JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Laws of Motion

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A stone of mass 900 g is tied to a string and moved in a vertical circle of radius 1 m making 10 rpm. The tension in the string, when the stone is at the lowest point is (if

π^{2} = 9.8

and

g = 9.8

m/s

^{2}

)

(A)97 N
(B)9.8 N✓ Correct
(C)8.82 N
(D)17.8 N

Answer: (B)

Solution

At the lowest point of a vertical circle, the string tension supports the weight and also supplies the centripetal force:

T = m g + m r ω^{2}

With

m = 0.9

kg,

r = 1

m, and angular speed

ω = \frac{2 π N}{60} = \frac{2 π ( 10 )}{60} = \frac{π}{3} rad/s

T = 0.9 \times 9.8 + 0.9 \times 1 \times (\frac{π}{3})^{2}

Using

π^{2} = 9.8

T = 8.82 + 0.9 \times \frac{9.8}{9} = 8.82 + 0.98 = 9.8 N

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Single correct option

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