149475
The temperature of two bodies $A$ and $B$ are respectively $727^{\circ} \mathrm{C}$ and $327^{\circ} \mathrm{C}$. The ratio $\mathrm{H}_{\mathrm{A}}: \mathrm{H}_{\mathrm{B}}$ of the rates of heat radiated by them is
1 $727: 327$
2 $5: 3$
3 $25: 9$
4 $625: 81$
Explanation:
D Given that, $\mathrm{T}_{\mathrm{A}}=273+727=1000 \mathrm{~K}$ $\mathrm{T}_{\mathrm{B}}=327+273=600 \mathrm{~K}$ We know that $\mathrm{H} \propto \mathrm{T}^{4}$ $\frac{\mathrm{H}_{\mathrm{A}}}{\mathrm{H}_{\mathrm{B}}}=\left(\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}\right)^{4}$ $\frac{\mathrm{H}_{\mathrm{A}}}{\mathrm{H}_{\mathrm{B}}}=\left[\frac{273+727}{273+327}\right]^{4}=\left(\frac{1000}{600}\right)^{4}=\frac{625}{81}$ $\mathrm{H}_{\mathrm{A}}: \mathrm{H}_{\mathrm{B}}=625: 81$
Manipal UGET-2017
Heat Transfer
149476
The surface area of a black body is $5 \times 10^{-4} \mathrm{~m}^{2}$ and its temperature is $727^{\circ} \mathrm{C}$. The energy radiated by it per minute is $\left(\sigma=5.670 \times 10^{-8}\right.$ $\mathbf{J} / \mathbf{m}^{2}-\mathbf{s}-\mathbf{K}^{4}$ )
149478
The amount of heat energy radiated by a metal at temperature $\mathbf{T}$ is $\mathbf{E}$. When the temperature is increased to $3 \mathrm{~T}$, energy radiated is
1 $81 \mathrm{E}$
2 $9 \mathrm{E}$
3 $3 \mathrm{E}$
4 $27 \mathrm{E}$
Explanation:
A Given that, $\mathrm{T}_{1}=\mathrm{T}$ $\mathrm{T}_{2}=3 \mathrm{~T}$ $\mathrm{E}_{1}=\mathrm{E}$ $\mathrm{E}_{2}=?$ According to Stefan's law $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}=\left(\frac{3 \mathrm{~T}}{\mathrm{~T}}\right)^{4}$ $\mathrm{E}_{2}=81 \mathrm{E}$
Manipal UGET-2009
Heat Transfer
149480
Two spheres $P$ and $Q$, of same colour having radii $8 \mathrm{~cm}$ and $2 \mathrm{~cm}$ are maintained at temperatures $127^{\circ} \mathrm{C}$ and $527^{\circ} \mathrm{C}$ respectively. The energy radiated by $P$ and $Q$ is
1 0.054
2 0.0034
3 1
4 2
Explanation:
C Given that $\mathrm{T}_{1}=127+273=400 \mathrm{~K}$ $\mathrm{T}_{2}=527+273=800 \mathrm{~K}$ According to Stefan's law $\mathrm{E}=\sigma \mathrm{AT}^{4}$ $\mathrm{E} \propto \mathrm{A}, \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{A}_{1} \mathrm{~T}_{1}^{4}}{\mathrm{~A}_{2} \mathrm{~T}_{2}^{4}}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{8^{2}}{2^{2}} \times \frac{400^{4}}{800^{4}}=1$
149475
The temperature of two bodies $A$ and $B$ are respectively $727^{\circ} \mathrm{C}$ and $327^{\circ} \mathrm{C}$. The ratio $\mathrm{H}_{\mathrm{A}}: \mathrm{H}_{\mathrm{B}}$ of the rates of heat radiated by them is
1 $727: 327$
2 $5: 3$
3 $25: 9$
4 $625: 81$
Explanation:
D Given that, $\mathrm{T}_{\mathrm{A}}=273+727=1000 \mathrm{~K}$ $\mathrm{T}_{\mathrm{B}}=327+273=600 \mathrm{~K}$ We know that $\mathrm{H} \propto \mathrm{T}^{4}$ $\frac{\mathrm{H}_{\mathrm{A}}}{\mathrm{H}_{\mathrm{B}}}=\left(\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}\right)^{4}$ $\frac{\mathrm{H}_{\mathrm{A}}}{\mathrm{H}_{\mathrm{B}}}=\left[\frac{273+727}{273+327}\right]^{4}=\left(\frac{1000}{600}\right)^{4}=\frac{625}{81}$ $\mathrm{H}_{\mathrm{A}}: \mathrm{H}_{\mathrm{B}}=625: 81$
