Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
        a soap bubble if radius 4 cm and surface tension 30 dyne / cm, is blown at the end of a tube of length 10 cm and internal radius 0.20 cm . if viscousity of air is 1.89 * 10^-4 poise, find the time taken by the bubble to be reduced to a radius of 2 cm.
2 years ago

Sunder Muthukumaran
28 Points
							Let the Volume flowing per unit time be Vt
To find the Vt ,
Use Poiseulle’s Formula  Vt =   $\frac{\Pi Pr^{4}}{8\eta L}$ (L – Length of tube , r – radius of tube . P – excess pressure on bubble )

Then ,
Find the difference in Volume ; $\Delta V$ = $V_{2}- V_{1}$
= $\frac{4\Pi }{3}((R_{2})^{3}-(R_{1})^{3})$
For excess pressure use ;

$\frac{4S}{R}=P$

Now for the time ,
By dividing the $\frac{\Delta V}{V_{t}}$   = Change in voume ÷ Rate of flow   = Time taken
( which is similar to where we divide x/v for t , that is distance / velocity = Time) , (Here velocity is rate of flow of volume , and X is Change in volume )
Use the given values and solve for time .
Regards ,
M,Sunder

2 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Mechanics

View all Questions »

### Course Features

• 101 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions