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. `alpha_(A)=alpha_(B)`
B. `alpha_(A)gtalpha_(B)`
C. `alpha_(A)ltalpha_(B)`
D. none of these

Answer

B

Explanation

In figure `(T_(2)-T_(1))R=(MR^(2))/2alpha_(A)`
`T_(1)-Mg=Ma_(A)`
`T_(2)=2Mg`
`a_(A)=Ralpha_(A)`
`alpha_(A)=(2g)/(3R)`

`(T_(2)-T_(1))xxR=(MR^(2))/2alpha_(B)`

`T_(1)-Mg=Ma_(B)`
`2Mg-T_(2)=2Ma_(B)`
`a_(B)=(2g)/(7R)`
so, `alpha_(A)=alpha_(B)`