Fill in the blanks by suitable conversion of units:

solution:

(a) $1 \mathrm{~kg}=10^{3} \mathrm{~g}$

$1 \mathrm{~m}^{2}=10^{4} \mathrm{~cm}^{2}$

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

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

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

$1 \mathrm{ly}=$ Speed of light $\times$ One year

$=\left(3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right) \times(365 \times 24 \times 60 \times 60 \mathrm{~s})$

$=9.46 \times 10^{15} \mathrm{~m}$

$\therefore 1 \mathrm{~m}=\frac{1}{9.46 \times 10^{15}}=1.057 \times 10^{-16} \mathrm{ly}$

(c) $1 \mathrm{~m}=10^{-3} \mathrm{~km}$

Again, $1 \mathrm{~s}=\frac{1}{3600} \mathrm{~h}$

$1 s^{-1}=3600 h^{-1}$

$1 s^{-2}=(3600)^{2} h^{-2}$

$\therefore 3 \mathrm{~m} \mathrm{~s}^{-2}=\left(3 \times 10^{-3} \mathrm{~km}\right) \times\left((3600)^{2} \mathrm{~h}^{-2}\right)=3.88 \times 10^{4} \mathrm{~km} \mathrm{~h}^{-2}$

(d) $1 \mathrm{~N}=1 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-2}$

$1 \mathrm{~kg}=10^{-3} \mathrm{~g}^{-1}$

$1 \mathrm{~m}^{3}=10^{6} \mathrm{~cm}^{3}$

$\therefore 6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}=6.67 \times 10^{-11} \times\left(1 \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-2}\right)\left(1 \mathrm{~m}^{2}\right)\left(1 \mathrm{~s}^{-2}\right)$

$=6.67 \times 10^{-11} \times\left(1 \mathrm{~kg} \times 1 \mathrm{~m}^{3} \times 1 \mathrm{~s}^{-2}\right)$

$=6.67 \times 10^{-11} \times\left(10^{-3} \mathrm{~g}^{-1}\right) \times\left(10^{6} \mathrm{~cm}^{3}\right) \times\left(1 \mathrm{~s}^{-2}\right)$

$=6.67 \times 10^{-8} \mathrm{~cm}^{3} \mathrm{~s}^{-2} \mathrm{~g}^{-1}$