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

Options

A. `sqrt2 Mg`
B. `sqrt2 mg`
C. `sqrt((M+m)^(2)+m^(2)g)`
D. `sqrt((M+m)^(2)+M^(2)g)`

Answer

D

Explanation

Weight of the pulley mg acts downwards. Weight of the block Mg acts downwards. Total downward force. `F_(1)=(m+M)g`
Tension in the string T=Mg.
The force on the pulley by reaction of the clamp is equal to the resultant of `F_(1)` and T. Let it be F.
`F^(2)=T^(2)=F_(2)^(1)=(Mg)^(2)+[(m+M)g]^(2)`
`or F^(2)=[M^(2)+(M+m)^(2)]g^(2)`
`or F=sqrt((M+m)^(2)+M^(2)g)`