QBManager

Thermal Properties of Matter

146499 A silver wire has temperature coefficient of resistivity $4 \times 10^{- 3} /^{\circ} C$ and its resistance at $20^{\circ} C$ is $10 Ω$ . Neglecting any change in dimensions due to the change in temperature, its resistance at $40^{\circ} C$ is

1

0.8 Ω

2

1.8 Ω

3

9.2 Ω

4

10.8 Ω

5

11.6 Ω

Explanation:

D Given,
Coefficient of resistivity $= 4 \times 10^{- 3} /^{\circ} C$
$T_{1} = 20^{\circ} C$ ,
$T_{2} = 40^{\circ} C$
$Δ T = T_{2} - T_{1} = 40 - 20 = 20^{\circ} C$
$R_{20^{\circ} C} = 10 Ω$
$R_{40^{\circ} C} =$ ?
We know that,
$R_{40^{\circ} C} = R_{20^{\circ} C} [1 + α Δ T]$
$R_{40}^{\circ} C = 10 [1 + 4 \times 10^{- 3} \times 20]$
$R_{40}^{\circ} C = 10 [1 + 80 \times 10^{- 3}]$
$R_{40}^{\circ} C = 10 [1.08]$
$R_{40}^{\circ} C = 10.8$

Kerala CEE - 2017

Thermal Properties of Matter

146500 A bubble of 8 mole of helium is submerged at a certain depth in water. The temperature of water increases by $30^{\circ} C$ . How much heat is added approximately to helium during expansion?

1

4000 J

2

3000 J

3

3500 J

4

4500 J

5

5000 J

Explanation:

B
Exp:
Heat added during expansion of Helium (for monoatomic gas) $= {nC}_{v} Δ T$ ,
Then,
$Q = 8 \times \frac{3}{2} \times 8.31 \times 30 [C_{v} = \frac{3}{2} R]$
$∴ Q = 2991.6 J ≃ 3000 J$

Kerala CEE - 2008

Thermal Properties of Matter

146501 The resistance of a wire at room temperature $30^{\circ} C$ is found to be $10 Ω$ . Now to increase the resistance by $10 %$ , the temperature of the wire must be [The temperature coefficient of resistance of the material of the wire is $0.002 /^{\circ} C]$

1

36^{\circ} C

2

83^{\circ} C

3

63^{\circ} C

4

33^{\circ} C

5

66^{\circ} C

Explanation:

B Given,
$R_{30}^{\circ} C = 10 Ω$
$T_{1} = 30^{\circ} C$
When resistance increases by $10 %$ at temperature $t$ ,
$Change in resistance (Δ R_{t}) = \frac{10}{100} \times 10 = 1 Ω$
$R_{t} = R_{30^{\circ} C} + Δ R_{t}$
$= 10 Ω + 1 Ω = 11 Ω$
Now,
Case - 1
When $T = 30^{\circ} C$
$R_{30}^{\circ} C = R_{0} [1 + α T_{1}]$
$10 Ω = R_{0} [1 + 30 α]$
Case -2
$When T = t^{\circ} C$
$R_{t} = R_{0} [1 + α T]$
$11 Ω = R_{0} [1 + α t] \dots . . (2)$
$Dividing eq^{n} (2) \div {eq}^{n} (1)$
$\frac{11}{10} = \frac{R_{0} [1 + α t]}{R_{0} [1 + 30 α]}$
$11 + 330 α = 10 + 10 α t$
$11 + 330 \times 0.002 = 10 + 10 \times 0.002 \times t$
$11.66 = 0.02 t + 10$
$0.02 t = 1.66$
$∴ t = 83^{\circ} C$

Kerala CEE 2007

Thermal Properties of Matter

146502 The volume of a metal sphere increases by $0.24 %$ when its temperature is raised by $40^{\circ} C$ . The coefficient of linear expansion of the metal is... $/^{\circ} C$ .

