-2}=\left(\dfrac{13^{-2}}{12^{-2}}\right)=\dfrac{\dfrac{1}{13^{2}}}{\dfrac{1}{12^{2}}}=\left\{\dfrac{1}{\dfrac{169}{144}}\right\}=\dfrac{144}{169}$ 


(iv) $\quad\left(3^{-1}+4^{-1}+5^{-1}\right)^{0}$

उत्तर: (iv) $\quad\left(3^{-1}+4^{-1}+5^{-1}\right)^{0}=1 \quad\left[a^{0}=1\right]$


(v) $\left\{\left(-\dfrac{2}{3}\right)^{-2}\right\}^{2}$

उत्तर: (v)$\left\{\left(-\dfrac{2}{3}\right)^{-2}\right\}^{2}=\left(-\dfrac{2}{3}\right)^{-2 \times 2}=\left(-\dfrac{2}{3}\right)^{-4}=\left(-\dfrac{2^{-4}}{3^{-4}}\right)=\left\{-\dfrac{\dfrac{1}{2^{4}}}{\dfrac{1}{3^{4}}}\right\}=$ $\left(-\dfrac{\dfrac{1}{16}}{\dfrac{1}{81}}\right)=\left(-\dfrac{81}{16}\right)$


4. मान ज्ञात कीजिए:

(i) $\dfrac{8^{-1} \times 5^{3}}{2^{-4}}$

उत्तर: (i) $\quad \dfrac{8^{-1} \times 5^{3}}{2^{-4}}=\dfrac{\left\{\left(\dfrac{1}{8^{1}}\right) \times 125\right\}}{\dfrac{1}{2^{4}}}=\dfrac{\dfrac{1}{8} \times 125}{\dfrac{1}{16}}=\dfrac{16 \times 125}{8}=2 \times 125=250$


(ii) $\left(5^{-1} \times 2^{-1}\right) \times 6^{-1}$

उत्तर:(ii) $\left(5^{-1} \times 2^{-1}\right) \times 6^{-1}-\left(5^{-1} \times 2^{-1} \times 6^{-1}\right)-(5 \times 2 \times 6)^{-1}$ $60^{-1}=\dfrac{1}{60}$


5. $m$ का मान ज्ञात कीजिए जिसके लिए $5^{m} \div 5^{-3}=5^{5}$

उत्तर: $5^{m} \div 5^{-3}=5^{5}$

तथा, $5^{m+3}=5^{5}$

,तथा $m+3=5$

तथा, $m=5-3=2$

तब, $m=2$ [proved]


6. मान ज्ञात कीजिए:

(i) $\left\{\left(\dfrac{1}{3}\right)^{-1}-\left(\dfrac{1}{4}\right)^{-1}\right\}^{-4}$

उत्तर:(i)$\left\{\left(\dfrac{1}{3}\right)^{-1}-\left(\dfrac{1}{4}\right)^{-1}\right\}^{-4}=\left\{\left(\dfrac{1}{3}-\dfrac{1}{4}\right)^{-1}\right\}^{-4}=\left\{\left(\dfrac{4-3}{12}\right)^{-1}\right\}^{-4}=$

$\left(\dfrac{1}{12}\right)^{\{(-1) \times(-4)\}}=\left(\dfrac{1}{12}\right)^{4}=\left(\dfrac{1^{4}}{12^{4}}\right)=\dfrac{1}{20746}$


(ii) $\left(\dfrac{5}{8}\right)^{-7} \times\left(\dfrac{8}{5}\right)^{-4}$

उत्तर: (ii) $\left(\dfrac{5}{8}\right)^{-7} \times\left(\dfrac{8}{5}\right)^{-4}=\dfrac{\dfrac{1}{5^{7}}}{\dfrac{1}{8^{7}}} \times \dfrac{\dfrac{1}{8^{4}}}{\dfrac{1}{5^{4}}}=\dfrac{8^{7}}{5^{7}} \times \dfrac{5^{4}}{8^{4}}=\dfrac{8^{7}}{8^{4}} \times \dfrac{5^{4}}{5^{7}}=8^{3} \times 5^{-3}=$ $512 \times \dfrac{1}{5^{3}}=512 \times \dfrac{1}{125}=4.096$


7. सरल कीजिए :

(i) $\quad \dfrac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}}(t \neq 0)$

उत्तर: (i) $\dfrac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}}=\dfrac{25 \times \dfrac{1}{t^{4}}}{\dfrac{1}{5^{3}} \times 10 \times \dfrac{1}{t^{8}}}=\dfrac{25 \times 5^{3} \times t^{8}}{\left(10 \times t^{4}\right)}=\dfrac{25 \times 125 \times t^{4}}{10}=\dfrac{3125 \times t^{4}}{10}$


