A uniform sphere of weight W and radius 5, cm | vedclass

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(D) Let the center of the sphere be $O$ and the point where the string is attached to the wall be $P$. Let the point of contact between the sphere and

Vedclass · Accepted Answer

$\frac{13}{12}\,W$

Vedclass · Answer

(D) Let the center of the sphere be $O$ and the point where the string is attached to the wall be $P$. Let the point of contact between the sphere and the wall be $Q$. The radius of the sphere is $r = 5\, cm$. The length of the string is $l = 8\, cm$. In the right-angled triangle formed by the center of the sphere $O$,the point of contact $Q$,and the attachment point $P$,the hypotenuse $OP = l + r = 8 + 5 = 13\, cm$. The base $OQ = r = 5\, cm$. Using the Pythagorean theorem,the vertical distance $PQ = \sqrt{OP^2 - OQ^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\, cm$. Let $	heta$ be the angle the string makes with the vertical wall. Then $\cos 	heta = \frac{PQ}{OP} = \frac{12}{13}$. For the sphere to be in equilibrium,the vertical component of the tension $T$ must balance the weight $W$ of the sphere. Thus,$T \cos 	heta = W$. Substituting the value of $\cos 	heta$,we get $T 	imes \frac{12}{13} = W$,which implies $T = \frac{13}{12} W$.

$A$ uniform sphere of weight $W$ and radius $5\, cm$ is held by a string of length $8\, cm$ attached to a smooth vertical wall as shown in the figure. The tension in the string is:

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