Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

Multiple choice physics option b: engineering physics conversion of heat into work: heat engine and it's efficiency engines and cycles refrigerators and heat pumps

A heat engine is supplied with 250 kJ/s of heat at a constant fixed temperature of $227^0C$; the heat is rejected at $27^0C$, the cycle is reversible, then what amount of heat is rejected?

  1. 24kJ/s

  2. 223kJ/s

  3. 150kJ/s

  4. none of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The temperature in kelvin scales are $T _1=273+27=300K,T _2=273+237=500K$

$T _1$is temperature of sink And$T _2$ is temperature of source hence by efficiency we get

$\eta=1-\dfrac{T _1}{T _2}=1-\dfrac{Q _1}{Q _2}$


$Q _1=\dfrac{T _1}{T _2}*Q _2=\dfrac{300}{500}*250=150kW$
Multiple choice physics option b: engineering physics conversion of heat into work: heat engine and it's efficiency engines and cycles refrigerators and heat pumps

An engine working on Carnot cycle rejects 40% of absorbed heat from the source, while the sink temperature is maintained at $27^0C$, then what is the source temperature (in $^0C$)?

  1. 477

  2. 346

  3. 564

  4. none of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\eta =\dfrac { 60 }{ 100 } $

${ T } _{ 2 }={ 27 }^{ o }C$
${ T } _{ 2 }={ 300 }^{ o }K$
${ T } _{ 1 }=?$

$\eta =\dfrac { { T } _{ 1 }-{ T } _{ 2 } }{ { T } _{ 1 } } $

Here, ${ T } _{ 1 }=Source\quad temperature$
${ T } _{ 2 }=sink\quad temperature$
$\eta =Efficiency$
$\eta =100-rejecion$

$\therefore \dfrac { 60 }{ 100 } =\dfrac { { T } _{ 1 }-{ T } _{ 2 } }{ { T } _{ 1 } } $

$\therefore 0.6=\dfrac { { T } _{ 1 }-300 }{ { T } _{ 1 } } $

$\therefore { T } _{ 1 }\left( 1-0.6 \right) =300$

${ T } _{ 1 }=\dfrac { 300 }{ 0.4 } =\dfrac { 3000 }{ 4 } ={ 750 }^{ o }K$

$\therefore { T } _{ 1 }={ \left( 750-273 \right)  }^{ o }C$
$\therefore { T } _{ 1 }={ 477 }^{ o }C$
Multiple choice physics option b: engineering physics conversion of heat into work: heat engine and it's efficiency engines and cycles refrigerators and heat pumps

One reversible heat engine operates between 1600 K and $T _2$ K, and another reversible heat engine operates between $T _2$ K and 400 K. If both the engines have the same heat input and output, then the temperature $T _2$ must be equal to:

  1. 600

  2. 800

  3. 650

  4. 675
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

${ T } _{ 1 }=1600K$


${ T } _{ 2 }={ T } _{ 2 }K$


${ T } _{ 1 }^{ 1 }={ T } _{ 2 }K$

${ T } _{ 2 }^{ 1 }=400K$

Same input and same output.Then the efficiency is same.
$\eta =\dfrac { { T } _{ 1 }-{ T } _{ 2 } }{ { T } _{ 1 } } $

$\therefore \dfrac { 1600-{ T } _{ 2 } }{ 1600 } =\dfrac { { T } _{ 2 }-400 }{ { T } _{ 2 } } $

$=1600{ T } _{ 2 }-{ T } _{ 2 }^{ 2 }=1600{ T } _{ 2 }-640000$

$\therefore { T } _{ 2 }^{ 2 }=640000$
$\therefore { T } _{ 2 }=800K$

Multiple choice physics option b: engineering physics conversion of heat into work: heat engine and it's efficiency engines and cycles refrigerators and heat pumps

The freezer in a refrigerator is located at the top section so that;

  1. The entire chamber of the refrigerator is cooled quickly due to convection

  2. The motor is not heated

  3. The heat gained from the environment is high

  4. The heat gained from the environment is low
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The freezer in a refrigerator is located at the top section so that the entire chamber of the refrigerator is cooled quickly due to convection.  
The correct option is A.

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

How can resolving power of the instrument be increased?

  1. use UV light

  2. immerse in oil

  3. use IR light

  4. use one more lens.
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

Resolving power for the instrument is found to be $\dfrac{\mu\sin\theta}{0.61\lambda}$ , UV light has short wavelength, hence higher resolving power. Oil is optically denser than air, that is, its $\mu$ is greater than that of air. Thus immersing in oil would increase the resolving power.

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

The ability of an optical instruments to show the images of two adjacent point objects as separate is called :

  1. dispersive power

  2. magnifying power

  3. resolving power

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

By definition, resolving power of an optical instrument is its ability to show two closely adjacent point (closely spaced) as distinct as possible.

Multiple choice physics wave optics resolving power of optical instruments resolution of optical instruments diffraction

Two lenses of focal lengths $+ 100 cm$ and $+ 5 cm$ are used to prepare an astronomical telescope. The minimum tube length will be : (final image is at $\displaystyle \infty $)

  1. $95 cm$

  2. $100 cm$

  3. $105 cm$

  4. $500 cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

the length of telescope =focal  length  of  object $(-f _0)$ +focal  length  of  eyepiece  $(f _e)$$=100+5=105cm$