CBSE Class 11 Physics Kinematics Objective Questions

OBJECTIVES
CLASS 12 PHYSICS : KINEMATICS

1. A body is projected vertically upwards with a velocity v. It returns to the point from which it was projected after some time. The average velocity and speed for the total time of flight are respectively.
(a) \( \frac{v}{2} \), v
(b) 0, \( \frac{v}{2} \)
(c) 0, 0
(d) \( \frac{v}{2} \),0.

Answer: (b) 0, \( \frac{v}{2} \)

1. The body returns to the starting point, so the displacement is zero and the average velocity is 0.
2. Time of flight = 2v/g and the total distance = 2 × v²/(2g) = v²/g.
3. Average speed = (v²/g)/(2v/g) = v/2.
4. Hence, the correct option is (b).

Teacher's Note:
a) Average velocity uses displacement; average speed uses total path length.
b) A return to the starting point always gives zero average velocity.

2. A body is thrown vertically upwards with a velocity u. Which of the following graphs represents the variation of velocity with time correctly ?
(a) a V-shaped graph in which v falls along a straight line to the t-axis and then rises along another straight line
(b) two straight lines which start at different points on the left, one above and one below the t-axis, and come together at a point on the t-axis
(c) two straight lines which start at the origin O and spread out to the right, one lying above the t-axis and one on the right of O below the t-axis
(d) a single straight line sloping downwards, starting above the t-axis and crossing it

Answer: (d) a single straight line sloping downwards, starting above the t-axis and crossing it

1. Take upward as positive. The velocity at time t is v = u - gt.
2. This is a straight line with negative slope -g that starts at +u when t = 0.
3. It crosses the t-axis at the top of the flight (t = u/g) and becomes negative on the way down.
4. Hence, the correct option is (d).

Teacher's Note:
a) The slope of a v-t graph is the acceleration, which is a constant -g for the whole flight.
b) A V-shaped graph would show speed, not velocity.

3. A ball is thrown vertically upwards and its velocity v varies with time t as shown in the figure. Which of the graphs (a), (b), (c), (d) shows the correct curve of displacement versus time
(a) s is constant, a horizontal straight line
(b) s starts from the origin and rises slowly at first, then more and more steeply
(c) s starts from the origin, rises steeply at first and then flattens out
(d) s is a straight line through the origin with a positive slope

Answer: (c) s starts from the origin, rises steeply at first and then flattens out

1. The v-t graph is a straight line falling from a positive value to zero, so the ball is moving upward with decreasing speed.
2. The slope of the s-t graph is the velocity. So the s-t graph must rise with a slope that starts large and decreases to zero.
3. This is a curve through the origin that is steep at first and flattens out.
4. Hence, the correct option is (c).

Teacher's Note:
a) The slope of a displacement-time graph gives the velocity.
b) A straight line in the s-t graph would mean constant velocity.

4. A particle moving with a constant acceleration from A to B in the straight line AB has velocities u and v at A and B respectively. If C is the mid-point of AB then the velocity of particle while passing C will be
(a) \( \sqrt{\frac{v^2 + u^2}{2}} \)
(b) \( \frac{v+u}{2} \)
(c) \( \frac{v−u}{2} \)
(d) \( \frac{\left(\frac{1}{v}+\frac{1}{u}\right)}{2} \)

Answer: (a) \( \sqrt{\frac{v^2 + u^2}{2}} \)

1. Let AB = 2s and the acceleration be a. Then v² = u² + 2a(2s) and v_C² = u² + 2as.
2. From the first, 2as = (v² - u²)/2.
3. v_C² = u² + (v² - u²)/2 = (u² + v²)/2.
4. Hence, v_C = √((u² + v²)/2) and the correct option is (a).

Teacher's Note:
a) The velocity at the mid-point is the root mean square of u and v, not their ordinary average.
b) The average (u + v)/2 is the velocity at the mid-time.

