Power Class 11 Physics Notes for NEET | Complete Revision Guide

 - Dr.Sanjaykumar Pawar  

Power Physics Notes PDF | NEET Work, Energy and Power Chapter

Power in Physics – Complete NEET Notes with Formulas, Units, Examples and Quick Revision

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Complete Class 11 Physics Notes

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Work, Energy and Power · Unit IV

Power — Complete Question Bank

A full CBSE Class 11 Physics practice set on Power: MCQs, assertion–reason, case studies, match-the-columns, numericals, and a quick answer key — click any answer to reveal it.

Class 11 · CBSE Chapter: Work, Energy & Power 100+ Questions Answer Key Included

Jump to a section

1. A · Multiple Choice Questions
2. B · Very Short Answer (1 mark)
3. C · Short Answer (2 marks)
4. D · Long Answer (3–5 marks)
5. E · Assertion & Reason
6. F · Fill in the Blanks
7. G · Case Study Based
8. H · Statement Based
9. I · Match the Columns
10. J · True or False
11. K · Numerical Problems
12. Answer Key (Quick Reference)
13. Exam Tips

Section A Multiple Choice Questions

Tap "Show Answer" under any question to reveal the solution.

Level · Easy

Q1.Power is defined as:

Easy

• (a) Total work done
• (b) Rate of doing work
• (c) Energy stored in a body
• (d) Force applied per unit area

Show Answer

Answer: (b) Rate of doing work

Q2.The SI unit of power is:

Easy

• (a) Joule
• (b) Newton
• (c) Watt
• (d) Pascal

Show Answer

Answer: (c) Watt

Q3.1 horsepower is equal to:

Easy

• (a) 500 W
• (b) 746 W
• (c) 1000 W
• (d) 980 W

Show Answer

Answer: (b) 746 W

Q4.A machine does 1000 J of work in 10 s. Its power is:

Easy

• (a) 10 W
• (b) 100 W
• (c) 1000 W
• (d) 10000 W

Show Answer

Answer: (b) 100 W

Q5.The dimensional formula of power is:

Easy

• (a) [ML²T⁻²]
• (b) [ML²T⁻³]
• (c) [MLT⁻²]
• (d) [M²L²T⁻²]

Show Answer

Answer: (b) [ML²T⁻³]

Q6.Power is a:

Easy

• (a) Vector quantity
• (b) Scalar quantity
• (c) Neither scalar nor vector
• (d) Tensor quantity

Show Answer

Answer: (b) Scalar quantity

Q7.A pump lifts 200 kg of water to a height of 5 m in 10 s (g = 10 m/s²). The power of the pump is:

Easy

• (a) 100 W
• (b) 500 W
• (c) 1000 W
• (d) 2000 W

Show Answer

Answer: (c) 1000 W

Q8.If force and velocity are perpendicular to each other, the power is:

Easy

• (a) Maximum
• (b) Minimum
• (c) Zero
• (d) Infinite

Show Answer

Answer: (c) Zero

Level · Medium

Q9.A car engine exerts a force of 500 N while moving at a constant speed of 20 m/s. The power of the engine in hp is approximately:

Medium

• (a) 10.7 hp
• (b) 13.4 hp
• (c) 15.2 hp
• (d) 20.0 hp

Show Answer

Answer: (b) 13.4 hpP = Fv = 500 × 20 = 10000 W = 10000/746 ≈ 13.4 hp

Q10.The position of a particle is given by x = 3t² + 2t (x in m, t in s). A force of 6 N acts on it. The power at t = 2 s is:

Medium

• (a) 72 W
• (b) 84 W
• (c) 96 W
• (d) 108 W

Show Answer

Answer: (b) 84 Wv = dx/dt = 6t + 2. At t = 2s, v = 14 m/s. P = Fv = 6 × 14 = 84 W

Q11.A body of mass 2 kg is moved by a force of 10 N at constant velocity of 5 m/s at 60° to the direction of force. The power is:

Medium

• (a) 50 W
• (b) 25 W
• (c) 43.3 W
• (d) 0 W

Show Answer

Answer: (b) 25 WP = Fv cosθ = 10 × 5 × cos60° = 50 × 0.5 = 25 W

Q12.An engine of power 2 kW can do how much work in 1 minute?

Medium

• (a) 120 J
• (b) 2000 J
• (c) 120000 J
• (d) 20000 J

Show Answer

Answer: (c) 120000 JW = P × t = 2000 × 60 = 120000 J

Q13.A force F acts on a body moving with velocity v. If the angle between F and v is 120°, the power is:

Medium

• (a) Fv
• (b) Fv/2
• (c) Zero
• (d) −Fv/2

Show Answer

Answer: (d) −Fv/2P = Fv cos120° = Fv × (−1/2) = −Fv/2

Q14.The power of a pump that can lift 5000 kg of water per minute to a height of 20 m is: (g = 10 m/s²)

Medium

• (a) 16.67 kW
• (b) 10 kW
• (c) 100 kW
• (d) 1.67 kW

Show Answer

Answer: (a) 16.67 kWP = mgh/t = (5000 × 10 × 20)/60 = 1,000,000/60 ≈ 16,667 W ≈ 16.67 kW

Q15.A man of mass 60 kg climbs up a staircase carrying a load of 20 kg. If the total height gained is 10 m in 20 s, the average power is: (g = 10 m/s²)

Medium

• (a) 200 W
• (b) 300 W
• (c) 400 W
• (d) 800 W

Show Answer

Answer: (c) 400 WTotal mass = 80 kg. P = mgh/t = (80 × 10 × 10)/20 = 400 W

Level · Hard

Q16.A particle of mass m moves along a circular path of radius r with uniform speed v. The power delivered by the centripetal force is:

Hard

• (a) mv²/r
• (b) mv³/r
• (c) Zero
• (d) mv²r

Show Answer

Answer: (c) ZeroCentripetal force is always perpendicular to velocity (θ = 90°), so P = Fv cos90° = 0

Q17.The power delivered to a body moving in a straight line is given by P = 3t² + 2t (in watts). The work done in the first 2 seconds is:

Hard

• (a) 10 J
• (b) 12 J
• (c) 14 J
• (d) 16 J

Show Answer

Answer: (d) 16 JW = ∫P dt = ∫(3t² + 2t)dt = t³ + t². At t = 2: W = 8 + 8 = 16 J

Q18.A body of mass 1 kg is thrown vertically upward with initial velocity 20 m/s. The instantaneous power due to gravity at t = 1 s is: (g = 10 m/s²)

Hard

• (a) 100 W
• (b) −100 W
• (c) 200 W
• (d) −200 W

Show Answer

Answer: (b) −100 Wv = u − gt = 20 − 10 = 10 m/s (upward). F = mg = 10 N (downward). P = Fv cos180° = 10 × 10 × (−1) = −100 W

Q19.A pump motor is rated at 5 hp. How many kilograms of water can it raise in 1 minute through a height of 10 m? (g = 10 m/s², 1 hp = 746 W)

Hard

• (a) 1492 kg
• (b) 2238 kg
• (c) 2984 kg
• (d) 3730 kg

Show Answer

Answer: (b) 2238 kgP = 5 × 746 = 3730 W. W = P × t = 3730 × 60 = 223,800 J. m = W/(gh) = 223,800/(10×10) = 2238 kg

Q20.A vehicle of mass m accelerates uniformly from rest to velocity v in time t. The instantaneous power delivered by the engine at time t/2 is:

Hard

• (a) mv²/2t
• (b) mv²/4t
• (c) mv²/t
• (d) 3mv²/4t

Show Answer

Answer: (a) mv²/2ta = v/t (constant). At t/2, instantaneous velocity v′ = a(t/2) = v/2. F = ma = mv/t. P = F·v′ = (mv/t)(v/2) = mv²/2t.

Section B Very Short Answer Questions (1 Mark Each)

Q1.Define power.

Show Answer

Power is defined as the rate of doing work or the rate of energy transfer. Mathematically, P = W/t

Q2.Write the SI unit of power.

Show Answer

Watt (W), where 1 W = 1 J/s.

Q3.What is 1 horsepower in watts?

Show Answer

1 hp = 746 W.

Q4.Is power a scalar or vector quantity? Why?

