JEE Physics Laws Of Motion MCQs Set 23

Answer: (b) \( \dfrac{6mg}{5\sqrt2} \)

Explanation:
1. For P (moving up the smooth face): \( T-mg\sin45^\circ=ma \).
2. \( T=m\left(\dfrac{g}{5\sqrt2}+\dfrac{g}{\sqrt2}\right)=\dfrac{6mg}{5\sqrt2} \).

Teacher's Note:
1. Use the block on the smooth face because no friction acts on it.
2. The friction on the rough face does not enter the answer.

Answer: (b) Only (1) and (3)

Explanation:
1. Let the string tension at \( m_1 \) be \( T_2 \) and at \( m_2 \) be \( T_1 \), with \( T_2-T_1=10 \).
2. \( 40-T_2=4a \) and \( T_1-20=2a \), giving \( 10=6a \), so \( a=\tfrac53 \) m/s² and \( T_1=\tfrac{70}3 \) N.
3. The tension differs on the two sides of the slit, and after interchange the acceleration is again 5/3 m/s², not 10/3.

Teacher's Note:
1. Friction at the slit makes the string tension unequal on the two sides.
2. Check statement (4) by redoing the equations with the friction direction reversed.

Answer: (b) \( g\sin\alpha\left(1+\dfrac Mm\right) \) down the incline

Explanation:
1. The plank is at rest, so the friction from the man on it is \( Mg\sin\alpha \), up the incline.
2. So the friction on the man is \( Mg\sin\alpha \) down the incline.
3. \( ma=mg\sin\alpha+Mg\sin\alpha \), so \( a=g\sin\alpha(1+M/m) \) down the incline.

Teacher's Note:
1. The supports are frictionless, so the man's friction alone balances the plank's weight component.
2. Newton's third law gives the opposite force on the man.

Answer: (a) 12.5 m

Explanation:
1. Retardation = \( \mu g=1 \) m/s².
2. \( s=\dfrac{v^2}{2\mu g}=\dfrac{25}{2}=12.5 \) m.

Teacher's Note:
1. The mass does not appear in the answer.
2. Use \( v^2=u^2+2as \) with a negative acceleration.

Answer: (d) 2 s

Explanation:
1. \( a_A=g(\sin45^\circ-0.2\cos45^\circ)=\dfrac{8}{\sqrt2} \) and \( a_B=\dfrac{7}{\sqrt2} \) m/s².
2. Relative acceleration = \( \dfrac1{\sqrt2} \) m/s².
3. \( \sqrt2=\tfrac12\cdot\dfrac1{\sqrt2}t^2 \), so \( t^2=4 \) and t = 2 s.

Teacher's Note:
1. The relative motion has constant relative acceleration.
2. A catches up with B when the relative displacement equals the initial gap.

Answer: (b) 3.2 m/s²

Explanation:
1. Check if they move together: the pulley forces on the 15 kg block cancel, so the external horizontal force on the system is F, giving \( a=\dfrac{80}{25}=3.2 \) m/s².
2. Friction needed on the 10 kg block: \( 80-10(3.2)=48 \) N, less than the limit 60 N, so there is no slipping.
3. Hence the acceleration of the 10 kg block is 3.2 m/s².

Teacher's Note:
1. Test the no-slip assumption first by comparing the friction required with 0.6 × 100 N.
2. Friction is the only force that moves the 15 kg block.

Answer: (a) 20 m/s²

Explanation:
1. Normal reaction N = ma, and friction must balance the weight: \( \mu N=mg \).
2. \( \mu ma=mg \), so \( a=g/\mu=20 \) m/s².

Teacher's Note:
1. A smaller coefficient of friction needs a larger acceleration.
2. The mass cancels.

Answer: (c) \( \dfrac{mg}{\sin\theta+\mu\cos\theta} \)

Explanation:
1. Normal reaction \( N=F\cos\theta \).
2. For the least F the block is on the point of slipping down, so friction acts up: \( F\sin\theta+\mu F\cos\theta=mg \).
3. \( F=\dfrac{mg}{\sin\theta+\mu\cos\theta} \).

Teacher's Note:
1. At minimum F friction is limiting and acts upward.
2. The component F sinθ and friction together carry the weight.

Answer: (d) 2 m/s² towards the right

Explanation:
1. The string tension is 80 N, pulling the 10 kg block to the right; the net horizontal string force on the 15 kg block's pulleys is 80 N to the left.
2. With no slipping the acceleration would be zero, needing 80 N of friction, more than the limit 60 N, so slipping occurs.
3. 10 kg block: \( 80-60=10a \), so \( a=2 \) m/s² to the right.

Teacher's Note:
1. Slipping is certain once the required friction exceeds 60 N.
2. The 15 kg block moves the other way, at 4/3 m/s².

Answer: (c) remains unchanged

Explanation:
1. The coefficient of friction depends only on the nature of the two surfaces.
2. Friction doubles with N, but the ratio \( f/N \) stays the same.

Teacher's Note:
1. Friction force is proportional to N, the coefficient is the constant of proportionality.
2. It does not depend on the area of contact either.

Answer: (c) \( \dfrac{kt-\mu m_2g}{m_2} \)

Explanation:
1. After slipping, kinetic friction \( \mu m_2g \) acts backward on the bar.
2. \( m_2a_2=kt-\mu m_2g \).

