Q.29. Two blocks (P and Q) with respectively masses 2 kg and 1.5 kg are joined by a massless thread.

Q: What is the correct answer for JEE Main Physics Q29?

The correct answer is B. Sol. Let the acceleration of the cube be $a_0 = g/2$ towards the right. Consider the system from a non-inertial frame of reference moving with the cube (S). In this frame, a pseudo force acts on blocks P and Q. For block P (mass $m_P = 2 \text{ kg}$): The pseudo force is $m_P a_0 = m_P g/2$ acting to the left. The tension in the string is $T$. The block P is stationary relative to the cube, so $T

Q: Q.29. Two blocks (P and Q) with respectively masses 2 kg and 1.5 kg are joined by a massless thread. These blocks are mounted on a frictionless pulley which is fixed on the edge of a cube (S), as…

The correct answer is B. Sol. Let the acceleration of the cube be $a_0 = g/2$ towards the right. Consider the system from a non-inertial frame of reference moving with the cube (S). In this frame, a pseudo force acts on blocks P and Q. For block P (mass $m_P = 2 \text{ kg}$): The pseudo force is $m_P a_0 = m_P g/2$ acting to the left. The tension in the string is $T$. The block P is stationary relative to the cube, so $T

Question

Q.29. Two blocks (P and Q) with respectively masses 2 kg and 1.5 kg are joined by a massless thread. These blocks are mounted on a frictionless pulley which is fixed on the edge of a cube (S), as shown in the figure below. Block P is positioned on the top surface which has no friction and block Q is in contact with side-surface, having coefficient friction . The cube (S) moves towards the right with acceleration of , where  is gravitational acceleration. During this movement the block P and Q remain stationary. The value of  is ________ (take ).

FirstInTest · Accepted Answer

Sol. Let the acceleration of the cube be $a_0 = g/2$ towards the right. Consider the system from a non-inertial frame of reference moving with the cube (S). In this frame, a pseudo force acts on blocks P and Q. For block P (mass $m_P = 2 	ext{ kg}$): The pseudo force is $m_P a_0 = m_P g/2$ acting to the left. The tension in the string is $T$. The block P is stationary relative to the cube, so $T = m_P a_0 = 2(g/2) = g$. For block Q (mass $m_Q = 1.5 	ext{ kg}$): The pseudo force is $m_Q a_0 = m_Q g/2$ acting downwards (due to the cube accelerating horizontally, the pseudo force on Q is horizontal, but Q is on a vertical surface. The pseudo force is $m_Q a_0$ acting to the left. The normal force from the vertical surface on Q is $N = m_Q a_0 = 1.5(g/2) = 0.75g$. The gravitational force on Q is $m_Q g = 1.5g$ acting downwards. The tension in the string is $T$ acting upwards. The frictional force $f = \mu N$ acts upwards, opposing the tendency to slide down. Since block Q is stationary, the forces in the vertical direction must balance: $T + f = m_Q g$. We found $T = g$. So, $g + \mu N = 1.5g$. $g + \mu (0.75g) = 1.5g$. Divide by $g$: $1 + 0.75\mu = 1.5$. $0.75\mu = 0.5$. $\mu = \frac{0.5}{0.75} = \frac{2}{3} \approx 0.666...$. The closest option is 0.67.

Answer

33

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67

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1

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5

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