Show Answer

Power is a scalar quantity because it has only magnitude and no direction. It is the ratio of two scalars (work and time).

Q5.Write the dimensional formula of power.

Show Answer

[M¹L²T⁻³]

Q6.A force acts perpendicular to the velocity of a body. What is the power?

Show Answer

Zero, because P = Fv cos90° = 0.

Q7.What is the relation between power, force, and velocity?

Show Answer

P = Fv (when force and velocity are in the same direction).

Q8.A machine does no work. Can it have power?

Show Answer

No. Since P = W/t, if W = 0, then P = 0.

Q9.What is the commercial unit of power?

Show Answer

Horsepower (hp).

Q10.Write the expression for instantaneous power.

Show Answer

P = dW/dt or P = F·v (dot product of force and instantaneous velocity).

Section C Short Answer Questions (2 Marks Each)

Q1.Distinguish between average power and instantaneous power.

Show Answer

Average Power	Instantaneous Power
Defined as total work divided by total time	Defined as power at a particular instant
Formula: Pavg = W/t	Formula: Pinst = F·v
Used when total work and time are given	Used when force or velocity varies with time

Q2.Derive the relation P = Fv.

Show Answer

We know power P = W/t. Work done W = F × s (for constant force in direction of displacement). Therefore, P = (F × s)/t = F × (s/t) = F × v. Hence, P = Fv

Q3.A pump delivers 1000 liters of water per minute to a tank at a height of 20 m. Find the power of the pump. (Density of water = 1000 kg/m³, g = 10 m/s²)

Show Answer

Volume per second = 1000/60 = 50/3 L/s = (50/3) × 10⁻³ m³/s. Mass per second = ρ × volume/s = 1000 × (50/3) × 10⁻³ = 50/3 kg/s. Power = (m/t) × gh = (50/3) × 10 × 20 = 10000/3 ≈ 3333.33 W ≈ 3.33 kW

Q4.Show that the power delivered by the centripetal force in uniform circular motion is zero.

Show Answer

In uniform circular motion, the centripetal force is always directed towards the center, while the velocity is always tangential. So θ = 90° always. P = Fv cosθ = Fv cos90° = 0.

Q5.The power of an engine is 5 kW. How much work can it do in 10 minutes?

Show Answer

Given: P = 5 kW = 5000 W, t = 10 min = 600 s. W = P × t = 5000 × 600 = 3,000,000 J = 3 × 10⁶ J

Q6.A car of mass 1000 kg moves up an incline of 1 in 20 at a constant speed of 10 m/s. Find the power of the engine. (g = 10 m/s², neglect friction)

Show Answer

Slope = 1/20, so sinθ = 1/20. Force required F = mg sinθ = 1000 × 10 × (1/20) = 500 N. Power P = Fv = 500 × 10 = 5000 W = 5 kW

Q7.Why is the power of a body moving with constant velocity on a frictionless horizontal surface zero?

Show Answer

On a frictionless horizontal surface, no external force is required to maintain constant velocity (Newton's first law). Since F = 0, power P = Fv = 0 × v = 0.

Q8.A 2 kW motor pump is used to pump water from a well 10 m deep. How much water can be pumped per minute? (g = 10 m/s²)

Show Answer

P = 2 kW = 2000 W, t = 60 s. W = P × t = 2000 × 60 = 120,000 J. m = W/(gh) = 120,000/(10 × 10) = 1200 kg

Section D Long Answer Questions (3–5 Marks Each)

Q1.(a) Define power and derive its SI unit. (2 marks)
(b) A pump can throw 8000 kg of water per minute to a height of 15 m. Calculate the power of the pump. (g = 9.8 m/s²) (3 marks)

Show Answer

(a) Power is defined as the rate of doing work. If W is the work done in time t, then P = W/t. The SI unit of work is Joule (J) and time is second (s). Therefore, SI unit of power = J/s = Watt (W). 1 Watt = 1 Joule per second.

(b) Given: m = 8000 kg (per minute), h = 15 m, t = 60 s, g = 9.8 m/s². Work done per minute = mgh = 8000 × 9.8 × 15 = 1,176,000 J. Power P = W/t = 1,176,000/60 = 19,600 W = 19.6 kW

Q2.(a) Derive the expression for instantaneous power. (2 marks)
(b) The position of a body of mass 2 kg is given by x = 2t³ + 3t² + 5, where x is in meters and t in seconds. A constant force of 12 N acts on the body in the direction of motion. Calculate the power delivered at t = 2 s. (3 marks)

Show Answer

(a) Consider a small amount of work dW done in a small time interval dt. Instantaneous power P = dW/dt. Since dW = F·ds, we get P = (F·ds)/dt = F·(ds/dt) = F·v. Therefore, P = F·v (dot product of force and instantaneous velocity).

(b) x = 2t³ + 3t² + 5 → v = dx/dt = 6t² + 6t. At t = 2 s: v = 6(4) + 6(2) = 24 + 12 = 36 m/s. Power P = F × v = 12 × 36 = 432 W

Q3.(a) Prove that power can also be expressed as the scalar product of force and velocity. (2 marks)
(b) An engine of power 10 hp is used to pump water from a well 8 m deep. How many kilograms of water can be pumped in 1 hour? (1 hp = 746 W, g = 9.8 m/s²) (3 marks)

Show Answer

(a) Work done by a force F in displacing a body by ds is dW = F·ds. Power is rate of doing work: P = dW/dt = (F·ds)/dt = F·(ds/dt) = F·v. Hence, P = F·v.

(b) P = 10 hp = 10 × 746 = 7460 W. t = 1 hour = 3600 s. Total work W = P × t = 7460 × 3600 = 26,856,000 J. m = W/(gh) = 26,856,000/(9.8 × 8) = 26,856,000/78.4 ≈ 342,551 kg

Q4.(a) What is the difference between kilowatt and kilowatt-hour? (2 marks)
(b) A family uses a 2 kW electric heater for 4 hours daily. Calculate the energy consumed in 30 days in kWh and joules. (3 marks)

Show Answer

(a)

Kilowatt (kW)	Kilowatt-hour (kWh)
Unit of power	Unit of energy
1 kW = 1000 W	1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J
Measures rate of energy consumption	Measures total energy consumed

(b) Power P = 2 kW, daily usage t = 4 h. Daily energy = 2 × 4 = 8 kWh. 30-day energy = 8 × 30 = 240 kWh. In joules: 240 kWh = 240 × 3.6 × 10⁶ = 8.64 × 10⁸ J

Section E Assertion and Reason Questions

Choose: (a) Both true, Reason correctly explains Assertion  |  (b) Both true, Reason does NOT explain Assertion  |  (c) Assertion true, Reason false  |  (d) Assertion false, Reason true

Q1.Assertion: Power is a scalar quantity.
Reason: Power is the ratio of two scalar quantities, work and time.

Show Answer

Answer: (a) Both true and Reason correctly explains Assertion.

Q2.Assertion: When a body moves in a circular path with uniform speed, the power delivered by the centripetal force is zero.
Reason: The centripetal force is always perpendicular to the velocity of the body.

Show Answer

Answer: (a) Both true and Reason correctly explains Assertion.

Q3.Assertion: A body moving with constant velocity on a frictionless horizontal surface has zero power.
Reason: No force is required to maintain constant velocity on a frictionless surface.

Show Answer

Answer: (a) Both true and Reason correctly explains Assertion.

Q4.Assertion: 1 horsepower is equal to 1000 watts.
Reason: Horsepower is the commercial unit of power.

Show Answer

Answer: (c) Assertion is false but Reason is true. (1 hp = 746 W, not 1000 W)

Q5.Assertion: The instantaneous power of a body can be negative.
Reason: Power is always positive because it is the rate of doing work.

Show Answer

Answer: (c) Assertion is true but Reason is false. (Power can be negative when force opposes motion, e.g., friction doing negative work)

Q6.Assertion: Average power is always equal to instantaneous power.
Reason: Average power is calculated over a time interval while instantaneous power is at a specific instant.

Show Answer

Answer: (d) Assertion is false but Reason is true.

Q7.Assertion: The power delivered by gravity to a body thrown vertically upward is negative.
Reason: The gravitational force acts opposite to the direction of motion during upward journey.

