A light rigid wire of length 1 ,m is attached | vedclass

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(B) Given: Length of the rigid wire $l = 1 \,m$, mass of the ball $m = 500 \,g = 0.5 \,kg$, initial velocity at the lowest point $u = 6 \,m/s$. Let $v

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$6$

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(B) Given: Length of the rigid wire $l = 1 \,m$, mass of the ball $m = 500 \,g = 0.5 \,kg$, initial velocity at the lowest point $u = 6 \,m/s$. Let $v$ be the velocity of the ball when the wire makes an angle $	heta = 60^{\circ}$ with the upward vertical.Using the law of conservation of energy between the lowest point (let it be $P$) and the point $C$ where the wire makes $60^{\circ}$ with the upward vertical:Total energy at $P$ = Total energy at $C$$\frac{1}{2}mu^2 + 0 = \frac{1}{2}mv^2 + mgh$where $h$ is the height of point $C$ above the lowest point. From the geometry, $h = l + l \cos 60^{\circ} = l(1 + \cos 60^{\circ}) = 1(1 + 0.5) = 1.5 \,m$.Substituting the values:$\frac{1}{2} 	imes (6)^2 = \frac{1}{2}v^2 + g 	imes 1.5$$18 = \frac{1}{2}v^2 + 10 	imes 1.5$$18 = \frac{1}{2}v^2 + 15$$\frac{1}{2}v^2 = 3 \implies v^2 = 6 \,m^2/s^2$.The radial component of acceleration (centripetal acceleration) is given by $a_r = \frac{v^2}{l}$.$a_r = \frac{6}{1} = 6 \,m/s^2$.

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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