146612
The two metal rods $A$ and $B$ are having their initial lengths in the ratio 2:3 and coefficient of linear expansion in the ratio $3: 4$. When they are heated through same temperature difference, the ratio of linear expansions is:
1 $1: 2$
2 $2: 3$
3 $3: 4$
4 $4: 3$
Explanation:
A Given that, Ratio of initial length $=\frac{2}{3}$ Ratio of linear expansion $=3 / 4$ As we know that, $\frac{\Delta l_{1}}{\Delta l_{2}}=\frac{l_{1} \alpha_{1}}{l_{2} \alpha_{2}}$ $=\frac{2}{3} \cdot \frac{3}{4}$ $\frac{\Delta l_{1}}{\Delta l_{2}}=\frac{1}{2}$ The ratio of linear expansion is 1:2.
AP EAMCET(Medical)-2000
Thermal Properties of Matter
146613
Density of a substance at $0^{\circ} \mathrm{C}$ is $10 \mathrm{~g} / \mathrm{cc}$ and at $100^{\circ} \mathrm{C}$ its density is $9.7 \mathrm{~g} / \mathrm{cc}$. The coefficient of linear expansion of the substance is :
1 $10^{-4}$
2 $3 \times 10^{-4}$
3 $19.7 \times 10^{-3}$
4 $10^{-3}$
Explanation:
A Given that, Density of a substance at $0^{\circ} \mathrm{C}=10 \mathrm{~g} / \mathrm{cc}$ Density of substance at $100^{\circ} \mathrm{C}=9.7 \mathrm{~g} / \mathrm{cc}$ As we know that, Coefficient of volume expansion, $\gamma=\frac{\Delta \rho}{\rho \cdot \Delta \mathrm{T}}$ Temperature difference, $\Delta \mathrm{T}=100-0$ $=100^{\circ} \mathrm{C}$ $\gamma=\frac{\rho_{0}-\rho_{100}}{\rho_{0} \cdot\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)}$ $=\frac{10-9.7}{10(100-0)}$ $=\frac{0.3}{10^{3}}$ $\gamma=3 \times 10^{-4} /{ }^{\circ} \mathrm{C}$ Coefficient of linear expansion, $\alpha=\frac{\gamma}{3}$ $=\frac{3 \times 10^{-4}}{3}$ $=10^{-4} /{ }^{\circ} \mathrm{C}$
AP EAMCET(Medical)-1998
Thermal Properties of Matter
146614
A $2 \mathrm{~m}$ long $\mathrm{Al}$ pipe at $27^{\circ} \mathrm{C}$ is heated until it is $0.0024 \mathrm{~m}$, at $77^{\circ} \mathrm{C}$. The coefficient of linear expansion of $\mathrm{Al}$ is
1 $2.4 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
2 $1.4 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
3 $2.4 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
4 $1.4 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
Explanation:
A Given that, Length of pipe $(l)=2 \mathrm{~m}$. Change in length $(\Delta l)=0.0024 \mathrm{~m}$ Temperature Variation $(\Delta \mathrm{T})=(77-27)^{\circ} \mathrm{C}$ $=50^{\circ} \mathrm{C}$ We know that, $\Delta l=l \alpha \Delta \mathrm{T}$ $0.0024=2 \times \alpha \times(77-27)$ $0.0024=2 \alpha \times 50$ $2.4 \times 10^{-3}=100 \alpha$ $\alpha=\frac{2.4 \times 10^{-3}}{100}$ $=2.4 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
EAMCET-1992
Thermal Properties of Matter
