Question

If the unit of length and mass be doubled, then the numerical value w.r.t present value of the universal gravitation constant $G$ will become

• A. $\frac{1}{2}$ times
• B. $\frac{1}{4}$ times
• C. $8$ times
• D. $\frac{1}{3}$ times

Answer 1

Answer: (B)

As, $F=\frac{GM_{1}M_{2}}{r^{2}}$
$\Rightarrow \left[G\right]=\left[\frac{Fr^{2}}{M^{2}}\right]=\left(\frac{\left[MLT^{-2}\right]\left[L^{2}\right]}{\left[M^{2}\right]}\right)=\left[M^{-1}L^{3}T^{-2}\right]$
Now, $n_{1}u_{1}=n_{2}u_{2}$
Therefore, $n_{2}=\left(n_{1}\right)\left(\frac{u_{1}}{u_{2}}\right)=\left(n_{1}\right)\left(\frac{M_{1}^{-1}L^{3}_{1}T^{-2}_{1}}{M_{2}^{-1}L^{3}_{3}T^{-2}_{3}}\right)=\left(\frac{n_{1}}{4}\right)$