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Heat Transfer

149323 If a metallic sphere gets cooled from $62^{\circ} C$ to $50^{\circ} C$ in $10 \min$ and in the next $10 \min$ gets cooled to $42^{\circ} C$ , then the temperature of the surroundings is

1

30^{\circ} C

2

36^{\circ} C

3

26^{\circ} C

4

20^{\circ} C

Explanation:

C According to Newton's law cooling,
$\frac{θ_{1} - θ_{2}}{t} = K (\frac{θ_{1} + θ_{2}}{2} - θ_{o})$
Case-I
$θ_{1} = 62^{\circ} C, θ_{2} = 50^{\circ} C, t = 10 minute,$
$\frac{62 - 50}{10} = K [\frac{62 + 50}{2} - θ_{o}]$
$\frac{12}{10} = K [56 - θ_{o}]$
Case-II
$θ_{1} = 50^{\circ} C, θ_{2} = 42^{\circ} C, t = 10 minute,$
$\frac{(50 - 42)}{10} = K [\frac{50 + 42}{2} - θ_{o}]$
$\frac{8}{10} = K [46 - θ_{o}]$
On dividing equation (i) and (ii), we get
$\frac{12}{10} \times \frac{10}{8} = \frac{K (56 - θ_{o})}{K (46 - θ_{o})}$
$\frac{3}{2} = \frac{[56 - θ_{o}]}{[46 - θ_{o}]}$
$138 - 3 θ_{o} = 112 - 2 θ_{o}$
$138 - 112 = - 2 θ_{o} + 3 θ_{o}$
$θ_{o} = 26^{\circ} C$

UPSEE - 2010

Heat Transfer

149324 A metal rod of length $2 m$ has cross - sectional areas $2 A$ and $A$ as shown in figure. The two ends are maintained at temperatures $100^{\circ} C$ and $70^{\circ} C$ . The temperature of middle point $C$ is

1

80^{\circ} C

2

85^{\circ} C

3

90^{\circ} C

4

95^{\circ} C

Explanation:

C Here, $K_{1} = K_{2}, l_{1} = l_{2} = 1 m, A_{1} = 2 A_{2} = 2 A$ Also, $T_{1} = 100^{\circ} C$ and $T_{2} = 70^{\circ} C$
Using, $\frac{Δ Q}{Δ t} = \frac{KA (Δ T)}{l}$
Let temperature at $C$ be $T_{C}$ . then
${(\frac{Δ Q}{Δ t})}_{BC} = {(\frac{Δ Q}{Δ t})}_{CA}$
$∴ \frac{K.2A. (100 - T_{C})}{1} = \frac{KA (T_{C} - 70)}{1}$
$200 - 2 T_{C} = T_{C} - 70$
$3 T_{C} = 270$
$T_{C} = 90^{\circ} C$

UPSEE - 2010

Heat Transfer

149325 Two rods of the same length and diameter having thermal conductivities $K_{1}$ and $K_{2}$ are joined in parallel. The equivalent thermal conductivity of the combination is

1

\frac{K_{1} K_{2}}{K_{1} + K_{2}}

2

K_{1} + K_{2}

3

\frac{K_{1} + K_{2}}{2}

4

\sqrt{K_{1} K_{2}}

Explanation:

C
Thermal resistance, $R_{1} = \frac{L}{K_{1} A}, R_{2} = \frac{L}{K_{2} A}$
In series, equivalent thermal resistance conductivity-
$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$\frac{1}{\frac{L}{2 {AK}_{eq}}} = \frac{1}{(\frac{L}{{AK}_{1}})} + \frac{1}{(\frac{L}{{AK}_{2}})}$
$\frac{2 A K_{eq}}{L} = \frac{A K_{1}}{L} = \frac{A K_{2}}{L}$
$2 K_{eq} = K_{1} + K_{2}$
$K_{eq} = \frac{K_{1} + K_{2}}{2}$

UPSEE - 2009

Heat Transfer

149326 If $K$ denotes coefficient of thermal conductivity, $d$ the density and $c$ the specific heat, the unit of $X$ , where $X = K / d c$ will be

1

cm \sec^{- 1}

2

{cm}^{2} \sec^{- 2}

3

cm \sec

4

{cm}^{2} \sec^{- 1}

Explanation:

D Given,
The units of given quantities in SI are:
$[K] = J (mKs)^{- 1}$
$[d] = m / v = kg m^{- 3}$
$[c] = J (kgK)^{- 1}$
$[X] = \frac{[K]}{[d] [c]} = \frac{J (mks)^{- 1}}{kg (m)^{- 3} (J (kgk))^{- 1}}$
$= m^{2} s^{- 1} or cm s^{- 1} .$

