TS EAMCET · Physics · Laws of Motion
Two touching blocks 1 and 2 are placed on an inclined plane forming an angle \(60^{\circ}\) with the horizontal. The masses are \(m_1\) and \(m_2\) and the coefficient of friction between the inclined plane and the two blocks are \(1.5 \mu\) and \(1.0 \mu\), respectively. The force of reaction between the blocks during the motion is ( \(g=\) acceleration due to gravity)
- A \(\left(m_2-m_1\right) \mu g\)
- B \(\left(m_2+m_1\right) \mu g\)
- C \(\frac{1}{2} \frac{m_1 m_2}{m_1+m_2} \mu g\)
- D \(\frac{1}{4} \frac{m_1 m_2}{m_1+m_2} \mu \mathrm{g}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{4} \frac{m_1 m_2}{m_1+m_2} \mu \mathrm{g}\)
Step-by-step Solution
Detailed explanation
According to the question, there are two blocks of masses \(m_1\) and \(m_2\) which are kept in contact on inclined plane of inclination \(60^{\circ}\) as shown in the figure, Coefficient of friction between the surface of plane and surface of first block, \(\mu_1=1.5 \mu\)…
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