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Thermal Properties of Matter

146453 A horizontal fire hose with a nozzle of crosssectional area $\frac{5}{\sqrt{21}} \times 10^{- 3} m^{2}$ delivers a cubic meter of water in $10 s$ . What will be the maximum possible increase in the temperature of water while it hits a rigid wall (neglecting the effect of gravity)?

1

1^{\circ} C

2

{0.1}^{\circ} C

3

10^{\circ} C

4

{0.01}^{\circ} C

Explanation:

A Given that, $A = \frac{5}{\sqrt{21}} \times 10^{- 3} m^{2}$

Rate of heat transfer, $\frac{dQ}{dt} = A \times V$ ?
$\frac{1}{10} = A \times V$
$V = \frac{1}{10 A}$
Kinetic energy change into heat
$\frac{1}{2} {mV}^{2} = mS Δ T$
$Δ T = \frac{V^{2}}{2 S} = {(\frac{1}{10 A})}^{2} \times \frac{1}{2 S}$
$= \frac{1}{2} \times \frac{21}{100 \times 25 \times 10^{- 6}} \times \frac{1}{\frac{4.18}{10^{- 3}}}$
$Δ T = 1^{\circ} C$

WB JEE 2019

Thermal Properties of Matter

146454 Two black bodies $A$ and $B$ have equal surface areas and are maintained at temperatures $27^{\circ} C$ and $177^{\circ} C$ respectively. What will be the ratio of the thermal energy radiated per second by $A$ to that by $B$ ?

1

4 : 9

2

2 : 3

3

16 : 81

4

27 : 177

Explanation:

C According to the question-
Area of both bodies $A$ and $B$ are equal then
Temperature of body $A = 27^{\circ} C = 300 K$
Temperature of body $B = 177^{\circ} C = 450 K$
According to Stefan -Boltzman law, thermal energy radiated per second by a body-
$Q = σ {AT}^{4}$
Where, $A =$ Area
$T = Temperature$
$σ = Stefan-Boltzman's constant$
So, the ratio of thermal energy radiated per second by $A$ to that by $B$ is
$\frac{Q_{1}}{Q_{2}} = \frac{σ A}{σ A} {(\frac{T_{1}}{T_{2}})}^{4}$
Putting these value, we get-
$\frac{Q_{1}}{Q_{2}} = {(\frac{300}{450})}^{4}$
$\frac{Q_{1}}{Q_{2}} = \frac{16}{81}$
Ratio of thermal energy radiated per second
$Q_{1} : Q_{2} = 16 : 81$

WB JEE 2019

Thermal Properties of Matter

146456 A wire has resistance of $3.1 Ω$ at $30^{\circ} C$ and 4.5 $Ω$ at $100^{\circ} C$ . The temperature coefficient of resistance of the wire is

1

0.0012^{\circ} C^{- 1}

2

0.0024^{\circ} C^{- 1}

3

0.0032^{\circ} C^{- 1}

4

0.0064^{\circ} C^{- 1}

Explanation:

D Given that,
$R_{1} = 3.1 Ω T_{1} = 30^{\circ} C$
$R_{2} = 4.5 Ω T_{2} = 100^{\circ} C$
Resistance of wire at any temperature-
$R_{2} = R_{1} [1 + α (T_{2} - T_{1})]$
$4.5 = 3.1 [1 + α (100 - 30)]$
$4.5 = 3.1 [1 + 70 α]$
$\frac{4.5}{3.1} = 1 + 70 α$
$1.4516 = 1 + 70 α$
$0.4515 = 70 α$
$α = \frac{0.4515}{70}$
$α = {0.0064}^{\circ} C^{- 1}$

TS EAMCET (Engg.)-2017

Thermal Properties of Matter

146457 A sphere at $600 K$ is losing heat due to radiation. At this temperature its rate of cooling is $R$ . The rate of cooling of this sphere at $400 K$ is (temperature of surroundings is 200 K)

1

\frac{3}{16} R

2

\frac{8}{27} R

3

\frac{16}{3} R

4

7 R

Explanation:

A According to Stefan - Boltzman Law-
$E = σ T^{4}$
$R = \frac{dT}{dt} \propto (T^{4} - T_{o}^{4})$
$\frac{R_{1}}{R_{2}} = \frac{T_{1}^{4} - T_{o}^{4}}{T_{2}^{4} - T_{o}^{4}}$
$\frac{R_{1}}{R_{2}} = \frac{[(600)^{4} - (200)^{4}]}{[(400)^{4} - (200)^{4}]}$
$\frac{R_{1}}{R_{2}} = \frac{100^{4} [(6)^{4} - (2)^{4}]}{100^{4} [(4)^{4} - (2)^{4}]}$
$\frac{R_{1}}{R_{2}} = \frac{1280}{240}$
$\frac{R_{1}}{R_{2}} = \frac{16}{3}$
$R_{2} = \frac{3}{16} R_{1} or \frac{3}{16} R$

TS EAMCET(Medical)-2015

Thermal Properties of Matter

146458 Two identical shaped metallic spheres $A$ and $B$ made up of same material of mass ' $m$ ' and ' $4 m$ ' are heated to attain a temperature $T_{1}$ and then they are placed in a container maintained at temperature $T_{2} (T_{2} < T_{1})$ . The spheres are thermally insulated from each other. If $R$ is the rate of change of temperature, then the ratio $R_{A} & R_{B}$ is

1

\frac{1}{4}

2

{(\frac{1}{4})}^{\frac{1}{3}}

3

{(\frac{1}{4})}^{\frac{2}{3}}

4

(4)^{\frac{1}{3}}

Explanation:

D Given that, $M_{1} = m, M_{2} = 4 m$
Heat loss as per Stephan's boltzman law
$E = e σ {AT}^{4}$
$E = e σ T^{4} (4 π r^{2})$
We know that,
Volume of sphere $(V) = 4 / 3 π R^{3}$
$V = \frac{mass}{Density}$
$\frac{4}{3} π r^{3} = \frac{m}{ρ}$
$r = K (m)^{1 / 3}$
Therefore, $\frac{r_{1}}{r_{2}} = {(\frac{m_{1}}{m_{2}})}^{\frac{1}{3}}$
$\frac{E_{1}}{E_{2}} = {(\frac{r_{1}}{r_{2}})}^{2} = {(\frac{m_{1}}{m_{2}})}^{2 / 3}$
Rate of Cooling $(R) = \frac{E}{mS} = \frac{AE}{m}$
Hence $\frac{R_{A}}{R_{B}} = {(\frac{m_{2}}{m_{1}})}^{\frac{1}{3}} = {(\frac{4}{1})}^{\frac{1}{3}}$

Shift-II]

Thermal Properties of Matter

146453 A horizontal fire hose with a nozzle of crosssectional area $\frac{5}{\sqrt{21}} \times 10^{- 3} m^{2}$ delivers a cubic meter of water in $10 s$ . What will be the maximum possible increase in the temperature of water while it hits a rigid wall (neglecting the effect of gravity)?

1

1^{\circ} C

2

{0.1}^{\circ} C

3

10^{\circ} C

4

{0.01}^{\circ} C

Explanation:

A Given that, $A = \frac{5}{\sqrt{21}} \times 10^{- 3} m^{2}$

Rate of heat transfer, $\frac{dQ}{dt} = A \times V$ ?
$\frac{1}{10} = A \times V$
$V = \frac{1}{10 A}$
Kinetic energy change into heat
$\frac{1}{2} {mV}^{2} = mS Δ T$
$Δ T = \frac{V^{2}}{2 S} = {(\frac{1}{10 A})}^{2} \times \frac{1}{2 S}$
$= \frac{1}{2} \times \frac{21}{100 \times 25 \times 10^{- 6}} \times \frac{1}{\frac{4.18}{10^{- 3}}}$
$Δ T = 1^{\circ} C$

WB JEE 2019

Thermal Properties of Matter

146454 Two black bodies $A$ and $B$ have equal surface areas and are maintained at temperatures $27^{\circ} C$ and $177^{\circ} C$ respectively. What will be the ratio of the thermal energy radiated per second by $A$ to that by $B$ ?

1

4 : 9

2

2 : 3

3

16 : 81

4

27 : 177

Explanation:

C According to the question-
Area of both bodies $A$ and $B$ are equal then
Temperature of body $A = 27^{\circ} C = 300 K$
Temperature of body $B = 177^{\circ} C = 450 K$
According to Stefan -Boltzman law, thermal energy radiated per second by a body-
$Q = σ {AT}^{4}$
Where, $A =$ Area
$T = Temperature$
$σ = Stefan-Boltzman's constant$
So, the ratio of thermal energy radiated per second by $A$ to that by $B$ is
$\frac{Q_{1}}{Q_{2}} = \frac{σ A}{σ A} {(\frac{T_{1}}{T_{2}})}^{4}$
Putting these value, we get-
$\frac{Q_{1}}{Q_{2}} = {(\frac{300}{450})}^{4}$
$\frac{Q_{1}}{Q_{2}} = \frac{16}{81}$
Ratio of thermal energy radiated per second
$Q_{1} : Q_{2} = 16 : 81$

WB JEE 2019

Thermal Properties of Matter

146456 A wire has resistance of $3.1 Ω$ at $30^{\circ} C$ and 4.5 $Ω$ at $100^{\circ} C$ . The temperature coefficient of resistance of the wire is

1

0.0012^{\circ} C^{- 1}

2

0.0024^{\circ} C^{- 1}

3

0.0032^{\circ} C^{- 1}

4

0.0064^{\circ} C^{- 1}

Explanation:

D Given that,
$R_{1} = 3.1 Ω T_{1} = 30^{\circ} C$
$R_{2} = 4.5 Ω T_{2} = 100^{\circ} C$
Resistance of wire at any temperature-
$R_{2} = R_{1} [1 + α (T_{2} - T_{1})]$
$4.5 = 3.1 [1 + α (100 - 30)]$
$4.5 = 3.1 [1 + 70 α]$
$\frac{4.5}{3.1} = 1 + 70 α$
$1.4516 = 1 + 70 α$
$0.4515 = 70 α$
$α = \frac{0.4515}{70}$
$α = {0.0064}^{\circ} C^{- 1}$

TS EAMCET (Engg.)-2017

Thermal Properties of Matter

146457 A sphere at $600 K$ is losing heat due to radiation. At this temperature its rate of cooling is $R$ . The rate of cooling of this sphere at $400 K$ is (temperature of surroundings is 200 K)

1

\frac{3}{16} R

2

\frac{8}{27} R

3

\frac{16}{3} R

4

7 R

Explanation:

A According to Stefan - Boltzman Law-
$E = σ T^{4}$
$R = \frac{dT}{dt} \propto (T^{4} - T_{o}^{4})$
$\frac{R_{1}}{R_{2}} = \frac{T_{1}^{4} - T_{o}^{4}}{T_{2}^{4} - T_{o}^{4}}$
$\frac{R_{1}}{R_{2}} = \frac{[(600)^{4} - (200)^{4}]}{[(400)^{4} - (200)^{4}]}$
$\frac{R_{1}}{R_{2}} = \frac{100^{4} [(6)^{4} - (2)^{4}]}{100^{4} [(4)^{4} - (2)^{4}]}$
$\frac{R_{1}}{R_{2}} = \frac{1280}{240}$
$\frac{R_{1}}{R_{2}} = \frac{16}{3}$
$R_{2} = \frac{3}{16} R_{1} or \frac{3}{16} R$

TS EAMCET(Medical)-2015

Thermal Properties of Matter

146458 Two identical shaped metallic spheres $A$ and $B$ made up of same material of mass ' $m$ ' and ' $4 m$ ' are heated to attain a temperature $T_{1}$ and then they are placed in a container maintained at temperature $T_{2} (T_{2} < T_{1})$ . The spheres are thermally insulated from each other. If $R$ is the rate of change of temperature, then the ratio $R_{A} & R_{B}$ is

1

\frac{1}{4}

2

{(\frac{1}{4})}^{\frac{1}{3}}

3

{(\frac{1}{4})}^{\frac{2}{3}}

4

(4)^{\frac{1}{3}}

Explanation:

D Given that, $M_{1} = m, M_{2} = 4 m$
Heat loss as per Stephan's boltzman law
$E = e σ {AT}^{4}$
$E = e σ T^{4} (4 π r^{2})$
We know that,
Volume of sphere $(V) = 4 / 3 π R^{3}$
$V = \frac{mass}{Density}$
$\frac{4}{3} π r^{3} = \frac{m}{ρ}$
$r = K (m)^{1 / 3}$
Therefore, $\frac{r_{1}}{r_{2}} = {(\frac{m_{1}}{m_{2}})}^{\frac{1}{3}}$
$\frac{E_{1}}{E_{2}} = {(\frac{r_{1}}{r_{2}})}^{2} = {(\frac{m_{1}}{m_{2}})}^{2 / 3}$
Rate of Cooling $(R) = \frac{E}{mS} = \frac{AE}{m}$
Hence $\frac{R_{A}}{R_{B}} = {(\frac{m_{2}}{m_{1}})}^{\frac{1}{3}} = {(\frac{4}{1})}^{\frac{1}{3}}$

Shift-II]

Thermal Properties of Matter

146453 A horizontal fire hose with a nozzle of crosssectional area $\frac{5}{\sqrt{21}} \times 10^{- 3} m^{2}$ delivers a cubic meter of water in $10 s$ . What will be the maximum possible increase in the temperature of water while it hits a rigid wall (neglecting the effect of gravity)?

1

1^{\circ} C

2

{0.1}^{\circ} C

3

10^{\circ} C

4

{0.01}^{\circ} C

Explanation:

A Given that, $A = \frac{5}{\sqrt{21}} \times 10^{- 3} m^{2}$

Rate of heat transfer, $\frac{dQ}{dt} = A \times V$ ?
$\frac{1}{10} = A \times V$
$V = \frac{1}{10 A}$
Kinetic energy change into heat
$\frac{1}{2} {mV}^{2} = mS Δ T$
$Δ T = \frac{V^{2}}{2 S} = {(\frac{1}{10 A})}^{2} \times \frac{1}{2 S}$
$= \frac{1}{2} \times \frac{21}{100 \times 25 \times 10^{- 6}} \times \frac{1}{\frac{4.18}{10^{- 3}}}$
$Δ T = 1^{\circ} C$

WB JEE 2019

Thermal Properties of Matter

146454 Two black bodies $A$ and $B$ have equal surface areas and are maintained at temperatures $27^{\circ} C$ and $177^{\circ} C$ respectively. What will be the ratio of the thermal energy radiated per second by $A$ to that by $B$ ?

1

4 : 9

2

2 : 3

3

16 : 81

4

27 : 177

Explanation:

C According to the question-
Area of both bodies $A$ and $B$ are equal then
Temperature of body $A = 27^{\circ} C = 300 K$
Temperature of body $B = 177^{\circ} C = 450 K$
According to Stefan -Boltzman law, thermal energy radiated per second by a body-
$Q = σ {AT}^{4}$
Where, $A =$ Area
$T = Temperature$
$σ = Stefan-Boltzman's constant$
So, the ratio of thermal energy radiated per second by $A$ to that by $B$ is
$\frac{Q_{1}}{Q_{2}} = \frac{σ A}{σ A} {(\frac{T_{1}}{T_{2}})}^{4}$
Putting these value, we get-
$\frac{Q_{1}}{Q_{2}} = {(\frac{300}{450})}^{4}$
$\frac{Q_{1}}{Q_{2}} = \frac{16}{81}$
Ratio of thermal energy radiated per second
$Q_{1} : Q_{2} = 16 : 81$

WB JEE 2019

Thermal Properties of Matter

146456 A wire has resistance of $3.1 Ω$ at $30^{\circ} C$ and 4.5 $Ω$ at $100^{\circ} C$ . The temperature coefficient of resistance of the wire is

1

0.0012^{\circ} C^{- 1}

2

0.0024^{\circ} C^{- 1}

3

0.0032^{\circ} C^{- 1}

4

0.0064^{\circ} C^{- 1}

Explanation:

D Given that,
$R_{1} = 3.1 Ω T_{1} = 30^{\circ} C$
$R_{2} = 4.5 Ω T_{2} = 100^{\circ} C$
Resistance of wire at any temperature-
$R_{2} = R_{1} [1 + α (T_{2} - T_{1})]$
$4.5 = 3.1 [1 + α (100 - 30)]$
$4.5 = 3.1 [1 + 70 α]$
$\frac{4.5}{3.1} = 1 + 70 α$
$1.4516 = 1 + 70 α$
$0.4515 = 70 α$
$α = \frac{0.4515}{70}$
$α = {0.0064}^{\circ} C^{- 1}$

