Question

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. What is the determine the area of ​​cross section of the wire.​

Don't spam !!

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 :

•Length of the wire = L

•Mass of the object = m

•Wire growth = l

•Young's modules = Y

•Acceleration of gravity = g

★ 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

We know the formula of Young modules.

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

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

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

So,

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

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

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

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