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JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
Two blocks of mass \(M_1 = 20\,kg\) and \(M_2 = 12\,kg\) are connected by a metal rod of mass \(8\,kg.\) The system is pulled vertically up by applying a force of \(480\,N\) as shown. The tension at the mid-point of the rod is ........ \(N\)
- A \(144\)
- B \(96\)
- C \(240\)
- D \(192\)
Answer & Solution
Correct Answer
(D) \(192\)
Step-by-step Solution
Detailed explanation
\begin{array}{l} Acceleration\,produced\,in\,upward\,direction\\ a = \frac{F}{{{M_1} + {M_2} + Mass\,of\,metal\,rod}}\\ = \frac{{480}}{{20 + 12 + 8}} = 12\,m{s^{ - 2}}\\ Tension\,at\,the\,mid\,{\rm{point}}\,\\ T = \left( {{M_2} + \frac{{Mass\,of\,rod}}{2}} \right)a\\ \,\,\,\, =…
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