Question

A weight of 20 kg is suspended from a wire. The area of cross-section of the wire is 1 mm' and when
stretched, the length is exactly 6 m. When the weight is removed, the length reduces to 5.995 m.
Calculate the Young's modulus (Y) of the material of the wire.​

Answer 1

Answer:

2.4x

${10}^{11}$

Explanation:

$y = fl \div ae$

=200x6÷

$1200 \div {10}^{ - 6 } \times 5 \times {10}^{ - 3}$

Answer 2

The Young's modulus of the material of the wire is $2.398\times10^{10}\ N/m^2$

Explanation:

Given that,

Weight = 20 kg

Area of cross section = 1 mm²

Stretched length = 6 m

Removed length = 5.995 m

We need to calculate the Young's modulus of the wire

Using formula of young's modulus

$Y=\dfrac{stress}{strain}$

$Y=\dfrac{mg}{A}\times\dfrac{l}{\Delta l}$

Put the value into the formula

$Y=\dfrac{20}{1\times10^{-6}}\times\dfrac{5.995}{5\times10^{-3}}$

$Y=2.398\times10^{10}\ N/m^2$

Hence, The Young's modulus of the material of the wire is $2.398\times10^{10}\ N/m^2$

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Topic : young's modulus

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