QBManager

Thermal Properties of Matter

146459 If $K_{1}$ and $K_{2}$ are the thermal conductivities $L_{1}$ $L_{2}$ are the lengths and $A_{1}$ and $A_{2}$ are the cross sectional areas of steel and cooper rods respectively such that $\frac{K_{2}}{K_{1}} = 9, \frac{A_{1}}{A_{2}} = 2, \frac{L_{1}}{L_{2}} = 2$ . Then, for the arrangement as shown in the figure. The value of temperature $T$ of the steelcopper junction in the steady state will be:

1

18^{\circ} C

2

14^{\circ} C

3

45^{\circ} C

4

150^{\circ} C

Explanation:

C
Considering the junction, we can write the heat equation as,
$450 - T = \frac{dQ}{dt} \times \frac{L_{1}}{K_{1} A_{1}}$
and $T - 0 = \frac{dQ}{dt} \times \frac{L_{2}}{K_{2} A_{2}}$
Dividing equation (i) by (ii), we get-
$\frac{450 - T}{T} = \frac{K_{2} A_{2} L_{1}}{K_{1} A_{1} L_{2}} = 9 \times \frac{1}{2} \times 2 = 9$
$450 - T = 9 T$
$T = 45^{\circ} C$

Shift-I]

Thermal Properties of Matter

146460 A standard resistance coil marked $2 Ω$ is found to have a resistance of $2.118 Ω$ at $30^{\circ} C$ the temperature at which marking is correct is (temperature coefficient of resistant of the material of the coil is 0.0042 per degree Celsius)

1

{15.05}^{\circ} C

2

{15.07}^{\circ} C

3

{15.09}^{\circ} C

4

{15.06}^{\circ} C

Explanation:

D Given, that,
$R_{1} = 2 Ω, α = 0.0042 /^{\circ} C, R_{2} = 2.118 Ω, T_{2} = 30^{\circ} C$ We know that,
$\frac{R_{1}}{R_{2}} = \frac{1 + α T_{1}}{1 + α T_{2}}$
$\frac{2}{2.118} = \frac{1 + 0.0042 T_{1}}{1 + 0.0042 \times 30}$
$0.94428 = \frac{1 + 0.0042 T_{1}}{1 + 0.126}$
$0.94428 \times 1.126 = 1 + 0.0042 T_{1}$
$1.06326 = 1 + 0.0042 T_{1}$
$T_{1} = \frac{0.06326}{0.0042}$
$T_{1} = {15.06}^{\circ} C$

Assam CEE-2016

Thermal Properties of Matter

146461 The resistance of a wire at $0^{\circ} C$ is $20 Ω$ . If the temperature coefficient of the resistance is $5 \times 10^{- 3}^{\circ} C^{- 1}$ . The temperature at which the resistance will be double of that at $0^{\circ} C$ is

1

10^{\circ} C

2

200^{\circ} C

3

250^{\circ} C

4

300^{\circ} C

Explanation:

B
Exp:
Given that,
Resistance of wire $(R_{o}) = 20 Ω$
Temperature coefficient of the resistance $(α) = 5$ $\times 10^{- 3}^{\circ} C^{- 1}$
We know that, $R = R_{o} (1 + α . T)$
According to the question -
$2 R_{o} = R_{o} (1 + α T)$
$T = \frac{1}{α}$
$T = \frac{1}{5 \times 10^{- 3}}$
$T = 200^{\circ} C$

Shift-I]

Thermal Properties of Matter

146462 When a tyre pumped to a pressure $3.3375 atm$ at $27^{\circ} C$ suddenly bursts, find its final temperature $(γ = 1.5)$

1

27^{\circ} C

2

- 27^{0} C

3

- 0^{0} C

4

- 73^{\circ} C

Explanation:

D Given that, $P_{1} = 3.3375 atm$
$T_{1} = 27^{\circ} C = (273 + 27) K = 300 K, P_{2} = 1 atm, γ = 1.5$ Final temperature,
$T_{2} = T_{1} {(\frac{P_{2}}{P_{1}})}^{\frac{γ - 1}{γ}}$
$T_{2} = 300 {(\frac{1}{3.3375})}^{\frac{1.5 - 1}{1.5}} = 300 {(\frac{1}{3.3375})}^{\frac{1}{3}}$
$T_{2} = 200.7462 K$
$T_{2} = (200.746 - 273)^{\circ} C$
$T_{2} = - {72.25}^{\circ} C \approx - 73^{\circ} C$

Shift-I]

