Past NEET Papers

Hint

In first conditions;
Given, the initial temperature of the cup of coffee,\(T_i=90^{\circ} \mathrm{C}\)
The final temperature of the cup of coffee, \(T_f=80^{\circ} \mathrm{C}\)
The time taken to drop the temperature \(90^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\) is \(\mathrm{t}\).
The temperature of the surrounding, \(T_0=20^{\circ} \mathrm{C}\)
Using the Newton’s law of cooling, rate of cooling \(=\frac{d T}{d t}=K\left[\frac{T_i+T_f}{2}-T_0\right]\)
Substituting the values in the above equation, we get
\(
\begin{aligned}
\frac{90-80}{t} & =K\left[\frac{90+80}{2}-20\right] \\
\Rightarrow \quad \frac{10}{t} & =K[65] \Rightarrow K=\frac{2}{13 t}
\end{aligned}
\)
In second conditions;
The initial temperature of the cup of coffee,\(T_i^{\prime}=80^{\circ} \mathrm{C}\)
The final temperature of the cup of coffee, \(T_f^{\prime}=60^{\circ} \mathrm{C}\)
Using the Newton’s law of cooling, rate of cooling \(=\frac{d T}{d t}=K\left[\frac{T_1^{\prime}+T_f^{\prime}}{2}-T_0\right]\)
Substituting the values in the above equation, we get
\(
\begin{aligned}
\frac{80-60}{t_1} & =\frac{2}{13 t}\left[\frac{60+80}{2}-20\right] \\
\frac{20}{t_1} & =\frac{2}{13 t}[50] \Rightarrow t_1=\frac{13}{5} t
\end{aligned}
\)

Hint

Since, heat required, \(Q=m c \Delta T\)
\(
=\left(\frac{4}{3} \pi r^3 \cdot \rho\right) c \Delta T \quad\left[\because m=V_{\text {sphere }} \rho\right]
\)
Since, \(\pi, \rho, c\) and \(T\) are constants.
\(
\begin{aligned}
\Rightarrow \quad Q & \propto r^3 \text { or } \frac{Q_1}{Q_2}=\frac{r_1^3}{r_2^3} \\
& =\left(\frac{r_1}{r_2}\right)^3=\left(\frac{1.5 r_2}{r_2}\right)^3=\frac{27}{8}
\end{aligned}
\)

Hint

According to Wein’s displacement law,
\(
\lambda=\frac{b}{T}
\)
i.e., \(\quad \lambda \propto \frac{1}{T} \dots(i)\)
We know that,
\(
\lambda_{\text {blish }}<\lambda_{\text {yellowish }}<\lambda_{\text {reddish }}
\)
Hence, using Eq. (i), we have
\(
T_A>T_C>T_B
\)

Hint

Due to change in temperature, the thermal strain produced in a rod of length \(L\) is given by
\(
\begin{aligned}
& \frac{\Delta L}{L}=\alpha \Delta T \\
& \Rightarrow \quad \Delta L=L \alpha \Delta T \\
& \text { wherel }=\text { original length of rod and } \\
& \alpha=\text { coefficient of liner expansion of solid } \\
& \text { rod }
\end{aligned}
\)
As the change in length \((\Delta l)\) of the given two rods of copper and aluminium are independent of temperature change, i.e. \(\Delta T\) is same for both copper and aluminium.
\(
\begin{aligned}
L_{C u} \alpha_{C u} & =L_{A l} \alpha_{A l} \dots(i)\\
\text { Here, } \quad \alpha_{C u} & =1.7 \times 10^{-5} \mathrm{~K}^{-1} \\
& \alpha_{A l}=2.2 \times 10^{-5} \mathrm{~K}^{-1} \\
\text { and } \quad L_{C u} & =88 \mathrm{~cm}
\end{aligned}
\)
Substituting the given values in Eq. (i), we get
\(
L_{A l}=\frac{L_{C u} \alpha_{C u}}{\alpha_{A l}}=\frac{1.7 \times 10^{-5} \times 88}{2.2 \times 10^{-5}} \simeq 68 \mathrm{~cm}
\)

Hint

According to Newton’s law of cooling
\(
\frac{\left(T_1-T_2\right)}{t}=K\left(\frac{T_1+T_2}{2}-T_0\right)
\)
\(
\begin{aligned}
& \frac{(80-70)}{12}=K(75-25) \dots(i)\\
& \frac{(70-60)}{t^{\prime}}=K(65-25) \dots(ii)\\
& \Rightarrow t^{\prime}=12 \times=15 \mathrm{~min}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \mathrm{Q}=\mathrm{KA}\left(\frac{\Delta \mathrm{T}}{\Delta \mathrm{x}}\right) \mathrm{dt} \\
& \mathrm{Now}, \mathrm{Q}=\mathrm{mL} \\
& \mathrm{mL}=\mathrm{KA}\left(\frac{\Delta \mathrm{T}}{\Delta \mathrm{x}}\right) \mathrm{dt} \\
& \mathrm{mL}=\frac{\mathrm{KA}[0-(-26)] \mathrm{dt}}{\mathrm{x}} \\
& (\rho \mathrm{dx}) \mathrm{L}=\mathrm{KA} \frac{[0-(-26)]}{\mathrm{x}} \mathrm{dt} \\
& \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{26 \mathrm{~K}}{\mathrm{x} \rho \mathrm{L}}
\end{aligned}
\)

