1
JEE Main 2025 (Online) 2nd April Morning Shift
MCQ (Single Correct Answer)
+4
-1

A river is flowing from west to east direction with speed of $9 \mathrm{~km} \mathrm{~h}^{-1}$. If a boat capable of moving at a maximum speed of $27 \mathrm{~km} \mathrm{~h}^{-1}$ in still water, crosses the river in half a minute, while moving with maximum speed at an angle of $150^{\circ}$ to direction of river flow, then the width of the river is :

A
112.5 m
B
75 m
C
300 m
D
$112.5 \times \sqrt{3} \mathrm{~m}$
2
JEE Main 2025 (Online) 29th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
Two projectiles are fired with same initial speed from same point on ground at angles of $(45^\circ - \alpha)$ and $(45^\circ + \alpha)$, respectively, with the horizontal direction. The ratio of their maximum heights attained is :
A

$ \frac{1+\sin\alpha}{1-\sin\alpha} $

B

$ \frac{1+\sin2\alpha}{1-\sin2\alpha} $

C

$ \frac{1-\tan\alpha}{1+\tan\alpha} $

D

$ \frac{1-\sin2\alpha}{1+\sin2\alpha} $

3
JEE Main 2025 (Online) 24th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The position vector of a moving body at any instant of time is given as $\overrightarrow{\mathrm{r}}=\left(5 \mathrm{t}^2 \hat{i}-5 \mathrm{t} \hat{j}\right) \mathrm{m}$. The magnitude and direction of velocity at $t=2 s$ is,

A
$5 \sqrt{17} \mathrm{~m} / \mathrm{s}$, making an angle of $\tan ^{-1} 4$ with - ve Y axis
B
$5 \sqrt{15} \mathrm{~m} / \mathrm{s}$, making an angle of $\tan ^{-1} 4$ with + ve $X$ axis
C
$5 \sqrt{17} \mathrm{~m} / \mathrm{s}$, making an angle of $\tan ^{-1} 4$ with + ve $X$ axis
D
$5 \sqrt{15} \mathrm{~m} / \mathrm{s}$, making an angle of $\tan ^{-1} 4$ with $-$ ve $Y$ axis
4
JEE Main 2025 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

An object of mass ' m ' is projected from origin in a vertical xy plane at an angle $45^{\circ}$ with the $\mathrm{x}-$ axis with an initial velocity $\mathrm{v}_0$. The magnitude and direction of the angular momentum of the object with respect to origin, when it reaches at the maximum height, will be [ g is acceleration due to gravity]

A
$\frac{m v_o{ }^3}{2 \sqrt{2} g}$ along negative $z$-axis
B
$\frac{m v_o^3}{2 \sqrt{2} g}$ along positive $z$-axis
C
$\frac{m v_o^3}{4 \sqrt{2} g}$ along positive $z$-axis
D
$\frac{m v_o^3}{4 \sqrt{2} g}$ along negative z-axis
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