JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
A hanging mass \(M\) is connected to a four times bigger mass by using a string-pulley arrangement. as shown in the figure. The bigger mass is placed on a horizontal ice-slab and being pulled by \(2\,Mg\) force. In this situation. tension in the string is \(\frac{x}{5}\) \(Mg\) for \(x =\) Neglect mass of the string and friction of the block (bigger mass) with ice slab. (Given \(g=\) acceleration due to gravity)
- A \(2\)
- B \(7\)
- C \(6\)
- D \(8\)
Answer & Solution
Correct Answer
(C) \(6\)
Step-by-step Solution
Detailed explanation
Using \(\overrightarrow{ F }_{\text {net }}=\mu \vec{a}\) \(2 Mg - T =4 Ma\) \(T - Mg = Ma\) _______________ \(\Rightarrow a =\frac{ g }{5}\) \(T = Mg + Ma = Mg +\frac{ Mg }{5}=\frac{6}{5}\,Mg\)
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