Question

The length of two wires made of different materials is equal. One is 2 mm in diameter and the other is 4mm in diameter. Pulling both by equal force causes the length of the first wire to double that of the second wire. Compare their Young modulus.

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Answer 1

Given :

• The length of two wires made of different materials is equal.
• One is 2 mm in diameter and the other is 4 mm in diameter.
• Pulling both by equal force causes the length of the first wire to double that of the second wire.

To find :

• Compare their Young's modulus.

Formula Used :

${\boxed{\large{\purple{\bold{Y=\dfrac{FL}{Al}}}}}}$

• Terms identification :-
• Y refers the young's modules.
• F refers force.
• L refers the length of wire.
• A refers the area of cross section.
• l refers the wire growth.

Solution :

✪The length of two wires made of different materials is equal.✪

So,

$\sf{Length\:of\: first\: wire=Length\:of\: second\:wire=L\:mm}$

• Diameter of first wire = 2 mm.

• Radius of first wire = 1 mm.

• Diameter of second wire = 4 mm.

• Radius of second wire = 2 mm.

✪ Force of two wires are equal ✪

So,

$\sf{Force\:of\: first\: wire=Force\:of\: second\:wire=F}$

Consider,

• Growth of 2 nd wire = l mm.
• Growth of 1st wire = 2l mm.

Let the young's modulus of 1st wire be $\sf{Y_1}$ and the young's modulus of 2nd wire be $\sf{Y_2}$.

➪ In case of 1st wire ,

$\sf{Y_1=\frac{FL}{3.14\times\:1^2\times\:2l}}$

$\implies\sf{Y_1=\frac{FL}{3.14\times\:2l}}$

➪ In case of 2nd wire ,

$\sf{Y_2=\frac{FL}{3.14\times\:2^2\times\:l}}$

$\implies\sf{Y_2=\frac{FL}{3.14\times\:4l}}$

★ Now compare their young modulus ★

$\sf{Y_1 : Y_2}$

$\implies\sf{\frac{FL}{3.14\times\:2l}:\frac{FL}{3.14\times\:4l}}$

$\implies\sf{\frac{FL}{3.14\times\:2l}\times\frac{3.14\times\:4l}{FL}}$

$\implies\sf{\frac{2}{1}}$

$\implies\sf{2:1}$

Therefore, the ratio of their young's modulus is 2:1 .