Physics

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

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

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

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 $\left(\frac{5}{2}\right)=2500J\mathit{}$

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During driving temperature of the gas increases while volume remains constant. So according charle's law, at constant volume V.

Pressure is directly proportional to temperature. Therefore pressure of gas increases.

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Explanation- according to ohm's law V= IR

I= 6/6 = 1A

I= AneV_d  or V_d= i/neA

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Yes during adiabatic compression the temperature of a gas increases while no heat is

In adiabatic compression dQ=0

From the first law of thermodynamics dU= dQ-dW

dU=-dW

in compression work is done on the gas i.e work done is negative

dU=positive

hence internal energy of the gas increases due to which its temperature increases.

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

If a refrigerator's doors is kept open, then room will become hot, because amount of heat removed would be less than the amount of heat released in the room.