NEET Test Series from KOTA - 10 Papers In MS WORD
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Heat Transfer
149577
Two spheres of the same material have radii 1 $\mathrm{m}$ and $4 \mathrm{~m}$ and temperatures $4000 \mathrm{~K}$ and 2000 $K$ respectively. The ratio of energy radiated per second by the first sphere to the second is
1 $1: 1$
2 $16: 1$
3 $4: 1$
4 $1: 9$
Explanation:
A Given, Radius of $1^{\text {st }}$ sphere $=1 \mathrm{~m}$ Radius of $2^{\text {nd }}$ sphere $=4 \mathrm{~m}$ Temperature of $1^{\text {st }}$ sphere $=4000 \mathrm{~K}$ Temperature of $2^{\text {nd }}$ sphere $=2000 \mathrm{~K}$ We know that, Energy radiated per second $=\sigma \mathrm{T}^{4} \mathrm{~A}$ $\therefore \quad \frac{\text { Energy radiated by } 1^{\text {st }} \text { sphere }\left(\mathrm{E}_{1}\right)}{\text { Energy radiated by } 2^{\text {nd }} \operatorname{sphere}\left(\mathrm{E}_{2}\right)}=\frac{\sigma \mathrm{T}_{1}^{4} \mathrm{~A}_{1}}{\sigma \mathrm{T}_{2}^{4} \mathrm{~A}_{2}}$ $=\left(\frac{4000}{2000}\right)^{4} \times \frac{4 \pi \times 1^{2}}{4 \pi \times 4^{2}}$ $=2^{4} \times \frac{1}{16}$ $=\frac{16}{16}$ $\therefore \quad \frac{\text { Energy radiated by } 1^{\text {st }} \text { sphere }\left(\mathrm{E}_{1}\right)}{\text { Energy radiated by } 2^{\text {nd }} \operatorname{sphere}\left(\mathrm{E}_{2}\right)}=1: 1$
AMU-2015
Heat Transfer
149578
Stars $S_{1}$ and $S_{2}$ emit maximum energy at wavelengths $5000 \AA$ and $50 \mu \mathrm{m}$, respectively. The surface temperature of $S_{1}$ is $6000 \mathrm{~K}$. Find the surface temperature of $S_{2}$
1 $90 \mathrm{~K}$
2 $80 \mathrm{~K}$
3 $70 \mathrm{~K}$
4 $60 \mathrm{~K}$
Explanation:
D Given, Wavelength of $\operatorname{star}\left(\mathrm{S}_{1}\right)=\lambda_{1}=5000 \AA=5000 \times 10^{-10}$ Wavelength of $\operatorname{star}\left(\mathrm{S}_{2}\right)=\lambda_{2}=50 \mu \mathrm{m}=50 \times 10^{-6}$ Temperature of $\operatorname{star}\left(\mathrm{T}_{1}\right)=6000 \mathrm{~K}$ We know that, Wein's displacement Law $\lambda \mathrm{T}=$ Constant $\therefore \quad \lambda_{1} \mathrm{~T}_{1}=\lambda_{2} \mathrm{~T}_{2}$ $5000 \times 10^{-10} \times 6000=50 \times 10^{-6} \times \mathrm{T}_{2}$ $\mathrm{T}_{2}=\frac{5000 \times 10^{-10} \times 6000}{50 \times 10^{-6}}$ $\mathrm{T}_{2}=60 \mathrm{~K}$
AMU-2014
Heat Transfer
149579
The following figure shows the Maxwell's speed distribution plots at four different temperatures $T_{1}, T_{2}, T_{3}$ and $T_{4}$ Which of the following gives the correct relation between temperatures?
