JEE Physics Laws Of Motion MCQs Set 15

Answer: (b) Statement I is true but Statement II is false

Explanation:
1. Statement I is true: for uniform velocity the net force is zero, so the tension equals the weight.
2. Going down with increasing speed means the acceleration is downward: \( mg-N=ma\Rightarrow N=m(g-a)\lt mg \). The floor force is less than the weight, so Statement II is false.

Teacher's Note:
1. An apparent weight greater than the true weight needs an upward acceleration.
2. Uniform velocity means zero acceleration.

Answer: (b) \( 4(\sqrt3+1) \) N

Explanation:
1. Driving forces: \( 4g\sin60^\circ=20\sqrt3\ \text{N} \) and \( 1g\sin30^\circ=5\ \text{N} \).
2. Acceleration \( a=\frac{20\sqrt3-5}{5}=4\sqrt3-1\ \text{m/s}^2 \).
3. For the 1 kg block: \( T-5=1\times a\Rightarrow T=5+4\sqrt3-1=4(\sqrt3+1)\ \text{N} \).

Teacher's Note:
1. The heavier block on the steeper face pulls the lighter block up its incline.
2. Apply Newton's second law to either block.

Answer: (c) 300 N

Explanation:
1. Take torques about the hinge C.
2. Clockwise: weight of the load \( 80\times1.0=80 \) N m and weight of the rod \( 20\times0.5=10 \) N m, total 90 N m.
3. Anticlockwise: vertical component of tension at B \( T\sin30^\circ\times0.6=0.3T \).
4. \( 0.3T=90\Rightarrow T=300\ \text{N} \).

Teacher's Note:
1. Choose the hinge as the pivot so that the unknown hinge force has no torque.
2. The weight of a uniform rod acts at its centre.

Answer: (a) 0.33

Explanation:
1. Pushing up: \( F_1=mg(\sin45^\circ+\mu\cos45^\circ) \). Preventing sliding down: \( F_2=mg(\sin45^\circ-\mu\cos45^\circ) \).
2. With \( F_1=2F_2 \) and \( \sin45^\circ=\cos45^\circ \): \( 1+\mu=2(1-\mu)\Rightarrow3\mu=1\Rightarrow\mu=0.33 \).

Teacher's Note:
1. Friction acts down the plane in the first case and up the plane in the second.
2. At 45° the sine and cosine are equal, which simplifies the algebra.

Answer: (b) \( \frac1{2\sqrt3} \)

Explanation:
1. \( a=g\sin30^\circ-\mu g\cos30^\circ=\frac g4 \).
2. \( \frac12-\mu\frac{\sqrt3}2=\frac14\Rightarrow\mu=\frac{1}{2\sqrt3} \).

Teacher's Note:
1. Net acceleration down the plane = g sinθ - μ g cosθ.
2. μ must be less than tan 30° = 0.577 for the block to slide.

Answer: (b) 5 N

Explanation:
1. After the force ceases the speed is constant: \( v=\frac{50}{10}=5\ \text{m/s} \).
2. During the force \( a=\frac{5}{20}=0.25\ \text{m/s}^2 \), so \( F=ma=20\times0.25=5\ \text{N} \).

Teacher's Note:
1. The speed gained during the first 20 s is the constant speed in the next 10 s.
2. Newton's second law gives the force.

Answer: (d) \( 1-\frac1{n^2} \)

Explanation:
1. Smooth: \( a_1=g\sin45^\circ \). Rough: \( a_2=g\sin45^\circ(1-\mu) \) since sin 45° = cos 45°.
2. For the same distance \( t\propto\frac1{\sqrt a} \): \( n^2=\frac{a_1}{a_2}=\frac1{1-\mu} \).
3. \( \mu=1-\frac1{n^2} \).

Teacher's Note:
1. Use s = at²/2 for the same length of incline.
2. The ratio of the squares of the times is the inverse ratio of the accelerations.

Answer: (a) 20 N

Explanation:
1. Vertical equilibrium at the knot: \( T\cos30^\circ=mg=10\sqrt3 \).
2. \( T=\frac{10\sqrt3}{\sqrt3/2}=20\ \text{N} \).

Teacher's Note:
1. The angle of 30° is measured from the vertical wall, so the vertical component is T cos 30°.
2. The horizontal force F balances the horizontal component T sin 30° = 10 N.

Answer: (c) 800 N

Explanation:
1. The downward component of the push is \( 200\sin30^\circ=100\ \text{N} \).
2. \( N=mg+F\sin30^\circ=700+100=800\ \text{N} \).

Teacher's Note:
1. A downward-directed push adds to the weight; a pull upward would reduce the normal reaction.
2. Resolve the force along the vertical.

Answer: (a) 20 m/s

Explanation:
1. Tension provides the centripetal force: \( T=\frac{mv^2}{r} \).
2. \( v_{max}=\sqrt{\frac{T r}{m}}=\sqrt{\frac{400\times1}{1}}=20\ \text{m/s} \).

Teacher's Note:
1. Gravity is ignored for a horizontal circle with the string in the plane.
2. The maximum tension corresponds to the maximum speed.

Answer: (d) 0.4

Explanation:
1. Retardation \( a=\frac{20}{5}=4\ \text{m/s}^2 \).
2. \( \mu g=a\Rightarrow\mu=\frac4{10}=0.4 \).

Teacher's Note:
1. Friction is the only horizontal force, so a = μ g.
2. The mass cancels out.

Answer: (c) 0.50

Explanation:
1. \( s=\frac12at^2\Rightarrow a=\frac{2\times50}{100}=1\ \text{m/s}^2 \).
2. \( F-\mu mg=ma\Rightarrow30-50\mu=5\Rightarrow\mu=0.5 \).

Teacher's Note:
1. The block starts from rest.
2. Net force = applied force - kinetic friction.

Answer: (b) 25.2 N

Explanation:
1. Normal reaction \( N=mg-F\sin30^\circ=100-0.5F \).
2. Just starting to move: \( F\cos30^\circ=\mu_sN=0.25(100-0.5F) \).
3. \( F(0.866+0.125)=25\Rightarrow F=25.2\ \text{N} \).

Teacher's Note:
1. The upward component of F reduces the normal reaction and therefore the friction.
2. Use the limiting value of static friction.