A small ball of mass 1 kg is suspended through a light string of length 1m. A constant force of magnitude 20 N is in always acting on the ball.
The minimum velocity of ball at lowest point to complete the vertical circle is given in column II
Match the column I with column II and mark the correct option from the given codes
Column I Column II
A) If force is directed downward P) $\sqrt{50} m / s$
B) If force directed upward Q) $\sqrt{150} m / s$
C) If force is directed horizontally R) $\sqrt{10} m / s$
D) If force is zero S) $\sqrt{30 \sqrt{5} + 20} m / s$

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$A \to Q; B \to R; C \to S; D \to P$

$A \to Q; B \to R; C \to P; D \to S$

$A \to Q; B \to R; C \to Q; D \to P$

$A \to Q; B \to R; C \to P; D \to P$

answer is D.

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Detailed Solution

Force minimum at the lowest point

$v = \sqrt{5 g l} = \sqrt{5 \times 10 \times 1} = \sqrt{50} m / s$

Force is downward $w_{g} + w_{F} + w_{T} = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$

$- 20 - 40 = 15 - \frac{1}{2} \times 1 \times U^{2}; U = \sqrt{150} m / s$

Force is upward $20 - 10 = \frac{m u^{2}}{l} \Rightarrow u = \sqrt{10} m / s$

For the bob to pass through that point,

the centripetal force must be equal to the effective acceleration due to gravity $\frac{v^{2}}{r} = \sqrt{g^{2} + {(\frac{F}{m})}^{2}} = 10 \sqrt{5}$ .....(1)

From work energy theorem

$\frac{1}{2} {mu}^{2} - \frac{1}{2} {mv}^{2} = mgr (1 + sinθ) + Frcosθ$

on solving we get, $u = \sqrt{30 \sqrt{5} + 20} m / s$

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Column I	Column II
A) If force is directed downward	P) $\sqrt{50} m / s$
B) If force directed upward	Q) $\sqrt{150} m / s$
C) If force is directed horizontally	R) $\sqrt{10} m / s$
D) If force is zero	S) $\sqrt{30 \sqrt{5} + 20} m / s$