146615
A glass vessel just holds $50 \mathrm{~g}$ of a liquid at $0^{\circ} \mathrm{C}$. If the coefficient of linear expansion of glass is $8 \times 10^{-4} /{ }^{\circ} \mathrm{C}$. The mass of the liquid, it holds at $50^{\circ} \mathrm{C}$ is
1 $46 \mathrm{~g}$
2 $48 \mathrm{~g}$
3 $56 \mathrm{~g}$
4 $42 \mathrm{~g}$
Explanation:
C Given that, $\text { Mass of vessel }\left(\mathrm{M}_{1}\right)=50 \mathrm{~g}$ Coefficient of Linear Expansion $(\alpha)=8 \times 10^{-4} /{ }^{\circ} \mathrm{C}$ Change in temperature, $(\Delta \mathrm{T})=(50-0)^{\circ} \mathrm{C}$ $=50^{\circ} \mathrm{C}$ We know that, Coefficient of volume expansion, $\gamma=\frac{\Delta \mathrm{V}}{\mathrm{V}_{1} \Delta \mathrm{T}}$ $\gamma=\frac{\mathrm{V}_{1}-\mathrm{V}_{2}}{\mathrm{~V}_{1}(\Delta \mathrm{T})}$ $\mathrm{V}_{2}=\mathrm{V}_{1}(1+\gamma \Delta \mathrm{T})$ $\frac{\mathrm{M}_{2}}{\rho}=\frac{\mathrm{M}_{1}}{\rho}(1+3 \alpha \cdot \Delta \mathrm{T})\left(\mathrm{M}_{2}=\text { mass of liquid }\right)$ $\mathrm{M}_{2}=50\left[1+\left(3 \times 8 \times 10^{-4} \times 50\right)\right]$ $\quad=56.0 \mathrm{~g}$
EAMCET-1996
Thermal Properties of Matter
146616
Coefficient of real expansion of mercury is 0.18 $\times 10^{-3} /{ }^{\circ} \mathrm{C}$. If the density of mercury at $0^{\circ} \mathrm{C}$ is $13.6 \mathrm{~g} / \mathrm{cc}$, its density at $473 \mathrm{~K}$ will be
1 $13.11 \mathrm{~g} \mathrm{cc}^{-1}$
2 $13.65 \mathrm{~g} \mathrm{cc}^{-1}$
3 $13.51 \mathrm{~g} \mathrm{cc}^{-1}$
4 $13.22 \mathrm{~g} \mathrm{cc}^{-1}$
Explanation:
A Given that, Coefficient of real expansion of mercury $(\gamma)$ $=0.18 \times 10^{-3} /{ }^{\circ} \mathrm{C}$ Density of mercury $\left(\mathrm{d}_{0}\right)=13.6 \mathrm{~g} / \mathrm{cc}$ Temperature difference $(\Delta \mathrm{T})=200^{\circ} \mathrm{C}$ Coefficient of volume expansion, $\Delta \mathrm{V}=\mathrm{V}(\gamma \Delta \mathrm{T}+1)$ $d=\frac{d_{0}}{(1+\gamma \Delta T)}$ $d=\frac{13.6}{\left(1+0.18 \times 10^{-3} \times 200\right)}$ $d=13.1274 \mathrm{~g} / \mathrm{cc}$
146612
The two metal rods $A$ and $B$ are having their initial lengths in the ratio 2:3 and coefficient of linear expansion in the ratio $3: 4$. When they are heated through same temperature difference, the ratio of linear expansions is:
1 $1: 2$
2 $2: 3$
3 $3: 4$
4 $4: 3$
Explanation:
A Given that, Ratio of initial length $=\frac{2}{3}$ Ratio of linear expansion $=3 / 4$ As we know that, $\frac{\Delta l_{1}}{\Delta l_{2}}=\frac{l_{1} \alpha_{1}}{l_{2} \alpha_{2}}$ $=\frac{2}{3} \cdot \frac{3}{4}$ $\frac{\Delta l_{1}}{\Delta l_{2}}=\frac{1}{2}$ The ratio of linear expansion is 1:2.
