A) 144 N
B) 96 N
C) 240 N
D) 192 N
Correct Answer: D
Solution :
[d] Acceleration produced in upward direction \[a=\frac{F}{{{M}_{1}}+{{M}_{2}}+\text{ }\text{Mass of metal rod}}\] =\[\frac{480}{20+12+8}=12\text{ m}{{\text{s}}^{-2}}\] Tensions at the mid-point \[T=\left( {{M}_{2}}+\frac{\text{Mass of rod}}{2} \right)a\] \[=\left( 12+4 \right)\times 12=192N\]You need to login to perform this action.
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