149323
If a metallic sphere gets cooled from $62^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in $10 \mathrm{~min}$ and in the next $10 \mathrm{~min}$ gets cooled to $42^{\circ} \mathrm{C}$, then the temperature of the surroundings is
1 $30^{\circ} \mathrm{C}$
2 $36^{\circ} \mathrm{C}$
3 $26^{\circ} \mathrm{C}$
4 $20^{\circ} \mathrm{C}$
Explanation:
C According to Newton's law cooling, $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=\mathrm{K}\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta_{\mathrm{o}}\right)$ Case-I $\theta_{1}=62^{\circ} \mathrm{C}, \quad \theta_{2}=50^{\circ} \mathrm{C}, \quad \mathrm{t}=10 \text { minute, }$ $\frac{62-50}{10}=\mathrm{K}\left[\frac{62+50}{2}-\theta_{\mathrm{o}}\right]$ $\frac{12}{10}=\mathrm{K}\left[56-\theta_{\mathrm{o}}\right]$ Case-II $\theta_{1}=50^{\circ} \mathrm{C}, \quad \theta_{2}=42^{\circ} \mathrm{C}, \quad \mathrm{t}=10 \text { minute, }$ $\frac{(50-42)}{10}=\mathrm{K}\left[\frac{50+42}{2}-\theta_{\mathrm{o}}\right]$ $\frac{8}{10}=\mathrm{K}\left[46-\theta_{\mathrm{o}}\right]$ On dividing equation (i) and (ii), we get $\frac{12}{10} \times \frac{10}{8}=\frac{K\left(56-\theta_{\mathrm{o}}\right)}{\mathrm{K}\left(46-\theta_{\mathrm{o}}\right)}$ $\frac{3}{2}=\frac{\left[56-\theta_{\mathrm{o}}\right]}{\left[46-\theta_{\mathrm{o}}\right]}$ $138-3 \theta_{\mathrm{o}}=112-2 \theta_{\mathrm{o}}$ $138-112=-2 \theta_{\mathrm{o}}+3 \theta_{\mathrm{o}}$ $\theta_{\mathrm{o}}=26^{\circ} \mathrm{C}$
UPSEE - 2010
Heat Transfer
149324
A metal rod of length $2 \mathrm{~m}$ has cross - sectional areas $2 \mathrm{~A}$ and $\mathrm{A}$ as shown in figure. The two ends are maintained at temperatures $100^{\circ} \mathrm{C}$ and $70^{\circ} \mathrm{C}$. The temperature of middle point $\mathrm{C}$ is
1 $80^{\circ} \mathrm{C}$
2 $85^{\circ} \mathrm{C}$
3 $90^{\circ} \mathrm{C}$
4 $95^{\circ} \mathrm{C}$
Explanation:
C Here, $\mathrm{K}_{1}=\mathrm{K}_{2}, l_{1}=l_{2}=1 \mathrm{~m}, \mathrm{~A}_{1}=2 \mathrm{~A}_{2}=2 \mathrm{~A}$ Also, $\mathrm{T}_{1}=100^{\circ} \mathrm{C}$ and $\mathrm{T}_{2}=70^{\circ} \mathrm{C}$ Using, $\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}=\frac{\mathrm{KA}(\Delta \mathrm{T})}{l}$ Let temperature at $\mathrm{C}$ be $\mathrm{T}_{\mathrm{C}}$. then $\left(\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}\right)_{\mathrm{BC}}=\left(\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}\right)_{\mathrm{CA}}$ $\therefore \quad \frac{\text { K.2A. }\left(100-\mathrm{T}_{\mathrm{C}}\right)}{1}=\frac{\mathrm{KA}\left(\mathrm{T}_{\mathrm{C}}-70\right)}{1}$ $200-2 \mathrm{~T}_{\mathrm{C}}=\mathrm{T}_{\mathrm{C}}-70$ $3 \mathrm{~T}_{\mathrm{C}}=270$ $\mathrm{T}_{\mathrm{C}}=90^{\circ} \mathrm{C}$
