×

#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

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 liquid soap film in shape of a plane loop has an initial area 0.05 m2. If its area is slowly doubled then the increase in its surface potential energy from its initial value will be (Surface tension of liquid = 0.2 N/m).

```
9 years ago

509 Points
```							dE  = Tds
T = surface tension , ds change in area
=(0.05)2(0.2) = 2*10-3 joules
```
9 years ago
SATYAM JAISWAL
15 Points
```							U= Ts×Area
=2×0.2×0.05= 2×10-2J

This is correct answer . In this question we take a change in area and we multiplied by 2 because soap drop has two layer and surface tension is given so easily you will get answer.

Regards,
Satyam Jaiswal

```
2 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Discuss with colleagues and IITians

View all Questions »

### Course Features

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

### Course Features

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