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Thermal Properties of Matter

146594 Consider a compound slab consisting of two different materials having equal thickneses and thermal conductivities $K$ and $2 K$ , respectively. The equivalent thermal conductivity of the slab is

1

3 K

2

\frac{4}{3} K

3

\frac{2}{3} K

4

\sqrt{2} K

Explanation:

B
Exp:
We know that, thermal resistance of a material is given as-
$R = \frac{l}{KA}$
Where, $A =$ Area
$K = Thermal conductivity$
$l = Length of the material$
Therefore, the equivalent thermal conductivity of the slab,
$\frac{2 l}{K_{eq} A} = \frac{l}{KA} + \frac{l}{2 KA}$
$\frac{2 l}{K_{eq} A} = \frac{l}{A} (\frac{1}{K} + \frac{1}{2 K})$
$\frac{2}{K_{eq}} = \frac{2 + 1}{2 K}$
Or $\frac{K_{eq}}{2} = \frac{2 K}{3}$
$K_{eq} = \frac{4}{3} K$

AIPMT-2003

Thermal Properties of Matter

146595 The two ends of a rod of length $L$ and a uniform cross-sectional area $A$ are kept at two temperatures $T_{1}$ and $T_{2} (T_{1} > T_{2})$ . The rate of heat transfer ' $\frac{dQ}{dt}$ ' through the rod in a steady state is given by

1

\frac{dQ}{dt} = \frac{KL (T_{1} - T_{2})}{A}

2

\frac{dQ}{dt} = \frac{K (T_{1} - T_{2})}{LA}

3

\frac{dQ}{dt} = KLA (T_{1} - T_{2})

4

\frac{dQ}{dt} = \frac{KA (T_{1} - T_{2})}{L}

Explanation:

D Thermal conductivity :-The rate of heat transferred by conduction through a unit cross-section area of a material when a temperature difference exists perpendicular to the area is known as thermal conductivity.
$(K) = \frac{QL}{A Δ t}$
Thermal Resistance 'R' $\Rightarrow R = \frac{L}{K.A}$
Where, $Q =$ Amount of heat transferred
$A =$ Area of material
$L =$ Distance between two planes
$Δ T =$ Temperature difference

The temperature difference $(Δ T) = (T_{1} - T_{2})$
$[∵ T_{1} > T_{2}]$
$∵$ Rate of heat transfer $(\frac{dQ}{dt}) = \frac{Δ T}{R}$
$\frac{dQ}{dt} = \frac{Δ T}{L / KA}$
$\frac{dQ}{dt} = \frac{KA (T_{1} - T_{2})}{L}$

AIPMT-2009

Thermal Properties of Matter

146596 Two rods $A$ and $B$ of different materials are welded together as shown in figure. Their thermal conductivities are $K_{1}$ and $K_{2}$ . The thermal conductivity of the composite rod will be

1

\frac{K_{1} + K_{2}}{2}

2

\frac{3 (K_{1} + K_{2})}{2}

3

K_{1} + K_{2}

4

2 (K_{1} + K_{2})

Explanation:

A Given that, thermal conductivities are $K_{1}$ and $K_{2}$ .
We know that, thermal resistance $(R) = \frac{d}{KA}$
$R_{1} = \frac{d}{K_{1} A}$
$R_{2} = \frac{d}{K_{2} A}$
Equivalent thermal resistance of $K_{1}$ and $K_{2}$ are paralld, are-
$R_{eq} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}$
Putting the value of these, we get-
$\frac{d}{K_{e q} (2 A)} = \frac{\frac{d}{K_{1} A} \times \frac{d}{K_{2} A}}{\frac{d}{K_{1} A} + \frac{d}{K_{2} A}}$
$\frac{d}{K_{e q} (2 A)} = \frac{\frac{d^{2}}{A^{2}} \times (\frac{1}{K_{1} K_{2}})}{\frac{d}{A} (\frac{1}{K_{1}} + \frac{1}{K_{2}})}$
$\frac{1}{2 K_{e q}} = \frac{1 / K_{1} K_{2}}{\frac{K_{2} + K_{1}}{K_{1} K_{2}}}$
$\frac{1}{2 K_{e q}} = \frac{1}{K_{2} + K_{1}}$
$Or K_{eq} = \frac{K_{1} + K_{2}}{2}$

