JEE Physics Motion In A Straight Line MCQs Set 20

Answer: (c) at 3h/4 from the ground

Explanation:
1. The drop in time t is ½gt², so in T/2 the drop is ½g(T/2)² = ¼ of ½gT² = h/4.
2. The height above the ground is h − h/4 = 3h/4.

Teacher's Note:
1. The distance dropped through from rest is proportional to t².
2. Mass does not matter in free fall.

Answer: (b) \( \dfrac{4u^2}{g} \)

Explanation:
1. Whether thrown up or down, the stone passes the top again (moving down) with speed u, so use v² = u² + 2gh with u as the initial speed.
2. 9u² = u² + 2gh, so h = 4u²/g.

Teacher's Note:
1. The direction of the initial velocity does not matter for v² = u² + 2gh.
2. Both up and down throws of the same speed reach the ground with the same speed.

Answer: (a) \( \dfrac{2v(n-1)}{n} \)

Explanation:
1. v = an, so a = v/n.
2. Last two seconds: ½an² − ½a(n − 2)² = ½a(4n − 4) = 2a(n − 1).
3. = 2v(n − 1)/n.

Teacher's Note:
1. Subtract the displacement in (n − 2) seconds from that in n seconds.
2. Express a using v and n.

Answer: (a) \( \dfrac{v}{g}\left[1+\sqrt{1+\dfrac{2gh}{v^2}}\right] \)

Explanation:
1. Taking up as positive, displacement is −h: −h = vt − ½gt².
2. ½gt² − vt − h = 0, so t = [v ± √(v² + 2gh)]/g.
3. Time must be positive, so take the plus sign: t = (v/g)[1 + √(1 + 2gh/v²)].

Teacher's Note:
1. Write the quadratic in t and pick the positive root.
2. The displacement is negative because the ground is below the start.

Answer: (a) 495 m

Explanation:
1. Upward positive: −h = 10(11) − ½(10)(11²) = 110 − 605 = −495.
2. So h = 495 m.

Teacher's Note:
1. The stone starts with the balloon's upward velocity.
2. Displacement to the ground is −h.

Answer: (d) \( \dfrac{15}{4} \) m

Explanation:
1. First drop: 5 = ½(10)t², so t = 1 s.
2. Two equal intervals fit in this time, so each interval is 0.5 s.
3. The second drop has dropped ½(10)(0.5)² = 1.25 m.
4. Height above ground = 5 − 1.25 = 3.75 = 15/4 m.

Teacher's Note:
1. The first drop has been falling for two intervals and the second drop for one.
2. Find the interval from the first drop's fall time.

Answer: (b) \( \sqrt{a} : \sqrt{b} \)

Explanation:
1. v² = 2gh, so v = √(2gh).
2. v₁ : v₂ = √a : √b.

Teacher's Note:
1. Speed on reaching the ground depends on the square root of the height.
2. Mass does not matter.

Answer: (a) The retardation of the particle is \( \dfrac{1}{4(x+1)^3} \)

Explanation:
1. (x + 1)² = t + 1, so x + 1 = √(t + 1).
2. v = dx/dt = 1/(2√(t+1)) = 1/(2(x+1)).
3. a = dv/dt = −1/(4(t+1)^{3/2}) = −1/(4(x+1)³).
4. The acceleration is negative, so the particle is retarding with magnitude 1/(4(x+1)³).

Teacher's Note:
1. Complete the square to get x in terms of t.
2. Differentiate twice and express the result in x.

Answer: (a) An ideal speedometer shows the magnitude of the instantaneous velocity

Explanation:
1. A speedometer reads instantaneous speed, equal to the magnitude of instantaneous velocity: true.
2. Displacement equals distance only for motion in one direction: (b) is false.
3. Average speed zero requires zero distance, so the speed must be zero throughout: (c) and (d) are false.

Teacher's Note:
1. Average speed is zero only when the particle never moves.
2. Displacement can be smaller than the distance.

Answer: (c) 3h

Explanation:
1. v² = 2gh. For velocity 2v: 4v² = 2gH, so H = 4h.
2. Further distance = 4h − h = 3h.

Teacher's Note:
1. The distance dropped through is proportional to the square of the velocity.
2. Subtract the distance already covered.

Answer: (a) (i), (ii) and (iv) only

Explanation:
1. Area under a-t gives Δv: the positive and negative triangles cancel, so v(20) = 0 and the average acceleration is zero: (i) true.
2. Here a = 10 − t, so v = ∫(10 − t)dt = 10t − t²/2, which is non-negative for 0 ≤ t ≤ 20, so the particle never reverses: (ii) true.
3. Since v ≥ 0 and not always zero, the displacement is positive: (iii) false.
4. Distance in 0 to 10 s = ∫(10t − t²/2)dt = 500 − 166.7 = 333.3 m and in 10 to 20 s is also 333.3 m for equal times: (iv) true.

