Two 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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$A$ metal ball initially at a pressure of $10^5 \ Pa$ is heated from $20^{\circ} C$ to $127^{\circ} C$ while keeping its volume constant. The coefficient of linear expansion of the metal is $10^{-5} \ {^{\circ} C}^{-1}$ and the bulk modulus of the metal is $2 \times 10^{11} \ N \ m^{-2}$. The final pressure inside the ball becomes:
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?
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View SolutionTwo metal rods $A$ and $B$ each of length $50 \ cm$ and diameter $4.0 \ mm$ are joined together at temperature $30^{\circ} C$. What is the change in length of the combined rod at $230^{\circ} C$ (in $mm$)? [Given linear expansion coefficients of rods $A$ and $B$ are respectively,$2.0 \times 10^{-5} /^{\circ} C$ and $1.0 \times 10^{-5} /^{\circ} C$]
$A$ sheet of steel is $40 \ cm$ long and $5 \ cm$ broad at $0^{\circ} C$. The surface area of the sheet increases by $1.4 \ cm^2$ at $100^{\circ} C$. The coefficient of linear expansion of steel is:
To increase the length of a metal rod by $0.4 \%$,the temperature of the rod is to be increased by (Coefficient of linear expansion of the metal $= 20 \times 10^{-6} \ {}^{\circ}C^{-1}$) (in $K$)
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