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Thermal Properties of Matter

146652 Two metallic spheres $P$ and $Q$ are made of same material have same smoothness but the weight of $P$ is 8 times of $Q$. If the two are heated to same temperature and left to cool, then the ratio of rate of cooling of $Q$ to that of $P$ is

1 4

2 8

3 2

4 1

AP EAMCET (22.04.2019) Shift-I

Thermal Properties of Matter

146653 A liquid of mass $250 \mathrm{~g}$ is kept warm in a vessel using an electric heater. The liquid is maintained at $57^{\circ} \mathrm{C}$ when the power supplied by the heater is $30 \mathrm{~W}$ and surrounding temperature is $27^{\circ} \mathrm{C}$. As the heater is switched off, it took $10 \mathrm{~s}$ time for the temperature of the liquid to fall from $47^{\circ} \mathrm{C}$ to $46.9^{\circ} \mathrm{C}$. The specific heat capacity of the liquid is

1 $8000 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$

2 $9000 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$

3 $6000 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$

4 $12000 \mathrm{Jkg}-1 \mathrm{~K}^{-1}$

AP EAMCET (21.04.2019) Shift-II

Thermal Properties of Matter

146654 A hammer of mass $200 \mathrm{~kg}$ strikes a steel block of mass $200 \mathrm{~g}$ with a velocity $8 \mathrm{~ms}^{-1}$. If $23 \%$ of the energy is utilized to heat the steel block, the rise in temperature, of the block is (specific heat capacity of steel, $=460 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$ )

1 $8 \mathrm{~K}$

2 $16 \mathrm{~K}$

3 $12 \mathrm{~K}$

4 $24 \mathrm{~K}$

AP EAMCET (21.04.2019) Shift-I

Thermal Properties of Matter

146655 A metal rod is subjected to cycles of magnetisation at the rate of $42 \mathrm{~Hz}$. Density of the metal is $6 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$ and its specific heat capacity is $0.1 \times 10^{-3} \mathrm{cal} \mathrm{kg}^{-1 \circ} \mathrm{C}^{-1}$. If the area of its B-H loop corresponds to energy density of $10^{-2} \mathbf{J m}^{-3}$, then the rise in its temperature in one minute is

1 $5^{\circ} \mathrm{C}$

2 $10^{\circ} \mathrm{C}$

3 $15^{\circ} \mathrm{C}$

4 $20^{\circ} \mathrm{C}$

AP EAMCET (20.04.2019) Shift-II

Thermal Properties of Matter

146656 One mole of a monoatomic ideal gas undergoes the process $A \rightarrow B$ in the given $p-V$ diagram. Specific heat capacity in the process is

1 $\frac{13 R}{3}$

2 $\frac{13 R}{6}$

3 $\frac{7 R}{3}$

4 $\frac{2 R}{3}$

Explanation:

