Static Friction — NEET Physics MCQs with Solutions
Free NEET Physics Static Friction MCQs with step-by-step solutions (32 questions). Part of Friction. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Static Friction · medium · numerical
The frictional force is -
A. Self adjustable ✓ Correct
B. Scalar quantity
C. Not self-adjustable
D. Equal to the limiting force
Solution: Static friction is a self-adjustable force: it adjusts itself to balance the applied force up to the limiting value.
Q2 — Static Friction · medium · numerical
A block is placed on a rough floor and a horizontal force F is applied on it. The force of friction f by the floor on the block is measured for different values of F and a graph is plotted between them. Consider the statements: (a) The graph is a straight line of slope 45°. (b) The graph is a straight line parallel to the F axis. (c) The graph is a straight line of slope 45° for small F and a straight line parallel to the F-axis for large F. (d) There is a small kink on the graph. Which of the following is correct?
A. c, d ✓ Correct
B. a, d
C. a, c
D. a, b
Solution: Up to the limiting force friction balances F, so the graph is a 45° straight line; a small kink appears as motion begins, after which f stays constant and the graph runs parallel to the F-axis. Hence statements (c) and (d) are correct.
Q3 — Static Friction · medium · numerical
A block A kept on an inclined surface just begins to slide if the inclination is 30°. The block is replaced by another block B and it is found that it just begins to slide if the inclination is 40° -
A. mass of A > mass of B
B. all the three are possible ✓ Correct
C. mass of A = mass of B
D. mass of A < mass of B
Solution: The angle of repose gives μ = tan θ and is independent of the block's mass, so the sliding angles reveal nothing about the masses — all three cases are possible.
Q4 — Static Friction · medium · numerical
It is easier to pull a body than to push, because-
A. the coefficient of friction is more in pushing than that in pulling
B. the body does not move forward when pushed
C. the friction force is more in pushing than that in pulling ✓ Correct
D. none of these
Solution: Pushing has a downward force component that raises the normal reaction and hence the friction, so the friction force is greater when pushing than when pulling.
Q5 — Static Friction · medium · numerical
The coefficient of static friction between two surfaces depends on -
A. The area of contact
B. Nature of surfaces ✓ Correct
C. The shape of the surfaces in contact
D. All of the above
Solution: The coefficient of static friction depends only on the nature of the surfaces in contact; it is independent of the contact area.
Q6 — Static Friction · medium · numerical
A box is lying on an inclined plane. If the box starts sliding when the angle of inclination is 60°, then the coefficient of static friction of the box and plane is-
A. 0.267
B. 0.176
C. 1.732 ✓ Correct
D. 2.732
Solution: At the angle of sliding, μ = tan θ = tan 60° = √3 ≈ 1.732.
Q7 — Static Friction · medium · numerical
A 20 kg block is initially at rest. A 75 N force is required to set the block in motion. After the motion, a force of 60 N is applied to keep the block moving with constant speed. The coefficient of static friction is-
A. 0.44
B. 0.52
C. 0.6
D. 0.35 ✓ Correct
Solution: The 75 N needed to start motion equals the limiting static friction, so μs = 75/(20 × g) ≈ 0.35.
Q8 — Static Friction · medium · numerical
A block of metal is lying on the floor of a bus. The maximum acceleration which can be given to the bus so that the block may remain at rest, will be-
A. μg ✓ Correct
B. μ²g
C. μg²
D. μ/g
Solution: The friction ma needed must not exceed the maximum friction μmg; the largest acceleration is when ma = μmg, giving a = μg.
Q9 — Static Friction · medium · numerical
A block of mass 0.1 kg is held against a wall by applying a horizontal force of 5 N on the block. If the coefficient of friction (μ) between the block and the wall is 0.5, the magnitude of frictional force acting on the block is (g = 9.8 m/s²)
A. 0.98 N ✓ Correct
B. 0.49 N
C. 2.5 N
D. 4.9 N
Solution: Maximum friction = μN = 0.5 × 5 = 2.5 N, but the weight mg = 0.1 × 9.8 = 0.98 N is smaller, so friction only balances the weight: f = 0.98 N.
Q10 — Static Friction · medium · numerical
A block of mass 2 kg rests on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block is (g = 9.8 m/s²):
A. 9.8 N ✓ Correct
B. 0.7 × 9.8 N
C. 9.8 × 7 N
D. 0.8 × 9.8 N
Solution: Since μ (0.7) > tan 30° (≈0.577) the block does not slide, so friction equals the gravity component along the incline: f = mg sin 30° = 2 × 9.8 × ½ = 9.8 N.
Q11 — Static Friction · medium · numerical
A block of mass 5 kg and surface area 2 m² just begins to slide down on an inclined plane when the angle of inclination is 30°. Keeping mass same, the surface area of the block is doubled. The angle at which it starts sliding down is:
A. 30° ✓ Correct
B. 60°
C. 15°
D. none
Solution: Friction depends only on the normal reaction, not on the contact area, so doubling the surface area leaves the sliding angle unchanged at 30°.