Teacher's Note:
1. The plank's acceleration is \( \mu m_2g/m_1 \) and constant.
2. The bar's acceleration keeps increasing with time.

Answer: (d) 49°

Explanation:
1. Length of slope \( s=\ell/\cos\alpha \), acceleration \( a=g(\sin\alpha-k\cos\alpha) \).
2. \( t^2=\dfrac{4\ell}{g[\sin2\alpha-k(1+\cos2\alpha)]} \), least when \( \tan2\alpha=-\dfrac1k \).
3. \( 2\alpha=180^\circ-82^\circ=98^\circ \), so \( \alpha\approx49^\circ \).

Teacher's Note:
1. Differentiate the denominator with respect to \( 2\alpha \).
2. The least time is for a steeper slope than 45° because friction slows a shallow slide.

Answer: (c) 1.0 s

Explanation:
1. At \( \alpha=49^\circ \): \( \sin2\alpha=0.990 \), \( \cos2\alpha=-0.139 \).
2. \( t^2=\dfrac{4(2.10)}{9.8[0.990-0.14(0.861)]}=\dfrac{8.4}{8.53}\approx0.985 \), so \( t\approx1.0 \) s.

Teacher's Note:
1. Substitute the optimum angle into the time formula.
2. The result is close to 1 s.

Answer: (c) 0.24

Explanation:
1. After sliding starts: \( 1.8t-\mu_k(1.5)(10)=1.5(1.2t-2.4) \).
2. Comparing the constant terms: \( -15\mu_k=-3.6 \), so \( \mu_k=0.24 \).

Teacher's Note:
1. Match coefficients of t on both sides to check the algebra.
2. Kinetic friction is constant once sliding begins.

Answer: (b) \( \sqrt5 \) s

Explanation:
1. Going down: \( a=g\sin37^\circ-\mu g\cos37^\circ=6-4=2 \) m/s².
2. \( 5=\tfrac12(2)t^2 \), so \( t=\sqrt5 \) s.

Teacher's Note:
1. The descent acceleration is smaller than the ascent retardation.
2. So the descent takes longer than the ascent, which lasts 1 s.

Answer: (d) Only (1) and (3)

Explanation:
1. Along the plane: \( f-mg\sin30^\circ=ma \), so \( f=5+1=6 \) N.
2. Normal reaction \( N=mg\cos30^\circ=5\sqrt3 \) N.
3. Contact force \( =\sqrt{6^2+(5\sqrt3)^2}=\sqrt{111}\approx10.5 \) N.

Teacher's Note:
1. The block moves up only because of friction.
2. Contact force combines both the normal reaction and the friction.

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

Explanation:
1. Assume A and B move together: \( 60-T=6a \), \( T-18-T'=9a \), \( T'-10=a \) (18 N is friction of table on B).
2. Adding the three equations: \( 60-18-10=16a \), so \( 32=16a \) and a = 2 m/s².
3. Then T = 48 N and the friction needed on A is \( T-12=36 \) N, less than the limit 42 N, so no slipping; T' = 12 N.

Teacher's Note:
1. Check slipping by comparing the friction needed with the limiting value 42 N.
2. The table friction on B is 0.2 times the total weight of A and B.

Answer: (a) 0.75

Explanation:
1. For the same distance, \( t\propto1/\sqrt a \), so doubling the time reduces the acceleration to one quarter.
2. \( g\sin45^\circ-\mu g\cos45^\circ=\tfrac14g\sin45^\circ \), so \( 1-\mu=\tfrac14 \).
3. \( \mu=0.75 \).

Teacher's Note:
1. At 45° the sine and cosine terms are equal, which simplifies the algebra.
2. Time is inversely proportional to the square root of acceleration for a given distance.

Answer: (d) \( \mu mg \)

Explanation:
1. The normal reaction is mg, so the limiting friction is \( \mu mg \).
2. The pull must at least equal this to start (or maintain) motion.

Teacher's Note:
1. A horizontal pull does not change the normal reaction.
2. A pull at an angle can be smaller, as the next type of problem shows.

Answer: (a) 0.33

Explanation:
1. Weight of the hanging part = \( \tfrac14Mg \) and the normal reaction on the table part = \( \tfrac34Mg \).
2. \( \tfrac14Mg=\mu\tfrac34Mg \), so \( \mu=\tfrac13 \approx0.33 \).

Teacher's Note:
1. Only the part on the table has friction.
2. The hanging weight is the driving force.

Answer: (b) \( \dfrac{mg}{\sin\theta} \)

Explanation:
1. With no friction the weight is balanced by the vertical component of F alone.
2. \( F\sin\theta=mg \), so \( F=\dfrac{mg}{\sin\theta} \).

Teacher's Note:
1. The wall's normal reaction balances \( F\cos\theta \) horizontally.
2. Zero friction is a special value between the minimum and maximum forces.

Answer: (a) \( t_a<t_d \)

Explanation:
1. Going up, gravity and friction both act down the plane; coming down, friction opposes gravity.
2. So the acceleration is larger going up, and for the same distance the ascent takes less time.

Teacher's Note:
1. On a smooth plane \( t_a=t_d \).
2. The kinetic energy lost to friction makes the return speed smaller.