IOE entrance Physics · Chapter 1
Mechanics
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85 questions in 7 syllabus topics · 36 are 2-mark questions.
1.1 Physical quantities, vectors and kinematics
12 questions
1. The dimensional formula of Planck's constant h is:
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Use E = hν and divide the dimensions of energy by those of frequency.
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Answer: D. [ML²T⁻¹]
From E = hν, h = E/ν = [ML²T⁻²]/[T⁻¹] = [ML²T⁻¹]. It has the same dimensions as angular momentum.
2. The velocity of a particle is given by v = At + B/(t + C), where t is time. The dimensions of B are:
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Quantities added together must have the same dimensions.
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Answer: B. [L]
C must have the dimensions of time, so B/(t + C) has dimensions [LT⁻¹] only if B = [LT⁻¹]×[T] = [L].
3. The acceleration due to gravity is found using g = 4π²l/T². If the percentage errors in measuring l and T are 1% and 2% respectively, the maximum percentage error in g is:
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Powers of a quantity multiply its percentage error; errors always add.
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Answer: D. 5%
Δg/g = Δl/l + 2(ΔT/T) = 1% + 2×2% = 5%.
4. Two forces, each of magnitude F, act on a particle. If their resultant also has magnitude F, the angle between the two forces is:
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Use R² = P² + Q² + 2PQ cosθ.
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Answer: D. 120°
F² = F² + F² + 2F²cosθ gives cosθ = −½, so θ = 120°.
5. Vector A = 2i + 3j − k and vector B = i − 2j + ak are perpendicular to each other. The value of a is:
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Perpendicular vectors have zero dot product.
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Answer: C. −4
A·B = (2)(1) + (3)(−2) + (−1)(a) = −4 − a = 0, so a = −4.
6. Three forces acting on a particle are represented in magnitude and direction by the three sides of a triangle taken in the same order. The particle will:
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What is the resultant of vectors forming a closed figure in order?
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Answer: A. remain in equilibrium
By the triangle (polygon) law, vectors represented by the sides of a closed polygon taken in order add to zero, so the net force is zero.
7. A vector of magnitude 10 units makes an angle of 30° with the positive x-axis in the x-y plane. Its x and y components are respectively:
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Component along an axis = magnitude × cosine of angle with that axis.
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Answer: D. 5√3 and 5
Ax = 10 cos30° = 5√3, Ay = 10 sin30° = 5.
8. A ball is thrown vertically upward with a speed of 30 m/s. The distance travelled by the ball in the first 4 s is (g = 10 m/s²):
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Find when the ball turns back; distance and displacement differ after that.
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Answer: D. 50 m
It rises for u/g = 3 s, covering 30²/(2×10) = 45 m, then falls for 1 s covering ½×10×1² = 5 m. Distance = 45 + 5 = 50 m (displacement is 40 m).
9. For a projectile at the highest point of its path (air resistance neglected):
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Which velocity component vanishes at the top, and does gravity ever switch off?
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Answer: C. velocity is horizontal and acceleration is g vertically downward
At the top only the vertical velocity becomes zero; the horizontal component u cosθ remains, and gravity still acts, so acceleration is g downward.
10. A ball is projected with a speed of 20 m/s at 30° above the horizontal. Its horizontal range is (g = 10 m/s²):
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R = u² sin2θ / g.
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Answer: A. 20√3 m
R = u² sin2θ/g = (400 × sin60°)/10 = 40 × (√3/2) = 20√3 m ≈ 34.6 m.
11. The horizontal range of a projectile is four times its maximum height. The angle of projection is:
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Divide R by H and simplify to a trigonometric ratio.
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Answer: B. 45°
R/H = (2u² sinθ cosθ/g)/(u² sin²θ/2g) = 4 cotθ. Setting 4 cotθ = 4 gives θ = 45°.
12. Two trains of lengths 100 m and 150 m run on parallel tracks in the same direction with speeds 15 m/s and 10 m/s. The time taken by the faster train to completely overtake the slower one is:
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Work in the frame of the slower train.
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Answer: B. 50 s
Relative speed = 15 − 10 = 5 m/s; relative distance to cover = 100 + 150 = 250 m; time = 250/5 = 50 s.
1.2 Newton’s laws of motion and friction
12 questions
13. Newton's first law of motion gives the concept of:
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What property makes a body resist change in its state of motion?
