Question

A mercury drop of radius 1.0 cm is sprayed into
106 droplets of equal sizes. The energy expended
in this process is (surface tension of mercury is
equal to 32 x 10-2 Nm-')
(1) 3.98 x 10-4 J (2) 8.46 x 104J
(3) 3.98 x 10-2 J (4) 8.46 x 10-2 J​

Answer 1

it's simple dude....

let the radius of each small drops be r

then

4/3π(.001)^3=10^6×4/3π(r)^3

1×10^-9=10^6×r^3

10^-9/10^6=r^3

now,r=10^-5

surface area of total sphere=4πr^2×10^6

=4×3.14×10^-10×10^6

=12.56×10^-4

energy expanded=12.55×10^-4×32×10^-2

=4.01×10^-4 or 3.9×10^-4(approx)

hope it helps you

be brainly ✌️

ask ur doubts here, I will surely help u......

Answer 2

Correct Question :

A mercury drop of radius 1.0 cm is sprayed into 10^6 droplets of equal sizes. The energy expended in this process is (surface tension of mercury is equal to 32 x 10^-2 N/m).

Theory :

${\red{\boxed{\large{\bold{Surface\:Energy}}}}}$

According to molecular theory of surface tension the molecules in the surface have some additional energy due to their position .This additional energy per unit area of the surface is called surface energy.

⇒ In case of Drops :

Work done = surface surface tension× change in area

Solution:

Given : Radius of big drop = 1 cm =0.01m

n = $10{}^6}$

suface tension ,T = 32 x 10^-2 N/m

________________________

Let R be the radius of the big drop and r be the radius of a small drop.

⇒Since the total volume remains conserved on the formation of small drops.

$\frac{4}{3}\pi \: R{}^{3} = \frac{4}{3}\pi \: r {}^{3} \times n$

$\implies \: R= n {}^{ \frac{1}{3} } r$

Energy expended = T× ΔA

$= T \times 4\pi(R{}^{2} - nr {}^{2})$

$= T \times 4\pi(R{}^{2} - n {}^{ \frac{1}{3} } R{}^{2} )$

$= T \times 4\pi \: R {}^{2} (1 - n {}^{ \frac{1}{3} } )$

Now put the values

$= 32 \times 10 {}^{ - 2} \times4 \times 3.14 \times 1 \times 10 {}^{ - 4} (1 - 10 {}^{2} )$

$= 32 \times 10 {}^{ - 2} \times 4 \times 3.14 \times 1 \times 10 {}^{ - 4} \times 99$

$= 39.8 \times 10 {}^{ - 4} \: J$