JEE Physics Laws Of Motion MCQs Set 18

Answer: (a) \( \frac{a\tau^3}{6} \)

Explanation:
1. \( p=\int_0^\tau at(\tau-t)\,dt=a\left[\frac{\tau t^2}{2}-\frac{t^3}{3}\right]_0^\tau \).
2. \( =a\tau^3\left(\frac12-\frac13\right)=\frac{a\tau^3}{6} \).

Teacher's Note:
1. Momentum gained equals the impulse.
2. Integrate the force over the time for which it acts.

Answer: (c) \( \frac{a\tau^4}{12m} \)

Explanation:
1. Velocity \( v=\frac1m\int_0^t at(\tau-t)\,dt=\frac am\left(\frac{\tau t^2}{2}-\frac{t^3}{3}\right) \).
2. \( x=\int_0^\tau v\,dt=\frac am\left(\frac{\tau^4}{6}-\frac{\tau^4}{12}\right)=\frac{a\tau^4}{12m} \).

Teacher's Note:
1. First find v(t), then integrate again.
2. The upper limit is the time τ for which the force acts.

Answer: (d) \( \frac{4u}{3} \)

Explanation:
1. The string is slack until B has dropped as far as A has moved: \( \frac12gt^2=ut\Rightarrow t=\frac{2u}{g} \), when B's speed is 2u.
2. The jerk gives both particles the same speed v. Equal and opposite impulses J: \( 2m(v-u)=J \) and \( m(2u-v)=J \).
3. \( 2v-2u=2u-v\Rightarrow v=\frac{4u}{3} \).

Teacher's Note:
1. During the sudden jerk the weight of B is negligible compared with the impulsive tension.
2. The impulse on A and on B is the same since the string is massless.

Answer: (c) \( \frac{\sqrt{4u^2+6gl}}{3} \)

Explanation:
1. After the jerk the speed is \( v_0=\frac{4u}{3} \) and A has already covered \( ut=\frac{2u^2}{g} \), leaving \( l-\frac{2u^2}{g} \) to the pulley.
2. The common acceleration of the 3m system is \( \frac{mg}{3m}=\frac g3 \).
3. \( v^2=\frac{16u^2}{9}+2\cdot\frac g3\left(l-\frac{2u^2}{g}\right)=\frac{4u^2+6gl}{9} \).

Teacher's Note:
1. After the string tightens, B pulls both masses with acceleration g/3.
2. Use v² = v0² + 2as over the remaining distance.

Answer: (a) 0 and 0

Explanation:
1. The pulley is massless, so the tension in the string is \( T=\frac F2=49\ \text{N} \).
2. The weights are \( 196\ \text{N} \) (A) and \( 98\ \text{N} \) (B); the tension is smaller than both, so neither block leaves the floor.

Teacher's Note:
1. Tension is half the pulling force on a massless movable pulley.
2. A block lifts only if the tension exceeds its weight.

Answer: (a) 9.81 s

Explanation:
1. Tension in the string \( T=\frac F2=t \) (in newton).
2. m1 lifts when \( T=m_1g=9.81\ \text{N} \), i.e. at t = 9.81 s.

Teacher's Note:
1. A massless pulley has tension equal to half the applied force.
2. The lighter mass lifts first.

Answer: (a) 19.62 s

Explanation:
1. m2 lifts when \( T=m_2g=19.62\ \text{N} \), and T = t, so t = 19.62 s.

Teacher's Note:
1. The condition is the same as for m1 but with m2g.
2. Between 9.81 s and 19.62 s only m1 moves.

Answer: (c) 15.19 m/s² and 2.69 m/s²

Explanation:
1. At t = 25 s the tension is 25 N, which is more than both weights, so both masses accelerate upward.
2. \( a_1=\frac{25-9.81}{1}=15.19\ \text{m/s}^2 \), \( a_2=\frac{25-19.62}{2}=2.69\ \text{m/s}^2 \).

Teacher's Note:
1. Apply Newton's second law to each mass separately.
2. The same tension acts on both.

Answer: (d) A is false, but R is true.

Explanation:
1. Impulse is force multiplied by time, while momentum is mass multiplied by velocity; they are different physical quantities, so A is false.
2. Their units, N s and kg m/s, are the same, so R is true.

Teacher's Note:
1. Same units do not imply the same physical quantity.
2. Impulse equals the change in momentum.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. For zero net force the two forces must be equal and opposite, and for zero torque they must be collinear, so equilibrium is possible.
2. R states exactly the condition that makes A true.

Teacher's Note:
1. Equilibrium needs zero net force and zero net torque.
2. Two equal and opposite forces with different lines of action give a couple.

Answer: (d) A is false, but R is true.

Explanation:
1. Action and reaction act on different bodies, so they can never cancel each other on one body: A is false.
2. The block is at rest because its weight and the normal reaction (not an action-reaction pair) cancel, so the net force is zero: R is true.

Teacher's Note:
1. Newton's third-law pairs act on two different bodies.
2. Weight and normal reaction are balanced forces on the same body.

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

Explanation:
1. The dishes tend to stay at rest because of inertia, and friction acts for only a very short time, so A is true.
2. R is also a true statement, but inertia, not the third law, is the explanation.

Teacher's Note:
1. Newton's first law explains the dishes staying in place.
2. A true reason is not always the correct reason.

