Laws of Motion — Previous Year Questions (JEE Main 2026)

Both planes cover the same slope length $L$ starting from rest, so $L = \tfrac{1}{2}a t^2$, which gives $t \propto \dfrac{1}{\sqrt{a}}$.

Accelerations ($\theta = 45^\circ$, so $\sin\theta = \cos\theta$):

$$a_{\text{smooth}} = g\sin\theta, \qquad a_{\text{rough}} = g\sin\theta - \mu g\cos\theta$$

Time ratio (rough time is 50 % more, i.e. $t_{\text{rough}} = 1.5\,t_{\text{smooth}}$):

$$\frac{t_{\text{rough}}}{t_{\text{smooth}}} = \sqrt{\frac{a_{\text{smooth}}}{a_{\text{rough}}}} = 1.5 \implies \frac{a_{\text{smooth}}}{a_{\text{rough}}} = \frac{9}{4}$$

Solve for $\mu$ (dividing by $g\cos\theta$ with $\tan 45^\circ = 1$):

$$\frac{a_{\text{smooth}}}{a_{\text{rough}}} = \frac{\sin\theta}{\sin\theta - \mu\cos\theta} = \frac{1}{1-\mu} = \frac{9}{4}$$

$$1 - \mu = \frac{4}{9} \implies \mu = \frac{5}{9}$$

Answer: C ($\mu = 5/9$)

Mass $m = 100\ \text{g} = 0.1\ \text{kg}$. The body starts from rest at the origin.

Acceleration from $\vec{a} = \vec{F}/m$:

$$\vec{a} = \frac{5\hat{i} + 10\hat{j}}{0.1} = \left(50\hat{i} + 100\hat{j}\right)\ \text{m/s}^2$$

Displacement after $t = 2$ s, $\vec{s} = \tfrac{1}{2}\vec{a}\,t^2 = 2\vec{a}$:

$$\vec{s} = 2\left(50\hat{i} + 100\hat{j}\right) = \left(100\hat{i} + 200\hat{j}\right)\ \text{m}$$

Match components with $\left(2x\hat{i} + 5y\hat{j}\right)$:

$$2x = 100 \implies x = 50, \qquad 5y = 200 \implies y = 40$$$$x : y = 50 : 40 = 5 : 4$$

Answer: D ($5 : 4$)

Same slope length $L$ from rest, so $t \propto \dfrac{1}{\sqrt{a}}$.

Accelerations ($\theta = 45^\circ$):

$$a_{\text{rough}} = g\sin\theta - \mu g\cos\theta, \qquad a_{\text{smooth}} = g\sin\theta$$

Time ratio ($t_{\text{smooth}} = t/2$, so rough time is twice the smooth time):

$$\frac{t_{\text{rough}}}{t_{\text{smooth}}} = \sqrt{\frac{a_{\text{smooth}}}{a_{\text{rough}}}} = 2 \implies \frac{a_{\text{smooth}}}{a_{\text{rough}}} = 4$$

Solve for $\mu$ (with $\tan 45^\circ = 1$):

$$\frac{1}{1-\mu} = 4 \implies 1 - \mu = \frac{1}{4} \implies \mu = \frac{3}{4} = 0.75 = \frac{75}{100}$$

Hence $\alpha = 75$.

Answer: 75

Convert the speed: $v = 54\ \text{km/h} = 54 \times \dfrac{5}{18} = 15\ \text{m/s}$.

In the car’s (non-inertial) frame, the bob experiences a pseudo (centrifugal) force $\dfrac{mv^2}{r}$ horizontally outward and gravity $mg$ downward. At equilibrium the string tilts by angle $\theta$ from the vertical:

$$\tan\theta = \frac{mv^2/r}{mg} = \frac{v^2}{rg}$$

Substitute values:

$$\tan\theta = \frac{(15)^2}{20 \times 10} = \frac{225}{200} = 1.125$$$$\theta = \tan^{-1}(1.125)$$

Answer: C ($\tan^{-1}(1.125)$)