Hint

(d) From Wien’s law
\(
\begin{aligned}
& \lambda_{\max } \mathrm{T}=\text { constant } \\
& \text { i.e., } \lambda_{\max 1} \mathrm{~T}_1=\lambda_{\max _2} \mathrm{~T}_2 \\
& \Rightarrow \lambda_0 \mathrm{~T}_1=\frac{3 \lambda_0}{4} T_2 \\
& \Rightarrow \mathrm{T}_2=\frac{4}{3} T_1
\end{aligned}
\)
As we know that, from Stefan’s law, the power radiated by a body at temperature \(T\) is given as
\(
P=\sigma A e T^4
\)
Power radiated \(P \propto T^4\)
So, \(\frac{P_2}{P_1}=n=\left(\frac{T_2}{T_1}\right)^4=\left(\frac{4}{3}\right)^4=\frac{256}{81}\)

Hint

(d)
Radiated power of a black body,
\(
P=\sigma A T^4
\)
where, \(A=\) surface area of the body
\(T=\) temperature of the body
and \(\sigma=\) Stefan’s constant
When radius of the sphere is halved, new area,
\(
A^{\prime}=\frac{A}{4}
\)
\(\therefore\) Power radiated,
\(
\begin{aligned}
P^{\prime} & =\sigma\left(\frac{A}{4}\right)(2 T)^4=\frac{16}{4} \cdot\left(\sigma A T^4\right) \\
& =4 P=4 \times 450=1800 \text { watts }
\end{aligned}
\)

Hint

\(
\begin{aligned}
& R_1=\frac{d}{k_1 A} \\
& R_2=d K_2 A \\
& \operatorname{Req}=\frac{R_1 R_2}{R_1+R_2} \\
& \frac{d}{k e q(2 A)}=\frac{\left(\frac{d}{k_1 A}\right)\left(\frac{d}{k_2 A}\right)}{A}\left(\frac{1}{k_1}+\frac{1}{k_2}\right) \\
& \frac{1}{2 k e q}=\frac{\frac{1}{k_1 k_2}}{\frac{k_1+k_2}{k_1 k_2}}=\frac{1}{k_1+k_2} \\
& k e q=\frac{k_1+k_2}{2}
\end{aligned}
\)

Hint

(d) According to Wein’s displacement law, maximum amount of emitted radiation corresponding to \(\lambda_{\mathrm{m}}=\frac{\mathrm{b}}{\mathrm{T}}\)
\(
\lambda_{\mathrm{m}}=\frac{2.88 \times 10^6 \mathrm{nmK}}{5760 \mathrm{~K}}=500 \mathrm{~nm}
\)
\(
\text { From the graph } \mathrm{U}_1<\mathrm{U}_2>\mathrm{U}_3
\)

Hint

(b) According to the question, as conservation of energy, energy gained by the ice during its fall from height \(h\) is given by
\(
E=m g h
\)
As given, only one quarter of its energy is absorbed by the ice.
\(
\text { So, } \begin{aligned}
& \frac{m g h}{4}=m L \Rightarrow h=\frac{m L \times 4}{m g} \\
&= \frac{L \times 4}{g}=\frac{3.4 \times 10^5 \times 4}{10} \\
&=13.6 \times 10^4=136000 \mathrm{~m}=136 \mathrm{~km}
\end{aligned}
\)

Hint

According to question,
Coefficient of linear expression of brass
\(
=\alpha_1
\)
Coefficient of linear expression of steel
\(
=\alpha_2
\)
Length of brass and steel rods are \(l_1\) and \(l_2\) respectively.
As given difference increase in length \(\left(l_2^{\prime}-l_1^{\prime}\right)\) is same for all temperature.
So, \(l_2^{\prime}-l_1^{\prime}=l_2-l_1\)
\(\Rightarrow l_2\left(1+\alpha_2 \Delta t\right)-l_1\left(1+\alpha_1 \Delta t\right)=l_2-l_1\)
\(\Rightarrow l_2 \alpha_2=l_1 \alpha_1\)

Hint

Heat lost by 1st body \(=\) heat gained by 2nd body. Body at \(100^{\circ} \mathrm{C}\) temperature has greater heat capacity than body at \(0^{\circ} \mathrm{C}\) so final temperature will be closer to \(100^{\circ} \mathrm{C}\). So, \(T_c>50^{\circ} \mathrm{C}\)

