JEE Mains Previous Year Questions on Laws of Motion with Solutions
Welcome to this collection of previous year JEE Mains questions from the Laws of Motion chapter. Each question is accompanied by a detailed solution to help you understand the concepts and prepare effectively for your exams. Click the "Show Answer" button to reveal the solution for each question.
Question 1
A coin placed on a rotating table just slips when it is placed at a distance of 1 cm from the center. If the angular velocity of the table is halved, it will just slip when placed at a distance of _______ from the center: [11-Apr-2023 Shift 1]
Solution:
The coin slips when the centrifugal force equals the maximum static friction force.
Centrifugal force: \( F_c = m r \omega^2 \)
Static friction force: \( F_s = \mu m g \)
At the slipping point: \( m r \omega^2 = \mu m g \), so \( r \omega^2 = \mu g \).
First case: \( r_1 = 1 \) cm, \( \omega_1 = \omega \), thus \( 1 \cdot \omega^2 = \mu g \).
Second case: \( \omega_2 = \frac{\omega}{2} \), find \( r_2 \) such that \( r_2 \left( \frac{\omega}{2} \right)^2 = \mu g \).
Substitute \( \mu g = \omega^2 \): \( r_2 \cdot \frac{\omega^2}{4} = \omega^2 \).
Solve: \( \frac{r_2}{4} = 1 \Rightarrow r_2 = 4 \) cm.
Answer: 4 cm
Question 2
A body of mass 500 g moves along the x-axis such that its velocity varies with displacement \( x \) according to the relation \( v = 10 \sqrt{x} \) m/s. The force acting on the body is: [11-Apr-2023 Shift 2]
Solution:
Given: mass \( m = 500 \) g = \( 0.5 \) kg, velocity \( v = 10 \sqrt{x} \) m/s.
Force \( F = m a \), where \( a \) is acceleration.
Since \( v = 10 x^{1/2} \), compute acceleration using \( a = v \frac{dv}{dx} \).
\( \frac{dv}{dx} = \frac{d}{dx} (10 x^{1/2}) = 10 \cdot \frac{1}{2} x^{-1/2} = \frac{5}{\sqrt{x}} \).
Thus, \( a = v \cdot \frac{dv}{dx} = (10 \sqrt{x}) \cdot \frac{5}{\sqrt{x}} = 10 \cdot 5 = 50 \) m/s\(^2\).
Force: \( F = m a = 0.5 \cdot 50 = 25 \) N.
Answer: 25 N
Question 3
A uniform chain of 6 m length is placed on a table such that a part of its length is hanging over the edge of the table. The system is at rest. The coefficient of static friction between the chain and the surface of the table is 0.5. The maximum length of the chain hanging from the table is _____ m. [25-Jun-2022 Shift 1]
Solution:
Total length \( L = 6 \) m. Let the hanging length be \( x \), so length on table is \( 6 - x \).
Mass per unit length \( \mu \). Weight of hanging part: \( \mu x g \).
Normal force on table: \( N = \mu (6 - x) g \).
Maximum friction: \( f_{\text{max}} = \mu_s N = 0.5 \cdot \mu (6 - x) g \).
At equilibrium: \( f_{\text{max}} = \mu x g \).
\( 0.5 \cdot \mu (6 - x) g = \mu x g \). Cancel \( \mu g \): \( 0.5 (6 - x) = x \).
Solve: \( 3 - 0.5 x = x \Rightarrow 3 = 1.5 x \Rightarrow x = 2 \) m.
Answer: 2 m
Question 4
A disc with a flat small bottom beaker placed on it at a distance \( R \) from its center is revolving about an axis passing through the center and perpendicular to its plane with an angular velocity \( \omega \). The coefficient of static friction between the bottom of the beaker and the surface of the disc is \( \mu \). The beaker will revolve with the disc if: [25-Jun-2022 Shift 2]
Solution:
The beaker revolves if static friction provides the centripetal force.
Centripetal force required: \( m R \omega^2 \).
Maximum friction: \( \mu m g \).
Condition: \( m R \omega^2 \leq \mu m g \). Cancel \( m \): \( R \omega^2 \leq \mu g \).
Answer: \( R \omega^2 \leq \mu g \)
Question 5
A curve in a level road has a radius of 75 m. The maximum speed of a car turning this curved road can be 30 m/s without skidding. If the radius of the curved road is changed to 48 m and the coefficient of friction between the tyres and the road remains the same, then the maximum allowed speed would be _____ m/s. [25-Jun-2022 Shift 2]
Solution:
For a level road, \( v_{\text{max}} = \sqrt{\mu g r} \).
First case: \( r = 75 \) m, \( v_{\text{max}} = 30 \) m/s. So, \( 30^2 = \mu g \cdot 75 \Rightarrow 900 = \mu g \cdot 75 \Rightarrow \mu g = 12 \).
Second case: \( r = 48 \) m, \( v_{\text{max}}' = \sqrt{\mu g \cdot 48} = \sqrt{12 \cdot 48} = \sqrt{576} = 24 \) m/s.
Answer: 24 m/s
Question 6
One end of a massless spring of spring constant \( k \) and natural length \( l_0 \) is fixed while the other end is connected to a small object of mass \( m \) lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity \( \omega \) about an axis passing through the fixed end, then the elongation of the spring will be: [27-Jun-2022 Shift 2]
Solution:
Spring force provides centripetal force. Let elongation be \( x \), so total length is \( l_0 + x \).
