AP EAPCET 04-Jul-2022 Shift 2 Paper

FBD of the situation is given below
From Newton's law, ΣF= ma
Force on M1,
in x-direction, T=M1a........(i).
in y-direction, N1−w1=0......(ii)
Similarly, forces on M2,
in x-direction, M2gsin‌θ−T=M2a...(iii)
in y-direction, N2−m2g‌cos‌θ=0...(iv)
Adding to eq. (i) with eq. (ii), we get
M2gsin‌θ‌=M1a+M2a
⇒‌‌a‌=‌

M2gsin‌θ

M1+M2

According to question, when M1 is doubled and M2 is halved
a′=‌

‌ M2 2 gsin‌θ

2M1+‌ M2 2

=‌

M2gsin‌θ

4M1+M2

Now, ‌

a′

a

=‌

‌ M2gsin‌θ 4M1+M2

M2gsin‌θ M1+M2

⇒a′=(‌

M1+M2

4M1+M2

)a