The tension $T$ in the string shown in figure is ..................... $N$
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(a)
$m g \sin \theta=10(10) \sin 30^{\circ}=50 \,N$
Frictional force $=\mu m g \cos \theta=(0.7)$ (10) (10) $\frac{\sqrt{3}}{2}=35 \sqrt{3} \,N$
Frictional force is sufficient to oppose gravitational force. Tension will be zero.
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