A At lower temperature the molecules have less energy. hence, the molecular speeds are lower and distribution of molecules has smaller range. But as the temperature increases molecular speeds become higher. $\therefore \quad \mathrm{T}_{4}>\mathrm{T}_{3}>\mathrm{T}_{2}>\mathrm{T}_{1}$
AMU-2013
Heat Transfer
149580
$5 \%$ of the Power of $200 \mathrm{~W}$ bulb is converted into visible radiation. The average intensity of visible radiation at a distance of $1 \mathrm{~m}$ from the bulb is
1 $0.5 \mathrm{~W} / \mathrm{m}^{2}$
2 $0.8 \mathrm{~W} / \mathrm{m}^{2}$
3 $0.4 \mathrm{~W} / \mathrm{m}^{2}$
4 $2 \mathrm{~W} / \mathrm{m}^{2}$
Explanation:
B Given, Power $(\mathrm{P})=200 \mathrm{~W}$ distance $(\mathrm{d})=1 \mathrm{~m}$ We know that, Power of visible radiation $\mathrm{P}^{\prime}=5 \%$ of $\mathrm{P}$ $=\frac{200 \times 5}{100}$ $\mathrm{P}^{\prime} =10 \mathrm{~W}$ The intensity of radiation, $I=\frac{P^{\prime}}{4 \pi d^{2}}$ $\therefore \quad I=\frac{10}{4 \pi \times 1^{2}}$ $I=0.79 \mathrm{~W} / \mathrm{m}^{2}$ $I \simeq 0.8 \mathrm{~W} / \mathrm{m}^{2}$
149577
Two spheres of the same material have radii 1 $\mathrm{m}$ and $4 \mathrm{~m}$ and temperatures $4000 \mathrm{~K}$ and 2000 $K$ respectively. The ratio of energy radiated per second by the first sphere to the second is
1 $1: 1$
2 $16: 1$
3 $4: 1$
4 $1: 9$
Explanation:
A Given, Radius of $1^{\text {st }}$ sphere $=1 \mathrm{~m}$ Radius of $2^{\text {nd }}$ sphere $=4 \mathrm{~m}$ Temperature of $1^{\text {st }}$ sphere $=4000 \mathrm{~K}$ Temperature of $2^{\text {nd }}$ sphere $=2000 \mathrm{~K}$ We know that, Energy radiated per second $=\sigma \mathrm{T}^{4} \mathrm{~A}$ $\therefore \quad \frac{\text { Energy radiated by } 1^{\text {st }} \text { sphere }\left(\mathrm{E}_{1}\right)}{\text { Energy radiated by } 2^{\text {nd }} \operatorname{sphere}\left(\mathrm{E}_{2}\right)}=\frac{\sigma \mathrm{T}_{1}^{4} \mathrm{~A}_{1}}{\sigma \mathrm{T}_{2}^{4} \mathrm{~A}_{2}}$ $=\left(\frac{4000}{2000}\right)^{4} \times \frac{4 \pi \times 1^{2}}{4 \pi \times 4^{2}}$ $=2^{4} \times \frac{1}{16}$ $=\frac{16}{16}$ $\therefore \quad \frac{\text { Energy radiated by } 1^{\text {st }} \text { sphere }\left(\mathrm{E}_{1}\right)}{\text { Energy radiated by } 2^{\text {nd }} \operatorname{sphere}\left(\mathrm{E}_{2}\right)}=1: 1$
AMU-2015
Heat Transfer
149578
Stars $S_{1}$ and $S_{2}$ emit maximum energy at wavelengths $5000 \AA$ and $50 \mu \mathrm{m}$, respectively. The surface temperature of $S_{1}$ is $6000 \mathrm{~K}$. Find the surface temperature of $S_{2}$
1 $90 \mathrm{~K}$
2 $80 \mathrm{~K}$
3 $70 \mathrm{~K}$
4 $60 \mathrm{~K}$
Explanation:
D Given, Wavelength of $\operatorname{star}\left(\mathrm{S}_{1}\right)=\lambda_{1}=5000 \AA=5000 \times 10^{-10}$ Wavelength of $\operatorname{star}\left(\mathrm{S}_{2}\right)=\lambda_{2}=50 \mu \mathrm{m}=50 \times 10^{-6}$ Temperature of $\operatorname{star}\left(\mathrm{T}_{1}\right)=6000 \mathrm{~K}$ We know that, Wein's displacement Law $\lambda \mathrm{T}=$ Constant $\therefore \quad \lambda_{1} \mathrm{~T}_{1}=\lambda_{2} \mathrm{~T}_{2}$ $5000 \times 10^{-10} \times 6000=50 \times 10^{-6} \times \mathrm{T}_{2}$ $\mathrm{T}_{2}=\frac{5000 \times 10^{-10} \times 6000}{50 \times 10^{-6}}$ $\mathrm{T}_{2}=60 \mathrm{~K}$
AMU-2014
Heat Transfer
149579
The following figure shows the Maxwell's speed distribution plots at four different temperatures $T_{1}, T_{2}, T_{3}$ and $T_{4}$ Which of the following gives the correct relation between temperatures?
