$\therefore$ Radius of gyration $K_1=R_1$
For solid sphere
$I^{\prime}=\frac{2}{5} m^{\prime} R_2^2=m^{\prime} K_2^2$
$\therefore \text { Its radius of gyration }=K_2=\sqrt{\frac{2}{5}} R_2$
$\because K_1=K_2$
$\therefore R_1=\sqrt{\frac{2}{5}} R_2$
$\therefore \frac{R_1}{R_2}=\sqrt{\frac{2}{5}}$
$\therefore x=5$
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They use different lengths of the pendulum and /or record time for different number of oscillations. The observations are shown in the table.
Least count for length $=0.1 \mathrm{~cm}$
Least count for time $=0.1 \mathrm{~s}$
| Student | Length of the pendulum $(cm)$ | Number of oscillations $(n)$ | Total time for $(n)$ oscillations $(s)$ | Time period $(s)$ |
| $I.$ | $64.0$ | $8$ | $128.0$ | $16.0$ |
| $II.$ | $64.0$ | $4$ | $64.0$ | $16.0$ |
| $III.$ | $20.0$ | $4$ | $36.0$ | $9.0$ |
If $\mathrm{E}_{\mathrm{I}}, \mathrm{E}_{\text {II }}$ and $\mathrm{E}_{\text {III }}$ are the percentage errors in g, i.e., $\left(\frac{\Delta \mathrm{g}}{\mathrm{g}} \times 100\right)$ for students $\mathrm{I}, \mathrm{II}$ and III, respectively,