NEET-2017

Thermal Properties of Matter

146597 The coefficients of linear expansions of brass and steel are $α_{1}$ and $α_{2}$ respectively. When we take a brass rod of length $l_{1}$ and a steel rod of length $l_{2}$ at $0^{\circ} C$ , then the difference in their lengths $(l_{2} - l_{1})$ will remain the same at all temperatures, if

1

α_{1} l_{1} = α_{2} l_{2}

2

α_{1} l_{2} = α_{2} l_{1}

3

α_{1}^{2} l_{2} = α_{2}^{2} l_{1}

4

α_{1} l_{2}^{2} = α_{2} l_{1}^{2}

Explanation:

A Given, coefficient of linear expansion of brass and steel are $α_{1}$ and $α_{2}$ respectively we know that, coefficient of linear expansion is -
$α = \frac{Δ l}{l Δ t}$
$∵$ Length of brass rod $= l_{1} (1 + α_{1} T)$
Length of steel rod $= l_{2} (1 + α_{2} T)$
Difference in brass rod and steel rod,
$= l_{1} (1 + α_{1} T) - l_{2} (1 + α_{2} T)$
$= l_{1} + l_{1} α_{1} T - (l_{2} + l_{2} α_{2} T)$
$= (l_{1} - l_{2}) + T (l_{1} α_{1} - l_{2} α_{2})$
Thus, length will be independent of temperature only when coefficient of temperature will be equal to zero
$i.e l_{1} α_{1} - l_{2} α_{2} = 0$
$l_{1} α_{1} = l_{2} α_{2}$

MHT-CET.2020 NEET-2016 AIPMT-1999

Thermal Properties of Matter

146594 Consider a compound slab consisting of two different materials having equal thickneses and thermal conductivities $K$ and $2 K$ , respectively. The equivalent thermal conductivity of the slab is

1

3 K

2

\frac{4}{3} K

3

\frac{2}{3} K

4

\sqrt{2} K

Explanation:

B
Exp:
We know that, thermal resistance of a material is given as-
$R = \frac{l}{KA}$
Where, $A =$ Area
$K = Thermal conductivity$
$l = Length of the material$
Therefore, the equivalent thermal conductivity of the slab,
$\frac{2 l}{K_{eq} A} = \frac{l}{KA} + \frac{l}{2 KA}$
$\frac{2 l}{K_{eq} A} = \frac{l}{A} (\frac{1}{K} + \frac{1}{2 K})$
$\frac{2}{K_{eq}} = \frac{2 + 1}{2 K}$
Or $\frac{K_{eq}}{2} = \frac{2 K}{3}$
$K_{eq} = \frac{4}{3} K$

AIPMT-2003

Thermal Properties of Matter

146595 The two ends of a rod of length $L$ and a uniform cross-sectional area $A$ are kept at two temperatures $T_{1}$ and $T_{2} (T_{1} > T_{2})$ . The rate of heat transfer ' $\frac{dQ}{dt}$ ' through the rod in a steady state is given by

1

\frac{dQ}{dt} = \frac{KL (T_{1} - T_{2})}{A}

2

\frac{dQ}{dt} = \frac{K (T_{1} - T_{2})}{LA}

3

\frac{dQ}{dt} = KLA (T_{1} - T_{2})

4

\frac{dQ}{dt} = \frac{KA (T_{1} - T_{2})}{L}

Explanation:

D Thermal conductivity :-The rate of heat transferred by conduction through a unit cross-section area of a material when a temperature difference exists perpendicular to the area is known as thermal conductivity.
$(K) = \frac{QL}{A Δ t}$
Thermal Resistance 'R' $\Rightarrow R = \frac{L}{K.A}$
Where, $Q =$ Amount of heat transferred
$A =$ Area of material
$L =$ Distance between two planes
$Δ T =$ Temperature difference

