Motion in a Plane Class 11 Physics NCERT Solutions

Step-by-step solution

1. 1A scalar is specified by magnitude alone and has no direction; a vector has both magnitude and direction.
2. 2Scalars: volume, mass, speed, density, number of moles, angular frequency.
3. 3Vectors: acceleration, velocity, displacement, angular velocity.

Step-by-step solution

1. 1Work is the dot product of force and displacement; a dot product is always a scalar, so work is a scalar.
2. 2Current is described only by its magnitude — its direction is not taken into account, so it is a scalar.
3. 3All the remaining quantities (force, angular momentum, linear momentum, electric field, magnetic moment) have an associated direction and are vectors; average velocity and relative velocity are velocity quantities — vectors. Angular momentum and magnetic moment are vectors too.

Step-by-step solution

1. 1Impulse is the product of force and time (J = F × Δt).
2. 2Force is a vector quantity; multiplying a vector by the scalar time gives a vector along the force direction, so impulse is a vector.
3. 3Every other item (temperature, pressure, time, power, total path length, energy, gravitational potential, coefficient of friction, charge) has magnitude only and is a scalar.

Step-by-step solution

1. 1(a) Meaningful, but only if the two scalars represent the same physical quantity.
2. 2(b) Not meaningful — a scalar quantity cannot be added to a vector quantity.
3. 3(c) Meaningful — e.g., multiplying the force vector by the scalar time gives the vector impulse.
4. 4(d) Meaningful — a scalar can be multiplied with another scalar of the same or different dimensions.
5. 5(e) Meaningful, but only if the two vectors represent the same physical quantity.
6. 6(f) Meaningful — a component of a vector has the same dimensions as the vector itself, so they may be added.

Step-by-step solution

1. 1(a) True — the magnitude of a vector is a number (with a unit), so it is a scalar.
2. 2(b) False — each component of a vector is itself a vector (it has a direction along the axis).
3. 3(c) False — the total path length is a scalar and generally exceeds the magnitude of the displacement vector; the two are equal only for straight-line motion without reversal.
4. 4(d) True — since total path length ≥ magnitude of displacement, dividing both by the same time interval keeps the inequality: average speed ≥ magnitude of average velocity.
5. 5(e) True — three vectors not lying in one plane cannot be represented by the three sides of a triangle taken in order, so their vector sum cannot be the null vector.

Step-by-step solution

1. 1(a) Represent a and b as adjacent sides of a parallelogram; the diagonal gives |a + b|. In the triangle formed, each side is smaller than the sum of the other two: |a + b| ≤ |a| + |b|.
2. 2(b) In the same triangle, the third side exceeds the difference of the other two: |a + b| ≥ ||a| − |b||.
3. 3(c) Vector a − b is the third side of the triangle built on a and −b, so |a − b| ≤ |a| + |−b| = |a| + |b|.
4. 4(d) For a − b, the triangle inequality on the differences gives |a − b| ≥ ||a| − |b|| (taking moduli so both sides are non-negative).
5. 5Equality in (a) and (c) holds when the two vectors act along the same straight line and same sense; equality in (b) and (d) holds when they act along the same straight line in opposite directions.

Step-by-step solution

1. 1(a) Incorrect — a + b + c + d = 0 does not require each vector to be null; many non-zero combinations give a zero sum.
2. 2(b) Correct — rewrite as a + c = −(b + d); taking magnitudes, |a + c| = |−(b + d)| = |b + d|.
3. 3(c) Correct — a = −(b + c + d), so |a| = |b + c + d| ≤ |b| + |c| + |d|; hence |a| can never exceed that sum.
4. 4(d) Correct — a + (b + c) + d = 0 means (b + c) is the side that closes the triangle with a and d: (b + c) lies in the plane of a and d, or along their line if they are collinear.

Step-by-step solution

1. 1Displacement depends only on the initial and final positions: all three girls go from P to Q.
2. 2P and Q are diametrically opposite, so the displacement of each girl equals the diameter of the ground.
3. 3Diameter = 2 × radius = 2 × 200 = 400 m for every girl.
4. 4This equals the actual path length only for the girl who skates along the straight diameter path (girl B in Fig. 3.19).

