JEE Main 2019 (Online) 8th April Morning Slot | Center of Mass and Collision Question 79 | Physics

JEE Main 2019 (Online) 8th April Morning Slot

MCQ (Single Correct Answer)

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If 10²² gas molecules each of mass 10^–26 kg collide with a surface (perpendicular to it) elastically per second over an area 1 m² with a speed 10⁴ m/s, the pressure exerted by the gas molecules will be of the order of :

10⁸ N/m²

10¹⁶ N/m²

10⁴ N/m²

2 N/m²

JEE Main 2019 (Online) 12th January Morning Slot

MCQ (Single Correct Answer)

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The position vector of the centre of mass $$\overrightarrow r {\,_{cm}}\,$$ of an asymmetric uniform bar of negligible area of crosssection as shown in figure is :

JEE Main 2019 (Online) 12th January Morning Slot Physics - Center of Mass and Collision Question 81 English

JEE Main 2019 (Online) 12th January Morning Slot Physics - Center of Mass and Collision Question 81 English

$${\overrightarrow r _{cm}}\, = {{11} \over 8}L\,\,\widehat x + {8 \over 8}L\widehat y$$

$${\overrightarrow r _{cm}}\, = {5 \over 8}L\,\,\widehat x + {{13} \over 8}L\widehat y$$

$${\overrightarrow r _{cm}}\, = {{13} \over 8}L\,\,\widehat x + {5 \over 8}L\widehat y$$

$${\overrightarrow r _{cm}}\, = {3 \over 8}L\,\,\widehat x + {{11} \over 8}L\widehat y$$

JEE Main 2019 (Online) 12th January Morning Slot

MCQ (Single Correct Answer)

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A simple pendulum, made of a string of length $$\ell $$ and a bob of mass m, is released from a small angle $${{\theta _0}}$$. It strikes a block of mass M, kept on a horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle $${{\theta _1}}$$. Then M is given by :

$${m \over 2}\left( {{{{\theta _0} + {\theta _1}} \over {{\theta _0} - {\theta _1}}}} \right)$$

$${m \over 2}\left( {{{{\theta _0} - {\theta _1}} \over {{\theta _0} + {\theta _1}}}} \right)$$

$$m\left( {{{{\theta _0} + {\theta _1}} \over {{\theta _0} - {\theta _1}}}} \right)$$

$$m\left( {{{{\theta _0} - {\theta _1}} \over {{\theta _0} + {\theta _1}}}} \right)$$

JEE Main 2019 (Online) 11th January Evening Slot

MCQ (Single Correct Answer)

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A particle of mass m is moving in a straight line with momentum p. Starting at time t = 0, a force F = kt acts in the same direction on the moving particle during time interval T so that its momentum changes from p to 3p. Here k is a constant. The value of T is :

$$2\sqrt {{k \over p}} $$

$$2\sqrt {{p \over k}} $$

$$\sqrt {{{2p} \over 2}} $$

$$\sqrt {{{2k} \over p}} $$

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