Teacher's Note:
1. Area under an a-t graph is the change in velocity.
2. Integrate once more to check the sign of the velocity.

Answer: (a) (ii), (iii) and (iv) only

Explanation:
1. Zero acceleration at a single instant does not make velocity constant: (i) false.
2. A particle can be momentarily at rest with zero acceleration: (ii) true.
3. Zero acceleration over an interval means constant velocity, hence constant speed: (iii) true.
4. If the speed stays zero, velocity does not change, so a = 0: (iv) true.

Teacher's Note:
1. An instant is not an interval.
2. Constant velocity requires a = 0 for the whole interval.

Answer: (a) (i) and (iv) only

Explanation:
1. After one hour the tip is back at its starting point, so the displacement and average velocity are zero.
2. The tip travels a full circle, so the distance and average speed are not zero.

Teacher's Note:
1. Displacement depends on the end points; distance on the path.
2. Average velocity = displacement / time.

Answer: (a) (i), (ii) and (iv) only

Explanation:
1. (i) Opposite signs of velocity and acceleration mean retardation: true.
2. (ii) If x is positive and v negative, the particle moves toward the origin: true.
3. (iii) At the top of a throw v = 0 but a = g: false.
4. (iv) If v remains zero over an interval it does not change, so a = 0: true.

Teacher's Note:
1. Test statement (iii) with a vertical throw.
2. Use signs to judge direction.

Answer: (c) −4 m/s²

Explanation:
1. x = 2t − 2t², v = 2 − 4t, a = dv/dt = −4 m/s².

Teacher's Note:
1. Differentiate twice.
2. Constant acceleration, directed along negative x.

Answer: (c) \( \dfrac{1}{2} \) m

Explanation:
1. v = 2 − 4t = 0 at t = 1/2 s.
2. x = 2(1/2 − 1/4) = 1/2 m.

Teacher's Note:
1. Maximum position occurs where v = 0.
2. Substitute that time into x.

Answer: (c) It starts from the origin, goes up to x = 1/2 on the positive x-axis, then goes to the left

Explanation:
1. At t = 0, x = 0 and v = +2 m/s, so it moves right.
2. It stops at x = 1/2 at t = 0.5 s and then returns, passing the origin at t = 1 s and going on to negative x.
3. Only (c) describes this correctly.

Teacher's Note:
1. Check the sign of v at the start and where it reverses.
2. x becomes negative for t > 1 s.

Answer: (b) 1 m

Explanation:
1. It goes out to x = 1/2 m in the first 0.5 s, then returns to x = 0 at t = 1 s.
2. Distance = 1/2 + 1/2 = 1 m (displacement is zero).

Teacher's Note:
1. Split the path at the turning point.
2. Distance counts both legs.

Answer: (b) \( \dfrac{2n-1}{2n+1} \)

Explanation:
1. From rest, S_n = (a/2)(2n − 1).
2. S_{n+1} = (a/2)(2n + 1).
3. Ratio = (2n − 1)/(2n + 1).

Teacher's Note:
1. Use s_n = u + (a/2)(2n − 1) with u = 0.
2. Successive intervals are in the ratio of odd numbers.

Answer: (c) 55 m/s

Explanation:
1. The speed keeps increasing as long as a is positive, so it is maximum at t = 11 s.
2. Change in velocity = area under the a-t graph = ½ × 11 × 10 = 55.
3. Starting from rest, v_max = 55 m/s.

Teacher's Note:
1. Area under an a-t graph gives the change in velocity.
2. Speed peaks when acceleration reaches zero.

Answer: (a) 1 cm

Explanation:
1. Let the initial speed be u and the deceleration a. First stage: (u/2)² = u² − 2a(3), so a = u²/8.
2. Second stage: 0 = (u/2)² − 2a x, so x = u²/(8a) = 1 cm.

Teacher's Note:
1. Use v² = u² − 2ax for each stage.
2. The deceleration is the same in both stages.

Answer: (b) \( v_A = v_B \)

Explanation:
1. Ball A returns to the level of the building with speed u downward, the same as B's initial speed.
2. From then on both have the same speed and acceleration, so they reach the ground with equal speed.
3. Mass does not affect free-fall speeds.

Teacher's Note:
1. Use v² = u² + 2gh with the same u and h.
2. A ball thrown up passes the start again with the same speed.

Answer: (b) 1 : 4

Explanation:
1. Both stop after the same time t, so distance = (average speed) × t = (v/2) t.
2. Distances are in the ratio u/2 : 4u/2 = 1 : 4.

Teacher's Note:
1. Constant deceleration gives average speed v/2.
2. Same stopping time means different decelerations.

Answer: (b) \( 3t^2\sqrt{\alpha^2+\beta^2} \)

Explanation:
1. v_x = 3αt², v_y = 3βt².
2. Speed = √(v_x² + v_y²) = 3t²√(α² + β²).

Teacher's Note:
1. Differentiate each coordinate.
2. Combine components with Pythagoras.