Previous Year JEE Mains Questions on Laws of Motion with Solutions
Explore a collection of previous year JEE Mains questions from the Laws of Motion chapter. Each question comes with a detailed solution, interactive multiple-choice options, and a "Show Answer" button. Test your understanding by selecting an option and instantly see if you're correct!
Question 1
A force acts for 20 s on a body of mass 20 kg, starting from rest. After the force ceases, the body travels 50 m in the next 10 s. What is the value of the force? [29-Jan-2023 Shift 2]
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Detailed Solution:
Given:
- Mass, \( m = 20 \, \text{kg} \)
- Initial velocity, \( u = 0 \, \text{m/s} \) (starts from rest)
- Time of force application, \( t_1 = 20 \, \text{s} \)
- Distance traveled after force ceases, \( s = 50 \, \text{m} \) in \( t_2 = 10 \, \text{s} \)
After the force ceases, the body moves with constant velocity (no acceleration, assuming no other forces like friction). The velocity \( v \) at the end of 20 s becomes the constant velocity for the next 10 s.
Distance traveled at constant velocity:
\[ s = v \cdot t_2 \]
\[ 50 = v \cdot 10 \]
\[ v = \frac{50}{10} = 5 \, \text{m/s} \]
Now, this velocity \( v = 5 \, \text{m/s} \) is achieved after the force acts for 20 s. Using the equation of motion during the force application:
\[ v = u + a \cdot t_1 \]
\[ 5 = 0 + a \cdot 20 \]
\[ a = \frac{5}{20} = 0.25 \, \text{m/s}^2 \]
The force \( F \) is related to acceleration by Newton's second law:
\[ F = m \cdot a \]
\[ F = 20 \cdot 0.25 = 5 \, \text{N} \]
Final Answer: B. 5 N
Question 2
The time taken by an object to slide down a 45° rough inclined plane is \( n \) times that on a perfectly smooth 45° inclined plane. What is the coefficient of kinetic friction between the object and the inclined plane? [29-Jan-2023 Shift 2]
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Detailed Solution:
Given: Angle of inclination \( \theta = 45^\circ \), time on rough plane \( t_{\text{rough}} = n \cdot t_{\text{smooth}} \).
Smooth Plane:
Acceleration, \( a_{\text{smooth}} = g \sin \theta \), where \( \sin 45^\circ = \frac{\sqrt{2}}{2} \), so \( a_{\text{smooth}} = g \cdot \frac{\sqrt{2}}{2} \).
Starting from rest, distance \( s = \frac{1}{2} a t^2 \):
\[ s = \frac{1}{2} a_{\text{smooth}} t_{\text{smooth}}^2 \]
Rough Plane:
Net acceleration, \( a_{\text{rough}} = g \sin \theta - \mu_k g \cos \theta \).
For \( \theta = 45^\circ \), \( \sin \theta = \cos \theta = \frac{\sqrt{2}}{2} \), so:
\[ a_{\text{rough}} = g \cdot \frac{\sqrt{2}}{2} - \mu_k g \cdot \frac{\sqrt{2}}{2} = \frac{g \sqrt{2}}{2} (1 - \mu_k) \]
Distance: \( s = \frac{1}{2} a_{\text{rough}} t_{\text{rough}}^2 \), and \( t_{\text{rough}} = n t_{\text{smooth}} \).
Since \( s \) is the same:
\[ \frac{1}{2} a_{\text{smooth}} t_{\text{smooth}}^2 = \frac{1}{2} a_{\text{rough}} (n t_{\text{smooth}})^2 \]
\[ a_{\text{smooth}} = a_{\text{rough}} n^2 \]
\[ \frac{g \sqrt{2}}{2} = \frac{g \sqrt{2}}{2} (1 - \mu_k) n^2 \]
\[ 1 = (1 - \mu_k) n^2 \]
\[ 1 - \mu_k = \frac{1}{n^2} \]
\[ \mu_k = 1 - \frac{1}{n^2} \]
Final Answer: D. \({1 - \frac{1}{n^2}} \)
Question 3
A stone tied to a 180 cm long string makes 28 revolutions per minute in a horizontal circle. The magnitude of acceleration is \( \frac{1936}{x} \, \text{m/s}^2 \). Find \( x \). Take \( \pi = \frac{22}{7} \). [30-Jan-2023 Shift 2]
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Detailed Solution:
Given:
- Length of string, \( r = 180 \, \text{cm} = 1.8 \, \text{m} \)
- Revolutions per minute = 28
- \( \pi = \frac{22}{7} \)
Convert to revolutions per second: \( \frac{28}{60} = \frac{7}{15} \, \text{rev/s} \).
