Motion in a Plane — Class 11 Physics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 11 Physics chapter "Motion in a Plane" — 7 important questions with detailed answers for CBSE board exam preparation.

Motion in a Plane — Class 11 Physics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 11 Physics chapter "Motion in a Plane" — 7 important questions with detailed answers for CBSE board exam preparation.

✓ 100% Free ✓ No Login Needed ✓ NCERT / CBSE Aligned ✓ Download as PDF

TL;DR: Free step-by-step NCERT solutions for Class 11 Physics chapter "Motion in a Plane" — 7 important questions with detailed answers for CBSE board exam p…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated

🤖 Stuck on any question? Ask Syllab's free AI Tutor for a step-by-step explanation — instant, unlimited, no login.

NCERT Solutions for Class 11 Physics — Motion in a Plane. Step-by-step answers to all 7 textbook questions from this chapter, written for CBSE board preparation and free to use.

Motion in a Plane — All 7 Questions Solved

Q1. Define projectile motion. A ball is thrown horizontally from a cliff of height 45 m with initial velocity 20 m/s. Taking g = 10 m/s², find (a) time to reach ground (b) horizontal distance travelled (c) velocity when it hits the ground.

Projectile Motion Definition:
Projectile motion is the motion of an object that is launched into the air and moves under the influence of gravity alone (neglecting air resistance). The path followed by a projectile is called a trajectory, which is parabolic in shape.

Given:
Height h = 45 m
Initial horizontal velocity u_x = 20 m/s
Initial vertical velocity u_y = 0 m/s (thrown horizontally)
g = 10 m/s²

Part (a): Time to reach ground

Vertical motion: h = u_y × t + (1/2)gt²
45 = 0 + (1/2)(10)t²
45 = 5t²
t² = 9
t = 3 s

Part (b): Horizontal distance travelled

Horizontal motion (uniform velocity):
Range R = u_x × t
R = 20 × 3 = 60 m

Part (c): Velocity when hitting ground

Horizontal component (remains constant):
v_x = u_x = 20 m/s

Vertical component at t = 3 s:
v_y = u_y + gt = 0 + 10(3) = 30 m/s

Resultant velocity:
v = √(v_x² + v_y²) = √(20² + 30²) = √(400 + 900) = √1300 = 10√13 m/s ≈ 36.06 m/s

Angle with horizontal:
tan θ = v_y/v_x = 30/20 = 1.5
θ = tan⁻¹(1.5) ≈ 56.3° below horizontal

Answers:
(a) Time to reach ground = 3 s
(b) Horizontal distance = 60 m
(c) Final velocity = 10√13 m/s ≈ 36.06 m/s at 56.3° below horizontal

Q2. A stone is projected at an angle of 30° with the horizontal with an initial velocity of 40 m/s. Taking g = 10 m/s², find (a) maximum height (b) time of flight (c) range.

Given:
Angle of projection θ = 30°
Initial velocity u = 40 m/s
g = 10 m/s²

Resolve initial velocity:
u_x = u cos θ = 40 cos 30° = 40 × (√3/2) = 20√3 m/s
u_y = u sin θ = 40 sin 30° = 40 × (1/2) = 20 m/s

Part (a): Maximum height

At maximum height, v_y = 0
Using v_y² = u_y² - 2gH
0 = (20)² - 2(10)H
20H = 400
H = 20 m

Part (b): Time of flight

Time of flight T = 2u_y/g
T = 2(20)/10 = 40/10 = 4 s

Alternative: Using vertical motion y = u_y t - (1/2)gt²
When projectile returns to ground level, y = 0
0 = 20t - (1/2)(10)t²
0 = 20t - 5t²
0 = t(20 - 5t)
So t = 0 or t = 4 s
Time of flight = 4 s

Part (c): Range

Range R = u_x × T = u cos θ × (2u sin θ/g)
R = 20√3 × 4 = 80√3 m ≈ 138.6 m

Or using formula: R = u² sin 2θ/g
R = (40)² sin(2 × 30°)/10 = 1600 sin 60°/10 = 1600 × (√3/2)/10 = 800√3/10 = 80√3 m

Answers:
(a) Maximum height = 20 m
(b) Time of flight = 4 s
(c) Range = 80√3 m ≈ 138.6 m

Q3. Define uniform circular motion. A particle moves in a circle of radius 5 m with constant speed 10 m/s. Calculate (a) angular velocity (b) centripetal acceleration (c) period and frequency.