Thermal Properties of Matter

146459 If $K_{1}$ and $K_{2}$ are the thermal conductivities $L_{1}$ $L_{2}$ are the lengths and $A_{1}$ and $A_{2}$ are the cross sectional areas of steel and cooper rods respectively such that $\frac{K_{2}}{K_{1}} = 9, \frac{A_{1}}{A_{2}} = 2, \frac{L_{1}}{L_{2}} = 2$ . Then, for the arrangement as shown in the figure. The value of temperature $T$ of the steelcopper junction in the steady state will be:

1

18^{\circ} C

2

14^{\circ} C

3

45^{\circ} C

4

150^{\circ} C

Explanation:

C
Considering the junction, we can write the heat equation as,
$450 - T = \frac{dQ}{dt} \times \frac{L_{1}}{K_{1} A_{1}}$
and $T - 0 = \frac{dQ}{dt} \times \frac{L_{2}}{K_{2} A_{2}}$
Dividing equation (i) by (ii), we get-
$\frac{450 - T}{T} = \frac{K_{2} A_{2} L_{1}}{K_{1} A_{1} L_{2}} = 9 \times \frac{1}{2} \times 2 = 9$
$450 - T = 9 T$
$T = 45^{\circ} C$

Shift-I]

Thermal Properties of Matter

146460 A standard resistance coil marked $2 Ω$ is found to have a resistance of $2.118 Ω$ at $30^{\circ} C$ the temperature at which marking is correct is (temperature coefficient of resistant of the material of the coil is 0.0042 per degree Celsius)

1

{15.05}^{\circ} C

2

{15.07}^{\circ} C

3

{15.09}^{\circ} C

4

{15.06}^{\circ} C

Explanation:

D Given, that,
$R_{1} = 2 Ω, α = 0.0042 /^{\circ} C, R_{2} = 2.118 Ω, T_{2} = 30^{\circ} C$ We know that,
$\frac{R_{1}}{R_{2}} = \frac{1 + α T_{1}}{1 + α T_{2}}$
$\frac{2}{2.118} = \frac{1 + 0.0042 T_{1}}{1 + 0.0042 \times 30}$
$0.94428 = \frac{1 + 0.0042 T_{1}}{1 + 0.126}$
$0.94428 \times 1.126 = 1 + 0.0042 T_{1}$
$1.06326 = 1 + 0.0042 T_{1}$
$T_{1} = \frac{0.06326}{0.0042}$
$T_{1} = {15.06}^{\circ} C$

Assam CEE-2016

Thermal Properties of Matter

146461 The resistance of a wire at $0^{\circ} C$ is $20 Ω$ . If the temperature coefficient of the resistance is $5 \times 10^{- 3}^{\circ} C^{- 1}$ . The temperature at which the resistance will be double of that at $0^{\circ} C$ is

1

10^{\circ} C

2

200^{\circ} C

3

250^{\circ} C

4

300^{\circ} C

Explanation:

B
Exp:
Given that,
Resistance of wire $(R_{o}) = 20 Ω$
Temperature coefficient of the resistance $(α) = 5$ $\times 10^{- 3}^{\circ} C^{- 1}$
We know that, $R = R_{o} (1 + α . T)$
According to the question -
$2 R_{o} = R_{o} (1 + α T)$
$T = \frac{1}{α}$
$T = \frac{1}{5 \times 10^{- 3}}$
$T = 200^{\circ} C$

Shift-I]

Thermal Properties of Matter

146462 When a tyre pumped to a pressure $3.3375 atm$ at $27^{\circ} C$ suddenly bursts, find its final temperature $(γ = 1.5)$

1

27^{\circ} C

2

- 27^{0} C

3

- 0^{0} C

4

- 73^{\circ} C

Explanation:

D Given that, $P_{1} = 3.3375 atm$
$T_{1} = 27^{\circ} C = (273 + 27) K = 300 K, P_{2} = 1 atm, γ = 1.5$ Final temperature,
$T_{2} = T_{1} {(\frac{P_{2}}{P_{1}})}^{\frac{γ - 1}{γ}}$
$T_{2} = 300 {(\frac{1}{3.3375})}^{\frac{1.5 - 1}{1.5}} = 300 {(\frac{1}{3.3375})}^{\frac{1}{3}}$
$T_{2} = 200.7462 K$
$T_{2} = (200.746 - 273)^{\circ} C$
$T_{2} = - {72.25}^{\circ} C \approx - 73^{\circ} C$

Shift-I]

