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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.

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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 $\times$ I

2000W= 220V $\times$ 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 $\times$ 10^{-8 $\times \frac{10}{\pi \times {10}^{-6}}\times 81$} j/s = 4J/s approx.

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

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

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

Yes, aspirants can take admission at Ishan Institute of Pharmacy in D.Pharm and B.Pharm after successful completion of Class 12 from an open school which is recognised by the Ministry of education or any other appropriate government body. The minimum percentage required for admission is 45% for General, 42% for OBC and 40% for SC and ST.

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

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

$\begin{array}{l}\frac{6x+7}{x^2-9x+20}\\ Let6x+7=A\frac{d}{dx}\left(x^2–9x+20\right)+B\\ \Rightarrow 6x+7=A\left(2x–9\right)+B\\ Equatingthecoefficientofxandconstantterm,wehave\\ \begin{array}{l}2A = 6 and –9A+B= 7\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}\Rightarrow A=3 \Rightarrow B= 34\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}\therefore 6x+ 7 = 3\left(2x–9\right) + 34\hfill \end{array}\end{array}$

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 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 $\times$ I

2000W= 220V $\times$ 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 $\times$ 10^{-8 $\times \frac{10}{\pi \times {10}^{-6}}\times 81$} j/s = 4J/s approx.

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

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

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

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

$\begin{array}{l}Let5x–2=A\frac{d}{dx}\left(1+2x+3x^2\right)+B\\ \Rightarrow 5x–2=A\left(2+6x\right)+B\\ =2A+A6x+B \end{array}$

$\begin{array}{l}Equatingthecoefficientofxandconstanttermonbothsides,wehave\\ 5=6A; and 2A+B=–2 \\ \Rightarrow A= \frac{5}{6}\\ \Rightarrow B=–2–2A \\ B=\frac{-11}{3} \\ Therefore,\\ 5x-2=\frac{5}{6}(2+6x)+\left(\frac{-11}{3}\right)\\ \therefore I=\int \frac{5x-2}{1+2x+3x^2}dx\\ =\int \frac{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$6$}\right.(2+6x)+\left(\raisebox{1ex}{$-11$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)}{1+2x+3x^2}dx\\ =\frac{5}{6}\int \frac{(2+6x)}{1+2x+3x^2}dx-\frac{11}{3}\int \frac{1}{1+2x+3x^2}dx\\ Let,I_1=\int \frac{2+6x}{1+2x+3x^2}dxand{I}_2=\int \frac{1}{1+2x+3x^2}dx\\ \therefore I=\int \frac{5x-2}{1+2x+3x^2}dx\\ =\frac{5}{6}{I}_1-\frac{11}{3}{I}_2-----(1)\\ \begin{array}{l}Let 1 + 2x+ 3x^2=t\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}\Rightarrow \left(2+6x\right)dx=dt\hfill \end{array}\\ I_1=\int \frac{dt}{t}=log\left|t\right|\\ =log\left|1+2x+3x^2\right| \_\_\_\_\left(2\right)\\ I_2=\int \frac{1}{1+2x+3x^2}dx\\ Now,1+2x+3x^2=1+3\left(x^2+\frac{2}{3}x\right)\\ Therefore,1+3\left(x^2+\frac{2}{3}x\right)=1+3\left(x^2+\frac{2}{3}x+\frac{1}{9}-\frac{1}{9}\right)\end{array}$

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

Slope of the curve = f(V) , where V is the volume

Slope of P = f(V) curve at ((P_o, V₀ )= f(V_o)

Slope of adiabatic at (P_o, V₀ )= k(-Y)V_o^-1-Y =-YP_o/V_o

Now heat absorbed in the process P= f(V)

dQ=dU+dW= nC_vdT+pdV

pV=nRT

T= pV/nR

T= $\frac{1}{nR}\left[f\left(V\right)+V{f}^{\text{'}}\left(V\right)dV\right]$

$\frac{dQ}{dV}=$ nC_{v $\frac{dT}{dV}+p\frac{dV}{dV}=n{C}_V\frac{dT}{dV}+P$}

$\frac{C_V}{R}\left[f\right(V_o+V_0{f}^{\text{'}}\left(V\right)+f\left(V\right)]$

After solving we get

${\left[\frac{dQ}{dV}\right]}_{v=vo}=$ $\left[\frac{1}{Y-1}+1\right]f\left(V_o\right)+\frac{V_o{f}^{\text{'}{V}_o}}{Y-1}$

= $\frac{Y}{Y-1}{P}_o+\frac{V_o}{Y-1}{f}^{\text{'}}\left(V_o\right)$

Heat is absorbed where dQ/dV>0  when gas expands

Hence YP_o+V_of’(V_o)>0  or f’(V_o)>(-Y $\frac{P_o}{V_o}$ )

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