NCERT Solutions for Class 8 Maths Chapter 12 - In Hindi

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.

NCERT Solutions for Class 8 Maths Chapter 12 in Hindi medium PDF download are easily available on our official website (vedantu.com). Upon visiting the website, you have to register on the website with your phone number and email address. Then you will be able to download all the study materials of your preference in a click. You can also download the Class 8 Maths Exponents and Powers solution Hindi medium from Vedantu app as well by following the similar procedures, but you have to download the app from Google play store before doing that.

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.

These solutions are nothing but a compilation of all the answers to the questions of the textbook exercises. The answers/solutions are given in a stepwise format and very well researched by the subject matter experts who have relevant experience in this field. Relevant diagrams, graphs, illustrations are provided along with the answers wherever required. In nutshell, NCERT Solutions for Class 8 Maths in Hindi come really handy in exam preparation and quick revision as well prior to the final examinations.