(ii) $\dfrac{3^{-5} \times 10^{-5} \times 125}{5^{-7} \times 6^{-5}}$

उत्तर: (ii) $5^{2}=5^{3} \times 5^{2}=5^{5^{3}+2^{6^{5}}}=5^{5}$


प्रश्नावली 12.2

1. निम्नलखित संख्याओं को मानक रूप में व्यक्त कीजिए।

(i) $0.000000000085$

उत्तर: $0.000000000085=8.5 \times 10^{-11}$


(ii) $0.0000000000942$

उत्तर: $0 .0000000000942=9.42 \times 10^{-11}$


(iii) 6020000000000000

उत्तर: $6.02 \times 10^{15}$


(iv) $0.00000000837$

उत्तर: $0.00000000837=8.37 \times 10^{-9}$


(v) 31860000000

उत्तर: $31860000000=3.186 \times 10^{10}$


2. निम्नलिखत संख्याओं को सामान्य रूप में व्यक्त कीजिए।

(i) $\quad 3.02 \times 10^{-6}$

उत्तर: $3.02 \times 10^{-6}=0.00000302$


(ii) $\quad 4.5 \times 10^{4}$

उत्तर: $4 .5 \times 10^{4}=45000$


(iii) $3 \times 10-8$

उत्तर: $3 \times 10^{-8}=0.00000003$


(iv) $\quad 1.0001 \times 10^{9}$

उत्तर: $ \times 10^{9}=1000100000$


(v) $\quad 5.8 \times 10^{12}$

उत्तर: $5 .8 \times 10^{12}=5800000000000$


(vi) $\quad 3.61492 \times 10^{6}$

उत्तर: $3 .61492 \times 10^{6}=3614920$


3. निम्लिखित कथनों में को संख्या प्रकट हो रही है, उन्हें मानक रूप में प्रकट कीजिए।

(i) 1 माइक्रोन $1 / 1000000 m$ के बराबर होता है।

उत्तर:  $1 / 1000000=1 \times 10^{-6}$


(ii) एक इलेक्ट्रॉन आवेश $0.000,000,000,000,000,00016$ कुलंब होता है।

उत्तर:   $\times 10-{ }^{19}$


(iii) जीवाणु की माप $0.0000005 m$ है।

उत्तर:  $0 .0000005=5 \times 10^{-7}$


(iv) पौधों की कोशिकाओं की माप $0.00001275 m$ है।

उत्तर:   $0.00001275=1.275 \times 10^{-5}$


(v) मोटे कागज की मोटाई $0.07 mm$ है।

उत्तर:  $0.07=7 \times 10^{-2}$


4. एक ढेर में पांच किताबें है, जिनमें प्रत्येक की मोटाई $20 mm$ तथा पांच कागज की पत्रक

है। जिनमें प्रत्येक की मोटाई $0.016 mm$ है। इस ढेर की कुल मोटाई ज्ञात कीजिए।

उत्तर:  दिया गया 

एक पुस्तक की मोटाई $=20 mm$

5 पुस्तकों की मोटाई $=(5 \times 20) mm =100 mm$

एक कागज की मोटाई = $0.016 mm$

इसलिए 5 कागजों की मोटाई = $0.016 \times 5=0.080 mm$

एक ढेर की कुल मोटाई

$=5$ पुस्तकों की मोटाई $+5$ कागजों की मोटाई

$=(100+0.08) mm$

$=100.08 mm$

$=1.0008 \times 10^{2} mm$

ढेर की कुल मोटाई$=1.0008 \times 10^{2} mm$


NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers in Hindi

Chapter-wise NCERT Solutions are provided everywhere on the internet with an aim to help the students to gain a comprehensive understanding. Class 8 Maths Chapter 12 solution Hindi medium is created by our in-house experts keeping the understanding ability of all types of candidates in mind. NCERT textbooks and solutions are built to give a strong foundation to every concept. These NCERT Solutions for Class 8 Maths Chapter 12 in Hindi ensure a smooth understanding of all the concepts including the advanced concepts covered in the textbook.


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NCERT Solutions in Hindi medium have been created keeping those students in mind who are studying in a Hindi medium school. These NCERT Solutions for Class 8 Maths Exponents and Powers in Hindi medium pdf download have innumerable benefits as these are created in simple and easy-to-understand language. The best feature of these solutions is a free download option. Students of Class 8 can download these solutions at any time as per their convenience for self-study purpose.


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