Question

A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075 N/m. In this process the gain in surface energy will be ______.

Options

• 2.8 × 10^-4 J

• 1.5 × 10^-3 J

• 1.9 × 10^-4 J

• 9.4 × 10^-5 J

Answer 1

A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075 N/m. In this process the gain in surface energy will be 2.8 × 10^-4 J.

Explanation:

E_i = 0‾ A_i 0‾ = surface energy per unit area

= T a_i [∵ 0‾ = T] = T.4`pi"r"_"i"^2`

Now, V_i = V_f

⇒ `4/3pi"r"_"i"^3=64xx4/3pi"r"_"f"^3`

⇒ `"r"_"i"^3 = 64"r"_"f"^3`

⇒ r_i = 4r_F

So, `"E"_"f" =  0‾ "A"_"f" ="T"xx64xx4pi"r"_"f"^2=256"T"pi"r"_"i"^2/16=16pi"T""r"_"i"^2`

So, `Delta"E"="E"_"f"-"E"_"i"=12pi+"T""r"_"i"^2`

= 12π × 0.075 × 0.1²

= 2.82 × 10^-4 J