MHT CET 2026 11th April Evening Shift | MHT CET Year Wise Previous Years Questions

MHT CET 2026 11th April Evening Shift

MCQ (Single Correct Answer)

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In non uniform circular motion, the ratio of radial acceleration to tangential acceleration is (V is the velocity, r is the radius and $\alpha$ is the angular acceleration)

$\dfrac{\alpha r}{V}$

$\dfrac{V^2}{r^2\alpha}$

$\dfrac{r^2\alpha}{V^2}$

$\dfrac{r\alpha^2}{V^2}$

MHT CET 2026 11th April Evening Shift

MCQ (Single Correct Answer)

-0

A mass M moving with velocity V along X-axis collides and sticks to another mass 2M which is moving along Y-axis with velocity 3V. The velocity of the combination, after the collision is

$V\hat{i} + 2V\hat{j}$

$\dfrac{V}{3}\hat{i} + 2V\hat{j}$

$V\hat{i} + 3V\hat{j}$

$2V\hat{i} + 4V\hat{j}$

MHT CET 2026 11th April Evening Shift

MCQ (Single Correct Answer)

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The moment of inertia of a ring about an axis passing through its centre and perpendicular to its plane is I. It is rotating with angular velocity $\omega$. Another identical ring is gently placed on it so that their centres coincide. If both rings are rotating about the same axis then loss in kinetic energy is

$I\omega^2$

$\dfrac{I\omega^2}{2}$

$\dfrac{I\omega^2}{4}$

$\dfrac{I\omega^2}{8}$

MHT CET 2026 11th April Evening Shift

MCQ (Single Correct Answer)

-0

Let M and L be the mass and length of thin uniform rod respectively. In first case, axis of rotation is passing through centre and perpendicular to length of rod. In second case axis of rotation is passing through one end and perpendicular to length of rod. The ratio of radius of gyration in first case to second case is

$\dfrac{1}{4}$

$\dfrac{1}{2}$

$\dfrac{1}{8}$

$\dfrac{1}{6}$

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