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
C Thermal resistance, $\mathrm{R}_{1}=\frac{\mathrm{L}}{\mathrm{K}_{1} \mathrm{~A}}, \mathrm{R}_{2}=\frac{\mathrm{L}}{\mathrm{K}_{2} \mathrm{~A}}$ In series, equivalent thermal resistance conductivity- $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $\frac{1}{\frac{\mathrm{L}}{2 \mathrm{AK}_{\mathrm{eq}}}}=\frac{1}{\left(\frac{\mathrm{L}}{\mathrm{AK}_{1}}\right)}+\frac{1}{\left(\frac{\mathrm{L}}{\mathrm{AK}_{2}}\right)}$ $\frac{2 A K_{\text {eq }}}{L}=\frac{A K_{1}}{L}=\frac{A K_{2}}{L}$ $2 K_{\text {eq }}=K_{1}+K_{2}$ $K_{\text {eq }}=\frac{K_{1}+K_{2}}{2}$
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 $\mathrm{cm} \mathrm{sec}^{-1}$
2 $\mathrm{cm}^{2} \mathrm{sec}^{-2}$
3 $\mathrm{cm} \mathrm{sec}$
4 $\mathrm{cm}^{2} \mathrm{sec}^{-1}$
Explanation:
D Given, The units of given quantities in SI are: ${[\mathrm{K}]=\mathrm{J}(\mathrm{mKs})^{-1}}$ ${[\mathrm{~d}]=\mathrm{m} / \mathrm{v}=\mathrm{kg} \mathrm{m}^{-3}}$ ${[\mathrm{c}]=\mathrm{J}(\mathrm{kgK})^{-1}}$ ${[\mathrm{X}]=\frac{[\mathrm{K}]}{[\mathrm{d}][\mathrm{c}]}=\frac{\mathrm{J}(\mathrm{mks})^{-1}}{\mathrm{~kg}(\mathrm{~m})^{-3}(\mathrm{~J}(\mathrm{kgk}))^{-1}}}$ $=\mathrm{m}^{2} \mathrm{~s}^{-1} \text { or cm } \mathrm{s}^{-1} .$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Heat Transfer
149323
If a metallic sphere gets cooled from $62^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in $10 \mathrm{~min}$ and in the next $10 \mathrm{~min}$ gets cooled to $42^{\circ} \mathrm{C}$, then the temperature of the surroundings is
1 $30^{\circ} \mathrm{C}$
2 $36^{\circ} \mathrm{C}$
3 $26^{\circ} \mathrm{C}$
4 $20^{\circ} \mathrm{C}$
Explanation:
C According to Newton's law cooling, $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=\mathrm{K}\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta_{\mathrm{o}}\right)$ Case-I $\theta_{1}=62^{\circ} \mathrm{C}, \quad \theta_{2}=50^{\circ} \mathrm{C}, \quad \mathrm{t}=10 \text { minute, }$ $\frac{62-50}{10}=\mathrm{K}\left[\frac{62+50}{2}-\theta_{\mathrm{o}}\right]$ $\frac{12}{10}=\mathrm{K}\left[56-\theta_{\mathrm{o}}\right]$ Case-II $\theta_{1}=50^{\circ} \mathrm{C}, \quad \theta_{2}=42^{\circ} \mathrm{C}, \quad \mathrm{t}=10 \text { minute, }$ $\frac{(50-42)}{10}=\mathrm{K}\left[\frac{50+42}{2}-\theta_{\mathrm{o}}\right]$ $\frac{8}{10}=\mathrm{K}\left[46-\theta_{\mathrm{o}}\right]$ On dividing equation (i) and (ii), we get $\frac{12}{10} \times \frac{10}{8}=\frac{K\left(56-\theta_{\mathrm{o}}\right)}{\mathrm{K}\left(46-\theta_{\mathrm{o}}\right)}$ $\frac{3}{2}=\frac{\left[56-\theta_{\mathrm{o}}\right]}{\left[46-\theta_{\mathrm{o}}\right]}$ $138-3 \theta_{\mathrm{o}}=112-2 \theta_{\mathrm{o}}$ $138-112=-2 \theta_{\mathrm{o}}+3 \theta_{\mathrm{o}}$ $\theta_{\mathrm{o}}=26^{\circ} \mathrm{C}$
