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