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NEET PHYSICSRotational motionMedium

Question

A mass mm moves in a circle on a smooth horizontal plane with velocity v0v_0 at a radius R0R_0. The mass is attached to a string that passes through a smooth hole in the plane, as shown in the figure. The tension in the string is increased gradually and finally, mm moves in a circle of radius R02\frac{R_0}{2}. The final value of the kinetic energy is:

A

mv02m v_0^2

B

14mv02\frac{1}{4} m v_0^2

C

2mv022 m v_0^2

D

12mv02\frac{1}{2} m v_0^2

Step-by-Step Solution

The force on the mass is the tension in the string, which acts along the radius towards the centre (the hole). Since the line of action of the force passes through the centre, the torque about the centre is zero (τ=r×F=0\tau = r \times F = 0). According to the law of conservation of angular momentum, if the net external torque is zero, the angular momentum remains constant. Initial angular momentum, Li=mv0R0L_i = m v_0 R_0 Final angular momentum, Lf=mvf(R02)L_f = m v_f \left(\frac{R_0}{2}\right) Equating initial and final angular momentum: mv0R0=mvf(R02)m v_0 R_0 = m v_f \left(\frac{R_0}{2}\right) vf=2v0v_f = 2v_0 The final kinetic energy is given by: Kf=12mvf2=12m(2v0)2=12m(4v02)=2mv02K_f = \frac{1}{2} m v_f^2 = \frac{1}{2} m (2 v_0)^2 = \frac{1}{2} m (4 v_0^2) = 2 m v_0^2

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Rotational motion. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSRotational motioncirclesmoothhorizontalvelocityradius

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