Show Answer

Answer: (a) Both true and Reason correctly explains Assertion.

Q8.Assertion: A pump requires more power to lift the same amount of water to a greater height.
Reason: Power is directly proportional to the height through which water is lifted.

Show Answer

Answer: (a) Both true and Reason correctly explains Assertion.

Section F Fill in the Blanks

Q1.Power is defined as the ______ of doing work.

Show Answer

rate

Q2.The SI unit of power is ______, named after the scientist ______.

Show Answer

Watt, James Watt

Q3.1 horsepower = ______ watts.

Show Answer

746

Q4.The dimensional formula of power is ______.

Show Answer

[M¹L²T⁻³]

Q5.When force and velocity are perpendicular to each other, the power is ______.

Show Answer

zero

Q6.The commercial unit of power is ______.

Show Answer

horsepower (hp)

Q7.For a body moving with constant velocity on a frictionless surface, the power is ______.

Show Answer

zero

Q8.Instantaneous power is given by the scalar product of ______ and ______.

Show Answer

force, velocity

Q9.1 kilowatt = ______ watts.

Show Answer

1000

Q10.The power of a pump lifting mass m through height h in time t is given by P = ______.

Show Answer

mgh/t

Section G Case Study Based Questions

Case Study 1 · Hydroelectric Power Plant

A hydroelectric power plant uses water falling from a height to generate electricity. Water from a reservoir at a height of 100 m flows down through penstocks to turn turbines. The plant has a capacity to process 5000 kg of water per second. (Take g = 10 m/s²)

Q1.What is the power generated by the falling water?

Show Answer

P = mgh/t = (5000 × 10 × 100)/1 = 5,000,000 W = 5 MW

Q2.If the efficiency of the turbine-generator system is 80%, what is the actual electrical power output?

Show Answer

Actual power = 80% of 5 MW = 0.8 × 5 = 4 MW

Q3.How much energy is produced in 1 hour at this actual power output?

Show Answer

E = P × t = 4 MW × 1 h = 4 MWh = 4 × 10⁶ Wh = 1.44 × 10¹⁰ J

Case Study 2 · Electric Vehicle

An electric car of mass 1500 kg accelerates from rest to a speed of 30 m/s in 10 seconds. The motor delivers constant power during this time.

Q1.What is the acceleration of the car?

Show Answer

a = (v − u)/t = (30 − 0)/10 = 3 m/s²

Q2.What is the average power delivered by the motor during acceleration?

Show Answer

Work done = ΔKE = ½mv² = ½ × 1500 × 30² = 675,000 J. P_avg = W/t = 675,000/10 = 67,500 W = 67.5 kW

Q3.If the car maintains a constant speed of 30 m/s on a level road with a frictional force of 500 N, what power is required to overcome friction?

Show Answer

P = Fv = 500 × 30 = 15,000 W = 15 kW

Case Study 3 · Human Power Output

A person of mass 70 kg climbs a staircase of 50 steps, each 20 cm high, in 20 seconds. (g = 9.8 m/s²)

Q1.What is the total height climbed?

Show Answer

h = 50 × 0.20 = 10 m

Q2.What is the work done against gravity?

Show Answer

W = mgh = 70 × 9.8 × 10 = 6860 J

Q3.What is the average power output of the person?

Show Answer

P = W/t = 6860/20 = 343 W

Section H Statement Based Questions

Passage 1

"Power is the rate at which work is done. It is a scalar quantity. The SI unit of power is watt. When a force acts on a body in the direction of its motion, the power is given by P = Fv. If the force makes an angle θ with the velocity, then P = Fv cosθ."

Q1.Why is power called a scalar quantity?

Show Answer

Power is a scalar quantity because it has only magnitude and no direction. It is derived as the ratio of work (scalar) and time (scalar), and the dot product of two vectors (F·v) also yields a scalar.

Q2.A force of 20 N acts on a body moving with a velocity of 4 m/s. The force makes an angle of 60° with the direction of motion. Calculate the power.

Show Answer

P = Fv cosθ = 20 × 4 × cos60° = 80 × 0.5 = 40 W

Passage 2

"A pump is used to lift water from a well. The power of the pump depends on the mass of water lifted, the height to which it is lifted, and the time taken. The efficiency of the pump is defined as the ratio of useful power output to the total power input."

Q1.Write the expression for the power of a pump lifting water.

Show Answer

P = mgh/t

Q2.A pump of power 2 kW and efficiency 75% is used to lift water through 10 m. How much water can it lift in 1 minute? (g = 10 m/s²)

Show Answer

Useful power = 75% of 2 kW = 0.75 × 2000 = 1500 W. Work done in 1 min = 1500 × 60 = 90,000 J. m = W/(gh) = 90,000/(10 × 10) = 900 kg

Section I Match the Columns

Match the Following 1

Column A	Column B
(a) Unit of power	(p) [M¹L²T⁻³]
(b) Dimensional formula of power	(q) Joule
(c) Unit of work	(r) Watt
(d) 1 hp	(s) 746 W

Show Answer

(a) → (r)  |  (b) → (p)  |  (c) → (q)  |  (d) → (s)

Match the Following 2

Column A (Physical Quantity)	Column B (Expression)
(a) Average power	(p) F·v
(b) Instantaneous power	(q) W/t
(c) Power in lifting	(r) mgh/t
(d) Power when F ⊥ v	(s) Zero

Show Answer

(a) → (q)  |  (b) → (p)  |  (c) → (r)  |  (d) → (s)

Match the Following 3

Column A (Situation)	Column B (Power)
(a) Body moving with constant velocity on frictionless surface	(p) Positive
(b) Body thrown upward against gravity	(q) Negative
(c) Body falling freely under gravity	(r) Zero
(d) Body in uniform circular motion	(s) Variable

Show Answer

(a) → (r) Zero, as F = 0  |  (b) → (q) Negative, gravity opposes motion  |  (c) → (p) Positive, gravity aids motion  |  (d) → (r) Zero, centripetal force ⊥ velocity

Match the Following 4

Column A	Column B
(a) James Watt	(p) Unit of energy
(b) Joule	(q) Commercial unit of power
(c) Horsepower	(r) SI unit of power
(d) Kilowatt-hour	(s) Improved steam engine

Show Answer

(a) → (s)  |  (b) → (p)  |  (c) → (q)  |  (d) → (p) — kWh is a unit of energy, like Joule

Match the Following 5

Column A (Value)	Column B (Equivalent)
(a) 1 kW	(p) 3.6 × 10⁶ J
(b) 1 kWh	(q) 1000 W
(c) 1 hp	(r) 746 W
(d) 1 W	(s) 1 J/s

Show Answer

(a) → (q)  |  (b) → (p)  |  (c) → (r)  |  (d) → (s)

Section J True or False

Q1.Power and work have the same dimensions.

Show Answer

False. [Work] = [ML²T⁻²], [Power] = [ML²T⁻³]

Q2.1 kilowatt-hour is a unit of power.

Show Answer

False. It is a unit of energy.

Q3.The power of a body can be negative.

Show Answer

True. When force opposes motion (e.g., friction), power is negative.

Q4.A body in uniform circular motion has zero power due to centripetal force.

Show Answer

True. Centripetal force is perpendicular to velocity.

Q5.The SI unit of power is named after James Prescott Joule.

Show Answer

False. It is named after James Watt.

Section K Numerical Problems (With Detailed Solutions)

Q1.A pump lifts 1200 kg of water per minute to a height of 25 m. Calculate the power of the pump in kW. (g = 10 m/s²)

Show Solution

m = 1200 kg, h = 25 m, t = 60 s, g = 10 m/s². P = mgh/t = (1200 × 10 × 25)/60 = 300,000/60 = 5000 W = 5 kW

Q2.A car of mass 1000 kg moves on a level road at a constant speed of 20 m/s against a resistance of 400 N. Find the power of the engine in horsepower.

