QBManager

Thermal Properties of Matter

146533 A piece of metal weighs $45 g$ in air and $25 g$ in a liquid of density $1.5 \times 10^{3} kg - m^{- 3} kept$ at $30^{\circ} C$ . When the temperature of the liquid is raised to $40^{\circ} C$ , the metal piece weights $27 g$ . The density of liquid of $40^{\circ} C$ is $1.25 \times 10^{3} kg - m^{- 3}$ , the coefficient of linear expansion of metal is

1

1.3 \times 10^{- 3} /^{\circ} C

2

5.2 \times 10^{- 3} /^{\circ} C

3

2.6 \times 10^{- 3} /^{\circ} C

4

0.26 \times 10^{- 3} /^{\circ} C

Explanation:

C Given that, $M = 45 g$ (in air)
$m_{1} = 25 g$ (in liquid)
$m_{2} = 27 g$ ,
$ρ_{2} = 1.25 \times 10^{3} kg - m^{- 3}$
$T_{1} = 30^{\circ} C$
$T_{2} = 40^{\circ} C$
$T_{1} = 30^{\circ} C$ ; density of liquid, $ρ_{1} = 1.5 \times 10^{3} kg / m^{3}$
$V_{1} = \frac{M - m_{1}}{ρ_{1}} = \frac{(45 - 25)}{1.5 \times 10^{3}} \times 10^{- 3}$
$= 13.33 \times 10^{- 3} \times 10^{- 3} m^{3}$
$= 13.33 \times 10^{- 6} m^{3}$
$V_{2} = \frac{M - m_{2}}{ρ_{2}} = \frac{45 - 27}{1.25 \times 10^{3}} \times 10^{- 3}$
$= 14.4 \times 10^{- 6} m^{3}$
Coefficient of linear expansion
$γ = \frac{Δ V}{V_{1} \times Δ t}$
$= \frac{V_{2} - V_{1}}{V_{1} (T_{2} - T_{1})}$
$γ = \frac{(14.4 - 13.33) \times 10^{- 6}}{13.33 \times 10^{- 6} \times (40 - 30)}$
$γ = 8 \times 10^{- 3}^{\circ} C^{- 1}$
$γ γ = 3 α$
$α = \frac{γ}{3} = \frac{8 \times 10^{- 3}}{3}$
$α = 2.6 \times 10^{- 3}^{\circ} C^{- 1}$

VITEEE-2009

Thermal Properties of Matter

146534 A metal rod of length $L$ and cross-sectional area $A$ is heated through $T^{0} C$ . What is the force required to prevent the expansion of the rod lengthwise ?
( $Y$ = Young's modulus of material of the rod, $α$ = coefficient of linear expansion of the rod).

1

YA α T / (1 - α T)

2

YA α T / (1 + α T)

3

YA α / T (1 + α T)

4

YA α / (1 - α T)

Explanation:

B
Exp:
We know that, increasing length due to heating is given by
$Δ L = L_{0} α T$
$or, L = L_{0} (1 + α T)$
From the young modulus
$Y = \frac{F / A}{Δ L / L}$
$F = \frac{YA Δ L}{L}$
$F = \frac{YA L_{0} α T}{L_{0} (1 + α T)}$
$F = \frac{YA α T}{(1 + α T)}$

MHT-CET 2020

Thermal Properties of Matter

146535 A hole is drilled in a copper sheet. The diameter of the hole is $4.24 cm$ at ${27.0}^{\circ} C$ . What is the change in the diameter of the hole when the sheet is heated to $227^{\circ} C$ ? Coefficient of linear expansion of copper $= 1.70 \times 10^{- 5} K^{- 1}$

1

0.0144 cm

2

0.144 cm

3

1.44 cm

4

0.0014 cm

Explanation:

A Given, $L = 4.24 cm, T_{1} = 27^{\circ} C, T_{2} = 227^{\circ} C$ $Δ T = (T_{2} - T_{1}) = (227 - 27) = 200^{\circ} C$
We know that,
$Δ L = L α Δ T$
$Δ L = 4.24 \times 1.70 \times 10^{- 5} \times 200$
$Δ L = 0.0144 cm$

