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PHXI11:THERMAL PROPERTIES OF MATTER

366772 An isosceles triangle is formed with in a rod of length $l_{1}$ and coefficient of linear expansion $α_{1}$ as the base and two thin rods each of length $l_{2}$ and coefficient of linear expansions $α_{2}$ as the two sides. If the distance between the apex and the midpoint of the base remain unchanged as the temperature is varied, then
supporting img

1

\frac{l_{1}}{l_{2}} = 2 \sqrt{\frac{α_{2}}{α_{1}}}

2

\frac{l_{1}}{l_{2}} = \sqrt{\frac{α_{2}}{α_{1}}}

3

\frac{l_{1}}{l_{2}} = \frac{α_{2}}{α_{1}}

4

\frac{l_{1}}{l_{2}} = \frac{2 α_{2}}{α_{1}}

Explanation:

Let $l_{1}^{1} & l_{2}^{1}$ are new lengths of the rods after increasing temperature $Δ T$ as shown in the figure.
supporting img

As $h$ remains constant
$\sqrt{l_{2}^{2} - \frac{l_{1}^{2}}{4}} = \sqrt{l_{2}^{l^{2}} - \frac{l_{1}^{l^{2}}}{4}}$
square on both sides.
$\begin{aligned} l_{2}^{2} - \frac{l_{1}^{2}}{4} = l_{2}^{1^{2}} - \frac{l_{1}^{1^{2}}}{4} \\ l_{2}^{2} - \frac{l_{1}^{2}}{4} = l_{2}^{2} {(1 + α_{2} Δ T)}^{2} - \frac{l_{1}^{2} {(1 + α_{1} Δ T)}^{2}}{4} \\ \Rightarrow \frac{l_{1}}{l_{2}} = 2 \sqrt{\frac{α_{2}}{α_{1}}} \end{aligned}$

PHXI11:THERMAL PROPERTIES OF MATTER

366773 A hole is drilled in a copper sheet. The diameter of the hole is $4.24 c m$ at $27^{\circ} C$ . 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 is $1.70 \times 10^{- 5} /^{\circ} C$ .

1

1.44 \times 10^{- 2} c m

2

2.44 \times 10^{- 3} c m

3

1.44 \times 10^{- 2} m m

4

2.44 \times 10^{- 2} m m

Explanation:

Given, diameter of the hole
$(d_{1}) = 4.24 c m$
Initial temperature
$T_{1} = 27 + 273 = 300 K$
Final temperature $T_{2} = 227 + 273 = 500 K$
Coefficient of linear expansion
$(α) = 1.70 \times 10^{- 5} /^{\circ} C$
Coefficient of areal expansion $(β) = 2 α$
$= 3.40 \times 10^{- 5} /^{\circ} C$
Initial area of hole at $27^{\circ} C (A_{1}) = π r^{2} = \frac{π d_{1}^{2}}{4}$
$\frac{π}{4} (4.24)^{2} = 4.494 π c m^{2}$
Area of hole at $227^{\circ} C (A_{2}) = A_{1} (1 + β . Δ t)$
$\begin{aligned} = 4.494 π [1 + 3.40 \times 10^{- 5} \times (227 - 27)] \\ = 4.495 π \times 1.0068 \end{aligned}$
If diameter of hole becomes $d_{2} a t 227^{\circ} C$ , then
$A_{2} = \frac{π d_{2}^{2}}{4} 4.525 π = \frac{π d_{2}^{2}}{4}$
$o r d_{2} = 4.2544 c m$
Change in diameter $(Δ d) = d_{2} - d_{1}$
$= 1.44 \times 10^{- 2} c m$

NCERT Exempler

PHXI11:THERMAL PROPERTIES OF MATTER

366774 A bimetallic strip consists of metals $X$ and $Y$ . It is mounted rigidly at the base as shown. The metal $X$ has a higher coefficient of expansion compared to that for metal $Y$ . When bimetallic strip is placed in a hot bath:
supporting img