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Answer: A. inertia
The first law says a body stays at rest or in uniform motion unless an external force acts, which is the property of inertia (and a qualitative definition of force).
14. A ball of mass 0.2 kg hits a wall normally at 10 m/s and rebounds with the same speed. The magnitude of the impulse given to the ball is:
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Velocity reverses direction, so take signs into account.
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Answer: A. 4 N s
Impulse = change in momentum = 0.2 × 10 − (0.2 × (−10)) = 4 N s.
15. A man of mass 60 kg stands on a weighing scale in a lift that is accelerating downward at 2 m/s². The scale reads (g = 10 m/s²):
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For downward acceleration, the normal reaction is less than mg.
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Answer: A. 480 N
N = m(g − a) = 60 × (10 − 2) = 480 N.
16. Two masses of 3 kg and 2 kg hang from the ends of a light string passing over a smooth, light pulley. The tension in the string is (g = 10 m/s²):
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Find the acceleration first, then apply F = ma to one mass.
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Answer: A. 24 N
a = (3 − 2)g/(3 + 2) = 2 m/s². For the 2 kg mass: T − 20 = 2 × 2, so T = 24 N (also T = 2m₁m₂g/(m₁ + m₂)).
17. A block slides down a rough inclined plane of inclination 30°. If the coefficient of kinetic friction is 1/(2√3), the acceleration of the block is (g = 10 m/s²):
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Friction acts up the incline and equals μ mg cosθ.
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Answer: B. 2.5 m/s²
a = g(sinθ − μ cosθ) = 10(½ − (1/(2√3))(√3/2)) = 10(0.5 − 0.25) = 2.5 m/s².
18. A block placed on a plank just begins to slide when the plank is tilted to 30° with the horizontal. The coefficient of static friction is:
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Relate the angle of repose to μ.
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Answer: A. 1/√3
At the angle of repose, μs = tanθ = tan30° = 1/√3.
19. Which of the following statements about friction is NOT correct?
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Recall the laws of limiting friction.
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Answer: D. Limiting friction depends on the area of contact
By the laws of friction, limiting friction is independent of the apparent area of contact; the other statements are true.
20. A 2 kg block rests on a horizontal floor with μs = 0.4 and μk = 0.25. A horizontal force of 6 N is applied to it. The frictional force acting on the block is (g = 10 m/s²):
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Compare the applied force with limiting friction before using any formula.
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Answer: D. 6 N
Limiting friction = μs mg = 0.4 × 20 = 8 N. Since 6 N < 8 N the block does not move, so static friction just balances the applied force: 6 N.
21. A bomb at rest explodes into two fragments of masses 1 kg and 2 kg. If the 1 kg fragment moves at 6 m/s, the 2 kg fragment moves with:
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Total momentum before and after explosion is the same.
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Answer: A. 3 m/s opposite to the 1 kg fragment
Initial momentum is zero, so 1 × 6 = 2 × v gives v = 3 m/s, in the opposite direction.
22. The momentum of a particle varies with time as p = 3t² + 2t (SI units). The force on the particle at t = 2 s is:
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Force is the rate of change of momentum.
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Answer: A. 14 N
F = dp/dt = 6t + 2 = 6(2) + 2 = 14 N.
23. Two blocks of masses 4 kg and 6 kg are placed in contact on a smooth horizontal surface. A horizontal force of 20 N is applied to the 4 kg block. The force exerted by the 4 kg block on the 6 kg block is:
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Find the common acceleration, then isolate the block that is pushed only by the contact force.
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Answer: C. 12 N
Common acceleration a = 20/(4 + 6) = 2 m/s². The contact force alone accelerates the 6 kg block: N = 6 × 2 = 12 N.
24. A weight of 100 N hangs in equilibrium from two strings, each making an angle of 60° with the vertical. The tension in each string is:
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Resolve the tensions vertically; horizontal parts cancel.
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Answer: C. 100 N
Vertical balance: 2T cos60° = 100, so 2T × ½ = 100 and T = 100 N.
1.3 Work, energy and power
12 questions
25. The speed of a 2 kg body increases from 2 m/s to 4 m/s. The work done by the net force on it is:
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Net work equals change in kinetic energy.
