Question

calculate the potential energy of spherical Mercury drop of radius 0.01 m due to surface tension the surface tension of Mercury 0.46 Newton per metre

Answer 1

Answer:57.776 erg

Explanation:

We know that,

Work done is stored as P.E (Surface Energy)

Therefore, W = T X A

=0.46*4πr²

=0.46*4π(0.01)²

=0.00057775 Joules

=57.776 erg

Answer 2

Answer:

The answer is 57.776.

Explanation:

potential energy exists as stored energy that depends upon the relative position of different parts of a system. Spring has more additional potential energy when it stands compressed or stretched. A steel ball has more potential energy increased above the ground than it has after falling to Earth.

We know that,

Work done is stored as P.E (Surface Energy)

Therefore, $\mathrm{W}=\mathrm{T} \times \mathrm{A}$

$&=0.46*4 \pi r^{2}$

$&=0.46* 4 \pi(0.01)^{2}$

$&=0.00057775 \text { Joules }$

$&=57.776$ erg.

The answer is 57.776.

#SPJ2