Two particles of mass m each are tied at the | vedclass

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(B) Let the angle made by each segment of the string with the horizontal be $	heta$. When the separation between the particles is $2x$,the distance o

Vedclass · Accepted Answer

$\frac{F}{{2m}}\,\frac{x}{{\sqrt {{a^2} - {x^2}} }}$

Vedclass · Answer

(B) Let the angle made by each segment of the string with the horizontal be $	heta$. When the separation between the particles is $2x$,the distance of each particle from the center $P$ is $x$. The length of each half of the string is $a$. Thus,$\cos 	heta = \frac{x}{a}$ and $\sin 	heta = \frac{\sqrt{a^2 - x^2}}{a}$.Considering the equilibrium of the mid-point $P$ in the vertical direction,the force $F$ is balanced by the vertical components of the tension $T$ in the two segments of the string:$F = 2T \sin 	heta$$T = \frac{F}{2 \sin 	heta} = \frac{F}{2 (\sqrt{a^2 - x^2} / a)} = \frac{Fa}{2 \sqrt{a^2 - x^2}}$The horizontal component of the tension $T$ provides the force for the acceleration $a_{acc}$ of the particles towards each other:$T \cos 	heta = m a_{acc}$$a_{acc} = \frac{T \cos 	heta}{m} = \frac{1}{m} \left( \frac{Fa}{2 \sqrt{a^2 - x^2}} ight) \left( \frac{x}{a} ight)$$a_{acc} = \frac{F}{2m} \frac{x}{\sqrt{a^2 - x^2}}$

List-$I$	List-$II$
$(A)$ $P(E_2)$	$(i)$ $1/4$
$(B)$ $P(E_1 \cup E_2)$	$(ii)$ $5/8$
$(C)$ $P(\bar{E}_1 / \bar{E}_2)$	$(iii)$ $1/8$
$(D)$ $P(E_1 / \bar{E}_2)$	$(iv)$ $1/2$
	$(v)$ $3/8$
	$(vi)$ $3/4$

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