Uniform Circular Motion:
Uniform circular motion is the motion of a body along a circular path at constant speed. Although the speed is constant, the velocity changes continuously in direction, so the body is accelerating toward the centre. This acceleration is called centripetal acceleration.

Given:
Radius r = 5 m
Linear speed v = 10 m/s

Part (a): Angular velocity

Angular velocity ω = v/r
ω = 10/5 = 2 rad/s

Part (b): Centripetal acceleration

Centripetal acceleration a_c = v²/r
a_c = (10)²/5 = 100/5 = 20 m/s²

Alternative: a_c = ω²r = (2)² × 5 = 4 × 5 = 20 m/s²

Part (c): Period and frequency

Period T = 2π/ω = 2π/2 = π s ≈ 3.14 s

Alternative: T = 2πr/v = 2π(5)/10 = 10π/10 = π s

Frequency f = 1/T = 1/π Hz ≈ 0.318 Hz

Alternative: f = ω/2π = 2/(2π) = 1/π Hz

Answers:
(a) Angular velocity = 2 rad/s
(b) Centripetal acceleration = 20 m/s² (directed toward centre)
(c) Period = π s ≈ 3.14 s; Frequency = 1/π Hz ≈ 0.318 Hz

Q4. Two projectiles are fired from the same point with the same initial speed u but at angles of 30° and 60° to the horizontal. Compare their ranges and maximum heights. Take g = 10 m/s².

Given:
Both projectiles: initial speed = u, g = 10 m/s²
Projectile 1: angle θ₁ = 30°
Projectile 2: angle θ₂ = 60°

Key observation: θ₂ = 90° - θ₁, so these are complementary angles.

Part 1: Range comparison

Range formula: R = u² sin 2θ/g

For projectile 1 (30°):
R₁ = u² sin(2 × 30°)/g = u² sin 60°/g = u² × (√3/2)/g

For projectile 2 (60°):
R₂ = u² sin(2 × 60°)/g = u² sin 120°/g = u² sin(180° - 60°)/g = u² sin 60°/g = u² × (√3/2)/g

Therefore: R₁ = R₂ = u²√3/(2g)

Conclusion: Both projectiles have the SAME range.

General Rule: Projectiles launched at complementary angles (θ and 90° - θ) have equal ranges.

Part 2: Maximum height comparison

Maximum height formula: H = (u sin θ)²/(2g) = u² sin² θ/(2g)

For projectile 1 (30°):
H₁ = u² sin² 30°/(2g) = u² × (1/2)²/(2g) = u² × (1/4)/(2g) = u²/(8g)

For projectile 2 (60°):
H₂ = u² sin² 60°/(2g) = u² × (√3/2)²/(2g) = u² × (3/4)/(2g) = 3u²/(8g)

Ratio: H₂/H₁ = [3u²/(8g)]/[u²/(8g)] = 3

Therefore: H₂ = 3H₁

Conclusion: The projectile at 60° reaches 3 times the maximum height of the one at 30°.

Part 3: Time of flight comparison

Time of flight: T = 2u sin θ/g

T₁ = 2u sin 30°/g = 2u × (1/2)/g = u/g
T₂ = 2u sin 60°/g = 2u × (√3/2)/g = u√3/g

Ratio: T₂/T₁ = √3

Conclusion: The time of flight for 60° is √3 times that for 30°.

Summary:
- Ranges: R₁ = R₂ (equal)
- Heights: H₂ = 3H₁ (60° angle gives 3 times greater height)
- Time of flight: T₂ = √3 × T₁ (60° angle takes √3 times longer)

Q5. A car is moving in a circular path of radius 100 m with a constant speed of 20 m/s. What is the centripetal force required if the mass of the car is 1000 kg? What provides this force?

Given:
Radius of circular path r = 100 m
Speed v = 20 m/s
Mass of car m = 1000 kg

Part 1: Calculate centripetal force

Centripetal force F_c = mv²/r

F_c = 1000 × (20)²/100
F_c = 1000 × 400/100
F_c = 1000 × 4
F_c = 4000 N = 4 kN

Alternative calculation using angular velocity:
ω = v/r = 20/100 = 0.2 rad/s
F_c = mω²r = 1000 × (0.2)² × 100 = 1000 × 0.04 × 100 = 4000 N

Part 2: What provides this force?

