Question

Calculate the workdone when a spherical drop of
mercury of radius 2mm, falls from some height and
breakes into a million droplets.
each of the
same size. The surface tension of mercury
is T = 0.5 N/m .​

Answer 1

Answer:

Answer is 2×0.5 just find it's Answer

Answer 2

The work done is $0.248\times10^{-2}\ J$

Explanation:

Given that,

Radius = 2 mm

Volume of the large drop $V=\dfrac{4}{3}\pi R^3$

Volume of the small drop $V'=\dfrac{4}{3}\pi r^3$

We need to calculate the radius of small drop

Using formula of volume

$V=V'$

Put the value of volume

$\dfrac{4}{3}\pi R^3=n\dfrac{4}{3}\pi r^3$

$R=10^{6}\times r^3$

$r=\dfrac{R}{100}$

We need to calculate the charge in surface area

$ds=4\pi(nr^2-R^2)$

$ds=4\pi(10^6\times\dfrac{R^2}{10^4}-R^2)$

$ds=99\times 4\pi(R)^2$

$ds=99\times4\pi\times(2\times10^{-3})^2$

$ds=4976.2\times10^{-6}\ m^2$

We need to calculate the work done

Using formula of work done

$W=T\cdot ds$

Where, ds = change in area

Put the value into the formula

$W=0.5\times4976.2\times10^{-6}$

$W=0.248\times10^{-2}\ J$

Hence, The work done is $0.248\times10^{-2}\ J$

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Topic : work done

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