CBSE Class 10 Physics Motion MCQs Set 36

Multiple Choice Questions (MCQs) for Class 10 Physics: Motion

Access targeted multiple-choice questions for Motion designed to align with the latest CBSE academic syllabus for Class 10 Physics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Practice Motion MCQs for Class 10 Physics

View or download the dedicated Motion MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Physics evaluations.

Multiple Choice Questions

Question 1: Which quantity has the unit g cm s-2?
(a) force
(b) momentum
(c) energy
(d) impulse
Show Answer & Explanation

Answer: (a) force

Explanation:
1. g cm s-2 = mass × acceleration, which is the dyne.
2. Mass × acceleration is force.

Teacher's Note:
1. g cm s-2 is the CGS unit of force (dyne).
2. Its SI counterpart is kg m s-2 (newton).

Question 2: If the dimensional formula of a physical quantity is [MLT-2], what is its SI unit?
(a) kg2 s-2
(b) kg m s-2
(c) g cm s
(d) kg m s
Show Answer & Explanation

Answer: (b) kg m s-2

Explanation:
1. [M] gives kg, [L] gives m and [T-2] gives s-2.
2. So the unit is kg m s-2.

Teacher's Note:
1. Replace M, L, T by kg, m, s.
2. kg m s-2 is the newton.

Question 3: Which of the following equations is dimensionally incorrect (s = displacement, u, v = velocities, a = acceleration, t = time)?
(a) v2 = as
(b) s = ut
(c) s = at
(d) s = at2
Show Answer & Explanation

Answer: (c) s = at

Explanation:
1. For s = at: left side has dimension L, right side has LT-2 × T = LT-1.
2. The dimensions differ, so s = at is incorrect.
3. The other three have matching dimensions on both sides (for v2 = as, L2T-2 on both sides).

Teacher's Note:
1. Check both sides of an equation for the same dimensions.
2. Numerical factors such as ½ have no dimensions.

Question 4: The thimble of a screw gauge (100 divisions on the head scale) is given 40 complete rotations and the tip of the screw advances 20 mm. When the studs are in contact, the 85th division of the head scale is on the index line. What is the zero correction?
(a) 0.075 mm
(b) 0.75 mm
(c) -0.075 mm
(d) -0.75 mm
Show Answer & Explanation

Answer: (a) 0.075 mm

Explanation:
1. Pitch = 20 mm / 40 = 0.5 mm, so least count = 0.5 / 100 = 0.005 mm.
2. The 85th division at the index means the reading is 15 divisions short of 100, so the zero error = -15 × 0.005 = -0.075 mm.
3. Zero correction = +0.075 mm.

Teacher's Note:
1. Zero correction has the opposite sign to zero error.
2. Pitch divided by the number of head divisions gives the least count.

Question 5: What is the acceleration of a body that is initially moving at 10 m s-1 and attains 20 m s-1 in 2 s?
(a) 2 m s-2
(b) 10 m s-2
(c) 5 m s-2
(d) 1 m s-2
Show Answer & Explanation

Answer: (c) 5 m s-2

Explanation:
1. a = (v - u)/t = (20 - 10)/2 = 5 m s-2.

Teacher's Note:
1. Acceleration is the rate of change of velocity.
2. Use a = (v - u)/t.

Question 6: A body starts with a velocity of 2 m s-1 and moves with an acceleration of 5 m s-2. What distance does it travel in the 5th second?
(a) 18 m
(b) 24.5 m
(c) 27.5 m
(d) 20 m
Show Answer & Explanation

Answer: (b) 24.5 m

Explanation:
1. Distance in the nth second: sn = u + a(n - ½).
2. s5 = 2 + 5 × 4.5 = 24.5 m.

Teacher's Note:
1. sn is the distance covered during the nth second only.
2. Do not confuse it with distance in n seconds.

Question 7: An object dropped from the top of a building travels 25 m in the last second of its motion. What is the total time of fall? (g = 10 m s-2)
(a) 4.5 s
(b) 2.5 s
(c) 3 s
(d) 6.5 s
Show Answer & Explanation

Answer: (c) 3 s

Explanation:
1. sn = u + g(n - ½) gives 25 = 0 + 10(n - ½).
2. n - ½ = 2.5, so n = 3 s.

