Physics

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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.

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

Explanation- according to ohm's law V= IR

I= 6/6 = 1A

I= AneV_d  or V_d= i/neA

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

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.

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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.

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For path1

Heat Q₁= 1000J

Work done =W₁

For path 2

Work done W₂= W₁-100

As change in internal energy is same

dU=Q₁-W₁=Q₂-W₂

1000-W₁=Q₂-W₁+100

Q₂= 1000-100= 900J

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

Explanation – power consumption in a day i.e in 5  = 10 units

Power consumption per hour = 2 units

Power consumption = 2 units =2KW= 2000J/s

Also power =V $*$ I

2000W= 220V $*$ l or l= 9A approx.

R= $\frac{\rho l}{A}$

Power consumption in first current carrying wire

P= I²R

$\frac{\rho l}{A}$ l²= 1.7 $*$ 10^{-8 $*\frac{10}{\pi *{10}^{-6}}*81$} j/s = 4J/s approx.

Loss due to joule heating in first wire = $\frac{4}{2000}*$ 100=0.2%

Power loss in Al wire =1.6 $*$ 4= 6.4J/s

Fractional loss due to joule heating in second wire = $\frac{6.4}{2000}*$ 100= 0.32%

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

Yes this is possible when the entire heat supplied to the system is utilised in expansion.

So its working against the surroundings.

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

(a) Initially the piston is in equilibrium P_i=P_a

(b) On supplying heat , the gas expands from V_o to V_i

so increase in volume of the gas =V_i-V_o

as the piston is of unit cross sectional area hence extension in the spring

x= $\frac{V_i-V_0}{area}=V_i-V_0$

force exerted by the spring on the piston= F= kx= K(V_i - V_o)

hence final pressure =P_f =P_a +kx

= P_a+K $*$ ( $V_i-V_0$ )

(c) From first law of thermodynamics

dQ=dU+dW

dU=C_v(T-T_o) = C_v(T-T_o)

T= $\frac{P_f{V}_1}{R}=\left[\frac{P_a+K\left(V_i-V_0\right)}{R}\right]\frac{V_1}{R}$

Work done by the gas =pdV+ increase in PE of the spring

= P_a(V₁-V_o) + $\frac{1}{2}k$ x²

dQ=dU+dW

= C_v(T-T_o)+P_a(V-V_o)+ $\frac{1}{2}k$ x²

= C_v(T-T_o)+P_a(v-V_o)+1/2 ( $V_i-V_0$ )²

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

Explanation – power consumption in a day i.e in 5  = 10 units

Power consumption per hour = 2 units

Power consumption = 2 units =2KW= 2000J/s

Also power =V $*$ I

2000W= 220V $*$ l or l= 9A approx.

R= $\frac{\rho l}{A}$

Power consumption in first current carrying wire

P= I²R

$\frac{\rho l}{A}$ l²= 1.7 $*$ 10^{-8 $*\frac{10}{\pi *{10}^{-6}}*81$} j/s = 4J/s approx.

Loss due to joule heating in first wire = $\frac{4}{2000}*$ 100=0.2%

Power loss in Al wire =1.6 $*$ 4= 6.4J/s

Fractional loss due to joule heating in second wire = $\frac{6.4}{2000}*$ 100= 0.32%