A At lower temperature the molecules have less energy. hence, the molecular speeds are lower and distribution of molecules has smaller range. But as the temperature increases molecular speeds become higher. $\therefore \quad \mathrm{T}_{4}>\mathrm{T}_{3}>\mathrm{T}_{2}>\mathrm{T}_{1}$
AMU-2013
Heat Transfer
149580
$5 \%$ of the Power of $200 \mathrm{~W}$ bulb is converted into visible radiation. The average intensity of visible radiation at a distance of $1 \mathrm{~m}$ from the bulb is
1 $0.5 \mathrm{~W} / \mathrm{m}^{2}$
2 $0.8 \mathrm{~W} / \mathrm{m}^{2}$
3 $0.4 \mathrm{~W} / \mathrm{m}^{2}$
4 $2 \mathrm{~W} / \mathrm{m}^{2}$
Explanation:
B Given, Power $(\mathrm{P})=200 \mathrm{~W}$ distance $(\mathrm{d})=1 \mathrm{~m}$ We know that, Power of visible radiation $\mathrm{P}^{\prime}=5 \%$ of $\mathrm{P}$ $=\frac{200 \times 5}{100}$ $\mathrm{P}^{\prime} =10 \mathrm{~W}$ The intensity of radiation, $I=\frac{P^{\prime}}{4 \pi d^{2}}$ $\therefore \quad I=\frac{10}{4 \pi \times 1^{2}}$ $I=0.79 \mathrm{~W} / \mathrm{m}^{2}$ $I \simeq 0.8 \mathrm{~W} / \mathrm{m}^{2}$
149577
Two spheres of the same material have radii 1 $\mathrm{m}$ and $4 \mathrm{~m}$ and temperatures $4000 \mathrm{~K}$ and 2000 $K$ respectively. The ratio of energy radiated per second by the first sphere to the second is
1 $1: 1$
2 $16: 1$
3 $4: 1$
4 $1: 9$
Explanation:
A Given, Radius of $1^{\text {st }}$ sphere $=1 \mathrm{~m}$ Radius of $2^{\text {nd }}$ sphere $=4 \mathrm{~m}$ Temperature of $1^{\text {st }}$ sphere $=4000 \mathrm{~K}$ Temperature of $2^{\text {nd }}$ sphere $=2000 \mathrm{~K}$ We know that, Energy radiated per second $=\sigma \mathrm{T}^{4} \mathrm{~A}$ $\therefore \quad \frac{\text { Energy radiated by } 1^{\text {st }} \text { sphere }\left(\mathrm{E}_{1}\right)}{\text { Energy radiated by } 2^{\text {nd }} \operatorname{sphere}\left(\mathrm{E}_{2}\right)}=\frac{\sigma \mathrm{T}_{1}^{4} \mathrm{~A}_{1}}{\sigma \mathrm{T}_{2}^{4} \mathrm{~A}_{2}}$ $=\left(\frac{4000}{2000}\right)^{4} \times \frac{4 \pi \times 1^{2}}{4 \pi \times 4^{2}}$ $=2^{4} \times \frac{1}{16}$ $=\frac{16}{16}$ $\therefore \quad \frac{\text { Energy radiated by } 1^{\text {st }} \text { sphere }\left(\mathrm{E}_{1}\right)}{\text { Energy radiated by } 2^{\text {nd }} \operatorname{sphere}\left(\mathrm{E}_{2}\right)}=1: 1$
AMU-2015
Heat Transfer
149578
Stars $S_{1}$ and $S_{2}$ emit maximum energy at wavelengths $5000 \AA$ and $50 \mu \mathrm{m}$, respectively. The surface temperature of $S_{1}$ is $6000 \mathrm{~K}$. Find the surface temperature of $S_{2}$
1 $90 \mathrm{~K}$
2 $80 \mathrm{~K}$
3 $70 \mathrm{~K}$
4 $60 \mathrm{~K}$
Explanation:
D Given, Wavelength of $\operatorname{star}\left(\mathrm{S}_{1}\right)=\lambda_{1}=5000 \AA=5000 \times 10^{-10}$ Wavelength of $\operatorname{star}\left(\mathrm{S}_{2}\right)=\lambda_{2}=50 \mu \mathrm{m}=50 \times 10^{-6}$ Temperature of $\operatorname{star}\left(\mathrm{T}_{1}\right)=6000 \mathrm{~K}$ We know that, Wein's displacement Law $\lambda \mathrm{T}=$ Constant $\therefore \quad \lambda_{1} \mathrm{~T}_{1}=\lambda_{2} \mathrm{~T}_{2}$ $5000 \times 10^{-10} \times 6000=50 \times 10^{-6} \times \mathrm{T}_{2}$ $\mathrm{T}_{2}=\frac{5000 \times 10^{-10} \times 6000}{50 \times 10^{-6}}$ $\mathrm{T}_{2}=60 \mathrm{~K}$
AMU-2014
Heat Transfer
149579
The following figure shows the Maxwell's speed distribution plots at four different temperatures $T_{1}, T_{2}, T_{3}$ and $T_{4}$ Which of the following gives the correct relation between temperatures?
A At lower temperature the molecules have less energy. hence, the molecular speeds are lower and distribution of molecules has smaller range. But as the temperature increases molecular speeds become higher. $\therefore \quad \mathrm{T}_{4}>\mathrm{T}_{3}>\mathrm{T}_{2}>\mathrm{T}_{1}$
AMU-2013
Heat Transfer
149580
$5 \%$ of the Power of $200 \mathrm{~W}$ bulb is converted into visible radiation. The average intensity of visible radiation at a distance of $1 \mathrm{~m}$ from the bulb is
1 $0.5 \mathrm{~W} / \mathrm{m}^{2}$
2 $0.8 \mathrm{~W} / \mathrm{m}^{2}$
3 $0.4 \mathrm{~W} / \mathrm{m}^{2}$
4 $2 \mathrm{~W} / \mathrm{m}^{2}$
Explanation:
B Given, Power $(\mathrm{P})=200 \mathrm{~W}$ distance $(\mathrm{d})=1 \mathrm{~m}$ We know that, Power of visible radiation $\mathrm{P}^{\prime}=5 \%$ of $\mathrm{P}$ $=\frac{200 \times 5}{100}$ $\mathrm{P}^{\prime} =10 \mathrm{~W}$ The intensity of radiation, $I=\frac{P^{\prime}}{4 \pi d^{2}}$ $\therefore \quad I=\frac{10}{4 \pi \times 1^{2}}$ $I=0.79 \mathrm{~W} / \mathrm{m}^{2}$ $I \simeq 0.8 \mathrm{~W} / \mathrm{m}^{2}$
149577
Two spheres of the same material have radii 1 $\mathrm{m}$ and $4 \mathrm{~m}$ and temperatures $4000 \mathrm{~K}$ and 2000 $K$ respectively. The ratio of energy radiated per second by the first sphere to the second is
1 $1: 1$
2 $16: 1$
3 $4: 1$
4 $1: 9$
Explanation:
A Given, Radius of $1^{\text {st }}$ sphere $=1 \mathrm{~m}$ Radius of $2^{\text {nd }}$ sphere $=4 \mathrm{~m}$ Temperature of $1^{\text {st }}$ sphere $=4000 \mathrm{~K}$ Temperature of $2^{\text {nd }}$ sphere $=2000 \mathrm{~K}$ We know that, Energy radiated per second $=\sigma \mathrm{T}^{4} \mathrm{~A}$ $\therefore \quad \frac{\text { Energy radiated by } 1^{\text {st }} \text { sphere }\left(\mathrm{E}_{1}\right)}{\text { Energy radiated by } 2^{\text {nd }} \operatorname{sphere}\left(\mathrm{E}_{2}\right)}=\frac{\sigma \mathrm{T}_{1}^{4} \mathrm{~A}_{1}}{\sigma \mathrm{T}_{2}^{4} \mathrm{~A}_{2}}$ $=\left(\frac{4000}{2000}\right)^{4} \times \frac{4 \pi \times 1^{2}}{4 \pi \times 4^{2}}$ $=2^{4} \times \frac{1}{16}$ $=\frac{16}{16}$ $\therefore \quad \frac{\text { Energy radiated by } 1^{\text {st }} \text { sphere }\left(\mathrm{E}_{1}\right)}{\text { Energy radiated by } 2^{\text {nd }} \operatorname{sphere}\left(\mathrm{E}_{2}\right)}=1: 1$
AMU-2015
Heat Transfer
149578
Stars $S_{1}$ and $S_{2}$ emit maximum energy at wavelengths $5000 \AA$ and $50 \mu \mathrm{m}$, respectively. The surface temperature of $S_{1}$ is $6000 \mathrm{~K}$. Find the surface temperature of $S_{2}$
1 $90 \mathrm{~K}$
2 $80 \mathrm{~K}$
3 $70 \mathrm{~K}$
4 $60 \mathrm{~K}$
Explanation:
D Given, Wavelength of $\operatorname{star}\left(\mathrm{S}_{1}\right)=\lambda_{1}=5000 \AA=5000 \times 10^{-10}$ Wavelength of $\operatorname{star}\left(\mathrm{S}_{2}\right)=\lambda_{2}=50 \mu \mathrm{m}=50 \times 10^{-6}$ Temperature of $\operatorname{star}\left(\mathrm{T}_{1}\right)=6000 \mathrm{~K}$ We know that, Wein's displacement Law $\lambda \mathrm{T}=$ Constant $\therefore \quad \lambda_{1} \mathrm{~T}_{1}=\lambda_{2} \mathrm{~T}_{2}$ $5000 \times 10^{-10} \times 6000=50 \times 10^{-6} \times \mathrm{T}_{2}$ $\mathrm{T}_{2}=\frac{5000 \times 10^{-10} \times 6000}{50 \times 10^{-6}}$ $\mathrm{T}_{2}=60 \mathrm{~K}$
AMU-2014
Heat Transfer
149579
The following figure shows the Maxwell's speed distribution plots at four different temperatures $T_{1}, T_{2}, T_{3}$ and $T_{4}$ Which of the following gives the correct relation between temperatures?
A At lower temperature the molecules have less energy. hence, the molecular speeds are lower and distribution of molecules has smaller range. But as the temperature increases molecular speeds become higher. $\therefore \quad \mathrm{T}_{4}>\mathrm{T}_{3}>\mathrm{T}_{2}>\mathrm{T}_{1}$
AMU-2013
Heat Transfer
149580
$5 \%$ of the Power of $200 \mathrm{~W}$ bulb is converted into visible radiation. The average intensity of visible radiation at a distance of $1 \mathrm{~m}$ from the bulb is
1 $0.5 \mathrm{~W} / \mathrm{m}^{2}$
2 $0.8 \mathrm{~W} / \mathrm{m}^{2}$
3 $0.4 \mathrm{~W} / \mathrm{m}^{2}$
4 $2 \mathrm{~W} / \mathrm{m}^{2}$
Explanation:
B Given, Power $(\mathrm{P})=200 \mathrm{~W}$ distance $(\mathrm{d})=1 \mathrm{~m}$ We know that, Power of visible radiation $\mathrm{P}^{\prime}=5 \%$ of $\mathrm{P}$ $=\frac{200 \times 5}{100}$ $\mathrm{P}^{\prime} =10 \mathrm{~W}$ The intensity of radiation, $I=\frac{P^{\prime}}{4 \pi d^{2}}$ $\therefore \quad I=\frac{10}{4 \pi \times 1^{2}}$ $I=0.79 \mathrm{~W} / \mathrm{m}^{2}$ $I \simeq 0.8 \mathrm{~W} / \mathrm{m}^{2}$