5. An aeroplane is flying in a horizontal direction with a velocity of 900 km/h and at a height of 1960 m. When it is vertically above the point A on the ground, a body is dropped from it. The body strikes the ground at point B. The distance AB will be (take g = 9.8 m/s²)
(a) 3.33 km
(b) 5 km
(c) 10 km
(d) 18 km

Answer: (b) 5 km

1. Horizontal velocity = 900 km/h = 900 × (5/18) = 250 m/s.
2. Time of fall: 1960 = (1/2)(9.8)t², so t² = 400 and t = 20 s.
3. AB = 250 × 20 = 5000 m = 5 km.
4. Hence, the correct option is (b).

Teacher's Note:
a) The dropped body keeps the horizontal velocity of the aeroplane.
b) Time of fall depends only on the height.

6. To a man walking at the rate of 3 km/h the rain appears to fall vertically. When he increases his speed to 6 km/h it appears to meet him at an angle of 45º with vertical from the front. The actual speed of rain is
(a) 3 km/hr
(b) 4 km/hr
(c) 3√2 km/h
(d) 2√3 km/h

Answer: (c) 3√2 km/h

1. When the man walks at 3 km/h the rain appears vertical, so the horizontal component of the rain velocity is 3 km/h in the direction of his walk.
2. At 6 km/h the horizontal component of the rain relative to him is 3 - 6 = -3 km/h (towards him), and the vertical component is v_y.
3. The relative velocity makes 45° with the vertical, so v_y = 3 km/h.
4. Actual speed of the rain = √(3² + 3²) = 3√2 km/h.
5. Hence, the correct option is (c).

Teacher's Note:
a) Use relative velocity: v_rain,man = v_rain - v_man.
b) The vertical component of the rain is the same in every frame.

7. A body is released from the top of a tower of height h metre, it takes t second to reach the ground. Where is the body at the time \( \frac{t}{2} \) ?
(a) at h/2 meters from the ground
(b) at h/4 from the ground
(c) depends on the mass and velocity of the body
(d) at 3h/4 from the ground

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

1. For a free fall from rest, the distance fallen is proportional to t².
2. In time t the body falls h. In time t/2 it falls h × (1/2)² = h/4.
3. Height above the ground = h - h/4 = 3h/4.
4. Hence, the correct option is (d).

Teacher's Note:
a) The distance fallen in half the time is one-fourth, not half.
b) The mass has no effect on free fall.

8. If a body travels half of its total path in the last second of its fall from rest then the total time of its fall will be
(a) 1.82 s
(b) 3.4 s
(c) 3.41
(d) 3.73

Answer: (c) 3.41

1. Let the total time be T. Total height = (1/2)gT².
2. The distance fallen in (T - 1) s is (1/2)g(T - 1)², and the last second covers half the path: (1/2)gT² - (1/2)g(T - 1)² = (1/4)gT².
3. So (T - 1)² = T²/2, that is, T - 1 = T/√2.
4. T = 1/(1 - 1/√2) = √2/(√2 - 1) = √2(√2 + 1) = 2 + √2 = 3.41 s.
5. Hence, the correct option is (c).

Teacher's Note:
a) Half the height is covered in the first (T - 1) seconds.
b) Option (c) is given in seconds, as the others are.

9. A particle moves along the positive branch of the curve y = x²/2 with x governed by x = t²/2 where x and y are measured in metre and t in second. At t = 2 s, the velocity of the particle is
(a) 2i - 4j m/s
(b) 2i + 4j m/s
(c) 4i + 2j m/s
(d) 4i - 2j m/s

Answer: (b) 2i + 4j m/s

1. x = t²/2, so v_x = dx/dt = t = 2 m/s at t = 2 s, and x = 2 m.
2. y = x²/2, so v_y = dy/dt = x(dx/dt) = 2 × 2 = 4 m/s.
3. Velocity = (2i + 4j) m/s.
4. Hence, the correct option is (b).

Teacher's Note:
a) Use the chain rule: dy/dt = (dy/dx)(dx/dt).
b) Both components are positive on the positive branch of the curve.

10. A balloon is ascending at the rate of 12 m/s. When it is at a height of 65 m from the ground, a packet is dropped from it. The packet will reach the ground after time of about (take g = 10 m/s²)
(a) 10 s
(b) 2.5 s
(c) 5 s
(d) 7.5 s

Answer: (c) 5 s

1. The packet leaves with the velocity of the balloon, 12 m/s upward. Take the upward direction as positive.
2. The displacement is -65 m: -65 = 12t - 5t², so 5t² - 12t - 65 = 0.
3. t = [12 + √(144 + 1300)]/10 = (12 + 38)/10 = 5 s.
4. Hence, the correct option is (c).