AP EAMCET(Medical)-2000
Thermal Properties of Matter
146613
Density of a substance at $0^{\circ} \mathrm{C}$ is $10 \mathrm{~g} / \mathrm{cc}$ and at $100^{\circ} \mathrm{C}$ its density is $9.7 \mathrm{~g} / \mathrm{cc}$. The coefficient of linear expansion of the substance is :
1 $10^{-4}$
2 $3 \times 10^{-4}$
3 $19.7 \times 10^{-3}$
4 $10^{-3}$
Explanation:
A Given that, Density of a substance at $0^{\circ} \mathrm{C}=10 \mathrm{~g} / \mathrm{cc}$ Density of substance at $100^{\circ} \mathrm{C}=9.7 \mathrm{~g} / \mathrm{cc}$ As we know that, Coefficient of volume expansion, $\gamma=\frac{\Delta \rho}{\rho \cdot \Delta \mathrm{T}}$ Temperature difference, $\Delta \mathrm{T}=100-0$ $=100^{\circ} \mathrm{C}$ $\gamma=\frac{\rho_{0}-\rho_{100}}{\rho_{0} \cdot\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)}$ $=\frac{10-9.7}{10(100-0)}$ $=\frac{0.3}{10^{3}}$ $\gamma=3 \times 10^{-4} /{ }^{\circ} \mathrm{C}$ Coefficient of linear expansion, $\alpha=\frac{\gamma}{3}$ $=\frac{3 \times 10^{-4}}{3}$ $=10^{-4} /{ }^{\circ} \mathrm{C}$
AP EAMCET(Medical)-1998
Thermal Properties of Matter
146614
A $2 \mathrm{~m}$ long $\mathrm{Al}$ pipe at $27^{\circ} \mathrm{C}$ is heated until it is $0.0024 \mathrm{~m}$, at $77^{\circ} \mathrm{C}$. The coefficient of linear expansion of $\mathrm{Al}$ is
1 $2.4 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
2 $1.4 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
3 $2.4 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
4 $1.4 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
Explanation:
A Given that, Length of pipe $(l)=2 \mathrm{~m}$. Change in length $(\Delta l)=0.0024 \mathrm{~m}$ Temperature Variation $(\Delta \mathrm{T})=(77-27)^{\circ} \mathrm{C}$ $=50^{\circ} \mathrm{C}$ We know that, $\Delta l=l \alpha \Delta \mathrm{T}$ $0.0024=2 \times \alpha \times(77-27)$ $0.0024=2 \alpha \times 50$ $2.4 \times 10^{-3}=100 \alpha$ $\alpha=\frac{2.4 \times 10^{-3}}{100}$ $=2.4 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
EAMCET-1992
Thermal Properties of Matter
146615
A glass vessel just holds $50 \mathrm{~g}$ of a liquid at $0^{\circ} \mathrm{C}$. If the coefficient of linear expansion of glass is $8 \times 10^{-4} /{ }^{\circ} \mathrm{C}$. The mass of the liquid, it holds at $50^{\circ} \mathrm{C}$ is
1 $46 \mathrm{~g}$
2 $48 \mathrm{~g}$
3 $56 \mathrm{~g}$
4 $42 \mathrm{~g}$
Explanation:
C Given that, $\text { Mass of vessel }\left(\mathrm{M}_{1}\right)=50 \mathrm{~g}$ Coefficient of Linear Expansion $(\alpha)=8 \times 10^{-4} /{ }^{\circ} \mathrm{C}$ Change in temperature, $(\Delta \mathrm{T})=(50-0)^{\circ} \mathrm{C}$ $=50^{\circ} \mathrm{C}$ We know that, Coefficient of volume expansion, $\gamma=\frac{\Delta \mathrm{V}}{\mathrm{V}_{1} \Delta \mathrm{T}}$ $\gamma=\frac{\mathrm{V}_{1}-\mathrm{V}_{2}}{\mathrm{~V}_{1}(\Delta \mathrm{T})}$ $\mathrm{V}_{2}=\mathrm{V}_{1}(1+\gamma \Delta \mathrm{T})$ $\frac{\mathrm{M}_{2}}{\rho}=\frac{\mathrm{M}_{1}}{\rho}(1+3 \alpha \cdot \Delta \mathrm{T})\left(\mathrm{M}_{2}=\text { mass of liquid }\right)$ $\mathrm{M}_{2}=50\left[1+\left(3 \times 8 \times 10^{-4} \times 50\right)\right]$ $\quad=56.0 \mathrm{~g}$
