TS EAMCET PYQs for Motion in a Plane with Solutions: Practice TS EAMCET Previous Year Questions

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Yashodeep Mahurkar

Updated on - Jan 4, 2026

Motion in a Plane is an important topic in the Physics section in TS EAMCET exam. Practising this topic will increase your score overall and make your conceptual grip on TS EAMCET exam stronger.

This article gives you a full set of TS EAMCET PYQs for Motion in a Plane with explanations for effective preparation. Practice of TS EAMCET Physics PYQs including Motion in a Plane questions regularly will improve accuracy, speed, and confidence in the TS EAMCET 2026 exam.

TS EAMCET PYQs for Motion in a Plane with Solutions

1.
A cannon on a cliff 55 m above the ground fires a shell with a velocity of $50\hat{i}+50\hat{j}$ ms\textsuperscript{−1}. The displacement vector of the shell when it hits the ground is (Acceleration due to gravity = 10 ms\textsuperscript{−2})
- $(550\hat{i}-55\hat{j})$\,m
- $(550\hat{i}-500\hat{j})$\,m
- $(500\hat{i}-55\hat{j})$\,m
- $(500\hat{i}-550\hat{j})$\,m
View Solution
2.
The time period of a simple pendulum on the surface of the earth is $ T $. The time period of the same pendulum at a height of 1280 km from the surface of the earth is
- $ 1.5T $
- $ 1.2T $
- $ 2T $
- $ 2.4T $
View Solution
3.
The minimum velocity of a projectile is 50% of its maximum velocity. If the minimum velocity of the projectile is $10~\text{m s}^{-1}$ then the time of flight of the projectile is (Acceleration due to gravity = $10~\text{m s}^{-2}$
- 4 $\sqrt{3}$ sec
- 3 $\sqrt{3}$ sec
- 2 $\sqrt{3}$ sec
- $\sqrt{3}$ sec
View Solution
4.
If a body projected with a velocity of 19.6 ms⁻¹ reaches a maximum height of 9.8 m, then the range of the projectile is (Neglect air resistance)
- 19.6 m
- 39.2 m
- 78.4 m
- 9.8 m
View Solution
5.
A body thrown vertically upwards with certain velocity from the ground reaches a maximum height $ H $. The ratio of the times at which the body is at a height of $ \frac{H}{2} $ is:
- $ 3:2 $
- $ \sqrt{3} : \sqrt{2} $
- $ (\sqrt{3} - 1) : (\sqrt{3} + 1) $
- $\mathbf{(\sqrt{2} -1) : (\sqrt{2} +1)}$
  \
View Solution
6.
A 20-ton truck is traveling along a curved path of radius 240 m. If the center of gravity of the truck above the ground is 2 m and the distance between its wheels is 1.5 m, the maximum speed of the truck with which it can travel without toppling over is:
(Acceleration due to gravity $ g = 10 $ ms$^{-2}$)
- $ 43 $ ms$^{-1}$
- $ 40 $ ms$^{-1}$
- $ 38 $ ms$^{-1}$
- $ 30 $ ms$^{-1}$
View Solution
7.
A cannon placed on a cliff at a height of 375 m fires a cannonball with a velocity of 100 m/s$^{-1}$ at an angle of 30Â° above the horizontal. The horizontal distance between the cannon and the target is:
(Acceleration due to gravity $ g = 10 $ ms$^{-2}$)
- $ 750\sqrt{3} $ m
- $ 500\sqrt{3} $ m
- $ 250\sqrt{3} $ m
- $ 750 $ m
View Solution
8.
y = a sin(βx + γt)wherex and t represent displacement and time, respectively. Then, the dimensional formula for β— γis:
- $\mathbf{[M^0L^1T^{-1}]}$
- $\mathbf{[M^0L^1T^{0}]}$
- $\mathbf{[M^1L^1T^{-1}]}$
- $\mathbf{[M^1L^0T^{-1}]}$
View Solution
9.
The theory currently accepted as proper framework for explaining microscopic phenomena is
- Classical physics
- Quantum theory
- Wave theory
- Electromagnetic theory
View Solution
10.
A cannon fires two similar shells one after the other each with a velocity of 100 m/s at angles 60$^\circ$ and 30$^\circ$ respectively with the horizontal such that they both hit the target at the same time. The time interval between the firing of the two shells is (Acceleration due to gravity = 10 m/s²).
- 6.57 s
- 7.32 s
- 8.28 s
- 3.14 s
View Solution

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TS EAMCET PYQs for Motion in a Plane with Solutions: Practice TS EAMCET Previous Year Questions