Step-by-step solution

1. 1(a) The cyclist ends where he started (back at O), so the net displacement is zero.
2. 2(b) Average velocity = net displacement / total time = 0 / (10 min) = 0.
3. 3(c) Total path length = OP + arc PQ + QO = 1 + (1/4)(2π × 1) + 1 = 2 + π/2 ≈ 3.570 km.
4. 4Time = 10 min = 10/60 h = 1/6 h.
5. 5Average speed = 3.570 / (1/6) = 21.42 km h⁻¹.

Step-by-step solution

1. 1Turning left by 60° after each 500 m side traces a regular hexagon of side 500 m.
2. 2Third turn (vertex S): the two opposite vertices are two diameters apart through the centre — displacement = PS = 500 + 500 = 1000 m; total path length = 3 × 500 = 1500 m.
3. 3Sixth turn: the motorist is back at the starting point P — displacement = 0; total path length = 6 × 500 = 3000 m.
4. 4Eighth turn (vertex R): displacement PR = √(500² + 500² + 2·500·500·cos 60°) = √(500² + 500² + 500²) = 500√3 ≈ 866.03 m, directed at 30° to the first side PQ; total path length = 8 × 500 = 4000 m.
5. 5Summary — third: 1000 m vs 1500 m, sixth: 0 vs 3000 m, eighth: 866.03 m at 30° vs 4000 m.

Step-by-step solution

1. 1(a) Total distance = 23 km, time = 28 min = 28/60 h.
2. 2Average speed = 23/(28/60) = 23 × 60/28 ≈ 49.29 km h⁻¹.
3. 3(b) Displacement = straight-line distance between hotel and station = 10 km.
4. 4Magnitude of average velocity = 10/(28/60) ≈ 21.43 km h⁻¹.
5. 5The two quantities are not equal because the path taken (23 km) is not a straight line — average speed uses path length, average velocity uses displacement.

Step-by-step solution

1. 1Maximum height h = u² sin²θ / 2g. The ball must not exceed h = 25 m with u = 40 m s⁻¹.
2. 225 = (40)² sin²θ / (2 × 9.8), so sin²θ = 25 × 2 × 9.8 / 1600 = 0.30625.
3. 3sin θ ≈ 0.5534, hence θ ≈ 33.60°.
4. 4Range R = u² sin 2θ / g = 1600 × sin 67.2° / 9.8 = 1600 × 0.922 / 9.8 ≈ 150.53 m.

Step-by-step solution

1. 1Maximum horizontal range occurs at θ = 45°: R = u² sin 2θ/g = u²/g × sin 90° = u²/g.
2. 2Given R = 100 m, u²/g = 100.
3. 3The maximum height is reached when the ball is thrown vertically upward, where v = 0 at the top.
4. 4Using v² − u² = −2gH: H = u²/2g = (1/2)(u²/g) = (1/2)(100) = 50 m.

Step-by-step solution

1. 1Length of string = radius r = 80 cm = 0.8 m.
2. 2Frequency ν = 14/25 Hz; angular frequency ω = 2πν = 2 × (22/7) × (14/25) = 88/25 rad s⁻¹.
3. 3Centripetal acceleration a_c = ω²r = (88/25)² × 0.8 ≈ 9.91 m s⁻².
4. 4Direction: acceleration is always along the string, radially towards the centre of the circle.

Step-by-step solution

1. 1r = 1 km = 1000 m; v = 900 km h⁻¹ = 900 × (5/18) = 250 m s⁻¹.
2. 2Centripetal acceleration a_c = v²/r = (250)²/1000 = 62.5 m s⁻².
3. 3With g = 9.8 m s⁻², a_c/g = 62.5/9.8 ≈ 6.38.
4. 4So a_c ≈ 6.38 g — the aircraft feels a centripetal acceleration about 6.4 times the acceleration due to gravity.

Step-by-step solution

1. 1(a) False — the net acceleration points radially inward only for uniform circular motion. If the speed also changes, a tangential acceleration component exists, so the net acceleration is not purely radial.
2. 2(b) True — at any point of the path, the particle moves tangentially, so the velocity vector is always tangent to the path at that point.
3. 3(c) True — in uniform circular motion the acceleration always points toward the centre and its direction keeps changing symmetrically; the average of these vectors over one full cycle is the null vector.

