The length of the seconds pendulum is \(1 \mathrm{~m}\) on earth. If the mass and diameter of the planet is 1.5 times that of the earth, the length of the seconds pendulum on the planet will be nearly

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