(C) For vertical circular motion:$1$. If $v_0 = \sqrt{5 g \ell}$,the bob completes the circle. At the lowest point,$T - mg = \frac{mv_0^2}{\ell} = 5mg

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(C) For vertical circular motion:$1$. If $v_0 = \sqrt{5 g \ell}$,the bob completes the circle. At the lowest point,$T - mg = \frac{mv_0^2}{\ell} = 5mg$,so $T = 6mg$. Thus,$(P) \rightarrow (1)$.$2$. If $v_0 = \sqrt{g \ell}$,since $v_0 $3$. If $v_0 = 2 \sqrt{g \ell}$,since $\sqrt{2 g \ell} $4$. If $v_0 = 3 \sqrt{g \ell}$,at the highest point,$v^2 = v_0^2 - 4g\ell = 9g\ell - 4g\ell = 5g\ell$. Then $T + mg = \frac{mv^2}{\ell} = 5mg$,so $T = 4mg$. Thus,$(S) \rightarrow (4)$.Therefore,the correct matching is $P-1, Q-3, R-2, S-4$.

Class 11 Physics Work, Energy, Power and Collision Vertical Circular Motion

Difficult

In vertical circular motion of a bob,match the entries of List-$I$ with entries of List-$II$. Here,$v_0$ is the velocity of the bob at the lowest point.

List-$I$ (Speed at lowest point) List-$II$ (Possible situation)

$(P) v_0 = \sqrt{5 g \ell}$ $(1)$ Tension at lowest point $= 6 mg$

$(Q) v_0 = \sqrt{g \ell}$ $(2)$ String will slack for a finite time

$(R) v_0 = 2 \sqrt{g \ell}$ $(3)$ Bob will oscillate

$(S) v_0 = 3 \sqrt{g \ell}$ $(4)$ Tension at highest point $= 4 mg$

A
$P-1, Q-2, R-3, S-4$
B
$P-4, Q-2, R-3, S-1$
C
$P-1, Q-3, R-2, S-4$
D
$P-4, Q-3, R-2, S-1$

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Work, Energy, Power and Collision

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Vertical Circular Motion

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List-$I$ (Speed at lowest point)	List-$II$ (Possible situation)
$(P) v_0 = \sqrt{5 g \ell}$	$(1)$ Tension at lowest point $= 6 mg$
$(Q) v_0 = \sqrt{g \ell}$	$(2)$ String will slack for a finite time
$(R) v_0 = 2 \sqrt{g \ell}$	$(3)$ Bob will oscillate
$(S) v_0 = 3 \sqrt{g \ell}$	$(4)$ Tension at highest point $= 4 mg$

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