Hint

Newton’s law of cooling
\(
\begin{aligned}
& \frac{\mathrm{T}_1-\mathrm{T}_2}{\mathrm{t}}=\mathrm{k}\left(\frac{\mathrm{T}_1+\mathrm{T}_2}{2}-\mathrm{T}\right) \\
& \frac{3 \mathrm{~T}-2 \mathrm{~T}}{10}=\mathrm{k}\left(\frac{5 \mathrm{~T}-2 \mathrm{~T}}{2}\right) \\
& \Rightarrow \frac{\mathrm{T}}{10}=\mathrm{k}\left(\frac{3 \mathrm{~T}}{2}\right)—-(\mathrm{i}) \\
& \Rightarrow \frac{2 \mathrm{~T}-\mathrm{T}^{\prime}}{10}=\mathrm{k}\left(\frac{2 \mathrm{~T}+\mathrm{T}^{\prime}}{2}-\mathrm{T}\right) \\
& \Rightarrow \frac{2 \mathrm{~T}-\mathrm{T}^{\prime}}{10}=\mathrm{k}\left(\frac{\mathrm{T}^{\prime}}{2}\right)—–(\mathrm{ii})
\end{aligned}
\)
By solving (i) and (ii)
\(
\mathrm{T}^{\prime}=\frac{3}{2} \mathrm{~T}
\)

Alternate:

According to Newton’s law of cooling,
\(
\begin{aligned}
& \Delta T=\Delta T_0 e^{-\lambda t} \\
& \Rightarrow 3 T-2 T=(3 T-T) e^{-\lambda \times 10} \dots(i) \\
& \text { Again for next } 10 \text { minutes } \\
& T^{\prime}-T=(2 T) \times e^{-\lambda(20)} \dots(ii) \\
& \text { From Eqs. (i) and (ii), we get } \\
& T^{\prime}-T=(2 T)\left(e^{-\lambda \times 10}\right)^2=(2 T)\left(\frac{1}{2}\right)^2 \\
& =\frac{T}{2} \\
& \therefore \quad T^{\prime}=T+\frac{T}{2}=\frac{3 T}{2} \\
&
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \rho=\rho_0(1+Y \Delta T) \\
& \rho-\rho_0=\rho_0 Y \Delta T \\
& \text { fraction change } \\
& \frac{\rho-\rho_0}{\rho_0}=Y \Delta T=5 \times 10^{-4} \times 40 \\
& =200 \times 10^{-4}=0.020
\end{aligned}
\)

Hint

(a) From Wein’s displacement law
\(\lambda_{\mathrm{m}} \times \mathrm{T}=\mathrm{constant}\)
\(\mathrm{P}\) – max. intensity is at violet
\(\Rightarrow \lambda_{\mathrm{m}}\) is minimum \(\Rightarrow\) temp maximum
\(\mathrm{R}\) – max. intensity is at green
\(\Rightarrow \lambda_{\mathrm{m}}\) is moderate \(\Rightarrow\) temp moderate
\(\mathrm{Q}-\max _{\mathrm{max}}\). intensity is at red \(\Rightarrow \lambda_{\mathrm{m}}\) is maximum \(\Rightarrow\)
temp. minimum i.e., \(\mathrm{T}_{\mathrm{p}}>\mathrm{T}_{\mathrm{R}}>\mathrm{T}_{\mathrm{Q}}\)

Hint

Here, \(\quad \Delta T_1=110-100=10^{\circ} \mathrm{C}\)
\(
\begin{aligned}
& \frac{d \emptyset_1}{d t}=4 \mathrm{~J} / \mathrm{s} \Rightarrow \Delta T_2=210-200=10^0 \mathrm{C} \\
& \frac{d \emptyset_2}{d t}=?
\end{aligned}
\)
As the rate of heat flow is directly proportional to the temperature difference and the temperature difference in both the cases is same i.e. \(10^{\circ} \mathrm{C}\). So, the same rate of heat will flow in the second case.
Hence, \(\quad \frac{d O_2}{d t}=4 \mathrm{~J} / \mathrm{s}\)

Hint

(d) According to the principle of calorimetry.
Heat lost \(=\) Heat gained
\(
\begin{aligned}
& \mathrm{mL}_{\mathrm{v}}+\mathrm{ms}_{\mathrm{w}} \Delta \theta=\mathrm{m}_{\mathrm{w}} \mathrm{s}_{\mathrm{w}} \Delta \theta \\
& \Rightarrow \mathrm{m} \times 540+\mathrm{m} \times 1 \times(100-80) = 20 \times 1 \times(80-10) \\
& \Rightarrow \mathrm{m}=2.5 \mathrm{~g}
\end{aligned}
\)
Therefore total mass of water at \(80^{\circ} \mathrm{C}\)
\(
=(20+2.5) \mathrm{g}=22.5 \mathrm{~g}
\)

