JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Laws of Motion

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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:

(A) $6.28 \times 1 0^{- 3} N$
(B)0.0314 N
(C) $9.859 \times 1 0^{- 2} N$
(D) $9.859 \times 1 0^{- 4} N$ ✓ Correct

Answer: (D)

Solution

The vertical side walls of the groove supply the horizontal force that keeps the block on its circle. One round in

T = 40

s gives

ω = \frac{2 π}{T} = \frac{π}{20} rad s^{- 1}

The required centripetal force is

N = m ω^{2} r = 0.2 \times (\frac{π}{20})^{2} \times 0.2 = 0.04 \times (\frac{3.14}{20})^{2}

The side walls therefore push on the block with

N \approx 9.859 \times 1 0^{- 4}

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Single correct option

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