1 It will bend towards the right

2 It will bend towards the left

3 It will neither bend nor shrink

4 It will not bend but shrink

Explanation:

The strip on heating bends in the form of an arc with more expandable metal $(X)$ on outer side. So it bends towards right

PHXI11:THERMAL PROPERTIES OF MATTER

366775 As the temperature is increased, the time period of a pendulum.

1 Increase as the effective length increases even though its centre of mass still remains at the centre of the bob.

2 Decreases as its effective length increases even through its centre of mass still remains at the centre of the bob.

3 Increases as its effective length increases due to shifting of centre of mass below below centre of the bob

4 Decreases as its effective length increases remains same but centre of the mass shifts above the centre of the bob.

Explanation:

With increase in temperature, the effective length $(l)$ of simple pendulum increases even through its centre of mass still remains at the centre of the bob. Time period, $T = 2 π \sqrt{l / g}$ or $T \propto \sqrt{l}$ . $T$ increases as temperature increases

NCERT Exempler

PHXI11:THERMAL PROPERTIES OF MATTER

366772 An isosceles triangle is formed with in a rod of length $l_{1}$ and coefficient of linear expansion $α_{1}$ as the base and two thin rods each of length $l_{2}$ and coefficient of linear expansions $α_{2}$ as the two sides. If the distance between the apex and the midpoint of the base remain unchanged as the temperature is varied, then
supporting img

1

\frac{l_{1}}{l_{2}} = 2 \sqrt{\frac{α_{2}}{α_{1}}}

2

\frac{l_{1}}{l_{2}} = \sqrt{\frac{α_{2}}{α_{1}}}

3

\frac{l_{1}}{l_{2}} = \frac{α_{2}}{α_{1}}

4

\frac{l_{1}}{l_{2}} = \frac{2 α_{2}}{α_{1}}

Explanation:

Let $l_{1}^{1} & l_{2}^{1}$ are new lengths of the rods after increasing temperature $Δ T$ as shown in the figure.
supporting img

As $h$ remains constant
$\sqrt{l_{2}^{2} - \frac{l_{1}^{2}}{4}} = \sqrt{l_{2}^{l^{2}} - \frac{l_{1}^{l^{2}}}{4}}$
square on both sides.
$\begin{aligned} l_{2}^{2} - \frac{l_{1}^{2}}{4} = l_{2}^{1^{2}} - \frac{l_{1}^{1^{2}}}{4} \\ l_{2}^{2} - \frac{l_{1}^{2}}{4} = l_{2}^{2} {(1 + α_{2} Δ T)}^{2} - \frac{l_{1}^{2} {(1 + α_{1} Δ T)}^{2}}{4} \\ \Rightarrow \frac{l_{1}}{l_{2}} = 2 \sqrt{\frac{α_{2}}{α_{1}}} \end{aligned}$

PHXI11:THERMAL PROPERTIES OF MATTER

366773 A hole is drilled in a copper sheet. The diameter of the hole is $4.24 c m$ at $27^{\circ} C$ . 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 is $1.70 \times 10^{- 5} /^{\circ} C$ .

1

1.44 \times 10^{- 2} c m

2

2.44 \times 10^{- 3} c m

3

1.44 \times 10^{- 2} m m

4

2.44 \times 10^{- 2} m m

Explanation:

Given, diameter of the hole
$(d_{1}) = 4.24 c m$
Initial temperature
$T_{1} = 27 + 273 = 300 K$
Final temperature $T_{2} = 227 + 273 = 500 K$
Coefficient of linear expansion
$(α) = 1.70 \times 10^{- 5} /^{\circ} C$
Coefficient of areal expansion $(β) = 2 α$
$= 3.40 \times 10^{- 5} /^{\circ} C$
Initial area of hole at $27^{\circ} C (A_{1}) = π r^{2} = \frac{π d_{1}^{2}}{4}$
$\frac{π}{4} (4.24)^{2} = 4.494 π c m^{2}$
Area of hole at $227^{\circ} C (A_{2}) = A_{1} (1 + β . Δ t)$
$\begin{aligned} = 4.494 π [1 + 3.40 \times 10^{- 5} \times (227 - 27)] \\ = 4.495 π \times 1.0068 \end{aligned}$
If diameter of hole becomes $d_{2} a t 227^{\circ} C$ , then
$A_{2} = \frac{π d_{2}^{2}}{4} 4.525 π = \frac{π d_{2}^{2}}{4}$
$o r d_{2} = 4.2544 c m$
Change in diameter $(Δ d) = d_{2} - d_{1}$
$= 1.44 \times 10^{- 2} c m$

