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Equilibrium of a Particle and Common Forces in Mechanics — NEET Physics PYQ MCQs with Solutions

Free NEET Physics PYQ Equilibrium of a Particle and Common Forces in Mechanics MCQs with step-by-step solutions (12 questions). Part of Laws of Motion. Practise online on Prepizo — no login needed.

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Questions with solutions

Q1 — Equilibrium of a Particle and Common Forces in Mechanics · medium
Two bodies of mass 4 kg and 6 kg are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity $g$ is
A. $g/2$
B. $g/5$  ✓ Correct
C. $g/10$
D. $g$
Solution: For this Atwood system with $m_1=4$ kg and $m_2=6$ kg, $a=\left(\dfrac{m_2-m_1}{m_1+m_2}\right)g=\dfrac{6-4}{4+6}\times g=\dfrac{g}{5}$.
Q2 — Equilibrium of a Particle and Common Forces in Mechanics · medium
A block of mass $m$ is placed on a smooth inclined wedge $ABC$ of inclination $\theta$ as shown in the figure. The wedge is given an acceleration $a$ towards the right. The relation between $a$ and $\theta$ for the block to remain stationary on the wedge is
A. $a=g\cos\theta$
B. $a=\dfrac{g}{\sin\theta}$
C. $a=\dfrac{g}{\text{cosec}\,\theta}$
D. $a=g\tan\theta$  ✓ Correct
Solution: In the wedge's frame a pseudo force $ma$ acts on the block towards the left. For equilibrium, resolving the normal reaction $R$: $R\sin\theta=ma$ and $R\cos\theta=mg$. Dividing gives $\tan\theta=\dfrac{a}{g}$, so $a=g\tan\theta$.
Q3 — Equilibrium of a Particle and Common Forces in Mechanics · medium
Two blocks $A$ and $B$ of masses $3m$ and $m$ respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of $A$ and $B$ immediately after the string is cut, are respectively
A. $g,\ \dfrac{g}{3}$
B. $\dfrac{g}{3},\ g$  ✓ Correct
C. $g,\ g$
D. $\dfrac{g}{3},\ \dfrac{g}{3}$
Solution: Initially the system is in equilibrium, so the spring force $kx=4mg$. The instant the string is cut, the spring still pulls A up with $4mg$. For A ($3m$): $F_{net}=4mg-3mg=mg\Rightarrow a_A=\dfrac{g}{3}$. For B ($m$): only its own weight acts, so $a_B=g$. Hence $\dfrac{g}{3},\,g$.
Q4 — Equilibrium of a Particle and Common Forces in Mechanics · medium
Three blocks $A$, $B$ and $C$ of masses 4 kg, 2 kg and 1 kg respectively, are in contact on a frictionless surface, as shown. If a force of 14 N is applied on the 4 kg block, then the contact force between $A$ and $B$ is
A. 2 N
B. 6 N  ✓ Correct
C. 8 N
D. 18 N
Solution: Total mass $M=4+2+1=7$ kg, so $a=\dfrac{F}{M}=\dfrac{14}{7}=2\ \text{m/s}^2$. For block A: $F-F'=m_A a\Rightarrow F'=14-4\times2=6$ N, which is the contact force between A and B.
Q5 — Equilibrium of a Particle and Common Forces in Mechanics · medium
A person of mass 60 kg is inside a lift of mass 940 kg and presses the button on control panel. The lift starts moving upwards with an acceleration $1.0$ m/s$^2$. If $g = 10$ m/s$^2$, the tension in the supporting cable is
A. 9680 N
B. 11000 N  ✓ Correct
C. 1200 N
D. 8600 N
Solution: Total mass $m = $ mass of lift $+$ mass of person $= 940 + 60 = 1000$ kg. From the free body diagram, $T - mg = ma \Rightarrow T = m(g+a) = 1000(10+1) = 11000$ N.
Q6 — Equilibrium of a Particle and Common Forces in Mechanics · medium
The mass of a lift is 2000 kg. When the tension in the supporting cable is 28000 N, then its acceleration is
