QBManager

Thermodynamics

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 E

2

6 E

3

3 E

4

E

Explanation:

A $E_{1} = σ \times π R_{1}^{2} \times T_{1}^{4}$
$E_{2} = σ \times π R_{2}^{2} \times T_{2}^{4}$
$∴ \frac{E_{2}}{E_{1}} = {(\frac{R_{2}}{R_{1}})}^{2} {(\frac{T_{2}}{T_{1}})}^{4} = {(\frac{1}{3})}^{2} (3)^{4} = \frac{1}{9} \times 81$
$E_{2} = 9 E_{1}$
$E_{2} = 9 E$
A black sphere has radius ' $R$ ' whose rate of radiation $\frac{R}{3}$ and temperature ' $T$ '. If radius is made $\frac{R}{3}$ and temp ' $3 T$ ', the rate of radiation will be $9 E$ .

MHT-CET 2020

Thermodynamics

148309 An electric kettle takes $4 A$ current at $220 V$ . How much time will it take to boil $1 kg$ of water from temperature $20^{\circ} C$ ? The temperature of boiling water is $100^{\circ} C$

1

12.6 \min

2

4.2 \min

3

6.3 \min

4

8.4 \min

Explanation:

C Given,
$m = 1000 gm$
$S = 1 cal / g^{\circ} C$
We know that
$Q = mS Δ T = 1000 \times 1 \times (100 - 20)$
$Q = 1000 \times 80 cal$
Heat produced in time $t$ is
$H = VIt$
$H = 220 \times 4 \times t Joule$
$H = \frac{220 \times 4 \times t}{4.18} cal$
Equating (i) and (ii)
$∴ \frac{200 \times 4 \times t}{4.18} = 1000 \times 80$
$t = \frac{1000 \times 80 \times 4.18}{220 \times 4}$
$t = 380 \sec = 6.3 \min$

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} PV

3

2 PV

4 PV

Explanation:

D According to question-
The work done during a cycle is equal to the area under a P-V diagram.

$W = \frac{1}{2} \times AC \times BC$
$W = \frac{1}{2} \times (3 V - V) \times (2 P - P)$
$W = \frac{1}{2} \times 2 V \times P$
$W = PV$

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 $PV = nRT$
$PV =$ constant $= K$ \{temp remain constant $}$
Pressure decrease by $20 % P_{2} = 0.8 P$
So,
$PV = (0.8 P) V_{2}$
$V_{2} = \frac{PV}{0.8 P} = \frac{V}{0.8} = 1.25 V$
So, volume of gas increases by $25 %$

J and K CET- 2007

Thermodynamics

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 E

2

6 E

3

3 E

4

E

Explanation:

A $E_{1} = σ \times π R_{1}^{2} \times T_{1}^{4}$
$E_{2} = σ \times π R_{2}^{2} \times T_{2}^{4}$
$∴ \frac{E_{2}}{E_{1}} = {(\frac{R_{2}}{R_{1}})}^{2} {(\frac{T_{2}}{T_{1}})}^{4} = {(\frac{1}{3})}^{2} (3)^{4} = \frac{1}{9} \times 81$
$E_{2} = 9 E_{1}$
$E_{2} = 9 E$
A black sphere has radius ' $R$ ' whose rate of radiation $\frac{R}{3}$ and temperature ' $T$ '. If radius is made $\frac{R}{3}$ and temp ' $3 T$ ', the rate of radiation will be $9 E$ .

MHT-CET 2020

Thermodynamics

148309 An electric kettle takes $4 A$ current at $220 V$ . How much time will it take to boil $1 kg$ of water from temperature $20^{\circ} C$ ? The temperature of boiling water is $100^{\circ} C$

1

12.6 \min

2

4.2 \min

3

6.3 \min

4

8.4 \min

Explanation:

C Given,
$m = 1000 gm$
$S = 1 cal / g^{\circ} C$
We know that
$Q = mS Δ T = 1000 \times 1 \times (100 - 20)$
$Q = 1000 \times 80 cal$
Heat produced in time $t$ is
$H = VIt$
$H = 220 \times 4 \times t Joule$
$H = \frac{220 \times 4 \times t}{4.18} cal$
Equating (i) and (ii)
$∴ \frac{200 \times 4 \times t}{4.18} = 1000 \times 80$
$t = \frac{1000 \times 80 \times 4.18}{220 \times 4}$
$t = 380 \sec = 6.3 \min$

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} PV

3

2 PV

4 PV

Explanation:

D According to question-
The work done during a cycle is equal to the area under a P-V diagram.

