Q128 / 155 · WBJEE 2020 · Physics

The bob of a swinging seconds pendulum (one whose time period is $2 \mathrm{~s}$ ) has a small speed $\mathrm{v}_{0}$ at its lowest point. Its height from this lowest point $2.25 \mathrm{~s}$ after passing through it is given by

  • A
    $\frac{\mathrm{v}_{0}^{2}}{2 \mathrm{~g}}$
  • B
    $\frac{\mathrm{v}_{0}^{2}}{\mathrm{~g}}$
  • Correct answer
    C
    $\frac{v_{0}^{2}}{4 g}$
  • D
    $\frac{9 v_{0}^{2}}{4 g}$

Solution

Hint: $T=2$ sec. at $\mathrm{t}=\mathrm{T}+\mathrm{T} / 8$ $\mathrm{t}=\mathrm{T}+\mathrm{T} / 8 \quad \mathrm{v}=\mathrm{A} \omega \cos \omega \mathrm{t}=\mathrm{v}_{0} / \sqrt{2}$ By Mechanical energy conservation: $\left(\frac{v_{0}}{\sqrt{2}}\right)^{2}=v_{0}^{2}-2 g h \quad h=\frac{v_{0}^{2}}{4 g}$

Hence, the correct answer is option: 'C'.

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