Question

T) UT
A drop of liquid of diameter 0.28 cm breaks into 125
droplets of equal radii. The change in surface energy
is nearly (Surface tension = 7.5 x 10-2 N/m)
(1) Zero
(2) 2 w
(3) 4.6 W
(4) 7.5 u​

Answer 1

Answer:

Explanation:

E=4piR^2T (n^1/3-1)

=4×3.14×(14×10^-4)^2×7.5×10^-2×(125^1/3-1)

=7.38×10^-6

=approximately 7.5uj

Answer 2

Answer:

The change in surface energy is 7.5 μJ.

(4) is correct option.

Explanation:

Given that,

Diameter = 0.28 cm

Number of drops = 125

Surface tension $T=7.5 \times10^{-2}\ N/m$

We need to calculate the change in surface energy

Using formula of energy

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

Put the value into the formula

$E=4\pi\times\dfrac{(0.28\times10^{-2})^2}{4}\times7.5 \times10^{-2}(125^{\dfrac{1}{3}}-1)$

$E=0.00000738\ J$

$E=7.4\times10^{-6}\ J$

$E=7.4\ \mu J\approx 7.5\ \mu J$

Hence, The change in surface energy is 7.5 μJ.