B
Exp:
From figure -
original image
$\mathrm{V}_{0}$
$\mathrm{CB}=6 \mathrm{P}_{0}-3 \mathrm{P}_{0}=3 \mathrm{P}_{0}$
$\mathrm{AE}=3 \mathrm{P}_{0}, \mathrm{ED}=\mathrm{AC}=4 \mathrm{~V}_{0}$
Given, 1 mole of an ideal monoatomic gas undergoes the process $A B$. So, number of mole $n=1$
Work done in part $\mathrm{A}$ to $\mathrm{B}\left(\mathrm{W}_{\mathrm{A} \rightarrow \mathrm{B}}\right)=$ Area under $\mathrm{P}-\mathrm{V}$ diagram
$\mathrm{W}_{\mathrm{A} \rightarrow \mathrm{B}}=$ Area of $\triangle \mathrm{ACB}+$ Area of $\square \mathrm{ACDE}$
$=\frac{1}{2} \times \mathrm{AC} \times \mathrm{CB}+\mathrm{AE} \times \mathrm{ED}$
$=\frac{1}{2} \times 4 \mathrm{~V}_{0} \times 3 \mathrm{P}_{0}+3 \mathrm{P}_{0} \times 4 \mathrm{~V}_{0}$
$=6 \mathrm{P}_{0} \mathrm{~V}_{0}+12 \mathrm{P}_{0} \mathrm{~V}_{0}=18 \mathrm{P}_{0} \mathrm{~V}_{0}$
Now internal energy change $\Delta \mathrm{U}_{\mathrm{A} \rightarrow \mathrm{B}}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$
$=1 \times \frac{3}{2} \mathrm{R}\left(\mathrm{T}_{\mathrm{B}}-\mathrm{T}_{\mathrm{A}}\right)$
$\left\{\because\right.$ As the gas is monoatomic that's why $\left.\mathrm{C}_{\mathrm{v}}=\frac{3}{2} \mathrm{R}\right\}$
At the point $\mathrm{B}, \mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}=\mathrm{RT}_{\mathrm{B}}\left(\because \mathrm{P}_{\mathrm{B}}=6 \mathrm{P}_{0}, \mathrm{~V}_{\mathrm{B}}=5 \mathrm{~V}_{0}\right)$
$\mathrm{T}_{\mathrm{B}}=\frac{6 \mathrm{P}_{0} \cdot 5 \mathrm{~V}_{0}}{\mathrm{R}}$
$\mathrm{T}_{\mathrm{B}}=\frac{30 \mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}$
Similarly, $\mathrm{T}_{\mathrm{A}}=\frac{3 \mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}$
$\text { So, } \Delta \mathrm{U}_{\mathrm{A} \rightarrow \mathrm{B}} =\frac{3}{2} \mathrm{R} \times\left(\frac{30 \mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}-\frac{3 \mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}\right)$
$\Delta \mathrm{U}_{\mathrm{A} \rightarrow \mathrm{B}} =\frac{3}{2} \mathrm{R} \times \frac{\mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}(30-3)$
$=\frac{81}{2} \mathrm{P}_{0} \mathrm{~V}_{0}$
Thus, from $1^{\text {st }}$ law of thermodynamics -
$\Delta \mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}} =\Delta \mathrm{W}_{\mathrm{A} \rightarrow \mathrm{B}}+\Delta \mathrm{U}_{\mathrm{A} \rightarrow \mathrm{B}}$
$=18 \mathrm{P}_{0} \mathrm{~V}_{0}+\frac{81}{2} \mathrm{P}_{0} \mathrm{~V}_{0}$
$\mathrm{DQ}_{\mathrm{A} \rightarrow \mathrm{B}} =\frac{117}{2} \mathrm{P}_{0} \mathrm{~V}_{0} \text { (Putting from (i) and (ii)) }$
$\because$ Molar heat capacity $\mathrm{C}=\frac{\Delta \mathrm{Q}}{\mathrm{n} \cdot \Delta \mathrm{T}}$
$\mathrm{C}=\frac{\frac{117}{2} \mathrm{P}_{0} \mathrm{~V}_{0}}{1 \times\left(\frac{30 \mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{R}}-\frac{3 \mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{R}}\right)}$
$\mathrm{C}=\frac{\frac{117}{2} \mathrm{P}_{0} \mathrm{~V}_{0}}{\frac{27 \mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{R}}}$
$\mathrm{C}=\frac{13 \mathrm{R}}{6}$

AP EAMCET (20.04.2019) Shift-II

Thermal Properties of Matter

146652 Two metallic spheres $P$ and $Q$ are made of same material have same smoothness but the weight of $P$ is 8 times of $Q$. If the two are heated to same temperature and left to cool, then the ratio of rate of cooling of $Q$ to that of $P$ is

1 4

2 8

3 2

4 1

AP EAMCET (22.04.2019) Shift-I

Thermal Properties of Matter

146653 A liquid of mass $250 \mathrm{~g}$ is kept warm in a vessel using an electric heater. The liquid is maintained at $57^{\circ} \mathrm{C}$ when the power supplied by the heater is $30 \mathrm{~W}$ and surrounding temperature is $27^{\circ} \mathrm{C}$. As the heater is switched off, it took $10 \mathrm{~s}$ time for the temperature of the liquid to fall from $47^{\circ} \mathrm{C}$ to $46.9^{\circ} \mathrm{C}$. The specific heat capacity of the liquid is

1 $8000 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$

2 $9000 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$

3 $6000 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$

4 $12000 \mathrm{Jkg}-1 \mathrm{~K}^{-1}$

AP EAMCET (21.04.2019) Shift-II

Thermal Properties of Matter

146654 A hammer of mass $200 \mathrm{~kg}$ strikes a steel block of mass $200 \mathrm{~g}$ with a velocity $8 \mathrm{~ms}^{-1}$. If $23 \%$ of the energy is utilized to heat the steel block, the rise in temperature, of the block is (specific heat capacity of steel, $=460 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$ )

1 $8 \mathrm{~K}$

2 $16 \mathrm{~K}$

3 $12 \mathrm{~K}$

4 $24 \mathrm{~K}$

AP EAMCET (21.04.2019) Shift-I

Thermal Properties of Matter

146655 A metal rod is subjected to cycles of magnetisation at the rate of $42 \mathrm{~Hz}$. Density of the metal is $6 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$ and its specific heat capacity is $0.1 \times 10^{-3} \mathrm{cal} \mathrm{kg}^{-1 \circ} \mathrm{C}^{-1}$. If the area of its B-H loop corresponds to energy density of $10^{-2} \mathbf{J m}^{-3}$, then the rise in its temperature in one minute is

