Surface tension of a soap bubble is 2.0 times | vedclass

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

Question

(C) soap bubble has two surfaces (inner and outer),so its total surface area is $A = 2 	imes (4 \pi R^2) = 8 \pi R^2$.The work done $W$ in increasing

Vedclass · Accepted Answer

$18.48$

Vedclass · Answer

(C) soap bubble has two surfaces (inner and outer),so its total surface area is $A = 2 	imes (4 \pi R^2) = 8 \pi R^2$.The work done $W$ in increasing the radius from $R_1$ to $R_2$ is equal to the change in surface energy:$W = T 	imes \Delta A = T 	imes (A_2 - A_1) = T 	imes 8 \pi (R_2^2 - R_1^2)$.Given:$T = 2.0 	imes 10^{-2} \; N m^{-1}$$R_1 = 3.5 \; cm = 3.5 	imes 10^{-2} \; m$$R_2 = 7.0 \; cm = 7.0 	imes 10^{-2} \; m$Substituting the values:$W = 2.0 	imes 10^{-2} 	imes 8 	imes \frac{22}{7} 	imes [(7.0 	imes 10^{-2})^2 - (3.5 	imes 10^{-2})^2]$$W = 16 	imes 10^{-2} 	imes \frac{22}{7} 	imes (49 - 12.25) 	imes 10^{-4}$$W = 16 	imes 10^{-2} 	imes \frac{22}{7} 	imes 36.75 	imes 10^{-4}$$W = 16 	imes 10^{-2} 	imes 22 	imes 5.25 	imes 10^{-4}$$W = 18.48 	imes 10^{-4} \; J$.

Surface tension of a soap bubble is $2.0 \times 10^{-2} \; N m^{-1}$. Work done to increase the radius of the soap bubble from $3.5 \; cm$ to $7 \; cm$ will be $......... \times 10^{-4} \; J$ [Take $\pi = \frac{22}{7}$]

Similar Questions

$A$ liquid drop having surface energy $E$ is sprayed into $512$ droplets of same size. The final surface energy is (in $E$)

Work done in increasing the size of a soap bubble from a radius of $3 \ cm$ to $5 \ cm$ in millijoules is nearly (surface tension of soap solution $= 0.03 \ Nm^{-1}$) (in $\pi$)

$A$ soap bubble of radius $R$ is blown. After heating a solution,a second bubble of radius $2R$ is blown. The work required to blow the $2^{nd}$ bubble in comparison to that required for the $1^{st}$ bubble is

$A$ soap bubble of radius $r$ is blown up to form a bubble of radius $2r$ under isothermal conditions. If $T$ is the surface tension of the soap solution, the energy spent in the blowing is: (in $\pi T r^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module