Figure shown two pulley arrangments for lifting a mass `m` . In case-1, the mass is lifting by attaching a mass 2m while in case-2 the mass is lifted by pulling the other end with a downward force `F=2mg` . If `a_(a)` and `a_(b)` are the accelerations of the two masses then (Assumme string is massless and pulley is ideal).

Options

A. `f_(a)=f_(b)`
B. `f_(a)=f_(b)//2`
C. `f_(a)=f_(b)//3`
D. `f_(a)=2f_(b)`

Answer

C

Explanation

`f_(a)=((2m-m)/(2m+m))g=g/3`
`f_(b)=(2mg-mg)/m=g`
So, `f_(a)=f_(b)//3`