146403
The surface temperature of the stars is determined using
1 Plank's law
2 Wien's displacement law
3 Rayleigh-Jeans law
4 Kirchhoff's law
Explanation:
B According to the Wien's displacement law, black body radiation has different peaks of temperature at wavelengths that are inversely proportional to temperature. $\lambda_{\mathrm{m}}=\frac{\mathrm{b}}{\mathrm{T}}$ or $\lambda_{\mathrm{m}} \mathrm{T}=\mathrm{b}$ Where, $b=$ Wien's constant $=2.8977 \times 10^{-3} \mathrm{mK}$ $\mathrm{T}=$ Temperature (in Kelvin) It is used for determining the temperature of hot radiant objects. So, Wien's displacement law is used to determine the surface of temperature of stars, moon, sun or celestial bodies.
Thermal Properties of Matter
146404
The resistance of a thermometer is $100 \Omega$ at the triple point of water $(273 \mathrm{~K})$ and is $300 \Omega$ at the melting point of gold $(\sim 873 \mathrm{~K})$. The temperature at which the resistance of the thermometer is $200 \Omega$ is
1 $273 \mathrm{~K}$
2 $373 \mathrm{~K}$
3 $473 \mathrm{~K}$
4 $573 \mathrm{~K}$
Explanation:
D Given, At $273 \mathrm{~K}$ or $0^{\circ} \mathrm{C}, \mathrm{R}_{0}=100 \Omega$ At $873 \mathrm{~K}$ or $600^{\circ} \mathrm{C}, \mathrm{R}_{600}=300 \Omega$ Assume that at $\theta \mathrm{K}$ temperature, $\mathrm{R}_{\theta}=200 \Omega$ We know that, $\theta =\frac{\mathrm{R}_{\theta}-\mathrm{R}_{0}}{\mathrm{R}_{600}-\mathrm{R}_{0}}\left(\mathrm{~T}_{2}-\mathrm{T}_{1}\right)$ $\theta =\frac{200-100}{300-100}(600-0)$ $\theta =\frac{100}{200} \times 600$ $\theta =300^{\circ} \mathrm{C}$ $\theta =(300+273) \mathrm{K}=573 \mathrm{~K}$
Shift-1]
Thermal Properties of Matter
146405
A cup of coffee cools from $150^{\circ} \mathrm{F}$ to $144^{\circ} \mathrm{F}$ in 1 min in a room temperature at $72^{\circ} \mathrm{F}$. How much time with the coffee take to cool from $110^{\circ} \mathrm{F}$ to $104^{\circ} \mathrm{F}$ in the same room?
1 $1.55 \mathrm{~min}$
2 $2.14 \mathrm{~min}$
3 $2.89 \mathrm{~min}$
4 $3.35 \mathrm{~min}$
Explanation:
B Exp: Case-1 Initial temperature of coffee $\left(T_{1}\right)=150^{\circ} \mathrm{F}$ Temperature of coffee after $1 \mathrm{~min}\left(\mathrm{~T}_{2}\right)=144^{\circ} \mathrm{F}$ Temperature of room $\left(T_{R}\right)=72^{\circ} \mathrm{F}$ and $\mathrm{t}_{1}=1 \mathrm{~min}$ According to Newton's law of cooling- $\frac{\mathrm{Q}_{1}-\mathrm{Q}_{2}}{\mathrm{t}_{1}}=\mathrm{k}\left(\frac{\mathrm{Q}_{1}+\mathrm{Q}_{2}}{2}-\mathrm{Q}_{\mathrm{R}}\right)$ $\mathrm{mS} \frac{\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}{\mathrm{t}_{1}}=\mathrm{k}\left(\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}-\mathrm{T}_{\mathrm{R}}\right)$ $\mathrm{mS} \frac{150-144}{1}=\mathrm{k}\left(\frac{150+144}{2}-72\right)$ $6 \mathrm{mS}=75 \mathrm{k}$ Case-2 Initial temperature of coffee $\left(\mathrm{T}_{1}\right)=110^{\circ} \mathrm{F}$ Temperature of coffee after $\mathrm{t}_{2} \min \left(\mathrm{T}_{2}\right)=104^{\circ} \mathrm{F}$ and $\mathrm{t}_{2}=$ ? According to Newton's law of cooling- $\mathrm{mS}\left(\frac{\mathrm{T}_{2}-\mathrm{T}_{1}}{\mathrm{t}_{2}}\right)=\mathrm{k}\left(\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}-\mathrm{T}_{\mathrm{R}}\right)$ $\mathrm{mS} \frac{(110-104)}{\mathrm{t}_{2}}=\mathrm{k}\left(\frac{110+104}{2}-72\right)$ $\frac{6 \mathrm{mS}}{\mathrm{t}_{2}}=35 \mathrm{k}$ Dividing equation (i) from equation (ii), we get $\frac{6 \mathrm{mS}}{\frac{6 \mathrm{mS}}{\mathrm{t}_{2}}}=\frac{75 \mathrm{k}}{35 \mathrm{k}}$ $\therefore \quad \mathrm{t}_{2}=\frac{15}{7}=2.14 \mathrm{~min}$
Shift-2]
Thermal Properties of Matter
146406
$30^{\circ} \mathrm{C}$ temperature in Fahrenheit scale is
1 $68^{\circ} \mathrm{F}$
2 $86^{\circ} \mathrm{F}$
3 $100^{\circ} \mathrm{F}$
4 $48.5^{\circ} \mathrm{F}$
Explanation:
B Given, temperature $=30^{\circ} \mathrm{C}$ We know that, $\mathrm{F}=\frac{9}{5} \mathrm{C}+32$ $\mathrm{~F}=\frac{9}{5} \times 30+32$ $\mathrm{~F}=54+32$ $\mathrm{~F}=86^{\circ} \mathrm{F}$
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Thermal Properties of Matter
146403
The surface temperature of the stars is determined using
1 Plank's law
2 Wien's displacement law
3 Rayleigh-Jeans law
4 Kirchhoff's law
Explanation:
B According to the Wien's displacement law, black body radiation has different peaks of temperature at wavelengths that are inversely proportional to temperature. $\lambda_{\mathrm{m}}=\frac{\mathrm{b}}{\mathrm{T}}$ or $\lambda_{\mathrm{m}} \mathrm{T}=\mathrm{b}$ Where, $b=$ Wien's constant $=2.8977 \times 10^{-3} \mathrm{mK}$ $\mathrm{T}=$ Temperature (in Kelvin) It is used for determining the temperature of hot radiant objects. So, Wien's displacement law is used to determine the surface of temperature of stars, moon, sun or celestial bodies.
Thermal Properties of Matter
146404
The resistance of a thermometer is $100 \Omega$ at the triple point of water $(273 \mathrm{~K})$ and is $300 \Omega$ at the melting point of gold $(\sim 873 \mathrm{~K})$. The temperature at which the resistance of the thermometer is $200 \Omega$ is
1 $273 \mathrm{~K}$
2 $373 \mathrm{~K}$
3 $473 \mathrm{~K}$
4 $573 \mathrm{~K}$
Explanation:
D Given, At $273 \mathrm{~K}$ or $0^{\circ} \mathrm{C}, \mathrm{R}_{0}=100 \Omega$ At $873 \mathrm{~K}$ or $600^{\circ} \mathrm{C}, \mathrm{R}_{600}=300 \Omega$ Assume that at $\theta \mathrm{K}$ temperature, $\mathrm{R}_{\theta}=200 \Omega$ We know that, $\theta =\frac{\mathrm{R}_{\theta}-\mathrm{R}_{0}}{\mathrm{R}_{600}-\mathrm{R}_{0}}\left(\mathrm{~T}_{2}-\mathrm{T}_{1}\right)$ $\theta =\frac{200-100}{300-100}(600-0)$ $\theta =\frac{100}{200} \times 600$ $\theta =300^{\circ} \mathrm{C}$ $\theta =(300+273) \mathrm{K}=573 \mathrm{~K}$
Shift-1]
Thermal Properties of Matter
146405
A cup of coffee cools from $150^{\circ} \mathrm{F}$ to $144^{\circ} \mathrm{F}$ in 1 min in a room temperature at $72^{\circ} \mathrm{F}$. How much time with the coffee take to cool from $110^{\circ} \mathrm{F}$ to $104^{\circ} \mathrm{F}$ in the same room?