UPSEE - 2008

Heat Transfer

149323 If a metallic sphere gets cooled from $62^{\circ} C$ to $50^{\circ} C$ in $10 \min$ and in the next $10 \min$ gets cooled to $42^{\circ} C$ , then the temperature of the surroundings is

1

30^{\circ} C

2

36^{\circ} C

3

26^{\circ} C

4

20^{\circ} C

Explanation:

C According to Newton's law cooling,
$\frac{θ_{1} - θ_{2}}{t} = K (\frac{θ_{1} + θ_{2}}{2} - θ_{o})$
Case-I
$θ_{1} = 62^{\circ} C, θ_{2} = 50^{\circ} C, t = 10 minute,$
$\frac{62 - 50}{10} = K [\frac{62 + 50}{2} - θ_{o}]$
$\frac{12}{10} = K [56 - θ_{o}]$
Case-II
$θ_{1} = 50^{\circ} C, θ_{2} = 42^{\circ} C, t = 10 minute,$
$\frac{(50 - 42)}{10} = K [\frac{50 + 42}{2} - θ_{o}]$
$\frac{8}{10} = K [46 - θ_{o}]$
On dividing equation (i) and (ii), we get
$\frac{12}{10} \times \frac{10}{8} = \frac{K (56 - θ_{o})}{K (46 - θ_{o})}$
$\frac{3}{2} = \frac{[56 - θ_{o}]}{[46 - θ_{o}]}$
$138 - 3 θ_{o} = 112 - 2 θ_{o}$
$138 - 112 = - 2 θ_{o} + 3 θ_{o}$
$θ_{o} = 26^{\circ} C$

UPSEE - 2010

Heat Transfer

149324 A metal rod of length $2 m$ has cross - sectional areas $2 A$ and $A$ as shown in figure. The two ends are maintained at temperatures $100^{\circ} C$ and $70^{\circ} C$ . The temperature of middle point $C$ is

1

80^{\circ} C

2

85^{\circ} C

3

90^{\circ} C

4

95^{\circ} C

Explanation:

C Here, $K_{1} = K_{2}, l_{1} = l_{2} = 1 m, A_{1} = 2 A_{2} = 2 A$ Also, $T_{1} = 100^{\circ} C$ and $T_{2} = 70^{\circ} C$
Using, $\frac{Δ Q}{Δ t} = \frac{KA (Δ T)}{l}$
Let temperature at $C$ be $T_{C}$ . then
${(\frac{Δ Q}{Δ t})}_{BC} = {(\frac{Δ Q}{Δ t})}_{CA}$
$∴ \frac{K.2A. (100 - T_{C})}{1} = \frac{KA (T_{C} - 70)}{1}$
$200 - 2 T_{C} = T_{C} - 70$
$3 T_{C} = 270$
$T_{C} = 90^{\circ} C$

UPSEE - 2010

Heat Transfer

149325 Two rods of the same length and diameter having thermal conductivities $K_{1}$ and $K_{2}$ are joined in parallel. The equivalent thermal conductivity of the combination is

1

\frac{K_{1} K_{2}}{K_{1} + K_{2}}

2

K_{1} + K_{2}

3

\frac{K_{1} + K_{2}}{2}

4

\sqrt{K_{1} K_{2}}

Explanation:

C
Thermal resistance, $R_{1} = \frac{L}{K_{1} A}, R_{2} = \frac{L}{K_{2} A}$
In series, equivalent thermal resistance conductivity-
$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$\frac{1}{\frac{L}{2 {AK}_{eq}}} = \frac{1}{(\frac{L}{{AK}_{1}})} + \frac{1}{(\frac{L}{{AK}_{2}})}$
$\frac{2 A K_{eq}}{L} = \frac{A K_{1}}{L} = \frac{A K_{2}}{L}$
$2 K_{eq} = K_{1} + K_{2}$
$K_{eq} = \frac{K_{1} + K_{2}}{2}$

UPSEE - 2009

Heat Transfer

149326 If $K$ denotes coefficient of thermal conductivity, $d$ the density and $c$ the specific heat, the unit of $X$ , where $X = K / d c$ will be

1

cm \sec^{- 1}

2

{cm}^{2} \sec^{- 2}

3

cm \sec

4

{cm}^{2} \sec^{- 1}

Explanation:

D Given,
The units of given quantities in SI are:
$[K] = J (mKs)^{- 1}$
$[d] = m / v = kg m^{- 3}$
$[c] = J (kgK)^{- 1}$
$[X] = \frac{[K]}{[d] [c]} = \frac{J (mks)^{- 1}}{kg (m)^{- 3} (J (kgk))^{- 1}}$
$= m^{2} s^{- 1} or cm s^{- 1} .$