NCERT Exempler

PHXI11:THERMAL PROPERTIES OF MATTER

366774 A bimetallic strip consists of metals $X$ and $Y$ . It is mounted rigidly at the base as shown. The metal $X$ has a higher coefficient of expansion compared to that for metal $Y$ . When bimetallic strip is placed in a hot bath:
supporting img

1 It will bend towards the right

2 It will bend towards the left

3 It will neither bend nor shrink

4 It will not bend but shrink

Explanation:

The strip on heating bends in the form of an arc with more expandable metal $(X)$ on outer side. So it bends towards right

PHXI11:THERMAL PROPERTIES OF MATTER

366775 As the temperature is increased, the time period of a pendulum.

1 Increase as the effective length increases even though its centre of mass still remains at the centre of the bob.

2 Decreases as its effective length increases even through its centre of mass still remains at the centre of the bob.

3 Increases as its effective length increases due to shifting of centre of mass below below centre of the bob

4 Decreases as its effective length increases remains same but centre of the mass shifts above the centre of the bob.

Explanation:

With increase in temperature, the effective length $(l)$ of simple pendulum increases even through its centre of mass still remains at the centre of the bob. Time period, $T = 2 π \sqrt{l / g}$ or $T \propto \sqrt{l}$ . $T$ increases as temperature increases

NCERT Exempler

PHXI11:THERMAL PROPERTIES OF MATTER

366772 An isosceles triangle is formed with in a rod of length $l_{1}$ and coefficient of linear expansion $α_{1}$ as the base and two thin rods each of length $l_{2}$ and coefficient of linear expansions $α_{2}$ as the two sides. If the distance between the apex and the midpoint of the base remain unchanged as the temperature is varied, then
supporting img

1

\frac{l_{1}}{l_{2}} = 2 \sqrt{\frac{α_{2}}{α_{1}}}

2

\frac{l_{1}}{l_{2}} = \sqrt{\frac{α_{2}}{α_{1}}}

3

\frac{l_{1}}{l_{2}} = \frac{α_{2}}{α_{1}}

4

\frac{l_{1}}{l_{2}} = \frac{2 α_{2}}{α_{1}}

Explanation:

Let $l_{1}^{1} & l_{2}^{1}$ are new lengths of the rods after increasing temperature $Δ T$ as shown in the figure.
supporting img

As $h$ remains constant
$\sqrt{l_{2}^{2} - \frac{l_{1}^{2}}{4}} = \sqrt{l_{2}^{l^{2}} - \frac{l_{1}^{l^{2}}}{4}}$
square on both sides.
$\begin{aligned} l_{2}^{2} - \frac{l_{1}^{2}}{4} = l_{2}^{1^{2}} - \frac{l_{1}^{1^{2}}}{4} \\ l_{2}^{2} - \frac{l_{1}^{2}}{4} = l_{2}^{2} {(1 + α_{2} Δ T)}^{2} - \frac{l_{1}^{2} {(1 + α_{1} Δ T)}^{2}}{4} \\ \Rightarrow \frac{l_{1}}{l_{2}} = 2 \sqrt{\frac{α_{2}}{α_{1}}} \end{aligned}$

PHXI11:THERMAL PROPERTIES OF MATTER

366773 A hole is drilled in a copper sheet. The diameter of the hole is $4.24 c m$ at $27^{\circ} C$ . 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 is $1.70 \times 10^{- 5} /^{\circ} C$ .

