#### Thank you for registering.

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

Click to Chat

1800-1023-196

+91 7353221155

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.

sivakrishna
37 Points
7 years ago
2.3×10?5
sireesh
11 Points
7 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
sivakrishna
37 Points
7 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
MTAMBIRAJ
37 Points
7 years ago
given Initial length (l1) = 40.125 final length (l2) = 40.148 temp (t1) =20 temp (t2) =45 we had know that coeffecient of l.e =(L2-L1)/L1*(T2-T1) =(40.148-40125)/40.125*(25) =(0.023)/1003.125 =2.29*10^-5
Bhumika
13 Points
2 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