TS EAMCET (Engg.)-2017

Thermal Properties of Matter

146457 A sphere at $600 K$ is losing heat due to radiation. At this temperature its rate of cooling is $R$ . The rate of cooling of this sphere at $400 K$ is (temperature of surroundings is 200 K)

1

\frac{3}{16} R

2

\frac{8}{27} R

3

\frac{16}{3} R

4

7 R

Explanation:

A According to Stefan - Boltzman Law-
$E = σ T^{4}$
$R = \frac{dT}{dt} \propto (T^{4} - T_{o}^{4})$
$\frac{R_{1}}{R_{2}} = \frac{T_{1}^{4} - T_{o}^{4}}{T_{2}^{4} - T_{o}^{4}}$
$\frac{R_{1}}{R_{2}} = \frac{[(600)^{4} - (200)^{4}]}{[(400)^{4} - (200)^{4}]}$
$\frac{R_{1}}{R_{2}} = \frac{100^{4} [(6)^{4} - (2)^{4}]}{100^{4} [(4)^{4} - (2)^{4}]}$
$\frac{R_{1}}{R_{2}} = \frac{1280}{240}$
$\frac{R_{1}}{R_{2}} = \frac{16}{3}$
$R_{2} = \frac{3}{16} R_{1} or \frac{3}{16} R$

TS EAMCET(Medical)-2015

Thermal Properties of Matter

146458 Two identical shaped metallic spheres $A$ and $B$ made up of same material of mass ' $m$ ' and ' $4 m$ ' are heated to attain a temperature $T_{1}$ and then they are placed in a container maintained at temperature $T_{2} (T_{2} < T_{1})$ . The spheres are thermally insulated from each other. If $R$ is the rate of change of temperature, then the ratio $R_{A} & R_{B}$ is

1

\frac{1}{4}

2

{(\frac{1}{4})}^{\frac{1}{3}}

3

{(\frac{1}{4})}^{\frac{2}{3}}

4

(4)^{\frac{1}{3}}

Explanation:

D Given that, $M_{1} = m, M_{2} = 4 m$
Heat loss as per Stephan's boltzman law
$E = e σ {AT}^{4}$
$E = e σ T^{4} (4 π r^{2})$
We know that,
Volume of sphere $(V) = 4 / 3 π R^{3}$
$V = \frac{mass}{Density}$
$\frac{4}{3} π r^{3} = \frac{m}{ρ}$
$r = K (m)^{1 / 3}$
Therefore, $\frac{r_{1}}{r_{2}} = {(\frac{m_{1}}{m_{2}})}^{\frac{1}{3}}$
$\frac{E_{1}}{E_{2}} = {(\frac{r_{1}}{r_{2}})}^{2} = {(\frac{m_{1}}{m_{2}})}^{2 / 3}$
Rate of Cooling $(R) = \frac{E}{mS} = \frac{AE}{m}$
Hence $\frac{R_{A}}{R_{B}} = {(\frac{m_{2}}{m_{1}})}^{\frac{1}{3}} = {(\frac{4}{1})}^{\frac{1}{3}}$

Shift-II]

Thermal Properties of Matter

146453 A horizontal fire hose with a nozzle of crosssectional area $\frac{5}{\sqrt{21}} \times 10^{- 3} m^{2}$ delivers a cubic meter of water in $10 s$ . What will be the maximum possible increase in the temperature of water while it hits a rigid wall (neglecting the effect of gravity)?

