Question 72: Properties of Matter

Question

The Young's modulus of a steel wire of length $$6 \mathrm{~m}$$ and cross-sectional area $$3 \mathrm{~mm}^{2}$$, is $$2 	imes 10^{11}~\mathrm{N} / \mathrm{m}^{2}$$. 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 $$=10 \mathrm{~m} / \mathrm{s}^{2}$$) :

Accepted Answer

The elongation of the wire can be calculated using the formula for stress and strain. The stress in the wire is given by:
$$\sigma = \frac{mg}{A}$$
where m is the mass of the block (4 kg), g is the acceleration due to gravity on the planet (1/4 of its value on the earth, or 2.5 m/s
2
), and A is the cross-sectional area of the wire (3 mm
2
).
The strain in the wire is given by:
$$\epsilon = \frac{\Delta L}{L}$$
where ÃÂL is the elongation of the wire and L is the original length of the wire (6 m).
Using Hooke's law, which states that stress is proportional to strain, we can find the elongation of the wire:
$$\sigma = Y\epsilon$$
where Y is the Young's modulus of the wire (2 $$ 	imes $$ 10
11
N/m
2
).
Combining the above equations, we can find the elongation of the wire:
$$\epsilon = \frac{\sigma}{Y} = \frac{mg}{A Y} = \frac{4 	imes 2.5}{3 	imes 10^{-6} 	imes 2 	imes 10^{11}} = \frac{5}{3 	imes 10^{-6} 	imes 10^{11}} = \frac{5}{3 	imes 10^{5}}$$
$$ 	herefore $$ $$\frac{\Delta L}{L}$$ $$= \frac{5}{3 	imes 10^{5}}$$
$$ \Rightarrow $$ $$\Delta L = {{5 	imes 6} \over {3 	imes {{10}^5}}}$$ = $${1 \over {{{10}^4}}}$$ = 0.1 mm
So, the elongation of the wire is 0.1 mm.

Answer

0.1 cm

Answer

1 cm

Answer

0.1 mm

Answer

1 mm

Properties of Matter

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