If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is x/2 times its original time period. Then the value of x is :

For a simple pendulum the time period of small‑angle oscillations is given by

$T=\; 2\pi \sqrt{L/g}$

where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity. The mass of the bob does not appear in this formula, so changing the mass does not affect the period.

Let the original length be $L$ and the original period be $T$. After the changes we have

$L\prime \; =\; \backslash tfrac\{1\}\{2\}\backslash ,L$

The new period $T\prime$ becomes

$T��\; =\; 2\pi \sqrt{L\prime /g}=\; 2\pi \sqrt{\backslash tfrac\{L\}\{2g\}}$

Factor out the original period:

$T\prime \; =\; 2\pi \; \backslash sqrt\{\backslash tfrac\{1\}\{2\}\}\; \backslash sqrt\{L/g\}\; =\; \backslash sqrt\{\backslash tfrac\{1\}\{2\}\}\; \backslash ,\; (2\pi \; \backslash sqrt\{L/g\})\; =\; \backslash tfrac\{1\}\{\backslash sqrt\{2\}\}\; \backslash ,\; \backslash mi\{T\}$

The problem states that the new period can be expressed as $\backslash tfrac\{x\}\{2\}\backslash ,\; \backslash mi\{T\}$. Equate this to the expression we derived:

$\backslash tfrac\{x\}\{2\}\backslash ,\; \backslash mi\{T\}\; =\; \backslash tfrac\{1\}\{\backslash sqrt\{2\}\}\backslash ,\; \backslash mi\{T\}$

Cancel $\backslash mi\{T\}$ and solve for $x$:

$\backslash tfrac\{x\}\{2\}\; =\; \backslash tfrac\{1\}\{\backslash sqrt\{2\}\}\; \backslash ;\backslash ;\backslash Rightarrow\backslash ;\backslash ;\; x\; =\; \backslash tfrac\{2\}\{\backslash sqrt\{2\}\}\; =\; \backslash sqrt\{2\}$

Hence the value of $x$ is √2, which matches the provided correct option.