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A mecury drop of radius 1 cm is sprayed ...

A mecury drop of radius 1 cm is sprayed into `10^(5)` droplets of equal size. Calculate the increase in surface energy if surface tension of mercury is `35xx10^(-3)N//m`.

Text Solution

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The correct Answer is:
A, C, D

`(4)/(3) pi R^(3)=10^(6)((4)/(3) pi r^(3))`
`implies r=(10^(-2))R=10^(-2)cm`
`A_(i)=4piR^(2)`
`A_(f)=(10^(6))(4 pi r^(2))`
`Delta A =A_(f)-A_(i)`
`:. Delta U =(T)(Delta A)`.
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Knowledge Check

  • A mercury drop of radius 1.0 cm is sprayed into 10^(6) droplets of equal sizes. The energy expended in this process is (Given, surface tension of mercury is 32 xx 10^(-2) Nm^(-1) )

    A
    `3.98 xx 10^(-4) J`
    B
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    D
    `8.46 xx 10^(-2) J`
  • A mercury drop of radius 1 cm is sprayed into 10^(6) drops of equal size. The energy expressed in joule is (surface tension of Mercury is 460 xx 10^(–3) N/m)

    A
    `0. 0 57`
    B
    `5. 7`
    C
    `5. 7 xx 10 ^(-4)`
    D
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  • A mercury drop of radius 1 cm is sprayed into 10^(6) drops of equil size. The energy expended in joule is (surface tension of mercury is (460xx10^(-3)N//m)

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