The variation of lengths of two metal rods A and B with change in temperature are as shown in the figure. The ratio of (αB/αA) is (Here αA= Linear coefficient of thermal expansion of rod A, αB= Linear coefficient of thermal expansion of rod B )

Question

The variation of lengths of two metal rods A and B with change in temperature are as shown in the figure. The ratio of (αB/αA) is (Here αA= Linear coefficient of thermal expansion of rod A, αB= Linear coefficient of thermal expansion of rod B ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

l=l0(1+α Δ t) ⇒ l-l0=l0 α Δ T ⇒ (l0 αA Δ T/l0 αB Δ T)=(104-100/106-100)=(4/6) ⇒ (αA/αB)=(2/3)

Q. The variation of lengths of two metal rods $A$ and $B$ with change in temperature are as shown in the figure. The ratio of $\frac{α _{B}}{α _{A}}$ is
(Here $α_{A} =$ Linear coefficient of thermal expansion of rod $A, α_{B} =$ Linear coefficient of thermal expansion of rod $B$ )

$l = l_{0} (1 + α Δ t)$
$\Rightarrow l - l_{0} = l_{0} α Δ T$
$\Rightarrow \frac{l _{0} α _{A} Δ T}{l _{0} α _{B} Δ T} = \frac{104 - 100}{106 - 100} = \frac{4}{6}$
$\Rightarrow \frac{α _{A}}{α _{B}} = \frac{2}{3}$

Q. The variation of lengths of two metal rods A and B with change in temperature are as shown in the figure. The ratio of αA​αB​​ is (Here αA​= Linear coefficient of thermal expansion of rod A,αB​= Linear coefficient of thermal expansion of rod B )

l=l0​(1+αΔt) ⇒l−l0​=l0​αΔT ⇒l0​αB​ΔTl0​αA​ΔT​=106−100104−100​=64​ ⇒αB​αA​​=32​

Q. The variation of lengths of two metal rods $A$ and $B$ with change in temperature are as shown in the figure. The ratio of $\frac{α _{B}}{α _{A}}$ is
(Here $α_{A} =$ Linear coefficient of thermal expansion of rod $A, α_{B} =$ Linear coefficient of thermal expansion of rod $B$ )

$l = l_{0} (1 + α Δ t)$
$\Rightarrow l - l_{0} = l_{0} α Δ T$
$\Rightarrow \frac{l _{0} α _{A} Δ T}{l _{0} α _{B} Δ T} = \frac{104 - 100}{106 - 100} = \frac{4}{6}$
$\Rightarrow \frac{α _{A}}{α _{B}} = \frac{2}{3}$