1

1.44 \times 10^{- 2} c m

2

2.44 \times 10^{- 3} c m

3

1.44 \times 10^{- 2} m m

4

2.44 \times 10^{- 2} m m

Explanation:

Given, diameter of the hole
$(d_{1}) = 4.24 c m$
Initial temperature
$T_{1} = 27 + 273 = 300 K$
Final temperature $T_{2} = 227 + 273 = 500 K$
Coefficient of linear expansion
$(α) = 1.70 \times 10^{- 5} /^{\circ} C$
Coefficient of areal expansion $(β) = 2 α$
$= 3.40 \times 10^{- 5} /^{\circ} C$
Initial area of hole at $27^{\circ} C (A_{1}) = π r^{2} = \frac{π d_{1}^{2}}{4}$
$\frac{π}{4} (4.24)^{2} = 4.494 π c m^{2}$
Area of hole at $227^{\circ} C (A_{2}) = A_{1} (1 + β . Δ t)$
$\begin{aligned} = 4.494 π [1 + 3.40 \times 10^{- 5} \times (227 - 27)] \\ = 4.495 π \times 1.0068 \end{aligned}$
If diameter of hole becomes $d_{2} a t 227^{\circ} C$ , then
$A_{2} = \frac{π d_{2}^{2}}{4} 4.525 π = \frac{π d_{2}^{2}}{4}$
$o r d_{2} = 4.2544 c m$
Change in diameter $(Δ d) = d_{2} - d_{1}$
$= 1.44 \times 10^{- 2} c m$

NCERT Exempler

PHXI11:THERMAL PROPERTIES OF MATTER

366774 A bimetallic strip consists of metals $X$ and $Y$ . It is mounted rigidly at the base as shown. The metal $X$ has a higher coefficient of expansion compared to that for metal $Y$ . When bimetallic strip is placed in a hot bath:
supporting img

1 It will bend towards the right

2 It will bend towards the left

3 It will neither bend nor shrink

4 It will not bend but shrink

Explanation:

The strip on heating bends in the form of an arc with more expandable metal $(X)$ on outer side. So it bends towards right

PHXI11:THERMAL PROPERTIES OF MATTER

366775 As the temperature is increased, the time period of a pendulum.

1 Increase as the effective length increases even though its centre of mass still remains at the centre of the bob.

2 Decreases as its effective length increases even through its centre of mass still remains at the centre of the bob.

3 Increases as its effective length increases due to shifting of centre of mass below below centre of the bob

4 Decreases as its effective length increases remains same but centre of the mass shifts above the centre of the bob.

Explanation:

With increase in temperature, the effective length $(l)$ of simple pendulum increases even through its centre of mass still remains at the centre of the bob. Time period, $T = 2 π \sqrt{l / g}$ or $T \propto \sqrt{l}$ . $T$ increases as temperature increases

NCERT Exempler

PHXI11:THERMAL PROPERTIES OF MATTER

366772 An isosceles triangle is formed with in a rod of length $l_{1}$ and coefficient of linear expansion $α_{1}$ as the base and two thin rods each of length $l_{2}$ and coefficient of linear expansions $α_{2}$ as the two sides. If the distance between the apex and the midpoint of the base remain unchanged as the temperature is varied, then
supporting img

1

\frac{l_{1}}{l_{2}} = 2 \sqrt{\frac{α_{2}}{α_{1}}}

2

\frac{l_{1}}{l_{2}} = \sqrt{\frac{α_{2}}{α_{1}}}

3

\frac{l_{1}}{l_{2}} = \frac{α_{2}}{α_{1}}

4

\frac{l_{1}}{l_{2}} = \frac{2 α_{2}}{α_{1}}

Explanation:

Let $l_{1}^{1} & l_{2}^{1}$ are new lengths of the rods after increasing temperature $Δ T$ as shown in the figure.
supporting img