The centripetal force is provided by the frictional force between the tyres and the road surface.

When the car moves along a curved path, the road (or a banking system) exerts a normal force or friction force on the car toward the centre of the circular path. This force is directed radially inward and causes the change in direction of the car's velocity.

For a horizontal circular path:
- Friction force from the road provides the centripetal force
- Maximum friction = μN = μmg (where μ is coefficient of friction)

For a banked circular path:
- Normal force and friction together provide the centripetal force
- Optimum banking angle θ is given by: tan θ = v²/(rg)

Answer: The centripetal force required is 4000 N or 4 kN, directed toward the center of the circular path. This force is provided by the friction between the car's tyres and the road surface.

Q6. Define relative velocity. A boat moving at 10 m/s in still water wants to cross a river flowing at 5 m/s. The river is 400 m wide. If the boat is directed perpendicular to the river current, find (a) the velocity of boat relative to ground (b) time to cross (c) how far downstream the boat will be when it reaches the opposite bank.

Relative Velocity Definition:
Relative velocity is the velocity of an object as observed from a reference frame that is itself moving. It is the vector difference between the velocities of two objects.

Given:
Boat velocity in still water v_b = 10 m/s
River current velocity v_r = 5 m/s
River width = 400 m
Boat directed perpendicular to river

Part (a): Velocity of boat relative to ground

When boat is directed perpendicular to the current:
- Velocity component perpendicular to river (across) = v_b = 10 m/s
- Velocity component along river (downstream) = v_r = 5 m/s

Resultant velocity v_bg = √(v_b² + v_r²)
v_bg = √(10² + 5²) = √(100 + 25) = √125 = 5√5 m/s ≈ 11.18 m/s

Direction: tan θ = v_r/v_b = 5/10 = 0.5
θ = tan⁻¹(0.5) ≈ 26.57° downstream from perpendicular

Part (b): Time to cross

Time to cross = (River width)/(velocity perpendicular to river)
Time t = 400/10 = 40 s

Part (c): Distance downstream

While crossing, the boat is also carried downstream by the current.
Downstream distance = v_r × t = 5 × 40 = 200 m

Alternative representation:
The boat's position when it reaches opposite bank:
- Perpendicular distance = 400 m
- Downstream distance = 200 m
- Actual path length = √(400² + 200²) = √(160000 + 40000) = √200000 = 200√5 m

Answers:
(a) Velocity relative to ground = 5√5 m/s ≈ 11.18 m/s at 26.57° downstream from perpendicular
(b) Time to cross = 40 s
(c) Boat is 200 m downstream when it reaches opposite bank

Q7. A particle is moving in a circle with angular velocity ω = 5 rad/s. If the radius of the circle is 2 m, calculate (a) linear velocity (b) angular displacement in 10 seconds (c) number of revolutions in 10 seconds.

Given:
Angular velocity ω = 5 rad/s (constant)
Radius r = 2 m
Time t = 10 s

Part (a): Linear velocity

Linear velocity v = ωr
v = 5 × 2 = 10 m/s

Part (b): Angular displacement in 10 seconds

For constant angular velocity:
Angular displacement θ = ωt
θ = 5 × 10 = 50 radians

Alternative expression:
θ = 50 rad = 50/(2π) revolutions = 25/π revolutions ≈ 7.96 revolutions

Part (c): Number of revolutions in 10 seconds

Since 1 complete revolution = 2π radians
Number of revolutions = θ/(2π) = 50/(2π) = 25/π ≈ 7.96 revolutions

Alternative method:
Frequency f = ω/(2π) = 5/(2π) Hz
Number of revolutions = f × t = [5/(2π)] × 10 = 50/(2π) = 25/π

Or using period:
Period T = 2π/ω = 2π/5 s ≈ 1.256 s
Number of revolutions = t/T = 10/[2π/5] = 50/(2π) = 25/π ≈ 7.96 revolutions

Answers:
(a) Linear velocity = 10 m/s
(b) Angular displacement = 50 radians
(c) Number of revolutions = 25/π ≈ 7.96 revolutions (or about 8 complete revolutions)

Motion in a Plane — Practice MCQs with Answers

Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.