1

1^{\circ} C

2

{0.1}^{\circ} C

3

10^{\circ} C

4

{0.01}^{\circ} C

Explanation:

A Given that, $A = \frac{5}{\sqrt{21}} \times 10^{- 3} m^{2}$

Rate of heat transfer, $\frac{dQ}{dt} = A \times V$ ?
$\frac{1}{10} = A \times V$
$V = \frac{1}{10 A}$
Kinetic energy change into heat
$\frac{1}{2} {mV}^{2} = mS Δ T$
$Δ T = \frac{V^{2}}{2 S} = {(\frac{1}{10 A})}^{2} \times \frac{1}{2 S}$
$= \frac{1}{2} \times \frac{21}{100 \times 25 \times 10^{- 6}} \times \frac{1}{\frac{4.18}{10^{- 3}}}$
$Δ T = 1^{\circ} C$

WB JEE 2019

Thermal Properties of Matter

146454 Two black bodies $A$ and $B$ have equal surface areas and are maintained at temperatures $27^{\circ} C$ and $177^{\circ} C$ respectively. What will be the ratio of the thermal energy radiated per second by $A$ to that by $B$ ?

1

4 : 9

2

2 : 3

3

16 : 81

4

27 : 177

Explanation:

C According to the question-
Area of both bodies $A$ and $B$ are equal then
Temperature of body $A = 27^{\circ} C = 300 K$
Temperature of body $B = 177^{\circ} C = 450 K$
According to Stefan -Boltzman law, thermal energy radiated per second by a body-
$Q = σ {AT}^{4}$
Where, $A =$ Area
$T = Temperature$
$σ = Stefan-Boltzman's constant$
So, the ratio of thermal energy radiated per second by $A$ to that by $B$ is
$\frac{Q_{1}}{Q_{2}} = \frac{σ A}{σ A} {(\frac{T_{1}}{T_{2}})}^{4}$
Putting these value, we get-
$\frac{Q_{1}}{Q_{2}} = {(\frac{300}{450})}^{4}$
$\frac{Q_{1}}{Q_{2}} = \frac{16}{81}$
Ratio of thermal energy radiated per second
$Q_{1} : Q_{2} = 16 : 81$

WB JEE 2019

Thermal Properties of Matter

146456 A wire has resistance of $3.1 Ω$ at $30^{\circ} C$ and 4.5 $Ω$ at $100^{\circ} C$ . The temperature coefficient of resistance of the wire is

1

0.0012^{\circ} C^{- 1}

2

0.0024^{\circ} C^{- 1}

3

0.0032^{\circ} C^{- 1}

4

0.0064^{\circ} C^{- 1}

Explanation:

D Given that,
$R_{1} = 3.1 Ω T_{1} = 30^{\circ} C$
$R_{2} = 4.5 Ω T_{2} = 100^{\circ} C$
Resistance of wire at any temperature-
$R_{2} = R_{1} [1 + α (T_{2} - T_{1})]$
$4.5 = 3.1 [1 + α (100 - 30)]$
$4.5 = 3.1 [1 + 70 α]$
$\frac{4.5}{3.1} = 1 + 70 α$
$1.4516 = 1 + 70 α$
$0.4515 = 70 α$
$α = \frac{0.4515}{70}$
$α = {0.0064}^{\circ} C^{- 1}$

TS EAMCET (Engg.)-2017

Thermal Properties of Matter

146457 A sphere at $600 K$ is losing heat due to radiation. At this temperature its rate of cooling is $R$ . The rate of cooling of this sphere at $400 K$ is (temperature of surroundings is 200 K)

1

\frac{3}{16} R

2

\frac{8}{27} R

3

\frac{16}{3} R

4

7 R

Explanation:

A According to Stefan - Boltzman Law-
$E = σ T^{4}$
$R = \frac{dT}{dt} \propto (T^{4} - T_{o}^{4})$
$\frac{R_{1}}{R_{2}} = \frac{T_{1}^{4} - T_{o}^{4}}{T_{2}^{4} - T_{o}^{4}}$
$\frac{R_{1}}{R_{2}} = \frac{[(600)^{4} - (200)^{4}]}{[(400)^{4} - (200)^{4}]}$
$\frac{R_{1}}{R_{2}} = \frac{100^{4} [(6)^{4} - (2)^{4}]}{100^{4} [(4)^{4} - (2)^{4}]}$
$\frac{R_{1}}{R_{2}} = \frac{1280}{240}$
$\frac{R_{1}}{R_{2}} = \frac{16}{3}$
$R_{2} = \frac{3}{16} R_{1} or \frac{3}{16} R$

