The co-efficient of linear expansion of iron is 11-180 of volume coefficient of expansion of mercury which is 18- 10-5-0C- An iron rod is 10m long at 27 0 C- The length of the rod will be decreased by 1-1 mm when the temperature of the rod changes by

The correct option is B 10 ^0 C α_rod=11180× 18× 10^ 5/^∘C=1.1× 10^ 5/^∘C Change in length of rod ,Δ l=l_0αΔθ 1.1× 10^ 3=10× 1.1× 10^ 5×Δθ ...

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Standard XII

Physics

Thermal Expansion

The co-effici...

Question

The co-efficient of linear expansion of iron is 11/180 of volume coefficient of expansion of mercury which is $18 \times 10^{- 5 / 0} C$ . An iron rod is 10m long at 27 $^{0}$ C. The length of the rod will be decreased by 1.1 mm when the temperature of the rod changes by

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Solution

The correct option is B 10 $^{0}$ C
$α_{r o d} = \frac{11}{180} \times 18 \times 10^{- 5} /^{\circ} C = 1.1 \times 10^{- 5} /^{\circ} C$
Change in length of rod $, Δ l = l_{0} α Δ θ$
$⟹ 1.1 \times 10^{- 3} = 10 \times 1.1 \times 10^{- 5} \times Δ θ$
$⟹ Δ θ = 10^{\circ} C$

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The co-efficient of linear expansion of iron is 11/180 of volume coefficient of expansion of mercury which is 18×10−5/0C. An iron rod is 10m long at 270 C. The length of the rod will be decreased by 1.1 mm when the temperature of the rod changes by

The correct option is B 100Cαrod=11180×18×10−5/∘C=1.1×10−5/∘CChange in length of rod,Δl=l0αΔθ⟹1.1×10−3=10×1.1×10−5×Δθ⟹Δθ=10∘C

The co-efficient of linear expansion of iron is 11/180 of volume coefficient of expansion of mercury which is $18 \times 10^{- 5 / 0} C$ . An iron rod is 10m long at 27 $^{0}$ C. The length of the rod will be decreased by 1.1 mm when the temperature of the rod changes by

The correct option is B 10 $^{0}$ C
$α_{r o d} = \frac{11}{180} \times 18 \times 10^{- 5} /^{\circ} C = 1.1 \times 10^{- 5} /^{\circ} C$
Change in length of rod $, Δ l = l_{0} α Δ θ$
$⟹ 1.1 \times 10^{- 3} = 10 \times 1.1 \times 10^{- 5} \times Δ θ$
$⟹ Δ θ = 10^{\circ} C$