Hint

(a) Let the temperature of surroundings be \(\theta_0\)
By Newton’s law of cooling
\(
\begin{aligned}
& \frac{\theta_1-\theta_2}{\mathrm{t}}=\mathrm{k}\left[\frac{\theta_1+\theta_2}{2}-\theta_0\right] \\
& \Rightarrow \frac{70-60}{5}=\mathrm{k}\left[\frac{70+60}{2}-\theta_0\right] \\
& \Rightarrow 2=\mathrm{k}\left[65-\theta_0\right] \dots(i)\\
& \text { Similarly, } \frac{60-54}{5}=\mathrm{k}\left[\frac{60+54}{2}-\theta_0\right] \\
& \Rightarrow \frac{6}{5}=\mathrm{k}\left[57-\theta_0\right] \dots(ii)
\end{aligned}
\)
By dividing (i) by (ii) we have
\(
\frac{10}{6}=\frac{65-\theta_0}{57-\theta_0} \Rightarrow \theta_0==45^{\circ} \mathrm{C}
\)

Hint

(a) Wein’s displacement law According to this law
\(
\lambda_{\max } \propto \frac{1}{T}
\)
or, \(\lambda_{\max } \times \mathrm{T}=\) constant
So, as the temperature increases \(\lambda\) decreases.

Hint

(b) Given
\(
\begin{array}{ll}
\mathrm{T}_1=20^{\circ} \mathrm{C}, & \mathrm{T}_2=40^{\circ} \mathrm{C} \\
\rho_{20}=998 \mathrm{~kg} / \mathrm{m}^3, & \rho_{40}=992 \mathrm{~kg} / \mathrm{m}^3
\end{array}
\)
\(
\begin{aligned}
& \text { As } \rho_{T 2}=\frac{\rho_{T 1}}{1+\gamma \wedge T}=\frac{\rho_{T 1}}{1+\gamma\left(T_2-T_1\right)} \\
& \therefore 992=\frac{998}{1+\gamma(40-20)} \Rightarrow 992=\frac{998}{1+20 \gamma} \\
& \Rightarrow 20 \gamma=\frac{998}{992}-1=\frac{6}{992} \\
& \Rightarrow \gamma=\frac{6}{992} \times \frac{1}{20}=3 \times 10^{-4} /{ }^{\circ} \mathrm{C}
\end{aligned}
\)

Hint

(b) \(Q_1=4 Q_2\) (Given)
\(
\Rightarrow \frac{K_1 A_1 \Delta t}{L}=4 \frac{K_2 A_2 \Delta t}{L} \Rightarrow K_1 A_1=4 K_2 A_2
\)

Hint

(a) Initially liquid oxygen will gain the temperature up to its boiling temperature then it changes its state liquid to gas. After this again its temperature will increase, so the corresponding graph will be graph (a).

Hint

(d) Stefan’s law for black body radiation \(Q=\sigma e A T^4\)
\(
T=\left[\frac{Q}{\sigma\left(4 \pi R^2\right)}\right]^{1 / 4}
\)
Here \(e=1\) and \(A=4 \pi R^2\)

Hint

Rate of heat given by steam \(=\) Rate of heat taken by ice
where \(K=\) Thermal conductivity of the slab
\(m=\) Mass of the ice
\(L=\) Latent heat of melting/fusion
\(A=\) Area of the slab
\(\frac{d Q}{d t}=\frac{K A(100-0)}{1}=m \frac{d L}{d t}\)
\(\frac{K \times 100 \times 0.36}{0.1}=\frac{4.8 \times 3.36 \times 10^5}{60 \times 60}\)
\(K=1.24 \mathrm{~J} / \mathrm{m} / \mathrm{s} /{ }^{\circ} \mathrm{C}\)

Hint

(c)
Change in entropy is given by \(\Delta S=\frac{m l}{T}=\frac{1000 \times 80}{273}=293 \mathrm{cal} \mathrm{K}^{-1}\)

Hint

(b) The rate of heat flow is given by
\(
\begin{aligned}
& \frac{Q}{t}=\mathrm{K} . \text { A. } \frac{\Delta T}{\ell} \\
& \text { Area of Original rod } \mathrm{A}=\pi R^2 ; \\
& \text { Area of new rod } \mathrm{A}^{\prime}=\frac{\pi R^2}{4} \\
& \mathrm{~A}^{\prime}=\frac{\mathrm{A}}{4}
\end{aligned}
\)
Volume of original rod will be equal to the volume of new rod.
So, \(\mathrm{A}^{\prime} \ell^{\prime}=\mathrm{A} \ell\)
or,
\(
\begin{aligned}
& \therefore \pi R^2 \ell=\pi\left(\frac{R}{2}\right)^2 \ell^{\prime} \\
& \Rightarrow \frac{\ell^{\prime}}{\ell}=\frac{R^2}{\left(\frac{R^2}{4}\right)}=4 \\
& \therefore \quad \frac{Q^{\prime}}{Q}=\frac{A^{\prime}}{A} \frac{\ell}{\ell^{\prime}}=\frac{1}{4} \cdot \frac{1}{4}=\frac{1}{16}
\end{aligned}
\)
\(\therefore \quad Q^{\prime}=\frac{Q}{16}\)