MCQ 1. If two vectors $\vec{A}$ and $\vec{B}$ have magnitudes 3 units and 4 units respectively, and they are perpendicular to each other, what is the magnitude of their resultant vector $\vec{R} = \vec{A} + \vec{B}$?

  • A. 1 unit
  • B. 5 units
  • C. 7 units
  • D. 12 units

Answer: B. 5 units — For two perpendicular vectors, the magnitude of the resultant is given by the Pythagorean theorem: $|\vec{R}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2}$. So, $|\vec{R}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ units.

MCQ 2. Two vectors are given by $\vec{A} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{B} = \hat{i} - \hat{j} + 2\hat{k}$. What is their scalar product $\vec{A} \cdot \vec{B}$?

  • A. 2
  • B. -3
  • C. -1
  • D. 4

Answer: B. -3 — The scalar product $\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z$. Here, $\vec{A} \cdot \vec{B} = (2)(1) + (3)(-1) + (-1)(2) = 2 - 3 - 2 = -3$.

MCQ 3. For the vectors $\vec{A} = 2\hat{i}$ and $\vec{B} = 3\hat{j}$, what is the magnitude of their vector product $|\vec{A} \times \vec{B}|$?

  • A. 0
  • B. 5
  • C. 6
  • D. 12

Answer: C. 6 — The vector product $\vec{A} \times \vec{B} = (2\hat{i}) \times (3\hat{j}) = 6(\hat{i} \times \hat{j})$. Since $\hat{i} \times \hat{j} = \hat{k}$, then $\vec{A} \times \vec{B} = 6\hat{k}$. The magnitude is $|\vec{A} \times \vec{B}| = |6\hat{k}| = 6$.

MCQ 4. A projectile is launched with an initial velocity of $20 \text{ m/s}$ at an angle of $30^\circ$ above the horizontal. What is its approximate time of flight? (Take $g = 10 \text{ m/s}^2$)

  • A. 1 s
  • B. 2 s
  • C. 3 s
  • D. 4 s

Answer: B. 2 s — The time of flight $T = \frac{2u \sin\theta}{g}$. Given $u=20 \text{ m/s}$, $\theta=30^\circ$, $g=10 \text{ m/s}^2$. So, $T = \frac{2 \times 20 \times \sin30^\circ}{10} = \frac{40 \times 0.5}{10} = \frac{20}{10} = 2 \text{ s}$.

MCQ 5. A projectile is launched with an initial speed $u$ at an angle $\theta$ to the horizontal, reaching a maximum height $H$. If the launch angle is changed to $(90^\circ - \theta)$ while keeping the initial speed $u$ constant, what is the new maximum height $H'$?

  • A. $H$
  • B. $H \tan^2\theta$
  • C. $H \cot^2\theta$
  • D. Cannot be determined without $u$ and $\theta$

Answer: C. $H \cot^2\theta$ — The maximum height is $H = \frac{u^2 \sin^2\theta}{2g}$. For an angle $(90^\circ - \theta)$, the new maximum height $H' = \frac{u^2 \sin^2(90^\circ - \theta)}{2g} = \frac{u^2 \cos^2\theta}{2g}$. Therefore, $H' = H \frac{\cos^2\theta}{\sin^2\theta} = H \cot^2\theta$.

MCQ 6. At the highest point of its trajectory, what is the vertical component of the velocity of a projectile launched at an angle to the horizontal?

  • A. Maximum
  • B. Minimum
  • C. Zero
  • D. Equal to horizontal component

Answer: C. Zero — At the highest point of its trajectory, a projectile momentarily stops moving upwards, so its vertical component of velocity becomes zero. The horizontal component remains constant (neglecting air resistance).

MCQ 7. A particle moves in a circle of radius $5 \text{ cm}$ with a constant speed. If it completes 2 revolutions in 5 seconds, what is its centripetal acceleration?