MHT-CET 2020

Thermal Properties of Matter

146536 A brass rod of length $50 cm$ and diameter 3.0 $mm$ is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at $250^{\circ} C$ , if the original lengths are at ${40.0}^{\circ} C$ ? (coefficient of linear expansion of brass $= 2.0 \times 10^{- 5} K^{- 1}$ , steel $=$ $1.2 \times 10^{- 5} K^{- 1}$ )

1

0.4365 cm

2

0.5525 cm

3

0.3465 cm

4

0.2465 cm

Explanation:

C Given,
$α_{b} = 2.0 \times 10^{- 5} K^{- 1}, α_{s} = 1.2 \times 10^{- 5} K^{- 1}$
$Δ T = (250^{\circ} C - 40^{\circ} C) = 210^{\circ} C$
$l_{1} = 50 cm, l_{2} = 50 cm$
$Δ l_{1} = l_{1} α_{b} Δ T$
$= 50 \times 2.0 \times 10^{- 5} \times 210$
$= 0.21 cm$
$Δ l_{2} = l_{2} α_{s} Δ T$
$= 50 \times 1.2 \times 10^{- 5} \times 210 = 0.126 cm$
$∴$ Change in length of the combined rod,
$Δ l = Δ l_{1} + Δ l_{2}$
$= 0.21 + 0.126$
$Δ l = 0.336 cm$
$Δ l \approx 0.3465 cm$

MHT-CET 2020

Thermal Properties of Matter

146537 The coefficient of volume expansion of glycerin is $49 \times 10^{- 5} K^{- 1}$ . What is the fractional change in its density for ${ar}^{\circ} C$ rise in temperature?

1 0.0145

2 0.0154

3 0.154

4

1.144 w

Explanation:

A Given,
Coefficient of volume expansion of glycerin $(α_{v})$
$= 49 \times 10^{- 5} K^{- 1}, Δ T = 30^{\circ} C$
We know that,
$Δ V = V α_{v} Δ T$
$V_{T_{2}} - V_{T_{1}} = V_{T_{1}} α_{v} Δ T (∵ ρ = \frac{m}{V})$
$\frac{m}{ρ_{T_{2}}} - \frac{m}{ρ_{T_{1}}} = \frac{m}{ρ_{T_{1}}} α_{v} Δ T$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}} ρ_{T_{1}}} = \frac{α_{v} Δ T}{ρ_{T_{1}}}$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}}} = α_{v} Δ T$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}}} = 49 \times 10^{- 5} \times 30$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}}} = 0.01470$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}}} \approx 0.0145$
$∴$ Fractional change in density $= 0.0145$

MHT-CET 2020

Thermal Properties of Matter

146533 A piece of metal weighs $45 g$ in air and $25 g$ in a liquid of density $1.5 \times 10^{3} kg - m^{- 3} kept$ at $30^{\circ} C$ . When the temperature of the liquid is raised to $40^{\circ} C$ , the metal piece weights $27 g$ . The density of liquid of $40^{\circ} C$ is $1.25 \times 10^{3} kg - m^{- 3}$ , the coefficient of linear expansion of metal is

1

1.3 \times 10^{- 3} /^{\circ} C

2

5.2 \times 10^{- 3} /^{\circ} C

3

2.6 \times 10^{- 3} /^{\circ} C

4

0.26 \times 10^{- 3} /^{\circ} C

Explanation:

C Given that, $M = 45 g$ (in air)
$m_{1} = 25 g$ (in liquid)
$m_{2} = 27 g$ ,
$ρ_{2} = 1.25 \times 10^{3} kg - m^{- 3}$
$T_{1} = 30^{\circ} C$
$T_{2} = 40^{\circ} C$
$T_{1} = 30^{\circ} C$ ; density of liquid, $ρ_{1} = 1.5 \times 10^{3} kg / m^{3}$
$V_{1} = \frac{M - m_{1}}{ρ_{1}} = \frac{(45 - 25)}{1.5 \times 10^{3}} \times 10^{- 3}$
$= 13.33 \times 10^{- 3} \times 10^{- 3} m^{3}$
$= 13.33 \times 10^{- 6} m^{3}$
$V_{2} = \frac{M - m_{2}}{ρ_{2}} = \frac{45 - 27}{1.25 \times 10^{3}} \times 10^{- 3}$
$= 14.4 \times 10^{- 6} m^{3}$
Coefficient of linear expansion
$γ = \frac{Δ V}{V_{1} \times Δ t}$
$= \frac{V_{2} - V_{1}}{V_{1} (T_{2} - T_{1})}$
$γ = \frac{(14.4 - 13.33) \times 10^{- 6}}{13.33 \times 10^{- 6} \times (40 - 30)}$
$γ = 8 \times 10^{- 3}^{\circ} C^{- 1}$
$γ γ = 3 α$
$α = \frac{γ}{3} = \frac{8 \times 10^{- 3}}{3}$
$α = 2.6 \times 10^{- 3}^{\circ} C^{- 1}$

