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Question

Physics Question on laws of motion

A string of negligible mass going over a clamped pulley of mass mm supports a block of mass MM as shown in the figure. The force on the pulley by the clamp is given by

A

2\sqrt{2} mgmg

B

4\sqrt{4} mgmg

C

(M+m)+mg\sqrt{\left(M+m\right)+m} \,g

D

(M+m)2+M2g\sqrt{\left(M+m\right)^{2}+M^{2}} \,g

Answer

(M+m)2+M2g\sqrt{\left(M+m\right)^{2}+M^{2}} \,g

Explanation

Solution

Weight of the pulley mgmg acts downwards. Weight of the block MgMg acts downwards.
Total downward force
F1F_{1} =(m+M)g=\left(m+M\right)g