1
JEE Main 2025 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A force $\mathrm{F}=\alpha+\beta \mathrm{x}^2$ acts on an object in the x -direction. The work done by the force is 5 J when the object is displaced by 1 m . If the constant $\alpha=1 \mathrm{~N}$ then $\beta$ will be

A
$15 \mathrm{~N} / \mathrm{m}^2$
B
$10 \mathrm{~N} / \mathrm{m}^2$
C
$12 \mathrm{~N} / \mathrm{m}^2$
D
$8 \mathrm{~N} / \mathrm{m}^2$
2
JEE Main 2025 (Online) 23rd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A ball having kinetic energy KE, is projected at an angle of $60^{\circ}$ from the horizontal. What will be the kinetic energy of ball at the highest point of its flight?

A
$\frac{(\mathrm{KE})}{2}$
B
$\frac{(\mathrm{KE})}{8}$
C
$\frac{(\mathrm{KE})}{4}$
D
$\frac{(\mathrm{KE})}{16}$
3
JEE Main 2025 (Online) 22nd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A force $\overrightarrow{\mathrm{F}}=2 \hat{i}+\mathrm{b} \hat{j}+\hat{k}$ is applied on a particle and it undergoes a displacement $\hat{i}-2 \hat{j}-\hat{k}$ What will be the value of $b$, if work done on the particle is zero.

A
$\frac{1}{3}$
B
$\frac{1}{2}$
C
0
D
2
4
JEE Main 2024 (Online) 9th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A particle of mass $$m$$ moves on a straight line with its velocity increasing with distance according to the equation $$v=\alpha \sqrt{x}$$, where $$\alpha$$ is a constant. The total work done by all the forces applied on the particle during its displacement from $$x=0$$ to $$x=\mathrm{d}$$, will be :

A
$$\frac{\mathrm{m}}{2 \alpha^2 \mathrm{~d}}$$
B
$$\frac{\mathrm{md}}{2 \alpha^2}$$
C
$$\frac{\mathrm{m} \alpha^2 \mathrm{~d}}{2}$$
D
$$2 \mathrm{~m} \alpha^2 \mathrm{~d}$$
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