As $h$ remains constant
$\sqrt{l_{2}^{2} - \frac{l_{1}^{2}}{4}} = \sqrt{l_{2}^{l^{2}} - \frac{l_{1}^{l^{2}}}{4}}$
square on both sides.
$\begin{aligned} l_{2}^{2} - \frac{l_{1}^{2}}{4} = l_{2}^{1^{2}} - \frac{l_{1}^{1^{2}}}{4} \\ l_{2}^{2} - \frac{l_{1}^{2}}{4} = l_{2}^{2} {(1 + α_{2} Δ T)}^{2} - \frac{l_{1}^{2} {(1 + α_{1} Δ T)}^{2}}{4} \\ \Rightarrow \frac{l_{1}}{l_{2}} = 2 \sqrt{\frac{α_{2}}{α_{1}}} \end{aligned}$

PHXI11:THERMAL PROPERTIES OF MATTER

366773 A hole is drilled in a copper sheet. The diameter of the hole is $4.24 c m$ at $27^{\circ} C$ . 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 is $1.70 \times 10^{- 5} /^{\circ} C$ .

1

1.44 \times 10^{- 2} c m

2

2.44 \times 10^{- 3} c m

3

1.44 \times 10^{- 2} m m

4

2.44 \times 10^{- 2} m m

Explanation:

Given, diameter of the hole
$(d_{1}) = 4.24 c m$
Initial temperature
$T_{1} = 27 + 273 = 300 K$
Final temperature $T_{2} = 227 + 273 = 500 K$
Coefficient of linear expansion
$(α) = 1.70 \times 10^{- 5} /^{\circ} C$
Coefficient of areal expansion $(β) = 2 α$
$= 3.40 \times 10^{- 5} /^{\circ} C$
Initial area of hole at $27^{\circ} C (A_{1}) = π r^{2} = \frac{π d_{1}^{2}}{4}$
$\frac{π}{4} (4.24)^{2} = 4.494 π c m^{2}$
Area of hole at $227^{\circ} C (A_{2}) = A_{1} (1 + β . Δ t)$
$\begin{aligned} = 4.494 π [1 + 3.40 \times 10^{- 5} \times (227 - 27)] \\ = 4.495 π \times 1.0068 \end{aligned}$
If diameter of hole becomes $d_{2} a t 227^{\circ} C$ , then
$A_{2} = \frac{π d_{2}^{2}}{4} 4.525 π = \frac{π d_{2}^{2}}{4}$
$o r d_{2} = 4.2544 c m$
Change in diameter $(Δ d) = d_{2} - d_{1}$
$= 1.44 \times 10^{- 2} c m$

NCERT Exempler

PHXI11:THERMAL PROPERTIES OF MATTER

366774 A bimetallic strip consists of metals $X$ and $Y$ . It is mounted rigidly at the base as shown. The metal $X$ has a higher coefficient of expansion compared to that for metal $Y$ . When bimetallic strip is placed in a hot bath:
supporting img

1 It will bend towards the right

2 It will bend towards the left

3 It will neither bend nor shrink

4 It will not bend but shrink

Explanation:

The strip on heating bends in the form of an arc with more expandable metal $(X)$ on outer side. So it bends towards right

PHXI11:THERMAL PROPERTIES OF MATTER

366775 As the temperature is increased, the time period of a pendulum.

1 Increase as the effective length increases even though its centre of mass still remains at the centre of the bob.

2 Decreases as its effective length increases even through its centre of mass still remains at the centre of the bob.

3 Increases as its effective length increases due to shifting of centre of mass below below centre of the bob

4 Decreases as its effective length increases remains same but centre of the mass shifts above the centre of the bob.

Explanation:

With increase in temperature, the effective length $(l)$ of simple pendulum increases even through its centre of mass still remains at the centre of the bob. Time period, $T = 2 π \sqrt{l / g}$ or $T \propto \sqrt{l}$ . $T$ increases as temperature increases

NCERT Exempler

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