VITEEE-2009

Thermal Properties of Matter

146534 A metal rod of length $L$ and cross-sectional area $A$ is heated through $T^{0} C$ . What is the force required to prevent the expansion of the rod lengthwise ?
( $Y$ = Young's modulus of material of the rod, $α$ = coefficient of linear expansion of the rod).

1

YA α T / (1 - α T)

2

YA α T / (1 + α T)

3

YA α / T (1 + α T)

4

YA α / (1 - α T)

Explanation:

B
Exp:
We know that, increasing length due to heating is given by
$Δ L = L_{0} α T$
$or, L = L_{0} (1 + α T)$
From the young modulus
$Y = \frac{F / A}{Δ L / L}$
$F = \frac{YA Δ L}{L}$
$F = \frac{YA L_{0} α T}{L_{0} (1 + α T)}$
$F = \frac{YA α T}{(1 + α T)}$

MHT-CET 2020

Thermal Properties of Matter

146535 A hole is drilled in a copper sheet. The diameter of the hole is $4.24 cm$ at ${27.0}^{\circ} C$ . What is the change in the diameter of the hole when the sheet is heated to $227^{\circ} C$ ? Coefficient of linear expansion of copper $= 1.70 \times 10^{- 5} K^{- 1}$

1

0.0144 cm

2

0.144 cm

3

1.44 cm

4

0.0014 cm

Explanation:

A Given, $L = 4.24 cm, T_{1} = 27^{\circ} C, T_{2} = 227^{\circ} C$ $Δ T = (T_{2} - T_{1}) = (227 - 27) = 200^{\circ} C$
We know that,
$Δ L = L α Δ T$
$Δ L = 4.24 \times 1.70 \times 10^{- 5} \times 200$
$Δ L = 0.0144 cm$

MHT-CET 2020

Thermal Properties of Matter

146536 A brass rod of length $50 cm$ and diameter 3.0 $mm$ is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at $250^{\circ} C$ , if the original lengths are at ${40.0}^{\circ} C$ ? (coefficient of linear expansion of brass $= 2.0 \times 10^{- 5} K^{- 1}$ , steel $=$ $1.2 \times 10^{- 5} K^{- 1}$ )

1

0.4365 cm

2

0.5525 cm

3

0.3465 cm

4

0.2465 cm

Explanation:

C Given,
$α_{b} = 2.0 \times 10^{- 5} K^{- 1}, α_{s} = 1.2 \times 10^{- 5} K^{- 1}$
$Δ T = (250^{\circ} C - 40^{\circ} C) = 210^{\circ} C$
$l_{1} = 50 cm, l_{2} = 50 cm$
$Δ l_{1} = l_{1} α_{b} Δ T$
$= 50 \times 2.0 \times 10^{- 5} \times 210$
$= 0.21 cm$
$Δ l_{2} = l_{2} α_{s} Δ T$
$= 50 \times 1.2 \times 10^{- 5} \times 210 = 0.126 cm$
$∴$ Change in length of the combined rod,
$Δ l = Δ l_{1} + Δ l_{2}$
$= 0.21 + 0.126$
$Δ l = 0.336 cm$
$Δ l \approx 0.3465 cm$

MHT-CET 2020

Thermal Properties of Matter

146537 The coefficient of volume expansion of glycerin is $49 \times 10^{- 5} K^{- 1}$ . What is the fractional change in its density for ${ar}^{\circ} C$ rise in temperature?

