Q55 / 270 · NDA 2023 (Phase 1) · General Ability

A particle is moving in a circle of radius $R$ with a constant speed $v$. Its average acceleration over the time when it moves over half the circle is :

  • A
    $\frac{v^2}{R}$
  • B
    $\frac{\pi v^2}{2 R}$
  • Correct answer
    C
    $\frac{2 v^2}{\pi R}$
  • D
    $0$

Solution

Change in momentum $=$ Final momentum - Initial momentum $=\mathrm{m} v-(-\mathrm{m} v)=2 \mathrm{~m} v$ Now, time taken $=\frac{\text { distance }}{\text { speed }}=\frac{\pi R}{v}$ $=\frac{2 \mathrm{~m} v}{\frac{\pi \mathrm{R}}{v}}=\frac{2 \mathrm{~m} v^2}{\pi \mathrm{R}}$ The average acceleration will be $\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{2 \mathrm{~m} v^2}{\pi \mathrm{Rm}}=\mathrm{a}=\frac{2 v^2}{\pi \mathrm{R}}$

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

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