Ncert Exercises & Solutions

2.2 Fill in the blanks by suitable conversion of units.

(a) 1 kg m² s^–2 = ....g cm²s^–2

(b) 1 m = ..... ly

(d) G = 6.67 × $10^{- 11}$ N m² (kg)^–2 = .... (cm)³ s^–2 g^–1.

(a) 1 kg = 10^{3} g

1 m^{2} = 10^{4} {cm}^{2}

1 kg m^{2} s^{- 2} = 1 kg \times 1 m^{2} \times 1 s^{- 2}

= 10^{3} g \times 10^{4} {cm}^{2} \times 1 s^{- 2} = 10^{7} g {cm}^{2} s^{- 2}

(b) A light-year is the total distance travelled by light in one year.

$\begin{array}{rcl} 1 ly & = & Speed of light \times One year \\ = & (3 \times 10^{8} m / s) \times (365 \times 24 \times 60 \times 60 s) \\ = & 9.46 \times 10^{15} m \\ \Rightarrow 1 m & = & \frac{1}{9.46 \times 10^{15}} = 1.057 \times 10^{- 16} ly \\ (c) 1 m & = & 10^{- 3} km \\ Again, 1 s & = & \frac{1}{3600} h \\ 1 s^{- 1} & = & 3600 h^{- 1} \\ 1 s^{- 2} & = & {(3600)}^{2} h^{- 2} \\ 3 m s^{- 2} & = & (3 \times 10^{- 3} km) \times {(3600)}^{2} h^{- 2} = 3.88 \times 10^{4} km h^{- 2} \\ (d) 1 N & = & 1 kg m s^{- 2} \\ 1 kg & = & 10^{3} g \\ 1 m^{3} & = & 10^{6} {cm}^{3} \\ So, 6.67 \times 10^{- 11} N m^{2} {kg}^{- 2} & = & 6.67 \times 10^{- 11} \times (1 kg m s^{- 2}) (1 m^{2}) (1 {kg}^{- 2}) \\ = & 6.67 \times 10^{- 11} \times (1 {kg}^{- 1} \times 1 m^{3} \times 1 s^{- 2}) \\ = & 6.67 \times 10^{- 11} \times (10^{- 3} g) \times (10^{6} {cm}^{3}) \times (1 s^{- 2}) \\ = & 6.67 \times 10^{- 8} {cm}^{3} s^{- 2} g^{- 1} \end{array}$

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