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Question

A sphere of radiusa and mass m rolls along horizontal plane with constant speedv0. It encounters an inclined plane at angle θ and climbs upward. Assuming that it rolls without slipping how far up the sphere will travel (along the incline)?


A

25v02gsinθ

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B

10v027gsinθ

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C

v025gsinθ

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D

7v0210gsinθ

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Solution

The correct option is D

7v0210gsinθ


Step 1: Given data and drawing the diagram

Radius=a

Mass =m

Speed=v0

Acceleration due to gravity =g

Let height sphere climb =h

Let length of incline =l

Angular velocity =ω=v0a

Shift 2 Feb 25 JEE Main 2021 Physics Papers With Solutions

Step 2: Find how far up the sphere will travel

Potential energy is the energy of object due to its position and it is given by

Potential energy =mgh

Kinetic energy is the energy of an object due to its motion and it is given by

Kinetic energy =12mv2

Friction produces heat, because the friction force carries out work when it acts. Heat due to friction is given by

Heat due to friction=12Iω2

From right triangle shown in figure,

sinθ=perpendicularhypotneussinθ=hlh=lsinθ

By law of conservation of energy, the energy can neither be created nor be destroyed, it can transfer from one form to another. So,

Potential energy = kinetic energy+ heat due to friction

mgh=12mv2+12Iω2mgh=12mv02+12×25ma2×v02a2gh=12v02+15v02h=710v02g

Putting value of h, we get

lsinθ=710v02gl=710v02gsinθ

Hence, option D is correct.


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