A drop of liquid of radius R=10^{-2} ,m havin | vedclass

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

Question

(B) The initial surface area of the large drop is $A_i = 4 \pi R^2$. The final surface area of $K$ small drops of radius $r$ is $A_f = K 	imes 4 \pi

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) The initial surface area of the large drop is $A_i = 4 \pi R^2$. The final surface area of $K$ small drops of radius $r$ is $A_f = K 	imes 4 \pi r^2$.Since the volume is conserved, $\frac{4}{3} \pi R^3 = K 	imes \frac{4}{3} \pi r^3$, which implies $r = R K^{-1/3}$.Substituting $r$ into the final area: $A_f = K 	imes 4 \pi (R K^{-1/3})^2 = 4 \pi R^2 K^{1/3}$.The change in surface energy is $\Delta U = S(A_f - A_i) = S 	imes 4 \pi R^2 (K^{1/3} - 1)$.Given $\Delta U = 10^{-3} \,J$, $R = 10^{-2} \,m$, and $S = \frac{0.1}{4 \pi} \,Nm^{-1}$:$10^{-3} = \left(\frac{0.1}{4 \pi}ight) 	imes 4 \pi 	imes (10^{-2})^2 	imes (K^{1/3} - 1)$.$10^{-3} = 0.1 	imes 10^{-4} 	imes (K^{1/3} - 1) = 10^{-5} 	imes (K^{1/3} - 1)$.$K^{1/3} - 1 = \frac{10^{-3}}{10^{-5}} = 10^2 = 100$.$K^{1/3} = 101 \approx 100 = 10^2$.$K = (10^2)^3 = 10^6$.Since $K = 10^\alpha$, we have $\alpha = 6$.

$A$ drop of liquid of radius $R=10^{-2} \,m$ having surface tension $S=\frac{0.1}{4 \pi} \,Nm^{-1}$ divides itself into $K$ identical drops. In this process, the total change in the surface energy is $\Delta U=10^{-3} \,J$. If $K=10^\alpha$, then the value of $\alpha$ is:

Similar Questions

$A$ soap bubble of radius $R$ is blown. After heating the solution,a second bubble of radius $2R$ is blown. The work required to blow the second bubble in comparison to that required for the first bubble is

$A$ mercury drop of radius $1\, cm$ is sprayed into $10^{6}$ drops of equal size. The energy expressed in joule is (surface tension of mercury is $460 \times 10^{-3}\, N/m$)

The work done in blowing a soap bubble of radius $0.2 \, m$,given that the surface tension of the soap solution is $60 \times 10^{-3} \, N/m$,is:

One thousand small water drops of equal radii combine to form a big drop. The ratio of final surface energy to total initial surface energy is

The work done in blowing a soap bubble of radius $R$ is $W_1$ at room temperature. Now the soap solution is heated. From the heated solution another soap bubble of radius $2R$ is blown and the work done is $W_2$. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module