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Answer: B. 12 J
By the work-energy theorem, W = ½m(v² − u²) = ½ × 2 × (16 − 4) = 12 J.
26. If the linear momentum of a body increases by 20%, its kinetic energy increases by:
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Write kinetic energy in terms of momentum.
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Answer: B. 44%
K = p²/2m, so K ∝ p². New K = (1.2)² K = 1.44 K, an increase of 44%.
27. A force F = 3x² N acts on a particle along the x-axis. The work done as the particle moves from x = 0 to x = 2 m is:
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For a variable force, W = ∫F dx.
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Answer: D. 8 J
W = ∫₀² 3x² dx = [x³]₀² = 8 J.
28. Which of the following is a non-conservative force?
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A conservative force does zero work round any closed path.
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Answer: C. Kinetic friction
Work done by friction depends on the path and is not zero over a closed path, so friction is non-conservative.
29. A spring of force constant 200 N/m is compressed by 10 cm and then released, pushing a 0.5 kg block on a smooth horizontal surface. The speed of the block when it leaves the spring is:
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Elastic potential energy converts fully into kinetic energy.
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Answer: B. 2 m/s
½kx² = ½ × 200 × (0.1)² = 1 J = ½ × 0.5 × v², so v² = 4 and v = 2 m/s.
30. A 1 kg block starts from rest at a height of 5 m and slides down a rough curved track, reaching the bottom with a speed of 8 m/s. The work done against friction is (g = 10 m/s²):
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Energy lost by the block equals work done against friction.
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Answer: C. 18 J
Loss of PE = mgh = 50 J; gain in KE = ½ × 1 × 64 = 32 J. Work done against friction = 50 − 32 = 18 J.
31. A crane lifts a load of 500 kg through a height of 10 m in 20 s at constant speed. The power developed is (g = 10 m/s²):
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Power = work done / time.
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Answer: D. 2.5 kW
P = mgh/t = (500 × 10 × 10)/20 = 2500 W = 2.5 kW.
32. A pump lifts 600 kg of water per minute to a height of 10 m. If the efficiency of the pump is 50%, its input power is (g = 10 m/s²):
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Find the useful power first, then divide by efficiency.
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Answer: D. 2 kW
Useful output = mgh/t = (600 × 10 × 10)/60 = 1000 W. Input = output/efficiency = 1000/0.5 = 2000 W = 2 kW.
33. A ball moving with speed v collides head-on elastically with an identical ball at rest. After the collision:
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Apply conservation of both momentum and kinetic energy with equal masses.
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Answer: A. the first ball stops and the second moves with speed v
For a one-dimensional elastic collision between equal masses, the velocities are exchanged.
34. A 2 kg body moving at 6 m/s strikes a 4 kg body at rest and the two stick together. The loss of kinetic energy in the collision is:
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Use momentum conservation to find the common velocity.
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Answer: D. 24 J
Common velocity v = (2 × 6)/6 = 2 m/s. Initial KE = ½ × 2 × 36 = 36 J; final KE = ½ × 6 × 4 = 12 J; loss = 24 J.
35. A ball dropped from a height of 10 m rebounds from the floor to a height of 2.5 m. The coefficient of restitution between the ball and floor is:
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Speed just before or after the bounce is √(2gh).
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Answer: D. 0.5
e = (speed after)/(speed before) = √(2gh₂)/√(2gh₁) = √(2.5/10) = 0.5.
36. The potential energy of a particle moving along the x-axis is U = 5x² − 10x (SI units). The force on the particle at x = 3 m is:
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For a conservative force, F = −dU/dx.
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Answer: C. −20 N
F = −dU/dx = −(10x − 10) = −(30 − 10) = −20 N.
1.4 Circular motion, gravitation and SHM
12 questions
37. A stone of mass 0.5 kg tied to a string is whirled in a horizontal circle of radius 1 m at 2 revolutions per second. The centripetal force on it is:
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Convert revolutions per second to angular velocity first.
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Answer: A. 8π² N
ω = 2πf = 4π rad/s. F = mω²r = 0.5 × 16π² × 1 = 8π² N (≈ 79 N).
38. A circular road of radius 40 m is to be banked so that a car at 20 m/s needs no friction to take the turn. The angle of banking is (g = 10 m/s²):
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tanθ = v²/rg for frictionless banking.
