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A copper block of mass 1kg slides down o...

A copper block of mass 1kg slides down one rough inclined plane of inclination `37^(@)` at a constant speed. Find the increase in the temperature of the block as it slides down temperature of the block as it slides down through 60cm assuming that the loss in mechanical energy goes into the copper block as thermal energy. (specific heat of copper=`420Jkg^(-1)K^(-1),g=10ms^(-2)`)

A

`6.6xx10^(-3).^@C`

B

`7.6xx10^(-3).^(@)C`

C

`8.6xx10^(-3).^(@)C`

D

`9.6xx10^(-3).^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
C
`(1)/(2)mv^(2)=mSDeltaT`, But `(1)/(2)mv^(2)=mgh`
`rArrV^(2)=2gl sin theta,(1)/(2)xx2gl sin theta=SDeltaT`
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Knowledge Check

  • A copper block of mass 5kg slides down along a rough inclined plane of inclination 30^(@)C with a constant speed. The increase in the temperature of the block as it slide down through 100cm assuming that the loss of mechanical energy goes into copper block as thermal energy. (specific heat of copper 420 Jkg^(-1)K^(-1),g=10ms^(-2))

    A
    `1.19xx10^(-3) .^(@)C`
    B
    `2.38xx10^(-3) .^(@)C`
    C
    `1.19xx10^(-2) .^(@)C`
    D
    `2.38xx10^(-2) .^(@)C`

  • A block of mass m slides down an inclined plane of inclination theta with uniform speed The coefficient of friction between the block and the plane is mu . The contact force between the block and the plane is .

    A
    `mg sin thetasqrt(1+mu^(2))`
    B
    `sqrt((mg sin theta)^(2)+(mumgcostheta)^(2))`
    C
    `mg sin theta`
    D
    `mg`

  • A block of mass m slides down an inclined plane of inclination theta with uniform speed. The coefficient of friction between the block and the plane is mu . The contact force between the block and the plane is

    A
    mg
    B
    `mg sin theta sqrt(1+mu^(2))`
    C
    `mg sin theta`
    D
    `sqrt((mg sin theta)^(2)+(mumg cos theta)^(2))`
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