Spring force: \( k x \). Centripetal force: \( m \omega^2 (l_0 + x) \).
Equate: \( k x = m \omega^2 (l_0 + x) \).
Solve: \( k x - m \omega^2 x = m \omega^2 l_0 \Rightarrow x (k - m \omega^2) = m \omega^2 l_0 \Rightarrow x = \frac{m \omega^2 l_0}{k - m \omega^2} \).
Answer: \( \frac{m \omega^2 l_0}{k - m \omega^2} \)
Question 7
A mass of 10 kg is suspended vertically by a rope of length 5 m from the roof. A force of 30 N is applied at the middle point of the rope in the horizontal direction. The angle made by the upper half of the rope with the vertical is \( \theta = \tan^{-1}(x \times 10^{-1}) \). The value of \( x \) is ____ (Given, \( g = 10 \) m/s\(^2\)). [27-Jun-2022 Shift 2]
Solution:
Weight of the mass: \( mg = 10 \times 10 = 100 \) N, acting downward.
The rope is divided into two segments by the horizontal force applied at the midpoint. Let:
- \( T_1 \): Tension in the upper half of the rope, making an angle \( \theta \) with the vertical.
- \( T_2 \): Tension in the lower half of the rope, making an angle \( \phi \) with the vertical.
For the mass (in equilibrium):
- Vertical force balance: \( T_2 \cos \phi = 100 \) N.
- Horizontal force balance: \( T_2 \sin \phi = 0 \), since no horizontal forces act on the mass.
Thus, \( \sin \phi = 0 \Rightarrow \phi = 0^\circ \), so the lower half is vertical, and \( T_2 = 100 \) N.
At the midpoint, where the 30 N horizontal force is applied, the forces are:
- \( T_1 \cos \theta \) upward (vertical component of upper tension).
- \( T_1 \sin \theta \) to the left (horizontal component, assuming the force is to the right).
- \( T_2 = 100 \) N downward (tension from the lower half).
- 30 N to the right (external force).
Vertical equilibrium at the midpoint:
\( T_1 \cos \theta - 100 = 0 \Rightarrow T_1 \cos \theta = 100 \) N.
Horizontal equilibrium at the midpoint:
\( - T_1 \sin \theta + 30 = 0 \Rightarrow T_1 \sin \theta = 30 \) N.
Divide the horizontal and vertical equations:
\( \frac{T_1 \sin \theta}{T_1 \cos \theta} = \frac{30}{100} \Rightarrow \tan \theta = 0.3 \).
Given \( \theta = \tan^{-1}(x \times 10^{-1}) \), so:
\( \tan \theta = x \times 10^{-1} = 0.1x \).
Thus, \( 0.1x = 0.3 \Rightarrow x = 3 \).
Answer: 3
Question 8
A block of mass 2 kg moving on a horizontal surface with speed of 4 m/s enters a rough surface ranging from \( x = 0.5 \) m to \( x = 1.5 \) m. The retarding force in this range is \( F = -k x \) where \( k = 12 \) N/m. The speed of the block as it just crosses the rough surface will be: [28-Jun-2022 Shift 2]
Solution:
Initial KE: \( \frac{1}{2} \cdot 2 \cdot 4^2 = 16 \) J.
Work done by force: \( W = \int_{0.5}^{1.5} (-12 x) \, dx = -12 \left[ \frac{x^2}{2} \right]_{0.5}^{1.5} = -12 \cdot \frac{1.5^2 - 0.5^2}{2} = -12 \cdot (1.125 - 0.125) = -12 \) J.
Final KE: \( 16 - 12 = 4 \) J.
\( \frac{1}{2} \cdot 2 \cdot v^2 = 4 \Rightarrow v^2 = 4 \Rightarrow v = 2 \) m/s.
Answer: 2.0 m/s
Question 9
A wire of length \( L \) is hanging from a fixed support. The length changes to \( L_1 \) and \( L_2 \) when masses 1 kg and 2 kg are suspended respectively from its free end. Then the value of \( L \) is equal to: [29-Jun-2022 Shift 1]
Solution:
Extension \( \delta = \frac{m g L}{A Y} \). Let \( c = \frac{g}{A Y} \).
\( L_1 = L + \frac{1 \cdot g L}{A Y} = L + c L \), \( L_2 = L + \frac{2 g L}{A Y} = L + 2 c L \).
So, \( L_2 - L = 2 (L_1 - L) \).
Solve: \( L_2 - L = 2 L_1 - 2 L \Rightarrow L_2 - 2 L_1 = -L \Rightarrow L = 2 L_1 - L_2 \).
Answer: \( 2 L_1 - L_2 \)
Question 10
A block of mass \( M \) placed inside a box descends vertically with acceleration \( a \). The block exerts a force equal to one-fourth of its weight on the floor of the box. The value of \( a \) will be: [29-Jun-2022 Shift 2]
Solution:
Normal force \( N = \frac{1}{4} M g \). Box accelerates downward with \( a \).
Inertial frame: \( M g - N = M a \).
Substitute: \( M g - \frac{1}{4} M g = M a \Rightarrow \frac{3}{4} M g = M a \Rightarrow a = \frac{3}{4} g \).
Answer: \( \frac{3}{4} g \)
.png)
0 Comments