Gujarati
Hindi
4-1.Newton's Laws of Motion
hard

A uniform sphere of weight $W$ and radius $3\,m$ is being held by a string of length $2\,m$ , attached to a frictionless wall as shown in the figure. The tension in the string will be

A

$5\,W/4$

B

$15\,W/4$

C

$15\,W/16$

D

$5\,W/3$

Solution

$\mathrm{W}=\mathrm{T} \sin \theta$

$\cos \theta=3 / 5 \Rightarrow \quad \theta=53^{\circ}$

$\mathrm{T}=\frac{\mathrm{W}}{\sin 53^{\circ}}=\frac{5}{4} \mathrm{W}$

Standard 11
Physics

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