1

2 \times 10^{- 5}

2

6 \times 10^{- 5}

3

18 \times 10^{- 5}

4

1.2 \times 10^{- 5}

5

2.1 \times 10^{- 5}

Explanation:

A Given,
Change in volume $(Δ V) = 0.24 %$
Temperature $(Δ T) = 40^{\circ} C$
We know that,
Change in volume, $Δ V = γ V Δ T$
$0.24 = γ \times 100 \times 40$
$γ = \frac{0.24}{100 \times 40}$
$γ = 6 \times 10^{- 5} /^{\circ} C$
Now,
$α = \frac{γ}{3}$
So,
$α = \frac{6 \times 10^{- 5}}{3} /^{\circ} C$
$α = 2 \times 10^{- 5} /^{\circ} C$

Kerala CEE 2005

Thermal Properties of Matter

146503 One junction of a certain thermoelectric couple is at a fixed temperature $T_{r}$ and the other junction is at temperature $T$ . The thermoelectromotive force for this is expressed by $E =$ $k (T - T_{r}) [T_{0} - \frac{1}{2} (T + T_{r})]$ . At temperature $T$ $= \frac{1}{2} T_{0}$ , the thermoelectric power is

1

\frac{1}{2} {kT}_{0}

2

{kT}_{0}

3

\frac{1}{2} {kT}_{0}^{2}

4

\frac{1}{2} k {(T_{0} - T_{r})}^{2}

Explanation:

A Given,
$E = k (T - T_{r}) [T_{0} - \frac{1}{2} (T + T_{r})]$
We know that,
Thermometric power, $S = \frac{dE}{dT}$
Differentiating the eq $^{n}$ (i) with respect to ' $t$ '
$\frac{dE}{dT} = k [T_{0} - T]$
Now, at $T = \frac{T_{0}}{2}$
$\frac{dE}{dT} = k [T_{0} - \frac{T_{0}}{2}]$
$∴ \frac{dE}{dT} = \frac{{kT}_{0}}{2}$

UPSEE - 2012

Thermal Properties of Matter

146499 A silver wire has temperature coefficient of resistivity $4 \times 10^{- 3} /^{\circ} C$ and its resistance at $20^{\circ} C$ is $10 Ω$ . Neglecting any change in dimensions due to the change in temperature, its resistance at $40^{\circ} C$ is

1

0.8 Ω

2

1.8 Ω

3

9.2 Ω

4

10.8 Ω

5

11.6 Ω

Explanation:

D Given,
Coefficient of resistivity $= 4 \times 10^{- 3} /^{\circ} C$
$T_{1} = 20^{\circ} C$ ,
$T_{2} = 40^{\circ} C$
$Δ T = T_{2} - T_{1} = 40 - 20 = 20^{\circ} C$
$R_{20^{\circ} C} = 10 Ω$
$R_{40^{\circ} C} =$ ?
We know that,
$R_{40^{\circ} C} = R_{20^{\circ} C} [1 + α Δ T]$
$R_{40}^{\circ} C = 10 [1 + 4 \times 10^{- 3} \times 20]$
$R_{40}^{\circ} C = 10 [1 + 80 \times 10^{- 3}]$
$R_{40}^{\circ} C = 10 [1.08]$
$R_{40}^{\circ} C = 10.8$

Kerala CEE - 2017

Thermal Properties of Matter

146500 A bubble of 8 mole of helium is submerged at a certain depth in water. The temperature of water increases by $30^{\circ} C$ . How much heat is added approximately to helium during expansion?

1

4000 J

2

3000 J

3

3500 J

4

4500 J

5

5000 J

Explanation:

B
Exp:
Heat added during expansion of Helium (for monoatomic gas) $= {nC}_{v} Δ T$ ,
Then,
$Q = 8 \times \frac{3}{2} \times 8.31 \times 30 [C_{v} = \frac{3}{2} R]$
$∴ Q = 2991.6 J ≃ 3000 J$

Kerala CEE - 2008

Thermal Properties of Matter

146501 The resistance of a wire at room temperature $30^{\circ} C$ is found to be $10 Ω$ . Now to increase the resistance by $10 %$ , the temperature of the wire must be [The temperature coefficient of resistance of the material of the wire is $0.002 /^{\circ} C]$

1

36^{\circ} C

2

83^{\circ} C

3

63^{\circ} C

4

33^{\circ} C

5

66^{\circ} C

Explanation:

