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Standard XII

Physics

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A thick unifo...

Question

A thick uniform rubber rope of density $1.5 g c m^{- 3}$ and Young’s modulus $5 \times 10^{6} N m^{- 7}$ has a length of 8m. When hung from the ceiling of a room, the increase in length of the rope due to its own weight will be

9.6 × 10⁻²m

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19.2 × 10⁻²m

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9.6 × 10⁻³m

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9.6 m

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Solution

The correct option is A
9.6 × 10⁻²m

$Y = \frac{\frac{F}{A}}{\frac{I}{L}} = \frac{F L}{A l} = \frac{(ρ A L) g L}{2 A l} = \frac{ρ g L^{2}}{2 l} \Rightarrow I = \frac{ρ g L^{2}}{2 Y} I = \frac{1.5 \times 10^{3} \times 8^{2}}{2 \times 5 \times 10^{6}} = 9.6 \times 10^{- 2} m$

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A thick uniform rubber rope of density 1.5gcm−3 and Young’s modulus 5×106Nm−7 has a length of 8m. When hung from the ceiling of a room, the increase in length of the rope due to its own weight will be

The correct option is A 9.6 × 10−2m Y=FAIL=FLAl=(ρAL)gL2Al=ρgL22l⇒I=ρgL22YI=1.5×103×822×5×106=9.6×10−2m

A thick uniform rubber rope of density $1.5 g c m^{- 3}$ and Young’s modulus $5 \times 10^{6} N m^{- 7}$ has a length of 8m. When hung from the ceiling of a room, the increase in length of the rope due to its own weight will be

The correct option is A
9.6 × 10⁻²m

$Y = \frac{\frac{F}{A}}{\frac{I}{L}} = \frac{F L}{A l} = \frac{(ρ A L) g L}{2 A l} = \frac{ρ g L^{2}}{2 l} \Rightarrow I = \frac{ρ g L^{2}}{2 Y} I = \frac{1.5 \times 10^{3} \times 8^{2}}{2 \times 5 \times 10^{6}} = 9.6 \times 10^{- 2} m$