Teacher's Note:
a) A dropped object keeps the velocity of its carrier at that instant.
b) Discard the negative root of the quadratic.

11. Two cars are moving in the same direction with a speed of 30 km/h. These are separated from each other by 5 km. Another third car moving in the opposite direction meets the two cars after an interval of 4 minutes. The speed of the third car is
(a) 45 km/h.
(b) 40 km/h
(c) 35 km/h
(d) 30 km/h.

Answer: (a) 45 km/h.

1. Let the speed of the third car be v. Its speed relative to each of the other cars is (v + 30) km/h.
2. The two cars are 5 km apart, so the third car meets them at an interval of 5/(v + 30) hour.
3. 5/(v + 30) = 4/60, so v + 30 = 75.
4. v = 45 km/h.
5. Hence, the correct option is (a).

Teacher's Note:
a) Treat the cars as moving points and use relative speed.
b) Convert 4 minutes to hours before using it.

12. The displacement of a particle undergoing rectilinear motion along the x-axis is given by x = 2t³ - 21t² + 60t + 6. The acceleration of the particle when its velocity is zero will be.
(a) 36 m/sec²
(b) 9 m/sec²
(c) -9m/sec²
(d) -18m/sec²

Answer: (d) -18m/sec²

1. Velocity v = dx/dt = 6t² - 42t + 60.
2. v = 0 gives t² - 7t + 10 = 0, so t = 2 s or t = 5 s.
3. Acceleration a = dv/dt = 12t - 42.
4. At t = 2 s, a = 24 - 42 = -18 m/s². (At t = 5 s, a = +18 m/s², which is not among the options.)
5. Hence, the correct option is (d).

Teacher's Note:
a) A particle can have zero velocity at more than one instant.
b) Only -18 m/s² appears among the options.

13. A passenger sitting by the window of a train moving with a velocity of 72 km/h observes for 10 seconds that a train moving with a velocity of 32.4 km/h completely passes by it in 10 seconds. The length of the second train is
(a) 110 m
(b) 145 m
(c) 210 m
(d) 290 m

Answer: (d) 290 m

1. The speed of the first train = 72 km/h = 20 m/s and that of the second = 32.4 km/h = 9 m/s.
2. The second train passes in the opposite direction, so the relative speed = 20 + 9 = 29 m/s.
3. Length of the second train = 29 × 10 = 290 m.
4. Hence, the correct option is (d).

Teacher's Note:
a) The direction of the second train is not stated; the options are based on opposite directions. For the same direction the length would be 11 × 10 = 110 m.
b) Convert km/h to m/s by multiplying by 5/18.

14. The points A, B, C and D lie in a vertical line such that AB = BC = CD. If a body falls from rest at A, then the times of descent through AB, BC and CD are in the ratio of
(a) 1 : √2:√3
(b) √2 : √3 : 1
(c) √3 : 1 : √2
(d) 1: (√2-1):(√3-√2)

Answer: (d) 1: (√2-1):(√3-√2)

1. Let AB = BC = CD = h. For a fall from rest, the time to fall a distance s is t = √(2s/g).
2. Times from A: to B = √(2h/g), to C = √(4h/g) = √2 × √(2h/g), to D = √3 × √(2h/g).
3. Time through AB = 1 unit, through BC = (√2 - 1) units and through CD = (√3 - √2) units.
4. Hence, the ratio is 1 : (√2 - 1) : (√3 - √2) and the correct option is (d).

Teacher's Note:
a) Subtract successive total times to find the time through each section.
b) The body moves faster in the lower sections, so the times decrease.

