A) zero
B) \[mga\]
C) \[mga\,\sin \,\theta \]
D) \[\frac{mga\,\,\sin \,\,\theta }{2}\]
Correct Answer: D
Solution :
Because the cubical block slides with a uniform velocity and does not topple. Hence, torque produced by weight = torque due to normal force on the block \[\therefore \] Torque due to normal face = torque due to weight = component of weight parallel to plane \[[O{{H}^{-}}]=2s=2\times 1.44\times {{10}^{-4}}=2.88\times {{10}^{-4}}\] perpendicular distance from lower face \[1\,d{{m}^{3}}=1\,L=1000\,mL\]You need to login to perform this action.
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