1 $1.55 \mathrm{~min}$
2 $2.14 \mathrm{~min}$
3 $2.89 \mathrm{~min}$
4 $3.35 \mathrm{~min}$
Explanation:
B Exp: Case-1 Initial temperature of coffee $\left(T_{1}\right)=150^{\circ} \mathrm{F}$ Temperature of coffee after $1 \mathrm{~min}\left(\mathrm{~T}_{2}\right)=144^{\circ} \mathrm{F}$ Temperature of room $\left(T_{R}\right)=72^{\circ} \mathrm{F}$ and $\mathrm{t}_{1}=1 \mathrm{~min}$ According to Newton's law of cooling- $\frac{\mathrm{Q}_{1}-\mathrm{Q}_{2}}{\mathrm{t}_{1}}=\mathrm{k}\left(\frac{\mathrm{Q}_{1}+\mathrm{Q}_{2}}{2}-\mathrm{Q}_{\mathrm{R}}\right)$ $\mathrm{mS} \frac{\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}{\mathrm{t}_{1}}=\mathrm{k}\left(\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}-\mathrm{T}_{\mathrm{R}}\right)$ $\mathrm{mS} \frac{150-144}{1}=\mathrm{k}\left(\frac{150+144}{2}-72\right)$ $6 \mathrm{mS}=75 \mathrm{k}$ Case-2 Initial temperature of coffee $\left(\mathrm{T}_{1}\right)=110^{\circ} \mathrm{F}$ Temperature of coffee after $\mathrm{t}_{2} \min \left(\mathrm{T}_{2}\right)=104^{\circ} \mathrm{F}$ and $\mathrm{t}_{2}=$ ? According to Newton's law of cooling- $\mathrm{mS}\left(\frac{\mathrm{T}_{2}-\mathrm{T}_{1}}{\mathrm{t}_{2}}\right)=\mathrm{k}\left(\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}-\mathrm{T}_{\mathrm{R}}\right)$ $\mathrm{mS} \frac{(110-104)}{\mathrm{t}_{2}}=\mathrm{k}\left(\frac{110+104}{2}-72\right)$ $\frac{6 \mathrm{mS}}{\mathrm{t}_{2}}=35 \mathrm{k}$ Dividing equation (i) from equation (ii), we get $\frac{6 \mathrm{mS}}{\frac{6 \mathrm{mS}}{\mathrm{t}_{2}}}=\frac{75 \mathrm{k}}{35 \mathrm{k}}$ $\therefore \quad \mathrm{t}_{2}=\frac{15}{7}=2.14 \mathrm{~min}$
Shift-2]
Thermal Properties of Matter
146406
$30^{\circ} \mathrm{C}$ temperature in Fahrenheit scale is
1 $68^{\circ} \mathrm{F}$
2 $86^{\circ} \mathrm{F}$
3 $100^{\circ} \mathrm{F}$
4 $48.5^{\circ} \mathrm{F}$
Explanation:
B Given, temperature $=30^{\circ} \mathrm{C}$ We know that, $\mathrm{F}=\frac{9}{5} \mathrm{C}+32$ $\mathrm{~F}=\frac{9}{5} \times 30+32$ $\mathrm{~F}=54+32$ $\mathrm{~F}=86^{\circ} \mathrm{F}$
146403
The surface temperature of the stars is determined using
1 Plank's law
2 Wien's displacement law
3 Rayleigh-Jeans law
4 Kirchhoff's law
Explanation:
B According to the Wien's displacement law, black body radiation has different peaks of temperature at wavelengths that are inversely proportional to temperature. $\lambda_{\mathrm{m}}=\frac{\mathrm{b}}{\mathrm{T}}$ or $\lambda_{\mathrm{m}} \mathrm{T}=\mathrm{b}$ Where, $b=$ Wien's constant $=2.8977 \times 10^{-3} \mathrm{mK}$ $\mathrm{T}=$ Temperature (in Kelvin) It is used for determining the temperature of hot radiant objects. So, Wien's displacement law is used to determine the surface of temperature of stars, moon, sun or celestial bodies.
