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PHXI11:THERMAL PROPERTIES OF MATTER

366776 A bimetal made of copper and iron strips welded together is straight at room temperature. It is held vertically so that the iron strip is towards the left hand and copper strip is towards right hand. The bimetal strip is then heated. The bimetal strip will $(α_{c u} > α_{F e})$

1 Remain straight

2 Bend towards right

3 Bend towards left

4 Have no change

PHXI11:THERMAL PROPERTIES OF MATTER

366777 The diameter of iron wheel is $1 c m$ . If its temperature is increased by $700^{\circ} C$ what is the increase in circumference of the wheel?
$(α = 12 \times 10^{- 6} /^{\circ} C)$

1

0.0264 c m

2

0.264 c m

3

2.64 c m

4

26.4 c m

Explanation:

Increase is circumference $= π D α Δ t$ where $D =$ diameter of the wheel
circumference, $C = π D$
Increase is circumference, $Δ C = (α Δ T) π D$
$= 12 \times 10^{- 6} \times 400 \times π \times 1 c m$
$= 0.0264 c m$

PHXI11:THERMAL PROPERTIES OF MATTER

366778 Two metal rods have coefficients of linear expansion $1.1 \times 10^{- 5} /^{\circ} C$ and $1.65 \times 10^{- 5} /^{\circ} C$ respectively. The difference in lengths is $10 c m$ at all temperatures. Their initial lengths must be respectively .

1

40 c m a n d 50 c m

2

40 c m a n d 30 c m

3

50 c m a n d 60 c m

4

30 c m a n d 20 c m

Explanation:

Given that the difference in lengths remain same
$Δ l_{1} = Δ l_{2}$
$l_{1} α_{1} Δ θ = l_{2} α_{2} Δ θ$
$\Rightarrow l_{1} α_{1} = l_{2} α_{2} (1)$
$l_{1} - l_{2} = 10 (G i v e n) (2)$
From eq's (1) and (2)
$\Rightarrow l_{1} - \frac{l_{1} α_{1}}{α_{2}} = 10$
$\Rightarrow l_{1} = \frac{10 α_{2}}{α_{2} - α_{1}} = \frac{10 \times 1.65 \times 10^{- 5}}{0.55 \times 10^{- 5}} = 30 c m$
$\Rightarrow l_{2} = 20 c m$ .

PHXI11:THERMAL PROPERTIES OF MATTER

366779 How much should the temperature of a brass rod be increased so as to increase its length by $1 %$ ? (Take, $α$ for brass is ${0.00002}^{\circ} C^{- 1}$ )

1

300^{\circ} C

2

400^{\circ} C

3

500^{\circ} C

4

550^{\circ} C

Explanation:

Here, $Δ T =$ ?, $\frac{Δ L}{L} = \frac{1}{100}$ ,
$a = {0.00002}^{\circ} C^{- 1}$
$A s Δ L = α L Δ T$
$∴ α Δ T = \frac{Δ L}{L}$
$o r Δ T = \frac{Δ L}{L α} = \frac{1}{100 \times 0.00002}$
$Δ T = \frac{10^{5}}{2 \times 10^{2}} = 500^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366780 A wheel is $80.3 c m$ in circumference. An iron tyre measures $80.0 c m$ around its inner face. If the coefficient of linear expansion for iron is $1.2 \times 10^{- 5}^{\circ} C^{- 1}$ , the temperature of the tyre must be raised by

1

417^{\circ} C

2

105^{\circ} C

3

223^{\circ} C

4

312^{\circ} C

Explanation:

Here, $Δ l = 80.3 - 80.0 = 0.3 c m$
$l = 80 c m, α = 1.2 \times 10^{- 6}^{\circ} C^{- 1}$
Rise in temperature $Δ T = \frac{Δ l}{l α}$
$Δ T = \frac{0.3}{80 \times 1.2 \times 10^{- 5}} = {312.5}^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366776 A bimetal made of copper and iron strips welded together is straight at room temperature. It is held vertically so that the iron strip is towards the left hand and copper strip is towards right hand. The bimetal strip is then heated. The bimetal strip will $(α_{c u} > α_{F e})$

1 Remain straight

2 Bend towards right

3 Bend towards left

4 Have no change

PHXI11:THERMAL PROPERTIES OF MATTER

366777 The diameter of iron wheel is $1 c m$ . If its temperature is increased by $700^{\circ} C$ what is the increase in circumference of the wheel?
$(α = 12 \times 10^{- 6} /^{\circ} C)$

1

0.0264 c m

2

0.264 c m

3

2.64 c m

4

26.4 c m

Explanation:

Increase is circumference $= π D α Δ t$ where $D =$ diameter of the wheel
circumference, $C = π D$
Increase is circumference, $Δ C = (α Δ T) π D$
$= 12 \times 10^{- 6} \times 400 \times π \times 1 c m$
$= 0.0264 c m$

PHXI11:THERMAL PROPERTIES OF MATTER

366778 Two metal rods have coefficients of linear expansion $1.1 \times 10^{- 5} /^{\circ} C$ and $1.65 \times 10^{- 5} /^{\circ} C$ respectively. The difference in lengths is $10 c m$ at all temperatures. Their initial lengths must be respectively .

