A pendulum made of a uniform wire of cross sectional area A has time period T. When an additional mass M is added to its bob, the time period changes to $$T_M$$. If the Young's modulus of the material of the wire is $$Y$$, then $$\frac{1}{Y}$$ is equal to: ($$g$$ = gravitational acceleration)

We have a simple pendulum whose bob is suspended by a uniform metallic wire of cross-sectional area $$A$$. Let the actual (already slightly stretched) length of this wire be $$L$$ when only the original bob is hanging. With this length the measured time period is

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

Now an additional mass $$M$$ is attached to the same bob. The extra load increases the tension in the wire by the amount

$$F = Mg\;.$$

Because the wire obeys Hooke’s law, this extra force produces an extra elongation $$\Delta L$$. For a wire of natural length $$L$$, Young’s modulus $$Y$$ is defined by the relation

$$Y = \dfrac{\text{Stress}}{\text{Strain}} = \dfrac{F/A}{\Delta L/L} = \dfrac{F\,L}{A\,\Delta L}\;.$$

Re-arranging, the extra extension is

$$\Delta L = \dfrac{F\,L}{A\,Y} = \dfrac{Mg\,L}{A\,Y}\;.$$

Because of this elongation, the new effective length of the pendulum becomes $$L + \Delta L$$ and the new time period is therefore

$$T_M = 2\pi \sqrt{\dfrac{L + \Delta L}{g}}\;.$$

Squaring the expressions for both time periods, we get

$$T^{2} = 4\pi^{2}\dfrac{L}{g}\;,\qquad T_M^{2} = 4\pi^{2}\dfrac{L + \Delta L}{g}\;.$$

Subtract the first equation from the second:

$$T_M^{2} - T^{2} = 4\pi^{2}\dfrac{(L + \Delta L) - L}{g} = 4\pi^{2}\dfrac{\Delta L}{g}\;.$$

Hence the extra elongation is also expressed as

$$\Delta L = \dfrac{g}{4\pi^{2}}\bigl(T_M^{2} - T^{2}\bigr)\;.$$

We now have two expressions for $$\Delta L$$, so we equate them:

$$\dfrac{Mg\,L}{A\,Y} = \dfrac{g}{4\pi^{2}}\bigl(T_M^{2} - T^{2}\bigr)\;.$$

The factor $$g$$ cancels from both sides, leaving

$$\dfrac{M\,L}{A\,Y} = \dfrac{1}{4\pi^{2}}\bigl(T_M^{2} - T^{2}\bigr)\;.$$

But from the first squared relation we already have $$L = \dfrac{g\,T^{2}}{4\pi^{2}}$$. Substitute this value of $$L$$ into the left-hand side:

$$\dfrac{M}{A\,Y}\left(\dfrac{g\,T^{2}}{4\pi^{2}}\right) = \dfrac{1}{4\pi^{2}}\bigl(T_M^{2} - T^{2}\bigr)\;.$$

Multiply both sides by $$4\pi^{2}$$ to clear the denominator:

$$\dfrac{M\,g\,T^{2}}{A\,Y} = T_M^{2} - T^{2}\;.$$

Solve for $$\dfrac{1}{Y}$$:

$$\dfrac{1}{Y} = \dfrac{T_M^{2} - T^{2}}{M\,g\,T^{2}}\,A = \left[\dfrac{T_M^{2}}{T^{2}} - 1\right]\dfrac{A}{M\,g}\;.$$

Writing the ratio compactly,

$$\dfrac{1}{Y} = \left[\left(\dfrac{T_M}{T}\right)^{2} - 1\right]\dfrac{A}{M\,g}\;.$$

This matches Option B in the given list.

Hence, the correct answer is Option B.