TS EAMCET(Medical)-2015

Thermal Properties of Matter

146458 Two identical shaped metallic spheres $A$ and $B$ made up of same material of mass ' $m$ ' and ' $4 m$ ' are heated to attain a temperature $T_{1}$ and then they are placed in a container maintained at temperature $T_{2} (T_{2} < T_{1})$ . The spheres are thermally insulated from each other. If $R$ is the rate of change of temperature, then the ratio $R_{A} & R_{B}$ is

1

\frac{1}{4}

2

{(\frac{1}{4})}^{\frac{1}{3}}

3

{(\frac{1}{4})}^{\frac{2}{3}}

4

(4)^{\frac{1}{3}}

Explanation:

D Given that, $M_{1} = m, M_{2} = 4 m$
Heat loss as per Stephan's boltzman law
$E = e σ {AT}^{4}$
$E = e σ T^{4} (4 π r^{2})$
We know that,
Volume of sphere $(V) = 4 / 3 π R^{3}$
$V = \frac{mass}{Density}$
$\frac{4}{3} π r^{3} = \frac{m}{ρ}$
$r = K (m)^{1 / 3}$
Therefore, $\frac{r_{1}}{r_{2}} = {(\frac{m_{1}}{m_{2}})}^{\frac{1}{3}}$
$\frac{E_{1}}{E_{2}} = {(\frac{r_{1}}{r_{2}})}^{2} = {(\frac{m_{1}}{m_{2}})}^{2 / 3}$
Rate of Cooling $(R) = \frac{E}{mS} = \frac{AE}{m}$
Hence $\frac{R_{A}}{R_{B}} = {(\frac{m_{2}}{m_{1}})}^{\frac{1}{3}} = {(\frac{4}{1})}^{\frac{1}{3}}$

Shift-II]

Thermal Properties of Matter

146453 A horizontal fire hose with a nozzle of crosssectional area $\frac{5}{\sqrt{21}} \times 10^{- 3} m^{2}$ delivers a cubic meter of water in $10 s$ . What will be the maximum possible increase in the temperature of water while it hits a rigid wall (neglecting the effect of gravity)?

1

1^{\circ} C

2

{0.1}^{\circ} C

3

10^{\circ} C

4

{0.01}^{\circ} C

Explanation:

A Given that, $A = \frac{5}{\sqrt{21}} \times 10^{- 3} m^{2}$

Rate of heat transfer, $\frac{dQ}{dt} = A \times V$ ?
$\frac{1}{10} = A \times V$
$V = \frac{1}{10 A}$
Kinetic energy change into heat
$\frac{1}{2} {mV}^{2} = mS Δ T$
$Δ T = \frac{V^{2}}{2 S} = {(\frac{1}{10 A})}^{2} \times \frac{1}{2 S}$
$= \frac{1}{2} \times \frac{21}{100 \times 25 \times 10^{- 6}} \times \frac{1}{\frac{4.18}{10^{- 3}}}$
$Δ T = 1^{\circ} C$

WB JEE 2019

Thermal Properties of Matter

146454 Two black bodies $A$ and $B$ have equal surface areas and are maintained at temperatures $27^{\circ} C$ and $177^{\circ} C$ respectively. What will be the ratio of the thermal energy radiated per second by $A$ to that by $B$ ?

1

4 : 9

2

2 : 3

3

16 : 81

4

27 : 177

Explanation:

C According to the question-
Area of both bodies $A$ and $B$ are equal then
Temperature of body $A = 27^{\circ} C = 300 K$
Temperature of body $B = 177^{\circ} C = 450 K$
According to Stefan -Boltzman law, thermal energy radiated per second by a body-
$Q = σ {AT}^{4}$
Where, $A =$ Area
$T = Temperature$
$σ = Stefan-Boltzman's constant$
So, the ratio of thermal energy radiated per second by $A$ to that by $B$ is
$\frac{Q_{1}}{Q_{2}} = \frac{σ A}{σ A} {(\frac{T_{1}}{T_{2}})}^{4}$
Putting these value, we get-
$\frac{Q_{1}}{Q_{2}} = {(\frac{300}{450})}^{4}$
$\frac{Q_{1}}{Q_{2}} = \frac{16}{81}$
Ratio of thermal energy radiated per second
$Q_{1} : Q_{2} = 16 : 81$

