148308
A black sphere has radius ' $R$ ' whose rate of radiation is ' $E$ ' at temperature ' $T$ '. If radius is made ' $\frac{R}{3}$ ' and temperature ' $3 T$ ', the rate of radiation will be
1 $9 \mathrm{E}$
2 $6 \mathrm{E}$
3 $3 \mathrm{E}$
4 $\mathrm{E}$
Explanation:
A $\mathrm{E}_{1} =\sigma \times \pi \mathrm{R}_{1}^{2} \times \mathrm{T}_{1}^{4}$ $\mathrm{E}_{2} =\sigma \times \pi \mathrm{R}_{2}^{2} \times \mathrm{T}_{2}^{4}$ $\therefore \quad \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}} =\left(\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}\right)^{2}\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}=\left(\frac{1}{3}\right)^{2}(3)^{4}=\frac{1}{9} \times 81$ $\mathrm{E}_{2} =9 \mathrm{E}_{1}$ $\mathrm{E}_{2} =9 \mathrm{E}$ A black sphere has radius ' $\mathrm{R}$ ' whose rate of radiation $\frac{\mathrm{R}}{3}$ and temperature ' $T$ '. If radius is made $\frac{\mathrm{R}}{3}$ and temp ' $3 \mathrm{~T}$ ', the rate of radiation will be $9 \mathrm{E}$.
MHT-CET 2020
Thermodynamics
148309
An electric kettle takes $4 \mathrm{~A}$ current at $220 \mathrm{~V}$. How much time will it take to boil $1 \mathrm{~kg}$ of water from temperature $20^{\circ} \mathrm{C}$ ? The temperature of boiling water is $100^{\circ} \mathrm{C}$
1 $12.6 \mathrm{~min}$
2 $4.2 \mathrm{~min}$
3 $6.3 \mathrm{~min}$
4 $8.4 \mathrm{~min}$
Explanation:
C Given, $\mathrm{m}=1000 \mathrm{gm}$ $\mathrm{S}=1 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C}$ We know that $\mathrm{Q}=\mathrm{mS} \Delta \mathrm{T}=1000 \times 1 \times(100-20)$ $\mathrm{Q}=1000 \times 80 \mathrm{cal}$ Heat produced in time $t$ is $\mathrm{H}=\mathrm{VIt}$ $\mathrm{H}=220 \times 4 \times \mathrm{t} \text { Joule }$ $\mathrm{H}=\frac{220 \times 4 \times \mathrm{t}}{4.18} \mathrm{cal}$ Equating (i) and (ii) $\therefore \quad \frac{200 \times 4 \times t}{4.18}=1000 \times 80$ $t=\frac{1000 \times 80 \times 4.18}{220 \times 4}$ $t=380 \mathrm{sec}=6.3 \mathrm{~min}$
VITEEE-2010
Thermodynamics
148311
An ideal gas is taken around ABCA as shown in the PV diagram. The work done during a cycle is:
1 Zero
2 $\frac{1}{2} \mathrm{PV}$
3 $2 \mathrm{PV}$
4 PV
Explanation:
D According to question- The work done during a cycle is equal to the area under a P-V diagram. $\mathrm{W}=\frac{1}{2} \times \mathrm{AC} \times \mathrm{BC}$ $\mathrm{W}=\frac{1}{2} \times(3 \mathrm{~V}-\mathrm{V}) \times(2 \mathrm{P}-\mathrm{P})$ $\mathrm{W}=\frac{1}{2} \times 2 \mathrm{~V} \times \mathrm{P}$ $\mathrm{W}=\mathrm{PV}$
Karnataka CET-2001
Thermodynamics
148312
Temperature remains constant, the pressure of gas is decreased by $20 \%$. The percentage change in volume is
1 increased by $20 \%$
2 decreased by $20 \%$
3 increased by $25 \%$
4 decreased by $25 \%$
Explanation:
C Ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PV}=$ constant $=\mathrm{K}$ \{temp remain constant $\}$ Pressure decrease by $20 \% \mathrm{P}_{2}=0.8 \mathrm{P}$ So, $\mathrm{PV}=(0.8 \mathrm{P}) \mathrm{V}_{2}$ $\mathrm{V}_{2}=\frac{\mathrm{PV}}{0.8 \mathrm{P}}=\frac{\mathrm{V}}{0.8}=1.25 \mathrm{~V}$ So, volume of gas increases by $25 \%$
148308
A black sphere has radius ' $R$ ' whose rate of radiation is ' $E$ ' at temperature ' $T$ '. If radius is made ' $\frac{R}{3}$ ' and temperature ' $3 T$ ', the rate of radiation will be
1 $9 \mathrm{E}$
2 $6 \mathrm{E}$
3 $3 \mathrm{E}$
4 $\mathrm{E}$
Explanation:
A $\mathrm{E}_{1} =\sigma \times \pi \mathrm{R}_{1}^{2} \times \mathrm{T}_{1}^{4}$ $\mathrm{E}_{2} =\sigma \times \pi \mathrm{R}_{2}^{2} \times \mathrm{T}_{2}^{4}$ $\therefore \quad \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}} =\left(\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}\right)^{2}\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}=\left(\frac{1}{3}\right)^{2}(3)^{4}=\frac{1}{9} \times 81$ $\mathrm{E}_{2} =9 \mathrm{E}_{1}$ $\mathrm{E}_{2} =9 \mathrm{E}$ A black sphere has radius ' $\mathrm{R}$ ' whose rate of radiation $\frac{\mathrm{R}}{3}$ and temperature ' $T$ '. If radius is made $\frac{\mathrm{R}}{3}$ and temp ' $3 \mathrm{~T}$ ', the rate of radiation will be $9 \mathrm{E}$.
