JEE Physics Laws Of Motion MCQs Set 32

Answer: (a) Only (1) and (4)

Explanation:
1. String constraint: \( u = v_B\cos\theta \Rightarrow v_B = \frac{u}{\cos\theta} \). The tension is less than mg for equal masses, so \( T\sin\theta2. The accelerations are not equal in general because the angle changes.

Teacher's Note:
1. Draw a free-body diagram and write \( F \) = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (b) 2 s

Explanation:
1. Acceleration along the groove \( = g\sin30^\circ\sin30^\circ = 2.5 m/s^{2} \) (gravity component down the plane times the sine of the groove angle).
2. \( 5 = \tfrac12(2.5)t^2 \Rightarrow t = 2 s \).

Teacher's Note:
1. Draw a free-body diagram and write \( F \) = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (d) 15 : 18 : 10

Explanation:
1. Tension in Atwood machine \( T = \frac{2m_1m_2g}{m_1+m_2} \): case \( (a) 2g \); \( (b) \frac{12g}{5} \); \( (c) \frac{4g}{3} \).
2. Extension is proportional to \( T \): \( 2:\frac{12}5:\frac43 = 30:36:20 = 15:18:10 \).

Teacher's Note:
1. Draw a free-body diagram and write \( F \) = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (c) 12 N

Explanation:
1. The block presses on the wedge with \( N = mg\cos37^\circ = 20 N \) perpendicular to the incline; its horizontal component \( N\sin37^\circ = 12 N \) must be balanced by the spring balance.

Teacher's Note:
1. Draw a free-body diagram and write \( F \) = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (c) 2 N

Explanation:
1. The groove walls supply the force in the \( y \) direction. Along the groove, friction alone must give the acceleration \( a_x = 2 m/s^{2} \), so \( f = ma_x = 2 N \).

Teacher's Note:
1. Resolve along and perpendicular to the groove.
2. Walls act perpendicular to the groove.

Answer: (d) zero

Explanation:
1. The limiting friction between \( B \) and the floor is \( 0.5\times100 = 50 N \), which is greater than \( 25 N \), so nothing moves. \( A \) has no tendency to slide on \( B \), so the friction between \( A \) and \( B \) is zero.

Teacher's Note:
1. Check whether the system moves at all.
2. No tendency to slip means no friction.

Answer: (b) 2.8 N

Explanation:
1. Limiting friction \( = 0.54\times20 = 10.8 N \), which is greater than \( 2.8 N \). The block stays at rest, so static friction equals the applied force, \( 2.8 N \).

Teacher's Note:
1. Static friction matches the applied force up to its limit.
2. Compare with \( \muN \) first.

Answer: (c) 1 kg

Explanation:
1. Constant velocity means zero net force: \( mg = m_Ag\sin37^\circ+\mu m_Ag\cos37^\circ = 6+4 = 10 N \).
2. So \( m = 1 \) kg.

Teacher's Note:
1. Draw a free-body diagram and write \( F \) = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (a) \( 2\tan\phi \)

Explanation:
1. Speed gained on the smooth half: \( v^2 = 2g\sin\phi\cdot\frac L2 \). On the rough half it falls to zero: \( v^2 = 2g(\mu\cos\phi-\sin\phi)\frac L2 \).
2. Equate: \( \sin\phi = \mu\cos\phi-\sin\phi \Rightarrow \mu = 2\tan\phi \).

Teacher's Note:
1. Draw a free-body diagram and write \( F \) = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (b) 10 m/s²

Explanation:
1. \( a = \frac{kx}{m} = \frac{15\times0.2}{0.3} = 10 m/s^{2} \).

Teacher's Note:
1. Draw a free-body diagram and write \( F \) = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (d) 1000 m

Explanation:
1. \( a = \mu g = 5 m/s^{2} \). \( s = \frac{v^2}{2a} = \frac{10000}{10} = 1000 m \).

Teacher's Note:
1. Draw a free-body diagram and write \( F \) = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (c) \( \frac{mF}{m+M} \)

Explanation:
1. The common acceleration is \( a = \frac{F}{m+M} \). The force on \( m \) is \( ma = \frac{mF}{m+M} \).

Teacher's Note:
1. Draw a free-body diagram and write \( F \) = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (c) 4.9 m/s² in the vertical direction

Explanation:
1. \( A's \) plane is at 60° to the horizontal: vertical acceleration \( g\sin^260^\circ = 7.35 m/s^{2} \). \( B's \) plane is at 30° to the horizontal: \( g\sin^230^\circ = 2.45 m/s^{2} \).
2. Relative vertical acceleration \( = 7.35-2.45 = 4.9 m/s^{2} \) (vertical).