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Answer: C. 45°
tanθ = v²/(rg) = 400/(40 × 10) = 1, so θ = 45°.
39. A conical pendulum has a string of length 0.2 m that makes an angle of 60° with the vertical. Its time period is (g = 10 m/s²):
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For a conical pendulum the effective length is the vertical height L cosθ.
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Answer: A. 0.2π s
T = 2π√(L cosθ/g) = 2π√((0.2 × 0.5)/10) = 2π√0.01 = 0.2π s.
40. At what depth below the Earth's surface is the acceleration due to gravity equal to its value at a height equal to the Earth's radius R above the surface?
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Use g(1 + h/R)⁻² above and g(1 − d/R) below.
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Answer: C. 3R/4
At height R: g' = g/(1 + 1)² = g/4. At depth d: g' = g(1 − d/R). Equating, 1 − d/R = ¼, so d = 3R/4.
41. A planet has 8 times the mass and twice the radius of the Earth. If the escape velocity from Earth is 11.2 km/s, that from the planet is:
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Escape velocity is proportional to √(M/R).
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Answer: B. 22.4 km/s
vₑ = √(2GM/R) ∝ √(M/R) = √(8/2) = 2, so vₑ = 2 × 11.2 = 22.4 km/s.
42. Which of the following statements about a geostationary satellite is NOT correct?
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Think about which plane its orbit must lie in.
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Answer: B. It can be kept stationary directly above Kathmandu
A geostationary satellite must orbit in the equatorial plane, so it can appear fixed only above points on the equator, not above Kathmandu (about 27° N).
43. The work required to raise a body of mass m from the Earth's surface to a height equal to the Earth's radius R is (g = acceleration due to gravity at the surface):
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Use U = −GMm/r, not mgh, for large heights.
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Answer: C. mgR/2
ΔU = −GMm/2R − (−GMm/R) = GMm/2R = mgR/2, using GM = gR². (mgh is valid only for h ≪ R.)
44. For a particle executing simple harmonic motion, at the mean position:
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Use a = −ω²x and v = ω√(A² − x²).
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Answer: A. velocity is maximum and acceleration is zero
Acceleration a = −ω²x is zero at x = 0, while speed ω√(A² − x²) is maximum there.
45. A particle executes SHM with amplitude 10 cm and period π s. Its speed when the displacement is 6 cm is:
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Find ω from the period, then use v = ω√(A² − x²).
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Answer: D. 16 cm/s
ω = 2π/T = 2 rad/s. v = ω√(A² − x²) = 2√(100 − 36) = 2 × 8 = 16 cm/s.
46. A mass on a spring oscillates with period T. The spring is cut into two equal halves and the same mass is attached to one half. The new period is:
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How does spring constant change when a spring is cut?
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Answer: C. T/√2
Halving a spring doubles its force constant (k' = 2k). Since T = 2π√(m/k), the new period is T/√2.
47. If the length of a simple pendulum is increased by 21%, its time period increases by:
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T ∝ √l; take the exact square root rather than halving the percentage.
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Answer: C. 10%
T ∝ √l, so T'/T = √1.21 = 1.1, an increase of 10%.
48. In forced oscillations, resonance occurs when:
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Resonance is a matching condition between two frequencies.
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Answer: D. the frequency of the driving force equals the natural frequency of the oscillator
At resonance the driving frequency matches the natural frequency, and the amplitude of the forced oscillation becomes maximum (limited only by damping).
1.5 Rotational dynamics
12 questions
49. The moment of inertia of a thin uniform ring of mass M and radius R about an axis through its centre and perpendicular to its plane is
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Where is all the mass located relative to the axis?
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Answer: C. MR²
Every mass element of the ring is at the same distance R from the axis, so I = ∑mR² = MR².
50. The radius of gyration of a uniform solid sphere of radius R about any diameter is
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Write I = Mk² and compare with the sphere's moment of inertia.
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Answer: B. R√(2/5)
I = (2/5)MR² = Mk², so k = R√(2/5) ≈ 0.63R. Note 2R/5 is k², not k.
51. A uniform disc of mass 2 kg rolls without slipping on a horizontal floor with its centre moving at 4 m/s. Its total kinetic energy is
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Add translational and rotational KE; for a disc I = ½MR² and ω = v/R.