Thermal Properties of Matter

146459 If $K_{1}$ and $K_{2}$ are the thermal conductivities $L_{1}$ $L_{2}$ are the lengths and $A_{1}$ and $A_{2}$ are the cross sectional areas of steel and cooper rods respectively such that $\frac{K_{2}}{K_{1}} = 9, \frac{A_{1}}{A_{2}} = 2, \frac{L_{1}}{L_{2}} = 2$ . Then, for the arrangement as shown in the figure. The value of temperature $T$ of the steelcopper junction in the steady state will be:

1

18^{\circ} C

2

14^{\circ} C

3

45^{\circ} C

4

150^{\circ} C

Explanation:

C
Considering the junction, we can write the heat equation as,
$450 - T = \frac{dQ}{dt} \times \frac{L_{1}}{K_{1} A_{1}}$
and $T - 0 = \frac{dQ}{dt} \times \frac{L_{2}}{K_{2} A_{2}}$
Dividing equation (i) by (ii), we get-
$\frac{450 - T}{T} = \frac{K_{2} A_{2} L_{1}}{K_{1} A_{1} L_{2}} = 9 \times \frac{1}{2} \times 2 = 9$
$450 - T = 9 T$
$T = 45^{\circ} C$

Shift-I]

Thermal Properties of Matter

146460 A standard resistance coil marked $2 Ω$ is found to have a resistance of $2.118 Ω$ at $30^{\circ} C$ the temperature at which marking is correct is (temperature coefficient of resistant of the material of the coil is 0.0042 per degree Celsius)

1

{15.05}^{\circ} C

2

{15.07}^{\circ} C

3

{15.09}^{\circ} C

4

{15.06}^{\circ} C

Explanation:

D Given, that,
$R_{1} = 2 Ω, α = 0.0042 /^{\circ} C, R_{2} = 2.118 Ω, T_{2} = 30^{\circ} C$ We know that,
$\frac{R_{1}}{R_{2}} = \frac{1 + α T_{1}}{1 + α T_{2}}$
$\frac{2}{2.118} = \frac{1 + 0.0042 T_{1}}{1 + 0.0042 \times 30}$
$0.94428 = \frac{1 + 0.0042 T_{1}}{1 + 0.126}$
$0.94428 \times 1.126 = 1 + 0.0042 T_{1}$
$1.06326 = 1 + 0.0042 T_{1}$
$T_{1} = \frac{0.06326}{0.0042}$
$T_{1} = {15.06}^{\circ} C$

Assam CEE-2016

Thermal Properties of Matter

146461 The resistance of a wire at $0^{\circ} C$ is $20 Ω$ . If the temperature coefficient of the resistance is $5 \times 10^{- 3}^{\circ} C^{- 1}$ . The temperature at which the resistance will be double of that at $0^{\circ} C$ is

1

10^{\circ} C

2

200^{\circ} C

3

250^{\circ} C

4

300^{\circ} C

Explanation:

B
Exp:
Given that,
Resistance of wire $(R_{o}) = 20 Ω$
Temperature coefficient of the resistance $(α) = 5$ $\times 10^{- 3}^{\circ} C^{- 1}$
We know that, $R = R_{o} (1 + α . T)$
According to the question -
$2 R_{o} = R_{o} (1 + α T)$
$T = \frac{1}{α}$
$T = \frac{1}{5 \times 10^{- 3}}$
$T = 200^{\circ} C$

Shift-I]

Thermal Properties of Matter

146462 When a tyre pumped to a pressure $3.3375 atm$ at $27^{\circ} C$ suddenly bursts, find its final temperature $(γ = 1.5)$

1

27^{\circ} C

2

- 27^{0} C

3

- 0^{0} C

4

- 73^{\circ} C

Explanation:

D Given that, $P_{1} = 3.3375 atm$
$T_{1} = 27^{\circ} C = (273 + 27) K = 300 K, P_{2} = 1 atm, γ = 1.5$ Final temperature,
$T_{2} = T_{1} {(\frac{P_{2}}{P_{1}})}^{\frac{γ - 1}{γ}}$
$T_{2} = 300 {(\frac{1}{3.3375})}^{\frac{1.5 - 1}{1.5}} = 300 {(\frac{1}{3.3375})}^{\frac{1}{3}}$
$T_{2} = 200.7462 K$
$T_{2} = (200.746 - 273)^{\circ} C$
$T_{2} = - {72.25}^{\circ} C \approx - 73^{\circ} C$

Shift-I]