B Given,
$R_{30}^{\circ} C = 10 Ω$
$T_{1} = 30^{\circ} C$
When resistance increases by $10 %$ at temperature $t$ ,
$Change in resistance (Δ R_{t}) = \frac{10}{100} \times 10 = 1 Ω$
$R_{t} = R_{30^{\circ} C} + Δ R_{t}$
$= 10 Ω + 1 Ω = 11 Ω$
Now,
Case - 1
When $T = 30^{\circ} C$
$R_{30}^{\circ} C = R_{0} [1 + α T_{1}]$
$10 Ω = R_{0} [1 + 30 α]$
Case -2
$When T = t^{\circ} C$
$R_{t} = R_{0} [1 + α T]$
$11 Ω = R_{0} [1 + α t] \dots . . (2)$
$Dividing eq^{n} (2) \div {eq}^{n} (1)$
$\frac{11}{10} = \frac{R_{0} [1 + α t]}{R_{0} [1 + 30 α]}$
$11 + 330 α = 10 + 10 α t$
$11 + 330 \times 0.002 = 10 + 10 \times 0.002 \times t$
$11.66 = 0.02 t + 10$
$0.02 t = 1.66$
$∴ t = 83^{\circ} C$

Kerala CEE 2007

Thermal Properties of Matter

146502 The volume of a metal sphere increases by $0.24 %$ when its temperature is raised by $40^{\circ} C$ . The coefficient of linear expansion of the metal is... $/^{\circ} C$ .

1

2 \times 10^{- 5}

2

6 \times 10^{- 5}

3

18 \times 10^{- 5}

4

1.2 \times 10^{- 5}

5

2.1 \times 10^{- 5}

Explanation:

A Given,
Change in volume $(Δ V) = 0.24 %$
Temperature $(Δ T) = 40^{\circ} C$
We know that,
Change in volume, $Δ V = γ V Δ T$
$0.24 = γ \times 100 \times 40$
$γ = \frac{0.24}{100 \times 40}$
$γ = 6 \times 10^{- 5} /^{\circ} C$
Now,
$α = \frac{γ}{3}$
So,
$α = \frac{6 \times 10^{- 5}}{3} /^{\circ} C$
$α = 2 \times 10^{- 5} /^{\circ} C$

Kerala CEE 2005

Thermal Properties of Matter

146503 One junction of a certain thermoelectric couple is at a fixed temperature $T_{r}$ and the other junction is at temperature $T$ . The thermoelectromotive force for this is expressed by $E =$ $k (T - T_{r}) [T_{0} - \frac{1}{2} (T + T_{r})]$ . At temperature $T$ $= \frac{1}{2} T_{0}$ , the thermoelectric power is

1

\frac{1}{2} {kT}_{0}

2

{kT}_{0}

3

\frac{1}{2} {kT}_{0}^{2}

4

\frac{1}{2} k {(T_{0} - T_{r})}^{2}

Explanation:

A Given,
$E = k (T - T_{r}) [T_{0} - \frac{1}{2} (T + T_{r})]$
We know that,
Thermometric power, $S = \frac{dE}{dT}$
Differentiating the eq $^{n}$ (i) with respect to ' $t$ '
$\frac{dE}{dT} = k [T_{0} - T]$
Now, at $T = \frac{T_{0}}{2}$
$\frac{dE}{dT} = k [T_{0} - \frac{T_{0}}{2}]$
$∴ \frac{dE}{dT} = \frac{{kT}_{0}}{2}$

UPSEE - 2012

Thermal Properties of Matter

146499 A silver wire has temperature coefficient of resistivity $4 \times 10^{- 3} /^{\circ} C$ and its resistance at $20^{\circ} C$ is $10 Ω$ . Neglecting any change in dimensions due to the change in temperature, its resistance at $40^{\circ} C$ is

1

0.8 Ω

2

1.8 Ω

3

9.2 Ω

4

10.8 Ω

5

11.6 Ω

Explanation:

D Given,
Coefficient of resistivity $= 4 \times 10^{- 3} /^{\circ} C$
$T_{1} = 20^{\circ} C$ ,
$T_{2} = 40^{\circ} C$
$Δ T = T_{2} - T_{1} = 40 - 20 = 20^{\circ} C$
$R_{20^{\circ} C} = 10 Ω$
$R_{40^{\circ} C} =$ ?
We know that,
$R_{40^{\circ} C} = R_{20^{\circ} C} [1 + α Δ T]$
$R_{40}^{\circ} C = 10 [1 + 4 \times 10^{- 3} \times 20]$
$R_{40}^{\circ} C = 10 [1 + 80 \times 10^{- 3}]$
$R_{40}^{\circ} C = 10 [1.08]$
$R_{40}^{\circ} C = 10.8$

Kerala CEE - 2017

Thermal Properties of Matter

146500 A bubble of 8 mole of helium is submerged at a certain depth in water. The temperature of water increases by $30^{\circ} C$ . How much heat is added approximately to helium during expansion?

1

4000 J

2

3000 J

3

3500 J

4

4500 J

5

5000 J

Explanation:

B
Exp:
Heat added during expansion of Helium (for monoatomic gas) $= {nC}_{v} Δ T$ ,
Then,
$Q = 8 \times \frac{3}{2} \times 8.31 \times 30 [C_{v} = \frac{3}{2} R]$
$∴ Q = 2991.6 J ≃ 3000 J$

Kerala CEE - 2008

Thermal Properties of Matter

146501 The resistance of a wire at room temperature $30^{\circ} C$ is found to be $10 Ω$ . Now to increase the resistance by $10 %$ , the temperature of the wire must be [The temperature coefficient of resistance of the material of the wire is $0.002 /^{\circ} C]$

1

36^{\circ} C

2

83^{\circ} C

3

63^{\circ} C

4

33^{\circ} C

5

66^{\circ} C

Explanation:

B Given,
$R_{30}^{\circ} C = 10 Ω$
$T_{1} = 30^{\circ} C$
When resistance increases by $10 %$ at temperature $t$ ,
$Change in resistance (Δ R_{t}) = \frac{10}{100} \times 10 = 1 Ω$
$R_{t} = R_{30^{\circ} C} + Δ R_{t}$
$= 10 Ω + 1 Ω = 11 Ω$
Now,
Case - 1
When $T = 30^{\circ} C$
$R_{30}^{\circ} C = R_{0} [1 + α T_{1}]$
$10 Ω = R_{0} [1 + 30 α]$
Case -2
$When T = t^{\circ} C$
$R_{t} = R_{0} [1 + α T]$
$11 Ω = R_{0} [1 + α t] \dots . . (2)$
$Dividing eq^{n} (2) \div {eq}^{n} (1)$
$\frac{11}{10} = \frac{R_{0} [1 + α t]}{R_{0} [1 + 30 α]}$
$11 + 330 α = 10 + 10 α t$
$11 + 330 \times 0.002 = 10 + 10 \times 0.002 \times t$
$11.66 = 0.02 t + 10$
$0.02 t = 1.66$
$∴ t = 83^{\circ} C$

Kerala CEE 2007

Thermal Properties of Matter

146502 The volume of a metal sphere increases by $0.24 %$ when its temperature is raised by $40^{\circ} C$ . The coefficient of linear expansion of the metal is... $/^{\circ} C$ .

1

2 \times 10^{- 5}

2

6 \times 10^{- 5}

3

18 \times 10^{- 5}

4

1.2 \times 10^{- 5}

5

2.1 \times 10^{- 5}

Explanation:

A Given,
Change in volume $(Δ V) = 0.24 %$
Temperature $(Δ T) = 40^{\circ} C$
We know that,
Change in volume, $Δ V = γ V Δ T$
$0.24 = γ \times 100 \times 40$
$γ = \frac{0.24}{100 \times 40}$
$γ = 6 \times 10^{- 5} /^{\circ} C$
Now,
$α = \frac{γ}{3}$
So,
$α = \frac{6 \times 10^{- 5}}{3} /^{\circ} C$
$α = 2 \times 10^{- 5} /^{\circ} C$

Kerala CEE 2005

Thermal Properties of Matter

146503 One junction of a certain thermoelectric couple is at a fixed temperature $T_{r}$ and the other junction is at temperature $T$ . The thermoelectromotive force for this is expressed by $E =$ $k (T - T_{r}) [T_{0} - \frac{1}{2} (T + T_{r})]$ . At temperature $T$ $= \frac{1}{2} T_{0}$ , the thermoelectric power is

