A block of 200 g mass moves with a uniform speed in a horizontal circular groove, with vertical side walls of radius 20 cm. If the block takes 40 s to complete one round, the normal force by the side walls of the groove is :

Circular Motion PYQ — JEE Main 2021 · 16 Mar · Shift 1 · Q47 with Solution

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Correct answer: C

The block moves in a horizontal circular groove with vertical side walls, so the normal force by the wall provides the required centripetal force.

For uniform circular motion,

v = \frac{2\pi r}{T}

Substitute the values:

v = \frac{2\pi(0.2)}{40} = \frac{0.4\pi}{40} = 0.01\pi\,\text{m/s}

So,

v = 0.0314\,\text{m/s}

The normal force $N$ by the side wall is:

N = \frac{mv^2}{r}

Substitute values:

N = \frac{0.2\,(0.0314)^2}{0.2}

Since $0.2$ cancels,

N = (0.0314)^2

Now,

(0.0314)^2 = 0.00098596 \approx 9.859\times 10^{-4}\,\text{N}

N = 9.859\times 10^{-4}\,\text{N}

This corresponds to Option C.

Stored correct answer: C

My derived answer: C

So, the derived answer agrees with the stored correct answer.