1 0.0145

2 0.0154

3 0.154

4

1.144 w

Explanation:

A Given,
Coefficient of volume expansion of glycerin $(α_{v})$
$= 49 \times 10^{- 5} K^{- 1}, Δ T = 30^{\circ} C$
We know that,
$Δ V = V α_{v} Δ T$
$V_{T_{2}} - V_{T_{1}} = V_{T_{1}} α_{v} Δ T (∵ ρ = \frac{m}{V})$
$\frac{m}{ρ_{T_{2}}} - \frac{m}{ρ_{T_{1}}} = \frac{m}{ρ_{T_{1}}} α_{v} Δ T$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}} ρ_{T_{1}}} = \frac{α_{v} Δ T}{ρ_{T_{1}}}$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}}} = α_{v} Δ T$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}}} = 49 \times 10^{- 5} \times 30$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}}} = 0.01470$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}}} \approx 0.0145$
$∴$ Fractional change in density $= 0.0145$

MHT-CET 2020

Thermal Properties of Matter

146533 A piece of metal weighs $45 g$ in air and $25 g$ in a liquid of density $1.5 \times 10^{3} kg - m^{- 3} kept$ at $30^{\circ} C$ . When the temperature of the liquid is raised to $40^{\circ} C$ , the metal piece weights $27 g$ . The density of liquid of $40^{\circ} C$ is $1.25 \times 10^{3} kg - m^{- 3}$ , the coefficient of linear expansion of metal is

1

1.3 \times 10^{- 3} /^{\circ} C

2

5.2 \times 10^{- 3} /^{\circ} C

3

2.6 \times 10^{- 3} /^{\circ} C

4

0.26 \times 10^{- 3} /^{\circ} C

Explanation:

C Given that, $M = 45 g$ (in air)
$m_{1} = 25 g$ (in liquid)
$m_{2} = 27 g$ ,
$ρ_{2} = 1.25 \times 10^{3} kg - m^{- 3}$
$T_{1} = 30^{\circ} C$
$T_{2} = 40^{\circ} C$
$T_{1} = 30^{\circ} C$ ; density of liquid, $ρ_{1} = 1.5 \times 10^{3} kg / m^{3}$
$V_{1} = \frac{M - m_{1}}{ρ_{1}} = \frac{(45 - 25)}{1.5 \times 10^{3}} \times 10^{- 3}$
$= 13.33 \times 10^{- 3} \times 10^{- 3} m^{3}$
$= 13.33 \times 10^{- 6} m^{3}$
$V_{2} = \frac{M - m_{2}}{ρ_{2}} = \frac{45 - 27}{1.25 \times 10^{3}} \times 10^{- 3}$
$= 14.4 \times 10^{- 6} m^{3}$
Coefficient of linear expansion
$γ = \frac{Δ V}{V_{1} \times Δ t}$
$= \frac{V_{2} - V_{1}}{V_{1} (T_{2} - T_{1})}$
$γ = \frac{(14.4 - 13.33) \times 10^{- 6}}{13.33 \times 10^{- 6} \times (40 - 30)}$
$γ = 8 \times 10^{- 3}^{\circ} C^{- 1}$
$γ γ = 3 α$
$α = \frac{γ}{3} = \frac{8 \times 10^{- 3}}{3}$
$α = 2.6 \times 10^{- 3}^{\circ} C^{- 1}$

VITEEE-2009

Thermal Properties of Matter

146534 A metal rod of length $L$ and cross-sectional area $A$ is heated through $T^{0} C$ . What is the force required to prevent the expansion of the rod lengthwise ?
( $Y$ = Young's modulus of material of the rod, $α$ = coefficient of linear expansion of the rod).

1

YA α T / (1 - α T)

2

YA α T / (1 + α T)

3

YA α / T (1 + α T)

4

YA α / (1 - α T)

Explanation:

B
Exp:
We know that, increasing length due to heating is given by
$Δ L = L_{0} α T$
$or, L = L_{0} (1 + α T)$
From the young modulus
$Y = \frac{F / A}{Δ L / L}$
$F = \frac{YA Δ L}{L}$
$F = \frac{YA L_{0} α T}{L_{0} (1 + α T)}$
$F = \frac{YA α T}{(1 + α T)}$