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

Explanation:
1. When pulling at an angle, the vertical component of the force lifts partly, reducing the normal reaction and so the friction; when pushing, the normal reaction increases. A is true.
2. R is also true, but the reason is the change in the normal reaction, not the nature of the surfaces.

Teacher's Note:
1. Friction depends on both μ and the normal reaction.
2. The same pair of surfaces is used in pushing and pulling.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Burnt-out stages are discarded so that the remaining rocket has less mass to accelerate, allowing the high orbital velocity to be reached.
2. A single stage cannot reach that velocity, which is exactly why multistage rockets are needed.

Teacher's Note:
1. Dropping empty tanks reduces the mass carried.
2. Orbital speed near the Earth is about 8 km/s.

Answer: (d) A is false, but R is true.

Explanation:
1. The apparent weight depends on the acceleration, not on the velocity. A lift moving upward at constant velocity gives the same reading as at rest, so A is false.
2. R is the correct expression for an upward acceleration.

Teacher's Note:
1. Apparent weight changes only when the lift accelerates.
2. Upward acceleration increases it, downward acceleration decreases it.

Answer: (a) \( \frac{F_1}{m} \)

Explanation:
1. For equilibrium the resultant of F2 and F3 is equal and opposite to F1.
2. Removing F1 leaves a net force of magnitude F1, so \( a=\frac{F_1}{m} \).

Teacher's Note:
1. The remaining forces always add up to the removed force reversed.
2. The perpendicularity is not needed for the answer.

Answer: (d) 2 N

Explanation:
1. Normal reaction \( N=10\ \text{N} \), so limiting friction \( =0.2\times10=2\ \text{N} \).
2. Just holding means friction equals the weight: weight = 2 N.

Teacher's Note:
1. 'Just hold' means friction is at its limiting value.
2. The pressing force is the normal reaction.

Answer: (d) 0.06

Explanation:
1. Deceleration \( a=\frac{6}{10}=0.6\ \text{m/s}^2 \).
2. \( \mu=\frac{a}{g}=\frac{0.6}{10}=0.06 \) (taking g = 10 m/s²).

Teacher's Note:
1. Friction is the only horizontal force.
2. The mass cancels out.

Answer: (c) \( g\tan\alpha \)

Explanation:
1. Forces on the block: weight mg and the normal reaction N. For a horizontal acceleration a: \( N\sin\alpha=ma \) and \( N\cos\alpha=mg \).
2. \( a=g\tan\alpha \).

Teacher's Note:
1. Resolve the normal reaction into horizontal and vertical parts.
2. The incline must accelerate in the direction in which it rises.

Answer: (b) 120 N

Explanation:
1. Treat A and B as a single system at rest: the only vertical forces are their total weight and the friction from the wall.
2. \( f=20+100=120\ \text{N} \) (upward).

Teacher's Note:
1. The wall is the only external support, so it carries the total weight.
2. Static friction need not equal the limiting value.

Answer: (b) \( -\frac{\alpha v^2}{M} \)

Explanation:
1. Momentum conservation with the dust gaining speed v: \( F=-v\frac{dM}{dt}=-\alpha v^2 \).
2. \( a=\frac FM=-\frac{\alpha v^2}{M} \).

Teacher's Note:
1. Dust initially at rest must be given speed v, taking momentum from the satellite.
2. The retarding force is v dM/dt.

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

Explanation:
1. Each half of the string makes an angle φ with the line joining the particles, where \( \cos\phi=\frac xa \) and \( \sin\phi=\frac{\sqrt{a^2-x^2}}{a} \).
2. Balance of the mid-point: \( F=2T\sin\phi\Rightarrow T=\frac{F}{2\sin\phi} \).
3. Acceleration of a particle: \( a=\frac{T\cos\phi}{m}=\frac{F}{2m}\cot\phi=\frac{F}{2m}\frac{x}{\sqrt{a^2-x^2}} \).

Teacher's Note:
1. The string tension acts along the string, so only its horizontal component moves the particle.
2. The mid-point of the string is massless, so the forces on it balance.

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

Explanation:
1. In the wire's frame the bead feels a pseudo force ma opposite to a, and its weight mg. At equilibrium the resultant is normal to the wire, so the tangent slope satisfies \( \frac{dy}{dx}=\frac{a}{g} \).
2. \( 2kx=\frac ag\Rightarrow x=\frac{a}{2gk} \).

Teacher's Note:
1. Use the frame of the wire; the effective gravity is tilted.
2. The slope of the tangent equals the ratio of the pseudo force to the weight.

Answer: (b) 5

Explanation:
1. \( mg(\sin45^\circ+\mu\cos45^\circ)=3mg(\sin45^\circ-\mu\cos45^\circ) \).
2. At 45° sin and cos cancel: \( 1+\mu=3-3\mu\Rightarrow\mu=\frac12 \), so N = 10μ = 5.

Teacher's Note:
1. Write both limiting-friction conditions and divide.
2. At 45° the trigonometric factors cancel.

Answer: (a) \( \frac{F_0}{mb}\left(1-e^{-bt}\right) \)

Explanation:
1. \( m\frac{dv}{dt}=F_0e^{-bt}\Rightarrow v=\frac{F_0}{m}\int_0^te^{-bt}\,dt \).
2. \( v=\frac{F_0}{mb}\left(1-e^{-bt}\right) \), which rises from zero and approaches F0/(mb).

Teacher's Note:
1. Integrate the acceleration with v = 0 at t = 0.
2. Option (b) would give a non-zero speed at t = 0.