EAMCET-1996
Thermal Properties of Matter
146616
Coefficient of real expansion of mercury is 0.18 $\times 10^{-3} /{ }^{\circ} \mathrm{C}$. If the density of mercury at $0^{\circ} \mathrm{C}$ is $13.6 \mathrm{~g} / \mathrm{cc}$, its density at $473 \mathrm{~K}$ will be
1 $13.11 \mathrm{~g} \mathrm{cc}^{-1}$
2 $13.65 \mathrm{~g} \mathrm{cc}^{-1}$
3 $13.51 \mathrm{~g} \mathrm{cc}^{-1}$
4 $13.22 \mathrm{~g} \mathrm{cc}^{-1}$
Explanation:
A Given that, Coefficient of real expansion of mercury $(\gamma)$ $=0.18 \times 10^{-3} /{ }^{\circ} \mathrm{C}$ Density of mercury $\left(\mathrm{d}_{0}\right)=13.6 \mathrm{~g} / \mathrm{cc}$ Temperature difference $(\Delta \mathrm{T})=200^{\circ} \mathrm{C}$ Coefficient of volume expansion, $\Delta \mathrm{V}=\mathrm{V}(\gamma \Delta \mathrm{T}+1)$ $d=\frac{d_{0}}{(1+\gamma \Delta T)}$ $d=\frac{13.6}{\left(1+0.18 \times 10^{-3} \times 200\right)}$ $d=13.1274 \mathrm{~g} / \mathrm{cc}$
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Thermal Properties of Matter
146612
The two metal rods $A$ and $B$ are having their initial lengths in the ratio 2:3 and coefficient of linear expansion in the ratio $3: 4$. When they are heated through same temperature difference, the ratio of linear expansions is:
1 $1: 2$
2 $2: 3$
3 $3: 4$
4 $4: 3$
Explanation:
A Given that, Ratio of initial length $=\frac{2}{3}$ Ratio of linear expansion $=3 / 4$ As we know that, $\frac{\Delta l_{1}}{\Delta l_{2}}=\frac{l_{1} \alpha_{1}}{l_{2} \alpha_{2}}$ $=\frac{2}{3} \cdot \frac{3}{4}$ $\frac{\Delta l_{1}}{\Delta l_{2}}=\frac{1}{2}$ The ratio of linear expansion is 1:2.
AP EAMCET(Medical)-2000
Thermal Properties of Matter
146613
Density of a substance at $0^{\circ} \mathrm{C}$ is $10 \mathrm{~g} / \mathrm{cc}$ and at $100^{\circ} \mathrm{C}$ its density is $9.7 \mathrm{~g} / \mathrm{cc}$. The coefficient of linear expansion of the substance is :
1 $10^{-4}$
2 $3 \times 10^{-4}$
3 $19.7 \times 10^{-3}$
4 $10^{-3}$
Explanation:
A Given that, Density of a substance at $0^{\circ} \mathrm{C}=10 \mathrm{~g} / \mathrm{cc}$ Density of substance at $100^{\circ} \mathrm{C}=9.7 \mathrm{~g} / \mathrm{cc}$ As we know that, Coefficient of volume expansion, $\gamma=\frac{\Delta \rho}{\rho \cdot \Delta \mathrm{T}}$ Temperature difference, $\Delta \mathrm{T}=100-0$ $=100^{\circ} \mathrm{C}$ $\gamma=\frac{\rho_{0}-\rho_{100}}{\rho_{0} \cdot\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)}$ $=\frac{10-9.7}{10(100-0)}$ $=\frac{0.3}{10^{3}}$ $\gamma=3 \times 10^{-4} /{ }^{\circ} \mathrm{C}$ Coefficient of linear expansion, $\alpha=\frac{\gamma}{3}$ $=\frac{3 \times 10^{-4}}{3}$ $=10^{-4} /{ }^{\circ} \mathrm{C}$
AP EAMCET(Medical)-1998
Thermal Properties of Matter
146614
A $2 \mathrm{~m}$ long $\mathrm{Al}$ pipe at $27^{\circ} \mathrm{C}$ is heated until it is $0.0024 \mathrm{~m}$, at $77^{\circ} \mathrm{C}$. The coefficient of linear expansion of $\mathrm{Al}$ is
1 $2.4 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
2 $1.4 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
3 $2.4 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
4 $1.4 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
