A machine is blowing spherical soap bubbles o | vedclass

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

Question

(D) For a soap bubble to float in air,the gravitational force must be equal to the buoyant force.Gravitational force = Buoyant force$g 	imes (	ext{m

Vedclass · Accepted Answer

$3.50$

Vedclass · Answer

(D) For a soap bubble to float in air,the gravitational force must be equal to the buoyant force.Gravitational force = Buoyant force$g 	imes (	ext{mass of helium} + 	ext{mass of soap film}) = 	ext{Weight of air displaced by bubble}$Let $r$ be the inner radius of the soap bubble and $t$ be the thickness of the film.The volume of helium is $V_{He} = \frac{4}{3} \pi r^3$.The volume of the soap film is $V_{film} \approx 4 \pi r^2 t$ (since $t \ll r$).The equation of equilibrium is:$\frac{4}{3} \pi r^3 ho_{He} g + (4 \pi r^2 t) ho_{soap} g = \frac{4}{3} \pi r^3 ho_{air} g$Dividing by $g$ and rearranging:$4 \pi r^2 t ho_{soap} = \frac{4}{3} \pi r^3 (ho_{air} - ho_{He})$$t = \frac{r}{3} \frac{(ho_{air} - ho_{He})}{ho_{soap}}$Given $r = 1 \, cm = 10^{-2} \, m$,$ho_{air} = 1.23 \, kg \, m^{-3}$,$ho_{He} = 0.18 \, kg \, m^{-3}$,and $ho_{soap} = 1000 \, kg \, m^{-3}$.$t = \frac{10^{-2}}{3} 	imes \frac{(1.23 - 0.18)}{1000}$$t = \frac{10^{-2}}{3} 	imes \frac{1.05}{1000} = \frac{1.05 	imes 10^{-5}}{3} = 0.35 	imes 10^{-5} \, m = 3.5 	imes 10^{-6} \, m$.Since $1 \, \mu m = 10^{-6} \, m$,the thickness $t = 3.50 \, \mu m$.

Similar Questions

The reading of a spring balance when a block is suspended from it in air is $60\,N$. This reading changes to $40\,N$ when the block is immersed in water. The specific density (relative density) of the block is:

$A$ container of large surface area is filled with a liquid of density $\rho$. $A$ cubical block of side edge $a$ and mass $M$ is floating in it with four-fifths of its volume submerged. If a coin of mass $m$ is placed gently on the top surface of the block,it is just submerged. $M$ is:

Determine the equation for the volume of the partially immersed part of a body floating in a fluid.

$A$ cubical block of wood of edge $10$ $cm$ and mass $0.92$ $kg$ floats in a tank of water with a layer of oil of relative density $0.6$ to a depth of $4$ $cm$ above the water. When the block attains equilibrium with four of its side edges vertical,which of the following is true?

$A$ body floats with $\frac{1}{n}$ of its volume outside of water. If the body is pushed to a depth $h$ inside the water and released, it will come to the surface after time $t$. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module