MHT-CET 2020
Thermodynamics
148309
An electric kettle takes $4 \mathrm{~A}$ current at $220 \mathrm{~V}$. How much time will it take to boil $1 \mathrm{~kg}$ of water from temperature $20^{\circ} \mathrm{C}$ ? The temperature of boiling water is $100^{\circ} \mathrm{C}$
1 $12.6 \mathrm{~min}$
2 $4.2 \mathrm{~min}$
3 $6.3 \mathrm{~min}$
4 $8.4 \mathrm{~min}$
Explanation:
C Given, $\mathrm{m}=1000 \mathrm{gm}$ $\mathrm{S}=1 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C}$ We know that $\mathrm{Q}=\mathrm{mS} \Delta \mathrm{T}=1000 \times 1 \times(100-20)$ $\mathrm{Q}=1000 \times 80 \mathrm{cal}$ Heat produced in time $t$ is $\mathrm{H}=\mathrm{VIt}$ $\mathrm{H}=220 \times 4 \times \mathrm{t} \text { Joule }$ $\mathrm{H}=\frac{220 \times 4 \times \mathrm{t}}{4.18} \mathrm{cal}$ Equating (i) and (ii) $\therefore \quad \frac{200 \times 4 \times t}{4.18}=1000 \times 80$ $t=\frac{1000 \times 80 \times 4.18}{220 \times 4}$ $t=380 \mathrm{sec}=6.3 \mathrm{~min}$
VITEEE-2010
Thermodynamics
148311
An ideal gas is taken around ABCA as shown in the PV diagram. The work done during a cycle is:
1 Zero
2 $\frac{1}{2} \mathrm{PV}$
3 $2 \mathrm{PV}$
4 PV
Explanation:
D According to question- The work done during a cycle is equal to the area under a P-V diagram. $\mathrm{W}=\frac{1}{2} \times \mathrm{AC} \times \mathrm{BC}$ $\mathrm{W}=\frac{1}{2} \times(3 \mathrm{~V}-\mathrm{V}) \times(2 \mathrm{P}-\mathrm{P})$ $\mathrm{W}=\frac{1}{2} \times 2 \mathrm{~V} \times \mathrm{P}$ $\mathrm{W}=\mathrm{PV}$
Karnataka CET-2001
Thermodynamics
148312
Temperature remains constant, the pressure of gas is decreased by $20 \%$. The percentage change in volume is
1 increased by $20 \%$
2 decreased by $20 \%$
3 increased by $25 \%$
4 decreased by $25 \%$
Explanation:
C Ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PV}=$ constant $=\mathrm{K}$ \{temp remain constant $\}$ Pressure decrease by $20 \% \mathrm{P}_{2}=0.8 \mathrm{P}$ So, $\mathrm{PV}=(0.8 \mathrm{P}) \mathrm{V}_{2}$ $\mathrm{V}_{2}=\frac{\mathrm{PV}}{0.8 \mathrm{P}}=\frac{\mathrm{V}}{0.8}=1.25 \mathrm{~V}$ So, volume of gas increases by $25 \%$
148308
A black sphere has radius ' $R$ ' whose rate of radiation is ' $E$ ' at temperature ' $T$ '. If radius is made ' $\frac{R}{3}$ ' and temperature ' $3 T$ ', the rate of radiation will be
1 $9 \mathrm{E}$
2 $6 \mathrm{E}$
3 $3 \mathrm{E}$
4 $\mathrm{E}$
Explanation:
A $\mathrm{E}_{1} =\sigma \times \pi \mathrm{R}_{1}^{2} \times \mathrm{T}_{1}^{4}$ $\mathrm{E}_{2} =\sigma \times \pi \mathrm{R}_{2}^{2} \times \mathrm{T}_{2}^{4}$ $\therefore \quad \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}} =\left(\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}\right)^{2}\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}=\left(\frac{1}{3}\right)^{2}(3)^{4}=\frac{1}{9} \times 81$ $\mathrm{E}_{2} =9 \mathrm{E}_{1}$ $\mathrm{E}_{2} =9 \mathrm{E}$ A black sphere has radius ' $\mathrm{R}$ ' whose rate of radiation $\frac{\mathrm{R}}{3}$ and temperature ' $T$ '. If radius is made $\frac{\mathrm{R}}{3}$ and temp ' $3 \mathrm{~T}$ ', the rate of radiation will be $9 \mathrm{E}$.
MHT-CET 2020
Thermodynamics
148309
An electric kettle takes $4 \mathrm{~A}$ current at $220 \mathrm{~V}$. How much time will it take to boil $1 \mathrm{~kg}$ of water from temperature $20^{\circ} \mathrm{C}$ ? The temperature of boiling water is $100^{\circ} \mathrm{C}$
1 $12.6 \mathrm{~min}$
2 $4.2 \mathrm{~min}$
3 $6.3 \mathrm{~min}$
4 $8.4 \mathrm{~min}$
Explanation:
C Given, $\mathrm{m}=1000 \mathrm{gm}$ $\mathrm{S}=1 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C}$ We know that $\mathrm{Q}=\mathrm{mS} \Delta \mathrm{T}=1000 \times 1 \times(100-20)$ $\mathrm{Q}=1000 \times 80 \mathrm{cal}$ Heat produced in time $t$ is $\mathrm{H}=\mathrm{VIt}$ $\mathrm{H}=220 \times 4 \times \mathrm{t} \text { Joule }$ $\mathrm{H}=\frac{220 \times 4 \times \mathrm{t}}{4.18} \mathrm{cal}$ Equating (i) and (ii) $\therefore \quad \frac{200 \times 4 \times t}{4.18}=1000 \times 80$ $t=\frac{1000 \times 80 \times 4.18}{220 \times 4}$ $t=380 \mathrm{sec}=6.3 \mathrm{~min}$
VITEEE-2010
Thermodynamics
148311
An ideal gas is taken around ABCA as shown in the PV diagram. The work done during a cycle is:
1 Zero
2 $\frac{1}{2} \mathrm{PV}$
3 $2 \mathrm{PV}$
4 PV
Explanation:
D According to question- The work done during a cycle is equal to the area under a P-V diagram. $\mathrm{W}=\frac{1}{2} \times \mathrm{AC} \times \mathrm{BC}$ $\mathrm{W}=\frac{1}{2} \times(3 \mathrm{~V}-\mathrm{V}) \times(2 \mathrm{P}-\mathrm{P})$ $\mathrm{W}=\frac{1}{2} \times 2 \mathrm{~V} \times \mathrm{P}$ $\mathrm{W}=\mathrm{PV}$
Karnataka CET-2001
Thermodynamics
148312
Temperature remains constant, the pressure of gas is decreased by $20 \%$. The percentage change in volume is
1 increased by $20 \%$
2 decreased by $20 \%$
3 increased by $25 \%$
4 decreased by $25 \%$
Explanation:
C Ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PV}=$ constant $=\mathrm{K}$ \{temp remain constant $\}$ Pressure decrease by $20 \% \mathrm{P}_{2}=0.8 \mathrm{P}$ So, $\mathrm{PV}=(0.8 \mathrm{P}) \mathrm{V}_{2}$ $\mathrm{V}_{2}=\frac{\mathrm{PV}}{0.8 \mathrm{P}}=\frac{\mathrm{V}}{0.8}=1.25 \mathrm{~V}$ So, volume of gas increases by $25 \%$
148308
A black sphere has radius ' $R$ ' whose rate of radiation is ' $E$ ' at temperature ' $T$ '. If radius is made ' $\frac{R}{3}$ ' and temperature ' $3 T$ ', the rate of radiation will be
1 $9 \mathrm{E}$
2 $6 \mathrm{E}$
3 $3 \mathrm{E}$
4 $\mathrm{E}$
Explanation:
A $\mathrm{E}_{1} =\sigma \times \pi \mathrm{R}_{1}^{2} \times \mathrm{T}_{1}^{4}$ $\mathrm{E}_{2} =\sigma \times \pi \mathrm{R}_{2}^{2} \times \mathrm{T}_{2}^{4}$ $\therefore \quad \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}} =\left(\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}\right)^{2}\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}=\left(\frac{1}{3}\right)^{2}(3)^{4}=\frac{1}{9} \times 81$ $\mathrm{E}_{2} =9 \mathrm{E}_{1}$ $\mathrm{E}_{2} =9 \mathrm{E}$ A black sphere has radius ' $\mathrm{R}$ ' whose rate of radiation $\frac{\mathrm{R}}{3}$ and temperature ' $T$ '. If radius is made $\frac{\mathrm{R}}{3}$ and temp ' $3 \mathrm{~T}$ ', the rate of radiation will be $9 \mathrm{E}$.