UPSEE - 2010
Heat Transfer
149324
A metal rod of length $2 \mathrm{~m}$ has cross - sectional areas $2 \mathrm{~A}$ and $\mathrm{A}$ as shown in figure. The two ends are maintained at temperatures $100^{\circ} \mathrm{C}$ and $70^{\circ} \mathrm{C}$. The temperature of middle point $\mathrm{C}$ is
1 $80^{\circ} \mathrm{C}$
2 $85^{\circ} \mathrm{C}$
3 $90^{\circ} \mathrm{C}$
4 $95^{\circ} \mathrm{C}$
Explanation:
C Here, $\mathrm{K}_{1}=\mathrm{K}_{2}, l_{1}=l_{2}=1 \mathrm{~m}, \mathrm{~A}_{1}=2 \mathrm{~A}_{2}=2 \mathrm{~A}$ Also, $\mathrm{T}_{1}=100^{\circ} \mathrm{C}$ and $\mathrm{T}_{2}=70^{\circ} \mathrm{C}$ Using, $\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}=\frac{\mathrm{KA}(\Delta \mathrm{T})}{l}$ Let temperature at $\mathrm{C}$ be $\mathrm{T}_{\mathrm{C}}$. then $\left(\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}\right)_{\mathrm{BC}}=\left(\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}\right)_{\mathrm{CA}}$ $\therefore \quad \frac{\text { K.2A. }\left(100-\mathrm{T}_{\mathrm{C}}\right)}{1}=\frac{\mathrm{KA}\left(\mathrm{T}_{\mathrm{C}}-70\right)}{1}$ $200-2 \mathrm{~T}_{\mathrm{C}}=\mathrm{T}_{\mathrm{C}}-70$ $3 \mathrm{~T}_{\mathrm{C}}=270$ $\mathrm{T}_{\mathrm{C}}=90^{\circ} \mathrm{C}$
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
C Thermal resistance, $\mathrm{R}_{1}=\frac{\mathrm{L}}{\mathrm{K}_{1} \mathrm{~A}}, \mathrm{R}_{2}=\frac{\mathrm{L}}{\mathrm{K}_{2} \mathrm{~A}}$ In series, equivalent thermal resistance conductivity- $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $\frac{1}{\frac{\mathrm{L}}{2 \mathrm{AK}_{\mathrm{eq}}}}=\frac{1}{\left(\frac{\mathrm{L}}{\mathrm{AK}_{1}}\right)}+\frac{1}{\left(\frac{\mathrm{L}}{\mathrm{AK}_{2}}\right)}$ $\frac{2 A K_{\text {eq }}}{L}=\frac{A K_{1}}{L}=\frac{A K_{2}}{L}$ $2 K_{\text {eq }}=K_{1}+K_{2}$ $K_{\text {eq }}=\frac{K_{1}+K_{2}}{2}$
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 $\mathrm{cm} \mathrm{sec}^{-1}$
2 $\mathrm{cm}^{2} \mathrm{sec}^{-2}$
3 $\mathrm{cm} \mathrm{sec}$
4 $\mathrm{cm}^{2} \mathrm{sec}^{-1}$
Explanation:
D Given, The units of given quantities in SI are: ${[\mathrm{K}]=\mathrm{J}(\mathrm{mKs})^{-1}}$ ${[\mathrm{~d}]=\mathrm{m} / \mathrm{v}=\mathrm{kg} \mathrm{m}^{-3}}$ ${[\mathrm{c}]=\mathrm{J}(\mathrm{kgK})^{-1}}$ ${[\mathrm{X}]=\frac{[\mathrm{K}]}{[\mathrm{d}][\mathrm{c}]}=\frac{\mathrm{J}(\mathrm{mks})^{-1}}{\mathrm{~kg}(\mathrm{~m})^{-3}(\mathrm{~J}(\mathrm{kgk}))^{-1}}}$ $=\mathrm{m}^{2} \mathrm{~s}^{-1} \text { or cm } \mathrm{s}^{-1} .$