UPSEE - 2008

Heat Transfer

149323 If a metallic sphere gets cooled from $62^{\circ} C$ to $50^{\circ} C$ in $10 \min$ and in the next $10 \min$ gets cooled to $42^{\circ} C$ , then the temperature of the surroundings is

1

30^{\circ} C

2

36^{\circ} C

3

26^{\circ} C

4

20^{\circ} C

Explanation:

C According to Newton's law cooling,
$\frac{θ_{1} - θ_{2}}{t} = K (\frac{θ_{1} + θ_{2}}{2} - θ_{o})$
Case-I
$θ_{1} = 62^{\circ} C, θ_{2} = 50^{\circ} C, t = 10 minute,$
$\frac{62 - 50}{10} = K [\frac{62 + 50}{2} - θ_{o}]$
$\frac{12}{10} = K [56 - θ_{o}]$
Case-II
$θ_{1} = 50^{\circ} C, θ_{2} = 42^{\circ} C, t = 10 minute,$
$\frac{(50 - 42)}{10} = K [\frac{50 + 42}{2} - θ_{o}]$
$\frac{8}{10} = K [46 - θ_{o}]$
On dividing equation (i) and (ii), we get
$\frac{12}{10} \times \frac{10}{8} = \frac{K (56 - θ_{o})}{K (46 - θ_{o})}$
$\frac{3}{2} = \frac{[56 - θ_{o}]}{[46 - θ_{o}]}$
$138 - 3 θ_{o} = 112 - 2 θ_{o}$
$138 - 112 = - 2 θ_{o} + 3 θ_{o}$
$θ_{o} = 26^{\circ} C$

UPSEE - 2010

Heat Transfer

149324 A metal rod of length $2 m$ has cross - sectional areas $2 A$ and $A$ as shown in figure. The two ends are maintained at temperatures $100^{\circ} C$ and $70^{\circ} C$ . The temperature of middle point $C$ is

1

80^{\circ} C

2

85^{\circ} C

3

90^{\circ} C

4

95^{\circ} C

Explanation:

C Here, $K_{1} = K_{2}, l_{1} = l_{2} = 1 m, A_{1} = 2 A_{2} = 2 A$ Also, $T_{1} = 100^{\circ} C$ and $T_{2} = 70^{\circ} C$
Using, $\frac{Δ Q}{Δ t} = \frac{KA (Δ T)}{l}$
Let temperature at $C$ be $T_{C}$ . then
${(\frac{Δ Q}{Δ t})}_{BC} = {(\frac{Δ Q}{Δ t})}_{CA}$
$∴ \frac{K.2A. (100 - T_{C})}{1} = \frac{KA (T_{C} - 70)}{1}$
$200 - 2 T_{C} = T_{C} - 70$
$3 T_{C} = 270$
$T_{C} = 90^{\circ} C$

UPSEE - 2010

Heat Transfer

149325 Two rods of the same length and diameter having thermal conductivities $K_{1}$ and $K_{2}$ are joined in parallel. The equivalent thermal conductivity of the combination is

1

\frac{K_{1} K_{2}}{K_{1} + K_{2}}

2

K_{1} + K_{2}

3

\frac{K_{1} + K_{2}}{2}

4

\sqrt{K_{1} K_{2}}

Explanation:

C
Thermal resistance, $R_{1} = \frac{L}{K_{1} A}, R_{2} = \frac{L}{K_{2} A}$
In series, equivalent thermal resistance conductivity-
$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$\frac{1}{\frac{L}{2 {AK}_{eq}}} = \frac{1}{(\frac{L}{{AK}_{1}})} + \frac{1}{(\frac{L}{{AK}_{2}})}$
$\frac{2 A K_{eq}}{L} = \frac{A K_{1}}{L} = \frac{A K_{2}}{L}$
$2 K_{eq} = K_{1} + K_{2}$
$K_{eq} = \frac{K_{1} + K_{2}}{2}$

UPSEE - 2009

Heat Transfer

149326 If $K$ denotes coefficient of thermal conductivity, $d$ the density and $c$ the specific heat, the unit of $X$ , where $X = K / d c$ will be

1

cm \sec^{- 1}

2

{cm}^{2} \sec^{- 2}

3

cm \sec

4

{cm}^{2} \sec^{- 1}

Explanation:

D Given,
The units of given quantities in SI are:
$[K] = J (mKs)^{- 1}$
$[d] = m / v = kg m^{- 3}$
$[c] = J (kgK)^{- 1}$
$[X] = \frac{[K]}{[d] [c]} = \frac{J (mks)^{- 1}}{kg (m)^{- 3} (J (kgk))^{- 1}}$
$= m^{2} s^{- 1} or cm s^{- 1} .$

