Laws of Motion Previous Year Questions — JEE Main Physics
34 past-year Laws of Motion questions from JEE Main (Physics) — page 2.
Q14 (2025)
A block of mass 25 kg is pulled along a horizontal surface by a force at an angle $45^{circ}$ with the horizontal. The friction coefficient between the block and the surface is 0.25. The work done for a displacement of 5 m of the block with uniform velocity is:
- 970 J
- 735 J
- 245 J
- 490 J
### Related Formula
$$N + F\sin\theta = mg \implies N = mg - F\sin\theta$$
$$F\cos\theta = f_k = \mu_k N$$
$$W = F \cdot S \cdot \cos\theta$$
### Core Logic
Since the block moves with uniform velocity, horizontal acceleration is zero.
$$F\cos(45^\circ) = \mu [mg - F\sin(45^\circ)]$$
$$\frac{F}{\sqrt{2}} = 0.25 \left[25 \times 9.8 - \frac{F}{\sqrt{2}}\right]$$
$$\frac{F}{\sqrt{2}} = 61.25 - 0.25 \frac{F}{\sqrt{2}} \implies 1.25 \frac{F}{\sqrt{2}} = 61.25$$
### Step 1: Compute Force and Work Done
$$\frac{F}{\sqrt{2}} = \frac{61.25}{1.25} = 49 \implies F = 49\sqrt{2}\text{ N}$$
The work done by the external force over displacement $S=5\text{ m}$ is:
$$W = F S \cos(45^\circ) = (49\sqrt{2}) \times 5 \times \frac{1}{\sqrt{2}} = 245\text{ J}$$ {{SOL_IMG1}}
### Pattern Recognition
When uniform velocity is sustained, work done by the external pulling component perfectly matches the work consumed against internal friction dissipation ($W = f_k \cdot S$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q912 (2025)
A body of mass $m$ is suspended by two strings making angles $\theta{1}$ and $\theta_{2}$ with the horizontal ceiling with tensions $T_{1}$ and $T_{2}$ simultaneously. $T_{1}$ and $T_{2}$ are related by $T_{1}=\sqrt{3}T_{2}$ the angles \theta_{1} and \theta_{2} are
- $\theta_{1}=30^{\circ}, \theta_{2}=60^{\circ}\text{ with } T_{2}=\frac{3mg}{4}$
- $\theta_{1}=60^{\circ}, \theta_{2}=30^{\circ}\text{ with } T_{2}=\frac{mg}{2}$
- $\theta_{1}=45^{\circ}, \theta_{2}=45^{\circ}\text{ with } T_{2}=\frac{3mg}{4}$
- $\theta_{1}=30^{\circ}, \theta_{2}=60^{\circ}\text{ with } T_{2}=\frac{4mg}{5}$
### Related Formula
Horizontal equilibrium equation:
$$T_1 \cos\theta_1 = T_2 \cos\theta_2$$
Vertical balancing equation:
$$T_1 \sin\theta_1 + T_2 \sin\theta_2 = mg$$
### Core Logic
Substitute $T_1 = \sqrt{3}T_2$ into horizontal equilibrium:
$$\sqrt{3}T_2 \cos\theta_1 = T_2 \cos\theta_2 \implies \sqrt{3}\cos\theta_1 = \cos\theta_2$$
### Step 1: Audit Options Pattern Match
Check Option (B) where $\theta_1 = 60^{\circ}$ and $\theta_2 = 30^{\circ}$:
$$\sqrt{3}\cos(60^{\circ}) = \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}$$
$$\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$$
This matches the horizontal equilibrium condition $\sqrt{3}\cos\theta_1 = \cos\theta_2$.
### Step 2: Solve for T2
Substitute the angles into the vertical component balancing equation:
$$T_2 \left[\sqrt{3}\sin(60^{\circ}) + \sin(30^{\circ})\right] = mg$$
$$T_2 \left[\sqrt{3}\left(\frac{\sqrt{3}}{2}\right) + \frac{1}{2}\right] = mg$$
$$T_2 \left[\frac{3}{2} + \frac{1}{2}\right] = mg \implies T_2 (2) = mg \implies T_2 = \frac{mg}{2}$$
### Pattern Recognition
Lami's Theorem or standard rectangular resolution of forces. When tension is scaled by $\sqrt{3}$, it directly hints at complementary $30^{\circ}\text{--}60^{\circ}$ geometric alignments.
### Evaluation Rubric / Model Answer
Option B: $\theta_{1}=60^{\circ}, \theta_{2}=30^{\circ}\text{ with } T_{2}=\frac{mg}{2}$
### Chapter Mix
Class 11 Physics: Laws of Motion
Q22 (2025)
A string of length L is fixed at one end and carries a mass of M at the other end. The mass makes $\left(\frac{3}{\pi}\right)$ rotations per second about the vertical axis passing through end of the string as shown. The tension in the string is ____ ML.
### Related Formula
For a conical pendulum:
$$T \cos \theta = Mg$$
$$T \sin \theta = M\omega^2 R$$
where $R = L \sin \theta$.