MHT-CET 2020

Thermal Properties of Matter

146535 A hole is drilled in a copper sheet. The diameter of the hole is $4.24 cm$ at ${27.0}^{\circ} C$ . What is the change in the diameter of the hole when the sheet is heated to $227^{\circ} C$ ? Coefficient of linear expansion of copper $= 1.70 \times 10^{- 5} K^{- 1}$

1

0.0144 cm

2

0.144 cm

3

1.44 cm

4

0.0014 cm

Explanation:

A Given, $L = 4.24 cm, T_{1} = 27^{\circ} C, T_{2} = 227^{\circ} C$ $Δ T = (T_{2} - T_{1}) = (227 - 27) = 200^{\circ} C$
We know that,
$Δ L = L α Δ T$
$Δ L = 4.24 \times 1.70 \times 10^{- 5} \times 200$
$Δ L = 0.0144 cm$

MHT-CET 2020

Thermal Properties of Matter

146536 A brass rod of length $50 cm$ and diameter 3.0 $mm$ is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at $250^{\circ} C$ , if the original lengths are at ${40.0}^{\circ} C$ ? (coefficient of linear expansion of brass $= 2.0 \times 10^{- 5} K^{- 1}$ , steel $=$ $1.2 \times 10^{- 5} K^{- 1}$ )

1

0.4365 cm

2

0.5525 cm

3

0.3465 cm

4

0.2465 cm

Explanation:

C Given,
$α_{b} = 2.0 \times 10^{- 5} K^{- 1}, α_{s} = 1.2 \times 10^{- 5} K^{- 1}$
$Δ T = (250^{\circ} C - 40^{\circ} C) = 210^{\circ} C$
$l_{1} = 50 cm, l_{2} = 50 cm$
$Δ l_{1} = l_{1} α_{b} Δ T$
$= 50 \times 2.0 \times 10^{- 5} \times 210$
$= 0.21 cm$
$Δ l_{2} = l_{2} α_{s} Δ T$
$= 50 \times 1.2 \times 10^{- 5} \times 210 = 0.126 cm$
$∴$ Change in length of the combined rod,
$Δ l = Δ l_{1} + Δ l_{2}$
$= 0.21 + 0.126$
$Δ l = 0.336 cm$
$Δ l \approx 0.3465 cm$

MHT-CET 2020

Thermal Properties of Matter

146537 The coefficient of volume expansion of glycerin is $49 \times 10^{- 5} K^{- 1}$ . What is the fractional change in its density for ${ar}^{\circ} C$ rise in temperature?

1 0.0145

2 0.0154

3 0.154

4

1.144 w

Explanation:

A Given,
Coefficient of volume expansion of glycerin $(α_{v})$
$= 49 \times 10^{- 5} K^{- 1}, Δ T = 30^{\circ} C$
We know that,
$Δ V = V α_{v} Δ T$
$V_{T_{2}} - V_{T_{1}} = V_{T_{1}} α_{v} Δ T (∵ ρ = \frac{m}{V})$
$\frac{m}{ρ_{T_{2}}} - \frac{m}{ρ_{T_{1}}} = \frac{m}{ρ_{T_{1}}} α_{v} Δ T$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}} ρ_{T_{1}}} = \frac{α_{v} Δ T}{ρ_{T_{1}}}$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}}} = α_{v} Δ T$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}}} = 49 \times 10^{- 5} \times 30$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}}} = 0.01470$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}}} \approx 0.0145$
$∴$ Fractional change in density $= 0.0145$

MHT-CET 2020

Thermal Properties of Matter

146533 A piece of metal weighs $45 g$ in air and $25 g$ in a liquid of density $1.5 \times 10^{3} kg - m^{- 3} kept$ at $30^{\circ} C$ . When the temperature of the liquid is raised to $40^{\circ} C$ , the metal piece weights $27 g$ . The density of liquid of $40^{\circ} C$ is $1.25 \times 10^{3} kg - m^{- 3}$ , the coefficient of linear expansion of metal is

1

1.3 \times 10^{- 3} /^{\circ} C

2

5.2 \times 10^{- 3} /^{\circ} C

3

2.6 \times 10^{- 3} /^{\circ} C

4

0.26 \times 10^{- 3} /^{\circ} C

Explanation:

C Given that, $M = 45 g$ (in air)
$m_{1} = 25 g$ (in liquid)
$m_{2} = 27 g$ ,
$ρ_{2} = 1.25 \times 10^{3} kg - m^{- 3}$
$T_{1} = 30^{\circ} C$
$T_{2} = 40^{\circ} C$
$T_{1} = 30^{\circ} C$ ; density of liquid, $ρ_{1} = 1.5 \times 10^{3} kg / m^{3}$
$V_{1} = \frac{M - m_{1}}{ρ_{1}} = \frac{(45 - 25)}{1.5 \times 10^{3}} \times 10^{- 3}$
$= 13.33 \times 10^{- 3} \times 10^{- 3} m^{3}$
$= 13.33 \times 10^{- 6} m^{3}$
$V_{2} = \frac{M - m_{2}}{ρ_{2}} = \frac{45 - 27}{1.25 \times 10^{3}} \times 10^{- 3}$
$= 14.4 \times 10^{- 6} m^{3}$
Coefficient of linear expansion
$γ = \frac{Δ V}{V_{1} \times Δ t}$
$= \frac{V_{2} - V_{1}}{V_{1} (T_{2} - T_{1})}$
$γ = \frac{(14.4 - 13.33) \times 10^{- 6}}{13.33 \times 10^{- 6} \times (40 - 30)}$
$γ = 8 \times 10^{- 3}^{\circ} C^{- 1}$
$γ γ = 3 α$
$α = \frac{γ}{3} = \frac{8 \times 10^{- 3}}{3}$
$α = 2.6 \times 10^{- 3}^{\circ} C^{- 1}$

VITEEE-2009

Thermal Properties of Matter

146534 A metal rod of length $L$ and cross-sectional area $A$ is heated through $T^{0} C$ . What is the force required to prevent the expansion of the rod lengthwise ?
( $Y$ = Young's modulus of material of the rod, $α$ = coefficient of linear expansion of the rod).

1

YA α T / (1 - α T)

2

YA α T / (1 + α T)

3

YA α / T (1 + α T)

4

YA α / (1 - α T)

Explanation:

B
Exp:
We know that, increasing length due to heating is given by
$Δ L = L_{0} α T$
$or, L = L_{0} (1 + α T)$
From the young modulus
$Y = \frac{F / A}{Δ L / L}$
$F = \frac{YA Δ L}{L}$
$F = \frac{YA L_{0} α T}{L_{0} (1 + α T)}$
$F = \frac{YA α T}{(1 + α T)}$

MHT-CET 2020

Thermal Properties of Matter

146535 A hole is drilled in a copper sheet. The diameter of the hole is $4.24 cm$ at ${27.0}^{\circ} C$ . What is the change in the diameter of the hole when the sheet is heated to $227^{\circ} C$ ? Coefficient of linear expansion of copper $= 1.70 \times 10^{- 5} K^{- 1}$

1

0.0144 cm

2

0.144 cm

3

1.44 cm

4

0.0014 cm

Explanation:

A Given, $L = 4.24 cm, T_{1} = 27^{\circ} C, T_{2} = 227^{\circ} C$ $Δ T = (T_{2} - T_{1}) = (227 - 27) = 200^{\circ} C$
We know that,
$Δ L = L α Δ T$
$Δ L = 4.24 \times 1.70 \times 10^{- 5} \times 200$
$Δ L = 0.0144 cm$

MHT-CET 2020

Thermal Properties of Matter

146536 A brass rod of length $50 cm$ and diameter 3.0 $mm$ is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at $250^{\circ} C$ , if the original lengths are at ${40.0}^{\circ} C$ ? (coefficient of linear expansion of brass $= 2.0 \times 10^{- 5} K^{- 1}$ , steel $=$ $1.2 \times 10^{- 5} K^{- 1}$ )

1

0.4365 cm

2

0.5525 cm

3

0.3465 cm

4

0.2465 cm

Explanation:

C Given,
$α_{b} = 2.0 \times 10^{- 5} K^{- 1}, α_{s} = 1.2 \times 10^{- 5} K^{- 1}$
$Δ T = (250^{\circ} C - 40^{\circ} C) = 210^{\circ} C$
$l_{1} = 50 cm, l_{2} = 50 cm$
$Δ l_{1} = l_{1} α_{b} Δ T$
$= 50 \times 2.0 \times 10^{- 5} \times 210$
$= 0.21 cm$
$Δ l_{2} = l_{2} α_{s} Δ T$
$= 50 \times 1.2 \times 10^{- 5} \times 210 = 0.126 cm$
$∴$ Change in length of the combined rod,
$Δ l = Δ l_{1} + Δ l_{2}$
$= 0.21 + 0.126$
$Δ l = 0.336 cm$
$Δ l \approx 0.3465 cm$

MHT-CET 2020

Thermal Properties of Matter

146537 The coefficient of volume expansion of glycerin is $49 \times 10^{- 5} K^{- 1}$ . What is the fractional change in its density for ${ar}^{\circ} C$ rise in temperature?