Step-by-step solution

1. 1(a) Differentiate r with respect to t: v = dr/dt = 3.0 î − 4.0t ĵ (in m s⁻¹).
2. 2Differentiate again: a = dv/dt = −4.0 ĵ (in m s⁻²), a constant.
3. 3(b) At t = 2.0 s: v = 3.0 î − 4.0(2.0) ĵ = 3.0 î − 8.0 ĵ.
4. 4Magnitude |v| = √(3² + (−8)²) = √73 ≈ 8.54 m s⁻¹.
5. 5Direction: θ = tan⁻¹(v_y/v_x) = tan⁻¹(−8/3) ≈ −69.45° — the negative sign means the direction is 69.45° below the x-axis.

Step-by-step solution

1. 1Acceleration a = d v/d t = 8.0 î + 2.0 ĵ. Integrating, v(t) = 8.0t î + 2.0t ĵ + u with u = 10.0 ĵ at t = 0.
2. 2Integrating again from the origin gives r(t) = 4.0t² î + (10.0t + t²) ĵ.
3. 3(a) x = 4.0t² = 16 ⇒ t = 2 s.
4. 4At t = 2 s: y = 10(2) + (2)² = 24 m.
5. 5(b) v at t = 2 s: v = 8(2) î + 2(2) ĵ + 10 ĵ = 16 î + 14 ĵ.
6. 6Speed = √(16² + 14²) = √452 ≈ 21.26 m s⁻¹.

Step-by-step solution

1. 1For P = î + ĵ: components P_x = P_y = 1, so |P| = √(1² + 1²) = √2.
2. 2Direction of î + ĵ: tan θ = 1/1 = 1 ⇒ θ = 45° with the x-axis.
3. 3For Q = î − ĵ: |Q| = √2 and θ = tan⁻¹(−1/1) = −45° with the x-axis.
4. 4Angle between A = 2 î + 3 ĵ and î + ĵ is θ′ = 56.31° − 45° = 11.31°, where tan θ of A is 3/2 ⇒ 56.31°.
5. 5Component of A along (î + ĵ)/√2 = |A| cos θ′ = √13 × cos 11.31° ≈ 3.54 = 5/√2.
6. 6Angle between A and (î − ĵ) is θ″ = 56.31° + 45° = 101.31°; the component along (î − ĵ)/√2 = √13 cos 101.31° ≈ −0.71 = −1/√2.

Step-by-step solution

1. 1(a) False — this linear average is valid only when acceleration is constant; the motion here is arbitrary.
2. 2(b) True — by definition, average velocity is the total displacement divided by the time interval.
3. 3(c) False — requires constant acceleration, which need not hold for arbitrary motion.
4. 4(d) False — this is the constant-acceleration position formula, not valid for arbitrary, non-uniform acceleration.
5. 5(e) True — by definition, average acceleration is the change in velocity divided by the time interval.

Step-by-step solution

1. 1(a) False — energy is a scalar yet is not conserved in inelastic collisions.
2. 2(b) False — temperature is a scalar but can take negative values.
3. 3(c) False — total path length is a scalar and yet it has the dimension of length.
4. 4(d) False — gravitational potential is a scalar and yet varies from point to point in space.
5. 5(e) True — a scalar is defined so that its value is the same for observers with differently oriented axes; a vector's components change with orientation, but a scalar does not.

Step-by-step solution

1. 1The aircraft is at height OR = 3400 m, and the two positions P and Q subtend ∠POQ = 30° at the observer.
2. 2In triangle PRO, tan 15° = PR/OR, so PR = OR tan 15° = 3400 × tan 15°.
3. 3Triangle PRO is similar to RQO and PR = RQ, so PQ = PR + RQ = 2 × 3400 × tan 15°.
4. 4PQ = 6800 × 0.268 ≈ 1822.4 m covered in 10 s.
5. 5Speed = 1822.4/10 ≈ 182.24 m s⁻¹.