Thermal Properties of Matter
146404
The resistance of a thermometer is $100 \Omega$ at the triple point of water $(273 \mathrm{~K})$ and is $300 \Omega$ at the melting point of gold $(\sim 873 \mathrm{~K})$. The temperature at which the resistance of the thermometer is $200 \Omega$ is
1 $273 \mathrm{~K}$
2 $373 \mathrm{~K}$
3 $473 \mathrm{~K}$
4 $573 \mathrm{~K}$
Explanation:
D Given, At $273 \mathrm{~K}$ or $0^{\circ} \mathrm{C}, \mathrm{R}_{0}=100 \Omega$ At $873 \mathrm{~K}$ or $600^{\circ} \mathrm{C}, \mathrm{R}_{600}=300 \Omega$ Assume that at $\theta \mathrm{K}$ temperature, $\mathrm{R}_{\theta}=200 \Omega$ We know that, $\theta =\frac{\mathrm{R}_{\theta}-\mathrm{R}_{0}}{\mathrm{R}_{600}-\mathrm{R}_{0}}\left(\mathrm{~T}_{2}-\mathrm{T}_{1}\right)$ $\theta =\frac{200-100}{300-100}(600-0)$ $\theta =\frac{100}{200} \times 600$ $\theta =300^{\circ} \mathrm{C}$ $\theta =(300+273) \mathrm{K}=573 \mathrm{~K}$
Shift-1]
Thermal Properties of Matter
146405
A cup of coffee cools from $150^{\circ} \mathrm{F}$ to $144^{\circ} \mathrm{F}$ in 1 min in a room temperature at $72^{\circ} \mathrm{F}$. How much time with the coffee take to cool from $110^{\circ} \mathrm{F}$ to $104^{\circ} \mathrm{F}$ in the same room?
1 $1.55 \mathrm{~min}$
2 $2.14 \mathrm{~min}$
3 $2.89 \mathrm{~min}$
4 $3.35 \mathrm{~min}$
Explanation:
B Exp: Case-1 Initial temperature of coffee $\left(T_{1}\right)=150^{\circ} \mathrm{F}$ Temperature of coffee after $1 \mathrm{~min}\left(\mathrm{~T}_{2}\right)=144^{\circ} \mathrm{F}$ Temperature of room $\left(T_{R}\right)=72^{\circ} \mathrm{F}$ and $\mathrm{t}_{1}=1 \mathrm{~min}$ According to Newton's law of cooling- $\frac{\mathrm{Q}_{1}-\mathrm{Q}_{2}}{\mathrm{t}_{1}}=\mathrm{k}\left(\frac{\mathrm{Q}_{1}+\mathrm{Q}_{2}}{2}-\mathrm{Q}_{\mathrm{R}}\right)$ $\mathrm{mS} \frac{\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}{\mathrm{t}_{1}}=\mathrm{k}\left(\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}-\mathrm{T}_{\mathrm{R}}\right)$ $\mathrm{mS} \frac{150-144}{1}=\mathrm{k}\left(\frac{150+144}{2}-72\right)$ $6 \mathrm{mS}=75 \mathrm{k}$ Case-2 Initial temperature of coffee $\left(\mathrm{T}_{1}\right)=110^{\circ} \mathrm{F}$ Temperature of coffee after $\mathrm{t}_{2} \min \left(\mathrm{T}_{2}\right)=104^{\circ} \mathrm{F}$ and $\mathrm{t}_{2}=$ ? According to Newton's law of cooling- $\mathrm{mS}\left(\frac{\mathrm{T}_{2}-\mathrm{T}_{1}}{\mathrm{t}_{2}}\right)=\mathrm{k}\left(\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}-\mathrm{T}_{\mathrm{R}}\right)$ $\mathrm{mS} \frac{(110-104)}{\mathrm{t}_{2}}=\mathrm{k}\left(\frac{110+104}{2}-72\right)$ $\frac{6 \mathrm{mS}}{\mathrm{t}_{2}}=35 \mathrm{k}$ Dividing equation (i) from equation (ii), we get $\frac{6 \mathrm{mS}}{\frac{6 \mathrm{mS}}{\mathrm{t}_{2}}}=\frac{75 \mathrm{k}}{35 \mathrm{k}}$ $\therefore \quad \mathrm{t}_{2}=\frac{15}{7}=2.14 \mathrm{~min}$