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Answer: B. 24 J
KE = ½mv² + ½Iω² = ½mv² + ½(½mR²)(v/R)² = ¾mv² = ¾ × 2 × 16 = 24 J.
52. A force of 20 N is applied at a point 0.5 m from the hinge of a door, the force making an angle of 30° with the line joining the hinge to that point. The torque about the hinge is
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τ = rF sin θ, with θ the angle between r and F.
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Answer: C. 5 N m
τ = rF sin θ = 0.5 × 20 × sin 30° = 5 N m.
53. A disc of moment of inertia 0.2 kg m² spins freely at 10 rad/s about a vertical axis. A second disc of moment of inertia 0.3 kg m², initially at rest, is gently dropped coaxially on it and they rotate together. The kinetic energy lost is
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Conserve angular momentum first, then compare kinetic energies.
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Answer: D. 6 J
Angular momentum is conserved: 0.2 × 10 = 0.5ω, so ω = 4 rad/s. KE before = ½ × 0.2 × 100 = 10 J; after = ½ × 0.5 × 16 = 4 J. Loss = 6 J.
54. Particles of mass 2 kg and 3 kg are placed on the x-axis at x = 0 and x = 5 m respectively. The centre of mass is at x =
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Use the mass-weighted average of positions.
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Answer: C. 3 m
x_cm = (2 × 0 + 3 × 5)/(2 + 3) = 15/5 = 3 m.
55. The centre of gravity of a body coincides with its centre of mass when
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Think about what makes weight distribution differ from mass distribution.
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Answer: A. the gravitational field over the body is uniform
The centre of gravity is the point where the resultant weight acts; it coincides with the centre of mass only if g is the same at every part of the body.
56. A thin uniform rod of mass 3 kg and length 2 m rotates about an axis perpendicular to it passing through a point 0.5 m from one end. Its moment of inertia about this axis is
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Use the parallel axis theorem; find the distance of the axis from the centre.
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Answer: A. 1.75 kg m²
I_cm = ML²/12 = 3 × 4/12 = 1 kg m². The axis is 1 − 0.5 = 0.5 m from the centre, so I = 1 + 3 × 0.5² = 1.75 kg m².
57. A ring, a hollow sphere, a disc and a solid sphere are released together from rest at the top of the same rough incline and roll without slipping. Which reaches the bottom first?
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Compare k²/R² for each body.
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Answer: B. Solid sphere
Acceleration a = g sin θ/(1 + k²/R²). k²/R² is smallest for the solid sphere (2/5), so it accelerates most and arrives first.
58. A solid sphere starts from rest and rolls without slipping down an incline through a vertical height of 7 m. Its speed at the bottom is (g = 10 m/s²)
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Energy conservation including rotational KE with k²/R² = 2/5.
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Answer: B. 10 m/s
mgh = ½mv²(1 + k²/R²) = ½mv²(7/5), so v = √(10gh/7) = √(10 × 10 × 7/7) = 10 m/s.
59. A constant torque of 10 N m acts on a flywheel of moment of inertia 2 kg m² initially at rest. Its rotational kinetic energy after 4 s is
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Find α from τ = Iα, then ω after 4 s.
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Answer: D. 400 J
α = τ/I = 5 rad/s², ω = αt = 20 rad/s, KE = ½Iω² = ½ × 2 × 400 = 400 J.
60. If the Earth (treated as a uniform sphere) suddenly contracted to half its present radius with no change in mass, the length of a day would become about
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Angular momentum is conserved; how does I depend on R?
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Answer: B. 6 h
L = Iω is conserved and I ∝ R², so I becomes I/4 and ω becomes 4ω. The period becomes 24/4 = 6 h.
1.6 Elasticity
13 questions
61. To what depth below the surface of a sea should a rubber ball be taken so that its volume decreases by 1%? Bulk modulus of rubber is 9 × 10⁸ N m⁻². (Take density of sea water 1000 kg/m³ and g = 10 m/s².)
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Pressure needed = B × (ΔV/V), then h = P/(ρg).
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Answer: D. 1 km
The exact depth is about 900 m; “1 km” is the nearest option.
ΔV/V = P/B, so P = 0.01 × 9 × 10⁸ = 9 × 10⁶ Pa. Depth h = P/(ρg) = 9 × 10⁶/(1000 × 10) = 900 m, which is about 1 km.