Teacher's Note:
1. For a freely falling body u = 0.
2. Distance in the last second equals sn for n = total time.

Question 8: An object is thrown vertically upward with a velocity of 15 m s-1. After how many seconds does it strike the ground? (g = 10 m s-2)
(a) 6 s
(b) 5 s
(c) 2 s
(d) 3 s
Show Answer & Explanation

Answer: (d) 3 s

Explanation:
1. Time of flight T = 2u/g = 2 × 15/10 = 3 s.

Teacher's Note:
1. Time of ascent equals time of descent.
2. T = 2u/g for a body returning to its starting level.

Question 9: An object is projected horizontally with a velocity of 10 m s-1 from the top of a tower 50 m high. What horizontal distance does it travel? (g = 10 m s-2)
(a) 10 m
(b) \( 10\sqrt{10} \) m
(c) \( 12\sqrt{2} \) m
(d) \( 10\sqrt{2} \) m
Show Answer & Explanation

Answer: (b) \( 10\sqrt{10} \) m

Explanation:
1. Vertical motion: 50 = ½ × 10 × t2, so t = \( \sqrt{10} \) s.
2. Horizontal distance = 10 × \( \sqrt{10} \) = \( 10\sqrt{10} \) m.

Teacher's Note:
1. Horizontal and vertical motions are independent.
2. Time of fall depends only on the height.

Question 10: If a vertically projected body takes 3 s to reach its maximum height, what time does it take to reach half of the maximum height on the way up? (g = 10 m s-2)
(a) 0.88 s
(b) 1.52 s
(c) 2.15 s
(d) 3.27 s
Show Answer & Explanation

Answer: (a) 0.88 s

Explanation:
1. u = g t = 30 m s-1, so Hmax = 900/20 = 45 m and half of it is 22.5 m.
2. 22.5 = 30t - 5t2 gives t2 - 6t + 4.5 = 0, so t = 3 - \( \sqrt{4.5} \) = 0.88 s (the smaller root is the upward journey).

Teacher's Note:
1. A quadratic in t gives two times: going up and coming down.
2. Choose the smaller one for the upward journey.

Question 11: A particle is projected at 10 m s-1 making 45° with the horizontal on a planet where the weight of the body is 200 N. The weight of the same body on the Earth is 150 N. What is the maximum height reached? (g on Earth = 10 m s-2)
(a) 2.563 m
(b) 1.875 m
(c) 16.232 m
(d) 2.255 m
Show Answer & Explanation

Answer: (b) 1.875 m

Explanation:
1. gp/ge = 200/150, so gp = 40/3 m s-2.
2. H = u2sin2θ/(2gp) = 100 × ½ / (80/3) = 1.875 m.

Teacher's Note:
1. Weight is proportional to g for the same mass.
2. H = u2sin2θ/2g.

Question 12: An object is thrown vertically upwards with a velocity of 20 m s-1. After how many seconds will its displacement be zero? (g = 10 m s-2)
(a) 2 s
(b) 4 s
(c) 8 s
(d) 16 s
Show Answer & Explanation

Answer: (b) 4 s

Explanation:
1. Displacement is zero when the object returns to the starting point.
2. T = 2u/g = 2 × 20/10 = 4 s.

Teacher's Note:
1. Zero displacement means back at the start.
2. Do not confuse it with zero velocity (at the top).

Question 13: Which quantity has the unit kg m2 s-2?
(a) power
(b) energy
(c) force
(d) momentum
Show Answer & Explanation

Answer: (b) energy

Explanation:
1. kg m2 s-2 = (kg m s-2) × m = force × displacement.
2. That is work or energy (the joule).

Teacher's Note:
1. The joule is the unit of energy.
2. Power is J s-1.

Question 14: If the dimensional formula of a physical quantity is [MLT-1], what is its CGS unit?
(a) g cm s-1
(b) g cm s-2
(c) g cm s
(d) g-1 s cm
Show Answer & Explanation

Answer: (a) g cm s-1

Explanation:
1. In CGS, M, L and T are replaced by g, cm and s.
2. So the unit is g cm s-1.