1

40 c m a n d 50 c m

2

40 c m a n d 30 c m

3

50 c m a n d 60 c m

4

30 c m a n d 20 c m

Explanation:

Given that the difference in lengths remain same
$Δ l_{1} = Δ l_{2}$
$l_{1} α_{1} Δ θ = l_{2} α_{2} Δ θ$
$\Rightarrow l_{1} α_{1} = l_{2} α_{2} (1)$
$l_{1} - l_{2} = 10 (G i v e n) (2)$
From eq's (1) and (2)
$\Rightarrow l_{1} - \frac{l_{1} α_{1}}{α_{2}} = 10$
$\Rightarrow l_{1} = \frac{10 α_{2}}{α_{2} - α_{1}} = \frac{10 \times 1.65 \times 10^{- 5}}{0.55 \times 10^{- 5}} = 30 c m$
$\Rightarrow l_{2} = 20 c m$ .

PHXI11:THERMAL PROPERTIES OF MATTER

366779 How much should the temperature of a brass rod be increased so as to increase its length by $1 %$ ? (Take, $α$ for brass is ${0.00002}^{\circ} C^{- 1}$ )

1

300^{\circ} C

2

400^{\circ} C

3

500^{\circ} C

4

550^{\circ} C

Explanation:

Here, $Δ T =$ ?, $\frac{Δ L}{L} = \frac{1}{100}$ ,
$a = {0.00002}^{\circ} C^{- 1}$
$A s Δ L = α L Δ T$
$∴ α Δ T = \frac{Δ L}{L}$
$o r Δ T = \frac{Δ L}{L α} = \frac{1}{100 \times 0.00002}$
$Δ T = \frac{10^{5}}{2 \times 10^{2}} = 500^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366780 A wheel is $80.3 c m$ in circumference. An iron tyre measures $80.0 c m$ around its inner face. If the coefficient of linear expansion for iron is $1.2 \times 10^{- 5}^{\circ} C^{- 1}$ , the temperature of the tyre must be raised by

1

417^{\circ} C

2

105^{\circ} C

3

223^{\circ} C

4

312^{\circ} C

Explanation:

Here, $Δ l = 80.3 - 80.0 = 0.3 c m$
$l = 80 c m, α = 1.2 \times 10^{- 6}^{\circ} C^{- 1}$
Rise in temperature $Δ T = \frac{Δ l}{l α}$
$Δ T = \frac{0.3}{80 \times 1.2 \times 10^{- 5}} = {312.5}^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366776 A bimetal made of copper and iron strips welded together is straight at room temperature. It is held vertically so that the iron strip is towards the left hand and copper strip is towards right hand. The bimetal strip is then heated. The bimetal strip will $(α_{c u} > α_{F e})$

1 Remain straight

2 Bend towards right

3 Bend towards left

4 Have no change

PHXI11:THERMAL PROPERTIES OF MATTER

366777 The diameter of iron wheel is $1 c m$ . If its temperature is increased by $700^{\circ} C$ what is the increase in circumference of the wheel?
$(α = 12 \times 10^{- 6} /^{\circ} C)$

1

0.0264 c m

2

0.264 c m

3

2.64 c m

4

26.4 c m

Explanation:

Increase is circumference $= π D α Δ t$ where $D =$ diameter of the wheel
circumference, $C = π D$
Increase is circumference, $Δ C = (α Δ T) π D$
$= 12 \times 10^{- 6} \times 400 \times π \times 1 c m$
$= 0.0264 c m$

PHXI11:THERMAL PROPERTIES OF MATTER

366778 Two metal rods have coefficients of linear expansion $1.1 \times 10^{- 5} /^{\circ} C$ and $1.65 \times 10^{- 5} /^{\circ} C$ respectively. The difference in lengths is $10 c m$ at all temperatures. Their initial lengths must be respectively .

1

40 c m a n d 50 c m

2

40 c m a n d 30 c m

3

50 c m a n d 60 c m

4

30 c m a n d 20 c m

Explanation:

Given that the difference in lengths remain same
$Δ l_{1} = Δ l_{2}$
$l_{1} α_{1} Δ θ = l_{2} α_{2} Δ θ$
$\Rightarrow l_{1} α_{1} = l_{2} α_{2} (1)$
$l_{1} - l_{2} = 10 (G i v e n) (2)$
From eq's (1) and (2)
$\Rightarrow l_{1} - \frac{l_{1} α_{1}}{α_{2}} = 10$
$\Rightarrow l_{1} = \frac{10 α_{2}}{α_{2} - α_{1}} = \frac{10 \times 1.65 \times 10^{- 5}}{0.55 \times 10^{- 5}} = 30 c m$
$\Rightarrow l_{2} = 20 c m$ .