1 $5^{\circ} \mathrm{C}$

2 $10^{\circ} \mathrm{C}$

3 $15^{\circ} \mathrm{C}$

4 $20^{\circ} \mathrm{C}$

AP EAMCET (20.04.2019) Shift-II

Thermal Properties of Matter

146656 One mole of a monoatomic ideal gas undergoes the process $A \rightarrow B$ in the given $p-V$ diagram. Specific heat capacity in the process is

1 $\frac{13 R}{3}$

2 $\frac{13 R}{6}$

3 $\frac{7 R}{3}$

4 $\frac{2 R}{3}$

Explanation:

B
Exp:
From figure -
original image
$\mathrm{V}_{0}$
$\mathrm{CB}=6 \mathrm{P}_{0}-3 \mathrm{P}_{0}=3 \mathrm{P}_{0}$
$\mathrm{AE}=3 \mathrm{P}_{0}, \mathrm{ED}=\mathrm{AC}=4 \mathrm{~V}_{0}$
Given, 1 mole of an ideal monoatomic gas undergoes the process $A B$. So, number of mole $n=1$
Work done in part $\mathrm{A}$ to $\mathrm{B}\left(\mathrm{W}_{\mathrm{A} \rightarrow \mathrm{B}}\right)=$ Area under $\mathrm{P}-\mathrm{V}$ diagram
$\mathrm{W}_{\mathrm{A} \rightarrow \mathrm{B}}=$ Area of $\triangle \mathrm{ACB}+$ Area of $\square \mathrm{ACDE}$
$=\frac{1}{2} \times \mathrm{AC} \times \mathrm{CB}+\mathrm{AE} \times \mathrm{ED}$
$=\frac{1}{2} \times 4 \mathrm{~V}_{0} \times 3 \mathrm{P}_{0}+3 \mathrm{P}_{0} \times 4 \mathrm{~V}_{0}$
$=6 \mathrm{P}_{0} \mathrm{~V}_{0}+12 \mathrm{P}_{0} \mathrm{~V}_{0}=18 \mathrm{P}_{0} \mathrm{~V}_{0}$
Now internal energy change $\Delta \mathrm{U}_{\mathrm{A} \rightarrow \mathrm{B}}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$
$=1 \times \frac{3}{2} \mathrm{R}\left(\mathrm{T}_{\mathrm{B}}-\mathrm{T}_{\mathrm{A}}\right)$
$\left\{\because\right.$ As the gas is monoatomic that's why $\left.\mathrm{C}_{\mathrm{v}}=\frac{3}{2} \mathrm{R}\right\}$
At the point $\mathrm{B}, \mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}=\mathrm{RT}_{\mathrm{B}}\left(\because \mathrm{P}_{\mathrm{B}}=6 \mathrm{P}_{0}, \mathrm{~V}_{\mathrm{B}}=5 \mathrm{~V}_{0}\right)$
$\mathrm{T}_{\mathrm{B}}=\frac{6 \mathrm{P}_{0} \cdot 5 \mathrm{~V}_{0}}{\mathrm{R}}$
$\mathrm{T}_{\mathrm{B}}=\frac{30 \mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}$
Similarly, $\mathrm{T}_{\mathrm{A}}=\frac{3 \mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}$
$\text { So, } \Delta \mathrm{U}_{\mathrm{A} \rightarrow \mathrm{B}} =\frac{3}{2} \mathrm{R} \times\left(\frac{30 \mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}-\frac{3 \mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}\right)$
$\Delta \mathrm{U}_{\mathrm{A} \rightarrow \mathrm{B}} =\frac{3}{2} \mathrm{R} \times \frac{\mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}(30-3)$
$=\frac{81}{2} \mathrm{P}_{0} \mathrm{~V}_{0}$
Thus, from $1^{\text {st }}$ law of thermodynamics -
$\Delta \mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}} =\Delta \mathrm{W}_{\mathrm{A} \rightarrow \mathrm{B}}+\Delta \mathrm{U}_{\mathrm{A} \rightarrow \mathrm{B}}$
$=18 \mathrm{P}_{0} \mathrm{~V}_{0}+\frac{81}{2} \mathrm{P}_{0} \mathrm{~V}_{0}$
$\mathrm{DQ}_{\mathrm{A} \rightarrow \mathrm{B}} =\frac{117}{2} \mathrm{P}_{0} \mathrm{~V}_{0} \text { (Putting from (i) and (ii)) }$
$\because$ Molar heat capacity $\mathrm{C}=\frac{\Delta \mathrm{Q}}{\mathrm{n} \cdot \Delta \mathrm{T}}$
$\mathrm{C}=\frac{\frac{117}{2} \mathrm{P}_{0} \mathrm{~V}_{0}}{1 \times\left(\frac{30 \mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{R}}-\frac{3 \mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{R}}\right)}$
$\mathrm{C}=\frac{\frac{117}{2} \mathrm{P}_{0} \mathrm{~V}_{0}}{\frac{27 \mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{R}}}$
$\mathrm{C}=\frac{13 \mathrm{R}}{6}$

