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 uniform chain of length L and mass M is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If g is the acceleration due to gravity, work required to pull the hanging part into the table is: 	MgL	MgL/3	MgL/9	MgL/18`
one year ago

Abhishek Anand
40 Points
```							As L/3 length is variable. So we assume it `x`.Then mass of hanging chain be M(x)/L. This is also a variable force by which we act on chain for take whole chain on table. As this force is variable so work done is W=(int)Fdx, where (int=integration).    Then equation be  W=(int) {Mg(x)/L}dx  where integration is from 0 to L/3       ,After integration, W=Mg(x^2/2)/L    ,Then put limit 0 to L/3 , we get MgL/18 , which is our answer ....Option D is correct.
```
one year ago
Abhishek Anand
40 Points
```							As L/3 length is variable. So we assume it `x`.Then mass of hanging chain be M(x)/L. Force is Mg(x)/L which is also a variable force by which we act on chain for take whole chain on table. As this force is variable so work done is W=(int)Fdx, where (int=integration). Then equation be W=(int) {Mg(x)/L}dx where integration is from 0 to L/3 ,After integration, W=Mg(x^2/2)/L ,Then put limit 0 to L/3 , we get MgL/18 , which is our answer ....Option D is correct.
```
one year 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