Q12 — Static Friction · medium · numerical
A 60 kg body is pushed horizontally with just enough force to start it moving across a floor and the same force continues to act afterwards. The coefficient of static friction and sliding friction are 0.5 and 0.4 respectively. The acceleration of the body is (g = 10 m/s²):
A. 4.9 m/s²
B. 1 m/s² ✓ Correct
C. 6 m/s²
D. 3.92 m/s²
Solution: The applied force equals the limiting static friction μs·mg; once moving the net force is (μs − μk)mg, so a = (μs − μk)g = (0.5 − 0.4) × 10 = 1 m/s².
Q13 — Static Friction · medium · numerical
The blocks A and B are arranged as shown in the figure: block A (mass 10 kg) rests on a horizontal surface and is connected over a frictionless pulley to a hanging block B. The coefficient of friction between block A and the horizontal surface is 0.20. The minimum mass of B to start the motion will be-
A. 5 kg
B. 2 kg ✓ Correct
C. 0.2 kg
D. 10 kg
Solution: For motion to just start, the hanging block's weight must equal the limiting friction on A: mB·g = μ·mA·g, so mB = 0.20 × 10 = 2 kg.
Q14 — Static Friction · medium · numerical
In the case of horse pulling a cart, the force that causes the horse to move forward is the force that:
A. the cart exerts on the horse
B. the ground exerts on the horse ✓ Correct
C. the horse exerts on the cart
D. the horse exerts on the ground
Solution: The horse pushes back on the ground; by Newton's third law the ground pushes forward on the horse, and this forward reaction from the ground moves the horse forward.
Q15 — Static Friction · medium · numerical
Block A of mass 4 kg rests on top of block B of mass 6 kg, which rests on a horizontal surface. There is no friction between block B and the horizontal surface. The coefficient of friction between the blocks is 0.2. If g = 10 m/s², the maximum horizontal force F that can be applied on block B without any relative motion between A and B is
A. 20 N ✓ Correct
B. 40 N
C. 100 N
D. 60 N
Solution: Friction between the blocks accelerates A, so the maximum common acceleration is a = μg = 0.2 × 10 = 2 m/s². Then F = (mA + mB)a = (4 + 6) × 2 = 20 N.
Q16 — Static Friction · medium · numerical
A body of mass M is kept on a rough horizontal surface (friction coefficient = μ). A person is trying to pull the body by applying a horizontal force but the body is not moving. The force by the surface on the body is F, where-
A. Mg ≥ F ≥ Mg√(1+μ²)
B. F = Mg
C. Mg ≤ F ≤ Mg√(1+μ²) ✓ Correct
D. F = μMg
Solution: The contact force is the resultant of the normal reaction (Mg) and friction (0 to μMg): its minimum is Mg (friction zero) and maximum is Mg√(1+μ²), so Mg ≤ F ≤ Mg√(1+μ²).
Q17 — Static Friction · medium · numerical
In a situation the contact force by a rough horizontal surface on a body placed on it has constant magnitude. If the angle between this force and the vertical is decreased, the frictional force between the surface and the body will-
A. may increase or decrease
B. decrease ✓ Correct
C. remain the same
D. increase
Solution: The contact force is the resultant of the vertical normal reaction and the horizontal friction; decreasing its angle with the vertical decreases its horizontal component, i.e. the friction decreases.
Q18 — Static Friction · medium · numerical
An inclined plane is inclined at an angle θ with the horizontal. A body of mass m rests on it. If the coefficient of friction is μ, then the minimum force that has to be applied to the inclined plane to make the body just move up the inclined plane is-
A. μmg cos θ – mg sin θ
B. μmg cos θ
C. μmg cos θ + mg sin θ ✓ Correct
D. mg sin θ
Solution: To just push the body up the incline the applied force must overcome both the gravity component and friction: F = mg sin θ + μmg cos θ.
Q19 — Static Friction · medium · numerical
A block W is held against a vertical wall by applying a horizontal force F. The minimum value of F needed to hold the block is (if μ < 1)
A. Less than W
B. Greater than W ✓ Correct
C. Equal to W
D. Data is insufficient
Solution: Friction holds the block: f = W and f ≤ μF, so F ≥ W/μ. Since μ < 1, W/μ > W, meaning the required F is greater than W.
Q20 — Static Friction · medium · numerical
A block of mass 20 kg is kept on a rough inclined plane inclined at 60°. If the angle of repose is 30°, then what should be the value of Fext (applied along the incline) so that the block does not move over the inclined plane?
A. Both (a) & (b) ✓ Correct
B. 120 N
C. 200 N
D. 110 N
Solution: With μ = tan 30° on the 60° incline, the force along the incline may range from mg sin θ − μmg cos θ ≈ 120 N (to prevent sliding down) up to mg sin θ + μmg cos θ ≈ 200 N (before sliding up); both 120 N and 200 N are valid limiting values.