### Core Logic
Substituting $R = L \sin \theta$ into the centripetal equation: {{SOL_IMG1}}
$$T \sin \theta = M \omega^2 (L \sin \theta)$$
$$T = M \omega^2 L$$
Given rotational frequency: $f = \frac{3}{\pi}\ \mathrm{rev/s}$
Angular velocity:
$$\omega = 2\pi f = 2\pi \left(\frac{3}{\pi}\right) = 6\ \mathrm{rad/s}$$
Substituting $\omega$ back into the simplified tension equation:
$$T = M (6)^2 L = 36\ ML$$
### Pattern Recognition
In a conical pendulum, the horizontal projection of tension provides the exact centripetal force. The $\sin\theta$ terms cancel out, making string tension independent of the semi-vertical angle.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q1135 (2025)
A car of mass 'm' moves on a banked road having radius 'r' and banking angle $\theta$ To avoid slipping from banked road, the maximum permissible speed of the car is $v_{0}.$ The coefficient of friction $\mu$ between the wheels of the car and the banked road is :-
- $\mu=\frac{v_{0}^{2}+rg~\tan~\theta}{rg-v_{0}^{2}\tan~\theta}$
- $\mu=\frac{v_{0}^{2}+rg~\tan~\theta}{rg+v_{0}^{2}\tan~\theta}$
- $\mu=\frac{v_{0}^{2}-rg~\tan~\theta}{rg+v_{0}^{2}\tan~\theta}$
- $\mu=\frac{v_{0}^{2}-rg~\tan~\theta}{rg-v_{0}^{2}\tan~\theta}$
### Related Formula
The maximum velocity limit preventing outer slide breakout on a rough banked plane profile is given by standard centrifugal force equations:
$$v_{0} = \sqrt{rg \left(\frac{\tan\theta + \mu}{1 - \mu\tan\theta}\right)}$$
### Core Logic
By writing out the components shown in the free body layout {{SOL_IMG1}} :
$$N\sin\theta + f\cos\theta = \frac{mv_{0}^{2}}{r}$$
$$N\cos\theta - f\sin\theta = mg$$
### Step 1: Isolate Coefficient of Friction
Squaring the velocity boundary relation gives :
$$\frac{v_{0}^{2}}{rg} = \frac{\tan\theta + \mu}{1 - \mu\tan\theta}$$
Cross multiply to isolate the variable terms :
$$v_{0}^{2} - \mu v_{0}^{2}\tan\theta = rg\tan\theta + \mu rg$$
$$v_{0}^{2} - rg\tan\theta = \mu(rg + v_{0}^{2}\tan\theta)$$
$$\mu = \frac{v_{0}^{2} - rg\tan\theta}{rg + v_{0}^{2}\tan\theta}$$
### Pattern Recognition
Maximum limits always balance with plus signs in the numerator fraction. Simply rearrange that template expression directly to map the targeted parameter.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q20 (2025)
A balloon and its content having mass M is moving up with an acceleration 'a'. The mass that must be released from the content so that the balloon starts moving up with an acceleration '3a' will be : (Take 'g' as acceleration due to gravity) [cite: 174, 175]
- $\frac{3\mathrm{Ma}}{2\mathrm{a} - \mathrm{g}}$
- $\frac{3 \mathrm{Ma}}{2 \mathrm{a} + \mathrm{g}}$
- $\frac{2 \mathrm{Ma}}{3 \mathrm{a} + \mathrm{g}}$
- $\frac{2 \mathrm{Ma}}{3 \mathrm{a} - \mathrm{g}}$
### Related Formula
By Newton's second law of motion, the net upward force acting on an accelerating balloon system is given by:
$$F_{\text{buoyant}} - m_{\text{total}} g = m_{\text{total}} a$$
### Core Logic
Let $F$ be the constant buoyant force acting upward on the balloon.
**Case 1 (Initial upward acceleration)** :
$$F - M g = M a \implies F = M(g + a) \quad \text{}$$
**Case 2 (After releasing mass $x$)** :
The new total mass becomes $(M - x)$, and its acceleration increases to $3a$:
$$F - (M - x)g = (M - x)3a \quad \text{}$$
Substitute the value of $F$ from Case 1 into Case 2 [cite: 836, 839]:
$$M(g + a) - (M - x)g = (M - x)3a$$
$$M g + M a - M g + x g = 3 M a - 3 x a \quad \text{}$$
$$M a + x g = 3 M a - 3 x a$$
$$x(g + 3a) = 2 M a$$
$$x = \frac{2 M a}{3a + g} \quad \text{}$$
### Step 1: Visual Context
The free-body force layout for both accelerating phases is shown below:
{{SOL_IMG1}}
{{SOL_IMG2}}
### Pattern Recognition
Since the upward buoyant force is completely determined by the balloon's volume, it remains constant. Expressing this constant force in terms of the initial conditions allows you to quickly solve for mass changes when acceleration states vary.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
