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A Sphere pure rolls on a rough inclined ...

A Sphere pure rolls on a rough inclined plane with initial velocity 2.8 m/s. Find the maximum distance on the inclined plane

A

2.74 m

B

5.48 m

C

1.38 m

D

3.2 m

Text Solution

Verified by Experts

The correct Answer is:
A

`a=(g sin theta)/( 1+k^(2)//r^(2))`
`I=(2)/(5)mr^(2)=mk^(2)`
`(k^(2))/(r^(2))=(2)/(5)`
`a=(g sin theta)/( 1+2//5)`
`2.8^(2)=2(gxx(1)/(2))/((7)/(2)) xxS`
`S=2.8^(2) xx(7)/(20)`
`S=2.744m`
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  • a particle is projected down on inclined plane with a velocity of 21m/s at an angle of 60^@ with the horizontal its range on the inclined plane inclined at an angle of 30^@ with the horizontal is

    A
    21dm
    B
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  • a particle is projected down on inclined plane with a velocity of 21m/s at an angle of 60^@ with the horizontal its range on the inclined plane inclined at an angle of 30^@ with the horizontal is

    A
    21dm
    B
    2.1dm
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    30dm
    D
    6dm
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    A
    `3m`
    B
    `4m`
    C
    `5m`
    D
    zero
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