Explanation:
A Given that, Length of pipe $(l)=2 \mathrm{~m}$. Change in length $(\Delta l)=0.0024 \mathrm{~m}$ Temperature Variation $(\Delta \mathrm{T})=(77-27)^{\circ} \mathrm{C}$ $=50^{\circ} \mathrm{C}$ We know that, $\Delta l=l \alpha \Delta \mathrm{T}$ $0.0024=2 \times \alpha \times(77-27)$ $0.0024=2 \alpha \times 50$ $2.4 \times 10^{-3}=100 \alpha$ $\alpha=\frac{2.4 \times 10^{-3}}{100}$ $=2.4 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
EAMCET-1992
Thermal Properties of Matter
146615
A glass vessel just holds $50 \mathrm{~g}$ of a liquid at $0^{\circ} \mathrm{C}$. If the coefficient of linear expansion of glass is $8 \times 10^{-4} /{ }^{\circ} \mathrm{C}$. The mass of the liquid, it holds at $50^{\circ} \mathrm{C}$ is
1 $46 \mathrm{~g}$
2 $48 \mathrm{~g}$
3 $56 \mathrm{~g}$
4 $42 \mathrm{~g}$
Explanation:
C Given that, $\text { Mass of vessel }\left(\mathrm{M}_{1}\right)=50 \mathrm{~g}$ Coefficient of Linear Expansion $(\alpha)=8 \times 10^{-4} /{ }^{\circ} \mathrm{C}$ Change in temperature, $(\Delta \mathrm{T})=(50-0)^{\circ} \mathrm{C}$ $=50^{\circ} \mathrm{C}$ We know that, Coefficient of volume expansion, $\gamma=\frac{\Delta \mathrm{V}}{\mathrm{V}_{1} \Delta \mathrm{T}}$ $\gamma=\frac{\mathrm{V}_{1}-\mathrm{V}_{2}}{\mathrm{~V}_{1}(\Delta \mathrm{T})}$ $\mathrm{V}_{2}=\mathrm{V}_{1}(1+\gamma \Delta \mathrm{T})$ $\frac{\mathrm{M}_{2}}{\rho}=\frac{\mathrm{M}_{1}}{\rho}(1+3 \alpha \cdot \Delta \mathrm{T})\left(\mathrm{M}_{2}=\text { mass of liquid }\right)$ $\mathrm{M}_{2}=50\left[1+\left(3 \times 8 \times 10^{-4} \times 50\right)\right]$ $\quad=56.0 \mathrm{~g}$
EAMCET-1996
Thermal Properties of Matter
146616
Coefficient of real expansion of mercury is 0.18 $\times 10^{-3} /{ }^{\circ} \mathrm{C}$. If the density of mercury at $0^{\circ} \mathrm{C}$ is $13.6 \mathrm{~g} / \mathrm{cc}$, its density at $473 \mathrm{~K}$ will be
1 $13.11 \mathrm{~g} \mathrm{cc}^{-1}$
2 $13.65 \mathrm{~g} \mathrm{cc}^{-1}$
3 $13.51 \mathrm{~g} \mathrm{cc}^{-1}$
4 $13.22 \mathrm{~g} \mathrm{cc}^{-1}$
Explanation:
A Given that, Coefficient of real expansion of mercury $(\gamma)$ $=0.18 \times 10^{-3} /{ }^{\circ} \mathrm{C}$ Density of mercury $\left(\mathrm{d}_{0}\right)=13.6 \mathrm{~g} / \mathrm{cc}$ Temperature difference $(\Delta \mathrm{T})=200^{\circ} \mathrm{C}$ Coefficient of volume expansion, $\Delta \mathrm{V}=\mathrm{V}(\gamma \Delta \mathrm{T}+1)$ $d=\frac{d_{0}}{(1+\gamma \Delta T)}$ $d=\frac{13.6}{\left(1+0.18 \times 10^{-3} \times 200\right)}$ $d=13.1274 \mathrm{~g} / \mathrm{cc}$
146612
The two metal rods $A$ and $B$ are having their initial lengths in the ratio 2:3 and coefficient of linear expansion in the ratio $3: 4$. When they are heated through same temperature difference, the ratio of linear expansions is:
1 $1: 2$
2 $2: 3$
3 $3: 4$
4 $4: 3$
Explanation:
A Given that, Ratio of initial length $=\frac{2}{3}$ Ratio of linear expansion $=3 / 4$ As we know that, $\frac{\Delta l_{1}}{\Delta l_{2}}=\frac{l_{1} \alpha_{1}}{l_{2} \alpha_{2}}$ $=\frac{2}{3} \cdot \frac{3}{4}$ $\frac{\Delta l_{1}}{\Delta l_{2}}=\frac{1}{2}$ The ratio of linear expansion is 1:2.
