question_answer

A cubical block of mass m and edge a slides down a rough inclined plane of inclination \[\theta \] with a uniform velocity. The torque of the normal force on the block about its centre has a magnitude

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 \[1\,d{{m}^{3}}=1\,L=1000\,mL\]