149323
If a metallic sphere gets cooled from $62^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in $10 \mathrm{~min}$ and in the next $10 \mathrm{~min}$ gets cooled to $42^{\circ} \mathrm{C}$, then the temperature of the surroundings is
1 $30^{\circ} \mathrm{C}$
2 $36^{\circ} \mathrm{C}$
3 $26^{\circ} \mathrm{C}$
4 $20^{\circ} \mathrm{C}$
Explanation:
C According to Newton's law cooling, $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=\mathrm{K}\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta_{\mathrm{o}}\right)$ Case-I $\theta_{1}=62^{\circ} \mathrm{C}, \quad \theta_{2}=50^{\circ} \mathrm{C}, \quad \mathrm{t}=10 \text { minute, }$ $\frac{62-50}{10}=\mathrm{K}\left[\frac{62+50}{2}-\theta_{\mathrm{o}}\right]$ $\frac{12}{10}=\mathrm{K}\left[56-\theta_{\mathrm{o}}\right]$ Case-II $\theta_{1}=50^{\circ} \mathrm{C}, \quad \theta_{2}=42^{\circ} \mathrm{C}, \quad \mathrm{t}=10 \text { minute, }$ $\frac{(50-42)}{10}=\mathrm{K}\left[\frac{50+42}{2}-\theta_{\mathrm{o}}\right]$ $\frac{8}{10}=\mathrm{K}\left[46-\theta_{\mathrm{o}}\right]$ On dividing equation (i) and (ii), we get $\frac{12}{10} \times \frac{10}{8}=\frac{K\left(56-\theta_{\mathrm{o}}\right)}{\mathrm{K}\left(46-\theta_{\mathrm{o}}\right)}$ $\frac{3}{2}=\frac{\left[56-\theta_{\mathrm{o}}\right]}{\left[46-\theta_{\mathrm{o}}\right]}$ $138-3 \theta_{\mathrm{o}}=112-2 \theta_{\mathrm{o}}$ $138-112=-2 \theta_{\mathrm{o}}+3 \theta_{\mathrm{o}}$ $\theta_{\mathrm{o}}=26^{\circ} \mathrm{C}$
UPSEE - 2010
Heat Transfer
149324
A metal rod of length $2 \mathrm{~m}$ has cross - sectional areas $2 \mathrm{~A}$ and $\mathrm{A}$ as shown in figure. The two ends are maintained at temperatures $100^{\circ} \mathrm{C}$ and $70^{\circ} \mathrm{C}$. The temperature of middle point $\mathrm{C}$ is
1 $80^{\circ} \mathrm{C}$
2 $85^{\circ} \mathrm{C}$
3 $90^{\circ} \mathrm{C}$
4 $95^{\circ} \mathrm{C}$
Explanation:
C Here, $\mathrm{K}_{1}=\mathrm{K}_{2}, l_{1}=l_{2}=1 \mathrm{~m}, \mathrm{~A}_{1}=2 \mathrm{~A}_{2}=2 \mathrm{~A}$ Also, $\mathrm{T}_{1}=100^{\circ} \mathrm{C}$ and $\mathrm{T}_{2}=70^{\circ} \mathrm{C}$ Using, $\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}=\frac{\mathrm{KA}(\Delta \mathrm{T})}{l}$ Let temperature at $\mathrm{C}$ be $\mathrm{T}_{\mathrm{C}}$. then $\left(\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}\right)_{\mathrm{BC}}=\left(\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}\right)_{\mathrm{CA}}$ $\therefore \quad \frac{\text { K.2A. }\left(100-\mathrm{T}_{\mathrm{C}}\right)}{1}=\frac{\mathrm{KA}\left(\mathrm{T}_{\mathrm{C}}-70\right)}{1}$ $200-2 \mathrm{~T}_{\mathrm{C}}=\mathrm{T}_{\mathrm{C}}-70$ $3 \mathrm{~T}_{\mathrm{C}}=270$ $\mathrm{T}_{\mathrm{C}}=90^{\circ} \mathrm{C}$