15. A body is thrown up in a lift with a velocity u relative to a lift and its time of flight is found to be t. The acceleration with which the lift is moving up will be.
(a) \( \frac{u−gt}{2} \)
(b) \( \frac{2u−gt}{t} \)
(c) \( \frac{u²−gt}{t} \)
(d) \( \frac{u²}{gt} \)

Answer: (b) \( \frac{2u−gt}{t} \)

1. Let the lift move up with acceleration a. In the frame of the lift the effective acceleration due to gravity is g + a.
2. Time of flight t = 2u/(g + a), so g + a = 2u/t.
3. a = 2u/t - g = (2u - gt)/t.
4. Hence, the correct option is (b).

Teacher's Note:
a) Use the pseudo force in the lift frame; the velocity u is measured relative to the lift.
b) Check: if a = 0, then t = 2u/g, as expected.

16. A body falls from a certain height. Two seconds later another body falls from the same height. How long after the beginning of motion of the first body is the distance between the bodies twice the distance at the moment the second starts to fall ?
(a) 3 s
(b) 10 s
(c) 15 s
(d) 20 s

Answer: (a) 3 s

1. When the second body starts, the first has fallen (1/2)g(2)² = 2g metres, so the distance between them is 2g.
2. After a time t from the start of the first, the distance between them = (1/2)gt² - (1/2)g(t - 2)² = 2g(t - 1).
3. Setting 2g(t - 1) = 2 × 2g gives t - 1 = 2, so t = 3 s.
4. Hence, the correct option is (a).

Teacher's Note:
a) Write both distances using the same time origin.
b) The separation grows linearly with time, so doubling is quick to find.

17. A body falls freely from a height of 50 m. Simultaneously, another body is thrown from the surface of earth with a certain initial velocity. The two bodies meet at a height of 10 m. What is the initial velocity of the second body ?
(a) 4.9 m/s
(b) 9.8 m/s
(c) 17.5 m/s
(d) 19.6 m/s

Answer: (c) 17.5 m/s

1. Take g = 9.8 m/s². The first body falls 50 - 10 = 40 m, so 40 = (1/2)(9.8)t², t² = 8.16 and t = 2.86 s.
2. The second body rises 10 m in the same time: 10 = ut - (1/2)gt² = ut - 40.
3. So ut = 50 and u = 50/2.86 = 17.5 m/s.
4. Hence, the correct option is (c).

Teacher's Note:
a) Both bodies move for the same time before meeting.
b) In the frame of the falling body the other body moves in a straight line, so u × t equals the initial separation of 50 m.

18. The height y and distance x along the horizontal for a body projected in the vertical plane are given by y = 8 t - 5t² and x = 6 t
The initial velocity of the projected body is
(a) 8 m/s
(b) 9 m/s
(c) 10 m/s
(d) \( \frac{10}{3} \) m/s

Answer: (c) 10 m/s

1. v_y = dy/dt = 8 - 10t, so the initial vertical velocity is 8 m/s.
2. v_x = dx/dt = 6 m/s.
3. Initial speed = √(6² + 8²) = √100 = 10 m/s.
4. Hence, the correct option is (c).

Teacher's Note:
a) Differentiate both coordinates and put t = 0.
b) The 6-8-10 combination is a Pythagorean triple.

19. A particle is projected vertically upwards and it reaches the maximum height H in time T seconds. The height of the particle at any time t will be
(a) g (t - T)²
(b) H- \( \frac{1}{2} \) g (t - T)²
(c) \( \frac{1}{2} \) g (t - T)²
(d) H - g (t - T)

Answer: (b) H- \( \frac{1}{2} \) g (t - T)²

1. At time T the particle is at the highest point and its velocity is zero.
2. From the top, the particle behaves as if falling from rest for a time (t - T), so it is below the top by (1/2)g(t - T)².
3. Height = H - (1/2)g(t - T)².
4. Hence, the correct option is (b).

Teacher's Note:
a) The same formula works for t less than T, because the motion is symmetric about the top.
b) Check: at t = T the height is H.

