- Home
- Standard 11
- Physics
10-1.Thermometry, Thermal Expansion and Calorimetry
hard
The coefficient of linear expansion of brass and steel are ${\alpha _1}$ and ${\alpha _2}$. If we take a brass rod of length ${l_1}$ and steel rod of length ${l_2}$ at $0°C$, their difference in length $({l_2} - {l_1})$ will remain the same at a temperature if
A
${\alpha _1}{l_2} = {\alpha _2}{l_1}$
B
${\alpha _1}l_2^2 = {\alpha _2}l_1^2$
C
$\alpha _1^2{l_1} = \alpha _2^2{l_2}$
D
${\alpha _1}{l_1} = {\alpha _2}{l_2}$
Solution
(d) ${L_2} = {l_2}(1 + {\alpha _2}\Delta \theta )$ and ${L_1} = {l_1}(1 + {\alpha _1}\Delta \theta )$
$ \Rightarrow ({L_2} – {L_1}) = ({l_2} – {l_1}) + \Delta \theta ({l_2}{\alpha _2} – {l_1}{\alpha _1})$
Now $({L_2} – {L_1}) = ({l_2} – {l_1})$ so, ${l_2}{\alpha _2} – {l_1}{\alpha _1} = 0$
Standard 11
Physics