UPSEE - 2008

Heat Transfer

149323 If a metallic sphere gets cooled from $62^{\circ} C$ to $50^{\circ} C$ in $10 \min$ and in the next $10 \min$ gets cooled to $42^{\circ} C$ , then the temperature of the surroundings is

1

30^{\circ} C

2

36^{\circ} C

3

26^{\circ} C

4

20^{\circ} C

Explanation:

C According to Newton's law cooling,
$\frac{θ_{1} - θ_{2}}{t} = K (\frac{θ_{1} + θ_{2}}{2} - θ_{o})$
Case-I
$θ_{1} = 62^{\circ} C, θ_{2} = 50^{\circ} C, t = 10 minute,$
$\frac{62 - 50}{10} = K [\frac{62 + 50}{2} - θ_{o}]$
$\frac{12}{10} = K [56 - θ_{o}]$
Case-II
$θ_{1} = 50^{\circ} C, θ_{2} = 42^{\circ} C, t = 10 minute,$
$\frac{(50 - 42)}{10} = K [\frac{50 + 42}{2} - θ_{o}]$
$\frac{8}{10} = K [46 - θ_{o}]$
On dividing equation (i) and (ii), we get
$\frac{12}{10} \times \frac{10}{8} = \frac{K (56 - θ_{o})}{K (46 - θ_{o})}$
$\frac{3}{2} = \frac{[56 - θ_{o}]}{[46 - θ_{o}]}$
$138 - 3 θ_{o} = 112 - 2 θ_{o}$
$138 - 112 = - 2 θ_{o} + 3 θ_{o}$
$θ_{o} = 26^{\circ} C$

UPSEE - 2010

Heat Transfer

149324 A metal rod of length $2 m$ has cross - sectional areas $2 A$ and $A$ as shown in figure. The two ends are maintained at temperatures $100^{\circ} C$ and $70^{\circ} C$ . The temperature of middle point $C$ is

1

80^{\circ} C

2

85^{\circ} C

3

90^{\circ} C

4

95^{\circ} C

Explanation:

C Here, $K_{1} = K_{2}, l_{1} = l_{2} = 1 m, A_{1} = 2 A_{2} = 2 A$ Also, $T_{1} = 100^{\circ} C$ and $T_{2} = 70^{\circ} C$
Using, $\frac{Δ Q}{Δ t} = \frac{KA (Δ T)}{l}$
Let temperature at $C$ be $T_{C}$ . then
${(\frac{Δ Q}{Δ t})}_{BC} = {(\frac{Δ Q}{Δ t})}_{CA}$
$∴ \frac{K.2A. (100 - T_{C})}{1} = \frac{KA (T_{C} - 70)}{1}$
$200 - 2 T_{C} = T_{C} - 70$
$3 T_{C} = 270$
$T_{C} = 90^{\circ} C$

UPSEE - 2010

Heat Transfer

149325 Two rods of the same length and diameter having thermal conductivities $K_{1}$ and $K_{2}$ are joined in parallel. The equivalent thermal conductivity of the combination is

1

\frac{K_{1} K_{2}}{K_{1} + K_{2}}

2

K_{1} + K_{2}

3

\frac{K_{1} + K_{2}}{2}

4

\sqrt{K_{1} K_{2}}

Explanation:

C
Thermal resistance, $R_{1} = \frac{L}{K_{1} A}, R_{2} = \frac{L}{K_{2} A}$
In series, equivalent thermal resistance conductivity-
$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$\frac{1}{\frac{L}{2 {AK}_{eq}}} = \frac{1}{(\frac{L}{{AK}_{1}})} + \frac{1}{(\frac{L}{{AK}_{2}})}$
$\frac{2 A K_{eq}}{L} = \frac{A K_{1}}{L} = \frac{A K_{2}}{L}$
$2 K_{eq} = K_{1} + K_{2}$
$K_{eq} = \frac{K_{1} + K_{2}}{2}$

UPSEE - 2009

Heat Transfer

149326 If $K$ denotes coefficient of thermal conductivity, $d$ the density and $c$ the specific heat, the unit of $X$ , where $X = K / d c$ will be

1

cm \sec^{- 1}

2

{cm}^{2} \sec^{- 2}

3

cm \sec

4

{cm}^{2} \sec^{- 1}

Explanation:

D Given,
The units of given quantities in SI are:
$[K] = J (mKs)^{- 1}$
$[d] = m / v = kg m^{- 3}$
$[c] = J (kgK)^{- 1}$
$[X] = \frac{[K]}{[d] [c]} = \frac{J (mks)^{- 1}}{kg (m)^{- 3} (J (kgk))^{- 1}}$
$= m^{2} s^{- 1} or cm s^{- 1} .$

UPSEE - 2008

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