Fluid Mechanics

JEE Main level, numerical answer, medium

110 of 192

Question 110 of 192

A soap bubble is blown to a diameter of 7 cm.36960erg7 \mathrm{~cm} .36960 \mathrm{erg} of work is done in blowing it further. If surface tension of soap solution is 40dyne/cm40 \mathrm{dyne} / \mathrm{cm} then the new radius is______ cm\mathrm{cm} Take (π=227)\left(\pi=\frac{22}{7}\right)

Show answer and solution

Answer: 7

ω=ΔU=SΔA36960erg=40 dyne cm8π[(r)2−(72)2]cm2r=7 cm\begin{aligned} & \omega=\Delta \mathrm{U}=\mathrm{S} \Delta \mathrm{A} \\ & 36960 \mathrm{erg}=\frac{40 \text { dyne }}{\mathrm{cm}} 8 \pi\left[(\mathrm{r})^2-\left(\frac{7}{2}\right)^2\right] \mathrm{cm}^2 \\ & \mathrm{r}=7 \mathrm{~cm}\end{aligned}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Fluid mechanics simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Fluid Mechanics (Surface tension), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.