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:

The correct option is √2.

Step-by-step derivation:

The time period of oscillation ($T$) of a simple pendulum is given by the formula:
$T=2\pi \sqrt{\frac{l}{g}}$
where:
• $l$ is the length of the pendulum,
• $g$ is the acceleration due to gravity.

From this formula, we can make two key observations:
1. The time period is directly proportional to the square root of the length of the pendulum ($T\propto \sqrt{l}$).
2. The time period is completely independent of the mass of the bob ($m$).

Let the original mass be $m_1$, the original length be $l_1=l$, and the original time period be $T_1$:
$T_1=2\pi \sqrt{\frac{l}{g}}$

According to the given conditions:
• The mass of the bob is increased to thrice its original mass ($m_2=3m_1$). Since the time period does not depend on the mass, this change has no effect.
• The length of the pendulum is halved ($l_2=\frac{l}{2}$).

Now, let's write the expression for the new time period ($T_2$):
$T_2=2\pi \sqrt{\frac{l_2}{g}}=2\pi \sqrt{\frac{l}{2g}}$

We can factor out the constant terms to relate the new time period to the original one:
$T_2=\frac{1}{\sqrt{2}}\left(2\pi \sqrt{\frac{l}{g}}\right)=\frac{1}{\sqrt{2}}{T}_1$

We are given that the new time period is $\frac{x}{2}$ times the original time period:
$T_2=\frac{x}{2}{T}_1$

Comparing the two expressions for $T_2$:
$\frac{x}{2}=\frac{1}{\sqrt{2}}$

Solving for $x$ by multiplying both sides by 2:
$x=\frac{2}{\sqrt{2}}=\sqrt{2}$

Therefore, the value of $x$ is √2.