Physics · Work, energy and power · chapter 5 of 14

Class 11 Physics Chapter 5: Work, Energy and Power MCQs with answers

10 multiple-choice questions on Work, Energy and Power, each with its answer and the reason, written from the NCERT chapter. For school exams on the CBSE pattern, JEE Main and NEET, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.

Question 1
Two vectors A⃗ and B⃗ have magnitudes 4 units and 6 units, and the angle between them is 120°. The scalar product A⃗·B⃗ is
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Answer: (b) −12

A⃗·B⃗ = AB cos θ = 4 × 6 × cos 120° = 24 × (−½) = −12. Taking cos 120° as +½ gives 12; leaving out cos θ gives 24.

Question 2
The scalar product (2î − ĵ + 3k̂)·(î + 4ĵ + k̂) is
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Answer: (a) 1

Using î·î = ĵ·ĵ = k̂·k̂ = 1 and the cross terms zero: (2)(1) + (−1)(4) + (3)(1) = 2 − 4 + 3 = 1. Adding the magnitudes of the products gives 9.

Question 3
For what value of p are the vectors (pî + 2ĵ − k̂) and (3î − pĵ + 4k̂) perpendicular to each other?
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Answer: (c) 4

Perpendicular vectors have zero scalar product: 3p − 2p − 4 = 0, so p = 4. A sign slip in the k̂ term gives −4.

Question 4
The component of A⃗ = 4î + 2ĵ + 4k̂ along the direction of B⃗ = 2î − ĵ + 2k̂ is
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Answer: (d) 14/3

Component of A⃗ along B⃗ = A⃗·B⃗ / B = (8 − 2 + 8)/√(4 + 1 + 4) = 14/3. Not dividing by B gives 14; dividing by B² = 9 gives 14/9, and by A = 6 gives 7/3.

Question 5
In which of the following does the named force do zero work?
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Answer: (c) The normal force of the floor on a box sliding along a level floor

The normal force on the box is vertical while its displacement is horizontal: θ = 90°, cos 90° = 0, so W = 0. Friction (θ = 180°), gravity on the falling coconut (θ = 0°) and the bullock's pull all act along or against the displacement.

Question 6
A porter lowers a suitcase from a rack to the floor at constant speed. Which statement about the work done on the suitcase by gravity (Wg) and by the porter's hands (Wp) is correct?
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Answer: (b) Wg is positive, Wp is negative, and Wg + Wp = 0

Gravity acts downward along the downward displacement (θ = 0°): Wg > 0. The porter's force is upward, opposite to the displacement (θ = 180°): Wp < 0. The speed is constant, so ΔK = 0 and by the work-energy theorem Wg + Wp = 0.

Question 7
A child pulls a toy cart with a string that makes 60° with the horizontal. The tension in the string is 10 N and the cart moves 5 m along the level floor. The work done by the tension is
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Answer: (d) 25 J

W = Fd cos θ = 10 × 5 × cos 60° = 25 J. Using sin 60° gives 43.3 J; leaving out the angle gives 50 J.

Question 8
The dimensions of the spring constant k are
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Answer: (a) [MT⁻²]

k = force/extension = [MLT⁻²]/[L] = [MT⁻²]; its unit is N m⁻¹. [MLT⁻²] is force and [ML²T⁻²] is energy.

Question 9
A 2 kW geyser is used for 45 minutes every day. The electrical energy it uses in 30 days is
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Answer: (c) 45 kWh

Energy = power × time = 2 kW × 0.75 h × 30 = 45 kWh. Using 45 h instead of 0.75 h each day gives 2700 kWh; using 1 h each day gives 60 kWh.

Question 10
The kinetic energy of a car of mass 1500 kg moving at 72 km h⁻¹ is
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Answer: (b) 300 kJ

v = 72 km h⁻¹ = 20 m s⁻¹, so K = ½mv² = ½ × 1500 × 20² = 3.0 × 10⁵ J = 300 kJ. Keeping v in km h⁻¹ gives 3888 kJ; leaving out the ½ gives 600 kJ.

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