AP EAMCET(Medical)-2000
Thermal Properties of Matter
146613
Density of a substance at $0^{\circ} \mathrm{C}$ is $10 \mathrm{~g} / \mathrm{cc}$ and at $100^{\circ} \mathrm{C}$ its density is $9.7 \mathrm{~g} / \mathrm{cc}$. The coefficient of linear expansion of the substance is :
1 $10^{-4}$
2 $3 \times 10^{-4}$
3 $19.7 \times 10^{-3}$
4 $10^{-3}$
Explanation:
A Given that, Density of a substance at $0^{\circ} \mathrm{C}=10 \mathrm{~g} / \mathrm{cc}$ Density of substance at $100^{\circ} \mathrm{C}=9.7 \mathrm{~g} / \mathrm{cc}$ As we know that, Coefficient of volume expansion, $\gamma=\frac{\Delta \rho}{\rho \cdot \Delta \mathrm{T}}$ Temperature difference, $\Delta \mathrm{T}=100-0$ $=100^{\circ} \mathrm{C}$ $\gamma=\frac{\rho_{0}-\rho_{100}}{\rho_{0} \cdot\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)}$ $=\frac{10-9.7}{10(100-0)}$ $=\frac{0.3}{10^{3}}$ $\gamma=3 \times 10^{-4} /{ }^{\circ} \mathrm{C}$ Coefficient of linear expansion, $\alpha=\frac{\gamma}{3}$ $=\frac{3 \times 10^{-4}}{3}$ $=10^{-4} /{ }^{\circ} \mathrm{C}$
AP EAMCET(Medical)-1998
Thermal Properties of Matter
146614
A $2 \mathrm{~m}$ long $\mathrm{Al}$ pipe at $27^{\circ} \mathrm{C}$ is heated until it is $0.0024 \mathrm{~m}$, at $77^{\circ} \mathrm{C}$. The coefficient of linear expansion of $\mathrm{Al}$ is
1 $2.4 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
2 $1.4 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
3 $2.4 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
4 $1.4 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
Explanation:
A Given that, Length of pipe $(l)=2 \mathrm{~m}$. Change in length $(\Delta l)=0.0024 \mathrm{~m}$ Temperature Variation $(\Delta \mathrm{T})=(77-27)^{\circ} \mathrm{C}$ $=50^{\circ} \mathrm{C}$ We know that, $\Delta l=l \alpha \Delta \mathrm{T}$ $0.0024=2 \times \alpha \times(77-27)$ $0.0024=2 \alpha \times 50$ $2.4 \times 10^{-3}=100 \alpha$ $\alpha=\frac{2.4 \times 10^{-3}}{100}$ $=2.4 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
EAMCET-1992
Thermal Properties of Matter
146615
A glass vessel just holds $50 \mathrm{~g}$ of a liquid at $0^{\circ} \mathrm{C}$. If the coefficient of linear expansion of glass is $8 \times 10^{-4} /{ }^{\circ} \mathrm{C}$. The mass of the liquid, it holds at $50^{\circ} \mathrm{C}$ is
1 $46 \mathrm{~g}$
2 $48 \mathrm{~g}$
3 $56 \mathrm{~g}$
4 $42 \mathrm{~g}$
Explanation:
C Given that, $\text { Mass of vessel }\left(\mathrm{M}_{1}\right)=50 \mathrm{~g}$ Coefficient of Linear Expansion $(\alpha)=8 \times 10^{-4} /{ }^{\circ} \mathrm{C}$ Change in temperature, $(\Delta \mathrm{T})=(50-0)^{\circ} \mathrm{C}$ $=50^{\circ} \mathrm{C}$ We know that, Coefficient of volume expansion, $\gamma=\frac{\Delta \mathrm{V}}{\mathrm{V}_{1} \Delta \mathrm{T}}$ $\gamma=\frac{\mathrm{V}_{1}-\mathrm{V}_{2}}{\mathrm{~V}_{1}(\Delta \mathrm{T})}$ $\mathrm{V}_{2}=\mathrm{V}_{1}(1+\gamma \Delta \mathrm{T})$ $\frac{\mathrm{M}_{2}}{\rho}=\frac{\mathrm{M}_{1}}{\rho}(1+3 \alpha \cdot \Delta \mathrm{T})\left(\mathrm{M}_{2}=\text { mass of liquid }\right)$ $\mathrm{M}_{2}=50\left[1+\left(3 \times 8 \times 10^{-4} \times 50\right)\right]$ $\quad=56.0 \mathrm{~g}$