Q21 — Static Friction · medium · numerical
The coefficient of static friction, μs, between block A of mass 2 kg and the table is 0.2. Block A rests on the table and is connected over a smooth, massless pulley to a hanging block B. What would be the maximum mass of block B so that the two blocks do not move? (g = 10 m/s²)
A. 0.2 kg
B. 4.0 kg
C. 0.4 kg ✓ Correct
D. 2.0 kg
Solution: The hanging block's weight equals the tension, which equals the limiting friction on A: M·g = μs·m·g, so M = μs·m = 0.2 × 2 = 0.4 kg.
Q22 — Static Friction · medium · numerical
A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is
A. 2 N ✓ Correct
B. 2 kg
C. 50 N
D. 100 N
Solution: The block is held by friction f = μN = μF = 0.2 × 10 = 2 N, which must equal the block's weight, so the weight is 2 N.
Q23 — Static Friction · medium · numerical
A block rests on a rough inclined plane making an angle of 30° with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is (take g = 10 m/s²)
A. 4.0
B. 2.0 ✓ Correct
C. 1.6
D. 2.5
Solution: At rest the friction balances the gravity component along the incline: f = mg sin 30°, so 10 = m × 10 × ½, giving m = 2.0 kg.
Q24 — Static Friction · medium · numerical
A block of mass 20 kg is acted upon by a force F = 30 N at an angle 53° with the horizontal in the downward direction. The coefficient of friction between the block and the horizontal surface is 0.2. The friction force acting on the block by the ground is (g = 10 m/s²)
A. 40.0 N
B. 30.0 N
C. 18.0 N ✓ Correct
D. 44.8 N
Solution: Maximum friction = μ(mg + F sin 53°) = 0.2(200 + 30 × 0.8) = 44.8 N. The applied horizontal component F cos 53° = 30 × 0.6 = 18 N is less than this, so the actual friction is 18 N.
Q25 — Static Friction · medium · numerical
A block of mass m lying on a rough horizontal plane is acted upon by a horizontal force P and another force Q inclined at an angle θ to the vertical. The block will remain in equilibrium if the coefficient of friction between it and the surface is:
A. (P + Q cos θ)/(mg + Q sin θ)
B. (P sin θ + Q)/(mg – Q cos θ)
C. (P + Q sin θ)/(mg + Q cos θ) ✓ Correct
D. (P cos θ + Q)/(mg – Q sin θ)
Solution: Normal reaction N = mg + Q cos θ, and horizontal balance gives P + Q sin θ = μN, so μ = (P + Q sin θ)/(mg + Q cos θ).
Q26 — Static Friction · medium · numerical
In the arrangement shown, a 5 kg block is placed on a rough table (μ = 0.4) with a 3 kg mass connected over a pulley at one end and a mass m connected over a pulley at the other end. The range of mass m for which the system will remain in equilibrium is
A. 1 kg to 5 kg ✓ Correct
B. 3 kg to 5 kg
C. 1 kg to 3 kg
D. Any value greater than 8 kg
Solution: Friction on the 5 kg block = μR = 0.4 × 5g = 2g. With the 3 kg side giving T = 3g, equilibrium holds from mg = 3g − 2g (m = 1 kg) to mg = 3g + 2g (m = 5 kg), i.e. 1 kg to 5 kg.
Q27 — Static Friction · medium · numerical
Two masses A and B of 10 kg and 5 kg respectively are connected with a string passing over a frictionless pulley fixed at the corner of a table (A rests on the table, B hangs). The coefficient of static friction of A with the table is 0.2. The minimum mass of C that may be placed on A to prevent it from moving is
A. 5 kg
B. 15 kg ✓ Correct
C. 10 kg
D. 12 kg
Solution: To prevent A from moving, the hanging weight must not exceed the friction: mB·g = μ(mA + mC)g, so mC = mB/μ − mA = 5/0.2 − 10 = 15 kg.
Q28 — Static Friction · medium · numerical
A block of mass m is at rest relative to a stationary wedge of mass M (wedge angle θ). The coefficient of friction between block and wedge is μ. The wedge is now pulled horizontally with acceleration 'a'. Then the minimum magnitude of 'a' for the friction between block and wedge to be zero is:
A. g cot θ ✓ Correct
B. μ g cot θ
C. g tan θ
D. μ g tan θ
Solution: For the friction to be zero the normal reaction alone must supply the forces normal to the plane: mg cos θ = ma sin θ, giving a = g cot θ.
Q29 — Static Friction · medium · numerical
A uniform rope of length l lies on a table. If the coefficient of friction is μ, then the maximum length l₁ of the part of this rope which can overhang from the edge of the table without sliding down is
A. μl/(μ-1)
B. l/(μ+1)
C. μl/(μ+1) ✓ Correct
D. l/μ
Solution: The overhanging weight is balanced by friction on the part on the table: (l₁/l)mg = μ((l − l₁)/l)mg, giving l₁ = μl/(μ+1).
Q30 — Static Friction · medium · numerical
A heavy uniform chain lies on a horizontal table-top. If the coefficient of friction between the chain and table surface is 0.25, then the maximum fraction of length of the chain that can hang over one edge of the table is
A. 15%
B. 25%
C. 20% ✓ Correct
D. 35%
Solution: The hanging fraction satisfies x/l = μ/(1+μ) = 0.25/1.25 = 0.2, i.e. 20%.