5 Marks Questions

Question 15 Marks

Write all the facts given in the standard

Answer

All the facts in the standard form are as follows.
1. The distance from the Earth to the Sun is 149600000000 m i.e. $1.496 \times 10^{11} m$.
2. The speed of light is $300000000 m / sec$ i.e. $3 \times 10^8 m / sec$.
3. Thickness of Class VIII Mathematics book is 20 mm i.e. $2 \times 10^1 mm$
4. The average diameter of a red blood cell is $0.000007=\frac{7}{1000000}=7 \times 10^{-6} mm$
5. The thickness of human hair is in the range of 0.005 cm i.e. $5 \times 10^{-3} cm$
6. The distance of Moon from the Earth is 384467000 m i.e. $3.84467 \times 10^8 m$.
7. The size of plant cell is 0.00001275 m i.e. $1.275 \times 10^{-5} m$.
8. Average radius of the Sun is 695000 km ie. $695 \times 10^5 km$.
9. Mass of the propellant in a space shuttle solid rocket booster is 503600 kg i.e. $5.036 \times 10^5 kg$.
10. Thickness of a piece of paper is 0.0016 cm i.e. $1.6 \times 10^{-3} cm$.
11. Diameter of a wire in a computer chip is 0.000003 mi.e. $3 \times 10^{-6} m$.
12. The height of Mount Everest is 8848 m i.e. $8.848 \times 10^3 m$.

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Question 25 Marks

Express the number appearing in the following statements in standard form.
(i) 1 micron is equal to $\frac{1}{1000000} m$.
(ii) Charge of an electron is 0.00000000000000000016 coulomb.
(iii) Size of a bacteria is 0.0000005 m .
(iv) Size of a plant cell is 0.00001275 m .
(v) Thickness of a thick paper is 0.07 mm .

Answer

(i) Here, $\frac{1}{1000000}=\frac{1}{10^6}$
$
=1 \times 10^{-6} m \quad\left[\because \frac{1}{a^m}=a^{-m}\right]
$
Hence, 1 micron $=1 \times 10^{-6} m$.
(ii) Here, 0.00000000000000000016 coulomb.
$
\begin{array}{l}
=\frac{16}{100000000000000000000} \\
=\frac{16}{10^{20}}=\frac{16 \times 10}{10^{20}}
\end{array}
$
$\begin{array}{l}=16 \times 10 \times 10^{-20} \quad\left[\because \frac{1}{a^m}=a^{-m}\right] \\ =16 \times 10^{1-20} \\ =16 \times 10^{-19} \text { coulomb }\left[\because a^m \times a^n=a^{m+n}\right]\end{array}$
Hence, the charge of an electron is $16 \times 10^{-19}$ coulomb.
(iii)
$
\begin{aligned}
\text { Here, } 0.0000005 & =\frac{5}{10000000} \\
& =\frac{5}{10^7}=5 \times 10^{-7} m \quad\left[\because \frac{1}{a^m}=a^{-m}\right]
\end{aligned}
$
Hence, the size of a bacteria is $5 \times 10^{-7} m$.
(iv)
$\begin{aligned} \text { Here, } 0.00001275 & =\frac{1275}{100000000} \\ & =\frac{1275}{10^8}=\frac{1.275 \times 1000}{10^8}\end{aligned}$
$\begin{array}{l}=1.275 \times 10^3 \times 10^{-6} \quad\left[\because \frac{1}{a^m}=a^{-m}\right] \\ =1.275 \times 10^{3-8} \\ =1.275 \times 10^{-5} m\left[\because a^m \times a^n=a^{m+n}\right]\end{array}$
Hence, the size of a plant cell is $1.275 \times 10^{-5} m$.
(v) Here, $0.07=\frac{7}{100}=\frac{7}{10^2}=7 \times 10^{-2} mm$
Hence, the thickness of a thick paper is $7 \times 10^{-2} mm$.

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Question 35 Marks

Express the following numbers in usual form.
(i) $3.02 \times 10^{-6}$
(ii) $4.5 \times 10^4$
(iii) $3 \times 10^{-8}$
(iv) $1.0001 \times 10^9$
(v) $5.8 \times 10^{12}$
(vi) $3.61492 \times 10^6$

Answer

(i) We have,
$
\begin{array}{rlr}
3.02 \times 10^{-6} & =\frac{3.02}{10^6}=\frac{3.02}{1000000} & \\
& =0.00000302 & {\left[\because a^{-m}=\frac{1}{a^m}\right]}
\end{array}
$
(ii) We have,
$
4.5 \times 10^4=4.5 \times 10000=45000
$
(iii) We have,
$
3 \times 10^{-8}=\frac{3}{10^8}=\frac{3}{100000000}
$
$=0.00000003 \quad\left[\because a^{-m}=\frac{1}{a^m}\right]$
(iv) We have,
$
1.0001 \times 10^9=1.0001 \times 1000000000=1000100000
$
(v) We have,
$
\begin{aligned}
5.8 \times 10^{12} & =5.8 \times 1000000000000 \\
& =5800000000000
\end{aligned}
$
(vi) We have,
$
\begin{aligned}
3.61492 \times 10^6 & =3.61492 \times 1000000 \\
& =3614920
\end{aligned}
$

