A thick rope of rubber of density $1.5 \times 10^{3} k g m^{- 3}$ and Young’s modulus $5 \times 10^{6} N m^{- 2},$ 8m in length is hung from the ceiling of a room, the increase in its length due to its own weight is

9.6 X 10^–2 m

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19.2 X 10^–2 m

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9.6 X 10^–3 m

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9.6 m

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Solution

The correct option is A 9.6 X 10^–2 m
If A is the area of cross section, d be the density of material and l is length of rope
then mass of rope m=Ald
As the weight of the rope acts on the mid point of the rope
$Y = \frac{m g}{A} \times \frac{l}{△ l}$
$△ l = \frac{m g l}{2 A Y} = \frac{A l p g l}{2 A Y} = \frac{g p l^{2}}{2 Y}$
or $△ l = \frac{9.8 \times 1.5 \times 10^{3} \times 8^{2}}{2 \times 5 \times 10^{6}} = 9.6 \times 10^{- 2} m$

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