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JEE Mains · Physics · STD 11 - Newtons laws of motion
- JEE Mains 2013
- Hard
- Numerical
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 ........ \(N\)

Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(192\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
Acceleration\,produced\,in\,upward\,direction\\
a = \frac{F}{{{M_1} + {M_2} + Mass\,of\,metal\,rod}}\\
= \frac{{480}}{{20 + 12 + 8}} = 12\,m{s^{ - 2}}\\
Tension\,at\,the\,mid\,{\rm{point}}\,\\
T = \left( {{M_2} + \frac{{Mass\,of\,rod}}{2}} \right)a\\
\,\,\,\, = \left( {12 + 4} \right) \times 12 = 192N
\end{array}\)
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\\
\,\,\,\, = \left( {12 + 4} \right) \times 12 = 192N
\end{array}\)
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