1 0.0145

2 0.0154

3 0.154

4

1.144 w

Explanation:

A Given,
Coefficient of volume expansion of glycerin $(α_{v})$
$= 49 \times 10^{- 5} K^{- 1}, Δ T = 30^{\circ} C$
We know that,
$Δ V = V α_{v} Δ T$
$V_{T_{2}} - V_{T_{1}} = V_{T_{1}} α_{v} Δ T (∵ ρ = \frac{m}{V})$
$\frac{m}{ρ_{T_{2}}} - \frac{m}{ρ_{T_{1}}} = \frac{m}{ρ_{T_{1}}} α_{v} Δ T$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}} ρ_{T_{1}}} = \frac{α_{v} Δ T}{ρ_{T_{1}}}$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}}} = α_{v} Δ T$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}}} = 49 \times 10^{- 5} \times 30$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}}} = 0.01470$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}}} \approx 0.0145$
$∴$ Fractional change in density $= 0.0145$

MHT-CET 2020

Thermal Properties of Matter

146533 A piece of metal weighs $45 g$ in air and $25 g$ in a liquid of density $1.5 \times 10^{3} kg - m^{- 3} kept$ at $30^{\circ} C$ . When the temperature of the liquid is raised to $40^{\circ} C$ , the metal piece weights $27 g$ . The density of liquid of $40^{\circ} C$ is $1.25 \times 10^{3} kg - m^{- 3}$ , the coefficient of linear expansion of metal is

1

1.3 \times 10^{- 3} /^{\circ} C

2

5.2 \times 10^{- 3} /^{\circ} C

3

2.6 \times 10^{- 3} /^{\circ} C

4

0.26 \times 10^{- 3} /^{\circ} C

Explanation:

C Given that, $M = 45 g$ (in air)
$m_{1} = 25 g$ (in liquid)
$m_{2} = 27 g$ ,
$ρ_{2} = 1.25 \times 10^{3} kg - m^{- 3}$
$T_{1} = 30^{\circ} C$
$T_{2} = 40^{\circ} C$
$T_{1} = 30^{\circ} C$ ; density of liquid, $ρ_{1} = 1.5 \times 10^{3} kg / m^{3}$
$V_{1} = \frac{M - m_{1}}{ρ_{1}} = \frac{(45 - 25)}{1.5 \times 10^{3}} \times 10^{- 3}$
$= 13.33 \times 10^{- 3} \times 10^{- 3} m^{3}$
$= 13.33 \times 10^{- 6} m^{3}$
$V_{2} = \frac{M - m_{2}}{ρ_{2}} = \frac{45 - 27}{1.25 \times 10^{3}} \times 10^{- 3}$
$= 14.4 \times 10^{- 6} m^{3}$
Coefficient of linear expansion
$γ = \frac{Δ V}{V_{1} \times Δ t}$
$= \frac{V_{2} - V_{1}}{V_{1} (T_{2} - T_{1})}$
$γ = \frac{(14.4 - 13.33) \times 10^{- 6}}{13.33 \times 10^{- 6} \times (40 - 30)}$
$γ = 8 \times 10^{- 3}^{\circ} C^{- 1}$
$γ γ = 3 α$
$α = \frac{γ}{3} = \frac{8 \times 10^{- 3}}{3}$
$α = 2.6 \times 10^{- 3}^{\circ} C^{- 1}$

VITEEE-2009

Thermal Properties of Matter

146534 A metal rod of length $L$ and cross-sectional area $A$ is heated through $T^{0} C$ . What is the force required to prevent the expansion of the rod lengthwise ?
( $Y$ = Young's modulus of material of the rod, $α$ = coefficient of linear expansion of the rod).