EAMCET-1996
Thermal Properties of Matter
146616
Coefficient of real expansion of mercury is 0.18 $\times 10^{-3} /{ }^{\circ} \mathrm{C}$. If the density of mercury at $0^{\circ} \mathrm{C}$ is $13.6 \mathrm{~g} / \mathrm{cc}$, its density at $473 \mathrm{~K}$ will be
1 $13.11 \mathrm{~g} \mathrm{cc}^{-1}$
2 $13.65 \mathrm{~g} \mathrm{cc}^{-1}$
3 $13.51 \mathrm{~g} \mathrm{cc}^{-1}$
4 $13.22 \mathrm{~g} \mathrm{cc}^{-1}$
Explanation:
A Given that, Coefficient of real expansion of mercury $(\gamma)$ $=0.18 \times 10^{-3} /{ }^{\circ} \mathrm{C}$ Density of mercury $\left(\mathrm{d}_{0}\right)=13.6 \mathrm{~g} / \mathrm{cc}$ Temperature difference $(\Delta \mathrm{T})=200^{\circ} \mathrm{C}$ Coefficient of volume expansion, $\Delta \mathrm{V}=\mathrm{V}(\gamma \Delta \mathrm{T}+1)$ $d=\frac{d_{0}}{(1+\gamma \Delta T)}$ $d=\frac{13.6}{\left(1+0.18 \times 10^{-3} \times 200\right)}$ $d=13.1274 \mathrm{~g} / \mathrm{cc}$
146612
The two metal rods $A$ and $B$ are having their initial lengths in the ratio 2:3 and coefficient of linear expansion in the ratio $3: 4$. When they are heated through same temperature difference, the ratio of linear expansions is:
1 $1: 2$
2 $2: 3$
3 $3: 4$
4 $4: 3$
Explanation:
A Given that, Ratio of initial length $=\frac{2}{3}$ Ratio of linear expansion $=3 / 4$ As we know that, $\frac{\Delta l_{1}}{\Delta l_{2}}=\frac{l_{1} \alpha_{1}}{l_{2} \alpha_{2}}$ $=\frac{2}{3} \cdot \frac{3}{4}$ $\frac{\Delta l_{1}}{\Delta l_{2}}=\frac{1}{2}$ The ratio of linear expansion is 1:2.
AP EAMCET(Medical)-2000
Thermal Properties of Matter
146613
Density of a substance at $0^{\circ} \mathrm{C}$ is $10 \mathrm{~g} / \mathrm{cc}$ and at $100^{\circ} \mathrm{C}$ its density is $9.7 \mathrm{~g} / \mathrm{cc}$. The coefficient of linear expansion of the substance is :
1 $10^{-4}$
2 $3 \times 10^{-4}$
3 $19.7 \times 10^{-3}$
4 $10^{-3}$
Explanation:
A Given that, Density of a substance at $0^{\circ} \mathrm{C}=10 \mathrm{~g} / \mathrm{cc}$ Density of substance at $100^{\circ} \mathrm{C}=9.7 \mathrm{~g} / \mathrm{cc}$ As we know that, Coefficient of volume expansion, $\gamma=\frac{\Delta \rho}{\rho \cdot \Delta \mathrm{T}}$ Temperature difference, $\Delta \mathrm{T}=100-0$ $=100^{\circ} \mathrm{C}$ $\gamma=\frac{\rho_{0}-\rho_{100}}{\rho_{0} \cdot\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)}$ $=\frac{10-9.7}{10(100-0)}$ $=\frac{0.3}{10^{3}}$ $\gamma=3 \times 10^{-4} /{ }^{\circ} \mathrm{C}$ Coefficient of linear expansion, $\alpha=\frac{\gamma}{3}$ $=\frac{3 \times 10^{-4}}{3}$ $=10^{-4} /{ }^{\circ} \mathrm{C}$
AP EAMCET(Medical)-1998
Thermal Properties of Matter
146614
