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.You need to login to perform this action.
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