JEE Main 2019 (Online) 10th April Morning Slot | Center of Mass and Collision Question 71 | Physics

JEE Main 2019 (Online) 10th April Morning Slot

MCQ (Single Correct Answer)

-1

A thin disc of mass M and radius R has mass per unit area $$\sigma $$(r) = kr² where r is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is :

$${{M{R^2}} \over 3}$$

$${{M{R^2}} \over 6}$$

$${{2M{R^2}} \over 3}$$

$${{M{R^2}} \over 2}$$

JEE Main 2019 (Online) 10th April Morning Slot

MCQ (Single Correct Answer)

-1

A particle of mass m is moving along a trajectory given by
x = x₀ + a cos$$\omega $$₁t
y = y₀ + b sin$$\omega $$₂t
The torque, acting on the particle about the origin, at t = 0 is :

Zero

+my₀a $$\omega _1^2$$$$\widehat k$$

$$ - m\left( {{x_0}b\omega _2^2 - {y_0}a\omega _1^2} \right)\widehat k$$

m (–x₀b + y₀a) $$\omega _1^2$$$$\widehat k$$

JEE Main 2019 (Online) 10th April Morning Slot

MCQ (Single Correct Answer)

-1

Two particles, of masses M and 2M, moving, as shown, with speeds of 10 m/s and 5 m/s, collide elastically at the origin. After the collision, they move along the indicated directions with speeds u₁ and u₂, respectively. The values of u₁ and u₂ are nearly : JEE Main 2019 (Online) 10th April Morning Slot Physics - Center of Mass and Collision Question 71 English

JEE Main 2019 (Online) 10th April Morning Slot Physics - Center of Mass and Collision Question 71 English

6.5 m/s and 3.2 m/s

6.5 m/s and 6.3 m/s

3.2 m/s and 6.3 m/s

3.2 m/s and 12.6 m/s

JEE Main 2019 (Online) 9th April Evening Slot

MCQ (Single Correct Answer)

-1

A particle of mass 'm' is moving with speed '2v' and collides with a mass '2m' moving with speed 'v' in the same direction. After collision, the first mass is stopped completely while the second one splits into two particles each of mass 'm', which move at angle 45° with respect to the origianl direction. The speed of each of the moving particle will be :-

2 $$\sqrt2$$v

v / (2 $$\sqrt2$$ )

v / $$\sqrt2$$

$$\sqrt2$$v

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