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JIPMER 2016
Exam
JIPMER
Year
2016
Subject
Physics
Chapter
Laws of Motion

JIPMER 2016 · Physics · Laws of Motion

A light string passes over a frictionless pulley. To one of its...

A light string passes over a frictionless pulley. To one of its ends a mass of 8 kg is attached. To its other end two masses of 7 kg each are attached. The acceleration of the system will be A light string passes over a frictionless pulley. To one of its ends a mass of 8 kg is attached. To its other end two

Options

  1. A10.2 g
  2. B5.10 g
  3. C20.36 g
  4. D0.27 g

Correct answer

D. 0.27 g

Step-by-step solution

From free body diagram of body AA, A light string passes over a frictionless pulley. To one of its ends a mass of 8 kg is attached. To its other end two m1a=T1−m1gm_1 a=T_1-m_1 g....(i) For body BB, m2a=m2g+T2−T1m_2 \mathrm{a}=m_2 g+T_2-T_1...(ii) For body C, m3a=m3g−T2m_3 a=m_3 g-T_2....(iii) On solving Eqs. (i), (ii) and (iii), we get a=[(m2+m3)−m1]gm1+m2+m3a=\frac{\left[\left(m_2+m_3\right)-m_1\right] g}{m_1+m_2+m_3} As, \quadm1=8 kg,m2=m3=7 kga=622 g=0.27 g\begin{aligned} m_1 & =8 \mathrm{~kg}, m_2=m_3=7 \mathrm{~kg} \\ a & =\frac{6}{22} \mathrm{~g}=0.27 \mathrm{~g}\end{aligned}$$
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