WB JEE 2019

Thermal Properties of Matter

146456 A wire has resistance of $3.1 Ω$ at $30^{\circ} C$ and 4.5 $Ω$ at $100^{\circ} C$ . The temperature coefficient of resistance of the wire is

1

0.0012^{\circ} C^{- 1}

2

0.0024^{\circ} C^{- 1}

3

0.0032^{\circ} C^{- 1}

4

0.0064^{\circ} C^{- 1}

Explanation:

D Given that,
$R_{1} = 3.1 Ω T_{1} = 30^{\circ} C$
$R_{2} = 4.5 Ω T_{2} = 100^{\circ} C$
Resistance of wire at any temperature-
$R_{2} = R_{1} [1 + α (T_{2} - T_{1})]$
$4.5 = 3.1 [1 + α (100 - 30)]$
$4.5 = 3.1 [1 + 70 α]$
$\frac{4.5}{3.1} = 1 + 70 α$
$1.4516 = 1 + 70 α$
$0.4515 = 70 α$
$α = \frac{0.4515}{70}$
$α = {0.0064}^{\circ} C^{- 1}$

TS EAMCET (Engg.)-2017

Thermal Properties of Matter

146457 A sphere at $600 K$ is losing heat due to radiation. At this temperature its rate of cooling is $R$ . The rate of cooling of this sphere at $400 K$ is (temperature of surroundings is 200 K)

1

\frac{3}{16} R

2

\frac{8}{27} R

3

\frac{16}{3} R

4

7 R

Explanation:

A According to Stefan - Boltzman Law-
$E = σ T^{4}$
$R = \frac{dT}{dt} \propto (T^{4} - T_{o}^{4})$
$\frac{R_{1}}{R_{2}} = \frac{T_{1}^{4} - T_{o}^{4}}{T_{2}^{4} - T_{o}^{4}}$
$\frac{R_{1}}{R_{2}} = \frac{[(600)^{4} - (200)^{4}]}{[(400)^{4} - (200)^{4}]}$
$\frac{R_{1}}{R_{2}} = \frac{100^{4} [(6)^{4} - (2)^{4}]}{100^{4} [(4)^{4} - (2)^{4}]}$
$\frac{R_{1}}{R_{2}} = \frac{1280}{240}$
$\frac{R_{1}}{R_{2}} = \frac{16}{3}$
$R_{2} = \frac{3}{16} R_{1} or \frac{3}{16} R$

TS EAMCET(Medical)-2015

Thermal Properties of Matter

146458 Two identical shaped metallic spheres $A$ and $B$ made up of same material of mass ' $m$ ' and ' $4 m$ ' are heated to attain a temperature $T_{1}$ and then they are placed in a container maintained at temperature $T_{2} (T_{2} < T_{1})$ . The spheres are thermally insulated from each other. If $R$ is the rate of change of temperature, then the ratio $R_{A} & R_{B}$ is

1

\frac{1}{4}

2

{(\frac{1}{4})}^{\frac{1}{3}}

3

{(\frac{1}{4})}^{\frac{2}{3}}

4

(4)^{\frac{1}{3}}

Explanation:

D Given that, $M_{1} = m, M_{2} = 4 m$
Heat loss as per Stephan's boltzman law
$E = e σ {AT}^{4}$
$E = e σ T^{4} (4 π r^{2})$
We know that,
Volume of sphere $(V) = 4 / 3 π R^{3}$
$V = \frac{mass}{Density}$
$\frac{4}{3} π r^{3} = \frac{m}{ρ}$
$r = K (m)^{1 / 3}$
Therefore, $\frac{r_{1}}{r_{2}} = {(\frac{m_{1}}{m_{2}})}^{\frac{1}{3}}$
$\frac{E_{1}}{E_{2}} = {(\frac{r_{1}}{r_{2}})}^{2} = {(\frac{m_{1}}{m_{2}})}^{2 / 3}$
Rate of Cooling $(R) = \frac{E}{mS} = \frac{AE}{m}$
Hence $\frac{R_{A}}{R_{B}} = {(\frac{m_{2}}{m_{1}})}^{\frac{1}{3}} = {(\frac{4}{1})}^{\frac{1}{3}}$

Shift-II]