A liquid drop of diameter D splits into 3375 | 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 $R = D/2$. Let the radius of each small drop be $r$.Since the volume remains constant,the volume of the large

Vedclass · Accepted Answer

$14 \pi D^2 S$

Vedclass · Answer

(A) Let the radius of the large drop be $R = D/2$. Let the radius of each small drop be $r$.Since the volume remains constant,the volume of the large drop equals the sum of the volumes of $n = 3375$ small drops:$\frac{4}{3} \pi R^3 = n 	imes \frac{4}{3} \pi r^3$$R^3 = 3375 r^3$Taking the cube root on both sides: $R = 3375^{1/3} r = 15r$,so $r = R/15$.The initial surface area is $A_i = 4 \pi R^2$.The final surface area is $A_f = n 	imes 4 \pi r^2 = 3375 	imes 4 \pi (R/15)^2 = 3375 	imes 4 \pi (R^2 / 225) = 15 	imes 4 \pi R^2 = 60 \pi R^2$.The change in surface area is $\Delta A = A_f - A_i = 60 \pi R^2 - 4 \pi R^2 = 56 \pi R^2$.The change in surface energy is $\Delta U = S 	imes \Delta A = S 	imes 56 \pi R^2$.Substituting $R = D/2$,we get $\Delta U = S 	imes 56 \pi (D/2)^2 = S 	imes 56 \pi (D^2 / 4) = 14 \pi D^2 S$.

$A$ liquid drop of diameter $D$ splits into $3375$ small identical drops. If $S$ is the surface tension of the liquid,then the change in the surface energy in the process is

Similar Questions

Two mercury drops,each with same radius $r$,merge to form a bigger drop. If $T$ is the surface tension of mercury,then the surface energy of the bigger drop is given by

$A$ spherical drop of liquid splits into $1000$ identical spherical drops. If $u_i$ is the surface energy of the original drop and $u_f$ is the total surface energy of the resulting drops (ignoring evaporation),and $\frac{u_f}{u_i} = \left(\frac{10}{x}\right)$,then the value of $x$ is $......$

If $T$ is the surface tension of a liquid,the energy needed to break a liquid drop of radius $R$ into $64$ drops is:

How much energy is required to double the radius of a soap bubble of radius $R$ and surface tension $T$ (in $\pi R^2 T$)?

$A$ mercury drop of $1\, cm$ radius is broken into ${10^6}$ small drops. The energy used will be (surface tension of mercury is $35 \times {10^{ - 3}} \, N/cm$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module