20. A balloon rises from rest with a constant acceleration g/8. A stone is released from it when it has risen to a height h. The time taken by the stone to reach the ground is
(a) \( \sqrt{\frac{h}{g}} \)
(b) \( \sqrt{\frac{2h}{g}} \)
(c) 2 \( \sqrt{\frac{h}{g}} \)
(d) 4 \( \sqrt{\frac{h}{g}} \)

Answer: (c) 2 \( \sqrt{\frac{h}{g}} \)

1. Velocity of the balloon at height h: v² = 2(g/8)h = gh/4, so v = (1/2)√(gh) upward.
2. The stone leaves with this velocity: -h = vt - (1/2)gt², that is, gt² - √(gh) t - 2h = 0.
3. t = [√(gh) + √(gh + 8gh)]/(2g) = [√(gh) + 3√(gh)]/(2g) = 2√(h/g).
4. Hence, the correct option is (c).

Teacher's Note:
a) The stone starts with the upward velocity of the balloon, not from rest.
b) Choose the positive root of the quadratic in t.

21. A particle, initially at rest, starts moving in a straight line with an acceleration a = 6t + 4 m/s². The distance covered by it in 3 s is
(a) 15 m
(b) 30 m
(c) 45 m
(d) 60 m

Answer: (c) 45 m

1. v = ∫a dt = 3t² + 4t (v = 0 at t = 0).
2. s = ∫v dt = t³ + 2t².
3. At t = 3 s, s = 27 + 18 = 45 m.
4. Hence, the correct option is (c).

Teacher's Note:
a) Integrate twice, using zero initial velocity and zero initial displacement.
b) Do not use s = (1/2)at², which holds only for constant acceleration.

22. The acceleration 'a' in m/s², of a particle is given by a = 3t² + 2t + 2 where 't' is the time. If the particle starts out with a velocity v = 2 m/s at t = 0, then the velocity at the end of 2 s is
(a) 12 m/s
(b) 14 m/s
(c) 16 m/s
(d) 18 m/s

Answer: (d) 18 m/s

1. v = ∫a dt = t³ + t² + 2t + C.
2. At t = 0, v = 2, so C = 2.
3. At t = 2 s, v = 8 + 4 + 4 + 2 = 18 m/s.
4. Hence, the correct option is (d).

Teacher's Note:
a) Do not forget the initial velocity as the constant of integration.

23. A body starts from rest with uniform acceleration. If its velocity after n seconds is v, then its displacement in the last two seconds is
(a) \( \frac{2v(n−1)}{n} \)
(b) \( \frac{v(n−1)}{n} \)
(c) \( \frac{v(n+1)}{n} \)
(d) \( \frac{2v(2n+1)}{n} \)

Answer: (a) \( \frac{2v(n−1)}{n} \)

1. The acceleration a = v/n.
2. Displacement in n seconds = (1/2)an², and in (n - 2) seconds = (1/2)a(n - 2)².
3. Displacement in the last two seconds = (1/2)a[n² - (n - 2)²] = (1/2)a(4n - 4) = 2a(n - 1).
4. = 2(v/n)(n - 1) = 2v(n - 1)/n.
5. Hence, the correct option is (a).

Teacher's Note:
a) Subtract the distance covered in the first (n - 2) seconds from the total distance.
b) Check with n = 2: the answer reduces to v, which is the total distance (1/2)(v/2)(4) = v.

24. A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 m/s. A pendulum bob is suspended from the roof of the car by a light rigid rod of length 1.00 m. The angle made by the rod with the track is
(a) 90º
(b) 30º
(c) 45º
(d) 60º

Answer: (c) 45º

1. In the frame of the car, the bob is acted on by gravity mg downward and the centrifugal force mv²/r horizontally outward.
2. v²/r = 100/10 = 10 m/s², which is about g (taking g = 10 m/s²).
3. tan θ = (v²/r)/g = 1, so the rod makes 45° with the vertical and therefore 45° with the horizontal track.
4. Hence, the correct option is (c).

Teacher's Note:
a) In a turning car, the effective gravity is the vector sum of g and the outward acceleration.
b) The length of the rod does not matter.

25. The distance x and y along the horizontal plane of a projectile is given by x = 6t and y = 8t-5t² in metres where t is in seconds. The angle with the horizontal at which the projectile is projected is
(a) tan⁻¹\( \frac{3}{4} \)
(b) tan⁻¹ \( \frac{4}{3} \)
(c) sin⁻¹\( \frac{3}{4} \)
(d) cos⁻¹\( \frac{3}{4} \)

Answer: (b) tan⁻¹ \( \frac{4}{3} \)

1. At t = 0, v_x = 6 m/s and v_y = 8 m/s.
2. tan θ = v_y/v_x = 8/6 = 4/3.
3. θ = tan⁻¹(4/3), which is about 53°.
4. Hence, the correct option is (b).

