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

A wheel is rolling on a plane surface. The speed of a particle on the highest point of the rim is $8 \mathrm{~m} / \mathrm{s}$. The speed of the particle on the rim of the wheel at the same level as the centre of wheel, will be :

A
$4 \sqrt{2} \mathrm{~m} / \mathrm{s}$
B
$8 \mathrm{~m} / \mathrm{s}$
C
$4 \mathrm{~m} / \mathrm{s}$
D
$8 \sqrt{2} \mathrm{~m} / \mathrm{s}$
2
JEE Main 2025 (Online) 2nd April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
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 10 English
A
$\pi r, 3 r$
B
$2 \mathrm{r}, 3 \pi \mathrm{r}$
C
$3 \pi \mathrm{r}, 2 \mathrm{r}$
D
$3 \pi r, \pi r$
3
JEE Main 2025 (Online) 29th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
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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

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

4
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
$\mu=\frac{v_o^2+r g \tan \theta}{r g-v_o^2 \tan \theta}$

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