Two blocks of mass \[{{M}_{1}}=20\,kg\] and \[{{\operatorname{M}}_{2}}=12\operatorname{K}\operatorname{g},\] are connected by a metal rod of mass 8 kg. The system is pulled vertically up applying a force of 480 N as shown. The tension at mid-point of the rod is: |
[JEE ONLINE 22-04-2013] |
A) \[144\] N
B) \[96\] N
C) \[240\] N
D) \[192\] N
Correct Answer: D
Solution :
[d] 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 |
\[\text{T =}\left( {{\text{M}}_{\text{2}}}\text{+}\frac{\text{Mass of rod}}{\text{2}} \right)\text{a}\] |
\[=(12+4)\times 12=192N\] |
[d] When the body has maximum speed then |
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