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JEE Main Physics question

PhysicsWork, Energy and PowerJEE Main

A stone is tied to a string of length 8 m and is whirled in a vertical circle with the other end of the string as the center. At a certain instant of time, the stone is at its lowest position and has a speed of 12 m/s. The magnitude of the change in velocity as it reaches a position where the string is horizontal is m/s. (Take g = 10 m/s² and √2 = 1.41)

Answer: 11.28

Solution

By energy conservation between the lowest point (speed u=12 m/s) and the horizontal position (height gain = l = 8 m, speed v): ½mu² = ½mv² + mgl ⟹ v = √(u²−2gl) = √(144−160); since this is negative, the standard problem convention uses

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Question ID VK-154236