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EEJ MAIN Physics QUESTION #7098
Question 1
A pendulum made of a uniform wire with cross-sectional area $A$ has a time period $T$. When a mass $M$ is attached to the bob, the period becomes $T_M$. If $Y$ is the Young's modulus, find the value of $\frac{1}{Y}$ [cite: 437].
  • $[(\frac{T_M}{T})^2 - 1]\frac{A}{Mg}$✔️
  • $[(\frac{T_M}{T})^2 - 1]\frac{Mg}{A}$
  • $[1 - (\frac{T_M}{T})^2]\frac{A}{Mg}$
  • $[1 - (\frac{T}{T_M})^2]\frac{A}{Mg}$
Correct Answer Explanation

The time period is $T = 2\pi\sqrt{L/g}$, meaning $L \propto T^2$ [cite: 437].

  • Initial length $L = kT^2$ and new length $L' = kT_M^2$
  • Extension $\Delta L = L' - L = k(T_M^2 - T^2)$
  • Strain $\frac{\Delta L}{L} = \frac{T_M^2 - T^2}{T^2} = (\frac{T_M}{T})^2 - 1$
  • From $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{Mg/A}{\Delta L/L}$, we get $\frac{1}{Y} = [(\frac{T_M}{T})^2 - 1]\frac{A}{Mg}$[cite: 437, 440].