Teacher's Note:
a) The initial speed is 10 m/s, so sin θ = 0.8 and cos θ = 0.6; options (c) and (d) do not fit.
b) The angle is measured from the horizontal, so use v_y/v_x.

26. The equation of projectile is y = √3 x - \( \frac{gx²}{2} \), The velocity of projection is
(a) 1.5 m/s
(b) 2 m/s
(c) 2.5 m/s
(d) 3 m/s

Answer: (b) 2 m/s

1. Compare with y = x tan θ - gx²/(2u² cos²θ).
2. tan θ = √3, so θ = 60° and cos²θ = 1/4.
3. gx²/(2u² cos²θ) = gx²/2, so u² cos²θ = 1 and u² = 4.
4. u = 2 m/s.
5. Hence, the correct option is (b).

Teacher's Note:
a) Match the coefficients of x and x² with the standard equation of the trajectory.
b) Check: u²/4 = 1 gives u = 2 m/s.

27. Bullets are fired with velocity u at a fixed angle to the horizontal in all the directions. The maximum area covered by the bullets is
(a) \( \frac{πu²}{g} \)
(b) \( \frac{π²u²}{g} \)
(c) \( \frac{π²u⁴}{g} \)
(d) \( \frac{πu⁴}{g²} \)

Answer: (d) \( \frac{πu⁴}{g²} \)

1. Bullets fired at the same angle in all horizontal directions land on a circle whose radius is the range R = u² sin 2θ/g.
2. Area = πR² = πu⁴ sin²2θ/g².
3. The area is maximum when sin 2θ = 1, that is, θ = 45°.
4. Maximum area = πu⁴/g².
5. Hence, the correct option is (d).

Teacher's Note:
a) The range is greatest at 45°, so the area covered is greatest then.
b) The area depends on R², giving u to the power 4.

28. The angle of projection, at which the horizontal range and maximum height of projectile are equal, is
(a) 45º
(b) θ = tan⁻¹ (0.25)
(c) θ = tan⁻¹ 4
(d) 60º.

Answer: (c) θ = tan⁻¹ 4

1. Range R = u² sin 2θ/g = 2u² sin θ cos θ/g. Maximum height H = u² sin²θ/(2g).
2. Setting R = H: 2 sin θ cos θ = sin²θ/2.
3. 4 cos θ = sin θ, so tan θ = 4.
4. Hence, θ = tan⁻¹ 4 and the correct option is (c).

Teacher's Note:
a) Cancel sin θ from both sides carefully; it is not zero.
b) The angle is about 76°.

29. From the top of a tower of height 40 m a ball is projected upwards with a speed of 20 m/sec at an angle of elevation of 30º. Then the ratio of the total time taken by the ball to hit the ground to its time of flight (time taken to come back to the same elevation) is (take g = 10 m/sec²)
(a) 2 : 1
(b) 3 : 1
(c) 3 : 2
(d) 4 : 1

Answer: (a) 2 : 1

1. The vertical component of the initial velocity = 20 sin 30° = 10 m/s.
2. Time of flight (to return to the same level) = 2 × 10/10 = 2 s.
3. Time to hit the ground: -40 = 10t - 5t², so t² - 2t - 8 = 0 and t = 4 s.
4. Ratio = 4 : 2 = 2 : 1.
5. Hence, the correct option is (a).

Teacher's Note:
a) The ball first rises, then falls past the height of the tower to the ground.
b) Choose the positive root of the quadratic.

30. Three particles, A, B, C are projected from the same point with same initial speeds making angles 30º, 45º and 60º respectively with the horizontal. Which of the following statement is correct ?
(a) A, B and C have unequal ranges
(b) Ranges of A and C are equal and less than that of B
(c) Ranges of A and C are equal and greater than that of B
(d) A, B and C have equal ranges.

