JEE Physics Laws Of Motion MCQs Set 25

Answer: (b) in the backward direction on the front wheel and in the forward direction on the rear wheel

Explanation:
1. The rear wheel is driven; it pushes the ground backward, so friction on it acts forward and moves the cycle.
2. The front wheel is pushed along by the frame and tends to slide forward relative to the ground, so friction on it acts backward.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Answer: (c) 9.8 N

Explanation:
1. The block is at rest, so static friction balances the component of weight along the incline.
2. \( f=mg\sin30^\circ=2\times9.8\times\tfrac12=9.8 \) N (the limiting friction \( 0.7\times2\times9.8\cos30^\circ \) is larger).

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Answer: (d) (a) and (c) only

Explanation:
1. Along the plane, the horizontal force gives \( F\cos\theta=\cos\theta \) N up the plane, and gravity gives \( mg\sin\theta=\sin\theta \) N down the plane.
2. At 45° they are equal, so no friction is needed: (a) correct. For \( \theta>45^\circ \), gravity wins and friction must act up the plane, towards Q: (c) correct. For \( \theta<45^\circ \), friction acts down the plane, towards P, so (d) is wrong.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Answer: (a) 5 N

Explanation:
1. The maximum possible friction is \( 0.6\times1\times10=6 \) N, which exceeds the needed \( ma=5 \) N, so the block does not slip.
2. \( f=ma=1\times5=5 \) N.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Answer: (b) in the direction of his motion

Explanation:
1. The foot pushes the ground backward, so by Newton's third law the ground pushes the person forward.
2. Static friction on the person is therefore along the direction of motion; it is what makes walking possible.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Answer: (c) 2 s

Explanation:
1. \( a_A=g(\sin45^\circ-0.2\cos45^\circ)=\dfrac{10}{\sqrt2}(0.8)=4\sqrt2 \) m s⁻².
2. \( a_B=\dfrac{10}{\sqrt2}(0.7)=3.5\sqrt2 \) m s⁻².
3. Relative: \( \tfrac12(a_A-a_B)t^2=\sqrt2\Rightarrow\tfrac12(0.5\sqrt2)t^2=\sqrt2\Rightarrow t=2 \) s.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Answer: (d) zero

Explanation:
1. Net driving force \( =2mg\sin45^\circ-mg\sin45^\circ=\dfrac{mg}{\sqrt2}\approx0.71\,mg \).
2. Maximum friction available: \( \tfrac23mg\cos45^\circ+\tfrac13(2m)g\cos45^\circ=0.94\,mg \), which is greater than the driving force.
3. So the system remains at rest: acceleration \( =0 \).

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Answer: (a) \( F=mg\sin\theta,\ \theta=\tan^{-1}\mu\ \text{above the horizontal} \)

Explanation:
1. Pull at angle \( \theta \) above the horizontal: \( F\cos\theta=\mu(mg-F\sin\theta) \), so \( F=\dfrac{\mu mg}{\cos\theta+\mu\sin\theta} \).
2. F is least when the denominator is maximum: \( \tan\theta=\mu \), giving \( F_{min}=\dfrac{\mu mg}{\sqrt{1+\mu^2}}=mg\sin\theta \).

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Answer: (b) 4.2 kg

Explanation:
1. Tension in the horizontal string \( T_2=\mu M_3g=0.25(4)(9.8)=9.8 \) N.
2. For \( M_2 \) moving up the incline: \( T_1=T_2+M_2g\sin37^\circ+\mu M_2g\cos37^\circ=9.8+23.52+7.84=41.16 \) N.
3. \( M_1g=T_1\Rightarrow M_1=4.2 \) kg.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Answer: (c) 1.3 m s⁻²

Explanation:
1. Along the plane: \( 6a=(4+2)g(0.6)-0.75(4)g(0.8)-0.25(2)g(0.8) \).
2. \( 6a=g(3.6-2.4-0.4)=0.8g\Rightarrow a=0.133g=1.3 \) m s⁻².
3. (The tension is then \( T=4a=5.2 \) N.)