PHXI11:THERMAL PROPERTIES OF MATTER

366779 How much should the temperature of a brass rod be increased so as to increase its length by $1 %$ ? (Take, $α$ for brass is ${0.00002}^{\circ} C^{- 1}$ )

1

300^{\circ} C

2

400^{\circ} C

3

500^{\circ} C

4

550^{\circ} C

Explanation:

Here, $Δ T =$ ?, $\frac{Δ L}{L} = \frac{1}{100}$ ,
$a = {0.00002}^{\circ} C^{- 1}$
$A s Δ L = α L Δ T$
$∴ α Δ T = \frac{Δ L}{L}$
$o r Δ T = \frac{Δ L}{L α} = \frac{1}{100 \times 0.00002}$
$Δ T = \frac{10^{5}}{2 \times 10^{2}} = 500^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366780 A wheel is $80.3 c m$ in circumference. An iron tyre measures $80.0 c m$ around its inner face. If the coefficient of linear expansion for iron is $1.2 \times 10^{- 5}^{\circ} C^{- 1}$ , the temperature of the tyre must be raised by

1

417^{\circ} C

2

105^{\circ} C

3

223^{\circ} C

4

312^{\circ} C

Explanation:

Here, $Δ l = 80.3 - 80.0 = 0.3 c m$
$l = 80 c m, α = 1.2 \times 10^{- 6}^{\circ} C^{- 1}$
Rise in temperature $Δ T = \frac{Δ l}{l α}$
$Δ T = \frac{0.3}{80 \times 1.2 \times 10^{- 5}} = {312.5}^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366776 A bimetal made of copper and iron strips welded together is straight at room temperature. It is held vertically so that the iron strip is towards the left hand and copper strip is towards right hand. The bimetal strip is then heated. The bimetal strip will $(α_{c u} > α_{F e})$

1 Remain straight

2 Bend towards right

3 Bend towards left

4 Have no change

PHXI11:THERMAL PROPERTIES OF MATTER

366777 The diameter of iron wheel is $1 c m$ . If its temperature is increased by $700^{\circ} C$ what is the increase in circumference of the wheel?
$(α = 12 \times 10^{- 6} /^{\circ} C)$

1

0.0264 c m

2

0.264 c m

3

2.64 c m

4

26.4 c m

Explanation:

Increase is circumference $= π D α Δ t$ where $D =$ diameter of the wheel
circumference, $C = π D$
Increase is circumference, $Δ C = (α Δ T) π D$
$= 12 \times 10^{- 6} \times 400 \times π \times 1 c m$
$= 0.0264 c m$

PHXI11:THERMAL PROPERTIES OF MATTER

366778 Two metal rods have coefficients of linear expansion $1.1 \times 10^{- 5} /^{\circ} C$ and $1.65 \times 10^{- 5} /^{\circ} C$ respectively. The difference in lengths is $10 c m$ at all temperatures. Their initial lengths must be respectively .

1

40 c m a n d 50 c m

2

40 c m a n d 30 c m

3

50 c m a n d 60 c m

4

30 c m a n d 20 c m

Explanation:

Given that the difference in lengths remain same
$Δ l_{1} = Δ l_{2}$
$l_{1} α_{1} Δ θ = l_{2} α_{2} Δ θ$
$\Rightarrow l_{1} α_{1} = l_{2} α_{2} (1)$
$l_{1} - l_{2} = 10 (G i v e n) (2)$
From eq's (1) and (2)
$\Rightarrow l_{1} - \frac{l_{1} α_{1}}{α_{2}} = 10$
$\Rightarrow l_{1} = \frac{10 α_{2}}{α_{2} - α_{1}} = \frac{10 \times 1.65 \times 10^{- 5}}{0.55 \times 10^{- 5}} = 30 c m$
$\Rightarrow l_{2} = 20 c m$ .

