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Step-by-Step Solution
Step 1: Write down the original time period
For a simple pendulum of length $l$, the time period $T$ is given by
$$
T = 2\pi \sqrt{\frac{l}{g}}.
$$
Step 2: Express the new time period with added mass
When an additional mass $M$ is added to the bob, the wire stretches by an amount $\Delta \ell$, increasing the length of the pendulum to $(l + \Delta \ell)$. Hence, the new time period $T_M$ becomes
$$
T_M = 2\pi \sqrt{\frac{l + \Delta \ell}{g}}.
$$
Step 3: Relate $T_M$ to $T$
Divide the new time period by the original time period:
$$
\frac{T_M}{T}
= \frac{2\pi \sqrt{\frac{l + \Delta \ell}{g}}}{2\pi \sqrt{\frac{l}{g}}}
= \sqrt{\frac{l + \Delta \ell}{l}}.
$$
Squaring both sides gives
$$
\left(\frac{T_M}{T}\right)^2
= \frac{l + \Delta \ell}{l}
= 1 + \frac{\Delta \ell}{l}.
$$
Step 4: Find the extension $\Delta \ell$ using Young’s modulus
For a wire of cross-sectional area $A$, Young’s modulus $Y$, under a load (tension) $Mg$:
$$
\Delta \ell = \frac{Mg\,l}{A\,Y}.
$$
Dividing by $l$,
$$
\frac{\Delta \ell}{l}
= \frac{Mg}{A\,Y}.
$$
Step 5: Substitute $\Delta \ell/l$ back into the time period ratio
Substituting $\Delta \ell/l = \frac{Mg}{A\,Y}$ into
$\left(\frac{T_M}{T}\right)^2 = 1 + \frac{\Delta \ell}{l},$ we get
$$
\left(\frac{T_M}{T}\right)^2
= 1 + \frac{Mg}{A\,Y}.
$$
This implies
$$
\left(\frac{T_M}{T}\right)^2 - 1
= \frac{Mg}{A\,Y}.
$$
Step 6: Isolate $1/Y$
From the above result, solve for $1/Y$:
$$
\frac{Mg}{A\,Y}
= \left(\frac{T_M}{T}\right)^2 - 1
\quad\Longrightarrow\quad
\frac{1}{Y}
= \left[\left(\frac{T_M}{T}\right)^2 - 1\right]\frac{A}{Mg}.
$$
Final Expression for $1/Y$
Therefore,
$$
\frac{1}{Y}
= \left[
\left(\frac{T_M}{T}\right)^2 - 1
\right]
\frac{A}{Mg}.
$$