Manipal UGET-2017
Heat Transfer
149476
The surface area of a black body is $5 \times 10^{-4} \mathrm{~m}^{2}$ and its temperature is $727^{\circ} \mathrm{C}$. The energy radiated by it per minute is $\left(\sigma=5.670 \times 10^{-8}\right.$ $\mathbf{J} / \mathbf{m}^{2}-\mathbf{s}-\mathbf{K}^{4}$ )
149478
The amount of heat energy radiated by a metal at temperature $\mathbf{T}$ is $\mathbf{E}$. When the temperature is increased to $3 \mathrm{~T}$, energy radiated is
1 $81 \mathrm{E}$
2 $9 \mathrm{E}$
3 $3 \mathrm{E}$
4 $27 \mathrm{E}$
Explanation:
A Given that, $\mathrm{T}_{1}=\mathrm{T}$ $\mathrm{T}_{2}=3 \mathrm{~T}$ $\mathrm{E}_{1}=\mathrm{E}$ $\mathrm{E}_{2}=?$ According to Stefan's law $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}=\left(\frac{3 \mathrm{~T}}{\mathrm{~T}}\right)^{4}$ $\mathrm{E}_{2}=81 \mathrm{E}$
Manipal UGET-2009
Heat Transfer
149480
Two spheres $P$ and $Q$, of same colour having radii $8 \mathrm{~cm}$ and $2 \mathrm{~cm}$ are maintained at temperatures $127^{\circ} \mathrm{C}$ and $527^{\circ} \mathrm{C}$ respectively. The energy radiated by $P$ and $Q$ is
1 0.054
2 0.0034
3 1
4 2
Explanation:
C Given that $\mathrm{T}_{1}=127+273=400 \mathrm{~K}$ $\mathrm{T}_{2}=527+273=800 \mathrm{~K}$ According to Stefan's law $\mathrm{E}=\sigma \mathrm{AT}^{4}$ $\mathrm{E} \propto \mathrm{A}, \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{A}_{1} \mathrm{~T}_{1}^{4}}{\mathrm{~A}_{2} \mathrm{~T}_{2}^{4}}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{8^{2}}{2^{2}} \times \frac{400^{4}}{800^{4}}=1$
149475
The temperature of two bodies $A$ and $B$ are respectively $727^{\circ} \mathrm{C}$ and $327^{\circ} \mathrm{C}$. The ratio $\mathrm{H}_{\mathrm{A}}: \mathrm{H}_{\mathrm{B}}$ of the rates of heat radiated by them is
1 $727: 327$
2 $5: 3$
3 $25: 9$
4 $625: 81$
Explanation:
D Given that, $\mathrm{T}_{\mathrm{A}}=273+727=1000 \mathrm{~K}$ $\mathrm{T}_{\mathrm{B}}=327+273=600 \mathrm{~K}$ We know that $\mathrm{H} \propto \mathrm{T}^{4}$ $\frac{\mathrm{H}_{\mathrm{A}}}{\mathrm{H}_{\mathrm{B}}}=\left(\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}\right)^{4}$ $\frac{\mathrm{H}_{\mathrm{A}}}{\mathrm{H}_{\mathrm{B}}}=\left[\frac{273+727}{273+327}\right]^{4}=\left(\frac{1000}{600}\right)^{4}=\frac{625}{81}$ $\mathrm{H}_{\mathrm{A}}: \mathrm{H}_{\mathrm{B}}=625: 81$
Manipal UGET-2017
Heat Transfer
149476
The surface area of a black body is $5 \times 10^{-4} \mathrm{~m}^{2}$ and its temperature is $727^{\circ} \mathrm{C}$. The energy radiated by it per minute is $\left(\sigma=5.670 \times 10^{-8}\right.$ $\mathbf{J} / \mathbf{m}^{2}-\mathbf{s}-\mathbf{K}^{4}$ )
149478
The amount of heat energy radiated by a metal at temperature $\mathbf{T}$ is $\mathbf{E}$. When the temperature is increased to $3 \mathrm{~T}$, energy radiated is
1 $81 \mathrm{E}$
2 $9 \mathrm{E}$
3 $3 \mathrm{E}$
4 $27 \mathrm{E}$
Explanation:
A Given that, $\mathrm{T}_{1}=\mathrm{T}$ $\mathrm{T}_{2}=3 \mathrm{~T}$ $\mathrm{E}_{1}=\mathrm{E}$ $\mathrm{E}_{2}=?$ According to Stefan's law $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}=\left(\frac{3 \mathrm{~T}}{\mathrm{~T}}\right)^{4}$ $\mathrm{E}_{2}=81 \mathrm{E}$