Hint

By Stefan’s law,
\(
\mathrm{E}=\sigma \mathrm{eAT}^4
\)
For black body, e \(=1\)
Energy, \(\mathrm{E}=\sigma\left(4 \pi \mathrm{r}^2\right) \mathrm{T}^4\)
Intensity, \(\mathrm{I}=\frac{\sigma\left(4 \pi \mathrm{r}^2 \mathrm{~T}^4\right)}{4 \pi \mathrm{R}^2}=\frac{\sigma r^2 \mathrm{~T}^4}{\mathrm{R}^2}\)

Hint

\(
\begin{aligned}
& \text { (b) According to Stefan’s law } \mathrm{E}=\sigma \mathrm{T}^4 \text {, } \\
& \mathrm{T}_1=500 \mathrm{~K} \\
& \mathrm{~T}_2=1000 \mathrm{~K} \\
& \frac{E_2}{E_1}=\left(\frac{T_2}{T_1}\right)^4=\left(\frac{1000}{500}\right)^4=16 \\
& \therefore \quad \mathrm{E}_2=16 \times 7=112 \mathrm{cal} / \mathrm{cm}^2 \mathrm{~s}
\end{aligned}
\)

Hint

For a rod of length \(L\) and area of cross-section \(A\) whose faces are maintained at temperatures \(T_1\) and \(T_2\) respectively. Then in steady state the rate of heat flowing from one face to the other face in time \(t\) is given by
\(
\frac{d Q}{d t}=\frac{K A\left(T_1-T_2\right)}{L}
\)

Hint

The relation between true scale and new scale of temperature is given by
\(
\begin{aligned}
& \left(\frac{t-L F P}{U F P-L F P}\right)_{\text {true }}=\left(\frac{t-L F P}{U F P-L F P}\right)_{\text {faulty }} \\
& \frac{39^{\circ} \mathrm{C}-0^{\circ} \mathrm{C}}{100^{\circ} \mathrm{C}-0^{\circ} \mathrm{C}}=\frac{t-39^{\circ} \mathrm{W}}{239^{\circ} \mathrm{W}-39^{\circ} \mathrm{W}} \\
\Rightarrow \quad & t=117^{\circ} \mathrm{W}
\end{aligned}
\)

Where L.F.P \(\Rightarrow\) Lower Fixed point
U.H.F. \(\Rightarrow\) Upper fixed point

Hint

(a) Heat required to raise the temperature of \(1 \mathrm{~kg}\) water from \(20^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) is given by \(\mathrm{Q}=\mathrm{ms} \Delta \theta=(1 \times 4200 \times 80) \mathrm{J}\)
Power of kettle \((P)=V I=(220 \times 4) \mathrm{W}\)
\(
\begin{aligned}
\therefore \text { Time taken } & =\frac{Q}{P}=\frac{1 \times 4200 \times 80}{220 \times 4} \\
& =381.81 \mathrm{sec}=6.36 \mathrm{~min}
\end{aligned}
\)

Hint

(a) According to Stefan’s law,
\(\mathrm{E} \propto T^4\)
\(\mathrm{E} \propto(t+273)^4 K\left[727^{\circ} \mathrm{C}=(727+273) K\right]\)
\(\mathrm{E} \propto(727+273)^4 K\)
\(\mathrm{E} \propto(1000)^4 K\)

Hint

(c) Power radiated by the sun at \(t^{\circ} \mathrm{C}\) \(=\sigma(t+273)^4 4 \pi r^2\)
Power received by a unit surface
\(
=\frac{\sigma(t+273)^4 4 \pi r^2}{4 \pi R^2}=\frac{r^2 \sigma(t+273)^4}{R^2}
\)

Hint

(d) Since \(T_n=\frac{T_i+T_c}{2}=\) Neutral temperature
\(
\begin{aligned}
T_n & =\frac{T_i+0}{2}=\frac{T_i}{2} \\
{\left[T_c\right.} & \left.=0^{\circ} \mathrm{C}=\text { temperature of cold junction }\right]
\end{aligned}
\)