The temperature difference $(Δ T) = (T_{1} - T_{2})$
$[∵ T_{1} > T_{2}]$
$∵$ Rate of heat transfer $(\frac{dQ}{dt}) = \frac{Δ T}{R}$
$\frac{dQ}{dt} = \frac{Δ T}{L / KA}$
$\frac{dQ}{dt} = \frac{KA (T_{1} - T_{2})}{L}$

AIPMT-2009

Thermal Properties of Matter

146596 Two rods $A$ and $B$ of different materials are welded together as shown in figure. Their thermal conductivities are $K_{1}$ and $K_{2}$ . The thermal conductivity of the composite rod will be

1

\frac{K_{1} + K_{2}}{2}

2

\frac{3 (K_{1} + K_{2})}{2}

3

K_{1} + K_{2}

4

2 (K_{1} + K_{2})

Explanation:

A Given that, thermal conductivities are $K_{1}$ and $K_{2}$ .
We know that, thermal resistance $(R) = \frac{d}{KA}$
$R_{1} = \frac{d}{K_{1} A}$
$R_{2} = \frac{d}{K_{2} A}$
Equivalent thermal resistance of $K_{1}$ and $K_{2}$ are paralld, are-
$R_{eq} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}$
Putting the value of these, we get-
$\frac{d}{K_{e q} (2 A)} = \frac{\frac{d}{K_{1} A} \times \frac{d}{K_{2} A}}{\frac{d}{K_{1} A} + \frac{d}{K_{2} A}}$
$\frac{d}{K_{e q} (2 A)} = \frac{\frac{d^{2}}{A^{2}} \times (\frac{1}{K_{1} K_{2}})}{\frac{d}{A} (\frac{1}{K_{1}} + \frac{1}{K_{2}})}$
$\frac{1}{2 K_{e q}} = \frac{1 / K_{1} K_{2}}{\frac{K_{2} + K_{1}}{K_{1} K_{2}}}$
$\frac{1}{2 K_{e q}} = \frac{1}{K_{2} + K_{1}}$
$Or K_{eq} = \frac{K_{1} + K_{2}}{2}$

NEET-2017

Thermal Properties of Matter

146597 The coefficients of linear expansions of brass and steel are $α_{1}$ and $α_{2}$ respectively. When we take a brass rod of length $l_{1}$ and a steel rod of length $l_{2}$ at $0^{\circ} C$ , then the difference in their lengths $(l_{2} - l_{1})$ will remain the same at all temperatures, if

1

α_{1} l_{1} = α_{2} l_{2}

2

α_{1} l_{2} = α_{2} l_{1}

3

α_{1}^{2} l_{2} = α_{2}^{2} l_{1}

4

α_{1} l_{2}^{2} = α_{2} l_{1}^{2}

Explanation:

A Given, coefficient of linear expansion of brass and steel are $α_{1}$ and $α_{2}$ respectively we know that, coefficient of linear expansion is -
$α = \frac{Δ l}{l Δ t}$
$∵$ Length of brass rod $= l_{1} (1 + α_{1} T)$
Length of steel rod $= l_{2} (1 + α_{2} T)$
Difference in brass rod and steel rod,
$= l_{1} (1 + α_{1} T) - l_{2} (1 + α_{2} T)$
$= l_{1} + l_{1} α_{1} T - (l_{2} + l_{2} α_{2} T)$
$= (l_{1} - l_{2}) + T (l_{1} α_{1} - l_{2} α_{2})$
Thus, length will be independent of temperature only when coefficient of temperature will be equal to zero
$i.e l_{1} α_{1} - l_{2} α_{2} = 0$
$l_{1} α_{1} = l_{2} α_{2}$