Teacher's Note:
1. [MLT-1] is the dimension of momentum.
2. Its SI unit is kg m s-1.

Question 15: Which of the following equations is dimensionally incorrect?
(a) v2 = 2as
(b) v = at
(c) v = st
(d) s = ½at2
Show Answer & Explanation

Answer: (c) v = st

Explanation:
1. For v = st: left side LT-1, right side L × T = LT, which do not match.
2. So v = st is incorrect.
3. Here v2 = 2as, v = at and s = ½at2 are all dimensionally correct.

Teacher's Note:
1. Compare the dimensions on both sides.
2. Speed = distance divided by time, not multiplied.

Question 16: When the thimble of a screw gauge (100 divisions on the head scale) is given 10 complete rotations, the tip advances 20 mm. When the studs are in contact, the 90th division of the head scale is on the index line. What is the zero error?
(a) 0.02 mm
(b) -0.2 mm
(c) 0.002 mm
(d) 0.2 mm
Show Answer & Explanation

Answer: (b) -0.2 mm

Explanation:
1. Pitch = 20/10 = 2 mm, least count = 2/100 = 0.02 mm.
2. The 90th division at the index is 10 divisions below 100, so the zero error = -10 × 0.02 = -0.2 mm.

Teacher's Note:
1. A reading just below 100 means a negative zero error.
2. Zero correction here would be +0.2 mm.

Question 17: A body moves with a uniform acceleration of 1 m s-2 and zero initial velocity. What is its velocity after 5 s?
(a) 1 m s-1
(b) 2 m s-1
(c) 10 m s-1
(d) 5 m s-1
Show Answer & Explanation

Answer: (d) 5 m s-1

Explanation:
1. v = u + at = 0 + 1 × 5 = 5 m s-1.

Teacher's Note:
1. Use v = u + at.
2. u = 0 because the body starts from rest.

Question 18: A body starting from rest travels 20 m in the 5th second. What is its acceleration?
(a) 4.4 m s-2
(b) 5.5 m s-2
(c) 6.6 m s-2
(d) 3.3 m s-2
Show Answer & Explanation

Answer: (a) 4.4 m s-2

Explanation:
1. s5 = u + a(n - ½) = a × 4.5 = 20.
2. a = 20/4.5 = 4.4 m s-2.

Teacher's Note:
1. Use the distance-in-nth-second formula.
2. u = 0 from rest.

Question 19: An object dropped from a cliff reaches the ground in 2 s. What distance does it travel in the last second? (g = 10 m s-2)
(a) 25 m
(b) 20 m
(c) 15 m
(d) 10 m
Show Answer & Explanation

Answer: (c) 15 m

Explanation:
1. s2 = 0 + 10 × (2 - ½) = 15 m.
2. Check: fall in 2 s = 20 m and in 1 s = 5 m, difference 15 m.

Teacher's Note:
1. Distance in the last second = total distance - distance in (n-1) seconds.

Question 20: A ball projected upwards returns to the initial position in 4 s. What is its initial velocity? (g = 10 m s-2)
(a) 20 m s-1
(b) 30 m s-1
(c) 40 m s-1
(d) 50 m s-1
Show Answer & Explanation

Answer: (a) 20 m s-1

Explanation:
1. T = 2u/g, so u = gT/2 = 10 × 4/2 = 20 m s-1.

Teacher's Note:
1. Time of ascent is half the total time.
2. u = g × time of ascent.

Question 21: An object is projected horizontally with a velocity of 20 m s-1 from a certain height and hits the ground in 3 s. What is the height above the ground? (g = 10 m s-2)
(a) 80 m
(b) 65 m
(c) 30 m
(d) 45 m
Show Answer & Explanation

Answer: (d) 45 m

Explanation:
1. Only vertical motion decides the fall time: h = ½gt2 = ½ × 10 × 9 = 45 m.

Teacher's Note:
1. The horizontal velocity does not affect the height.
2. Initial vertical velocity is zero.

