Question

Two wires of a given material have length L and nL while their areas of cross section are a and A/n^2 the ratio of Young's modulus for 2 wires would be what?.

Answer 1

Answer:

Applying a force F to stretch is applied on both wires simultaneously so we can directly add the elongation of each wire with force F, so,

Y = \dfrac{F/A}{\triangle L_{1} / L }Y=

△ L

1

​

/L

F/A

​

for wire with youngs modulus YY

2Y = \dfrac{F/A}{\triangle L_{2} / L } 2Y=

△ L

2

​

/L

F/A

​

for wire with youngs modulus 2Y2Y

\triangle L_{1} = \dfrac{F L}{Y A }△L

1

​

=

Y A

F L

​

\triangle L_{2} = \dfrac{F L}{2YA}△L

2

​

=

2YA

F L

​

\triangle L_{1} + \triangle L_{2} = \dfrac{F L}{Y A} + \dfrac{F L}{2YA} = \dfrac{3F L}{2YA}△L

1

​

+△ L

2

​

=

Y A

FL

​

+

2YA

F L

​

=

2YA

3F L

Answer 2

Answer:

yes it's my question also

but i didn't get the right answer