A. $30\ \text{ms}^{-2}$ downwards
B. $4\ \text{ms}^{-2}$ upwards  ✓ Correct
C. $4\ \text{ms}^{-2}$ downwards
D. $14\ \text{ms}^{-2}$ upwards
Solution: For the upward-accelerating lift the equation of motion is $R-mg=ma$, so $28000-2000\times10=2000a\Rightarrow a=\dfrac{8000}{2000}=4\ \text{ms}^{-2}$ upwards.
Q7 — Equilibrium of a Particle and Common Forces in Mechanics · medium
Three forces acting on a body are shown in the figure. To have the resultant force only along the $y$-direction, the magnitude of the minimum additional force needed is
A. $0.5$ N  ✓ Correct
B. $1.5$ N
C. $\dfrac{\sqrt{3}}{4}$ N
D. $\sqrt{3}$ N
Solution: Resolving, the net force along the $x$-axis $= -(1\cos60^\circ + 2\sin30^\circ) + 4\sin30^\circ = -\left(\dfrac{1}{2} + 1\right) + 2 = +\dfrac{1}{2}$ N. To make the resultant lie only along the $y$-direction, a minimum additional force of magnitude $0.5$ N is required to cancel this $x$-component.
Q8 — Equilibrium of a Particle and Common Forces in Mechanics · medium
A monkey of mass 20 kg is holding a vertical rope. The rope will not break, when a mass of 25 kg is suspended from it but will break, if the mass exceeds 25 kg. What is the maximum acceleration with which the monkey can climb up along the rope? (Take $g = 10$ m/s$^2$)
A. $25$ m/s$^2$
B. $2.5$ m/s$^2$  ✓ Correct
C. $5$ m/s$^2$
D. $10$ m/s$^2$
Solution: Maximum bearable tension in the rope $T = 25\times10 = 250$ N. For the climbing monkey, $T - mg = ma \Rightarrow a = \dfrac{T - mg}{m} = \dfrac{250 - 20\times10}{20} = \dfrac{50}{20} = 2.5$ m/s$^2$.
Q9 — Equilibrium of a Particle and Common Forces in Mechanics · medium
A man weighs 80 kg. He stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 5 m/s$^2$. What would be the reading on the scale? (Take $g = 10$ m/s$^2$)
A. $800$ N
B. $1200$ N  ✓ Correct
C. Zero
D. $400$ N
Solution: When the lift moves upwards, the scale reads $R = m(g+a) = 80(10+5) = 80\times15 = 1200$ N.
Q10 — Equilibrium of a Particle and Common Forces in Mechanics · medium
A lift of mass 1000 kg is moving upwards with an acceleration of 1 m/s$^2$. The tension developed in the string, which is connected to lift is ($g = 9.8$ m/s$^2$)
A. $9800$ N
B. $10800$ N  ✓ Correct
C. $11000$ N
D. $10000$ N
Solution: As the lift moves upwards, $T - mg = ma \Rightarrow T = m(g+a) = 1000(9.8+1) = 1000\times10.8 = 10800$ N.
Q11 — Equilibrium of a Particle and Common Forces in Mechanics · medium
Two masses $M_1 = 5$ kg, $M_2 = 10$ kg are connected at the ends of an inextensible string passing over a frictionless pulley as shown. When masses are released, then acceleration of masses will be
A. $g$
B. $\dfrac{g}{2}$
C. $\dfrac{g}{3}$  ✓ Correct
D. $\dfrac{g}{4}$
Solution: For the mass–pulley system, $a = \dfrac{M_2 - M_1}{M_1 + M_2}\,g = \dfrac{10-5}{5+10}\,g = \dfrac{5}{15}g = \dfrac{g}{3}$.
Q12 — Equilibrium of a Particle and Common Forces in Mechanics · medium
A mass of 1 kg is suspended by a thread. It is (1) lifted up with an acceleration $4.9$ m/s$^2$, (2) lowered with an acceleration $4.9$ m/s$^2$. The ratio of the tensions is
A. $3 : 1$  ✓ Correct
B. $1 : 3$
C. $1 : 2$
D. $2 : 1$
Solution: When lifted up, $T_1 = m(g+a) = 1(9.8+4.9) = 14.7$ N. When lowered, $T_2 = m(g-a) = 1(9.8-4.9) = 4.9$ N. Hence $\dfrac{T_1}{T_2} = \dfrac{14.7}{4.9} = \dfrac{3}{1}$, i.e. $3:1$.