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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

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

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

Kindly go through the solution

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

Height of stairs h= 10m

Energy produced by burning 1 kg of fat = 7000Kcal

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

Energy utilised in going up and down one time

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

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

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

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

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

$\begin{array}{l}I=\int \frac{dx}{x^2+2x+2}=\int \frac{dx}{(x^2+2x+1)+1}\\ =\int \frac{dx}{{(x+1)}^2+1}dx\\ \begin{array}{l}= \left[ta{n}^{–1}\left(x+1\right)\right] +C\hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}Hence, the correct Answer is \left(B\right).\hfill \end{array}\end{array}$

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

Kindly go through the solution

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

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

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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.

$\begin{array}{l}Let\left(x+3\right)=A\frac{d}{dx}\left(x^2–2x–5\right)+B\\ \Rightarrow \left(x+3\right)=A\left(2x–2\right)+B \\ Equatingthecoefficientofxandconstanttermonbothside,weget\\ Weget 2A=1 and –2A+B=3 \\ A= \frac{1}{2} B=4 \\ Therefore,\left(x+3\right)=\frac{1}{2}(2x-2)+4-----(1)\\ I=\int \frac{x+3}{x^2-2x-5}dx=\int \frac{\frac{1}{2}(2x-2)+4}{x^2-2x-5}dx\\ =\frac{1}{2}\int \frac{2x-2}{x^2-2x-5}dx+4\int \frac{1}{x^2-2x+5}dx\\ I=\int \frac{x+3}{x^2-2x-5}dx=\frac{1}{2}{I}_1+4I_2\\ I_1=\int \frac{2x-2}{x^2-2x-5}dx\\ \begin{array}{ll}Let{x}^2– 2x– 5 =t\hfill & \hfill \end{array}\end{array}$

$\begin{array}{l}\begin{array}{l}\Rightarrow \left(2x–2\right)dx=dt\hfill \end{array}\\ I_1=\int \frac{dt}{t}=log\left|t\right|=log\left|x^2-2x-5\right|-----(2)\\ I_2=\int \frac{1}{x^2-2x-5}dx=\int \frac{1}{\left(x^2-2x+1\right)-6}dx=\int \frac{1}{{(x-1)}^2-(\surd 6){}^2}dx\\ =\frac{1}{\mathrm{2\surd 6}}log\left(\frac{x-1-\surd 6}{x-1+\surd 6}\right)-----(3)\\ Substituting\left(2\right)and\left(3\right)in\left(1\right),weget\\ I=\frac{1}{2}log\left|x^2-2x-5\right|+\frac{4}{2}log\left|\frac{x-1-\surd 6}{x-1+\surd 6}\right|+C\end{array}$$\begin{array}{l}\end{array}$