The length of the seconds pendulum is $1 \mathrm{~m}$ on earth. If the mass and diameter of the planet is 1…
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
0.67 m
0.45 m
0.60 m
0.76 m
Solution
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 $\mathrm{T}=2 \mathrm{~s}$
$\begin{aligned} & \therefore \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}=2 \pi \sqrt{\frac{\ell^{\prime}}{\mathrm{g}^{\prime}}} \\ & \frac{\ell}{\mathrm{g}}=\frac{\ell^{\prime}}{\mathrm{g}^{\prime}}\end{aligned}$
$\ell^{\prime}=\ell \cdot \frac{\mathrm{g}^{\prime}}{\mathrm{g}}=1 \times \frac{1}{1.5}=\frac{2}{3}=0.67 \mathrm{~m}$
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