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  • The Young's modulus of a steel wire of length and cross-sectional area <img alt="3 \mat

The Young's modulus of a steel wire of length 6 \mathrm{~m} and cross-sectional area 3 \mathrm{~mm}^2, is 2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2. The wire is suspended from its support on a given planet. A block of mass 4 \mathrm{~kg} is attached to the free end of the wire. The acceleration due to gravity on the planet is \frac{1}{4} of its value on the earth. The elongation of wire is (Take g on the earth =10 \mathrm{~m} / \mathrm{s}^2 ):

Option: 1

0.1 cm


Option: 2

0.1 mm


Option: 3

1 cm


Option: 4

1 mm


Answers (1)

best_answer

As we know,
$$ \begin{gathered} Y=\frac{\text { stress }}{\text { strain }} \\ Y=\frac{\mathrm{FL}}{\mathrm{A} \Delta \mathrm{L}} \end{gathered}
Given: Y=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2
$$ \begin{array}{ll} \mathrm{L}=6 \mathrm{~m} & \mathrm{~g}_{\mathrm{p}}=\frac{\mathrm{g}}{4} \\ \mathrm{~A}=3 \mathrm{~mm}^2 \\ \mathrm{M}=4 \mathrm{~kg} \\ \mathrm{~F}=\mathrm{mg}_{\mathrm{p}} \\ \mathrm{F}=4 \times \frac{10}{4}=10 \mathrm{~N} \end{array}
Hence 2 \times 10^{11}=\frac{10 \times 6}{3 \times 10^{-6} \times \Delta \mathrm{L}}
$$ \Delta \mathrm{L}=0.1 \mathrm{~mm}

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avinash.dongre

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