Physics

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

Explanation- For the selection of metal for wiring in home the main criterion are: the availability, conductivity and the cost of the metal. The Cu wires or Al wires are used  for wiring in the home. The main considerations are involved in this process are cost of metal and good conductivity of metal.

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

Explanation- Metallic strips have negligible resistance and need not to be counted in the length l₁, of the null point of potentiometer. That's why the thick metallic strips are used in potentiometer. It is for the convenience of experimenter as he measures only their lengths along the straight segments each of lengths 1 m.

This is a multiple choice answer as classified in NCERT Exemplar

(a) Amount of sweat evaporated /minute = $\frac{\frac{sweatproduced}{minute}}{numberofcaloriesrequiredforevaporation\frac{}{kg}}$

 = $\frac{14.5*{10}^3}{580*{10}^3}=0.25kg$

This is a multiple choice answer as classified in NCERT Exemplar

(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.

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

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.

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

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

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

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