Hint

According to Wien’s law \(\lambda_m T=\) constant (say b)
where, \(\lambda_m\) is wavelength corresponding to maximum intensity of radiation and \(T\) is temperatures of the body in kelvin.
So for two different cases i.e. at two different temperature of body
\(
\therefore \quad \frac{\lambda_{m^{\prime}}^{\prime}}{\lambda_m}=\frac{T}{T^{\prime}}
\)
Given, \(\quad T=1227+273=1500 \mathrm{~K}\),
\(T^{\prime}=1227+1000+273\)
\(=2500 \mathrm{~K}\)
\(\lambda_m=5000\)
Hence, \(\lambda_m^{\prime}=\frac{1500}{2500} \times 5000=3000 \)

Hint

(b) From given option
(i) \(r=2 r_0, l=2 l_0\)
\(\therefore R=\frac{2 \ell_0}{K \pi\left(2 r_0\right)^2}=\frac{\ell_0}{2 K \pi r_0^2}\)
(ii) \(r=2 r_0, l=l_0\)
\(
\therefore R=\frac{\ell_0}{K \pi\left(2 r_0\right)^2}=\frac{\ell_0}{4 K \pi r_0^2}
\)
(iii) \(r=r_0, l=2 l_0\)
\(
\therefore R=\frac{2 \ell_0}{K \pi r_0^2}=\frac{2 \ell_0}{K \pi r_0^2}
\)
(iv) \(r=r_0, l=l_0\)
\(
\therefore R=\frac{\ell_0}{K \pi r_0^2}=\frac{\ell_0}{K \pi r_0^2}
\)
It is clear that for option (b) resistance is minimum, hence, heat flow will be maximum.
(i) Rate of heat flow is directly proportional to area
(ii) inversely proportional to length.
\(\therefore\) Heat flow will be maximum when \(r\) is maximum and \(\ell\) is minimum.

Note: We know that \(Q=\frac{T_H-T_L}{R}\) Also, Thermal resistance \(R=\frac{\ell}{K A}=\frac{\ell}{K \pi r^2}\)
Heat flow will be maximum when thermal resistance is minimum.

Hint

\(
\begin{aligned}
& \text { (c) } \theta_n=\frac{\theta_c+\theta_i}{2} \\
& \therefore \theta_c=2 \theta_n-\theta_i=2(300)-620=-20^{\circ} \mathrm{C}
\end{aligned}
\)

Alternate

As we know the temperature of inversion is higher than the neutral temperature by the same amount as the neutral temperature is higher than the temperature of cold junction.
Hence,
\(t_n-t_0=t_i-t_n \Rightarrow t_0=2 t_n-t_i\)
Given \(t_i=620^{\circ} \mathrm{C}, t_n=300^{\circ} \mathrm{C}\)
Hence, \(t_0=\) temperature of cold junction
\(2 \times 300-620=600-620=-20^{\circ} \mathrm{C}\).

Hint

(a) From Wein’s displacement law
\(
\begin{aligned}
& \lambda_m T=\text { constant } \\
& \Rightarrow \lambda_m \propto T^{-1}
\end{aligned}
\)

Hint

The heat radiation emitted by the human body is the infrared radiation. Their wavelength is of the order of \(7.9 \times 10^{-7} \mathrm{~m}\) to \(10^{-3} \mathrm{~m}\) which is of course the range of infrared region.

Note:

Not only the human body, but everything emits radiation. But the wavelength of this radiation depends on temperature.

The wavelength actually comes from the frequency of atom vibrations. If atom is hot it moves faster and generates higher frequencies of electromagnetic waves (shorter wavelength).

All atoms in your body have the temperature of your body and emit wavelength in your body frequency. All atoms outside your body also emit wavelength of its temperature. For human body, the wavelength lies in infrared region.

Hint

(a) In series, equivalent thermal conductivity
\(
\begin{aligned}
& K_{\text {eq }}=\frac{2 K_1 K_2}{K_1+K_2} \\
& \text { or, } K_{\text {eq }}=\frac{2 \times K \times 2 K}{K+2 K}=\frac{4}{3} K
\end{aligned}
\)

Note:
In electricity equivalent resistance in series grouping \(R_s=R_1+R_2+R_3 \ldots . .+R_n\)
In heat-thermal resistance in series \(R_s=R_1+R_2+\)
\(
\begin{aligned}
& R_3 \ldots . .+R_n \\
& \frac{\ell_1+\ell_2+\ldots . \ell_n}{K s}=\frac{\ell_1}{K_1 A}+\frac{\ell_2}{K_2 A}+\ldots . .+\frac{\ell_n}{K_n A}
\end{aligned}
\)
For \(n\) no. of rods of equal length
\(
K_s=\frac{n}{\frac{1}{K_1}+\frac{1}{K_2} \ldots . .+\frac{1}{K_n}}
\)