Answer: (b) Ranges of A and C are equal and less than that of B

1. R = u² sin 2θ/g.
2. For 30° and 60°: sin 60° = sin 120° = 0.866, so the ranges are equal.
3. For 45°: sin 90° = 1, which is the maximum.
4. So the ranges of A and C are equal and less than that of B.
5. Hence, the correct option is (b).

Teacher's Note:
a) Angles that add up to 90° give equal ranges.
b) The range is maximum at 45°.

31. A javelin thrown into air at an angle with the horizontal has a range of 200 m. If the time of flight is 5 second, then the horizontal component of velocity of the projectile at the highest point of the trajectory is
(a) 40 m/s
(b) 20 m/s
(c) 9.8 m/s
(d) 5 m/s

Answer: (a) 40 m/s

1. The horizontal component of velocity is constant throughout the flight.
2. Range = horizontal velocity × time of flight, so u_x = 200/5 = 40 m/s.
3. At the highest point the velocity is purely horizontal, so it equals 40 m/s.
4. Hence, the correct option is (a).

Teacher's Note:
a) At the highest point only the horizontal component remains.

32. A projectile is thrown with an initial velocity of (ai + bj) m/s. If the range of projectile is twice the maximum height reached by it, then
(a) b = a/2
(b) b = a
(c) b = 2 a
(d) b = 4 a

Answer: (c) b = 2 a

1. Range R = 2ab/g (time of flight 2b/g multiplied by horizontal velocity a).
2. Maximum height H = b²/(2g).
3. R = 2H gives 2ab/g = b²/g.
4. So b = 2a.
5. Hence, the correct option is (c).

Teacher's Note:
a) Horizontal component a and vertical component b give tan θ = b/a = 2.
b) R = 4H/tan θ is a useful general relation.

33. A projectile is projected with kinetic energy K. If it has the maximum possible horizontal range, then its kinetic energy at the highest point will be
(a) 0.25 K
(b) 0.5 K
(c) 0.75 K
(d) K

Answer: (b) 0.5 K

1. The maximum range is obtained at 45°.
2. At the highest point only the horizontal velocity u cos 45° remains.
3. KE at the top = (1/2)m(u cos 45°)² = K cos²45° = K × (1/2) = 0.5K, where K = (1/2)mu².
4. Hence, the correct option is (b).

Teacher's Note:
a) KE at the top = K cos²θ.
b) Half of the energy remains as kinetic energy at the top for a 45° projection.

34. A projectile has the maximum range of 500 m. If the projectile is now thrown up an inclined plane of 30º with the same speed, the distance covered by it along the inclined plane will be
(a) 250 m
(b) 500 m
(c) 750 m
(d) 1000 m

Answer: (b) 500 m

1. The maximum range on the level ground is u²/g = 500 m.
2. The projectile is thrown up along the inclined plane with the same speed u, so it moves along the plane.
3. The retardation along the plane = g sin 30° = g/2.
4. Distance = u²/(2 × g/2) = u²/g = 500 m.
5. Hence, the correct option is (b).

Teacher's Note:
a) The reading used is that the body is projected along the plane, not at an angle to it.
b) The component of gravity along the plane is g sin θ.

35. Two seconds after projection, a projectile is moving at 30º above the horizontal; after one more second it is moving horizontally. The initial speed of the projectile is (g = 10 m/s²)
(a) 10 √3 m/s
(b) 20 m/s
(c) 10 m/s
(d) 20 √3 m/s

Answer: (d) 20 √3 m/s

1. The projectile is moving horizontally at t = 3 s, which is the highest point, so the initial vertical velocity = 10 × 3 = 30 m/s.
2. At t = 2 s: v_y = 30 - 20 = 10 m/s and the direction is 30° above the horizontal, so v_x = 10/tan 30° = 10√3 m/s.
3. Initial speed = √(30² + (10√3)²) = √(900 + 300) = 20√3 m/s.
4. Hence, the correct option is (d).

Teacher's Note:
a) Use the fact that the velocity is horizontal at the top of the path.
b) The horizontal component never changes.