Show Solution

At constant speed, engine force = resistance = 400 N. P = Fv = 400 × 20 = 8000 W. In hp: 8000/746 ≈ 10.7 hp

Q3.The power of an engine is 5 kW. How much time will it take to lift a load of 500 kg to a height of 40 m? (g = 10 m/s²)

Show Solution

W = mgh = 500 × 10 × 40 = 200,000 J. P = 5000 W. t = W/P = 200,000/5000 = 40 s

Q4.A force F = (3i + 4j) N acts on a body moving with velocity v = (4i − 3j) m/s. Calculate the power.

Show Solution

P = F·v = (3)(4) + (4)(−3) = 12 − 12 = 0 W

Q5.A motor pump is rated at 3 hp. Calculate the maximum mass of water it can lift in 2 minutes through a height of 15 m. (1 hp = 746 W, g = 9.8 m/s²)

Show Solution

P = 3 × 746 = 2238 W. t = 120 s. W = P × t = 2238 × 120 = 268,560 J. m = W/(gh) = 268,560/(9.8 × 15) = 268,560/147 ≈ 1827 kg

Answer Key (Quick Reference — MCQs & Assertion-Reason)

Section	Q. No.	Answer
MCQ (Easy)	1	(b)
	2	(c)
	3	(b)
	4	(b)
	5	(b)
	6	(b)
	7	(c)
	8	(c)
MCQ (Medium)	9	(b)
	10	(b)
	11	(b)
	12	(c)
	13	(d)
	14	(a)
	15	(c)
MCQ (Hard)	16	(c)
	17	(d)
	18	(b)
	19	(b)
	20	(a)
Assertion-Reason	1	(a)
	2	(a)
	3	(a)
	4	(c)
	5	(c)
	6	(d)
	7	(a)
	8	(a)

Exam Tips for CBSE Class 11

1. Formula Sheet: Memorize P = W/t, P = Fv, P = Fv cosθ, and P = mgh/t
2. Unit Conversions: Always convert to SI units first (especially hp → W, min → s)
3. Sign Convention: Power is positive when force aids motion, negative when it opposes
4. Scalar Nature: Remember power is scalar — never add vectorially
5. Time Management: Most Power questions take 1–2 minutes max

Best of luck for your CBSE Class 11 exams — master these questions and you're set for full marks on Power. 🎓

NEET Physics Notes
Chapter: POWER

Definition:
Power is the rate at which work is done or energy is transferred.

Simple Meaning

Work tells us how much work is done, while Power tells us how fast the work is done.

Example

• Student A climbs 4 floors in 20 seconds.
• Student B climbs 4 floors in 40 seconds.

Both students do the same work, but Student A has more power because he completes the work in less time.

Average Power

P = W / t

Where

• P = Average Power (Watt)
• W = Work Done (Joule)
• t = Time Taken (Second)

More Work + Less Time = More Power

Instantaneous Power

The power at a particular instant of time is called Instantaneous Power.

P = dW / dt

Power in Terms of Force

We know

dW = F · dr

Therefore

P = dW / dt

Since

dr / dt = v

Hence

P = F · v

General Formula

P = Fv cosθ

Special Cases

Case 1: Force and Velocity in Same Direction

θ = 0°

P = Fv

Power is Maximum.

---

Case 2: Force Opposite to Velocity

θ = 180°

P = -Fv

Negative power means energy is removed from the body.

Example: Braking a moving car.

---

Case 3: Force Perpendicular to Velocity

θ = 90°

P = 0

Example: Uniform Circular Motion

The centripetal force changes only the direction of velocity, not its speed.

Nature of Power

Power is a Scalar Quantity.

Reason: It is obtained from the dot product of force and velocity.

Dimensions of Power

[ML²T⁻³]

SI Unit

Watt (W)

Named after James Watt, who improved the steam engine.

Definition of One Watt

1 Watt = 1 Joule / Second

A power of one watt means one joule of work is done every second.

Other Units

1 kW = 1000 W

1 hp = 746 W

Horsepower is commonly used for engines, cars and motorcycles.

Electrical Energy

Electrical appliances are rated in watts.

• 60 W Bulb
• 100 W Bulb
• 1000 W Heater

Higher wattage means the appliance consumes energy faster.

Kilowatt-hour (kWh)

Electricity bills are measured in kilowatt-hour (kWh).

Energy = Power × Time

Conversion

1 kWh = 1000 W × 3600 s

= 3.6 × 10⁶ J

Example

A 100 W bulb runs for 10 hours.

100 × 10 = 1000 Wh

= 1 kWh

A 100 W bulb used for 10 hours consumes 1 Unit of electricity.

Important Fact

1 Unit of Electricity = 1 kWh = 3.6 × 10⁶ Joules

Remember:
kWh is a unit of Energy, NOT Power.

Difference Between Power and Energy

Power	Energy
Rate of doing work	Capacity to do work
P = W / t	W = Pt
Unit = Watt (W)	Unit = Joule (J)
Scalar Quantity	Scalar Quantity
Measures speed of work	Measures amount of work

Graph Concept

Slope of Work-Time Graph = Power

NEET Formula Sheet

Formula	Expression
Average Power	P = W / t
Instantaneous Power	P = dW / dt
Power by Force	P = F · v
General Formula	P = Fv cosθ
Parallel Force	P = Fv
Perpendicular Force	P = 0
Opposite Force	P = -Fv
1 Watt	1 J/s
Horsepower	1 hp = 746 W
1 kWh	3.6 × 10⁶ J

NEET Quick Revision

• Power = Rate of doing work.
• Power = Speed of energy transfer.
• Power is a scalar quantity.
• P = Fv cosθ.
• Maximum power when θ = 0°.
• Zero power when θ = 90°.
• Negative power when θ = 180°.
• 1 Watt = 1 Joule/second.
• 1 Horsepower = 746 W.
• 1 Unit of electricity = 1 kWh = 3.6 × 10⁶ J.
• kWh is a unit of Energy, not Power.

Frequently Asked NEET Questions

Q1. What is Power?

Power is the rate at which work is done or energy is transferred.

Q2. Why is Power a Scalar Quantity?

Because Power is obtained from the dot product of Force and Velocity.

Q3. What is SI Unit of Power?

Watt (W).

Q4. Is kWh a unit of Power?

No. It is a unit of Energy.

Q5. What is the power when Force is perpendicular to Velocity?

P = 0

End of Chapter - POWER
Complete Beginner-Friendly NEET Physics Notes

Units and Measurements Class 11 Physics NCERT Solutions | Step-by-Step Answers, Short Notes & Mnemonics for NEET

- Dr.Sanjaykumar pawar 

Units and Measurements Class 11 Physics – Easy NCERT Solutions, Short Notes, Formulas, and NEET Revision Tricks

 Internal Links

Physics Class 11 Chapter 1 Physical World Notes

Measurement Errors Complete Notes

Significant Figures Explained

Vernier Calipers Complete Guide

Screw Gauge Complete Guide

Least Count Formula Explained

Dimensional Analysis Notes

NCERT Class 11 Physics Solutions

NEET Physics Short Notes

NEET Physics Formula Handbook

Class 11 Physics Important Questions

Physics MCQs with Solutions

NCERT Exemplar Physics Solutions

JEE Physics Revision Notes

Measurement Instruments Comparison

UNITS AND MEASUREMENTS

Question (a)

You are given a thread and a metre scale. How will you estimate the diameter of the thread?

Step-by-Step Answer

Step 1: Take a pencil or a cylindrical rod.

Step 2: Wind the thread closely around the pencil without leaving any gap.

Step 3: Make 20–50 turns of the thread.

Step 4: Measure the total length (L) of all turns using the metre scale.

Step 5: Count the total number of turns (N).

Diameter of Thread = Total Length (L) / Number of Turns (N)

Answer: Divide the total length of all turns by the number of turns to obtain the diameter of the thread.

Question (b)

A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?

Step-by-Step Answer

Step 1: Use the formula

Least Count = Pitch / Circular Scale Divisions

Step 2:

Least Count = 1.0 / 200 = 0.005 mm

Step 3: Increasing the number of divisions decreases the least count.

Step 4: However, accuracy cannot increase forever because of:

• Human error
• Instrument error
• Manufacturing limitations
• Backlash error

Final Answer:

No. Accuracy cannot be increased arbitrarily by increasing the number of divisions.

Question (c)

The mean diameter of a thin brass rod is to be measured using Vernier Calipers. Why is a set of 100 measurements more reliable than only 5 measurements?

