JEE Physics Laws Of Motion MCQs Set 29

Answer: (b) \( -\frac{\vec F_1}{m} \)

Explanation:
1. In equilibrium \( \vec F_1 + \vec F_2 + \dots + \vec F_n = 0 \).
2. After removing \( \vec F_1 \) the net force is \( \vec F_2+\dots+\vec F_n = -\vec F_1 \), so \( \vec a = -\vec F_1/m \).

Teacher's Note:
1. The remaining forces add to the negative of the removed force.
2. Apply Newton's second law to the net force.

Answer: (d) by spitting or sneezing

Explanation:
1. With no external horizontal force the momentum of you plus the expelled matter is conserved.
2. Throwing something horizontally gives you a recoil in the opposite direction.

Teacher's Note:
1. Internal forces cannot change total momentum, but recoil can move you.
2. Jumping, running or rolling all need friction.

Answer: (c) Newton's first law

Explanation:
1. Your body tends to continue its state of motion (or rest) when the bus changes its motion.
2. This property is inertia, described by the first law.

Teacher's Note:
1. Inertia explains tendencies to continue motion.
2. The bus changes state of motion but your body resists the change.

Answer: (c) force on the ground

Explanation:
1. Action and reaction act on different bodies, so they do not cancel.
2. The horse pushes the ground backward; the ground pushes the horse forward and this forward force, exceeding friction on the cart, accelerates the system.

Teacher's Note:
1. Reaction of the ground drives the horse.
2. Forces on different bodies never cancel each other.

Answer: (d) Newton's first law

Explanation:
1. The carpet is suddenly moved while the dust particles tend to remain at rest because of inertia.
2. So the dust separates from the carpet.

Teacher's Note:
1. Inertia of rest.
2. The same idea explains shaking a wet umbrella.

Answer: (b) decreases

Explanation:
1. The parallel component is \( mg\sin\theta \), which decreases as θ decreases to zero.

Teacher's Note:
1. Parallel component is mg sin θ and perpendicular component mg cos θ.
2. As θ falls, sin θ falls.

Answer: (c) 300 N, 1000 N

Explanation:
1. Constant velocity means zero net force in each direction.
2. Horizontal: P = 300 N. Vertical: 2S = 2000 N, so S = 1000 N.

Teacher's Note:
1. Zero acceleration means balanced forces.
2. The two S forces share the weight equally.

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

Explanation:
1. Horizontal balance: \( T_1\cos30^\circ = T_2\cos60^\circ \), so \( T_1 = T_2/\sqrt3 \).
2. Vertical balance: \( T_1\sin30^\circ + T_2\sin60^\circ = 50 \) gives \( T_2\cdot\frac{2}{\sqrt3} = 50 \), so \( T_2 = 25\sqrt3 \) N and \( T_1 = 25 \) N.

Teacher's Note:
1. Resolve into horizontal and vertical components.
2. The string at the larger angle with the horizontal carries more tension.

Answer: (b) 30 N

Explanation:
1. \( a = \frac{60}{10+20+30} = 1 \) m/s².
2. \( T_2 \) accelerates the first two blocks: \( T_2 = (10+20)\times1 = 30 \) N.

Teacher's Note:
1. Find the common acceleration first.
2. The tension pulls only the blocks behind it.

Answer: (d) 6400

Explanation:
1. Upward air force F: \( Mg - F = M(5) \), so \( F = 9600\times5 = 48000 \) N.
2. After the load is removed: \( F - (M-m)g = (M-m)\times5 \), so \( 48000 = 15(M-m) \), \( M-m = 3200 \) and \( m = 6400 \) kg.

Teacher's Note:
1. The upward force of the air is the same in both cases.
2. Write Newton's second law before and after.

Answer: (c) \( g\cot\theta \)

Explanation:
1. Block must stay in contact with the surface but exert no force on it: the wedge must retreat at the rate the block drops.
2. During time t the block falls \( \tfrac12gt^2 \) and the wedge moves \( \tfrac12at^2 \); contact along the incline requires \( \tan\theta = \frac{g}{a} \), so \( a = g\cot\theta \).

Teacher's Note:
1. Free fall means zero normal force.
2. Use the geometry of the incline for the displacement ratio.

Answer: (c) The string is perpendicular to the incline, so the tension is mg cos θ

Explanation:
1. The bob also accelerates down the plane at \( g\sin\theta \), the same as the trolley.
2. Along the incline \( mg\sin\theta \) alone gives that acceleration, so the string tension has no component along the incline: it is perpendicular to the plane with \( T = mg\cos\theta \).

Teacher's Note:
1. No relative acceleration means the string supplies only the perpendicular balance.
2. This is like an object sliding on a smooth incline.

Answer: (d) zero

Explanation:
1. Both bodies fall with acceleration g, so there is no relative acceleration.
2. The block feels weightlessness relative to the wedge: no normal force.

Teacher's Note:
1. Free fall gives apparent weightlessness.
2. Contact force vanishes when both accelerate at g.

Answer: (b) \( (M+m)g\tan\beta \)

Explanation:
1. For m to stay at rest on the smooth incline its acceleration must be \( a = g\tan\beta \).
2. Applying P to the whole system, \( P = (M+m)a = (M+m)g\tan\beta \).

Teacher's Note:
1. Both bodies share the same acceleration.
2. Use the incline normal to supply the horizontal force on m.

Answer: (c) \( \frac{mMg}{nm+M} \)

Explanation:
1. Common acceleration: \( a = \frac{Mg}{nm+M} \).
2. The string to the n-th (last) block pulls only that block: \( T = ma = \frac{mMg}{nm+M} \).

