Two bodies of masses m and M are attached to the two ends of a light string passing over a fixed ideal pulley `(M gt gt m)`. When the bodies are in motion, the tension in the string is approximately

Options

A. `(m-m) g`
B. `mg`
C. `2mg`
D. `((m)/(M ))mg`

Answer

C

Explanation

Let a be the commin acceleration of the system and T be the tension the string
The equation of motion for the block M is
`mg - T = Ma`
adding (I ) and (ii) , we get
` Mg - mg =ma +Ma`
` or a=((M-m)/(m+M)) g`
subtiting this value in (i) we get
` T= (2 Mm)/(M+m) g`
` :. M gt gt m`,m can be neglected as it is very small when compare to M
` therefore t=2 mg `