×

#### 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 uniform chain of mass 3m and length 3l is placed on a smooth pulley as shown. Due to slight disturbance right side chain has started to move down, then reaction of pulley when right side part of chain has length 3l/2

```
3 years ago

Arun
24735 Points
```							Weight of chain of length "x" =(M/Lx)g Amount of work done in moving dx to the topdW = F→. dx→       =F.dx   = (MLx)gdx The total amount of work done in moving the one third length of the hanging chain on the table will beW=∫L/3 - 0 MLxgdx = M / Lg ∫L / 3 - 0 xdx = M/ Lg[x^2 / 2] L/3 - 0 =MgL / 18 Hence the work required to pull the hanging part on the table is MgL / 18.
```
4 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on IIT JEE Entrance Exam

View all Questions »

### Course Features

• 731 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