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
C Thermal resistance, $\mathrm{R}_{1}=\frac{\mathrm{L}}{\mathrm{K}_{1} \mathrm{~A}}, \mathrm{R}_{2}=\frac{\mathrm{L}}{\mathrm{K}_{2} \mathrm{~A}}$ In series, equivalent thermal resistance conductivity- $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $\frac{1}{\frac{\mathrm{L}}{2 \mathrm{AK}_{\mathrm{eq}}}}=\frac{1}{\left(\frac{\mathrm{L}}{\mathrm{AK}_{1}}\right)}+\frac{1}{\left(\frac{\mathrm{L}}{\mathrm{AK}_{2}}\right)}$ $\frac{2 A K_{\text {eq }}}{L}=\frac{A K_{1}}{L}=\frac{A K_{2}}{L}$ $2 K_{\text {eq }}=K_{1}+K_{2}$ $K_{\text {eq }}=\frac{K_{1}+K_{2}}{2}$
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 $\mathrm{cm} \mathrm{sec}^{-1}$
2 $\mathrm{cm}^{2} \mathrm{sec}^{-2}$
3 $\mathrm{cm} \mathrm{sec}$
4 $\mathrm{cm}^{2} \mathrm{sec}^{-1}$
Explanation:
D Given, The units of given quantities in SI are: ${[\mathrm{K}]=\mathrm{J}(\mathrm{mKs})^{-1}}$ ${[\mathrm{~d}]=\mathrm{m} / \mathrm{v}=\mathrm{kg} \mathrm{m}^{-3}}$ ${[\mathrm{c}]=\mathrm{J}(\mathrm{kgK})^{-1}}$ ${[\mathrm{X}]=\frac{[\mathrm{K}]}{[\mathrm{d}][\mathrm{c}]}=\frac{\mathrm{J}(\mathrm{mks})^{-1}}{\mathrm{~kg}(\mathrm{~m})^{-3}(\mathrm{~J}(\mathrm{kgk}))^{-1}}}$ $=\mathrm{m}^{2} \mathrm{~s}^{-1} \text { or cm } \mathrm{s}^{-1} .$
149323
If a metallic sphere gets cooled from $62^{\circ} \mathrm{C}$ to $50^{\circ} \mathrm{C}$ in $10 \mathrm{~min}$ and in the next $10 \mathrm{~min}$ gets cooled to $42^{\circ} \mathrm{C}$, then the temperature of the surroundings is
1 $30^{\circ} \mathrm{C}$
2 $36^{\circ} \mathrm{C}$
3 $26^{\circ} \mathrm{C}$
4 $20^{\circ} \mathrm{C}$
Explanation:
C According to Newton's law cooling, $\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=\mathrm{K}\left(\frac{\theta_{1}+\theta_{2}}{2}-\theta_{\mathrm{o}}\right)$ Case-I $\theta_{1}=62^{\circ} \mathrm{C}, \quad \theta_{2}=50^{\circ} \mathrm{C}, \quad \mathrm{t}=10 \text { minute, }$ $\frac{62-50}{10}=\mathrm{K}\left[\frac{62+50}{2}-\theta_{\mathrm{o}}\right]$ $\frac{12}{10}=\mathrm{K}\left[56-\theta_{\mathrm{o}}\right]$ Case-II $\theta_{1}=50^{\circ} \mathrm{C}, \quad \theta_{2}=42^{\circ} \mathrm{C}, \quad \mathrm{t}=10 \text { minute, }$ $\frac{(50-42)}{10}=\mathrm{K}\left[\frac{50+42}{2}-\theta_{\mathrm{o}}\right]$ $\frac{8}{10}=\mathrm{K}\left[46-\theta_{\mathrm{o}}\right]$ On dividing equation (i) and (ii), we get $\frac{12}{10} \times \frac{10}{8}=\frac{K\left(56-\theta_{\mathrm{o}}\right)}{\mathrm{K}\left(46-\theta_{\mathrm{o}}\right)}$ $\frac{3}{2}=\frac{\left[56-\theta_{\mathrm{o}}\right]}{\left[46-\theta_{\mathrm{o}}\right]}$ $138-3 \theta_{\mathrm{o}}=112-2 \theta_{\mathrm{o}}$ $138-112=-2 \theta_{\mathrm{o}}+3 \theta_{\mathrm{o}}$ $\theta_{\mathrm{o}}=26^{\circ} \mathrm{C}$
