CBSE Class 9 Physics Force and Laws of Motion MCQs Set 16

Answer: (b) 1.23 m

Explanation:
1. The two strings and the beam form a triangle with sides 1.3, 2.0 and 2.0 m; the depth is the height of the triangle from the weight to the beam.
2. Distance along the beam from the end of the 1.3 m string: \( x=\frac{1.3^2-2.0^2+2.0^2}{2\times2.0}=0.4225 \) m.
3. Depth \( =\sqrt{1.3^2-0.4225^2}\approx1.23 \) m.

Teacher's Note:
1. Treat it as a triangle with three known sides.
2. The depth is the altitude from the weight to the beam.

Answer: (b) \( 4.8\times10^4 \) N

Explanation:
1. Downward acceleration: mg − T = ma.
2. T = m(g − a) = 6000 × 8 = \( 4.8\times10^4 \) N.

Teacher's Note:
1. Going down faster than you can be supported means the tension is less than the weight.
2. Use T = m(g − a).

Answer: (b) 30 N

Explanation:
1. Atwood machine: \( T=\frac{2m_1m_2}{m_1+m_2}g=\frac{2\times2\times6}{8}\times10=30 \) N.
2. Check: a = (6 − 2)(10)/8 = 5 m/s², T = 2(10 + 5) = 30 N.

Teacher's Note:
1. The tension is less than the heavier weight and more than the lighter weight.
2. Use T = 2m₁m₂g/(m₁ + m₂).

Answer: (a) 10 N

Explanation:
1. Each end of the balance is pulled by a tension equal to the weight of 1 kg, which is 10 N.
2. A spring balance reads the force at one end, not the sum of the two forces, so it reads 10 N.

Teacher's Note:
1. A balance reads the tension in it, not the total of both pulls.
2. Equal and opposite pulls of 10 N give a reading of 10 N.

Answer: (d) \( \frac{F-M_1a_1}{M_2} \)

Explanation:
1. On M₁: F − (spring force) = M₁a₁, so the spring force = F − M₁a₁.
2. On M₂ the only horizontal force is the spring force, so \( a_2=\frac{F-M_1a_1}{M_2} \).

Teacher's Note:
1. A spring can stretch or compress, so the two accelerations need not be equal.
2. Apply F = ma to each block separately.

Answer: (c) 49 m/s²

Explanation:
1. The pull is 10 kg-weight = 10 × 9.8 = 98 N.
2. a = F/m = 98/2 = 49 m/s².

Teacher's Note:
1. A reading in kg means kg-weight; multiply by g to get newtons.
2. Then use F = ma.

Answer: (b) Increasing speed at 20 m and decreasing speed at 50 m

Explanation:
1. At 20 m the reading (30 kg) is less than the true weight (32 kg), so the acceleration is downward: the lift is speeding up.
2. At 50 m the reading (36 kg) is more than 32 kg, so the acceleration is upward while the lift moves down: it is slowing down.

Teacher's Note:
1. Reading less than weight means downward acceleration; greater means upward acceleration.
2. Compare the direction of acceleration with the direction of motion (downward).

Answer: (c) 0.1 m/s

Explanation:
1. Work done = 5 × 10⁴ × 3 = 1.5 × 10⁵ J = ½ m v².
2. v² = 2 × 1.5 × 10⁵ / (3 × 10⁷) = 0.01, so v = 0.1 m/s.

Teacher's Note:
1. Use the work–energy theorem, or find a and use v² = 2as.
2. Check units of every quantity before substituting.

Answer: (b) the ground on the horse

Explanation:
1. The horse pushes the ground backward; by Newton's third law the ground pushes (friction) the horse forward.
2. Only an external force on the horse-cart system can accelerate it; this is the friction from the ground.

Teacher's Note:
1. A body cannot accelerate itself with internal forces.
2. The forward force on the horse comes from the ground, not from the cart.