$W = \frac{1}{2} \times AC \times BC$
$W = \frac{1}{2} \times (3 V - V) \times (2 P - P)$
$W = \frac{1}{2} \times 2 V \times P$
$W = PV$

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 $PV = nRT$
$PV =$ constant $= K$ \{temp remain constant $}$
Pressure decrease by $20 % P_{2} = 0.8 P$
So,
$PV = (0.8 P) V_{2}$
$V_{2} = \frac{PV}{0.8 P} = \frac{V}{0.8} = 1.25 V$
So, volume of gas increases by $25 %$

J and K CET- 2007

Thermodynamics

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 E

2

6 E

3

3 E

4

E

Explanation:

A $E_{1} = σ \times π R_{1}^{2} \times T_{1}^{4}$
$E_{2} = σ \times π R_{2}^{2} \times T_{2}^{4}$
$∴ \frac{E_{2}}{E_{1}} = {(\frac{R_{2}}{R_{1}})}^{2} {(\frac{T_{2}}{T_{1}})}^{4} = {(\frac{1}{3})}^{2} (3)^{4} = \frac{1}{9} \times 81$
$E_{2} = 9 E_{1}$
$E_{2} = 9 E$
A black sphere has radius ' $R$ ' whose rate of radiation $\frac{R}{3}$ and temperature ' $T$ '. If radius is made $\frac{R}{3}$ and temp ' $3 T$ ', the rate of radiation will be $9 E$ .

MHT-CET 2020

Thermodynamics

148309 An electric kettle takes $4 A$ current at $220 V$ . How much time will it take to boil $1 kg$ of water from temperature $20^{\circ} C$ ? The temperature of boiling water is $100^{\circ} C$

1

12.6 \min

2

4.2 \min

3

6.3 \min

4

8.4 \min

Explanation:

C Given,
$m = 1000 gm$
$S = 1 cal / g^{\circ} C$
We know that
$Q = mS Δ T = 1000 \times 1 \times (100 - 20)$
$Q = 1000 \times 80 cal$
Heat produced in time $t$ is
$H = VIt$
$H = 220 \times 4 \times t Joule$
$H = \frac{220 \times 4 \times t}{4.18} cal$
Equating (i) and (ii)
$∴ \frac{200 \times 4 \times t}{4.18} = 1000 \times 80$
$t = \frac{1000 \times 80 \times 4.18}{220 \times 4}$
$t = 380 \sec = 6.3 \min$

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} PV

3

2 PV

4 PV

Explanation:

D According to question-
The work done during a cycle is equal to the area under a P-V diagram.

$W = \frac{1}{2} \times AC \times BC$
$W = \frac{1}{2} \times (3 V - V) \times (2 P - P)$
$W = \frac{1}{2} \times 2 V \times P$
$W = PV$

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 $PV = nRT$
$PV =$ constant $= K$ \{temp remain constant $}$
Pressure decrease by $20 % P_{2} = 0.8 P$
So,
$PV = (0.8 P) V_{2}$
$V_{2} = \frac{PV}{0.8 P} = \frac{V}{0.8} = 1.25 V$
So, volume of gas increases by $25 %$

J and K CET- 2007

Thermodynamics

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 E

2

6 E

3

3 E

4

E

Explanation:

A $E_{1} = σ \times π R_{1}^{2} \times T_{1}^{4}$
$E_{2} = σ \times π R_{2}^{2} \times T_{2}^{4}$
$∴ \frac{E_{2}}{E_{1}} = {(\frac{R_{2}}{R_{1}})}^{2} {(\frac{T_{2}}{T_{1}})}^{4} = {(\frac{1}{3})}^{2} (3)^{4} = \frac{1}{9} \times 81$
$E_{2} = 9 E_{1}$
$E_{2} = 9 E$
A black sphere has radius ' $R$ ' whose rate of radiation $\frac{R}{3}$ and temperature ' $T$ '. If radius is made $\frac{R}{3}$ and temp ' $3 T$ ', the rate of radiation will be $9 E$ .

MHT-CET 2020

Thermodynamics

148309 An electric kettle takes $4 A$ current at $220 V$ . How much time will it take to boil $1 kg$ of water from temperature $20^{\circ} C$ ? The temperature of boiling water is $100^{\circ} C$

1

12.6 \min

2

4.2 \min

3

6.3 \min

4

8.4 \min

Explanation:

C Given,
$m = 1000 gm$
$S = 1 cal / g^{\circ} C$
We know that
$Q = mS Δ T = 1000 \times 1 \times (100 - 20)$
$Q = 1000 \times 80 cal$
Heat produced in time $t$ is
$H = VIt$
$H = 220 \times 4 \times t Joule$
$H = \frac{220 \times 4 \times t}{4.18} cal$
Equating (i) and (ii)
$∴ \frac{200 \times 4 \times t}{4.18} = 1000 \times 80$
$t = \frac{1000 \times 80 \times 4.18}{220 \times 4}$
$t = 380 \sec = 6.3 \min$

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} PV

3

2 PV

4 PV

Explanation:

D According to question-
The work done during a cycle is equal to the area under a P-V diagram.

$W = \frac{1}{2} \times AC \times BC$
$W = \frac{1}{2} \times (3 V - V) \times (2 P - P)$
$W = \frac{1}{2} \times 2 V \times P$
$W = PV$

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 $PV = nRT$
$PV =$ constant $= K$ \{temp remain constant $}$
Pressure decrease by $20 % P_{2} = 0.8 P$
So,
$PV = (0.8 P) V_{2}$
$V_{2} = \frac{PV}{0.8 P} = \frac{V}{0.8} = 1.25 V$
So, volume of gas increases by $25 %$

J and K CET- 2007

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