JEE Main 2020 (Online) 8th January Evening Slot | Rotational Motion Question 126 | Physics

A uniform sphere of mass 500 g rolls without slipping on a plane horizontal surface with its centre moving at a speed of 5.00 cm/s. Its kinetic energy is :

8.75 × 10^–3 J

1.13 × 10^–3 J

8.75 × 10^–4 J

6.25 × 10^–4 J

JEE Main 2020 (Online) 8th January Morning Slot

MCQ (Single Correct Answer)

-1

Consider a uniform rod of mass M = 4m and length $$\ell $$ pivoted about its centre. A mass m moving with velocity v making angle $$\theta = {\pi \over 4}$$ to the rod's long axis collides with one end of the rod and sticks to it. The angular speed of the rod-mass system just after the collision is :

$${{3\sqrt 2 } \over 7}{v \over \ell }$$

$${3 \over 7}{v \over \ell }$$

$${3 \over {7\sqrt 2 }}{v \over \ell }$$

$${4 \over 7}{v \over \ell }$$

JEE Main 2020 (Online) 7th January Evening Slot

MCQ (Single Correct Answer)

-1

Mass per unit area of a circular disc of radius $$a$$ depends on the distance r from its centre as $$\sigma \left( r \right)$$ = A + Br . The moment of inertia of the disc about the axis, perpendicular to the plane and assing through its centre is:

$$2\pi {a^4}\left( {{A \over 4} + {{aB} \over 5}} \right)$$

$$\pi {a^4}\left( {{A \over 4} + {{aB} \over 5}} \right)$$

$$2\pi {a^4}\left( {{{aA} \over 4} + {B \over 5}} \right)$$

$$2\pi {a^4}\left( {{A \over 4} + {B \over 5}} \right)$$

JEE Main 2020 (Online) 7th January Morning Slot

MCQ (Single Correct Answer)

-1

The radius of gyration of a uniform rod of length $$l$$, about an axis passing through a point $${l \over 4}$$ away from the centre of the rod, and perpendicular to it, is :

$${1 \over 8}l$$

$${1 \over 4}l$$

$$\sqrt {{7 \over {48}}} l$$

$$\sqrt {{3 \over 8}} l$$

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