An Overview of Ncert Solutions Class 8 Maths Chapter 12 In Hindi
In Ncert Solutions Class 8 Maths Chapter 12 In Hindi, you’ll learn all about exponents and powers—those special numbers that look like 23 or 5-2. This chapter shows you how to deal with big and small numbers easily using exponents, plus lots of tricks that make solving sums faster.
If you ever wondered how to compare tiny decimals or handle huge numbers with zeros, this is the chapter for you! The step-by-step NCERT Solutions from Vedantu, available in downloadable PDF format, help clear up doubts and build your confidence for every question. You can get the complete Class 8 Maths syllabus here.
These handy NCERT Solutions not only make your homework and exam prep easier, but also help you understand each concept in simple language. For more exercise solutions, find your Class 8 Maths NCERT Solutions at Vedantu.
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
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