1
COMEDK 2025 Evening Shift
MCQ (Single Correct Answer)
+1
-0
A point marked on a ring of radius 2 cm is in contact with a horizontal plane. Now the ring is rolled forward half a revolution along the positive X - direction. Then the angle made by the displacement vector of the point with the X - axis is:
A
$\theta=\boldsymbol{\operatorname { t a n }}^{-\mathbf{1}}\left(\frac{\mathbf{2}}{\mathbf{3} \boldsymbol{\pi}}\right)$
B
$\boldsymbol{\theta}=\boldsymbol{\operatorname { t a n }}^{-\mathbf{1}}\left(\frac{\mathbf{2}}{\boldsymbol{\pi}}\right)$
C
$\theta=\tan ^{-1}\left(\frac{2 \pi}{3}\right)$
D
$\theta=\cot ^{-1}\left(\frac{2}{\pi}\right)$
2
COMEDK 2025 Afternoon Shift
MCQ (Single Correct Answer)
+1
-0
A coin is placed on a disc rotating with an angular velocity $\omega$. The co-efficient of friction between the disc and the coin is $\mu$. The maximum distance of the coin from the centre of the disc up to which it will rotate with the disc is
A
$\sqrt{\frac{\mu}{\omega^2}}$
B
$\frac{\mu g}{\omega^2}$
C
$\sqrt{\frac{\mu g}{\omega^2}}$
D
$\frac{\mu g}{\omega}$
3
COMEDK 2024 Evening Shift
MCQ (Single Correct Answer)
+1
-0

A ball is moving in a circular path of radius $$5 \mathrm{~m}$$. If tangential acceleration at any instant is $$10 \mathrm{~ms}^{-2}$$ and the net acceleration makes an angle of $$30^{\circ}$$ with the centripetal acceleration, then, the instantaneous speed is

A
$$5.4 \mathrm{~ms}^{-1}$$
B
$$50 \sqrt{3} \mathrm{~ms}^{-1}$$
C
$$6.6 \mathrm{~ms}^{-1}$$
D
$$9.3 \mathrm{~ms}^{-1}$$
4
COMEDK 2024 Morning Shift
MCQ (Single Correct Answer)
+1
-0

A metal ball of $$20 \mathrm{~g}$$ is projected at an angle $$30^{\circ}$$ with the horizontal with an initial velocity $$10 \mathrm{~ms}^{-1}$$. If the mass and angle of projection are doubled keeping the initial velocity the same, the ratio of the maximum height attained in the former to the latter case is :

A
1 : 2
B
2 : 1
C
1 : 3
D
3 : 1
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