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A small block of mass 100 g is tied to a spring of spring constant 7.5 N/m and length 20 cm. The other end of spring is fixed at a particular point A. If the block moves in a circular path on a smooth horizontal surface with constant angular velocity 5 rad/s about point A, then tension in the spring is- (1) 0.75 N (2) 0.25 N (3) 0.50 N (4) 1.5 N

1. 0.75 N
2. 0.25 N
3. 0.50 N
4. 1.5 N

Correct answer: 0.75 N

Solution

The tension in the spring is equal to the centripetal force required to keep the block moving in a circular path. This force can be calculated using the formula F = mω²r, where m is the mass (0.1 kg), ω is the angular velocity (5 rad/s), and r is the length of the spring (0.2 m). Substituting these values gives a tension of 0.75 N.

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