A $2 \mathrm{~m}$ long $\mathrm{Al}$ pipe at $27^{\circ} \mathrm{C}$ is heated until it is $0.0024 \mathrm{~m}$, at $77^{\circ} \mathrm{C}$. The coefficient of linear expansion of $\mathrm{Al}$ is
1 $2.4 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
2 $1.4 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
3 $2.4 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
4 $1.4 \times 10^{-3} /{ }^{\circ} \mathrm{C}$
Explanation:
A Given that, Length of pipe $(l)=2 \mathrm{~m}$. Change in length $(\Delta l)=0.0024 \mathrm{~m}$ Temperature Variation $(\Delta \mathrm{T})=(77-27)^{\circ} \mathrm{C}$ $=50^{\circ} \mathrm{C}$ We know that, $\Delta l=l \alpha \Delta \mathrm{T}$ $0.0024=2 \times \alpha \times(77-27)$ $0.0024=2 \alpha \times 50$ $2.4 \times 10^{-3}=100 \alpha$ $\alpha=\frac{2.4 \times 10^{-3}}{100}$ $=2.4 \times 10^{-5} /{ }^{\circ} \mathrm{C}$
EAMCET-1992
Thermal Properties of Matter
146615
A glass vessel just holds $50 \mathrm{~g}$ of a liquid at $0^{\circ} \mathrm{C}$. If the coefficient of linear expansion of glass is $8 \times 10^{-4} /{ }^{\circ} \mathrm{C}$. The mass of the liquid, it holds at $50^{\circ} \mathrm{C}$ is
1 $46 \mathrm{~g}$
2 $48 \mathrm{~g}$
3 $56 \mathrm{~g}$
4 $42 \mathrm{~g}$
Explanation:
C Given that, $\text { Mass of vessel }\left(\mathrm{M}_{1}\right)=50 \mathrm{~g}$ Coefficient of Linear Expansion $(\alpha)=8 \times 10^{-4} /{ }^{\circ} \mathrm{C}$ Change in temperature, $(\Delta \mathrm{T})=(50-0)^{\circ} \mathrm{C}$ $=50^{\circ} \mathrm{C}$ We know that, Coefficient of volume expansion, $\gamma=\frac{\Delta \mathrm{V}}{\mathrm{V}_{1} \Delta \mathrm{T}}$ $\gamma=\frac{\mathrm{V}_{1}-\mathrm{V}_{2}}{\mathrm{~V}_{1}(\Delta \mathrm{T})}$ $\mathrm{V}_{2}=\mathrm{V}_{1}(1+\gamma \Delta \mathrm{T})$ $\frac{\mathrm{M}_{2}}{\rho}=\frac{\mathrm{M}_{1}}{\rho}(1+3 \alpha \cdot \Delta \mathrm{T})\left(\mathrm{M}_{2}=\text { mass of liquid }\right)$ $\mathrm{M}_{2}=50\left[1+\left(3 \times 8 \times 10^{-4} \times 50\right)\right]$ $\quad=56.0 \mathrm{~g}$
EAMCET-1996
Thermal Properties of Matter
146616
Coefficient of real expansion of mercury is 0.18 $\times 10^{-3} /{ }^{\circ} \mathrm{C}$. If the density of mercury at $0^{\circ} \mathrm{C}$ is $13.6 \mathrm{~g} / \mathrm{cc}$, its density at $473 \mathrm{~K}$ will be
1 $13.11 \mathrm{~g} \mathrm{cc}^{-1}$
2 $13.65 \mathrm{~g} \mathrm{cc}^{-1}$
3 $13.51 \mathrm{~g} \mathrm{cc}^{-1}$
4 $13.22 \mathrm{~g} \mathrm{cc}^{-1}$
Explanation:
A Given that, Coefficient of real expansion of mercury $(\gamma)$ $=0.18 \times 10^{-3} /{ }^{\circ} \mathrm{C}$ Density of mercury $\left(\mathrm{d}_{0}\right)=13.6 \mathrm{~g} / \mathrm{cc}$ Temperature difference $(\Delta \mathrm{T})=200^{\circ} \mathrm{C}$ Coefficient of volume expansion, $\Delta \mathrm{V}=\mathrm{V}(\gamma \Delta \mathrm{T}+1)$ $d=\frac{d_{0}}{(1+\gamma \Delta T)}$ $d=\frac{13.6}{\left(1+0.18 \times 10^{-3} \times 200\right)}$ $d=13.1274 \mathrm{~g} / \mathrm{cc}$