Step-by-Step Answer

Step 1: Every measurement contains small random errors.

Step 2: With only 5 readings, random errors affect the average more.

Step 3: With 100 readings, positive and negative errors cancel each other.

Step 4: Therefore, the mean value becomes closer to the true value.

More Measurements → Less Random Error → Better Mean Value

Short Notes (Quick Revision)

Topic	Key Point
Thread Diameter	Wind the thread around a pencil and use Length ÷ Turns.
Screw Gauge	Measures very small thickness like wire diameter.
Least Count	Least Count = Pitch ÷ Circular Divisions.
Vernier Calipers	Measures external diameter, internal diameter and depth.
Repeated Measurements	More readings reduce random errors.

Remember:

• More Readings = More Accuracy
• Smaller Least Count = Better Precision
• Screw Gauge is more accurate than Vernier Calipers.

Mnemonics (Easy Memory Tricks)

1. Instrument Order

Mnemonic:

Meter → Vernier → Screw

Sentence:

"My Very Smart Friend"

Word	Meaning
My	Meter Scale
Very	Vernier Calipers
Smart	Screw Gauge

2. Accuracy Order

Mnemonic:

Meter < Vernier < Screw

Sentence:

"Accuracy Climbs Up"

Instrument	Accuracy
Meter Scale	Low
Vernier Calipers	Medium
Screw Gauge	Highest

3. Formula Memory Trick

L ÷ N = Diameter

Mnemonic:

"Length Needs Number"

Formula	Meaning
L	Total Length
N	Number of Turns
L ÷ N	Diameter

4. NEET MCQ Trick

If Question Says	Remember
More Readings	Better Mean Value
Increase Circular Divisions	Accuracy improves only up to practical limits.
Least Count	Smaller LC → Better Precision

NCERT Class 11 Physics Chapter 1 Exercises Questions Only | Units and Measurements

 NCERT Units and Measurements Exercise Questions 

NCERT Class 11 Physics Chapter 1 – Units and Measurements Exercise Questions.

- Dr.Sanjaykumar Pawar 

EXERCISES

Note: In stating numerical answers, take care of significant figures.

1.1 Fill in the blanks.

(a) The volume of a cube of side 1 cm is equal to ______ m³.

(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to ______ (mm)².

(c) A vehicle moving with a speed of 18 km h⁻¹ covers ______ m in 1 s.

(d) The relative density of lead is 11.3. Its density is ______ g cm⁻³ or ______ kg m⁻³. 

Answer -1.1 

1.2 Fill in the blanks by suitable conversion of units.

(a) 1 kg m² s⁻² = ______ g cm² s⁻²

(b) 1 m = ______ ly

(c) 3.0 m s⁻² = ______ km h⁻²

(d) G = 6.67 × 10⁻¹¹ N m² kg⁻² = ______ cm³ s⁻² g⁻¹ 

Answer - 1.2

1.3 A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1 J = 1 kg m² s⁻². Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, and the unit of time is γ s. Show that a calorie has a magnitude 4.2 α⁻¹ β⁻² γ² in terms of the new units. 

Answer -  1.3

1.4 Explain this statement clearly:

“To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison.”

In view of this, reframe the following statements wherever necessary:

(a) Atoms are very small objects.

(b) A jet plane moves with great speed.

(c) The mass of Jupiter is very large.

(d) The air inside this room contains a large number of molecules.

(e) A proton is much more massive than an electron.

(f) The speed of sound is much smaller than the speed of light. 

Answer - 1.4

1.5 A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance? 

Answer - 1.5

1.6 Which of the following is the most precise device for measuring length?

(a) A vernier callipers with 20 divisions on the sliding scale.

(b) A screw gauge of pitch 1 mm and 100 divisions on the circular scale.

(c) An optical instrument that can measure length to within a wavelength of light. 

Answer - 1.6

1.7 A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate of the thickness of hair? 

Answer - 1.7

1.8 Answer the following:

(a) You are given a thread and a metre scale. How will you estimate the diameter of the thread?

(b) A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?

(c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only? 

Answer - 1.8

1.9 The photograph of a house occupies an area of 1.75 cm² on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m². What is the linear magnification of the projector-screen arrangement? 

Answer - 1.9

1.10 State the number of significant figures in the following:

(a) 0.007 m²

(b) 2.64 × 10²⁴ kg

(c) 0.2370 g cm⁻³

(d) 6.320 J

(e) 6.032 N m⁻²

(f) 0.0006032 m² 

Answer -1.10

1.11 The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures. 

Answer -1.11

1.12 The mass of a box measured by a grocer’s balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is:

(a) the total mass of the box?

(b) the difference in the masses of the pieces to correct significant figures? 

Answer - 1.12

1.13 A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m₀ of a particle in terms of its speed v and the speed of light c. A boy recalls the relation almost correctly but forgets where to put the constant c. He writes:

m/m₀ = (1 − v²)⁻¹ᐟ²

Guess where to put the missing c. 

Answer - 1.13

1.14 The unit of length convenient on the atomic scale is known as an angstrom (Å): 1 Å = 10⁻¹⁰ m. The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m³ of a mole of hydrogen atoms? 

Answer - 1.14

1.15 One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of the hydrogen molecule to be about 1 Å.) Why is this ratio so large? 

Answer - 1.15

1.16 Explain this common observation clearly: If you look out of the window of a fast-moving train, the nearby trees, houses, etc. seem to move rapidly in a direction opposite to the train’s motion, but the distant objects (hill tops, the Moon, the stars, etc.) seem to be stationary. 

Answer - 1.16

1.17 The Sun is a hot plasma (ionized matter) with its inner core at a temperature exceeding 10⁷ K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases? Check if your guess is correct from the following data:

Mass of the Sun = 2.0 × 10³⁰ kg

Radius of the Sun = 7.0 × 10⁸ m.  

Answer - 1.17

Internal Links

NCERT Class 11 Physics Chapter 1 Notes

NCERT Class 11 Physics Chapter 1 Solutions

SI Units and Dimensions Explained

Significant Figures and Errors Notes

Measurement of Length, Mass and Time

NCERT Class 11 Physics Chapter-wise Solutions

NCERT Class 11 Physics MCQs

NCERT Class 11 Physics Important Questions

CBSE Class 11 Physics Study Material

NCERT Class 11 Physics Previous Year Questions

Introduction to Units and Measurement Notes for NEET Physics

  

Introduction to Units and Measurement (NEET Level – Easy Notes) 

Introduction to Units and Measurement – Complete NEET Physics Mind Map and Quick Revision Notes.

- Dr.Sanjaykumar Pawar 

1. Measurement

• Measurement means finding the value of a physical quantity by comparing it with a standard quantity.
• Example: To measure the length of a table, we compare it with a standard unit like metre (m).

2. Physical Quantity

• A physical quantity is any quantity that can be measured.
• Examples: Length, mass, time, temperature, force, speed, etc.

3. Unit

• A unit is a fixed standard used to measure a physical quantity.
• It is internationally accepted and remains the same everywhere.
• Examples:
  • Length → metre (m)
  • Mass → kilogram (kg)
  • Time → second (s)

4. Why Do We Need Units?

• Units provide a common standard for measurement.
• Without units, measurements would be confusing and inconsistent.
• Example: Saying "the rod is 5" is incomplete. We must say "the rod is 5 metres long."

5. Result of Measurement

• Every measurement has two parts:
  1. Numerical value (number)
  2. Unit
• Example:
  • Length = 10 m
  • Here, 10 is the numerical value and m is the unit.

6. Number of Physical Quantities

• There are a very large number of physical quantities in physics.
• However, only a limited number of basic units are needed to express all of them.
• This is because many physical quantities are related to one another.

