First slide

Elastic Modulie

Question

A steel wire of length 4.5 m and cross-sectional area 3 x $10^{- 5}$ $m^{2}$ stretches by the same amount as a copper wire of length 3.5 m and cross sectional area of $4 \times 10^{- 5} m^{2}$ under a given load. The ratio of the Young's modulus of steel to that of copper is

Moderate

A

1.3
B

1.5
C

1.7
D

1.9

Solution

For copper wire, $L_{C} = 3.5 m,$
$A_{C} = 4 \times 10^{- 5} m^{2}$
For steel wire, LS = 4.5 m, $A_{S} = 3 \times 10^{- 5} m^{2}$
As Young's modulus, $Y = \frac{(\frac{F}{A})}{(\frac{∆ L}{L})}$
As applied force ,F and extension $∆ L$ are same for steel and copper wires
$∴ \frac{F}{∆ L} = \frac{Y_{S} A_{S}}{L_{S}} = \frac{Y_{C} A_{C}}{L_{C}}$
where the subscripts C and S steel refers to copper and steel respectively.
$∴ \frac{Y_{S}}{Y_{C}} = \frac{L_{S}}{L_{C}} \times \frac{A_{C}}{A_{S}} = \frac{(4.5 m) \times (4 \times 10^{- 5} m^{2})}{(3.5 m) \times (3 \times 10^{- 5} m^{2})} = 1.7$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App