MHT-CET.2020 NEET-2016 AIPMT-1999

Thermal Properties of Matter

146594 Consider a compound slab consisting of two different materials having equal thickneses and thermal conductivities $K$ and $2 K$ , respectively. The equivalent thermal conductivity of the slab is

1

3 K

2

\frac{4}{3} K

3

\frac{2}{3} K

4

\sqrt{2} K

Explanation:

B
Exp:
We know that, thermal resistance of a material is given as-
$R = \frac{l}{KA}$
Where, $A =$ Area
$K = Thermal conductivity$
$l = Length of the material$
Therefore, the equivalent thermal conductivity of the slab,
$\frac{2 l}{K_{eq} A} = \frac{l}{KA} + \frac{l}{2 KA}$
$\frac{2 l}{K_{eq} A} = \frac{l}{A} (\frac{1}{K} + \frac{1}{2 K})$
$\frac{2}{K_{eq}} = \frac{2 + 1}{2 K}$
Or $\frac{K_{eq}}{2} = \frac{2 K}{3}$
$K_{eq} = \frac{4}{3} K$

AIPMT-2003

Thermal Properties of Matter

146595 The two ends of a rod of length $L$ and a uniform cross-sectional area $A$ are kept at two temperatures $T_{1}$ and $T_{2} (T_{1} > T_{2})$ . The rate of heat transfer ' $\frac{dQ}{dt}$ ' through the rod in a steady state is given by

1

\frac{dQ}{dt} = \frac{KL (T_{1} - T_{2})}{A}

2

\frac{dQ}{dt} = \frac{K (T_{1} - T_{2})}{LA}

3

\frac{dQ}{dt} = KLA (T_{1} - T_{2})

4

\frac{dQ}{dt} = \frac{KA (T_{1} - T_{2})}{L}

Explanation:

D Thermal conductivity :-The rate of heat transferred by conduction through a unit cross-section area of a material when a temperature difference exists perpendicular to the area is known as thermal conductivity.
$(K) = \frac{QL}{A Δ t}$
Thermal Resistance 'R' $\Rightarrow R = \frac{L}{K.A}$
Where, $Q =$ Amount of heat transferred
$A =$ Area of material
$L =$ Distance between two planes
$Δ T =$ Temperature difference

The temperature difference $(Δ T) = (T_{1} - T_{2})$
$[∵ T_{1} > T_{2}]$
$∵$ Rate of heat transfer $(\frac{dQ}{dt}) = \frac{Δ T}{R}$
$\frac{dQ}{dt} = \frac{Δ T}{L / KA}$
$\frac{dQ}{dt} = \frac{KA (T_{1} - T_{2})}{L}$

AIPMT-2009

Thermal Properties of Matter

146596 Two rods $A$ and $B$ of different materials are welded together as shown in figure. Their thermal conductivities are $K_{1}$ and $K_{2}$ . The thermal conductivity of the composite rod will be

1

\frac{K_{1} + K_{2}}{2}

2

\frac{3 (K_{1} + K_{2})}{2}

3

K_{1} + K_{2}

4

2 (K_{1} + K_{2})

Explanation:

A Given that, thermal conductivities are $K_{1}$ and $K_{2}$ .
We know that, thermal resistance $(R) = \frac{d}{KA}$
$R_{1} = \frac{d}{K_{1} A}$
$R_{2} = \frac{d}{K_{2} A}$
Equivalent thermal resistance of $K_{1}$ and $K_{2}$ are paralld, are-
$R_{eq} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}$
Putting the value of these, we get-
$\frac{d}{K_{e q} (2 A)} = \frac{\frac{d}{K_{1} A} \times \frac{d}{K_{2} A}}{\frac{d}{K_{1} A} + \frac{d}{K_{2} A}}$
$\frac{d}{K_{e q} (2 A)} = \frac{\frac{d^{2}}{A^{2}} \times (\frac{1}{K_{1} K_{2}})}{\frac{d}{A} (\frac{1}{K_{1}} + \frac{1}{K_{2}})}$
$\frac{1}{2 K_{e q}} = \frac{1 / K_{1} K_{2}}{\frac{K_{2} + K_{1}}{K_{1} K_{2}}}$
$\frac{1}{2 K_{e q}} = \frac{1}{K_{2} + K_{1}}$
$Or K_{eq} = \frac{K_{1} + K_{2}}{2}$

