u = 72 kmph = 72 x 5/18 = 20 m/s

u = 20 m/s, v = 0, s = 200 m

v2= u2- 2as

2as = 400

a = 400/400 = 1 m/s2

For isothermal expansion PV = constant. Partially differentiating we have,

$V∆P+P∆V=0$

or, $\frac{∆P}{∆V}=-\frac{P}{V}$

For adiabatic expansion$P{V}^y=$constant. Partially differentiating we have,

$V^y∆P+Y{V}^{r-1}∆VP=0$

or, $\frac{∆P}{∆V}=-\frac{YP}{V}$

Thus, the slope at any point on the adiabatic curve is y times that for isothermal curve. Hence curve CD represents adiabatic expansion.

The correct choice is (1). From Ohm's law, the slope of the I–V graph gives the reciprocal of the resistance of the wire. Since the slope of the graph is smaller at temperature T₂, the resistance of the wire is greater at temperature T2 than at temperature T₁. Hence, T₂ is greater than T₁.

Using the formula for a spherical surface,

$\frac{{\mu }_a}{u}+\frac{{\mu }_z}{v}=\frac{{\mu }_z-{\mu }_a}{R}$

We have$\frac{1.0}{u}+\frac{1.5}{u}=\frac{1.5-1.0}{R}\left(Sincev=u\right)$

Which gives u = 5R. Hence the correct option is (1).

When a big drop breaks up into smaller drops, the total surface area of the smaller drops is more than the surface area of the big drop. The increase in the surface area can be brought about by supplying energy. Thus, a big drop has to absorb energy to break up into smaller drops. On the other hand, when smaller drops coalesce to form a big drop, there is a decrease in the surface area.

The balance condition of a metre bridge experiment,

$=\frac{R}{S}=\frac{l_1}{\left(100-l_1\right)}$

Hence :$R=R_1,S=R_2$

$\therefore \frac{R_1}{R_2}=\frac{l_1}{\left(100-l_1\right)}$

Ist Case;$\frac{R_1+10}{R_2}=\frac{50}{50}$

$\Rightarrow R_1+10=R_2$...........(1)

IInd Case;

$\Rightarrow R_2=\frac{40}{60}{R}_1$.............(2)

So, equations (1) and (2) give

$R_1+10=\frac{40}{60}{R}_1$

$\Rightarrow \frac{60}{40}{R}_1-R_1=10$

$\Rightarrow \frac{20}{40}{R}_1=10$

$\Rightarrow R_1=\frac{10\times 40}{20}$

$\therefore R_1=20\Omega$

Efficiency$=1-\frac{T_2}{T_1}$

$=1-\frac{400}{500}=\frac{1}{5}$

Work$=\frac{1}{5}\times heatprovided$

$=\frac{1}{5}\times 6\times {10}^{4}J$

$=1.2\times {10}^{4}J$

Weight of mass$m_2=6\times 10=60N.$The weight of$m_2$provides the tension. Thus,
T = 60 N
Opposing this force along the plane is the component$F_1=m,g\sin \theta$of the force m,g. Now$F_1=m,g\sin \theta =5\times 10\times \sin {30}^o=25N.$Since$F_1$is less than T and is, therefore, insufficient to balance T (see Fig.), the force of friction (f) down the plane is necessary to keep block$m_1$at the rest, the net force on$m_1$along the plane must be zero. Thus,

$T-m,g\sin {30}^o-f=0$
or,$f=T-m,g\sin {30}^o$
$=60-5\times 10\times \sin {30}^o$
$=60-25-35N$

Hence, the correct choice is (2).

Maximum efficiency of an engine working between temperatures T2and T1is given by the fraction of the heat absorbed by the engine, which can be converted into work.

Mathematically,

Efficiency, η = (T2– T1)/T2= 1/6

6 T2- 6 T1= T2

Therefore, T2= 1.2 T1………………….. (1)

Where T2is the source temperature and T1is the sink temperature

If the sink temperature (T1) is reduced by 62oC, then the efficiency gets doubled, i.e.

η = T2– (T1– 62)/T2= 2 x 1/6 = 2/6

6T2– 6T1+ 372 = 2T2

6T2– 2T2– 6T1+ 372 = 0

4T2– 6T1+ 372 = 0

Substituting the value of T2from equation (1), we get

4(1.2 T1) – 6T1+ 372 = 0

4.8T1– 6T1+ 372 = 0

372 = 1.2T1

T1= 310 K

Therefore, T2= 1.2 x 310 = 372 K

Hence, the temperatures of the source and the sink are 372 K and 310 K, respectively.

Source temperature = 372 - 273 = 99oC