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Answer: (d) 31.21 N

Explanation:
1. Friction \( f=\mu mg\cos30^\circ=\sqrt{\tfrac32}(20)\tfrac{\sqrt3}{2}=21.21 \) N.
2. Moving up uniformly: \( F=mg\sin30^\circ+f=10+21.21=31.21 \) N.
3. (For the downward case, \( F=f-mg\sin30^\circ=11.21 \) N.)

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Answer: (a) 5

Explanation:
1. Push up: \( F_1=mg(\sin45^\circ+\mu\cos45^\circ) \). Prevent sliding down: \( F_2=mg(\sin45^\circ-\mu\cos45^\circ) \).
2. \( F_1=3F_2\Rightarrow1+\mu=3(1-\mu)\Rightarrow\mu=0.5 \).
3. \( N=10\mu=5 \).

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Answer: (b) \( 10\ \text{m/s} \)

Explanation:
1. \( |\Delta\vec v|=2v\sin(\theta/2)=2(10)\sin30^\circ=10 \) m/s.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Answer: (c) 36

Explanation:
1. Maximum tension gives the centripetal force: \( T=m\omega^2l \).
2. \( \omega^2=\dfrac{324}{0.5\times0.5}=1296\Rightarrow\omega=36 \) rad/s.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Answer: (d) \( \sqrt{\dfrac{\mu}{\alpha}} \)

Explanation:
1. The rod's normal force supplies the tangential acceleration: \( N=mL\alpha \). Friction along the rod supplies the centripetal force.
2. Slipping starts when \( m\omega^2L=\mu N=\mu mL\alpha \), with \( \omega=\alpha t \).
3. \( \alpha^2t^2=\mu\alpha\Rightarrow t=\sqrt{\mu/\alpha} \).

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Answer: (a) \( 45^\circ \)

Explanation:
1. In the car's frame the bob is in equilibrium under gravity, tension and the pseudo force \( mv^2/r \).
2. \( \tan\theta=\dfrac{v^2}{rg}=\dfrac{100}{100}=1\Rightarrow\theta=45^\circ \).

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Answer: (b) (b) and (c) only

Explanation:
1. Along the string (radial direction): \( T-Mg\cos\theta=Mv^2/L \), so (b) is true.
2. Perpendicular to the string only the weight component acts: \( |a_T|=g\sin\theta \), so (c) is true.
3. (a) and (d) hold only for a bob at rest at the lowest point or momentarily at the extreme, so they are not true in general.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Answer: (c) greater than \( mg\cos20^\circ \)

Explanation:
1. At 20° the bob is moving, so \( T=mg\cos20^\circ+\dfrac{mv^2}{L} \) with \( v\neq0 \).
2. Hence \( T>mg\cos20^\circ \).

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Answer: (d) 36 N

Explanation:
1. \( m_2 \) is at \( 0.3-0.124=0.176 \) m. String tension \( T=m_2\omega^2r_2=5(100)(0.176)=88 \) N.
2. \( m_1 \) needs \( m_1\omega^2r_1=10(100)(0.124)=124 \) N towards the centre.
3. Friction \( f=124-88=36 \) N, directed towards the centre.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Answer: (a) 11.67 rad/s

Explanation:
1. Friction needed \( f=\omega^2(m_1r_1-m_2r_2)=\omega^2(1.24-0.88)=0.36\,\omega^2 \).
2. Slipping when \( f=\mu m_1g=49 \) N: \( \omega^2=136.1\Rightarrow\omega=11.67 \) rad/s.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.

Answer: (b) 0.1 m

Explanation:
1. No friction means the tension alone supplies the centripetal force of both: \( m_1\omega^2r_1=T=m_2\omega^2r_2 \), so \( 10r_1=5r_2 \).
2. With \( r_1+r_2=0.3 \): \( r_1=0.1 \) m.

Teacher's Note:
1. Draw a free body diagram and resolve the forces along and perpendicular to the surface before writing equations.
2. Check whether friction is static (adjusts to the need) or kinetic (equal to mu N) in the situation.