NEET-2017

Thermal Properties of Matter

146597 The coefficients of linear expansions of brass and steel are $α_{1}$ and $α_{2}$ respectively. When we take a brass rod of length $l_{1}$ and a steel rod of length $l_{2}$ at $0^{\circ} C$ , then the difference in their lengths $(l_{2} - l_{1})$ will remain the same at all temperatures, if

1

α_{1} l_{1} = α_{2} l_{2}

2

α_{1} l_{2} = α_{2} l_{1}

3

α_{1}^{2} l_{2} = α_{2}^{2} l_{1}

4

α_{1} l_{2}^{2} = α_{2} l_{1}^{2}

Explanation:

A Given, coefficient of linear expansion of brass and steel are $α_{1}$ and $α_{2}$ respectively we know that, coefficient of linear expansion is -
$α = \frac{Δ l}{l Δ t}$
$∵$ Length of brass rod $= l_{1} (1 + α_{1} T)$
Length of steel rod $= l_{2} (1 + α_{2} T)$
Difference in brass rod and steel rod,
$= l_{1} (1 + α_{1} T) - l_{2} (1 + α_{2} T)$
$= l_{1} + l_{1} α_{1} T - (l_{2} + l_{2} α_{2} T)$
$= (l_{1} - l_{2}) + T (l_{1} α_{1} - l_{2} α_{2})$
Thus, length will be independent of temperature only when coefficient of temperature will be equal to zero
$i.e l_{1} α_{1} - l_{2} α_{2} = 0$
$l_{1} α_{1} = l_{2} α_{2}$

MHT-CET.2020 NEET-2016 AIPMT-1999

Thermal Properties of Matter

146594 Consider a compound slab consisting of two different materials having equal thickneses and thermal conductivities $K$ and $2 K$ , respectively. The equivalent thermal conductivity of the slab is

1

3 K

2

\frac{4}{3} K

3

\frac{2}{3} K

4

\sqrt{2} K

Explanation:

B
Exp:
We know that, thermal resistance of a material is given as-
$R = \frac{l}{KA}$
Where, $A =$ Area
$K = Thermal conductivity$
$l = Length of the material$
Therefore, the equivalent thermal conductivity of the slab,
$\frac{2 l}{K_{eq} A} = \frac{l}{KA} + \frac{l}{2 KA}$
$\frac{2 l}{K_{eq} A} = \frac{l}{A} (\frac{1}{K} + \frac{1}{2 K})$
$\frac{2}{K_{eq}} = \frac{2 + 1}{2 K}$
Or $\frac{K_{eq}}{2} = \frac{2 K}{3}$
$K_{eq} = \frac{4}{3} K$

AIPMT-2003

Thermal Properties of Matter

146595 The two ends of a rod of length $L$ and a uniform cross-sectional area $A$ are kept at two temperatures $T_{1}$ and $T_{2} (T_{1} > T_{2})$ . The rate of heat transfer ' $\frac{dQ}{dt}$ ' through the rod in a steady state is given by

1

\frac{dQ}{dt} = \frac{KL (T_{1} - T_{2})}{A}

2

\frac{dQ}{dt} = \frac{K (T_{1} - T_{2})}{LA}

3

\frac{dQ}{dt} = KLA (T_{1} - T_{2})

4

\frac{dQ}{dt} = \frac{KA (T_{1} - T_{2})}{L}

Explanation:

D Thermal conductivity :-The rate of heat transferred by conduction through a unit cross-section area of a material when a temperature difference exists perpendicular to the area is known as thermal conductivity.
$(K) = \frac{QL}{A Δ t}$
Thermal Resistance 'R' $\Rightarrow R = \frac{L}{K.A}$
Where, $Q =$ Amount of heat transferred
$A =$ Area of material
$L =$ Distance between two planes
$Δ T =$ Temperature difference

