Length of an iron rod at 270C is 10 m -11 - 10-6

The correct option is B 7^0C l = 10 m #10; #10; T = 27^0c #10; #10; l_2 = l_1( l + αΔ T #10;) #10; #10; l_2 = l_1 + αl_1Δ ...

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Byju's Answer

Standard XII

Physics

Multiplication & Division of Errors

Length of an ...

Question

Length of an iron rod at $27^{0} C$ is $10 m$ $α = 11 \times 10^{- 6}$ . To decrease the length of the rod by 1.1 mm, a temperature of the rod should be decreased to:

0^{0} C

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7^{0} C

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17^{0} C

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10^{0} C

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Solution

The correct option is B $7^{0} C$
$l = 10 m$

$T = 27^{0} c$

$l_{2} = l_{1} (l + α Δ T)$

$l_{2} = l_{1} + α l_{1} Δ T$

$\frac{(l_{2} - l_{1})}{l_{1}} = α (T_{2} - T_{1})$

$\frac{1.1 \times 10^{- 3}}{10} = 11 \times 10^{- 6} (T_{2} - 27)$

$T_{2} = 7^{0} c$

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An Iron rod and a copper rod lie side by side. As the temperature is changed, the difference in the lengths of the rods remains constant at a value of 10 cm. Find the lengths at $0^{\circ} C$ . Coefficients of linear expansion of iron and copper are $1.1 \times 10^{- 6} /^{\circ} C$ and $1.7 \times 10^{- 5} /^{\circ} C$ respectively.

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Length of an iron rod at 270C is 10m α=11×10−6 . To decrease the length of the rod by 1.1 mm, a temperature of the rod should be decreased to:

The correct option is B 70Cl=10m T=270c l2=l1(l+αΔT) l2=l1+αl1ΔT (l2−l1)l1=α(T2−T1) 1.1×10−310=11×10−6(T2−27) T2=70c

Length of an iron rod at $27^{0} C$ is $10 m$ $α = 11 \times 10^{- 6}$ . To decrease the length of the rod by 1.1 mm, a temperature of the rod should be decreased to:

The correct option is B $7^{0} C$
$l = 10 m$

$T = 27^{0} c$

$l_{2} = l_{1} (l + α Δ T)$

$l_{2} = l_{1} + α l_{1} Δ T$

$\frac{(l_{2} - l_{1})}{l_{1}} = α (T_{2} - T_{1})$

$\frac{1.1 \times 10^{- 3}}{10} = 11 \times 10^{- 6} (T_{2} - 27)$

$T_{2} = 7^{0} c$