AP EAMCET · PHYSICS · Laws of Motion
One end of a light string is fixed to a clamp on the ground and the other end passes over a fixed frictionless pulley as shown in the figure. It makes an angle of \(30^{\circ}\) with the ground. The clamp can tolerate a vertical force of \(40 \mathrm{~N}\). If a monkey of mass \(5 \mathrm{~kg}\) were to climb up the rope, then the maximum acceleration in the upward direction with which it can climb safely is \(\left(g=10 \mathrm{~ms}^{-2}\right)\)
- A \(2 \mathrm{~ms}^{-2}\)
- B \(4 \mathrm{~ms}^{-2}\)
- C \(6 \mathrm{~ms}^{-2}\)
- D \(8 \mathrm{~ms}^{-2}\)
Answer & Solution
Correct Answer
(C) \(6 \mathrm{~ms}^{-2}\)
Step-by-step Solution
Detailed explanation
Let \(T\) is tension in the string. Maximum vertical force on clamp, \(T \sin 30^{\circ}=40\) \( \begin{array}{lrl} \Rightarrow & T \cdot \frac{1}{2}=40 \\ \Rightarrow & T=80 \mathrm{~N} \end{array} \) Let maximum acceleration of monkey for sate climbing is \(a\). Then, FBD of…
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