Teacher's Note:
1. Find the common acceleration of the whole train.
2. The end block is pulled only by one string.

Answer: (a) less than that in case II

Explanation:
1. Case I: \( a = \frac{2m-m}{2m+m}g = \frac g3 \).
2. Case II: \( a = \frac{2mg - mg}{m} = g \), so \( a_I < a_{II} \).

Teacher's Note:
1. A force of 2mg pulls harder than a hanging 2m mass would.
2. In case I the extra mass also has to be accelerated.

Answer: (b) 13 N (approximately)

Explanation:
1. System: \( a = \frac{(4-2)g}{6} = \frac g3 \) (B and C move down).
2. For C: \( T = m(g-a) = 2\times\frac{2g}{3} = 13.07 \) N, about 13 N.

Teacher's Note:
1. Treat B and C as one body for the acceleration.
2. Then isolate C.

Answer: (d) less than 6 kg

Explanation:
1. String tension \( T = \frac{2m_1m_2g}{m_1+m_2} = \frac{10}{6}g \).
2. Balance holds the pulley: reading \( = 2T = \frac{10g}{3} \), i.e. about 3.3 kg-wt, which is less than 6 kg (the total weight when at rest).

Teacher's Note:
1. Accelerating masses reduce the support force.
2. The balance reads 2T.

Answer: (c) 10 m/s²

Explanation:
1. Before: \( 2T\sin30^\circ = 20 \), so \( T = 20 \) N in each spring.
2. Immediately after: the left spring still pulls with 20 N at 30° to the ceiling. Net force = 20 N weight plus spring force; components: horizontal \( 20\cos30^\circ = 17.3 \) N, vertical \( 20 - 20\sin30^\circ = 10 \) N, net \( = 20 \) N, so \( a = 10 \) m/s².

Teacher's Note:
1. Spring force cannot change instantaneously.
2. Cut-string and cut-spring problems: other forces stay the same.

Answer: (c) 12000 N

Explanation:
1. Between 10 s and 12 s the deceleration is \( \frac{3.6}{2} = 1.8 \) m/s².
2. \( T = m(g-a) = 1500\times(9.8-1.8) = 12000 \) N.

Teacher's Note:
1. During retardation of an upward lift the tension is less than the weight.
2. Read the slope of the speed graph.

Answer: (d) 3g/4

Explanation:
1. \( N = M(g+a) = \frac{7Mg}{4} \).
2. So \( a = \frac{3g}{4} \).

Teacher's Note:
1. Apparent weight in an upward-accelerating lift is M(g + a).
2. Solve for a.

Answer: (b) The greater the mass of a body, the less its acceleration for a given force.

Explanation:
1. Only \( a = F/m \) holds universally.
2. Spring balance elongation depends on weight (location) and beam balance comparison depends on equal g on both pans, not an 'absolute' statement.

Teacher's Note:
1. Inertial mass appears in F = ma everywhere.
2. Weight-based statements depend on the local g.

Answer: (c) \( r\left[1-\frac{1}{\sqrt{1+\mu^2}}\right] \)

Explanation:
1. At the limit \( \mu = \tan\phi \), where φ is the angle of the radius with the vertical. Then \( \cos\phi = \frac{1}{\sqrt{1+\mu^2}} \).
2. Height above the bottom \( h = r - r\cos\phi = r\left[1-\frac1{\sqrt{1+\mu^2}}\right] \).

Teacher's Note:
1. Treat the bowl surface as a local incline of angle φ.
2. Insect stays if tan φ ≤ μ.

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

Explanation:
1. For small F the block is at rest and f = F (slope 45°).
2. When F exceeds the limiting value the block slides, f drops slightly to the kinetic value and stays constant (parallel to the F axis); the drop gives a small kink.

Teacher's Note:
1. Static friction is self-adjusting; kinetic is nearly constant.
2. Limiting friction is slightly more than kinetic friction.

Answer: (b) first remains constant, then decreases

Explanation:
1. Contact force = resultant of normal reaction and friction.
2. While the block is at rest it balances the weight, so it equals mg. After sliding starts, friction is kinetic and the resultant is less than mg and keeps decreasing as θ approaches 90° (it vanishes).

Teacher's Note:
1. Static equilibrium: contact force = weight.
2. At 90° the block falls freely and the contact force is zero.

Answer: (d) 9.8 N

Explanation:
1. \( \tan30^\circ = 0.577 < 0.7 \), so the block stays at rest.
2. Static friction equals the component along the plane: \( mg\sin30^\circ = 2\times9.8\times0.5 = 9.8 \) N.

Teacher's Note:
1. Compare tan θ with μ first.
2. When at rest, friction is not μN.

Answer: (c) upward and towards the left (up and backward)

Explanation:
1. Train decelerating means its acceleration points left (backward). The box stays at rest on the floor, so the friction on it points left.
2. Normal reaction is vertical up, friction is horizontal left, so R is up and to the left.

Teacher's Note:
1. Friction supplies the horizontal acceleration of the box.
2. The resultant reaction is the vector sum of N and f.

Answer: (c) \( \frac Fm(\cos\theta+\mu\sin\theta)-\mu g \)

Explanation:
1. Normal reaction \( N = mg - F\sin\theta \).
2. \( ma = F\cos\theta - \mu N = F\cos\theta - \mu mg + \mu F\sin\theta \), giving \( a = \frac Fm(\cos\theta+\mu\sin\theta)-\mu g \).

Teacher's Note:
1. The vertical component of F reduces the normal force and the friction.
2. Resolve vertically first.