Teacher's Note:
1. Draw a free-body diagram and write \( F \) = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (d) \( \frac16 \) m

Explanation:
1. The block is just stable where \( \tan\theta = \mu \): slope \( \frac{dy}{dx} = \frac{x^2}{2} = 0.5 \Rightarrow x = 1 \).
2. \( y = \frac{1}{6} m \).

Teacher's Note:
1. Draw a free-body diagram and write \( F \) = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (b) 120 N

Explanation:
1. Treat \( A \) and \( B \) as one system: the wall's friction must balance the total weight, \( 20+100 = 120 N \) (it is static, below the limit \( 0.15F \) since \( F \) is large).

Teacher's Note:
1. Static friction equals the load to be held.
2. Consider the combined system.

Answer: (c) g/2 and g

Explanation:
1. Before cutting the spring force is \( 3mg \). After cutting, the spring force is unchanged: \( A \) feels \( 3mg-2mg = mg \) upward, so \( a_A = \frac g2 \).
2. \( B \) is in free fall: \( a_B = g \).

Teacher's Note:
1. Draw a free-body diagram and write \( F \) = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (a) \( \frac{F}{2m}\frac{x}{\sqrt{a^2-x^2}} \)

Explanation:
1. Each half of the string has length a and makes angle φ with the pulling direction where \( \sin\phi = \frac xa \). Then \( F = 2T\cos\phi \).
2. A particle's acceleration is \( \frac{T\sin\phi}{m} = \frac{F}{2m}\tan\phi = \frac{F}{2m}\frac{x}{\sqrt{a^2-x^2}} \).

Teacher's Note:
1. Draw a free-body diagram and write \( F \) = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (b) \( \frac{a}{2gk} \)

Explanation:
1. At equilibrium (in the wire's frame) the bead sits where the tangent has \( \tan\theta = \frac ag \).
2. The slope \( \frac{dy}{dx} = 2kx = \frac ag \Rightarrow x = \frac{a}{2gk} \).

Teacher's Note:
1. Draw a free-body diagram and write \( F \) = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (d) 5

Explanation:
1. \( \sin45^\circ+\mu\cos45^\circ = 3(\sin45^\circ-\mu\cos45^\circ) \Rightarrow 4\mu = 2 \), so \( \mu = 0.5 \).
2. \( N = 10\mu = 5 \).

Teacher's Note:
1. Draw a free-body diagram and write \( F \) = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (a) Only (3) and (4)

Explanation:
1. If \( \mu_{1} = 0 \): torque about the foot gives \( N_1L\sin\theta = mg\frac L2\cos\theta \Rightarrow N_1\tan\theta = \frac{mg}{2} \) (4 true, 1 false).
2. With both friction forces at the limit \( N_1 = \mu_2N_2 \) and \( N_2+\mu_1N_1 = mg \Rightarrow N_2 = \frac{mg}{1+\mu_1\mu_2} \) (3 true). \( A \) ladder cannot be in equilibrium with \( \mu_{2} = 0 \).

Teacher's Note:
1. Draw a free-body diagram and write \( F \) = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (c) \( \frac{h}{\ell} = \frac{3\sqrt3}{16} \), \( f = \frac{16\sqrt3}{3} \) N

Explanation:
1. The stick makes 60° with the floor. Vertical: \( N+R\cos60^\circ = 16 \) with \( R = N \Rightarrow N = \frac{32}{3} N \). Horizontal: \( f = R\sin60^\circ = \frac{16\sqrt3}{3} N \).
2. Torque about the foot: \( 16\cdot\frac\ell2\cos60^\circ = R\,d \Rightarrow d = \frac{3\ell}{8} \), and \( h = d\sin60^\circ = \frac{3\sqrt3\ell}{16} \).

Teacher's Note:
1. Draw a free-body diagram and write \( F \) = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (c) 6.30 m

Explanation:
1. \( v_0 = \frac{J}{m} = 2.5 m/s \). \( x = \int_0^\tau v_0e^{-t/\tau}dt = v_0\tau(1-e^{-1}) \).
2. \( x = 2.5\times4\times0.63 = 6.30 m \).

Teacher's Note:
1. Draw a free-body diagram and write \( F \) = ma along each axis.
2. Check limiting friction before assuming a block moves.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Pulling the cloth quickly makes the friction act for a very short time, so the dishes gain almost no momentum (inertia); this is the reason.
2. Newton's third law is true, but it does not explain the effect.

Teacher's Note:
1. Statement 1 is true because of inertia.
2. The third law is correct but unrelated here.