×

#### 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
```
An aluminium rod has a breaking strain 0.2%. The minimum cross sectional area of the rod in m 2 in order to supporta load of 10 4 N is if young's modulus is 7×10 9 Nm -2
An aluminium rod has a breaking strain 0.2%. The minimum cross sectional area of the rod in m2 in order to supporta load of 104N is if young's modulus is 7×109Nm-2

```
4 years ago

Vikas TU
14146 Points
```							From Young’s Modulus formulae:Y =  Fl’/Alwhere l’ = 2l/1000putting all the values we get,7 x 10^9 = 10^4 * 2l/1000*A*lsolving we get,A = 1/1400 m^2.
```
4 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on General Physics

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

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