1
JEE Main 2025 (Online) 2nd April Evening Shift
MCQ (Single Correct Answer)
+4
-1
A sportsman runs around a circular track of radius $r$ such that he traverses the path $A B A B$. The distance travelled and displacement, respectively, are JEE Main 2025 (Online) 2nd April Evening Shift Physics - Circular Motion Question 3 English
A
$\pi r, 3 r$
B
$2 \mathrm{r}, 3 \pi \mathrm{r}$
C
$3 \pi \mathrm{r}, 2 \mathrm{r}$
D
$3 \pi r, \pi r$
2
JEE Main 2025 (Online) 29th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A body of mass ‘m’ connected to a massless and unstretchable string goes in vertical circle of radius ‘R’ under gravity g. The other end of the string is fixed at the center of circle. If velocity at top of circular path is $n\sqrt{ g R}$ , where, n ≥ 1, then ratio of kinetic energy of the body at bottom to that at top of the circle is :

A

$\frac{n^2 + 4}{n^2}$

B

$\frac{n + 4}{n}$

C

$\frac{n^2}{n^2 + 4}$

D

$\frac{n}{n + 4}$

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

A car of mass ' $m$ ' moves on a banked road having radius ' $r$ ' and banking angle $\theta$. To avoid slipping from banked road, the maximum permissible speed of the car is $v_0$. The coefficient of friction $\mu$ between the wheels of the car and the banked road is

A
$\mu=\frac{v_0^2+r g \tan \theta}{r g+v_0^2 \tan \theta}$
B
$\mu=\frac{v_0^2-r g \tan \theta}{\mathrm{rg}-\mathrm{v}_{\mathrm{o}}^2 \tan \theta}$
C
$\mu=\frac{v_0^2-r g \tan \theta}{r g+v_0^2 \tan \theta}$
D
$\mu=\frac{v_o^2+r g \tan \theta}{r g-v_o^2 \tan \theta}$
4
JEE Main 2025 (Online) 22nd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A body of mass 100 g is moving in circular path of radius 2 m on vertical plane as shown in figure. The velocity of the body at point $A$ is $10 \mathrm{~m} / \mathrm{s}$. The ratio of its kinetic energies at point B and C is :

JEE Main 2025 (Online) 22nd January Evening Shift Physics - Circular Motion Question 6 English

(Take acceleration due to gravity as $10 \mathrm{~m} / \mathrm{s}^2$)

A
$\frac{3-\sqrt{2}}{2}$
B
$\frac{2+\sqrt{3}}{3}$
C
$\frac{2+\sqrt{2}}{3}$
D
$\frac{3+\sqrt{3}}{2}$
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