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Question 45 Marks

Express the following numbers in standard form.
(i) 0.0000000000085
(ii) 0.00000000000942
(iii) 6020000000000000
(iv) 0.00000000837
(v) 31860000000

Answer

(i) We have, 0.0000000000085
$\begin{array}{ll}
=\frac{85}{10^{13}}=\frac{8.5 \times 10}{10^{13}} & \\
=8.5 \times 10 \times 10^{-13} & {\left[\because \frac{1}{a^m}=a^{-m}\right]} \\
=8.5 \times 10^{(1-13)} & \\
=8.5 \times 10^{-12} & {\left[\because a^m \times a^n=a^{m+m}\right]}
\end{array}
$
which is the required standard form.
(ii) Ans. $9.42 \times 10^{-12}$
(iii) We have, $6020000000000000=6.02 \times 10^{15}$ which is the required standard form.
(iv)Ans. $8.37 \times 10^{-9}$
(v) Ans. $3.186 \times 10^{10}$

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Question 55 Marks

Find the value of
(i) $\left(3^0+4^{-1}\right) \times 2^2$
(ii) $\left(2^{-1} \times 4^{-1}\right)+2^{-2}$
(iii) $\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2}$
(iv) $\left(3^{-1}+4^{-1}+5^{-1}\right)^0$
(v) $\left\{\left(\frac{-2}{3}\right)^{-2}\right\}^2$

Answer

(i) We have, $\left(3^0+4^{-1}\right) \times 2^2$
$
\begin{array}{l}
=\left(1+\frac{1}{4}\right) \times 4 \quad\left[\because a^0=1 \text { and } a^{-m}=\frac{1}{a^m}\right] \\
=\left(\frac{4+1}{4}\right) \times 4=\frac{5}{4} \times 4=5
\end{array}
$
(ii) We have, $\left(2^{-1} \times 4^{-1}\right)+2^{-2}$
$=\left(\frac{1}{2} \times \frac{1}{4}\right)+2^{-2} \quad\left[\because a^{-m}=\frac{1}{a^m}\right]$
$=\left(\frac{1}{8}\right)+\left(\frac{1}{2^2}\right)=\frac{1}{8}+\frac{1}{4}$
$=\frac{1}{8} \times \frac{4}{1}=\frac{1}{2} \quad\left[\because \frac{a}{b}+\frac{c}{d}=\frac{a}{b} \times \frac{d}{c}\right]$
(iii) Ans. 29
(iv) Ans. 1
(v)Ans. $\frac{81}{16}$

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Question 65 Marks

Simplify and express the result in power notation with positive exponent.
(i) $(-4)^5+(-4)^8$ $\qquad$ (ii) $\left(\frac{1}{2^3}\right)^2$ $\qquad$
(iii) $(-3)^4 \times\left(\frac{5}{3}\right)^4$ $\qquad$ (iv) $\left(3^{-7}+3^{-10}\right) \times 3^{-5}$ $\qquad$
(v) $2^{-3} \times(-7)^{-3}$

Answer

(i) We have, $(-4)^5+(-4)^8$
$
\begin{array}{l}
=\frac{(-4)^5}{(-4)^8}=\frac{1}{(-4)^8 \times(-4)^{-5}} \quad\left[\because a^m=\frac{1}{a^{-m}}\right] \\
=\frac{1}{(-4)^{8-5}}=\frac{1}{(-4)^3} \quad\left[\because a^m \times a^n=a^{m+n}\right]
\end{array}
$
which is the required form.
(ii) We have, $\left(\frac{1}{2^3}\right)^2$
$
\begin{array}{l}
=\frac{(1)^2}{\left(2^3\right)^2} \qquad{\left[\because\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\right]} \\=\frac{1}{2^6}\qquad{\left[\because\left(a^m\right)^n=a^{m \times n}\right]}
\end{array}
$
which is the required form.
(iii) We have, $(-3)^4 \times\left(\frac{5}{3}\right)^4$
$=(-1 \times 3)^4 \times\left(\frac{5}{3}\right)^4 \qquad[\because-a=-1 \times a]$
$=(-1)^4 \times 3^4 \times \frac{5^4}{3^4}$
$\left[\because(a \times b)^m=a^m \times b^m,\left(\frac{a}{b}\right)^m=\frac{a^m}{b^m}\right]$
$=1 \times 5^4=(5)^4 \qquad\left[\because(-1)^4=1\right]$
which is the required form.
(iv) Ans. $\frac{1}{3^2}$
(v) Ans. $\left(\frac{-1}{14}\right)^3$

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