AP EAMCET (20.04.2019) Shift-II

Thermal Properties of Matter

146652 Two metallic spheres $P$ and $Q$ are made of same material have same smoothness but the weight of $P$ is 8 times of $Q$. If the two are heated to same temperature and left to cool, then the ratio of rate of cooling of $Q$ to that of $P$ is

1 4

2 8

3 2

4 1

AP EAMCET (22.04.2019) Shift-I

Thermal Properties of Matter

146653 A liquid of mass $250 \mathrm{~g}$ is kept warm in a vessel using an electric heater. The liquid is maintained at $57^{\circ} \mathrm{C}$ when the power supplied by the heater is $30 \mathrm{~W}$ and surrounding temperature is $27^{\circ} \mathrm{C}$. As the heater is switched off, it took $10 \mathrm{~s}$ time for the temperature of the liquid to fall from $47^{\circ} \mathrm{C}$ to $46.9^{\circ} \mathrm{C}$. The specific heat capacity of the liquid is

1 $8000 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$

2 $9000 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$

3 $6000 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$

4 $12000 \mathrm{Jkg}-1 \mathrm{~K}^{-1}$

AP EAMCET (21.04.2019) Shift-II

Thermal Properties of Matter

146654 A hammer of mass $200 \mathrm{~kg}$ strikes a steel block of mass $200 \mathrm{~g}$ with a velocity $8 \mathrm{~ms}^{-1}$. If $23 \%$ of the energy is utilized to heat the steel block, the rise in temperature, of the block is (specific heat capacity of steel, $=460 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$ )

1 $8 \mathrm{~K}$

2 $16 \mathrm{~K}$

3 $12 \mathrm{~K}$

4 $24 \mathrm{~K}$

AP EAMCET (21.04.2019) Shift-I

Thermal Properties of Matter

146655 A metal rod is subjected to cycles of magnetisation at the rate of $42 \mathrm{~Hz}$. Density of the metal is $6 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$ and its specific heat capacity is $0.1 \times 10^{-3} \mathrm{cal} \mathrm{kg}^{-1 \circ} \mathrm{C}^{-1}$. If the area of its B-H loop corresponds to energy density of $10^{-2} \mathbf{J m}^{-3}$, then the rise in its temperature in one minute is

1 $5^{\circ} \mathrm{C}$

2 $10^{\circ} \mathrm{C}$

3 $15^{\circ} \mathrm{C}$

4 $20^{\circ} \mathrm{C}$

AP EAMCET (20.04.2019) Shift-II

Thermal Properties of Matter

146656 One mole of a monoatomic ideal gas undergoes the process $A \rightarrow B$ in the given $p-V$ diagram. Specific heat capacity in the process is

1 $\frac{13 R}{3}$

2 $\frac{13 R}{6}$

3 $\frac{7 R}{3}$

4 $\frac{2 R}{3}$

Explanation:

B
Exp:
From figure -
original image
$\mathrm{V}_{0}$
$\mathrm{CB}=6 \mathrm{P}_{0}-3 \mathrm{P}_{0}=3 \mathrm{P}_{0}$
$\mathrm{AE}=3 \mathrm{P}_{0}, \mathrm{ED}=\mathrm{AC}=4 \mathrm{~V}_{0}$
Given, 1 mole of an ideal monoatomic gas undergoes the process $A B$. So, number of mole $n=1$
Work done in part $\mathrm{A}$ to $\mathrm{B}\left(\mathrm{W}_{\mathrm{A} \rightarrow \mathrm{B}}\right)=$ Area under $\mathrm{P}-\mathrm{V}$ diagram
$\mathrm{W}_{\mathrm{A} \rightarrow \mathrm{B}}=$ Area of $\triangle \mathrm{ACB}+$ Area of $\square \mathrm{ACDE}$
$=\frac{1}{2} \times \mathrm{AC} \times \mathrm{CB}+\mathrm{AE} \times \mathrm{ED}$
$=\frac{1}{2} \times 4 \mathrm{~V}_{0} \times 3 \mathrm{P}_{0}+3 \mathrm{P}_{0} \times 4 \mathrm{~V}_{0}$
$=6 \mathrm{P}_{0} \mathrm{~V}_{0}+12 \mathrm{P}_{0} \mathrm{~V}_{0}=18 \mathrm{P}_{0} \mathrm{~V}_{0}$
Now internal energy change $\Delta \mathrm{U}_{\mathrm{A} \rightarrow \mathrm{B}}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$
$=1 \times \frac{3}{2} \mathrm{R}\left(\mathrm{T}_{\mathrm{B}}-\mathrm{T}_{\mathrm{A}}\right)$
$\left\{\because\right.$ As the gas is monoatomic that's why $\left.\mathrm{C}_{\mathrm{v}}=\frac{3}{2} \mathrm{R}\right\}$
At the point $\mathrm{B}, \mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}=\mathrm{RT}_{\mathrm{B}}\left(\because \mathrm{P}_{\mathrm{B}}=6 \mathrm{P}_{0}, \mathrm{~V}_{\mathrm{B}}=5 \mathrm{~V}_{0}\right)$
$\mathrm{T}_{\mathrm{B}}=\frac{6 \mathrm{P}_{0} \cdot 5 \mathrm{~V}_{0}}{\mathrm{R}}$
$\mathrm{T}_{\mathrm{B}}=\frac{30 \mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}$
Similarly, $\mathrm{T}_{\mathrm{A}}=\frac{3 \mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}$
$\text { So, } \Delta \mathrm{U}_{\mathrm{A} \rightarrow \mathrm{B}} =\frac{3}{2} \mathrm{R} \times\left(\frac{30 \mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}-\frac{3 \mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}\right)$
$\Delta \mathrm{U}_{\mathrm{A} \rightarrow \mathrm{B}} =\frac{3}{2} \mathrm{R} \times \frac{\mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}(30-3)$
$=\frac{81}{2} \mathrm{P}_{0} \mathrm{~V}_{0}$
Thus, from $1^{\text {st }}$ law of thermodynamics -
$\Delta \mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}} =\Delta \mathrm{W}_{\mathrm{A} \rightarrow \mathrm{B}}+\Delta \mathrm{U}_{\mathrm{A} \rightarrow \mathrm{B}}$
$=18 \mathrm{P}_{0} \mathrm{~V}_{0}+\frac{81}{2} \mathrm{P}_{0} \mathrm{~V}_{0}$
$\mathrm{DQ}_{\mathrm{A} \rightarrow \mathrm{B}} =\frac{117}{2} \mathrm{P}_{0} \mathrm{~V}_{0} \text { (Putting from (i) and (ii)) }$
$\because$ Molar heat capacity $\mathrm{C}=\frac{\Delta \mathrm{Q}}{\mathrm{n} \cdot \Delta \mathrm{T}}$
$\mathrm{C}=\frac{\frac{117}{2} \mathrm{P}_{0} \mathrm{~V}_{0}}{1 \times\left(\frac{30 \mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{R}}-\frac{3 \mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{R}}\right)}$
$\mathrm{C}=\frac{\frac{117}{2} \mathrm{P}_{0} \mathrm{~V}_{0}}{\frac{27 \mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{R}}}$
$\mathrm{C}=\frac{13 \mathrm{R}}{6}$

AP EAMCET (20.04.2019) Shift-II

Thermal Properties of Matter

146652 Two metallic spheres $P$ and $Q$ are made of same material have same smoothness but the weight of $P$ is 8 times of $Q$. If the two are heated to same temperature and left to cool, then the ratio of rate of cooling of $Q$ to that of $P$ is

1 4

2 8

3 2

4 1

AP EAMCET (22.04.2019) Shift-I

Thermal Properties of Matter

146653 A liquid of mass $250 \mathrm{~g}$ is kept warm in a vessel using an electric heater. The liquid is maintained at $57^{\circ} \mathrm{C}$ when the power supplied by the heater is $30 \mathrm{~W}$ and surrounding temperature is $27^{\circ} \mathrm{C}$. As the heater is switched off, it took $10 \mathrm{~s}$ time for the temperature of the liquid to fall from $47^{\circ} \mathrm{C}$ to $46.9^{\circ} \mathrm{C}$. The specific heat capacity of the liquid is

