Motion in Vertical Circle: NEET MCQs with Answers

Explanation: Take a point on the circle at angle θ from the lowest position. Along the string, T − mg cos θ = mv²/l, so T = mg cos θ + mv²/l. Both terms are largest at the lowest point: cos θ = +1 there, and the speed is greatest there because the stone has lost the most potential energy. In the figure A is the top, B the bottom, and C and D are at the level of the centre O. The tension is therefore maximum at B, so the string breaks at B.

Explanation: At the lowest point B the equation of motion is T_B = mv²/R + mg, whereas at the side points C and D, T = mv²/R, and at the top A, T_A = mv²/R − mg. Thus T_B is the largest tension. Increasing the speed raises every tension, but B remains the maximum, so the string breaks at B.

Explanation: Energy conservation between the lowest point and the horizontal position (a rise of l): ½mu² = ½mu′² + mgl ⟹ u′² = u² − 2gl. The velocity at the lowest point is horizontal while that at the horizontal position is vertical, so the two are mutually perpendicular. |Δv| = √(u² + u′²) = √(u² + u² − 2gl) = √(2(u² − gl)).

Explanation: At an angular position θ measured from the lowest point, T = mg cos θ + mv²/l. Both terms are largest at the lowest point: cos θ = 1 there, and by conservation of energy the speed v is also maximum there. Hence the tension peaks at the lowest point and the wire is most likely to snap there.

Explanation: At the lowest point the net upward force provides the centripetal force: T − mg = mv²/r T = mg + m(7gr)/r = mg + 7mg = 8mg. (Note that √(7gr) > √(5gr), so the mass does complete the full vertical circle.)

Explanation: A rigid rod can push as well as pull, so the bob need only reach the top with zero speed: V² = 4gl = 36, which gives V = 6 m s⁻¹. Momentum conservation with the rebound taken as negative gives 0.02v = 0.02(–50) + 2(6) = 11. Hence v = 550 m s⁻¹.

Explanation: At angle θ from the vertical the required centripetal force is mv²/R = 2mg(1 – cos θ), while gravity supplies mg cos θ towards the centre. Near A the speed is small, gravity supplies more than is needed, and the inner wall must push outward on the ball. Beyond cos θ = 2/3 more force is needed than gravity gives, so the outer wall must push inward.

Explanation: At the highest point the critical condition gives v² = gR. The point P lies R + R cos 60° = 3R/2 below the top. Energy conservation gives v_P² = gR + 2g(3R/2) = 4gR, so v_P = 2√(gR).

Explanation: The tension is greatest at the lowest point, where the speed is greatest. The ball falls l(1 – cos θ) = 0.4l, so v² = 0.8gl. Then T = mg + mv²/l = 1.8mg = 9 N.

Explanation: At the lowest point T = mg + mv²/l = 5 + 32 = 37 N. At the level of the centre v² = 64 – 2gl = 44, and the weight has no component along the string, so T = 0.5(44) = 22 N. At the top v² = 64 – 4gl = 24, and the weight helps, so T = 12 – 5 = 7 N.

Explanation: After turning through θ the ball has fallen R(1 – cos θ), so v² = 2gR(1 – cos θ). Contact is lost when the normal reaction becomes zero, which means mg cos θ = mv²/R. Substituting gives cos θ = 2(1 – cos θ), so cos θ = 2/3.

Explanation: When the string makes an angle θ with the vertical, the bob has fallen l cos θ, so v² = 2gl cos θ. The radial equation T − mg cos θ = mv²/l then gives T = 3mg cos θ. Setting T = 1.5mg gives cos θ = 0.5, so v² = 2 × 10 × 0.9 × 0.5 = 9 and v = 3 m s⁻¹.

Explanation: At angle θ from the vertical, energy conservation gives v² = 2gR(1 − cos θ). If N is the outward push of the tube on the ball, mg cos θ − N = mv²/R, so N = mg(3 cos θ − 2). N is positive for cos θ > 2/3, when the ball rests on the nearer wall, and negative beyond that, when the farther wall must supply the extra inward force. Unlike a ball on a sphere, the ball cannot lose contact, because the tube can push from either side.

Explanation: Energy conservation gives v² = u² − 4gL = 36 − 20 = 16, so the speed at the top is 4 m/s, and the string stays taut there because v² exceeds gL. The velocities at the bottom and top are opposite in direction, so the change in velocity has magnitude 6 + 4 = 10 m/s. The change in momentum is 0.5 × 10 = 5 kg m/s.

Explanation: Just completing the circle means the tension is zero at the top, so v_top² = gl. Falling through a height l to the horizontal position raises this to v² = gl + 2gl = 3gl. With the string horizontal, gravity is tangential and the tension alone supplies the centripetal force, so T = mv²/l = 3mg. This gives T = 3 × 0.2 × 10 = 6 N, independent of the string's length.