Angular velocity, \( \omega = 2\pi \cdot \frac{7}{15} = \frac{14 \cdot \frac{22}{7}}{15} = \frac{44}{15} \, \text{rad/s} \).
Centripetal acceleration, \( a_c = \omega^2 r \):
\[ \omega^2 = \left( \frac{44}{15} \right)^2 = \frac{1936}{225} \]
\[ a_c = \frac{1936}{225} \cdot 1.8 = \frac{1936}{225} \cdot \frac{9}{5} = \frac{1936 \cdot 9}{225 \cdot 5} = \frac{17424}{1125} \]
Simplify: \( \frac{17424}{1125} = \frac{1936 \cdot 9}{125 \cdot 9} = \frac{1936}{125} \).
Given \( a_c = \frac{1936}{x} \), so \( \frac{1936}{125} = \frac{1936}{x} \), thus \( x = 125 \).
Final Answer: 125
Question 4
A lift of mass 500 kg descends at 2 m/s. The cable slips, causing it to fall with a constant acceleration of 2 m/s² over 6 m. What is the kinetic energy at the end in kJ? [31-Jan-2023 Shift 1]
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Detailed Solution:
Given:
- Mass, \( M = 500 \, \text{kg} \)
- Initial velocity, \( u = 2 \, \text{m/s} \) (downward)
- Acceleration, \( a = 2 \, \text{m/s}^2 \) (downward)
- Distance, \( s = 6 \, \text{m} \)
Final velocity: \( v^2 = u^2 + 2 a s \)
\[ v^2 = 2^2 + 2 \cdot 2 \cdot 6 = 4 + 24 = 28 \]
\[ v = \sqrt{28} = 2\sqrt{7} \, \text{m/s} \]
Kinetic energy, \( KE = \frac{1}{2} M v^2 = \frac{1}{2} \cdot 500 \cdot 28 = 250 \cdot 28 = 7000 \, \text{J} = 7 \, \text{kJ} \).
Final Answer: 7 kJ
Question 5
A 1 kg stone is tied to a 1 m massless string with a breaking tension of 400 N. What is the maximum linear velocity without breaking the string in a horizontal plane? [31-Jan-2023 Shift 2]
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Detailed Solution:
Given:
- Mass, \( m = 1 \, \text{kg} \)
- Length (radius), \( r = 1 \, \text{m} \)
- Maximum tension, \( T_{\text{max}} = 400 \, \text{N} \)
Tension provides centripetal force: \( T = \frac{m v^2}{r} \).
\[ 400 = \frac{1 \cdot v^2}{1} \]
\[ v^2 = 400 \]
\[ v = 20 \, \text{m/s} \]
Final Answer: A. 20 m/s
Question 6
A 10 kg body moving at 20 m/s stops in 5 s due to friction. What is the coefficient of friction? (\( g = 10 \, \text{m/s}^2 \)) [31-Jan-2023 Shift 2]
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Detailed Solution:
Given:
- Mass, \( m = 10 \, \text{kg} \)
- Initial velocity, \( u = 20 \, \text{m/s} \)
- Final velocity, \( v = 0 \, \text{m/s} \)
- Time, \( t = 5 \, \text{s} \)
- \( g = 10 \, \text{m/s}^2 \)
Acceleration: \( v = u + a t \)
\[ 0 = 20 + a \cdot 5 \]
\[ a = -4 \, \text{m/s}^2 \]
Friction force, \( f = m |a| = 10 \cdot 4 = 40 \, \text{N} \).
