Motion in a Plane — Class 11 MCQs with Answers

Motion in a Plane — important questions & MCQs with answers (Class 11 Physics)

6 solved questions from this chapter's difficulty levels, each with its answer and explanation. Tap an option to check yourself.

0 of 6 answered

1. Q1 of 6Easy

   Which is a scalar?

   AVelocityBForceCDisplacementDMass
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   Answer: Mass

   Explanation: A scalar has magnitude only, no direction: mass = 5 kg needs no direction to specify it. Velocity, force and displacement all need direction too, so they are vectors, not scalars.

2. Q2 of 6Easy

   Maximum range of projectile (same speed) at angle:

   A30°B60°C45°D90°
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   Answer: 45°

   Explanation: Range R = u² sin2θ/g is maximum when sin2θ = 1, i.e., 2θ = 90°, so θ = 45°. At 30° or 60° sin2θ = √3/2 ≈ 0.87 < 1, giving a shorter range.

3. Q3 of 6Medium

   Object moves in a circle of radius 2 m with speed 4 m/s. Centripetal acceleration:

   A2 m/s²B4 m/s²C16 m/s²D8 m/s²
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   Answer: 8 m/s²

   Explanation: Centripetal acceleration a = v²/r = (4 m/s)²/(2 m) = 16/2 = 8 m/s², squaring the speed before dividing by the radius. Dividing 4/2 = 2 m/s² instead skips the required squaring of v.

4. Q4 of 6Medium

   Two perpendicular vectors of magnitude 3 and 4. Resultant:

   

   A7B5C1D12
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   Answer: 5

   Explanation: For perpendicular vectors, resultant magnitude follows Pythagoras: R = √(3² + 4²) = √25 = 5. Simply adding 3 + 4 = 7 ignores that they act at 90°, not along the same line.

5. Q5 of 6Hard

   In uniform circular motion, acceleration is:

   ADirected towards the centreBTangential onlyCZeroDRadial outward
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   Answer: Directed towards the centre

   Explanation: In uniform circular motion speed is constant but direction changes, producing centripetal acceleration a = v²/r directed toward the centre, not outward — that inward pull bends the path into a circle.

6. Q6 of 6Hard

   Stone tied to string moves in horizontal circle of radius 1 m at 2 rev/s. Centripetal acceleration:

   A8π² m/s²B4π² m/s²C16π² m/s²Dπ² m/s²
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   Answer: 16π² m/s²

   Explanation: Angular velocity ω = 2πf = 2π(2 rev/s) = 4π rad/s. Centripetal acceleration a = ω²r = (4π)²(1 m) = 16π² m/s². Forgetting to square ω before multiplying by r gives only 4π² instead.

Motion in a Plane notes: key concepts (Class 11 Physics)

Extending kinematics to two dimensions using vectors: addition by the triangle and parallelogram laws, resolution into components, the scalar and vector products, projectile motion as two independent one-dimensional motions, and uniform circular motion with its centripetal acceleration.

Scalars and vectors

A scalar is fully specified by magnitude alone; a vector needs both magnitude and direction and must obey the triangle law of addition.

Triangle and parallelogram law

Two vectors add head-to-tail along a triangle, or equivalently as the diagonal of the parallelogram they span. Both give the same resultant.

Resolution into components

Any vector can be split into perpendicular components, A cos θ along one axis and A sin θ along the other, which can then be handled independently.

Unit vector

A vector of magnitude one that carries only direction, written  = A/|A|. The Cartesian unit vectors î, ĵ and k̂ define the coordinate axes.

Scalar (dot) product

A·B = AB cos θ, which returns a scalar. It vanishes for perpendicular vectors, which is why work done by a centripetal force is always zero.

Vector (cross) product

A × B has magnitude AB sin θ and direction given by the right-hand rule. It vanishes for parallel vectors and is anti-commutative.

Projectile motion

Motion under gravity alone with an initial velocity that is not vertical. Its power lies in the horizontal and vertical motions being completely independent.

Independence of the two motions

Horizontally the velocity stays constant since no force acts; vertically the body accelerates at g. The same time variable links the two.

Parabolic trajectory

Eliminating time between the two component equations gives y as a quadratic in x — the path of any projectile in uniform gravity is a parabola.

Time of flight

The total time in the air, set entirely by the vertical component u sin θ. The horizontal velocity has no effect on how long the projectile stays up.

Maximum height

Reached when the vertical velocity component becomes zero. The horizontal velocity at that instant is still u cos θ, so the speed there is not zero.

Horizontal range

The horizontal distance covered over the full flight. It is maximum at a launch angle of 45°, and angles θ and (90° − θ) give the same range.

Uniform circular motion

Motion at constant SPEED on a circle. The velocity vector still changes direction continuously, so the motion is accelerated throughout.

Centripetal acceleration

Always directed towards the centre, with magnitude v²/r. It changes only the direction of velocity, never its magnitude.

Angular velocity

The rate at which the radius sweeps angle, ω = 2π/T. It links to linear speed through v = ωr, the bridge between linear and circular quantities.