7. Fundamental (Base) Quantities

• Fundamental quantities are basic physical quantities that do not depend on other quantities.
• Examples:
  • Length
  • Mass
  • Time
  • Electric current
  • Temperature
  • Amount of substance
  • Luminous intensity

8. Fundamental (Base) Units

• The units of fundamental quantities are called fundamental or base units.
• Examples:
  • Length → metre (m)
  • Mass → kilogram (kg)
  • Time → second (s)

9. Derived Quantities

• Quantities that can be expressed using fundamental quantities are called derived quantities.
• Examples:
  • Speed = Distance / Time
  • Force = Mass × Acceleration
  • Density = Mass / Volume

10. Derived Units

• Units obtained by combining base units are called derived units.
• Examples:
  • Speed → m/s
  • Force → kg·m/s² (newton, N)
  • Density → kg/m³

11. System of Units

• A complete collection of base units and derived units is called a system of units.
• It provides a standard method of measurement.

12. SI System (Most Important for NEET)

• SI stands for International System of Units.
• It is the globally accepted system of units.
• It contains 7 base units.

Fundamental Quantity	SI Unit	Symbol
Length	metre	m
Mass	kilogram	kg
Time	second	s
Electric current	ampere	A
Temperature	kelvin	K
Amount of substance	mole	mol
Luminous intensity	candela	cd

---

NEET Quick Revision Points

✅ Measurement = Comparison with a standard unit.

✅ Physical Quantity = Quantity that can be measured.

✅ Measurement result = Numerical value + Unit.

✅ Fundamental quantities are independent quantities.

✅ Fundamental units are units of fundamental quantities.

✅ Derived quantities depend on fundamental quantities.

✅ Derived units are combinations of base units.

✅ System of units = Collection of base units + derived units.

✅ SI system is the internationally accepted system of units.

One-Line Formula

Physical Quantity = Numerical Value × Unit

Example: Length = 5 m = 5 × metre. 

Below are CBSE Class 11 Physics (Units and Measurements – Introduction) exam-oriented questions with answers.

1. Multiple Choice Questions (MCQs)

Q1. Measurement of a physical quantity involves comparison with:

(a) Instrument
(b) Unit
(c) Formula
(d) Constant

Answer: (b) Unit

---

Q2. The internationally accepted standard for measurement is called:

(a) Scale
(b) Instrument
(c) Unit
(d) Quantity

Answer: (c) Unit

---

Q3. Which of the following is a fundamental quantity?

(a) Force
(b) Speed
(c) Length
(d) Density

Answer: (c) Length

---

Q4. Which of the following is a derived quantity?

(a) Mass
(b) Time
(c) Temperature
(d) Force

Answer: (d) Force

---

Q5. SI unit of length is:

(a) cm
(b) km
(c) metre
(d) inch

Answer: (c) metre

---

Q6. The SI unit of mass is:

(a) gram
(b) kilogram
(c) tonne
(d) pound

Answer: (b) kilogram

---

Q7. Which is NOT a fundamental quantity?

(a) Length
(b) Mass
(c) Time
(d) Speed

Answer: (d) Speed

---

Q8. A complete set of units is called:

(a) Physical quantity
(b) Standard
(c) System of units
(d) Measurement

Answer: (c) System of units

---

2. Very Short Answer Questions (1 Mark)

Q1. What is measurement?

Answer: Measurement is the comparison of a physical quantity with a standard unit.

---

Q2. Define unit.

Answer: A unit is a fixed standard used for measuring a physical quantity.

---

Q3. What are fundamental quantities?

Answer: Quantities that are independent and cannot be expressed in terms of other quantities.

---

Q4. Give one example of a derived quantity.

Answer: Force.

---

Q5. Write the SI unit of time.

Answer: Second (s).

---

3. Short Answer Questions (2–3 Marks)

Q1. What is a physical quantity? Give two examples.

Answer: A physical quantity is a quantity that can be measured and expressed by a number and a unit.

Examples:

1. Length
2. Mass

---

Q2. Differentiate between fundamental and derived quantities.

Fundamental Quantity	Derived Quantity
Independent quantity	Depends on fundamental quantities
Cannot be expressed in terms of other quantities	Can be expressed using fundamental quantities
Example: Length	Example: Speed

---

Q3. What are derived units? Give two examples.

Answer: Units obtained by combining fundamental units are called derived units.

Examples:

1. Speed = m/s
2. Force = kg m s⁻²

---

Q4. Why are units necessary?

Answer: Units are necessary because they:

• Provide a standard for measurement.
• Make communication of measurements clear.
• Allow comparison of physical quantities.

---

4. Long Answer Questions (5 Marks)

Q1. Explain fundamental quantities and derived quantities with examples.

Answer:

Fundamental Quantities

• These are basic physical quantities.
• They do not depend on other quantities.
• Examples: Length, Mass, Time.

Derived Quantities

• Quantities derived from fundamental quantities.
• Examples:
  • Speed = Distance/Time
  • Force = Mass × Acceleration
  • Density = Mass/Volume

Thus, all physical quantities can be expressed using fundamental quantities.

---

Q2. Explain the SI system of units.

Answer:

The SI system is the internationally accepted system of units.

The seven SI base units are:

Quantity	Unit	Symbol
Length	metre	m
Mass	kilogram	kg
Time	second	s
Electric Current	ampere	A
Temperature	kelvin	K
Amount of Substance	mole	mol
Luminous Intensity	candela	cd

The SI system ensures uniformity in measurements throughout the world.

---

5. Assertion and Reason Questions

Q1.

Assertion (A): Speed is a derived quantity.

Reason (R): Speed is obtained by dividing distance by time.

Answer: Both A and R are true and R is the correct explanation of A.

---

Q2.

Assertion (A): Length is a fundamental quantity.

Reason (R): Length can be expressed in terms of speed and time.

Answer: Assertion is true but Reason is false.

---

Q3.

Assertion (A): Force is a derived quantity.

Reason (R): Force depends on mass and acceleration.

Answer: Both A and R are true and R is the correct explanation.

---

6. Fill in the Blanks

1. Measurement involves comparison with a standard ________. Answer: unit

2. The SI unit of mass is ________. Answer: kilogram

3. The SI unit of length is ________. Answer: metre

4. Force is a ________ quantity. Answer: derived

5. A complete set of units is called a ________. Answer: system of units

6. Numerical value together with unit gives the result of a ________. Answer: measurement

---

7. Statement-Based Questions

Q1. Identify whether the statements are True or False.

(a) Every physical quantity has a numerical value and a unit. Answer: True

(b) Speed is a fundamental quantity. Answer: False

(c) Derived units are obtained from base units. Answer: True

(d) Kilogram is the SI unit of mass. Answer: True

---

Q2. Choose the correct statement.

(a) Force is a fundamental quantity. (b) Length is a derived quantity. (c) Mass is a fundamental quantity. (d) Speed is a fundamental quantity.

Answer: (c) Mass is a fundamental quantity.

---

8. Match the Following

Column A	Column B
(A) Length	(i) kg
(B) Mass	(ii) m
(C) Time	(iii) s
(D) Force	(iv) N

Answer:

A → ii

B → i

C → iii

D → iv

---

9. Case Study Questions

Case Study

A student measures the length of a table and finds it to be 2 m. Here, 2 is the numerical value and m is the unit. Measurement is the comparison of a physical quantity with a standard unit.

Questions

Q1. What is the measured physical quantity?

Answer: Length

---

Q2. What is the numerical value?

Answer: 2

---

Q3. What is the unit used?

Answer: Metre (m)

---

Q4. Is length a fundamental or derived quantity?

Answer: Fundamental quantity

---

Q5. Write the SI unit of length.

Answer: Metre (m)

---

Important CBSE Exam Questions

1. Define measurement and unit.
2. What is a physical quantity?
3. Differentiate between fundamental and derived quantities.
4. What are derived units? Give examples.
5. Explain the SI system of units.
6. List the seven SI base units.
7. Why are standard units necessary?
8. Write the difference between base units and derived units.