Normal force, \( N = m g = 10 \cdot 10 = 100 \, \text{N} \).
\[ f = \mu N \]
\[ 40 = \mu \cdot 100 \]
\[ \mu = 0.4 \]
Final Answer: D. 0.4
Question 7
A 5 kg block at rest on a rough table is pushed with 30 N, sliding 50 m in 10 s. What is the coefficient of kinetic friction? (\( g = 10 \, \text{m/s}^2 \)) [1-Feb-2023 Shift 1]
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Detailed Solution:
Given:
- Mass, \( m = 5 \, \text{kg} \)
- Force, \( F = 30 \, \text{N} \)
- Distance, \( s = 50 \, \text{m} \)
- Time, \( t = 10 \, \text{s} \)
- Initial velocity, \( u = 0 \)
- \( g = 10 \, \text{m/s}^2 \)
Acceleration: \( s = u t + \frac{1}{2} a t^2 \)
\[ 50 = 0 + \frac{1}{2} a \cdot 10^2 \]
\[ a = 1 \, \text{m/s}^2 \]
Net force: \( F - f = m a \)
\[ 30 - \mu_k \cdot 5 \cdot 10 = 5 \cdot 1 \]
\[ 30 - 50 \mu_k = 5 \]
\[ 50 \mu_k = 25 \]
\[ \mu_k = 0.5 \]
Final Answer: C. 0.50
Question 8
A 100 g block is tied to a 7.5 N/m spring of length 20 cm, fixed at point A. It moves in a circle on a smooth horizontal surface with \( \omega = 5 \, \text{rad/s} \) about A. What is the tension? [6-Apr-2023 Shift 1]
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Detailed Solution:
Given:
- Mass, \( m = 0.1 \, \text{kg} \)
- Spring constant, \( k = 7.5 \, \text{N/m} \)
- Natural length, \( l = 0.2 \, \text{m} \)
- Angular velocity, \( \omega = 5 \, \text{rad/s} \)
Tension = centripetal force: \( T = m \omega^2 r \).
Tension also: \( T = k (r - l) \).
\[ m \omega^2 r = k (r - l) \]
\[ 0.1 \cdot 25 \cdot r = 7.5 (r - 0.2) \]
\[ 2.5 r = 7.5 r - 1.5 \]
\[ r = 0.3 \, \text{m} \]
\[ T = 7.5 (0.3 - 0.2) = 0.75 \, \text{N} \]
Check: \( T = 0.1 \cdot 25 \cdot 0.3 = 0.75 \, \text{N} \).
Final Answer: A. 0.75 N
Question 9
A 500 g particle has velocity \( (2t \hat{i} + 3t^2 \hat{j}) \, \text{m/s} \). Force at \( t = 1 \, \text{s} \) is \( (\hat{i} + x \hat{j}) \, \text{N} \). Find \( x \). [8-Apr-2023 Shift 1]
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Detailed Solution:
Given:
- Mass, \( m = 0.5 \, \text{kg} \)
- Velocity, \( \vec{v} = 2t \hat{i} + 3t^2 \hat{j} \)
Acceleration: \( \vec{a} = \frac{d\vec{v}}{dt} = 2 \hat{i} + 6t \hat{j} \).
At \( t = 1 \): \( \vec{a} = 2 \hat{i} + 6 \hat{j} \).
Force: \( \vec{F} = m \vec{a} = 0.5 (2 \hat{i} + 6 \hat{j}) = \hat{i} + 3 \hat{j} \).
Given \( \vec{F} = \hat{i} + x \hat{j} \), so \( x = 3 \).
Final Answer: C. 3
Question 10
Assertion A: An electric fan continues to rotate after the current is switched off.
Reason R: The fan continues due to inertia of motion. Choose the correct option. [10-Apr-2023 Shift 2]
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Detailed Solution:
Assertion A is true: fans continue rotating due to inertia.
Reason R is true: inertia of motion (Newton’s first law) explains this.
R correctly explains A.
Final Answer: B. Both A and R are correct and R is the correct explanation of A
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