1

\frac{1}{2} {kT}_{0}

2

{kT}_{0}

3

\frac{1}{2} {kT}_{0}^{2}

4

\frac{1}{2} k {(T_{0} - T_{r})}^{2}

Explanation:

A Given,
$E = k (T - T_{r}) [T_{0} - \frac{1}{2} (T + T_{r})]$
We know that,
Thermometric power, $S = \frac{dE}{dT}$
Differentiating the eq $^{n}$ (i) with respect to ' $t$ '
$\frac{dE}{dT} = k [T_{0} - T]$
Now, at $T = \frac{T_{0}}{2}$
$\frac{dE}{dT} = k [T_{0} - \frac{T_{0}}{2}]$
$∴ \frac{dE}{dT} = \frac{{kT}_{0}}{2}$

UPSEE - 2012

Thermal Properties of Matter

146499 A silver wire has temperature coefficient of resistivity $4 \times 10^{- 3} /^{\circ} C$ and its resistance at $20^{\circ} C$ is $10 Ω$ . Neglecting any change in dimensions due to the change in temperature, its resistance at $40^{\circ} C$ is

1

0.8 Ω

2

1.8 Ω

3

9.2 Ω

4

10.8 Ω

5

11.6 Ω

Explanation:

D Given,
Coefficient of resistivity $= 4 \times 10^{- 3} /^{\circ} C$
$T_{1} = 20^{\circ} C$ ,
$T_{2} = 40^{\circ} C$
$Δ T = T_{2} - T_{1} = 40 - 20 = 20^{\circ} C$
$R_{20^{\circ} C} = 10 Ω$
$R_{40^{\circ} C} =$ ?
We know that,
$R_{40^{\circ} C} = R_{20^{\circ} C} [1 + α Δ T]$
$R_{40}^{\circ} C = 10 [1 + 4 \times 10^{- 3} \times 20]$
$R_{40}^{\circ} C = 10 [1 + 80 \times 10^{- 3}]$
$R_{40}^{\circ} C = 10 [1.08]$
$R_{40}^{\circ} C = 10.8$

Kerala CEE - 2017

Thermal Properties of Matter

146500 A bubble of 8 mole of helium is submerged at a certain depth in water. The temperature of water increases by $30^{\circ} C$ . How much heat is added approximately to helium during expansion?

1

4000 J

2

3000 J

3

3500 J

4

4500 J

5

5000 J

Explanation:

B
Exp:
Heat added during expansion of Helium (for monoatomic gas) $= {nC}_{v} Δ T$ ,
Then,
$Q = 8 \times \frac{3}{2} \times 8.31 \times 30 [C_{v} = \frac{3}{2} R]$
$∴ Q = 2991.6 J ≃ 3000 J$

Kerala CEE - 2008

Thermal Properties of Matter

146501 The resistance of a wire at room temperature $30^{\circ} C$ is found to be $10 Ω$ . Now to increase the resistance by $10 %$ , the temperature of the wire must be [The temperature coefficient of resistance of the material of the wire is $0.002 /^{\circ} C]$

1

36^{\circ} C

2

83^{\circ} C

3

63^{\circ} C

4

33^{\circ} C

5

66^{\circ} C

Explanation:

B Given,
$R_{30}^{\circ} C = 10 Ω$
$T_{1} = 30^{\circ} C$
When resistance increases by $10 %$ at temperature $t$ ,
$Change in resistance (Δ R_{t}) = \frac{10}{100} \times 10 = 1 Ω$
$R_{t} = R_{30^{\circ} C} + Δ R_{t}$
$= 10 Ω + 1 Ω = 11 Ω$
Now,
Case - 1
When $T = 30^{\circ} C$
$R_{30}^{\circ} C = R_{0} [1 + α T_{1}]$
$10 Ω = R_{0} [1 + 30 α]$
Case -2
$When T = t^{\circ} C$
$R_{t} = R_{0} [1 + α T]$
$11 Ω = R_{0} [1 + α t] \dots . . (2)$
$Dividing eq^{n} (2) \div {eq}^{n} (1)$
$\frac{11}{10} = \frac{R_{0} [1 + α t]}{R_{0} [1 + 30 α]}$
$11 + 330 α = 10 + 10 α t$
$11 + 330 \times 0.002 = 10 + 10 \times 0.002 \times t$
$11.66 = 0.02 t + 10$
$0.02 t = 1.66$
$∴ t = 83^{\circ} C$

Kerala CEE 2007

Thermal Properties of Matter

146502 The volume of a metal sphere increases by $0.24 %$ when its temperature is raised by $40^{\circ} C$ . The coefficient of linear expansion of the metal is... $/^{\circ} C$ .

