Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A 1 kg stone at the end of 1 m long string is whirled in a vertical circle at a constant speed of 4 m/s. The tension in the string is 6 N, when the stone is at (Take g = 10 m/s²):

Top of the circle

Bottom of the circle

Half way down

None of the above

Step-by-Step Solution

Identify the Dynamics: For a body moving in a vertical circle, the net radial force provides the centripetal acceleration. The forces involved are the tension ( $T$ ) in the string and the component of weight ( $mg$ ) along the radius [Source 73, 82].
Calculate Centripetal Force: $F_c = \frac{mv^2}{r}$ Given $m=1\text{ kg}$ , $v=4\text{ m/s}$ , $r=1\text{ m}$ . $F_c = \frac{1 \times (4)^2}{1} = 16\text{ N}$
Analyze Tension at Key Positions:

At the Top: Both tension and gravity point downwards (towards the center). $T_{top} + mg = \frac{mv^2}{r} \implies T_{top} = \frac{mv^2}{r} - mg$ $T_{top} = 16 - (1 \times 10) = 6\text{ N}$
At the Bottom: Tension points upwards (towards center), gravity downwards. $T_{bottom} - mg = \frac{mv^2}{r} \implies T_{bottom} = \frac{mv^2}{r} + mg$ $T_{bottom} = 16 + 10 = 26\text{ N}$

Conclusion: The given tension is 6 N, which matches the calculated tension at the top of the circle.

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