  • A. $3.2 \pi^2 \text{ cm/s}^2$
  • B. $1.6 \pi^2 \text{ cm/s}^2$
  • C. $0.8 \pi^2 \text{ cm/s}^2$
  • D. $0.4 \pi^2 \text{ cm/s}^2$

Answer: A. $3.2 \pi^2 \text{ cm/s}^2$ — The frequency $f = \frac{2 \text{ revolutions}}{5 \text{ s}} = 0.4 \text{ Hz}$. The angular velocity $\omega = 2\pi f = 2\pi (0.4) = 0.8\pi \text{ rad/s}$. Centripetal acceleration $a_c = \omega^2 r = (0.8\pi)^2 \times 5 = 0.64\pi^2 \times 5 = 3.2\pi^2 \text{ cm/s}^2$.

MCQ 8. In uniform circular motion, the centripetal acceleration is directed:

  • A. Along the tangent to the circle
  • B. Away from the center of the circle
  • C. Towards the center of the circle
  • D. Perpendicular to the plane of the circle

Answer: C. Towards the center of the circle — Centripetal acceleration is always directed radially inwards, towards the center of the circular path, responsible for changing the direction of the velocity vector.

MCQ 9. A man can swim at $4 \text{ km/h}$ in still water. He wants to cross a river $1 \text{ km}$ wide flowing at $3 \text{ km/h}$. If he heads perpendicular to the river flow, how long will it take him to cross the river?

  • A. 15 minutes
  • B. 20 minutes
  • C. 10 minutes
  • D. 25 minutes

Answer: A. 15 minutes — To cross the river, only the component of his velocity perpendicular to the river flow matters. His swimming speed in still water ($4 \text{ km/h}$) is this component. Time = Distance / Speed = $1 \text{ km} / 4 \text{ km/h} = 0.25 \text{ hours} = 15 \text{ minutes}$. The river flow only affects how far downstream he lands, not the time to cross.

MCQ 10. A force of $10 \text{ N}$ acts at an angle of $60^\circ$ with the horizontal. What is the magnitude of its vertical component?

  • A. $5 \text{ N}$
  • B. $5\sqrt{3} \text{ N}$
  • C. $10 \text{ N}$
  • D. $10\sqrt{3} \text{ N}$

Answer: B. $5\sqrt{3} \text{ N}$ — The vertical component of the force is $F_y = F \sin\theta$. Here, $F_y = 10 \sin60^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \text{ N}$.

MCQ 11. What is the unit vector in the direction of the vector $\vec{A} = 3\hat{i} - 4\hat{j}$?

  • A. $\frac{3\hat{i} - 4\hat{j}}{5}$
  • B. $3\hat{i} - 4\hat{j}$
  • C. $\frac{3\hat{i} + 4\hat{j}}{5}$
  • D. $\frac{-3\hat{i} + 4\hat{j}}{5}$

Answer: A. $\frac{3\hat{i} - 4\hat{j}}{5}$ — The magnitude of $\vec{A}$ is $|\vec{A}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$. The unit vector $\hat{A} = \frac{\vec{A}}{|\vec{A}|} = \frac{3\hat{i} - 4\hat{j}}{5}$.

MCQ 12. Which of the following statements about the vector product (cross product) is correct?

  • A. $\vec{A} \times \vec{B} = \vec{B} \times \vec{A}$
  • B. $\vec{A} \times \vec{B}$ is a scalar quantity
  • C. $\vec{A} \times \vec{B}$ is perpendicular to the plane containing $\vec{A}$ and $\vec{B}$
  • D. The magnitude of $\vec{A} \times \vec{A}$ is $|\vec{A}|^2$

Answer: C. $\vec{A} \times \vec{B}$ is perpendicular to the plane containing $\vec{A}$ and $\vec{B}$ — The cross product is anti-commutative ($\vec{A} \times \vec{B} = -\vec{B} \times \vec{A}$), it results in a vector, and this resultant vector is perpendicular to both $\vec{A}$ and $\vec{B}$ (hence perpendicular to the plane they form). The magnitude of $\vec{A} \times \vec{A}$ is zero, not $|\vec{A}|^2$.

MCQ 13. A projectile is thrown with initial velocity $u$ at an angle $\theta$ to the horizontal. If its range is $R$ and maximum height is $H$, which relation is correct?