Answer: (a) 6.0 N

Explanation:
1. Normal reaction N = mg − P = 25 − 9 = 16 N.
2. Maximum static friction = 0.4 × 16 = 6.4 N.
3. The applied horizontal force 6.0 N is less than 6.4 N, so the block stays at rest and friction equals 6.0 N.

Teacher's Note:
1. Compare the applied force with the limiting friction.
2. If the applied force is smaller, friction just equals the applied force.

Answer: (a) 2 tan φ

Explanation:
1. Let the length be L. Work–energy from top to bottom: mg L sin φ − μ mg cos φ (L/2) = 0.
2. So μ = 2 tan φ.

Teacher's Note:
1. Gain in kinetic energy over the smooth half is lost over the rough half.
2. Friction acts over only L/2.

Answer: (c) 7 N

Explanation:
1. Friction from the upper block: 0.1 × 1 × 10 = 1 N.
2. Friction from the floor: 0.2 × (1 + 2) × 10 = 6 N.
3. Minimum pull = 1 + 6 = 7 N.

Teacher's Note:
1. The floor supports the weight of both blocks.
2. Add the two friction forces opposing the motion.

Answer: (b) nmv

Explanation:
1. Each bullet loses momentum mv; n bullets hit each second.
2. Force = rate of change of momentum = n m v.

Teacher's Note:
1. The total number N does not matter; the rate n does.
2. Force = momentum lost per second.

Answer: (d) \( \frac{P^2}{2\mu m^2g} \)

Explanation:
1. Speed v = P/m and the retardation is a = μg.
2. Stopping distance \( s=\frac{v^2}{2a}=\frac{P^2}{2\mu m^2 g} \).

Teacher's Note:
1. Convert momentum to speed first.
2. Retardation due to friction is μg.

Answer: (a) 20 N

Explanation:
1. Horizontal: F cos 60° ≤ μN. Vertical: N = mg + F sin 60° = 10√3 + (√3/2)F.
2. F/2 ≤ (1/(2√3))(10√3 + (√3/2)F) = 5 + F/4.
3. So F/4 ≤ 5, F ≤ 20 N.

Teacher's Note:
1. The downward component of F increases the normal reaction.
2. At the limit, F cos θ = μN.

Answer: (c) 49 N

Explanation:
1. Maximum friction = μN = 0.5 × 100 = 50 N, which is more than the weight 5 × 9.8 = 49 N.
2. The block is at rest, so the friction equals the weight: 49 N.

Teacher's Note:
1. Friction adjusts up to its limit; it only needs to balance the weight.
2. Do not use μN unless the block is about to slip.

Answer: (b) 600 N

Explanation:
1. Let each person push with f. Uniform velocity: 2f = friction.
2. Three persons: 3f − friction = 1500 × 0.2 = 300, so f = 300 N and friction = 600 N.

Teacher's Note:
1. Uniform velocity means zero net force.
2. Write two equations and solve for f.

Answer: (d) Mg

Explanation:
1. The block is in equilibrium, so the total force by the plane (normal reaction + friction) balances the weight.
2. Its magnitude is Mg.

Teacher's Note:
1. Force by the plane is the resultant of normal reaction and friction.
2. Equilibrium means this resultant is equal and opposite to the weight.

Answer: (a) f increases linearly with t, then stays constant

Explanation:
1. Before sliding, static friction equals the applied force, so it grows linearly with time.
2. At the limit it equals μN; once sliding starts, kinetic friction = μN (the same value), which is constant.

Teacher's Note:
1. Static friction rises with the applied force up to μsN.
2. Kinetic friction is constant and here equals the limiting value.

Answer: (a) The resultant of the pushing force, weight of the toy, normal force by the ground on the toy and the frictional force is zero

Explanation:
1. The toy stays at rest, so the vector sum of all external forces on it is zero.
2. The normal force is not equal to the weight because F has a vertical component, and F is not balanced by friction alone, since F also has a vertical component.
3. Internal forces occur in equal and opposite pairs and cannot balance an external force.

Teacher's Note:
1. At rest means zero resultant external force.
2. Check each option by resolving F into components.