JEE Main 2026 (Online) 23rd January Evening Shift | Laws of Motion Question 11 | Physics

A block is sliding down on an inclined plane of slope $\theta$ and at an instant $t=0$ this block is given an upward momentum so that it starts moving up on the inclined surface with velocity $u$. The distance $(S)$ travelled by the block before its velocity become zero, is $\_\_\_\_$ .

(g = gravitational acceleration)

$\frac{2 u^2}{\mathrm{~g} \cos \theta}$

$\frac{u^2}{\sqrt{2} g \cos \theta}$

$\frac{u^2}{2 g \sin \theta}$

$\frac{u^2}{2 g \cos \theta}$

JEE Main 2026 (Online) 21st January Morning Shift

MCQ (Single Correct Answer)

-1

A 4 kg mass moves under the influence of a force $\vec{F}=\left(4 t^3 \hat{i}-3 t \hat{j}\right) \mathrm{N}$ where $t$ is the time in second. If mass starts from origin at $t=0$, the velocity and position after $t=2 \mathrm{~s}$ will be:

$\vec{v}=3 \hat{i}+\frac{3}{2} \hat{j} \quad \vec{r}=\frac{6}{5} \hat{i}+\hat{j}$

$\vec{v}=4 \hat{i}-\frac{3}{2} \hat{j} \quad \vec{r}=\frac{6}{5} \hat{i}-\hat{j}$

$\vec{v}=4 \hat{i}-\frac{3}{2} \hat{j} \quad \vec{r}=\frac{8}{5} \hat{i}-\hat{j}$

$\vec{v}=4 \hat{i}+\frac{5}{2} \hat{j} \quad \vec{r}=\frac{8}{5} \hat{i}+2 \hat{j}$

JEE Main 2025 (Online) 8th April Evening Shift

MCQ (Single Correct Answer)

-1

A body of mass 2 kg moving with velocity of $ \vec{v}_{in} = 3 \hat{i} + 4 \hat{j} \text{ ms}^{-1} $ enters into a constant force field of 6N directed along positive z-axis. If the body remains in the field for a period of $ \frac{5}{3} $ seconds, then velocity of the body when it emerges from force field is.

$ 3\hat{i} + 4\hat{j} + \sqrt{5} \hat{k} $

$ 4\hat{i} + 3\hat{j} + 5\hat{k} $

$ 3\hat{i} + 4\hat{j} - 5\hat{k} $

$ 3\hat{i} + 4\hat{j} + 5\hat{k} $

JEE Main 2025 (Online) 7th April Evening Shift

MCQ (Single Correct Answer)

-1

An object with mass 500 g moves along x-axis with speed $v = 4\sqrt{x}$ m/s. The force acting on the object is :

8 N

4 N

5 N

6 N