Physics

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(c) For the curve 1 volume is constant so it is isochoric process. But in curve 2 and 3 curve 2 is steeper so 2 is adiabatic and 3 is isothermal.

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process 1 isochoric and process 2 is isothermal .

Since, work done = area under P-V curve . here area under the pV curve 1 is more . so work done is more when the gas expands in isochoric process.

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Coefficient of $\beta =5$

T₁= 27+273=300K

Coefficient of performance $\beta =\frac{T_2}{300-T_2}$

1500-5T₂=T₂

6T₂=1500

T₂= 250K

T₂= 250-273=-23^oC

This is a short answer type question as classified in NCERT Exemplar

Temperature of the source is 27⁰C

T₁= 27+273= 300K

T₂= -3+273= 270K

Efficiency of heat engine = 1-T₂/T₁= 1-270/300=1/10

Efficiency of refrigerator  is 50% of a perfect engine

= 0.5 $*\eta$ = 1/20

Coefficient of performance of the refrigerator $\beta =\frac{Q_2}{W}=\frac{1-{\eta }^{\text{'}}}{\eta }$

= $\frac{1-1/20}{1/20}=\frac{19/20}{1/20}=19$

Q₂= $\beta W$ =19W

= 19 $*$ 1KW=19KW= 19kJ/s

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- relaxation time =mean free path/rms velocity of electron

Also $\rho$ = $\frac{1}{\sigma }$ = $\frac{m}{ne2t}$ relaxation time is inversely proportional to velocities

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For adiabatic change process we know

P₁V₁^y= P₂V₂^y

P (V+ $?V$ )^y = (P+ $?P$ )V^y

PV^y (1+ $\frac{?V}{V}$ ) ^y=p (1+ $\frac{?P}{P}$ )V^y

PV^y (1+ $\frac{?V}{V}$ ) $\approx$ PV^y (1+ $\frac{?V}{V}$ )

Y $\frac{?V}{V}=\frac{?P}{P}$

dV= $\frac{1}{Y}\frac{V}{p}dP$

hence work done increasing the pressure from P₁ to P₂

W= ${\int }_{p1}^{p2}pdV={\int }_{p2}^{p1}p\frac{1}{Y}\frac{V}{p}dp$

= $\frac{V}{Y}\left(P_2-P_1\right)$

W= $\frac{P_2-P_1}{Y}V$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation – In the circuit when an electron approaches a junction, in addition to the uniform E that faces it normally (which keep the drift velocity fixed), as drift velocity (v_d) is directly proportional to Electric field (E). That's why there are accumulation of charges on the surface of wires at the junction.
These produce additional electric fields. These fields alter the direction of momentum. Thus, the motion of a charge across junction is not momentum conserving

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Height of stairs h= 10m

Energy produced by burning 1 kg of fat = 7000Kcal

Energy produced by burning 5kg of fat = 5 $*7000=35000Kcal$

Energy utilised in going up and down one time

= mgh + $\frac{1}{2}mgh$ = $\frac{3}{2}mgh$

= $\frac{3}{2}*60*10*10$

= 9000J= 9000/4.2=3000/1.4cal

Number of times, the person has to go up and down the stairs

= $\frac{35*{10}^6}{\frac{3000}{1.4}}=\frac{3.5*1.4*{10}^6}{3000}$  = 16.3 $*{10}^3$ times

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- applying kirchhoff's junction rule I₁ = I+I₂

Applying kirchhoff's rule in outer loop containing 10V cell

10= IR+10I₁……………. (1)

Applying kirchhoff's rule in outer loop containing 2V cell

2= 5I₂-RI= 5 (I₁-I)-RI

4= 10I₁-10I-RI………… (2)

From 1 and 2

6=3RI+10I

2=I (R+10/3)

V= I (R+R_eff)

After comparing V=2V, R_eff= 10/3 ohm

Since effective internal resistance R_eff of two cells 10/3 ohm, being the parallel combination 5 ohm and 10 ohm . the equivalent circuit is

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Temperature of the source T₁= 500K and sink T₂= 300K

Work done W= 1000J

Efficiency of Carnot engine = 1-T₂/T₁= 1-300/500= 200/500= 2/5

Efficiency = W/Q₁

So Q₁= W/efficiency = 1000 $\left(\frac{5}{2}\right)=2500J\mathit{}$