62. Hooke's law states that, within the elastic limit,
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Think of the straight-line part of the stress–strain graph.
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Answer: A. stress is directly proportional to strain
Within the elastic (proportional) limit, stress/strain = constant, which is the modulus of elasticity.
63. The dimensional formula of Young's modulus is
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Strain has no dimensions.
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Answer: A. [ML⁻¹T⁻²]
Y = stress/strain; strain is dimensionless, so Y has dimensions of stress (force/area) = [MLT⁻²]/[L²] = [ML⁻¹T⁻²].
64. A wire of length 2 m and cross-sectional area 1 mm² hangs vertically and a mass of 10 kg is attached to its lower end. If Young's modulus of the material is 2 × 10¹¹ N/m², the extension is (g = 10 m/s²)
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Use Δl = FL/(AY) with A in m².
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Answer: A. 1 mm
Δl = FL/(AY) = (100 × 2)/(10⁻⁶ × 2 × 10¹¹) = 200/(2 × 10⁵) = 10⁻³ m = 1 mm.
65. The theoretical limits of Poisson's ratio for an isotropic material are
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Use Y = 2η(1 + σ) and Y = 3K(1 − 2σ) with all moduli positive.
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Answer: C. −1 to 0.5
From the relations among elastic moduli (e.g. Y = 3K(1 − 2σ) = 2η(1 + σ) with K, η positive), σ must lie between −1 and 0.5.
66. The elastic potential energy stored per unit volume of a stretched wire is
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It is the area of a triangle under the stress–strain line.
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Answer: D. ½ × stress × strain
Work per unit volume = area under the linear stress–strain graph = ½ × stress × strain.
67. A wire of length 1 m and cross-sectional area 2 mm² (Y = 2 × 10¹¹ N/m²) is stretched by 1 mm. The elastic energy stored in it is
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Find the force needed first, then U = ½ F Δl.
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Answer: A. 0.2 J
F = YAΔl/L = 2 × 10¹¹ × 2 × 10⁻⁶ × 10⁻³/1 = 400 N. U = ½FΔl = ½ × 400 × 10⁻³ = 0.2 J.
68. Compressibility of a material is defined as
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Which modulus deals with change in volume?
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Answer: D. the reciprocal of its bulk modulus
Compressibility = 1/K = −(ΔV/V)/ΔP, the fractional change in volume per unit pressure.
69. The bulk modulus of water is 2 × 10⁹ N/m². When the pressure on 1 litre of water is increased by 10⁷ N/m², the decrease in its volume is
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Use K = ΔP/(ΔV/V).
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Answer: C. 5 mL
ΔV/V = ΔP/K = 10⁷/(2 × 10⁹) = 5 × 10⁻³. ΔV = 5 × 10⁻³ × 1000 mL = 5 mL.
70. The top face of a rubber cube is displaced sideways relative to its fixed bottom face, changing its shape but not its volume. The elastic constant involved is
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Tangential force produces which kind of strain?
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Answer: B. modulus of rigidity
A tangential force producing a change of shape without volume change is shear; the relevant constant is the modulus of rigidity (shear modulus) η.
71. Two wires A and B of the same material have lengths in the ratio 1 : 2 and diameters in the ratio 2 : 1. When stretched by the same force, the ratio of their extensions (A : B) is
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Extension ∝ L/d² for the same force and material.
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Answer: B. 1 : 8
Δl = FL/(AY) ∝ L/d². Δl_A/Δl_B = (L_A/L_B)(d_B/d_A)² = (1/2)(1/2)² = 1/8.
72. A wire of length 2 m and diameter 1 mm is stretched so that its length increases by 1 mm and its diameter decreases by 1.5 × 10⁻⁴ mm. Poisson's ratio of the material is
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σ = lateral strain/longitudinal strain; use the right original dimensions.
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Answer: A. 0.30
Longitudinal strain = 1/2000 = 5 × 10⁻⁴. Lateral strain = 1.5 × 10⁻⁴/1 = 1.5 × 10⁻⁴. σ = 1.5 × 10⁻⁴/(5 × 10⁻⁴) = 0.30.
73. Steel is said to be more elastic than rubber because, for the same stress, steel
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Higher modulus means ... strain for a given stress.
