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

A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is :

A
$\frac{3}{4}$
B
$\frac{4}{3}$
C
$\frac{5}{2}$
D
$\frac{2}{5}$
2
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
3
JEE Main 2025 (Online) 24th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A photograph of a landscape is captured by a drone camera at a height of 18 km . The size of the camera film is $2 \mathrm{~cm} \times 2 \mathrm{~cm}$ and the area of the landscape photographed is $400 \mathrm{~km}^2$. The focal length of the lens in the drone camera is :

A
2.8 cm
B
0.9 cm
C
2.5 cm
D
1.8 cm
4
JEE Main 2025 (Online) 24th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

JEE Main 2025 (Online) 24th January Evening Shift Physics - Electrostatics Question 22 English

In the first configuration (1) as shown in the figure, four identical charges $\left(q_0\right)$ are kept at the corners A, B, C and D of square of side length ' $a$ '. In the second configuration (2), the same charges are shifted to mid points $G, E, H$ and $F$, of the square. If $K=\frac{1}{4 \pi \epsilon_0}$, the difference between the potential energies of configuration (2) and (1) is given by :

A
$\frac{K q_0^2}{a}(4-2 \sqrt{2})$
B
$\frac{K q_0^2}{a}(4 \sqrt{2}-2)$
C
$\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(3-\sqrt{2})$
D
$\frac{\mathrm{Kq}_0^2}{\mathrm{a}}(3 \sqrt{2}-2)$
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