Physics - Gravitation Question with Solution | TestHub

If $g$ is the acceleration due to gravity on earth's surface and g' on the planet, then $\begin{array}{rl}& \mathrm{g}=\frac{\mathrm{GM}}{{\mathrm{r}}^2}\text{ and }{\mathrm{g}}^{\prime }=\frac{\mathrm{G}\times 1.5\mathrm{M}}{(1.5{)}^2{\mathrm{r}}^2}=\frac{1}{1.5}\frac{\mathrm{GM}}{{\mathrm{r}}^2}\\ & \therefore {\mathrm{g}}^{\prime }=\frac{\mathrm{g}}{1.5}\end{array}$ For second's pendulum $\mathrm{T}=2\mathrm{s}$ $\begin{array}{rl}& \therefore \mathrm{T}=2\pi \sqrt{\frac{\ell }{\mathrm{g}}}=2\pi \sqrt{\frac{{\ell }^{\prime }}{{\mathrm{g}}^{\prime }}}\\ & \frac{\ell }{\mathrm{g}}=\frac{{\ell }^{\prime }}{{\mathrm{g}}^{\prime }}\end{array}$ ${\ell }^{\prime }=\ell \cdot \frac{{\mathrm{g}}^{\prime }}{\mathrm{g}}=1\times \frac{1}{1.5}=\frac{2}{3}=0.67\mathrm{m}$