Hint

Boltzmann corrected Stefan’s law and stated that the amount of radiations emitted by the body, not only depends upon the temperature of the body but also on the temperature of the surrounding. The radiated power by the body is given by
\(
P=\sigma\left(T^4-T_0^4\right) \quad \text {…(i) }
\)
where \(T_0\) is the absolute temperature of the surrounding and \(T\) is the temperature of body.
So for two different cases ratio of radiation power is given by
\(
\therefore \quad \frac{P_2}{P_1}=\left(\frac{T_2{ }^4-T_0{ }^4}{T_1{ }^4-T_0{ }^4}\right) \dots(ii)
\)
Here, \(P_1=60 \mathrm{~W}, T_1=727^{\circ} \mathrm{C}=1000 \mathrm{~K}\)
\(
\begin{aligned}
& T_0=227^{\circ} \mathrm{C}=500 \mathrm{~K} \\
& T_2=1227^{\circ} \mathrm{C}=1500 \mathrm{~K}
\end{aligned}
\)
Substituting in Eq. (ii), we get
\(
\begin{aligned}
P_2 & =\frac{(1500)^4-(500)^4}{(1000)^4-(500)^4} \times 60 \\
& =\frac{(500)^4}{(500)^4} \times\left[\frac{3^4-1}{2^4-1}\right] \times 60 \\
& =\frac{80}{15} \times 60=320 \mathrm{~W}
\end{aligned}
\)

Hint

(b) Rate of heat flow for one rod \(=\frac{K_1 A_1\left(T_1-T_2\right)}{d}(d \rightarrow\) Length \()\) Rate of heat flow for other rod \(=\frac{K_2 A_2\left(T_1-T_2\right)}{d}\)
In steady state, \(\frac{K_1 A_1\left(T_1-T_2\right)}{d}\) \(=\frac{K_2 A_2\left(T_1-T_2\right)}{d} \Rightarrow K_1 A_1=K_2 A_2\)

Hint

(d) According to Wein’s displacement law, the product of wavelength belonging to maximum intensity and temperature is constant i.e., \(\lambda_m T=\) constant.

Hint

Materials like black velvet or lamp black come close to ideal black bodies, but the best practical realization of an ideal black body is a small hole leading into a cavity maintained at constant temperature as this absorbs \(98 \%\) of the radiation incident on them. Cavity approxi- mating an ideal black body is shown in the figure. Radiation entering the cavity has little chance of leaving before it is completely absorbed.

Hint

Wien’s displacement law is given by
\(\begin{aligned} \lambda_m T & =\text { constant } \\ \lambda_m & =\text { maximum wavelength radiation } \\ T & =\text { temperature of the body }\end{aligned}\)
So for two different cases, i.e. at two different temperatures
\(
\text { or } \quad \begin{aligned}
\lambda_1 T_1 & =\lambda_2 T_2 \\
\lambda_2 & =\lambda_1\left(\frac{T_1}{T_2}\right)
\end{aligned}
\)
Given, \(T_1=2000 \mathrm{~K}, T_2=3000 \mathrm{~K}, \lambda_1=\lambda\)
\(
\therefore \quad \lambda_2=\lambda \times \frac{2000}{3000}=\frac{2}{3} \lambda
\)

Hint

\(
\begin{aligned}
& \text { Rate of heat flow }\left(\frac{Q}{t}\right)=\frac{k \pi r^2\left(\theta_1-\theta_2\right)}{L} \propto \frac{r^2}{L} \\
& \therefore \frac{Q_1}{Q_2}=\left(\frac{r_1}{r_2}\right)^2\left(\frac{l_2}{l_1}\right)=\left(\frac{1}{2}\right)^2 \times\left(\frac{2}{1}\right)=\frac{1}{2} \\
& \Rightarrow Q_2=2 Q_1 .
\end{aligned}
\)

Hint

(c) Natural convection
In convection, the temperature gradient exists in the vertical direction and not in the horizontal direction. So, up and down movement of particles takes place which depends on the weight or gravity.

Hint

Heat required by \(1 \mathrm{~g}\) ice at \(0^{\circ} \mathrm{C}\) to melt into \(\lg\) water at \(0^{\circ} \mathrm{C}\),
\(
\begin{aligned}
Q_1 & =m L(L=\text { latent heat of fusion }) \\
& =1 \times 80=80 \mathrm{cal} \quad(L=80 \mathrm{cal} / \mathrm{g})
\end{aligned}
\)
Heat required by \(1 \mathrm{~g}\) of water at \(0^{\circ} \mathrm{C}\) to boil at \(100^{\circ} \mathrm{C}\),
\(
\begin{aligned}
Q_2= & m c \Delta \theta \\
& \quad(c=\text { specific heat of water }) \\
& =1 \times 1(100-0) \quad\left(c=1 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C}\right) \\
& =100 \mathrm{cal}
\end{aligned}
\)
Thus, total heat required by \(1 \mathrm{~g}\) of ice to reach a temperature of \(100^{\circ} \mathrm{C}\),
\(
Q=Q_1+Q_2=80+100=180 \mathrm{cal}
\)
Heat available with \(1 \mathrm{~g}\) of steam to condense into \(\mathrm{lg}\) of water at \(100^{\circ} \mathrm{C}\), \(Q^{\prime}=m L^{\prime}\left(L^{\prime}=\right.\) latent heat of
vaporisation)
\(
\begin{aligned}
& =1 \times 536 \mathrm{cal} \quad\left(L^{\prime}=536 \mathrm{cal} / \mathrm{g}\right) \\
& =536 \mathrm{cal}
\end{aligned}
\)
Obviously, the whole steam will not be condensed and ice will attain temperature of \(100^{\circ} \mathrm{C}\). Thus, the temperature of mixture is \(100^{\circ} \mathrm{C}\).