Question 22: An object is projected vertically upwards from the top of a cliff of height h with initial velocity u. Arrange the steps to find the time to reach the ground.
(i) Determine the displacement and note the initial velocity.
(ii) Choose an equation of motion.
(iii) Substitute the values of initial velocity and displacement.
(iv) Assign signs to displacement and initial velocity according to the sign convention.
(v) Obtain the time taken to reach the ground.
(a) i ii iv iii v
(b) i ii iii iv v
(c) i iv ii iii v
(d) i iv iii ii v
Show Answer & Explanation

Answer: (c) i iv ii iii v

Explanation:
1. First find the displacement and initial velocity (i), then assign signs by the sign convention (iv).
2. Then choose the equation s = ut + ½at2 (ii), substitute the values (iii) and solve for t (v).
3. The order is i, iv, ii, iii, v.

Teacher's Note:
1. Signs must be fixed before substituting.
2. Displacement is -h when the object ends below the start.

Question 23: Match Column A with Column B.
(P) Ratio of maximum horizontal distances for bodies projected at 5 m s-1 with angles 15° and 45°
(Q) Two bodies projected with the same velocity have the ratio of times to reach maximum height 1 : \( \sqrt{2} \); ratio of the sines of their angles of projection
(R) Ratio of times of ascent of two bodies projected vertically up with the same velocity, one from the top of a tower and one from the ground
(p) 1 : \( \sqrt{2} \) (q) 1 : 2 (r) 1 : 1
(a) P → q; Q → p; R → r
(b) P → r; Q → p; R → q
(c) P → p; Q → q; R → r
(d) P → q; Q → r; R → p
Show Answer & Explanation

Answer: (a) P → q; Q → p; R → r

Explanation:
1. P: range ∝ sin2θ, so sin30° : sin90° = ½ : 1 = 1 : 2, which is (q).
2. Q: time to maximum height ∝ sinθ, so the ratio of sines is 1 : \( \sqrt{2} \), which is (p).
3. R: time of ascent = u/g is independent of height, so the ratio is 1 : 1, which is (r).

Teacher's Note:
1. Range ∝ sin2θ for the same speed.
2. Time of ascent = u sinθ / g.

Question 24: A vertically projected body reaches a maximum height of 20 m. How long does it take to travel the last 10 m of its upward journey? (g = 10 m s-2)
(a) 1.41 s
(b) 0.5 s
(c) 18 s
(d) 2 s
Show Answer & Explanation

Answer: (a) 1.41 s

Explanation:
1. The last 10 m below the top is like free fall from rest through 10 m.
2. 10 = ½ × 10 × t2, so t = \( \sqrt{2} \) = 1.41 s.

Teacher's Note:
1. Time going up through the last stretch equals time falling through it from the top.
2. Reverse the motion to simplify.

Free study material for Science

Multiple Choice Questions (MCQs) for Class 10 Physics Motion

Class 10 Physics Motion Objective Test Questions

Test your conceptual understanding of Motion with these targeted multiple-choice questions. Designed in alignment with the latest CBSE curriculum for Class 10 Physics, these problem sets build accuracy and prepare students for objective exams.

NCERT-Aligned Objective Questions and Solutions

Each question includes structured solution keys mapped directly to standard CBSE textbooks, helping students evaluate their reasoning and correct mistakes early in their revision.

Next Steps in Your Exam Preparation

Explore our broader library of printable assignments, chapter notes, and mock tests designed to support continuous revision and secure higher marks in CBSE assessments.

FAQs

Where can I access latest CBSE Class 10 Physics Motion MCQs Set 36?

You can get most exhaustive CBSE Class 10 Physics Motion MCQs Set 36 for free on StudiesToday.com. These MCQs for Class 10 Physics are updated for the 2026-27 academic session as per CBSE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Physics Class 10 material?

Yes, our CBSE Class 10 Physics Motion MCQs Set 36 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the CBSE paper is now competency-based.

How do practicing Physics MCQs help in scoring full marks in Class 10 exams?

By solving our CBSE Class 10 Physics Motion MCQs Set 36, Class 10 students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Physics.

Do you provide answers and explanations for CBSE Class 10 Physics Motion MCQs Set 36?

Yes, Physics MCQs for Class 10 have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused CBSE exams.

Can I practice these Physics Class 10 MCQs online?

Yes, you can also access online interactive tests for CBSE Class 10 Physics Motion MCQs Set 36 on StudiesToday.com as they provide instant answers and score to help you track your progress in Physics.