Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A particle of mass $10 \text{ g}$ moves along a circle of radius $6.4 \text{ cm}$ with a constant tangential acceleration. What is the magnitude of this acceleration, if the kinetic energy of the particle becomes equal to $8 \times 10^{-4} \text{ J}$ by the end of the second revolution after the beginning of the motion?

$0.15 \text{ m/s}^2$

$0.18 \text{ m/s}^2$

$0.2 \text{ m/s}^2$

$0.1 \text{ m/s}^2$

Step-by-Step Solution

Concept: According to the Work-Energy Theorem, the work done by the net force on a particle is equal to the change in its kinetic energy ( $W = \Delta K$ ) . In non-uniform circular motion, the tangential force $F_t$ does work over the distance covered.
Formulas:

Tangential Force: $F_t = m a_t$
Work Done: $W = F_t \cdot s = m a_t s$
Distance covered in $n$ revolutions: $s = n \times 2\pi R$

Given Values:

Mass ( $m$ ) = $10 \text{ g} = 10^{-2} \text{ kg}$
Radius ( $R$ ) = $6.4 \text{ cm} = 6.4 \times 10^{-2} \text{ m}$
Kinetic Energy ( $K$ ) = $8 \times 10^{-4} \text{ J}$ (Initial $K_i = 0$ )
Revolutions ( $n$ ) = 2

Calculation:

Distance ( $s$ ) = $2 \times 2\pi \times (6.4 \times 10^{-2}) = 4\pi (6.4 \times 10^{-2}) \text{ m}$
Applying Work-Energy Theorem: $m a_t s = K$ $(10^{-2}) a_t [4\pi (6.4 \times 10^{-2})] = 8 \times 10^{-4}$ $a_t = \frac{8 \times 10^{-4}}{10^{-2} \times 4\pi \times 6.4 \times 10^{-2}}$ $a_t = \frac{8 \times 10^{-4}}{4\pi \times 6.4 \times 10^{-4}} = \frac{8}{25.6 \pi}$ $a_t = \frac{1}{3.2 \pi} \approx \frac{1}{3.2 \times 3.1416} \approx \frac{1}{10.05}$ $a_t \approx 0.1 \text{ m/s}^2$

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