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Answer: D. develops a smaller strain
Elasticity is measured by the modulus: a larger Y means a smaller strain for the same stress. Steel's Y is far larger than rubber's.
1.7 Fluid mechanics
12 questions
74. A solid of volume 200 cm³ is completely immersed in water (density 1000 kg/m³). The buoyant force on it is (g = 10 m/s²)
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Weight of the water displaced; convert cm³ to m³.
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Answer: A. 2 N
Upthrust = Vρg = 200 × 10⁻⁶ × 1000 × 10 = 2 N.
75. A wooden block floats in water with 60% of its volume submerged. When placed in an oil of density 800 kg/m³, the fraction of its volume submerged will be
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For a floating body, fraction submerged = ρ_body/ρ_liquid.
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Answer: C. 75%
Density of wood = 0.6 × 1000 = 600 kg/m³. In oil, fraction submerged = 600/800 = 0.75 = 75%.
76. The dimensional formula of surface tension is
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Surface tension is force per unit length.
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Answer: C. [ML⁰T⁻²]
Surface tension = force/length, so [MLT⁻²]/[L] = [ML⁰T⁻²].
77. Water (surface tension 0.07 N/m, angle of contact 0°) rises in a clean glass capillary tube of radius 0.1 mm. The height of rise is (density of water 1000 kg/m³, g = 10 m/s²)
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Use h = 2T cos θ/(rρg) with r in metres.
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Answer: C. 14 cm
h = 2T cos θ/(rρg) = (2 × 0.07 × 1)/(10⁻⁴ × 1000 × 10) = 0.14/1 = 0.14 m = 14 cm.
78. The excess pressure inside a soap bubble of radius r, made from a solution of surface tension T, is
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Count the number of liquid–air surfaces.
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Answer: B. 4T/r
A soap bubble has two surfaces (inside and outside), so the excess pressure is 2 × 2T/r = 4T/r.
79. A small spherical raindrop falls through air with a terminal velocity of 2 cm/s. If 8 such identical drops coalesce to form a single drop, its terminal velocity becomes
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Find the new radius, then use v ∝ r².
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Answer: B. 8 cm/s
Volume ×8 means radius ×2. From Stokes' law, terminal velocity ∝ r², so v' = 4 × 2 = 8 cm/s.
80. The SI unit of coefficient of viscosity is
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Rearrange Newton's law of viscosity F = ηA dv/dx.
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Answer: D. N s m⁻²
From F = ηA(dv/dx), η = F/(A × dv/dx) with units N/(m² × s⁻¹) = N s m⁻² (pascal second).
81. Liquid flows steadily through a horizontal capillary tube at a volume rate Q. If the tube is replaced by another of the same length but half the radius, and the pressure difference across it is doubled, the new rate of flow is
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Use Poiseuille's equation: Q ∝ P r⁴/l.
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Answer: A. Q/8
Poiseuille: Q = πPr⁴/(8ηl) ∝ Pr⁴. New rate = Q × 2 × (1/2)⁴ = Q/8.
82. Which statement about Reynolds number is correct?
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Recall Re = ρvD/η.
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Answer: B. Flow is streamline for low values (below about 2000) of Reynolds number
Re = ρvD/η is dimensionless; it increases with density and decreases with viscosity. Low Re (below about 2000) means streamline flow.
83. Water flows at 2 m/s through a pipe of diameter 4 cm which narrows to a diameter of 2 cm. The speed of water in the narrow part is
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Equation of continuity; area depends on diameter squared.
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Answer: A. 8 m/s
A₁v₁ = A₂v₂ and A ∝ d², so v₂ = v₁(d₁/d₂)² = 2 × 4 = 8 m/s.
84. Water flows through a horizontal pipe. At point A its speed is 2 m/s and at point B it is 4 m/s. The pressure at A exceeds that at B by (density of water 1000 kg/m³)
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Use Bernoulli's equation with no height change.
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Answer: D. 6000 Pa
Bernoulli for a horizontal pipe: P_A − P_B = ½ρ(v_B² − v_A²) = ½ × 1000 × (16 − 4) = 6000 Pa.
85. As the temperature of a liquid increases, its surface tension generally
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What happens to cohesive forces when molecules move faster?
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Answer: C. decreases
Intermolecular cohesive forces weaken with rise in temperature, so surface tension decreases (becoming zero at the critical temperature).