JEE Physics Friction MCQs Set 01

Answer: (c) \( \mu = \tan\lambda \)

Explanation:
1. At limiting friction \( f = \mu R \), and the resultant of f and R makes angle \( \lambda \) with R.
2. \( \tan\lambda = f/R = \mu \).

Teacher's Note:
1. Angle of friction is the angle between normal reaction and the resultant contact force.
2. It equals the angle of repose.

Answer: (d) 0.1

Explanation:
1. Limiting friction \( = \mu mg \).
2. \( \mu = \dfrac{98}{100\times9.8} = 0.1 \).

Teacher's Note:
1. Force to just start motion equals the limiting friction.
2. Use \( g = 9.8\ \text{m/s}^2 \).

Answer: (b) is independent of the area of the surfaces in contact

Explanation:
1. Limiting friction \( = \mu_s R \) depends on the nature of surfaces and normal reaction only.
2. It does not depend on the apparent area of contact.

Teacher's Note:
1. Friction obeys the laws: independent of area (in the ideal model), proportional to normal reaction.

Answer: (a) limiting friction

Explanation:
1. Static friction adjusts up to a maximum value, called limiting friction, beyond which sliding starts.

Teacher's Note:
1. Static friction is self-adjusting, kinetic friction is nearly constant.

Answer: (c) not move

Explanation:
1. Limiting friction \( = 0.4\times10 = 4 \) N.
2. The applied force of 3 N is less than 4 N, so static friction (3 N) balances it and the block stays at rest.

Teacher's Note:
1. Compare the applied force with the limiting friction, not with the kinetic friction.

Answer: (a) 15 kg

Explanation:
1. Tension in the string \( = 5g \) (system at rest).
2. Limiting friction \( = 0.2(10+m_C)g \ge 5g \Rightarrow 10+m_C \ge 25 \).
3. \( m_C = 15 \) kg.

Teacher's Note:
1. Adding mass on A increases the normal reaction and hence the friction.

Answer: (d) All of the above

Explanation:
1. Ball bearings replace sliding by rolling, lubrication fills irregularities and polishing smooths them.
2. All three reduce friction.

Teacher's Note:
1. Rolling friction is much smaller than sliding friction.

Answer: (c) \( \dfrac{\mu l}{1+\mu} \)

Explanation:
1. Let mass per length be \( \lambda \). Hanging weight \( = \lambda l_1 g \); friction on the table part \( = \mu\lambda(l-l_1)g \).
2. Limiting case: \( l_1 = \mu(l-l_1) \Rightarrow l_1 = \dfrac{\mu l}{1+\mu} \).

Teacher's Note:
1. Only the portion lying on the table gives normal reaction.
2. The hanging part pulls the rest.

Answer: (c) Rolling friction is greater than sliding friction

Explanation:
1. Rolling friction is much smaller than sliding friction (this is why wheels were invented).
2. So statement (c) is false.

Teacher's Note:
1. Rolling < sliding < static (limiting) is the usual order.

Answer: (b) 49 N

Explanation:
1. Friction must balance weight: \( \mu F = mg \).
2. \( F = \dfrac{1\times9.8}{0.2} = 49 \) N.

Teacher's Note:
1. Here the normal reaction equals the applied force F.

Answer: (a) 20%

Explanation:
1. Fraction hanging \( x = \dfrac{\mu}{1+\mu} = \dfrac{0.25}{1.25} = 0.2 \).
2. That is 20%.

Teacher's Note:
1. Weight of hanging part = limiting friction of the part on the table.

Answer: (a) 2 kg

Explanation:
1. To start motion \( m_Bg = \mu m_Ag \).
2. \( m_B = 0.2\times10 = 2 \) kg.

Teacher's Note:
1. Tension equals the weight of B at the threshold.

Answer: (d) all of the above are possible

Explanation:
1. Kinetic friction on a sliding block opposes its motion, giving negative work.
2. Static friction on a body carried along by an accelerating truck acts forward, giving positive work; on a body with no displacement it does zero work.

Teacher's Note:
1. Sign of work depends on the angle between friction and displacement of its point of application.

Answer: (c) \( \dfrac{l}{L-l} \)

Explanation:
1. Hanging weight \( \propto l \); friction \( = \mu(L-l) \) in the same units.
2. Limiting case: \( l = \mu(L-l) \Rightarrow \mu = \dfrac{l}{L-l} \).

Teacher's Note:
1. Use mass per unit length to equate forces.

Answer: (b) slide upon each other

Explanation:
1. A lubricant layer reduces friction so the surfaces slide more easily past each other.

Teacher's Note:
1. Lubricants replace solid-solid contact by liquid layers.

Answer: (a) 0.38

Explanation:
1. Limiting friction \( = 75 \) N \( = \mu_s mg \).
2. \( \mu_s = \dfrac{75}{20\times9.8} = 0.38 \).

Teacher's Note:
1. Use the force that just starts motion for static friction, and the force for uniform motion for kinetic friction (\( \mu_k = 0.31 \)).

Answer: (c) 1100 N

Explanation:
1. Friction from A on B (opposing B): \( 0.2\times100\times10 = 200 \) N.
2. Friction from ground: \( 0.3\times(100+200)\times10 = 900 \) N.
3. \( F = 200+900 = 1100 \) N.

Teacher's Note:
1. The ground carries the weight of both blocks.
2. Two friction forces oppose the motion of B.

Answer: (b) 1.732

Explanation:
1. At the point of sliding \( \mu = \tan\theta = \tan60^\circ = 1.732 \).

Teacher's Note:
1. Coefficient of static friction \( = \tan \) of the angle of repose.

Answer: (a) 2.5 N

Explanation:
1. Limiting friction \( = 0.4\times2\times9.8 = 7.84 \) N, greater than 2.5 N.
2. The block does not move, so static friction equals the applied force, 2.5 N.

Teacher's Note:
1. Static friction matches the applied force until the limit is reached.

Answer: (c) Sand

Explanation:
1. Oil and graphite are lubricants, ball bearings give rolling.
2. Sand increases friction.

Teacher's Note:
1. Sand is spread on slippery roads to increase friction.

Answer: (a) 75 N

Explanation:
1. Wall is smooth, so the floor's normal reaction equals the ladder's weight, 250 N.
2. \( f_{max} = 0.3\times250 = 75 \) N.

Teacher's Note:
1. With a smooth wall, vertical equilibrium gives \( N = W \).

Answer: (b) 20 N

Explanation:
1. Maximum friction \( = 0.3\times100 = 30 \) N, which exceeds the weight 19.6 N.
2. The body is at rest, so friction equals weight \( \approx 20 \) N.

Teacher's Note:
1. Do not use \( \mu N \) when the body is at rest below the limit.

Answer: (a) 2 N

Explanation:
1. Weight = limiting friction \( = \mu N = 0.2\times10 = 2 \) N.

Teacher's Note:
1. For a block against a wall, N is the pressing force.

Answer: (d) 0.4 kg

Explanation:
1. \( m_Bg \le \mu m_Ag \Rightarrow m_B \le 0.2\times2 = 0.4 \) kg.

Teacher's Note:
1. Equilibrium holds up to the limiting friction.