1 $8000 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$

2 $9000 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$

3 $6000 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$

4 $12000 \mathrm{Jkg}-1 \mathrm{~K}^{-1}$

AP EAMCET (21.04.2019) Shift-II

Thermal Properties of Matter

146654 A hammer of mass $200 \mathrm{~kg}$ strikes a steel block of mass $200 \mathrm{~g}$ with a velocity $8 \mathrm{~ms}^{-1}$. If $23 \%$ of the energy is utilized to heat the steel block, the rise in temperature, of the block is (specific heat capacity of steel, $=460 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$ )

1 $8 \mathrm{~K}$

2 $16 \mathrm{~K}$

3 $12 \mathrm{~K}$

4 $24 \mathrm{~K}$

AP EAMCET (21.04.2019) Shift-I

Thermal Properties of Matter

146655 A metal rod is subjected to cycles of magnetisation at the rate of $42 \mathrm{~Hz}$. Density of the metal is $6 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$ and its specific heat capacity is $0.1 \times 10^{-3} \mathrm{cal} \mathrm{kg}^{-1 \circ} \mathrm{C}^{-1}$. If the area of its B-H loop corresponds to energy density of $10^{-2} \mathbf{J m}^{-3}$, then the rise in its temperature in one minute is

1 $5^{\circ} \mathrm{C}$

2 $10^{\circ} \mathrm{C}$

3 $15^{\circ} \mathrm{C}$

4 $20^{\circ} \mathrm{C}$

AP EAMCET (20.04.2019) Shift-II

Thermal Properties of Matter

146656 One mole of a monoatomic ideal gas undergoes the process $A \rightarrow B$ in the given $p-V$ diagram. Specific heat capacity in the process is

1 $\frac{13 R}{3}$

2 $\frac{13 R}{6}$

3 $\frac{7 R}{3}$

4 $\frac{2 R}{3}$

Explanation:

B
Exp:
From figure -
original image
$\mathrm{V}_{0}$
$\mathrm{CB}=6 \mathrm{P}_{0}-3 \mathrm{P}_{0}=3 \mathrm{P}_{0}$
$\mathrm{AE}=3 \mathrm{P}_{0}, \mathrm{ED}=\mathrm{AC}=4 \mathrm{~V}_{0}$
Given, 1 mole of an ideal monoatomic gas undergoes the process $A B$. So, number of mole $n=1$
Work done in part $\mathrm{A}$ to $\mathrm{B}\left(\mathrm{W}_{\mathrm{A} \rightarrow \mathrm{B}}\right)=$ Area under $\mathrm{P}-\mathrm{V}$ diagram
$\mathrm{W}_{\mathrm{A} \rightarrow \mathrm{B}}=$ Area of $\triangle \mathrm{ACB}+$ Area of $\square \mathrm{ACDE}$
$=\frac{1}{2} \times \mathrm{AC} \times \mathrm{CB}+\mathrm{AE} \times \mathrm{ED}$
$=\frac{1}{2} \times 4 \mathrm{~V}_{0} \times 3 \mathrm{P}_{0}+3 \mathrm{P}_{0} \times 4 \mathrm{~V}_{0}$
$=6 \mathrm{P}_{0} \mathrm{~V}_{0}+12 \mathrm{P}_{0} \mathrm{~V}_{0}=18 \mathrm{P}_{0} \mathrm{~V}_{0}$
Now internal energy change $\Delta \mathrm{U}_{\mathrm{A} \rightarrow \mathrm{B}}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$
$=1 \times \frac{3}{2} \mathrm{R}\left(\mathrm{T}_{\mathrm{B}}-\mathrm{T}_{\mathrm{A}}\right)$
$\left\{\because\right.$ As the gas is monoatomic that's why $\left.\mathrm{C}_{\mathrm{v}}=\frac{3}{2} \mathrm{R}\right\}$
At the point $\mathrm{B}, \mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}=\mathrm{RT}_{\mathrm{B}}\left(\because \mathrm{P}_{\mathrm{B}}=6 \mathrm{P}_{0}, \mathrm{~V}_{\mathrm{B}}=5 \mathrm{~V}_{0}\right)$
$\mathrm{T}_{\mathrm{B}}=\frac{6 \mathrm{P}_{0} \cdot 5 \mathrm{~V}_{0}}{\mathrm{R}}$
$\mathrm{T}_{\mathrm{B}}=\frac{30 \mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}$
Similarly, $\mathrm{T}_{\mathrm{A}}=\frac{3 \mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}$
$\text { So, } \Delta \mathrm{U}_{\mathrm{A} \rightarrow \mathrm{B}} =\frac{3}{2} \mathrm{R} \times\left(\frac{30 \mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}-\frac{3 \mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}\right)$
$\Delta \mathrm{U}_{\mathrm{A} \rightarrow \mathrm{B}} =\frac{3}{2} \mathrm{R} \times \frac{\mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}(30-3)$
$=\frac{81}{2} \mathrm{P}_{0} \mathrm{~V}_{0}$
Thus, from $1^{\text {st }}$ law of thermodynamics -
$\Delta \mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}} =\Delta \mathrm{W}_{\mathrm{A} \rightarrow \mathrm{B}}+\Delta \mathrm{U}_{\mathrm{A} \rightarrow \mathrm{B}}$
$=18 \mathrm{P}_{0} \mathrm{~V}_{0}+\frac{81}{2} \mathrm{P}_{0} \mathrm{~V}_{0}$
$\mathrm{DQ}_{\mathrm{A} \rightarrow \mathrm{B}} =\frac{117}{2} \mathrm{P}_{0} \mathrm{~V}_{0} \text { (Putting from (i) and (ii)) }$
$\because$ Molar heat capacity $\mathrm{C}=\frac{\Delta \mathrm{Q}}{\mathrm{n} \cdot \Delta \mathrm{T}}$
$\mathrm{C}=\frac{\frac{117}{2} \mathrm{P}_{0} \mathrm{~V}_{0}}{1 \times\left(\frac{30 \mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{R}}-\frac{3 \mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{R}}\right)}$
$\mathrm{C}=\frac{\frac{117}{2} \mathrm{P}_{0} \mathrm{~V}_{0}}{\frac{27 \mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{R}}}$
$\mathrm{C}=\frac{13 \mathrm{R}}{6}$