  • A. $R = 4H \cot\theta$
  • B. $R = H \cot\theta$
  • C. $R = 4H \tan\theta$
  • D. $R = H \tan\theta$

Answer: A. $R = 4H \cot\theta$ — Range $R = \frac{u^2 \sin(2\theta)}{g} = \frac{2u^2 \sin\theta \cos\theta}{g}$. Maximum height $H = \frac{u^2 \sin^2\theta}{2g}$. The ratio $\frac{R}{H} = \frac{2u^2 \sin\theta \cos\theta / g}{u^2 \sin^2\theta / 2g} = \frac{4 \cos\theta}{\sin\theta} = 4 \cot\theta$. Thus, $R = 4H \cot\theta$.

MCQ 14. A particle is performing uniform circular motion with radius $r$. If its linear speed is $v$ and angular speed is $\omega$, which relation is correct?

  • A. $v = \omega^2 r$
  • B. $v = \omega r^2$
  • C. $v = \omega r$
  • D. $v = r/\omega$

Answer: C. $v = \omega r$ — The relationship between linear speed $v$ and angular speed $\omega$ for a particle in circular motion of radius $r$ is $v = \omega r$.

MCQ 15. In projectile motion, neglecting air resistance, the acceleration of the projectile is:

  • A. Zero at the highest point
  • B. Constant and directed vertically downwards
  • C. Constant and directed horizontally
  • D. Increases as it moves upwards

Answer: B. Constant and directed vertically downwards — In projectile motion, the only acceleration acting on the projectile (ignoring air resistance) is due to gravity, which is constant ($g$) and always directed vertically downwards throughout the motion.

MCQ 16. The angle between vectors $\vec{A} = \hat{i} + \hat{j}$ and $\vec{B} = \hat{i} - \hat{j}$ is:

  • A. $0^\circ$
  • B. $45^\circ$
  • C. $90^\circ$
  • D. $180^\circ$

Answer: C. $90^\circ$ — The dot product $\vec{A} \cdot \vec{B} = (1)(1) + (1)(-1) = 1 - 1 = 0$. Since the dot product is zero, the vectors are perpendicular, meaning the angle between them is $90^\circ$.

MCQ 17. For a given initial speed, the maximum horizontal range of a projectile is achieved when the angle of projection is:

  • A. $0^\circ$
  • B. $30^\circ$
  • C. $45^\circ$
  • D. $90^\circ$

Answer: C. $45^\circ$ — The horizontal range $R = \frac{u^2 \sin(2\theta)}{g}$. For maximum range, $\sin(2\theta)$ must be maximum, which is 1. This occurs when $2\theta = 90^\circ$, so $\theta = 45^\circ$.

MCQ 18. If a particle completes 10 revolutions in 5 seconds in uniform circular motion, what is its period?

  • A. 0.5 s
  • B. 2 s
  • C. 10 s
  • D. 5 s

Answer: A. 0.5 s — Frequency $f = \frac{\text{Number of revolutions}}{\text{Time}} = \frac{10}{5} = 2 \text{ Hz}$. The period $T$ is the reciprocal of frequency, $T = 1/f = 1/2 = 0.5 \text{ s}$.

MCQ 19. The equation of trajectory of a projectile is given by $y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$. Which of the following represents the shape of the trajectory?

  • A. Straight line
  • B. Circle
  • C. Parabola
  • D. Ellipse

Answer: C. Parabola — The equation $y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$ is a quadratic equation in $x$ (of the form $y = Ax - Bx^2$). This is the standard equation of a parabola.

MCQ 20. Three vectors $\vec{A}$, $\vec{B}$, and $\vec{C}$ are such that $\vec{A} + \vec{B} + \vec{C} = 0$. If $\vec{A}$ and $\vec{B}$ are perpendicular to each other and $|\vec{A}|=3$, $|\vec{B}|=4$, what is the magnitude of $\vec{C}$?

  • A. 1
  • B. 5
  • C. 7
  • D. 25

Answer: B. 5 — From $\vec{A} + \vec{B} + \vec{C} = 0$, we have $\vec{C} = -(\vec{A} + \vec{B})$. The magnitude $|\vec{C}| = |-(\vec{A} + \vec{B})| = |\vec{A} + \vec{B}|$. Since $\vec{A}$ and $\vec{B}$ are perpendicular, $|\vec{A} + \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$.

How to Use These Solutions

Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.

Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.

All NCERT Solutions, Class 6–12 →