Class 11Physics1 Thermal Properties of Matter: Thermometry, Expansion and CalorimetryThermal Expansion for Solid
MediumTwo rods,one of aluminium and the other of steel,having initial lengths $L_1$ and $L_2$ are connected together to form a single rod of length $(L_1+L_2)$. The coefficients of linear expansion of aluminium and steel are $\alpha_1$ and $\alpha_2$ respectively. If the length of each rod increases by the same amount,when their temperatures are raised by $t^{\circ}C$,then the ratio $\frac{L_1}{L_1+L_2}$ will be
- A$\frac{\alpha_2}{\alpha_1}$
- B$\frac{\alpha_1}{\alpha_2}$
- C$\frac{\alpha_2}{(\alpha_1+\alpha_2)}$
- D$\frac{\alpha_1}{(\alpha_1+\alpha_2)}$
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Similar Questions
$A$ metal rod of length $20 \, cm$ expands by $0.075 \, cm$ when its temperature is raised from $0 \, ^\circ C$ to $100 \, ^\circ C$. Another rod $B$ of a different metal of the same length expands by $0.045 \, cm$ for the same rise in temperature. $A$ third rod of the same length is composed of two metals $A$ and $B$ joined together. This composite rod expands by $0.060 \, cm$ for the same temperature difference. The length of the portion of metal $A$ is ..... $cm$.
Difficult
View Solution$A$ sheet of steel at $20^{\circ} C$ has dimensions as shown in the figure. If the coefficient of linear expansion for steel is $10^{-5} {^{\circ} C}^{-1}$, what is the change in the area at $60^{\circ} C$ (in $\text{ cm}^2$)?
We would like to make a vessel whose volume does not change with temperature. We can use brass and iron $\left( {{\gamma _{{\text{brass}}}} = 6 \times {{10}^{ - 5}}/K} \right.$ and $\left. {{\gamma _{{\text{iron}}}} = 3.55 \times {{10}^{ - 5}}/K} \right)$ to create a volume of $100 \, cc$. How can you achieve this?
Difficult
View SolutionWhat must be the lengths of steel and copper rods at $0^o C$ for the difference in their lengths to be $10\,cm$ at any common temperature? $(\alpha_{steel}=1.2 \times 10^{-5} \;^o C^{-1})$ and $(\alpha_{copper} = 1.8 \times 10^{-5} \;^o C^{-1})$
Difficult
View Solution$A$ bar of iron is $10 \, cm$ at $20^{\circ}C$. At $19^{\circ}C$ it will be ($\alpha$ of iron $= 11 \times 10^{-6}/^{\circ}C$)
Medium
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