Shift-2]
Thermal Properties of Matter
146406
$30^{\circ} \mathrm{C}$ temperature in Fahrenheit scale is
1 $68^{\circ} \mathrm{F}$
2 $86^{\circ} \mathrm{F}$
3 $100^{\circ} \mathrm{F}$
4 $48.5^{\circ} \mathrm{F}$
Explanation:
B Given, temperature $=30^{\circ} \mathrm{C}$ We know that, $\mathrm{F}=\frac{9}{5} \mathrm{C}+32$ $\mathrm{~F}=\frac{9}{5} \times 30+32$ $\mathrm{~F}=54+32$ $\mathrm{~F}=86^{\circ} \mathrm{F}$
146403
The surface temperature of the stars is determined using
1 Plank's law
2 Wien's displacement law
3 Rayleigh-Jeans law
4 Kirchhoff's law
Explanation:
B According to the Wien's displacement law, black body radiation has different peaks of temperature at wavelengths that are inversely proportional to temperature. $\lambda_{\mathrm{m}}=\frac{\mathrm{b}}{\mathrm{T}}$ or $\lambda_{\mathrm{m}} \mathrm{T}=\mathrm{b}$ Where, $b=$ Wien's constant $=2.8977 \times 10^{-3} \mathrm{mK}$ $\mathrm{T}=$ Temperature (in Kelvin) It is used for determining the temperature of hot radiant objects. So, Wien's displacement law is used to determine the surface of temperature of stars, moon, sun or celestial bodies.
Thermal Properties of Matter
146404
The resistance of a thermometer is $100 \Omega$ at the triple point of water $(273 \mathrm{~K})$ and is $300 \Omega$ at the melting point of gold $(\sim 873 \mathrm{~K})$. The temperature at which the resistance of the thermometer is $200 \Omega$ is
1 $273 \mathrm{~K}$
2 $373 \mathrm{~K}$
3 $473 \mathrm{~K}$
4 $573 \mathrm{~K}$
Explanation:
D Given, At $273 \mathrm{~K}$ or $0^{\circ} \mathrm{C}, \mathrm{R}_{0}=100 \Omega$ At $873 \mathrm{~K}$ or $600^{\circ} \mathrm{C}, \mathrm{R}_{600}=300 \Omega$ Assume that at $\theta \mathrm{K}$ temperature, $\mathrm{R}_{\theta}=200 \Omega$ We know that, $\theta =\frac{\mathrm{R}_{\theta}-\mathrm{R}_{0}}{\mathrm{R}_{600}-\mathrm{R}_{0}}\left(\mathrm{~T}_{2}-\mathrm{T}_{1}\right)$ $\theta =\frac{200-100}{300-100}(600-0)$ $\theta =\frac{100}{200} \times 600$ $\theta =300^{\circ} \mathrm{C}$ $\theta =(300+273) \mathrm{K}=573 \mathrm{~K}$
Shift-1]
Thermal Properties of Matter
146405
A cup of coffee cools from $150^{\circ} \mathrm{F}$ to $144^{\circ} \mathrm{F}$ in 1 min in a room temperature at $72^{\circ} \mathrm{F}$. How much time with the coffee take to cool from $110^{\circ} \mathrm{F}$ to $104^{\circ} \mathrm{F}$ in the same room?