AP EAMCET (20.04.2019) Shift-II

Thermal Properties of Matter

146652 Two metallic spheres $P$ and $Q$ are made of same material have same smoothness but the weight of $P$ is 8 times of $Q$. If the two are heated to same temperature and left to cool, then the ratio of rate of cooling of $Q$ to that of $P$ is

1 4

2 8

3 2

4 1

AP EAMCET (22.04.2019) Shift-I

Thermal Properties of Matter

146653 A liquid of mass $250 \mathrm{~g}$ is kept warm in a vessel using an electric heater. The liquid is maintained at $57^{\circ} \mathrm{C}$ when the power supplied by the heater is $30 \mathrm{~W}$ and surrounding temperature is $27^{\circ} \mathrm{C}$. As the heater is switched off, it took $10 \mathrm{~s}$ time for the temperature of the liquid to fall from $47^{\circ} \mathrm{C}$ to $46.9^{\circ} \mathrm{C}$. The specific heat capacity of the liquid is

1 $8000 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$

2 $9000 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$

3 $6000 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$

4 $12000 \mathrm{Jkg}-1 \mathrm{~K}^{-1}$

AP EAMCET (21.04.2019) Shift-II

Thermal Properties of Matter

146654 A hammer of mass $200 \mathrm{~kg}$ strikes a steel block of mass $200 \mathrm{~g}$ with a velocity $8 \mathrm{~ms}^{-1}$. If $23 \%$ of the energy is utilized to heat the steel block, the rise in temperature, of the block is (specific heat capacity of steel, $=460 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$ )

1 $8 \mathrm{~K}$

2 $16 \mathrm{~K}$

3 $12 \mathrm{~K}$

4 $24 \mathrm{~K}$

AP EAMCET (21.04.2019) Shift-I

Thermal Properties of Matter

146655 A metal rod is subjected to cycles of magnetisation at the rate of $42 \mathrm{~Hz}$. Density of the metal is $6 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$ and its specific heat capacity is $0.1 \times 10^{-3} \mathrm{cal} \mathrm{kg}^{-1 \circ} \mathrm{C}^{-1}$. If the area of its B-H loop corresponds to energy density of $10^{-2} \mathbf{J m}^{-3}$, then the rise in its temperature in one minute is

1 $5^{\circ} \mathrm{C}$

2 $10^{\circ} \mathrm{C}$

3 $15^{\circ} \mathrm{C}$

4 $20^{\circ} \mathrm{C}$

AP EAMCET (20.04.2019) Shift-II

Thermal Properties of Matter

146656 One mole of a monoatomic ideal gas undergoes the process $A \rightarrow B$ in the given $p-V$ diagram. Specific heat capacity in the process is

1 $\frac{13 R}{3}$

2 $\frac{13 R}{6}$

3 $\frac{7 R}{3}$

4 $\frac{2 R}{3}$

Explanation:

B
Exp:
From figure -
original image
$\mathrm{V}_{0}$
$\mathrm{CB}=6 \mathrm{P}_{0}-3 \mathrm{P}_{0}=3 \mathrm{P}_{0}$
$\mathrm{AE}=3 \mathrm{P}_{0}, \mathrm{ED}=\mathrm{AC}=4 \mathrm{~V}_{0}$
Given, 1 mole of an ideal monoatomic gas undergoes the process $A B$. So, number of mole $n=1$
Work done in part $\mathrm{A}$ to $\mathrm{B}\left(\mathrm{W}_{\mathrm{A} \rightarrow \mathrm{B}}\right)=$ Area under $\mathrm{P}-\mathrm{V}$ diagram
$\mathrm{W}_{\mathrm{A} \rightarrow \mathrm{B}}=$ Area of $\triangle \mathrm{ACB}+$ Area of $\square \mathrm{ACDE}$
$=\frac{1}{2} \times \mathrm{AC} \times \mathrm{CB}+\mathrm{AE} \times \mathrm{ED}$
$=\frac{1}{2} \times 4 \mathrm{~V}_{0} \times 3 \mathrm{P}_{0}+3 \mathrm{P}_{0} \times 4 \mathrm{~V}_{0}$
$=6 \mathrm{P}_{0} \mathrm{~V}_{0}+12 \mathrm{P}_{0} \mathrm{~V}_{0}=18 \mathrm{P}_{0} \mathrm{~V}_{0}$
Now internal energy change $\Delta \mathrm{U}_{\mathrm{A} \rightarrow \mathrm{B}}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$
$=1 \times \frac{3}{2} \mathrm{R}\left(\mathrm{T}_{\mathrm{B}}-\mathrm{T}_{\mathrm{A}}\right)$
$\left\{\because\right.$ As the gas is monoatomic that's why $\left.\mathrm{C}_{\mathrm{v}}=\frac{3}{2} \mathrm{R}\right\}$
At the point $\mathrm{B}, \mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}=\mathrm{RT}_{\mathrm{B}}\left(\because \mathrm{P}_{\mathrm{B}}=6 \mathrm{P}_{0}, \mathrm{~V}_{\mathrm{B}}=5 \mathrm{~V}_{0}\right)$
$\mathrm{T}_{\mathrm{B}}=\frac{6 \mathrm{P}_{0} \cdot 5 \mathrm{~V}_{0}}{\mathrm{R}}$
$\mathrm{T}_{\mathrm{B}}=\frac{30 \mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}$
Similarly, $\mathrm{T}_{\mathrm{A}}=\frac{3 \mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}$
$\text { So, } \Delta \mathrm{U}_{\mathrm{A} \rightarrow \mathrm{B}} =\frac{3}{2} \mathrm{R} \times\left(\frac{30 \mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}-\frac{3 \mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}\right)$
$\Delta \mathrm{U}_{\mathrm{A} \rightarrow \mathrm{B}} =\frac{3}{2} \mathrm{R} \times \frac{\mathrm{P}_{0} \cdot \mathrm{V}_{0}}{\mathrm{R}}(30-3)$
$=\frac{81}{2} \mathrm{P}_{0} \mathrm{~V}_{0}$
Thus, from $1^{\text {st }}$ law of thermodynamics -
$\Delta \mathrm{Q}_{\mathrm{A} \rightarrow \mathrm{B}} =\Delta \mathrm{W}_{\mathrm{A} \rightarrow \mathrm{B}}+\Delta \mathrm{U}_{\mathrm{A} \rightarrow \mathrm{B}}$
$=18 \mathrm{P}_{0} \mathrm{~V}_{0}+\frac{81}{2} \mathrm{P}_{0} \mathrm{~V}_{0}$
$\mathrm{DQ}_{\mathrm{A} \rightarrow \mathrm{B}} =\frac{117}{2} \mathrm{P}_{0} \mathrm{~V}_{0} \text { (Putting from (i) and (ii)) }$
$\because$ Molar heat capacity $\mathrm{C}=\frac{\Delta \mathrm{Q}}{\mathrm{n} \cdot \Delta \mathrm{T}}$
$\mathrm{C}=\frac{\frac{117}{2} \mathrm{P}_{0} \mathrm{~V}_{0}}{1 \times\left(\frac{30 \mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{R}}-\frac{3 \mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{R}}\right)}$
$\mathrm{C}=\frac{\frac{117}{2} \mathrm{P}_{0} \mathrm{~V}_{0}}{\frac{27 \mathrm{P}_{0} \mathrm{~V}_{0}}{\mathrm{R}}}$
$\mathrm{C}=\frac{13 \mathrm{R}}{6}$

AP EAMCET (20.04.2019) Shift-II

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