TS EAMCET 10-Sep-2020 Shift 2 Solved Paper

Let block falls through a height h, then loss of PE of block appears in form of KE of block and KE of rotation of pulley.
⇒mgh=‌

1

2

mv2+‌

1

2

‌

MR2

2

⋅ω2
Here, m=M=2kg,
So, ‌‌gh=‌

v2

2

+‌

R2

4

ω2
As v=Rω,gh=‌

3

4

R2ω2
Differentiating with respect to time,
g⋅‌

dh

dt

‌‌=‌

3

4

R2⋅2ω⋅‌

dω

dt

gv‌‌=‌

3

4

R2×2×‌

v

R

⋅α
⇒‌‌g‌‌=‌

6

4

Rα⇒‌

4g

6

=Rα⇒‌‌Rα=‌

2g

3

‌ But ‌Rα=at‌‌=‌ tangential acceleration of pulley ‌
‌‌=‌ acceleration of block ‌
So, at=Rα=‌

2

3

g=‌

20

3

m∕s2