1 $1.55 \mathrm{~min}$
2 $2.14 \mathrm{~min}$
3 $2.89 \mathrm{~min}$
4 $3.35 \mathrm{~min}$
Explanation:
B Exp: Case-1 Initial temperature of coffee $\left(T_{1}\right)=150^{\circ} \mathrm{F}$ Temperature of coffee after $1 \mathrm{~min}\left(\mathrm{~T}_{2}\right)=144^{\circ} \mathrm{F}$ Temperature of room $\left(T_{R}\right)=72^{\circ} \mathrm{F}$ and $\mathrm{t}_{1}=1 \mathrm{~min}$ According to Newton's law of cooling- $\frac{\mathrm{Q}_{1}-\mathrm{Q}_{2}}{\mathrm{t}_{1}}=\mathrm{k}\left(\frac{\mathrm{Q}_{1}+\mathrm{Q}_{2}}{2}-\mathrm{Q}_{\mathrm{R}}\right)$ $\mathrm{mS} \frac{\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}{\mathrm{t}_{1}}=\mathrm{k}\left(\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}-\mathrm{T}_{\mathrm{R}}\right)$ $\mathrm{mS} \frac{150-144}{1}=\mathrm{k}\left(\frac{150+144}{2}-72\right)$ $6 \mathrm{mS}=75 \mathrm{k}$ Case-2 Initial temperature of coffee $\left(\mathrm{T}_{1}\right)=110^{\circ} \mathrm{F}$ Temperature of coffee after $\mathrm{t}_{2} \min \left(\mathrm{T}_{2}\right)=104^{\circ} \mathrm{F}$ and $\mathrm{t}_{2}=$ ? According to Newton's law of cooling- $\mathrm{mS}\left(\frac{\mathrm{T}_{2}-\mathrm{T}_{1}}{\mathrm{t}_{2}}\right)=\mathrm{k}\left(\frac{\mathrm{T}_{1}+\mathrm{T}_{2}}{2}-\mathrm{T}_{\mathrm{R}}\right)$ $\mathrm{mS} \frac{(110-104)}{\mathrm{t}_{2}}=\mathrm{k}\left(\frac{110+104}{2}-72\right)$ $\frac{6 \mathrm{mS}}{\mathrm{t}_{2}}=35 \mathrm{k}$ Dividing equation (i) from equation (ii), we get $\frac{6 \mathrm{mS}}{\frac{6 \mathrm{mS}}{\mathrm{t}_{2}}}=\frac{75 \mathrm{k}}{35 \mathrm{k}}$ $\therefore \quad \mathrm{t}_{2}=\frac{15}{7}=2.14 \mathrm{~min}$
Shift-2]
Thermal Properties of Matter
146406
$30^{\circ} \mathrm{C}$ temperature in Fahrenheit scale is
1 $68^{\circ} \mathrm{F}$
2 $86^{\circ} \mathrm{F}$
3 $100^{\circ} \mathrm{F}$
4 $48.5^{\circ} \mathrm{F}$
Explanation:
B Given, temperature $=30^{\circ} \mathrm{C}$ We know that, $\mathrm{F}=\frac{9}{5} \mathrm{C}+32$ $\mathrm{~F}=\frac{9}{5} \times 30+32$ $\mathrm{~F}=54+32$ $\mathrm{~F}=86^{\circ} \mathrm{F}$