1

YA α T / (1 - α T)

2

YA α T / (1 + α T)

3

YA α / T (1 + α T)

4

YA α / (1 - α T)

Explanation:

B
Exp:
We know that, increasing length due to heating is given by
$Δ L = L_{0} α T$
$or, L = L_{0} (1 + α T)$
From the young modulus
$Y = \frac{F / A}{Δ L / L}$
$F = \frac{YA Δ L}{L}$
$F = \frac{YA L_{0} α T}{L_{0} (1 + α T)}$
$F = \frac{YA α T}{(1 + α T)}$

MHT-CET 2020

Thermal Properties of Matter

146535 A hole is drilled in a copper sheet. The diameter of the hole is $4.24 cm$ at ${27.0}^{\circ} C$ . What is the change in the diameter of the hole when the sheet is heated to $227^{\circ} C$ ? Coefficient of linear expansion of copper $= 1.70 \times 10^{- 5} K^{- 1}$

1

0.0144 cm

2

0.144 cm

3

1.44 cm

4

0.0014 cm

Explanation:

A Given, $L = 4.24 cm, T_{1} = 27^{\circ} C, T_{2} = 227^{\circ} C$ $Δ T = (T_{2} - T_{1}) = (227 - 27) = 200^{\circ} C$
We know that,
$Δ L = L α Δ T$
$Δ L = 4.24 \times 1.70 \times 10^{- 5} \times 200$
$Δ L = 0.0144 cm$

MHT-CET 2020

Thermal Properties of Matter

146536 A brass rod of length $50 cm$ and diameter 3.0 $mm$ is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at $250^{\circ} C$ , if the original lengths are at ${40.0}^{\circ} C$ ? (coefficient of linear expansion of brass $= 2.0 \times 10^{- 5} K^{- 1}$ , steel $=$ $1.2 \times 10^{- 5} K^{- 1}$ )

1

0.4365 cm

2

0.5525 cm

3

0.3465 cm

4

0.2465 cm

Explanation:

C Given,
$α_{b} = 2.0 \times 10^{- 5} K^{- 1}, α_{s} = 1.2 \times 10^{- 5} K^{- 1}$
$Δ T = (250^{\circ} C - 40^{\circ} C) = 210^{\circ} C$
$l_{1} = 50 cm, l_{2} = 50 cm$
$Δ l_{1} = l_{1} α_{b} Δ T$
$= 50 \times 2.0 \times 10^{- 5} \times 210$
$= 0.21 cm$
$Δ l_{2} = l_{2} α_{s} Δ T$
$= 50 \times 1.2 \times 10^{- 5} \times 210 = 0.126 cm$
$∴$ Change in length of the combined rod,
$Δ l = Δ l_{1} + Δ l_{2}$
$= 0.21 + 0.126$
$Δ l = 0.336 cm$
$Δ l \approx 0.3465 cm$

MHT-CET 2020

Thermal Properties of Matter

146537 The coefficient of volume expansion of glycerin is $49 \times 10^{- 5} K^{- 1}$ . What is the fractional change in its density for ${ar}^{\circ} C$ rise in temperature?

1 0.0145

2 0.0154

3 0.154

4

1.144 w

Explanation:

A Given,
Coefficient of volume expansion of glycerin $(α_{v})$
$= 49 \times 10^{- 5} K^{- 1}, Δ T = 30^{\circ} C$
We know that,
$Δ V = V α_{v} Δ T$
$V_{T_{2}} - V_{T_{1}} = V_{T_{1}} α_{v} Δ T (∵ ρ = \frac{m}{V})$
$\frac{m}{ρ_{T_{2}}} - \frac{m}{ρ_{T_{1}}} = \frac{m}{ρ_{T_{1}}} α_{v} Δ T$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}} ρ_{T_{1}}} = \frac{α_{v} Δ T}{ρ_{T_{1}}}$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}}} = α_{v} Δ T$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}}} = 49 \times 10^{- 5} \times 30$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}}} = 0.01470$
$\frac{ρ_{T_{1}} - ρ_{T_{2}}}{ρ_{T_{2}}} \approx 0.0145$
$∴$ Fractional change in density $= 0.0145$

MHT-CET 2020

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