PHXI11:THERMAL PROPERTIES OF MATTER

366779 How much should the temperature of a brass rod be increased so as to increase its length by $1 %$ ? (Take, $α$ for brass is ${0.00002}^{\circ} C^{- 1}$ )

1

300^{\circ} C

2

400^{\circ} C

3

500^{\circ} C

4

550^{\circ} C

Explanation:

Here, $Δ T =$ ?, $\frac{Δ L}{L} = \frac{1}{100}$ ,
$a = {0.00002}^{\circ} C^{- 1}$
$A s Δ L = α L Δ T$
$∴ α Δ T = \frac{Δ L}{L}$
$o r Δ T = \frac{Δ L}{L α} = \frac{1}{100 \times 0.00002}$
$Δ T = \frac{10^{5}}{2 \times 10^{2}} = 500^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366780 A wheel is $80.3 c m$ in circumference. An iron tyre measures $80.0 c m$ around its inner face. If the coefficient of linear expansion for iron is $1.2 \times 10^{- 5}^{\circ} C^{- 1}$ , the temperature of the tyre must be raised by

1

417^{\circ} C

2

105^{\circ} C

3

223^{\circ} C

4

312^{\circ} C

Explanation:

Here, $Δ l = 80.3 - 80.0 = 0.3 c m$
$l = 80 c m, α = 1.2 \times 10^{- 6}^{\circ} C^{- 1}$
Rise in temperature $Δ T = \frac{Δ l}{l α}$
$Δ T = \frac{0.3}{80 \times 1.2 \times 10^{- 5}} = {312.5}^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366776 A bimetal made of copper and iron strips welded together is straight at room temperature. It is held vertically so that the iron strip is towards the left hand and copper strip is towards right hand. The bimetal strip is then heated. The bimetal strip will $(α_{c u} > α_{F e})$

1 Remain straight

2 Bend towards right

3 Bend towards left

4 Have no change

PHXI11:THERMAL PROPERTIES OF MATTER

366777 The diameter of iron wheel is $1 c m$ . If its temperature is increased by $700^{\circ} C$ what is the increase in circumference of the wheel?
$(α = 12 \times 10^{- 6} /^{\circ} C)$

1

0.0264 c m

2

0.264 c m

3

2.64 c m

4

26.4 c m

Explanation:

Increase is circumference $= π D α Δ t$ where $D =$ diameter of the wheel
circumference, $C = π D$
Increase is circumference, $Δ C = (α Δ T) π D$
$= 12 \times 10^{- 6} \times 400 \times π \times 1 c m$
$= 0.0264 c m$

PHXI11:THERMAL PROPERTIES OF MATTER

366778 Two metal rods have coefficients of linear expansion $1.1 \times 10^{- 5} /^{\circ} C$ and $1.65 \times 10^{- 5} /^{\circ} C$ respectively. The difference in lengths is $10 c m$ at all temperatures. Their initial lengths must be respectively .

1

40 c m a n d 50 c m

2

40 c m a n d 30 c m

3

50 c m a n d 60 c m

4

30 c m a n d 20 c m

Explanation:

Given that the difference in lengths remain same
$Δ l_{1} = Δ l_{2}$
$l_{1} α_{1} Δ θ = l_{2} α_{2} Δ θ$
$\Rightarrow l_{1} α_{1} = l_{2} α_{2} (1)$
$l_{1} - l_{2} = 10 (G i v e n) (2)$
From eq's (1) and (2)
$\Rightarrow l_{1} - \frac{l_{1} α_{1}}{α_{2}} = 10$
$\Rightarrow l_{1} = \frac{10 α_{2}}{α_{2} - α_{1}} = \frac{10 \times 1.65 \times 10^{- 5}}{0.55 \times 10^{- 5}} = 30 c m$
$\Rightarrow l_{2} = 20 c m$ .

PHXI11:THERMAL PROPERTIES OF MATTER

366779 How much should the temperature of a brass rod be increased so as to increase its length by $1 %$ ? (Take, $α$ for brass is ${0.00002}^{\circ} C^{- 1}$ )

1

300^{\circ} C

2

400^{\circ} C

3

500^{\circ} C

4

550^{\circ} C

Explanation:

Here, $Δ T =$ ?, $\frac{Δ L}{L} = \frac{1}{100}$ ,
$a = {0.00002}^{\circ} C^{- 1}$
$A s Δ L = α L Δ T$
$∴ α Δ T = \frac{Δ L}{L}$
$o r Δ T = \frac{Δ L}{L α} = \frac{1}{100 \times 0.00002}$
$Δ T = \frac{10^{5}}{2 \times 10^{2}} = 500^{\circ} C$

PHXI11:THERMAL PROPERTIES OF MATTER

366780 A wheel is $80.3 c m$ in circumference. An iron tyre measures $80.0 c m$ around its inner face. If the coefficient of linear expansion for iron is $1.2 \times 10^{- 5}^{\circ} C^{- 1}$ , the temperature of the tyre must be raised by

1

417^{\circ} C

2

105^{\circ} C

3

223^{\circ} C

4

312^{\circ} C

Explanation:

Here, $Δ l = 80.3 - 80.0 = 0.3 c m$
$l = 80 c m, α = 1.2 \times 10^{- 6}^{\circ} C^{- 1}$
Rise in temperature $Δ T = \frac{Δ l}{l α}$
$Δ T = \frac{0.3}{80 \times 1.2 \times 10^{- 5}} = {312.5}^{\circ} C$

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