MHT-CET 2020
Thermodynamics
148309
An electric kettle takes $4 \mathrm{~A}$ current at $220 \mathrm{~V}$. How much time will it take to boil $1 \mathrm{~kg}$ of water from temperature $20^{\circ} \mathrm{C}$ ? The temperature of boiling water is $100^{\circ} \mathrm{C}$
1 $12.6 \mathrm{~min}$
2 $4.2 \mathrm{~min}$
3 $6.3 \mathrm{~min}$
4 $8.4 \mathrm{~min}$
Explanation:
C Given, $\mathrm{m}=1000 \mathrm{gm}$ $\mathrm{S}=1 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C}$ We know that $\mathrm{Q}=\mathrm{mS} \Delta \mathrm{T}=1000 \times 1 \times(100-20)$ $\mathrm{Q}=1000 \times 80 \mathrm{cal}$ Heat produced in time $t$ is $\mathrm{H}=\mathrm{VIt}$ $\mathrm{H}=220 \times 4 \times \mathrm{t} \text { Joule }$ $\mathrm{H}=\frac{220 \times 4 \times \mathrm{t}}{4.18} \mathrm{cal}$ Equating (i) and (ii) $\therefore \quad \frac{200 \times 4 \times t}{4.18}=1000 \times 80$ $t=\frac{1000 \times 80 \times 4.18}{220 \times 4}$ $t=380 \mathrm{sec}=6.3 \mathrm{~min}$
VITEEE-2010
Thermodynamics
148311
An ideal gas is taken around ABCA as shown in the PV diagram. The work done during a cycle is:
1 Zero
2 $\frac{1}{2} \mathrm{PV}$
3 $2 \mathrm{PV}$
4 PV
Explanation:
D According to question- The work done during a cycle is equal to the area under a P-V diagram. $\mathrm{W}=\frac{1}{2} \times \mathrm{AC} \times \mathrm{BC}$ $\mathrm{W}=\frac{1}{2} \times(3 \mathrm{~V}-\mathrm{V}) \times(2 \mathrm{P}-\mathrm{P})$ $\mathrm{W}=\frac{1}{2} \times 2 \mathrm{~V} \times \mathrm{P}$ $\mathrm{W}=\mathrm{PV}$
Karnataka CET-2001
Thermodynamics
148312
Temperature remains constant, the pressure of gas is decreased by $20 \%$. The percentage change in volume is
1 increased by $20 \%$
2 decreased by $20 \%$
3 increased by $25 \%$
4 decreased by $25 \%$
Explanation:
C Ideal gas equation $\mathrm{PV}=\mathrm{nRT}$ $\mathrm{PV}=$ constant $=\mathrm{K}$ \{temp remain constant $\}$ Pressure decrease by $20 \% \mathrm{P}_{2}=0.8 \mathrm{P}$ So, $\mathrm{PV}=(0.8 \mathrm{P}) \mathrm{V}_{2}$ $\mathrm{V}_{2}=\frac{\mathrm{PV}}{0.8 \mathrm{P}}=\frac{\mathrm{V}}{0.8}=1.25 \mathrm{~V}$ So, volume of gas increases by $25 \%$