The temperature difference $(Δ T) = (T_{1} - T_{2})$
$[∵ T_{1} > T_{2}]$
$∵$ Rate of heat transfer $(\frac{dQ}{dt}) = \frac{Δ T}{R}$
$\frac{dQ}{dt} = \frac{Δ T}{L / KA}$
$\frac{dQ}{dt} = \frac{KA (T_{1} - T_{2})}{L}$

AIPMT-2009

Thermal Properties of Matter

146596 Two rods $A$ and $B$ of different materials are welded together as shown in figure. Their thermal conductivities are $K_{1}$ and $K_{2}$ . The thermal conductivity of the composite rod will be

1

\frac{K_{1} + K_{2}}{2}

2

\frac{3 (K_{1} + K_{2})}{2}

3

K_{1} + K_{2}

4

2 (K_{1} + K_{2})

Explanation:

A Given that, thermal conductivities are $K_{1}$ and $K_{2}$ .
We know that, thermal resistance $(R) = \frac{d}{KA}$
$R_{1} = \frac{d}{K_{1} A}$
$R_{2} = \frac{d}{K_{2} A}$
Equivalent thermal resistance of $K_{1}$ and $K_{2}$ are paralld, are-
$R_{eq} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}$
Putting the value of these, we get-
$\frac{d}{K_{e q} (2 A)} = \frac{\frac{d}{K_{1} A} \times \frac{d}{K_{2} A}}{\frac{d}{K_{1} A} + \frac{d}{K_{2} A}}$
$\frac{d}{K_{e q} (2 A)} = \frac{\frac{d^{2}}{A^{2}} \times (\frac{1}{K_{1} K_{2}})}{\frac{d}{A} (\frac{1}{K_{1}} + \frac{1}{K_{2}})}$
$\frac{1}{2 K_{e q}} = \frac{1 / K_{1} K_{2}}{\frac{K_{2} + K_{1}}{K_{1} K_{2}}}$
$\frac{1}{2 K_{e q}} = \frac{1}{K_{2} + K_{1}}$
$Or K_{eq} = \frac{K_{1} + K_{2}}{2}$

NEET-2017

Thermal Properties of Matter

146597 The coefficients of linear expansions of brass and steel are $α_{1}$ and $α_{2}$ respectively. When we take a brass rod of length $l_{1}$ and a steel rod of length $l_{2}$ at $0^{\circ} C$ , then the difference in their lengths $(l_{2} - l_{1})$ will remain the same at all temperatures, if

1

α_{1} l_{1} = α_{2} l_{2}

2

α_{1} l_{2} = α_{2} l_{1}

3

α_{1}^{2} l_{2} = α_{2}^{2} l_{1}

4

α_{1} l_{2}^{2} = α_{2} l_{1}^{2}

Explanation:

A Given, coefficient of linear expansion of brass and steel are $α_{1}$ and $α_{2}$ respectively we know that, coefficient of linear expansion is -
$α = \frac{Δ l}{l Δ t}$
$∵$ Length of brass rod $= l_{1} (1 + α_{1} T)$
Length of steel rod $= l_{2} (1 + α_{2} T)$
Difference in brass rod and steel rod,
$= l_{1} (1 + α_{1} T) - l_{2} (1 + α_{2} T)$
$= l_{1} + l_{1} α_{1} T - (l_{2} + l_{2} α_{2} T)$
$= (l_{1} - l_{2}) + T (l_{1} α_{1} - l_{2} α_{2})$
Thus, length will be independent of temperature only when coefficient of temperature will be equal to zero
$i.e l_{1} α_{1} - l_{2} α_{2} = 0$
$l_{1} α_{1} = l_{2} α_{2}$

MHT-CET.2020 NEET-2016 AIPMT-1999

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