JEE Main 2021 (Online) 25th February Evening Shift | Rotational Motion Question 62 | Physics

JEE Main 2021 (Online) 25th February Evening Shift

MCQ (Single Correct Answer)

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A sphere of radius 'a' and mass 'm' rolls along a horizontal plane with constant speed v₀. It encounters an inclined plane at angle $$\theta$$ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel?

JEE Main 2021 (Online) 25th February Evening Shift Physics - Rotational Motion Question 62 English

JEE Main 2021 (Online) 25th February Evening Shift Physics - Rotational Motion Question 62 English

$${{v_0^2} \over {2g\sin \theta }}$$

$${{7v_0^2} \over {10g\sin \theta }}$$

$${2 \over 5}{{v_0^2} \over {g\sin \theta }}$$

$${{v_0^2} \over {5g\sin \theta }}$$

JEE Main 2021 (Online) 24th February Evening Slot

MCQ (Single Correct Answer)

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A circular hole of radius $$\left( {{a \over 2}} \right)$$ is cut out of a circular disc of radius 'a' as shown in figure. The centroid of the remaining circular portion with respect to point 'O' will be :

JEE Main 2021 (Online) 24th February Evening Slot Physics - Rotational Motion Question 63 English

JEE Main 2021 (Online) 24th February Evening Slot Physics - Rotational Motion Question 63 English

$${1 \over 6}a$$

$${2 \over 3}a$$

$${5 \over 6}a$$

$${10 \over 11}a$$

JEE Main 2021 (Online) 24th February Morning Slot

MCQ (Single Correct Answer)

-1

Moment of inertia (M. I.) of four bodies, having same mass and radius, are reported as;

I₁ = M.I. of thin circular ring about its diameter,

I₂ = M.I. of circular disc about an axis perpendicular to disc and going through the centre,

I₃ = M.I. of solid cylinder about its axis and

I₄ = M.I. of solid sphere about its diameter.

Then :

I₁ = I₂ = I₃ > I₄

I₁ + I₃ < I₂ + I₄

I₁ = I₂ = I₃ < I₄

I₁ + I₂ = I₃ + $${5 \over 2}$$ I₄

JEE Main 2020 (Online) 6th September Evening Slot

MCQ (Single Correct Answer)

-1

The linear mass density of a thin rod AB of length L varies from A to B as
$$\lambda \left( x \right) = {\lambda _0}\left( {1 + {x \over L}} \right)$$, where x is the distance from A. If M is the mass of the rod then its moment of inertia about an axis passing through A and perpendicular to the rod is :

$${2 \over 5}M{L^2}$$

$${5 \over {12}}M{L^2}$$

$${7 \over {18}}M{L^2}$$

$${3 \over 7}M{L^2}$$

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