1

2 \times 10^{- 5}

2

6 \times 10^{- 5}

3

18 \times 10^{- 5}

4

1.2 \times 10^{- 5}

5

2.1 \times 10^{- 5}

Explanation:

A Given,
Change in volume $(Δ V) = 0.24 %$
Temperature $(Δ T) = 40^{\circ} C$
We know that,
Change in volume, $Δ V = γ V Δ T$
$0.24 = γ \times 100 \times 40$
$γ = \frac{0.24}{100 \times 40}$
$γ = 6 \times 10^{- 5} /^{\circ} C$
Now,
$α = \frac{γ}{3}$
So,
$α = \frac{6 \times 10^{- 5}}{3} /^{\circ} C$
$α = 2 \times 10^{- 5} /^{\circ} C$

Kerala CEE 2005

Thermal Properties of Matter

146503 One junction of a certain thermoelectric couple is at a fixed temperature $T_{r}$ and the other junction is at temperature $T$ . The thermoelectromotive force for this is expressed by $E =$ $k (T - T_{r}) [T_{0} - \frac{1}{2} (T + T_{r})]$ . At temperature $T$ $= \frac{1}{2} T_{0}$ , the thermoelectric power is

1

\frac{1}{2} {kT}_{0}

2

{kT}_{0}

3

\frac{1}{2} {kT}_{0}^{2}

4

\frac{1}{2} k {(T_{0} - T_{r})}^{2}

Explanation:

A Given,
$E = k (T - T_{r}) [T_{0} - \frac{1}{2} (T + T_{r})]$
We know that,
Thermometric power, $S = \frac{dE}{dT}$
Differentiating the eq $^{n}$ (i) with respect to ' $t$ '
$\frac{dE}{dT} = k [T_{0} - T]$
Now, at $T = \frac{T_{0}}{2}$
$\frac{dE}{dT} = k [T_{0} - \frac{T_{0}}{2}]$
$∴ \frac{dE}{dT} = \frac{{kT}_{0}}{2}$

UPSEE - 2012

Thermal Properties of Matter

146499 A silver wire has temperature coefficient of resistivity $4 \times 10^{- 3} /^{\circ} C$ and its resistance at $20^{\circ} C$ is $10 Ω$ . Neglecting any change in dimensions due to the change in temperature, its resistance at $40^{\circ} C$ is

1

0.8 Ω

2

1.8 Ω

3

9.2 Ω

4

10.8 Ω

5

11.6 Ω

Explanation:

D Given,
Coefficient of resistivity $= 4 \times 10^{- 3} /^{\circ} C$
$T_{1} = 20^{\circ} C$ ,
$T_{2} = 40^{\circ} C$
$Δ T = T_{2} - T_{1} = 40 - 20 = 20^{\circ} C$
$R_{20^{\circ} C} = 10 Ω$
$R_{40^{\circ} C} =$ ?
We know that,
$R_{40^{\circ} C} = R_{20^{\circ} C} [1 + α Δ T]$
$R_{40}^{\circ} C = 10 [1 + 4 \times 10^{- 3} \times 20]$
$R_{40}^{\circ} C = 10 [1 + 80 \times 10^{- 3}]$
$R_{40}^{\circ} C = 10 [1.08]$
$R_{40}^{\circ} C = 10.8$

Kerala CEE - 2017

Thermal Properties of Matter

146500 A bubble of 8 mole of helium is submerged at a certain depth in water. The temperature of water increases by $30^{\circ} C$ . How much heat is added approximately to helium during expansion?

