AP EAMCET Engineering 22 Sep 2020 Shift 2 Paper

The given situation is shown in the figure,We know that, torque, τ=Iα...(i)
If T is the tension in the string, then
τ=TR....(ii)
From Eqs. (i) and (ii), we get α=TR
Under no slipping condition,
α=‌

a

R

Hence, I⋅‌

a

R

=TR⇒T=‌

l⋅a

R2

For cylinder, I=‌

MR2

2

∴‌‌T=‌

MR2

2

⋅‌

a

R2

⇒T=‌

ma

2

....(iii)
For the motion of the block,
Mg−T=Ma
⇒‌‌Mg−‌

Ma

2

=Ma [from Eq. (iii)]
⇒‌‌Mg=‌

3Ma

2

⇒a=‌

2y

3