The coefficient of linear expansion of iron is 0-000011 - K - An iron rod is 10 metre long at 27 - C - The length of the rod will be decreased by 1-1 mm when the temperature of the rod changes to-

The correct option is C 17 ^ ∘ C [ Δ l=α l_ o Δ T; ⇒Δ T=Δ l /α l_ o = 1.1× 10 ^ 3 / 1.1× 10 ^ 6 × 10 =1 ...

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Standard XII

Physics

Area and Volume Expansion

The coefficie...

Question

The coefficient of linear expansion of iron is ${0.000011}^{\circ} K .$ An iron rod is $10$ metre long at $27^{\circ} C .$ The length of the rod will be decreased by $1.1 m m$ when the temperature of the rod changes to-

0^{\circ} C

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10^{\circ} C

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17^{\circ} C

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20^{\circ} C

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Solution

The correct option is C $17^{\circ} C$
$\begin{matrix} Δ l = α l_{o} Δ T \Rightarrow Δ T = \frac{Δ l}{α l_{o}} = \frac{1.1 \times 10^{- 3}}{1.1 \times 10^{- 6} \times 10} = 10 T = 27^{\circ} - 10^{\circ} = 17^{\circ} C A n s . (C) \end{matrix}$

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The coefficient of linear expansion of iron is 0.000011∘K. An iron rod is 10 metre long at 27∘C. The length of the rod will be decreased by 1.1mm when the temperature of the rod changes to-

The correct option is C 17∘CΔl=αloΔT⇒ΔT=Δlαlo=1.1×10−31.1×10−6×10=10T=27∘−10∘=17∘CAns.(C)

The coefficient of linear expansion of iron is ${0.000011}^{\circ} K .$ An iron rod is $10$ metre long at $27^{\circ} C .$ The length of the rod will be decreased by $1.1 m m$ when the temperature of the rod changes to-

The correct option is C $17^{\circ} C$
$\begin{matrix} Δ l = α l_{o} Δ T \Rightarrow Δ T = \frac{Δ l}{α l_{o}} = \frac{1.1 \times 10^{- 3}}{1.1 \times 10^{- 6} \times 10} = 10 T = 27^{\circ} - 10^{\circ} = 17^{\circ} C A n s . (C) \end{matrix}$