A liquid drop of diameter D breaks into 27 ti | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the radius of the large drop be $R = D/2$. The volume of the large drop is $V = \frac{4}{3}\pi R^3$.When it breaks into $n = 27$ smaller drops

Vedclass · Accepted Answer

$2\pi TD^2$

Vedclass · Answer

(A) Let the radius of the large drop be $R = D/2$. The volume of the large drop is $V = \frac{4}{3}\pi R^3$.When it breaks into $n = 27$ smaller drops of radius $r$,the volume remains conserved: $\frac{4}{3}\pi R^3 = n 	imes \frac{4}{3}\pi r^3$.Thus,$R^3 = 27r^3$,which gives $r = R/3$.The initial surface area is $A_i = 4\pi R^2$.The final surface area is $A_f = n 	imes 4\pi r^2 = 27 	imes 4\pi (R/3)^2 = 27 	imes 4\pi (R^2/9) = 12\pi R^2$.The change in surface area is $\Delta A = A_f - A_i = 12\pi R^2 - 4\pi R^2 = 8\pi R^2$.The change in energy (surface energy) is $\Delta E = T 	imes \Delta A = T 	imes 8\pi R^2$.Substituting $R = D/2$,we get $\Delta E = T 	imes 8\pi (D/2)^2 = T 	imes 8\pi (D^2/4) = 2\pi TD^2$.

$A$ liquid drop of diameter $D$ breaks into $27$ tiny drops. The change in energy is

Similar Questions

$A$ capacitor of capacity $4 \mu F$ is charged to $80 \ V$ and another capacitor of capacity $6 \mu F$ is charged to $30 \ V$. When they are connected,the energy lost by the $4 \mu F$ capacitor is: (in $mJ$)

In $\triangle ABC$,$(a+b+c)\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right)$ is equal to

What is the product formed when benzylamine reacts with nitrous acid?

$A$ transition metal $M$ forms a volatile chloride with a vapor density of $94.8$. If it contains $74.75\%$ chlorine,what is the formula of the metal chloride?

$A$ bullet of mass $m$ moving with velocity $v$ strikes a suspended wooden block of mass $M$. If the block rises to a height $h$,the initial velocity of the bullet will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module