Manipal UGET-2009
Heat Transfer
149480
Two spheres $P$ and $Q$, of same colour having radii $8 \mathrm{~cm}$ and $2 \mathrm{~cm}$ are maintained at temperatures $127^{\circ} \mathrm{C}$ and $527^{\circ} \mathrm{C}$ respectively. The energy radiated by $P$ and $Q$ is
1 0.054
2 0.0034
3 1
4 2
Explanation:
C Given that $\mathrm{T}_{1}=127+273=400 \mathrm{~K}$ $\mathrm{T}_{2}=527+273=800 \mathrm{~K}$ According to Stefan's law $\mathrm{E}=\sigma \mathrm{AT}^{4}$ $\mathrm{E} \propto \mathrm{A}, \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{A}_{1} \mathrm{~T}_{1}^{4}}{\mathrm{~A}_{2} \mathrm{~T}_{2}^{4}}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{8^{2}}{2^{2}} \times \frac{400^{4}}{800^{4}}=1$
149475
The temperature of two bodies $A$ and $B$ are respectively $727^{\circ} \mathrm{C}$ and $327^{\circ} \mathrm{C}$. The ratio $\mathrm{H}_{\mathrm{A}}: \mathrm{H}_{\mathrm{B}}$ of the rates of heat radiated by them is
1 $727: 327$
2 $5: 3$
3 $25: 9$
4 $625: 81$
Explanation:
D Given that, $\mathrm{T}_{\mathrm{A}}=273+727=1000 \mathrm{~K}$ $\mathrm{T}_{\mathrm{B}}=327+273=600 \mathrm{~K}$ We know that $\mathrm{H} \propto \mathrm{T}^{4}$ $\frac{\mathrm{H}_{\mathrm{A}}}{\mathrm{H}_{\mathrm{B}}}=\left(\frac{\mathrm{T}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{B}}}\right)^{4}$ $\frac{\mathrm{H}_{\mathrm{A}}}{\mathrm{H}_{\mathrm{B}}}=\left[\frac{273+727}{273+327}\right]^{4}=\left(\frac{1000}{600}\right)^{4}=\frac{625}{81}$ $\mathrm{H}_{\mathrm{A}}: \mathrm{H}_{\mathrm{B}}=625: 81$
Manipal UGET-2017
Heat Transfer
149476
The surface area of a black body is $5 \times 10^{-4} \mathrm{~m}^{2}$ and its temperature is $727^{\circ} \mathrm{C}$. The energy radiated by it per minute is $\left(\sigma=5.670 \times 10^{-8}\right.$ $\mathbf{J} / \mathbf{m}^{2}-\mathbf{s}-\mathbf{K}^{4}$ )
149478
The amount of heat energy radiated by a metal at temperature $\mathbf{T}$ is $\mathbf{E}$. When the temperature is increased to $3 \mathrm{~T}$, energy radiated is
1 $81 \mathrm{E}$
2 $9 \mathrm{E}$
3 $3 \mathrm{E}$
4 $27 \mathrm{E}$
Explanation:
A Given that, $\mathrm{T}_{1}=\mathrm{T}$ $\mathrm{T}_{2}=3 \mathrm{~T}$ $\mathrm{E}_{1}=\mathrm{E}$ $\mathrm{E}_{2}=?$ According to Stefan's law $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}=\left(\frac{3 \mathrm{~T}}{\mathrm{~T}}\right)^{4}$ $\mathrm{E}_{2}=81 \mathrm{E}$
Manipal UGET-2009
Heat Transfer
149480
Two spheres $P$ and $Q$, of same colour having radii $8 \mathrm{~cm}$ and $2 \mathrm{~cm}$ are maintained at temperatures $127^{\circ} \mathrm{C}$ and $527^{\circ} \mathrm{C}$ respectively. The energy radiated by $P$ and $Q$ is
1 0.054
2 0.0034
3 1
4 2
Explanation:
C Given that $\mathrm{T}_{1}=127+273=400 \mathrm{~K}$ $\mathrm{T}_{2}=527+273=800 \mathrm{~K}$ According to Stefan's law $\mathrm{E}=\sigma \mathrm{AT}^{4}$ $\mathrm{E} \propto \mathrm{A}, \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{A}_{1} \mathrm{~T}_{1}^{4}}{\mathrm{~A}_{2} \mathrm{~T}_{2}^{4}}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{8^{2}}{2^{2}} \times \frac{400^{4}}{800^{4}}=1$
