Laws of Motion Previous Year Questions — JEE Main Physics
34 past-year Laws of Motion questions from JEE Main (Physics) — page 7.
Q40 (2026)
A flexible chain of mass m hangs between two fixed points at the same level. The inclination of the chain with the horizontal at the two points of support is $30^{\circ}$ . Considering the equilibrium of each half of the chain, the tension of the chain at the lowest point is ____.
- $\frac{\sqrt{3}}{2} m g$
- $\frac{1}{2} m g$
- $m g$
- $\sqrt{3} m g$
### Related Formula
$$\sum F_y = 0 \implies T \sin\theta = \frac{mg}{2}$$
$$\sum F_x = 0 \implies T \cos\theta = T_0$$
### Core Logic
{{SOL_IMG1}}
Draw the Free Body Diagram (F.B.D) of half of the rope.
The forces acting on half the rope (mass $m/2$) are:
1. Weight $\frac{mg}{2}$ acting downwards.
2. Tension $T$ at the support point acting at $30^{\circ}$ to the horizontal.
3. Horizontal tension $T_0$ at the lowest point.
### Step 1: Equilibrium Equations
For vertical equilibrium:
$$T \sin 30^{\circ} = \frac{m}{2} g$$
For horizontal equilibrium:
$$T \cos 30^{\circ} = T_0$$
### Step 2: Solve for T_0
Dividing the vertical equation by the horizontal equation:
$$\tan 30^{\circ} = \frac{mg / 2}{T_0}$$
$$T_0 = \frac{mg}{2 \tan 30^{\circ}}$$
$$T_0 = \frac{mg}{2 (1/\sqrt{3})} = \frac{\sqrt{3}}{2} mg$$
### Pattern Recognition
When dealing with symmetrical hanging chains, always cut the chain at the lowest point. The tension at the lowest point is purely horizontal and is given by $T_0 = (W/2) / \tan\theta$, where $\theta$ is the angle at the supports.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q40 (2026)
A block of mass 5 kg is moving on an inclined plane which makes an angle of $30^{\circ}$ with the horizontal. Friction coefficient between the block and inclined plane surface is $\frac{\sqrt{3}}{2}$ . The force to be applied on the block so that the block will move down without acceleration is ____ N.
- $25$
- $12.5$
- $7.5$
- $15$
### Related Formula
$$f_k = \mu_k N = \mu_k mg \cos\theta$$
$$F_{\text{down}} = mg \sin\theta$$
$\text{Net force } = 0 \text{ for constant velocity}$
### Core Logic
For the block to move down at a constant velocity (zero acceleration), the sum of forces parallel to the incline must be zero.
{{SOL_IMG1}}
### Step 1: Force Balance Equation
Force pulling it down the incline = $mg \sin(30^{\circ})$
Friction acting up the incline (since block moves down) = $\mu mg \cos(30^{\circ})$
Let external force $F$ act down the incline.
$$mg \sin(30^{\circ}) + F = \mu mg \cos(30^{\circ})$$
### Step 2: Value Substitution
$$F = \mu mg \cos(30^{\circ}) - mg \sin(30^{\circ})$$
$$F = \left( \frac{\sqrt{3}}{2} \right) (5 \times 10) \left( \frac{\sqrt{3}}{2} \right) - (5 \times 10) \left( \frac{1}{2} \right)$$
### Step 3: Final Calculation
$$F = 50 \times \frac{3}{4} - 50 \times \frac{1}{2}$$
$$F = 37.5 - 25 = 12.5 \mathrm{~N}$$
Note: The PDF solution calculates this with F assigned up the incline and gets $F = -12.5\mathrm{N}$. The magnitude is $12.5\mathrm{N}$ downward on the incline.
### Pattern Recognition
If $\mu > \tan\theta$ (here $\frac{\sqrt{3}}{2} = 0.866$ and $\tan 30^\circ = \frac{1}{\sqrt{3}} \approx 0.577$), friction is stronger than gravity's pull. You must actively pull it down to keep it moving.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q45 (2026)
A particle of mass m falls from rest through a resistive medium having resistive force, $F = -kv$, where v is the velocity of the particle and k is a constant. Which of the following graphs represents velocity (v) versus time(t)?
