Two blocks, each having a mass M, rest on frictionless surfaces as shown in the figure. If the pulleys are light and frictionless and M on the incline is allowed to move down, then the tension in string will be

Options

A. `(2)/(3) Mg sin theta`
B. `(3)/(2) Mg sin theta`
C. `(Mg sin theta)/(2)`
D. `2Mg sin theta`

Answer

C

Explanation

( c) Acceleration of system , `a=(" Net pulling force")/(" Total mass")=(Mg sin theta)/(2M)`
`a=(1)/(2)g sin theta`
Now, the block on ground is moving due to tension.
Hence, `T=Ma=(Mg sin theta)/(2)`