- $T$
- $\dfrac{T}{2}$
- $\dfrac{T}{4}$
- $\dfrac{T}{8}$
- $\dfrac{T}{16}$

Solution :

Two circular hoops, X and Y, are hanging on nails in a wall. The mass of X is four times that of Y, and the diameter of X is also four times that of Y. If the period of small oscillations of X is $T$, the period of small oscillations of Y is:

**The answer is** $\dfrac{T}{2}$.

The key thing to note here is that the period of small oscillations of a circular hoop is the same as that of a normal pendulum with the mass where the centre of the hoop is.

Thus, the period is given by the formula (we'll disregard the constant of $2\pi$ at the front, since it will not matter to our calculations):

\[T=\sqrt{\frac{r}{g}}.\] For hoops X, we have \[T_x=\sqrt{\frac{r_x}{g}}.\] For hoop Y, we have \[T_y=\sqrt{\frac{r_y}{g}}.\] Since $r_y = \frac{1}{4}r_x$, we have: \[T_y=\sqrt{\frac{1}{4}\frac{r_x}{g}}=\frac{1}{2}\sqrt{\frac{r_x}{g}}=\frac{T}{2}.\]

- $T$
- $\dfrac{T}{2}$
- $\dfrac{T}{4}$
- $\dfrac{T}{8}$
- $\dfrac{T}{16}$

Solution :

The key thing to note here is that the period of small oscillations of a circular hoop is the same as that of a normal pendulum with the mass where the centre of the hoop is.

Thus, the period is given by the formula (we'll disregard the constant of $2\pi$ at the front, since it will not matter to our calculations):

\[T=\sqrt{\frac{r}{g}}.\] For hoops X, we have \[T_x=\sqrt{\frac{r_x}{g}}.\] For hoop Y, we have \[T_y=\sqrt{\frac{r_y}{g}}.\] Since $r_y = \frac{1}{4}r_x$, we have: \[T_y=\sqrt{\frac{1}{4}\frac{r_x}{g}}=\frac{1}{2}\sqrt{\frac{r_x}{g}}=\frac{T}{2}.\]

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