If work W is done in blowing a bubble of radi | vedclass

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

Question

(C) The work done $W$ in blowing a soap bubble of radius $R$ is given by the formula $W = T 	imes \Delta A$,where $T$ is the surface tension and $\De

Vedclass · Accepted Answer

$4W$

Vedclass · Answer

(C) The work done $W$ in blowing a soap bubble of radius $R$ is given by the formula $W = T 	imes \Delta A$,where $T$ is the surface tension and $\Delta A$ is the change in surface area.Since a soap bubble has two surfaces (inner and outer),the surface area $A = 2 	imes (4\pi R^2) = 8\pi R^2$.Thus,$W = 8\pi R^2 T$.From this expression,we can see that $W \propto R^2$.If the radius is increased from $R$ to $2R$,the new work done $W'$ will be:$W' = 8\pi (2R)^2 T = 8\pi (4R^2) T = 4 	imes (8\pi R^2 T) = 4W$.Therefore,the work done becomes $4W$.

If work $W$ is done in blowing a bubble of radius $R$ from a soap solution,then the work done in blowing a bubble of radius $2R$ from the same solution is

Similar Questions

$A$ bulb and a capacitor are connected in series to a source of alternating current. If its frequency is increased,while keeping the voltage of the source constant,then

$A$ body of mass $6 ~kg$ is under a force which causes displacement in it given by $s = \frac{t^2}{4}$ metres,where $t$ is time. The work done by the force in $2$ seconds is: (in $~J$)

$A$ particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement $x$ is proportional to

Magnifying power of a simple microscope is (when final image is formed at $D = 25 \ cm$ from eye)

The set $\{x \in R : [x - |x|] = 5\}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module