 INTRODUCTION TO UNITS AND MEASUREMENT

│

├── Measurement

│   ├── Comparison of physical quantity

│   ├── Compared with standard reference

│   └── Gives numerical value + unit

│

├── Physical Quantity

│   ├── Can be measured

│   ├── Examples

│   │   ├── Length

│   │   ├── Mass

│   │   ├── Time

│   │   ├── Temperature

│   │   └── Force

│

├── Unit

│   ├── Standard reference for measurement

│   ├── Arbitrarily chosen

│   └── Internationally accepted

│

├── Result of Measurement

│   ├── Numerical Value (Number)

│   └── Unit

│       └── Example: 10 m

│           ├── 10 → Number

│           └── m → Unit

│

├── Physical Quantities

│   ├── Very large in number

│   ├── Inter-related with each other

│   └── Need only limited basic units

│

├── Fundamental (Base) Quantities

│   ├── Independent quantities

│   ├── Length

│   ├── Mass

│   ├── Time

│   ├── Electric Current

│   ├── Temperature

│   ├── Amount of Substance

│   └── Luminous Intensity

│

├── Fundamental (Base) Units

│   ├── Units of fundamental quantities

│   ├── Metre (m)

│   ├── Kilogram (kg)

│   ├── Second (s)

│   ├── Ampere (A)

│   ├── Kelvin (K)

│   ├── Mole (mol)

│   └── Candela (cd)

│

├── Derived Quantities

│   ├── Depend on fundamental quantities

│   ├── Speed = Distance / Time

│   ├── Force = Mass × Acceleration

│   └── Density = Mass / Volume

│

├── Derived Units

│   ├── Combination of base units

│   ├── Speed → m s⁻¹

│   ├── Force → kg m s⁻² (N)

│   └── Density → kg m⁻³

│

└── System of Units

    ├── Collection of base units

    ├── Collection of derived units

    └── SI System

        ├── Internationally accepted

        └── Contains 7 base units

NEET FORMULA

Physical 

 = Numerical Value × Unit 

Internal Links

Physical Quantities and Their Types

SI Units and Dimensions

Errors in Measurement

Significant Figures

Dimensional Analysis

Motion in One Dimension

Vectors and Scalars

Basic Mathematics for Physics

Kinematics Complete Notes

NEET Physics Formula Sheet

Solving Problems in Mechanics NEET Notes | Free Body Diagram & Newton's Laws

 - Dr.Sanjaykumar pawar 

Understanding Free Body Diagrams is the first step toward solving mechanics problems in NEET Physics.

Solving Problems in Mechanics Class 11 Physics Notes for NEET

4.11 SOLVING PROBLEMS IN MECHANICS

│

├── Foundation

│   ├── Newton's First Law

│   ├── Newton's Second Law

│   └── Newton's Third Law

│

├── Mechanics Problems

│   ├── Single Body Problems

│   └── Multiple Body Systems

│       ├── Bodies interact with each other

│       ├── Gravity acts on each body

│       └── Forces act between bodies

│

├── Important Concepts

│   │

│   ├── System

│   │   ├── Chosen part for analysis

│   │   └── Newton's laws applied here

│   │

│   └── Environment

│       ├── Remaining bodies

│       ├── Surroundings

│       └── External force agencies

│

├── Problem Solving Steps

│   │

│   ├── Step 1: Draw Complete Diagram

│   │   ├── Bodies

│   │   ├── Strings

│   │   ├── Supports

│   │   └── Pulleys

│   │

│   ├── Step 2: Choose System

│   │   ├── Single body

│   │   └── Group of bodies

│   │

│   ├── Step 3: Draw Free Body Diagram (FBD)

│   │   ├── Show selected system only

│   │   ├── Show all external forces

│   │   └── Exclude surroundings

│   │

│   ├── Step 4: Include Known Information

│   │   ├── Force magnitudes

│   │   ├── Directions

│   │   ├── Tension direction

│   │   └── Unknowns as variables

│   │

│   └── Step 5: Repeat for Other Systems

│       └── Use Newton's Third Law

│

├── Free Body Diagram (FBD)

│   │

│   ├── Definition

│   │   └── Diagram showing all external forces

│   │

│   ├── Include

│   │   ├── Weight (mg)

│   │   ├── Normal Reaction (N)

│   │   ├── Tension (T)

│   │   ├── Friction (f)

│   │   └── Applied Forces

│   │

│   └── Do Not Include

│       ├── Surrounding objects

│       └── Forces exerted by system on environment

│

├── Newton's Third Law in FBD

│   │

│   ├── Force on A due to B = F

│   ├── Force on B due to A = -F

│   ├── Equal Magnitude

│   ├── Opposite Direction

│   └── Act on Different Bodies

│

├── Common Forces

│   │

│   ├── Weight

│   │   └── W = mg

│   │

│   ├── Normal Reaction

│   │   └── Perpendicular to surface

│   │

│   ├── Tension

│   │   └── Along string

│   │

│   └── Friction

│       └── Opposes motion

│

├── Newton's Laws Equations

│   │

│   ├── Equilibrium

│   │   └── ΣF = 0

│   │

│   ├── Motion

│   │   └── ΣF = ma

│   │

│   └── Weight

│       └── W = mg

│

├── NEET Tips

│   ├── Draw FBD first

│   ├── Identify all forces

│   ├── Choose convenient system

│   ├── Apply Newton's laws

│   └── Solve equations

│

└── Golden Rule

    └── "No FBD = No Proper Solution"

Class 11 Physics (CBSE)

Chapter: Laws of Motion

Topic: 4.11 Solving Problems in Mechanics

A. Multiple Choice Questions (MCQs)

1. The selected part of an assembly on which Newton's laws are applied is called:

(a) Environment (b) Force (c) System (d) Reaction

Answer: (c) System

---

2. A Free Body Diagram (FBD) shows:

(a) Only the body (b) Only the surroundings (c) The body and all external forces acting on it (d) Internal forces only

Answer: (c)

---

3. Which force must always be included in the FBD of an object near Earth?

(a) Tension (b) Friction (c) Weight (d) Spring force

Answer: (c)

---

4. The force exerted by a string on a body is called:

(a) Normal reaction (b) Tension (c) Friction (d) Weight

Answer: (b)

---

5. Newton's Third Law states:

(a) F = ma (b) ΣF = 0 (c) Every action has an equal and opposite reaction (d) Momentum is conserved

Answer: (c)

---

6. Normal reaction acts:

(a) Along the surface (b) Opposite to velocity (c) Perpendicular to the surface (d) Vertically upward always

Answer: (c)

---

7. In a Free Body Diagram, we should show:

(a) Forces acting on surroundings (b) Forces acting on the chosen body (c) Internal details of the body (d) Shape of surroundings

Answer: (b)

---

8. Friction acts:

(a) Along motion (b) Opposite to tendency of motion (c) Vertically upward (d) Downward

Answer: (b)

---

9. The first step in solving a mechanics problem is:

(a) Apply F = ma (b) Draw complete diagram (c) Calculate acceleration (d) Draw graph

Answer: (b)

---

10. Weight of a body is:

(a) N (b) mg (c) ma (d) mv

Answer: (b)

---

B. Very Short Answer Questions (1 Mark)

1. What is a system?

Answer: The selected object or group of objects chosen for analysis is called a system.

2. What is an environment?

Answer: Everything outside the system that exerts force on it is called the environment.

3. What is FBD?

Answer: A Free Body Diagram is a diagram showing a body and all external forces acting on it.

4. Write the formula for weight.

Answer: W = mg

5. Which force acts perpendicular to a surface?

Answer: Normal reaction.

6. What is the SI unit of force?

Answer: Newton (N)

7. Along which direction does tension act?

Answer: Along the string.

8. Which law gives F = ma?

Answer: Newton's Second Law.

9. What is the direction of weight?

Answer: Vertically downward.

10. Which force opposes motion?

Answer: Friction.

---

C. Short Answer Questions (2–3 Marks)

1. Define a Free Body Diagram.

Answer: A Free Body Diagram (FBD) is a diagram of a selected body showing all external forces acting on it. It helps in applying Newton's laws correctly.

---

2. Why is FBD important?

Answer:

1. It identifies all forces acting on a body.
2. It simplifies complex problems.
3. It helps apply Newton's laws accurately.

---

3. State Newton's Third Law.

Answer: For every action, there is an equal and opposite reaction. The two forces act on different bodies and are equal in magnitude but opposite in direction.

---

4. Differentiate between system and environment.

System	Environment
Chosen for study	Surroundings
Newton's laws applied	Exerts forces on system
Main object of analysis	External agencies

---

5. What forces act on a block resting on a table?

Answer:

1. Weight (mg) downward.
2. Normal reaction (N) upward.

---

D. Long Answer Questions (5 Marks)

1. Explain the steps involved in solving mechanics problems.

Answer:

The following steps should be followed:

1. Draw the complete diagram.
2. Choose a convenient system.
3. Draw the Free Body Diagram.
4. Include all known forces and directions.
5. Treat unknown forces as variables.
6. Apply Newton's laws.
7. Solve the equations obtained.