1

4000 J

2

3000 J

3

3500 J

4

4500 J

5

5000 J

Explanation:

B
Exp:
Heat added during expansion of Helium (for monoatomic gas) $= {nC}_{v} Δ T$ ,
Then,
$Q = 8 \times \frac{3}{2} \times 8.31 \times 30 [C_{v} = \frac{3}{2} R]$
$∴ Q = 2991.6 J ≃ 3000 J$

Kerala CEE - 2008

Thermal Properties of Matter

146501 The resistance of a wire at room temperature $30^{\circ} C$ is found to be $10 Ω$ . Now to increase the resistance by $10 %$ , the temperature of the wire must be [The temperature coefficient of resistance of the material of the wire is $0.002 /^{\circ} C]$

1

36^{\circ} C

2

83^{\circ} C

3

63^{\circ} C

4

33^{\circ} C

5

66^{\circ} C

Explanation:

B Given,
$R_{30}^{\circ} C = 10 Ω$
$T_{1} = 30^{\circ} C$
When resistance increases by $10 %$ at temperature $t$ ,
$Change in resistance (Δ R_{t}) = \frac{10}{100} \times 10 = 1 Ω$
$R_{t} = R_{30^{\circ} C} + Δ R_{t}$
$= 10 Ω + 1 Ω = 11 Ω$
Now,
Case - 1
When $T = 30^{\circ} C$
$R_{30}^{\circ} C = R_{0} [1 + α T_{1}]$
$10 Ω = R_{0} [1 + 30 α]$
Case -2
$When T = t^{\circ} C$
$R_{t} = R_{0} [1 + α T]$
$11 Ω = R_{0} [1 + α t] \dots . . (2)$
$Dividing eq^{n} (2) \div {eq}^{n} (1)$
$\frac{11}{10} = \frac{R_{0} [1 + α t]}{R_{0} [1 + 30 α]}$
$11 + 330 α = 10 + 10 α t$
$11 + 330 \times 0.002 = 10 + 10 \times 0.002 \times t$
$11.66 = 0.02 t + 10$
$0.02 t = 1.66$
$∴ t = 83^{\circ} C$

Kerala CEE 2007

Thermal Properties of Matter

146502 The volume of a metal sphere increases by $0.24 %$ when its temperature is raised by $40^{\circ} C$ . The coefficient of linear expansion of the metal is... $/^{\circ} C$ .

1

2 \times 10^{- 5}

2

6 \times 10^{- 5}

3

18 \times 10^{- 5}

4

1.2 \times 10^{- 5}

5

2.1 \times 10^{- 5}

Explanation:

A Given,
Change in volume $(Δ V) = 0.24 %$
Temperature $(Δ T) = 40^{\circ} C$
We know that,
Change in volume, $Δ V = γ V Δ T$
$0.24 = γ \times 100 \times 40$
$γ = \frac{0.24}{100 \times 40}$
$γ = 6 \times 10^{- 5} /^{\circ} C$
Now,
$α = \frac{γ}{3}$
So,
$α = \frac{6 \times 10^{- 5}}{3} /^{\circ} C$
$α = 2 \times 10^{- 5} /^{\circ} C$

Kerala CEE 2005

Thermal Properties of Matter

146503 One junction of a certain thermoelectric couple is at a fixed temperature $T_{r}$ and the other junction is at temperature $T$ . The thermoelectromotive force for this is expressed by $E =$ $k (T - T_{r}) [T_{0} - \frac{1}{2} (T + T_{r})]$ . At temperature $T$ $= \frac{1}{2} T_{0}$ , the thermoelectric power is

1

\frac{1}{2} {kT}_{0}

2

{kT}_{0}

3

\frac{1}{2} {kT}_{0}^{2}

4

\frac{1}{2} k {(T_{0} - T_{r})}^{2}

Explanation:

A Given,
$E = k (T - T_{r}) [T_{0} - \frac{1}{2} (T + T_{r})]$
We know that,
Thermometric power, $S = \frac{dE}{dT}$
Differentiating the eq $^{n}$ (i) with respect to ' $t$ '
$\frac{dE}{dT} = k [T_{0} - T]$
Now, at $T = \frac{T_{0}}{2}$
$\frac{dE}{dT} = k [T_{0} - \frac{T_{0}}{2}]$
$∴ \frac{dE}{dT} = \frac{{kT}_{0}}{2}$

UPSEE - 2012

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Question Bank Series

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