Question

The length of the wire is increased by 5 mm by hanging an object of mass 10 kg at one end of a 5m long wire. The young modulus of the wire component is 9.8×10¹¹dyn /cm². What is the area of cross section of the wire.

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Answer 1

Given :

• The length of the wire is increased by 5 mm by hanging an object of mass 10 kg at one end of a 5 m long wire.
• The young modulus of the wire component is 9.8×10¹¹ dyn/cm².

To find :

• Area of cross section of the wire.

Solution :

We know the formula of Young modulus.

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

Terms identification :-

• L refers the length of wire.
• M refers the mass of object.
• I refers the wire growth.
• Y refers the Young modulus.
• g refers the acceleration of gravity.
• A refers the area of cross section.

Here,

★ L = 5 m = 500 cm

★ m = 10 kg = 10000 g

★ l = 5 mm = 0.5 cm

★Y = 9.8 × 10¹¹ dyn/cm²

★ g = 9.8 m/s² = 980 cm/s²

${\large{\sf{\:\:\:\:\:\:\:\:\:\:\:\:\:\:Y=\dfrac{mgL}{Al}}}}$

$\implies\sf{9.8\times\:{10}^{11}=\frac{10000\times\:980\times\:500}{A\times\:0.5}}\\ \\ \implies\sf{A\times\:0.5\times\:9.8\times\:{10}^{11}=10000\times\:980\times\:500}\\ \\ \implies\sf{A=\frac{10000\times\:980\times\:500}{0.5\times\:9.8\times\:{10}^{11}}}\\ \\ \implies\sf{A=\frac{1}{100}}$

So,

${\boxed{\green{\large{\bold{A=\dfrac{1}{100}\:cm^2}}}}}$

For transferring into mm² , we need to multiply by 100.

$\sf{A=\frac{1}{100}\:cm^2}\\ \implies\sf{A=\frac{1}{100}\times\:100\:mm^2}\\ \implies\sf{A=1\:mm^2}$

${\boxed{\bold{Area\: of\: cross\:section=1\:mm^2}}}$

Therefore, the area of the cross section of the wire is 1 mm².