These steps help solve mechanics problems systematically.

---

2. Explain Free Body Diagram with an example.

Answer:

A Free Body Diagram is a diagram showing a body isolated from its surroundings along with all external forces acting on it.

Example: For a block resting on a table:

Forces acting:

• Weight (mg) downward
• Normal reaction (N) upward

Using equilibrium:

N = mg

The FBD helps in identifying forces and solving the problem correctly.

---

E. Assertion and Reason Questions

1.

Assertion (A): A Free Body Diagram contains all external forces acting on a body.

Reason (R): FBD helps in applying Newton's laws.

Answer: Both A and R are true and R is the correct explanation of A.

---

2.

Assertion (A): Weight acts vertically downward.

Reason (R): Weight is the gravitational force exerted by Earth.

Answer: Both A and R are true and R correctly explains A.

---

3.

Assertion (A): Action and reaction act on the same body.

Reason (R): Newton's Third Law states forces are equal and opposite.

Answer: Assertion is false but Reason is true.

---

4.

Assertion (A): Friction opposes relative motion.

Reason (R): Friction acts along the direction of motion.

Answer: Assertion is true but Reason is false.

---

5.

Assertion (A): Normal reaction acts perpendicular to a surface.

Reason (R): It is a contact force.

Answer: Both A and R are true but R is not the correct explanation.

---

F. Fill in the Blanks

1. The selected body for analysis is called a _______. Answer: System

2. The force due to gravity is called _______. Answer: Weight

3. The force exerted by a string is called _______. Answer: Tension

4. A diagram showing all external forces is called _______. Answer: Free Body Diagram

5. Weight of a body is equal to _______. Answer: mg

6. Newton's Second Law is represented by _______. Answer: F = ma

7. Friction opposes _______. Answer: Motion

8. Normal reaction acts _______ to the surface. Answer: Perpendicular

9. Newton's Third Law involves action and _______. Answer: Reaction

10. The SI unit of force is _______. Answer: Newton

---

G. Statement-Based Questions

Statement I:

A Free Body Diagram contains only the chosen body.

Statement II:

All external forces acting on the body must be shown.

(a) Both statements are true. (b) Both are false. (c) Statement I true, II false. (d) Statement I false, II true.

Answer: (a)

---

Statement I:

Action and reaction act on the same body.

Statement II:

Action and reaction are equal and opposite.

Answer: Statement I is false and Statement II is true.

---

H. Match the Columns

Column A	Column B
A. Weight	1. Along string
B. Tension	2. Opposes motion
C. Friction	3. mg
D. Normal Reaction	4. Perpendicular to surface

Answer:

A → 3

B → 1

C → 2

D → 4

---

I. Case Study Questions

Case Study

A block rests on a horizontal table. The block experiences its weight downward and the normal reaction upward. The block remains at rest.

Q1. Which force acts downward?

Answer: Weight (mg)

Q2. Which force acts upward?

Answer: Normal reaction

Q3. Why does the block remain at rest?

Answer: Net force is zero.

Q4. Which law explains equilibrium?

Answer: Newton's First Law.

Q5. What is the relation between N and mg?

Answer: N = mg

---

J. Competency-Based Questions

1.

A student forgets to include weight while drawing an FBD. How will this affect the solution?

Answer: The force analysis becomes incorrect and wrong answers may be obtained.

---

2.

Why should action and reaction never be shown on the same FBD?

Answer: Because they act on different bodies and belong to different free body diagrams.

---

Important CBSE Exam Tip

Always draw a neat Free Body Diagram before applying Newton's laws. Most mistakes in mechanics occur because of incorrect or incomplete FBDs.

 Internal Links

Newton's First Law of Motion – Complete Notes

Newton's Second Law of Motion – Formula and Examples

Newton's Third Law of Motion Explained

Force and Its Types in Physics

Free Body Diagram (FBD) Explained with Examples

Friction – Static and Kinetic Friction Notes

Tension in Strings and Pulley Problems

Laws of Motion Class 11 Notes

Work, Energy and Power NEET Notes

Important NEET Mechanics Questions and Solutions

Circular Motion NEET Notes

Centre of Mass and Momentum Notes

Gravitation Complete NEET Guide

Physics Formula Sheet for NEET

Class 11 Physics Chapter-wise Revision Notes

4.11 Solving Problems in Mechanics (NEET Notes)

Introduction

The three laws of motion are the foundation of mechanics. In many problems, more than one object is involved. These objects interact with each other through different forces such as tension, friction, normal reaction, and gravity.

To solve such problems systematically, we divide the arrangement into:

• System → The part chosen for study.
• Environment → Everything outside the system that exerts force on it.

Key Idea: Newton's laws can be applied to any chosen system if all external forces acting on it are considered.

1. System

The object or group of objects selected for analysis is called the system.

Example

• A single block on a table.
• Two connected blocks together.
• A pulley-block arrangement.

2. Environment

The surroundings that interact with the system are called the environment.

Example

• For a block on a table:
• System = Block
• Environment = Table, Earth, String, etc.

Free Body Diagram (FBD)

Definition: A Free Body Diagram (FBD) is a diagram showing only the chosen system and all external forces acting on it.

An FBD does not mean that the body is free from forces. It simply means the body is shown separately from its surroundings.

Steps for Solving Mechanics Problems

Step 1: Draw the Complete Diagram

• Draw all bodies.
• Show strings and pulleys.
• Show supports and surfaces.
• Understand the complete arrangement.

Step 2: Choose the System

Select a convenient object or group of objects for analysis.

Step 3: Draw the Free Body Diagram (FBD)

Separate the chosen system and show all external forces acting on it.

Step 4: Include Known Information

• Given forces
• Directions of forces
• Tension direction in strings
• Unknown forces represented by symbols

Step 5: Apply Newton's Laws

Use Newton's laws to determine unknown quantities.

Newton's First Law

ΣF = 0

If the net force acting on a body is zero, the body remains at rest or moves with constant velocity.

Newton's Second Law

ΣF = ma

The acceleration of a body is directly proportional to the net force acting on it.

Newton's Third Law

For every action, there is an equal and opposite reaction.

If object A exerts force F on object B, then object B exerts force -F on object A.

Using Newton's Third Law in FBD

Suppose two blocks A and B are in contact.

• Force by B on A = F
• Force by A on B = -F

These forces:

• Are equal in magnitude.
• Are opposite in direction.
• Act on different bodies.

Rules for Drawing Free Body Diagrams

Do's ✔

• Draw only the selected body.
• Show all external forces.
• Mark directions clearly.
• Label unknown forces.

Don'ts ✘

• Do not draw surrounding objects.
• Do not omit weight (mg).
• Do not omit normal reaction.
• Do not show action-reaction pairs on the same body.

Common Forces in Mechanics

Force	Symbol	Direction
Weight	W = mg	Vertically Downward
Normal Reaction	N	Perpendicular to Surface
Tension	T	Along the String
Friction	f	Opposite to Motion
Applied Force	F	Specified Direction

Example: Block on a Table

A block is resting on a horizontal table.

Forces acting on the block:

• Weight (mg) downward
• Normal reaction (N) upward

N ↑

[ BLOCK ]

mg ↓

Since the block is at rest:

N = mg

NEET Quick Tips

Always Follow:
1. Draw Diagram
2. Choose System
3. Draw FBD
4. Identify Forces
5. Apply Newton's Laws
6. Solve Equations

Common Mistakes in NEET

❌ Forgetting Weight (mg)
❌ Wrong Friction Direction
❌ Missing Normal Force
❌ Drawing Action-Reaction Pair on Same Body
❌ Solving Without FBD

Formula Box

W = mg

ΣF = 0

ΣF = ma

One-Line Summary

To solve any mechanics problem, first choose a system, draw its Free Body Diagram (FBD), identify all external forces, and then apply Newton's Laws of Motion.

⭐ Golden Rule for NEET: "No FBD = No Proper Solution."