- Graph 1
- Graph 2
- Graph 3
- Graph 4
### Related Formula
$$F_{\text{net}} = ma$$
$$mg - kv = m\frac{dv}{dt}$$
### Core Logic
Integrate the differential equation of motion to find how velocity depends on time. Initially, acceleration is $g$. As velocity increases, the resistive force $kv$ increases, reducing acceleration until it reaches zero (terminal velocity).
### Step 1: Setting up Integration
$$m \cdot \frac{dv}{dt} = mg - kv$$
$$\int_{0}^{v} \frac{dv}{mg - kv} = \int_{0}^{t} \frac{dt}{m}$$
### Step 2: Performing Integration
$$\left[ -\frac{1}{k} \ln(mg - kv) \right]_0^v = \frac{t}{m}$$
$$-\frac{1}{k} \ln\left(\frac{mg - kv}{mg}\right) = \frac{t}{m}$$
$$\ln\left(1 - \frac{kv}{mg}\right) = -\frac{kt}{m}$$
### Step 3: Final Velocity Equation
$$1 - \frac{kv}{mg} = e^{-kt/m}$$
$$v = \frac{mg}{k} \left(1 - e^{-kt/m}\right)$$
This shows an exponential increase that asymptotically approaches a terminal velocity $v_{\text{terminal}} = mg/k$. Option (2) represents this curve.
### Pattern Recognition
The equation $1 - e^{-t}$ always produces a curve starting at origin and flattening out horizontally (asymptotic approach to a limit).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Q40 (2026)
A small block of mass m slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration $a_{0}$ . The angle between the inclined plane and ground is $\theta$ and its base length is L. Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is ____.
{{IMG1}}
- $\sqrt{\frac{2L}{g\sin 2\theta - a_0(1 + \cos 2\theta)}}$
- $\sqrt{\frac{4L}{g\sin 2\theta - a_0(1 + \cos 2\theta)}}$
- $\sqrt{\frac{4L}{g\cos^2\theta - a_0\sin\theta\cos\theta}}$
- $\sqrt{\frac{2L}{g\sin\theta - a_0\cos\theta}}$
### Related Formula
$$s = ut + \frac{1}{2}at^2$$
$$F_{\text{pseudo}} = m a_0$$
### Core Logic
{{SOL_IMG1}}
We solve the problem from the non-inertial frame of reference of the inclined plane. A pseudo force $m a_0$ acts on the block towards the right.
Forces acting parallel to the incline (downward positive):
1. Component of gravity: $mg \sin \theta$
2. Component of pseudo force (pointing up the incline): $-ma_0 \cos \theta$
### Step 1: Calculate Effective Acceleration
Net force down the incline:
$$F_{\text{net}} = mg \sin \theta - m a_0 \cos \theta$$
Thus, the acceleration relative to the incline is:
$$a = g \sin \theta - a_0 \cos \theta$$
### Step 2: Distance Travelled
The base length of the incline is $L$. Thus, the total length of the inclined surface is:
$$S = \frac{L}{\cos \theta}$$
### Step 3: Calculate Time
Using the second equation of motion ($u=0$):
$$S = \frac{1}{2} a t^2$$
$$\frac{L}{\cos \theta} = \frac{1}{2} (g \sin \theta - a_0 \cos \theta) t^2$$
$$t^2 = \frac{2L}{\cos \theta (g \sin \theta - a_0 \cos \theta)}$$
$$t = \sqrt{ \frac{2L}{g \sin \theta \cos \theta - a_0 \cos^2 \theta} }$$
### Step 4: Trigonometric Identity Substitution
Multiply numerator and denominator by 2 inside the square root:
$$t = \sqrt{ \frac{4L}{2g \sin \theta \cos \theta - 2a_0 \cos^2 \theta} }$$
Using $\sin 2\theta = 2\sin\theta\cos\theta$ and $2\cos^2\theta = 1 + \cos 2\theta$:
$$t = \sqrt{ \frac{4L}{g \sin 2\theta - a_0(1 + \cos 2\theta)} }$$
### Pattern Recognition
Whenever an inclined plane accelerates horizontally, resolve the pseudo force $ma_0$ parallel and perpendicular to the incline. Multiply by 2 inside the root to simplify into double-angle formats if options demand it.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Laws of Motion
Class 11 Physics: Kinematics
