Q142 / 155 · WBJEE 2017 · Physics
The dimension of the universal constant of gravitation, $G$ is
- A$\left[\mathrm{ML}^{2} \mathrm{T}^{-1}\right]$
- B$\left[\mathrm{M}^{-1} \mathrm{L}^{3} \mathrm{T}^{-2}\right]$
- Correct answerC$\left[\mathrm{M}^{-1} \mathrm{L}^{2} \mathrm{T}^{-2}\right]$
- D$\left[\mathrm{ML}^{3} \mathrm{T}^{-2}\right]$
Solution
The dimensions of universal constant $G$, is obtained by Newton's law of gravitation
$F=\frac{G m_{1} m_{2}}{r^{2}}$
$\begin{array}{l}<br />G=\frac{F \times r^{2}}{m_{1} m_{2}} \\<br />F=\text { Force }<br />\end{array}$
$r=$ Distance $m_{1}$ and $m_{2}$ are masses
$\begin{aligned}<br />G &=\frac{\left[M^{1} L^{1} T^{-2}\right]\left[L^{2}\right]}{[M][M]} \\<br />&=\frac{\left[M^{\prime} L^{2} T^{-2}\right]}{\left[M^{2}\right]}=\left[M^{-1} L^{2} T^{-2}\right]<br />\end{aligned}$
$F=\frac{G m_{1} m_{2}}{r^{2}}$
$\begin{array}{l}<br />G=\frac{F \times r^{2}}{m_{1} m_{2}} \\<br />F=\text { Force }<br />\end{array}$
$r=$ Distance $m_{1}$ and $m_{2}$ are masses
$\begin{aligned}<br />G &=\frac{\left[M^{1} L^{1} T^{-2}\right]\left[L^{2}\right]}{[M][M]} \\<br />&=\frac{\left[M^{\prime} L^{2} T^{-2}\right]}{\left[M^{2}\right]}=\left[M^{-1} L^{2} T^{-2}\right]<br />\end{aligned}$
Hence, the correct answer is option: 'C'.