Thermal Properties of Matter

146459 If $K_{1}$ and $K_{2}$ are the thermal conductivities $L_{1}$ $L_{2}$ are the lengths and $A_{1}$ and $A_{2}$ are the cross sectional areas of steel and cooper rods respectively such that $\frac{K_{2}}{K_{1}} = 9, \frac{A_{1}}{A_{2}} = 2, \frac{L_{1}}{L_{2}} = 2$ . Then, for the arrangement as shown in the figure. The value of temperature $T$ of the steelcopper junction in the steady state will be:

1

18^{\circ} C

2

14^{\circ} C

3

45^{\circ} C

4

150^{\circ} C

Explanation:

C
Considering the junction, we can write the heat equation as,
$450 - T = \frac{dQ}{dt} \times \frac{L_{1}}{K_{1} A_{1}}$
and $T - 0 = \frac{dQ}{dt} \times \frac{L_{2}}{K_{2} A_{2}}$
Dividing equation (i) by (ii), we get-
$\frac{450 - T}{T} = \frac{K_{2} A_{2} L_{1}}{K_{1} A_{1} L_{2}} = 9 \times \frac{1}{2} \times 2 = 9$
$450 - T = 9 T$
$T = 45^{\circ} C$

Shift-I]

Thermal Properties of Matter

146460 A standard resistance coil marked $2 Ω$ is found to have a resistance of $2.118 Ω$ at $30^{\circ} C$ the temperature at which marking is correct is (temperature coefficient of resistant of the material of the coil is 0.0042 per degree Celsius)

1

{15.05}^{\circ} C

2

{15.07}^{\circ} C

3

{15.09}^{\circ} C

4

{15.06}^{\circ} C

Explanation:

D Given, that,
$R_{1} = 2 Ω, α = 0.0042 /^{\circ} C, R_{2} = 2.118 Ω, T_{2} = 30^{\circ} C$ We know that,
$\frac{R_{1}}{R_{2}} = \frac{1 + α T_{1}}{1 + α T_{2}}$
$\frac{2}{2.118} = \frac{1 + 0.0042 T_{1}}{1 + 0.0042 \times 30}$
$0.94428 = \frac{1 + 0.0042 T_{1}}{1 + 0.126}$
$0.94428 \times 1.126 = 1 + 0.0042 T_{1}$
$1.06326 = 1 + 0.0042 T_{1}$
$T_{1} = \frac{0.06326}{0.0042}$
$T_{1} = {15.06}^{\circ} C$

Assam CEE-2016

Thermal Properties of Matter

146461 The resistance of a wire at $0^{\circ} C$ is $20 Ω$ . If the temperature coefficient of the resistance is $5 \times 10^{- 3}^{\circ} C^{- 1}$ . The temperature at which the resistance will be double of that at $0^{\circ} C$ is

1

10^{\circ} C

2

200^{\circ} C

3

250^{\circ} C

4

300^{\circ} C

Explanation:

B
Exp:
Given that,
Resistance of wire $(R_{o}) = 20 Ω$
Temperature coefficient of the resistance $(α) = 5$ $\times 10^{- 3}^{\circ} C^{- 1}$
We know that, $R = R_{o} (1 + α . T)$
According to the question -
$2 R_{o} = R_{o} (1 + α T)$
$T = \frac{1}{α}$
$T = \frac{1}{5 \times 10^{- 3}}$
$T = 200^{\circ} C$

Shift-I]

Thermal Properties of Matter

146462 When a tyre pumped to a pressure $3.3375 atm$ at $27^{\circ} C$ suddenly bursts, find its final temperature $(γ = 1.5)$

1

27^{\circ} C

2

- 27^{0} C

3

- 0^{0} C

4

- 73^{\circ} C

Explanation:

D Given that, $P_{1} = 3.3375 atm$
$T_{1} = 27^{\circ} C = (273 + 27) K = 300 K, P_{2} = 1 atm, γ = 1.5$ Final temperature,
$T_{2} = T_{1} {(\frac{P_{2}}{P_{1}})}^{\frac{γ - 1}{γ}}$
$T_{2} = 300 {(\frac{1}{3.3375})}^{\frac{1.5 - 1}{1.5}} = 300 {(\frac{1}{3.3375})}^{\frac{1}{3}}$
$T_{2} = 200.7462 K$
$T_{2} = (200.746 - 273)^{\circ} C$
$T_{2} = - {72.25}^{\circ} C \approx - 73^{\circ} C$

Shift-I]

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: