 ×     #### 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 metal rod is 40.125cm long at 20.0°C and 40.148cm long at 45.0°C.calculate the average coefficient of linear expansion of the rod for this temperature range.

```
6 years ago

```							2.3×10?5
```
6 years ago
```							(40.148-40.125)=coefficint of linear expantion*40.125*(45-20)
(0.023)/(40.125*25)=coefficint of linear expantion
coefficint of linear expantion=0.000022928
therefore coefficint of linear expantion=0.000022928
```
6 years ago
```							given that
Initial length(L1)=40.125
Final length (L2)=40.148
(t1)=20
t2=45
we know that coefficient of linear expansion = (L2-L1)/L1*(t2-t1)
= (40.148-40.125)/40.125*(25)
= (0.023)/1003.125
=2.29*10^-5
```
6 years ago
```							given
Initial length (l1)    =  40.125
final length (l2)     =    40.148
temp (t1)                =20
temp  (t2)               =45
coeffecient  of  l.e =(L2-L1)/L1*(T2-T1)
=(40.148-40125)/40.125*(25)
=(0.023)/1003.125
=2.29*10^-5
```
6 years ago
```							A metal rod of 3 metre expanse by 1.0 CM when heated through 200° Celsius find the linear expansion coefficient of metal
```
one year ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Thermal Physics

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