36. From a point on the ground at a distance of 2 m from the foot of a vertical wall, a ball is thrown at an angle of 45º which just clears the top of the wall and then strikes the ground at a distance of 4 m from the foot of the wall on the other side. The height of the wall is
(a) 3 m
(b) \( \frac{2}{3} \) m
(c) \( \frac{3}{4} \) m
(d) \( \frac{4}{3} \) m

Answer: (d) \( \frac{4}{3} \) m

1. The total horizontal range R = 2 + 4 = 6 m.
2. For a projection at 45° the trajectory is y = x tan θ(1 - x/R) with tan 45° = 1, so y = x(1 - x/6).
3. At x = 2 m (the position of the wall), y = 2(1 - 1/3) = 4/3 m.
4. Hence, the height of the wall is 4/3 m and the correct option is (d).

Teacher's Note:
a) For a projection at 45°, the trajectory is y = x(1 - x/R), with R being the range.
b) The wall is 2 m from the starting point, not 4 m.

37. Two stones are projected with the same speed but making different angles with the horizontal. Their ranges are equal. If the angle of projection of one is π/3 and its maximum height is H, then the maximum height of the other is
(a) 3 H
(b) 2 H
(c) H/3
(d) H/2

Answer: (c) H/3

1. Equal ranges with the same speed mean that the angles are complementary: π/3 and π/6.
2. H = u² sin²θ/(2g). For θ = π/3, H = (u²/2g)(3/4).
3. For θ = π/6, H' = (u²/2g)(1/4).
4. H'/H = (1/4)/(3/4) = 1/3, so H' = H/3.
5. Hence, the correct option is (c).

Teacher's Note:
a) Complementary angles give the same range but different heights.
b) The ratio of heights is tan²θ-type: H'/H = tan²(30°) = 1/3.

38. The string of a pendulum of length l is displaced through 90º from the vertical and released. Then the minimum strength of the string in order to withstand the tension as the pendulum passes through the mean position is
(a) 3 m g
(b) 4 m g
(c) 5 m g
(d) 6 m g

Answer: (a) 3 m g

1. By energy conservation from the horizontal position to the lowest point: (1/2)mv² = mgl, so v² = 2gl.
2. At the mean position, T - mg = mv²/l = 2mg.
3. T = 3mg.
4. Hence, the string must withstand at least 3mg and the correct option is (a).

Teacher's Note:
a) The printed key gives (b), which is wrong; the correct tension at the lowest point is 3mg, not 4mg.
b) For a release from angle θ, the tension at the bottom is mg(3 - 2 cos θ).

39. A pendulum consisting of a small sphere of mass m, suspended by an inextensible and massless string of length l, is made to swing in a vertical plane. If the breaking strength of the string is 2 mg, then the maximum angular amplitude of the displacement from the vertical can be
(a) 45º
(b) 30º
(c) 60º
(d) 90º

Answer: (c) 60º

1. The tension at the lowest point for an amplitude θ is T = mg(3 - 2 cos θ).
2. The string does not break if T is at most 2mg: 3 - 2 cos θ ≤ 2, so cos θ ≥ 1/2.
3. θ ≤ 60°, so the maximum amplitude is 60°.
4. Hence, the correct option is (c).

Teacher's Note:
a) The tension is greatest at the mean position.
b) Amplitudes less than 60° are also safe, but 60° is the maximum.

40. The roadway of a bridge is in the form of a circle of radius 18 m. What is the greatest speed with which a motor cycle can cross the bridge without leaving the ground at the highest point ? (g = 9.8 m/s²)
(a) √(9.8) m/s
(b) √(18×9.8) m/s
(c) 18 × 9.8 m/s
(d) 18/9.8 m/s

Answer: (b) √(18×9.8) m/s

1. At the highest point of the bridge, the net downward force provides the centripetal force: mg - N = mv²/r.
2. The motorcycle just leaves the ground when N = 0, so mg = mv²/r.
3. v = √(rg) = √(18 × 9.8) = 13.3 m/s.
4. Hence, the correct option is (b).

Teacher's Note:
a) The printed key gives (a), which is wrong; the speed depends on both the radius and g, v = √(rg).
b) If the speed is higher, the normal force becomes zero and the motorcycle loses contact.