UPSEE - 2010
Heat Transfer
149324
A metal rod of length $2 \mathrm{~m}$ has cross - sectional areas $2 \mathrm{~A}$ and $\mathrm{A}$ as shown in figure. The two ends are maintained at temperatures $100^{\circ} \mathrm{C}$ and $70^{\circ} \mathrm{C}$. The temperature of middle point $\mathrm{C}$ is
1 $80^{\circ} \mathrm{C}$
2 $85^{\circ} \mathrm{C}$
3 $90^{\circ} \mathrm{C}$
4 $95^{\circ} \mathrm{C}$
Explanation:
C Here, $\mathrm{K}_{1}=\mathrm{K}_{2}, l_{1}=l_{2}=1 \mathrm{~m}, \mathrm{~A}_{1}=2 \mathrm{~A}_{2}=2 \mathrm{~A}$ Also, $\mathrm{T}_{1}=100^{\circ} \mathrm{C}$ and $\mathrm{T}_{2}=70^{\circ} \mathrm{C}$ Using, $\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}=\frac{\mathrm{KA}(\Delta \mathrm{T})}{l}$ Let temperature at $\mathrm{C}$ be $\mathrm{T}_{\mathrm{C}}$. then $\left(\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}\right)_{\mathrm{BC}}=\left(\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}\right)_{\mathrm{CA}}$ $\therefore \quad \frac{\text { K.2A. }\left(100-\mathrm{T}_{\mathrm{C}}\right)}{1}=\frac{\mathrm{KA}\left(\mathrm{T}_{\mathrm{C}}-70\right)}{1}$ $200-2 \mathrm{~T}_{\mathrm{C}}=\mathrm{T}_{\mathrm{C}}-70$ $3 \mathrm{~T}_{\mathrm{C}}=270$ $\mathrm{T}_{\mathrm{C}}=90^{\circ} \mathrm{C}$
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
C Thermal resistance, $\mathrm{R}_{1}=\frac{\mathrm{L}}{\mathrm{K}_{1} \mathrm{~A}}, \mathrm{R}_{2}=\frac{\mathrm{L}}{\mathrm{K}_{2} \mathrm{~A}}$ In series, equivalent thermal resistance conductivity- $\frac{1}{\mathrm{R}_{\mathrm{eq}}}=\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}$ $\frac{1}{\frac{\mathrm{L}}{2 \mathrm{AK}_{\mathrm{eq}}}}=\frac{1}{\left(\frac{\mathrm{L}}{\mathrm{AK}_{1}}\right)}+\frac{1}{\left(\frac{\mathrm{L}}{\mathrm{AK}_{2}}\right)}$ $\frac{2 A K_{\text {eq }}}{L}=\frac{A K_{1}}{L}=\frac{A K_{2}}{L}$ $2 K_{\text {eq }}=K_{1}+K_{2}$ $K_{\text {eq }}=\frac{K_{1}+K_{2}}{2}$
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 $\mathrm{cm} \mathrm{sec}^{-1}$
2 $\mathrm{cm}^{2} \mathrm{sec}^{-2}$
3 $\mathrm{cm} \mathrm{sec}$
4 $\mathrm{cm}^{2} \mathrm{sec}^{-1}$
Explanation:
D Given, The units of given quantities in SI are: ${[\mathrm{K}]=\mathrm{J}(\mathrm{mKs})^{-1}}$ ${[\mathrm{~d}]=\mathrm{m} / \mathrm{v}=\mathrm{kg} \mathrm{m}^{-3}}$ ${[\mathrm{c}]=\mathrm{J}(\mathrm{kgK})^{-1}}$ ${[\mathrm{X}]=\frac{[\mathrm{K}]}{[\mathrm{d}][\mathrm{c}]}=\frac{\mathrm{J}(\mathrm{mks})^{-1}}{\mathrm{~kg}(\mathrm{~m})^{-3}(\mathrm{~J}(\mathrm{kgk}))^{-1}}}$ $=\mathrm{m}^{2} \mathrm{~s}^{-1} \text { or cm } \mathrm{s}^{-1} .$
