## Pages

### Mechanical Properties of solids NCERT Solutions Class 11 Physics - Solved Exercise Question 9.1

Question 9.1:
A steel wire of length 4.7 m and cross-sectional area 3.0 × 10–5 m2 stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10–5 m2 under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
Solution:
Length of the steel wire, L1 = 4.7 m
Area of cross-section of the steel wire, A1 = 3.0 × 10–5 m2
Length of the copper wire, L2 = 3.5 m
Area of cross-section of the copper wire, A2 = 4.0 × 10–5 m2
Change in length = ΔL1 = ΔL2 = ΔL
Force applied in both the cases = F
Young’s modulus of the steel wire:
Y1 = (F1 / A1) (L1 / ΔL1)
= (F / 3 X 10-5) (4.7 / ΔL)     ....(i)
Young’s modulus of the copper wire:
Y2 = (F2 / A2) (L2 / ΔL2)
= (F / 4 X 10-5) (3.5 / ΔL)     ....(ii)
Dividing (i) by (ii), we get:
Y1 / Y2  =  (4.7 X 4 X 10-5) / (3 X 10-5 X 3.5)
= 1.79 : 1
The ratio of Young’s modulus of steel to that of copper is 1.79 : 1.