Question

Answer this question please

Answer 1

Answer:

Let the initial lengths of the rod at 0°C be L1 and

L2

Since the difference in lengths at all temperature

be X

Therefore L2-L1 = X

L 1 = 2*10-5= 20

L2 = 5*10-5 = 50

Therefore L1 ( 1 + A1 ∆) - L2 ( 1+ A2 ∆ )

= L1 A1 = L2 A2

=20L1 = 50 L2

Therefore L2-L1

= 50 L2 - 20 L1

= 30