Question

How much mass should be hung from a wire that is 3.14 m long and 1 mm long to increase 1 mm length?
[Young Modulus (Y)=2×10¹¹N/m²]

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Answer 1

Given :

• Length of the wire = 3.14 m
• Diameter of the wire = 1 mm
• Wire growth = 1 mm
• Young modules = 2 × 10¹¹ N/m²

To find :

• Mass of the object.

Solution :

We know the formula of Young modules.

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

✪ Terms identification :-

• Y refers the Young modules.
• m refers the mass of object.
• g refers the acceleration of gravity.
• L refers the length of wire.
• A refers the area of cross section.
• l refers the wire growth.

So,

★ L = 3.14 m

★ Y = 2 × 10¹¹ N/m²

★ g = 9.8 m/s²

★ l = 1 mm = 0.001 m

Diameter of the wire = 1 mm

Then,

Radius(r) = 0.5 mm = 0.0005 m

✪ Now find the area of cross section of the wire ✪

${\boxed{\large{\sf{A=\pi\:r^2}}}}$

A = 3.14 × (0.0005)² m²

→ A = (3.14 × 0.0005×0.0005) m²

✪Now substitute all values in the formula of Young modules✪

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

$\implies\sf{2\times\:{10}^{11}=\frac{m\times\:9.8\times\:3.14}{3.14\times\:0.0005\times\:0.0005\times\:0.001}}\\ \\ \implies\sf{9.8\:m=0.0005\times\:0.0005\times\:0.001\times\:2\times\:{10}^{11}}\\ \\ \implies\sf{m=\frac{0.0005\times\:0.0005\times\:0.001\times\:2\times\:{10}^{11}}{9.8}}\\ \\ \implies\sf{m=\frac{5\times\:5\times\:10}{49}}\\ \\ \implies\sf{m=5.1}$

${\purple{\boxed{\bold{m=5.1\: kg}}}}$

Therefore 5.1 kg should be hung from the wire that is 3.14 m long.