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, $F=\frac{G{M}_1{M}_2}{r^2}$
$\Rightarrow \left[G\right]=\left[\frac{F{r}^2}{M^2}\right]=\left(\frac{\left[ML{T}^{-2}\right]\left[L^2\right]}{\left[M^2\right]}\right)=\left[M^{-1}{L}^3{T}^{-2}\right]$
Now, $n_1{u}_1=n_2{u}_2$
Therefore, $n_2=\left(n_1\right)\left(\frac{u_1}{u_2}\right)=\left(n_1\right)\left(\frac{M_1^{-1}{L}_1^3{T}_1^{-2}}{M_2^{-1}{L}_3^3{T}_3^{-2}}\right)=\left(\frac{n_1}{4}\right)$