TS EAMCET PYQs for Motion in a Plane with Solutions

1.
A cannon on a cliff 55 m above the ground fires a shell with a velocity of \(50\hat{i}+50\hat{j}\) ms\textsuperscript{−1}. The displacement vector of the shell when it hits the ground is (Acceleration due to gravity = 10 ms\textsuperscript{−2})

2.
The time period of a simple pendulum on the surface of the earth is \( T \). The time period of the same pendulum at a height of 1280 km from the surface of the earth is

3.
The minimum velocity of a projectile is 50% of its maximum velocity. If the minimum velocity of the projectile is $10~\text{m s}^{-1}$ then the time of flight of the projectile is (Acceleration due to gravity = $10~\text{m s}^{-2}$

4.
If a body projected with a velocity of 19.6 ms⁻¹ reaches a maximum height of 9.8 m, then the range of the projectile is (Neglect air resistance)

5.
A body thrown vertically upwards with certain velocity from the ground reaches a maximum height \( H \). The ratio of the times at which the body is at a height of \( \frac{H}{2} \) is:

7.
A cannon placed on a cliff at a height of 375 m fires a cannonball with a velocity of 100 m/s\(^{-1}\) at an angle of 30Â° above the horizontal. The horizontal distance between the cannon and the target is:
(Acceleration due to gravity \( g = 10 \) ms\(^{-2}\))

8.
y = a sin(βx + γt)wherex and t represent displacement and time, respectively. Then, the dimensional formula for β— γis:

9.
The theory currently accepted as proper framework for explaining microscopic phenomena is

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TS EAMCET PYQs for Motion in a Plane with Solutions

1.A cannon on a cliff 55 m above the ground fires a shell with a velocity of \(50\hat{i}+50\hat{j}\) ms\textsuperscript{−1}. The displacement vector of the shell when it hits the ground is (Acceleration due to gravity = 10 ms\textsuperscript{−2})

2.The time period of a simple pendulum on the surface of the earth is \( T \). The time period of the same pendulum at a height of 1280 km from the surface of the earth is

3.The minimum velocity of a projectile is 50% of its maximum velocity. If the minimum velocity of the projectile is $10~\text{m s}^{-1}$ then the time of flight of the projectile is (Acceleration due to gravity = $10~\text{m s}^{-2}$

4.If a body projected with a velocity of 19.6 ms⁻¹ reaches a maximum height of 9.8 m, then the range of the projectile is (Neglect air resistance)

5.A body thrown vertically upwards with certain velocity from the ground reaches a maximum height \( H \). The ratio of the times at which the body is at a height of \( \frac{H}{2} \) is:

7.A cannon placed on a cliff at a height of 375 m fires a cannonball with a velocity of 100 m/s\(^{-1}\) at an angle of 30Â° above the horizontal. The horizontal distance between the cannon and the target is: (Acceleration due to gravity \( g = 10 \) ms\(^{-2}\))

8.y = a sin(βx + γt)wherex and t represent displacement and time, respectively. Then, the dimensional formula for β— γis:

9.The theory currently accepted as proper framework for explaining microscopic phenomena is

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1.
A cannon on a cliff 55 m above the ground fires a shell with a velocity of \(50\hat{i}+50\hat{j}\) ms\textsuperscript{−1}. The displacement vector of the shell when it hits the ground is (Acceleration due to gravity = 10 ms\textsuperscript{−2})

2.
The time period of a simple pendulum on the surface of the earth is \( T \). The time period of the same pendulum at a height of 1280 km from the surface of the earth is

3.
The minimum velocity of a projectile is 50% of its maximum velocity. If the minimum velocity of the projectile is $10~\text{m s}^{-1}$ then the time of flight of the projectile is (Acceleration due to gravity = $10~\text{m s}^{-2}$

4.
If a body projected with a velocity of 19.6 ms⁻¹ reaches a maximum height of 9.8 m, then the range of the projectile is (Neglect air resistance)

5.
A body thrown vertically upwards with certain velocity from the ground reaches a maximum height \( H \). The ratio of the times at which the body is at a height of \( \frac{H}{2} \) is:

7.
A cannon placed on a cliff at a height of 375 m fires a cannonball with a velocity of 100 m/s\(^{-1}\) at an angle of 30Â° above the horizontal. The horizontal distance between the cannon and the target is:
(Acceleration due to gravity \( g = 10 \) ms\(^{-2}\))

8.
y = a sin(βx + γt)wherex and t represent displacement and time, respectively. Then, the dimensional formula for β— γis:

9.
The theory currently accepted as proper framework for explaining microscopic phenomena is