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Properties of Matter question

2022 · 26 Jul · Shift 2 · Q62

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Properties of Matter
2022 · 26 Jul · Shift 2 · Q62

Properties of Matter question

2022 · 26 Jul · Shift 2 · Q62

JEE MainPhysicsProperties of MatterNumerical+4 / −1

A uniform heavy rod of mass

20 \mathrm{~kg}

, cross sectional area

0.4 \mathrm{~m}^{2}

and length

20 \mathrm{~m}

is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is

x \times 10^{-9} \mathrm{~m}

. The value of

x

is

\underline{\hspace{2cm}}

. (Given, young modulus Y = 2

\times

1011 Nm

-

2 and g = 10 ms

-

2)

Numerical answer

View written solutionFree

Correct answer: 25

Given data

Mass of rod: $M = 20\,\text{kg}$
Cross-sectional area: $A = 0.4\,\text{m}^2$
Length: $L = 20\,\text{m}$
Young's modulus: $Y = 2 \times 10^{11}\,\text{N/m}^2$
Acceleration due to gravity: $g = 10\,\text{m/s}^2$

We need elongation due to the rod's own weight.

Formula for elongation of a heavy uniform rod under its own weight

For a rod hanging vertically under its own weight,

\Delta L = \frac{MgL}{2AY}

This is because the tension varies from zero at the bottom to $Mg$ at the top, so the average tension is $\dfrac{Mg}{2}$ .

Substitute the values

\Delta L = \frac{20 \times 10 \times 20}{2 \times 0.4 \times 2 \times 10^{11}}

First, numerator:

20 \times 10 \times 20 = 4000

Denominator:

2 \times 0.4 \times 2 \times 10^{11} = 1.6 \times 10^{11}

So,

\Delta L = \frac{4000}{1.6 \times 10^{11}}

\Delta L = 2.5 \times 10^{-8}\,\text{m}

Match with the form $x \times 10^{-9}$ m

2.5 \times 10^{-8} = 25 \times 10^{-9}

Hence,

x = 25

Comparison with stored answer

Stored correct answer = $25$

Our derived answer matches the stored answer.

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