question_answer

In the figure shown, the pulleys and strings are massless. The acceleration of the block of mass 4 m just after the system is released from rest is \[(\theta ={{\sin }^{-1}}3/5)\]

A) \[\frac{2g}{5}\]downwards

B) \[\frac{2g}{5}\] upwards

C) \[\frac{5g}{11}\]downwards

D) \[\frac{5g}{11}\]upwards

Correct Answer: C

Solution :

[c] The FBD of blocks is as shown From Newton?s second law \[4mg-2T\cos \theta =4mA\text{ }.........\text{(i)}\] \[\text{and }T-mg=ma\text{         }.........\text{(ii)}\] \[\text{cos}\theta \text{=4/5 and from constraint we get}\] \[a=A\cos \theta \text{                   }.........\text{(iii)}\] Solving eq. (i), (ii) and (iii) We get acceleration of block of mass 4m, \[A=\frac{5g}{11}\]Downwards.