Hint

Apply Stefan’s law
According to Stefan’s law, the rate at which an object radiates energy is proportional to the fourth power of its absolute temperature, i.e.
\(E=\sigma T^4\) or \(E \propto T^4\) ( \(\sigma=\) Stefan’s constant)
Given, \(T_1=T, T_2=2 T, E_1=20 \mathrm{kcal} / \mathrm{m}^2 \mathrm{~min}\)
\(
\begin{aligned}
& \therefore \frac{20}{E_2}=\left(\frac{T}{2 T}\right)^4 \text { or } \frac{20}{E_2}=\frac{1}{16} \\
& \therefore \quad E_2=20 \times 16=320 \mathrm{kcal} / \mathrm{m}^2 \mathrm{~min}
\end{aligned}
\)

Hint

According to Stefan’s Law \(E=\sigma e A T^4\) \(E \propto T^4 ;\) so, \(E \propto(500)^4\)

Hint

(c) Let \(\theta_0\) be the temperature of the surrounding. Then
\(
\frac{80-75}{t_1}=k\left(\frac{80+75}{2}-\theta_0\right)
\)
or, \(\frac{5}{t_1}=k\left(77.5-\theta_0\right)\)
or, \(\quad t_1=\frac{5}{k\left(77.5-\theta_0\right)} \dots(1)\)
Similarly, \(t_2=\frac{5}{k\left(72.5-\theta_0\right)} \dots(2)\)
and \(t_3=\frac{5}{k\left(67.5-\theta_0\right)} \dots(3)\)
From (1), (2) \& (3), it is obvious that \(t_1<t_2<t_3\)

Hint

(b) Rate of cooling \(\propto\) temperature difference between system and surrounding.
As the temperature difference is halved, so the rate of cooling will also be halved.
So time taken will be doubled
\(
t=2 \times 5 \mathrm{sec} .=10 \mathrm{sec} \text {. }
\)

Hint

According to Stefan-Boltzmann law, amount of heat energy \((E)\) radiated per second by unit area of a body is directly proportional to the fourth power of absolute temperature (T) of the body i.e. \(E \propto T^4\) or \(E=\sigma T^4\)
If \(T\) is doubled, \(E\) becomes \((2)^4\) times (i.e. 16 times).

Hint

(c) Mercury thermometer is based on the principle of change of volume with rise of temperature and can measure temperatures ranging from \(-30^{\circ} \mathrm{C}\) to \(357^{\circ} \mathrm{C}\).

Hint

\(
\begin{aligned}
& \text { (c) Using } \frac{\mathrm{F}-32}{180}=\frac{\mathrm{C}}{100} \\
& \Rightarrow \frac{140-32}{180}=\frac{\mathrm{C}}{100} \\
& \Rightarrow \mathrm{C}=60
\end{aligned}
\)
Temperature of boiling water \(=100^{\circ} \mathrm{C}\)
We get, fall in temperature \(=100-60=40^{\circ} \mathrm{C}\)

Hint

(d) Thermal capacity \(=\mathrm{ms}=40 \times 0.2=8 \mathrm{cal} /{ }^{\circ} \mathrm{C}\) \(=4.2 \times 8 \mathrm{~J}=33.6\) joules \(/{ }^{\circ} \mathrm{C}\)

Hint

Let \(\theta\) be the temperature when thermal equilibrium has reached.
Heat gained by ice to be converted to water at \(\theta^{\circ} \mathrm{C}\)
\(
\begin{aligned}
& =m L+m \times s \times(\theta-0) \\
& =10 \times 80+10 \times 1 \times \theta
\end{aligned}
\)
Heat lost by tumbler and its contents
\(
=55 \times(40-\theta)
\)
Using principle of calorimetry that heat gained \(=\) heat lost
\(
\begin{aligned}
10 \times 80 & +10 \times 1 \times \theta=55 \times(40-\theta) \\
65 \theta & =2200-800=1400 \\
\theta & =\frac{1400}{65} \approx 21.5^{\circ} \mathrm{C} \approx 22^{\circ} \mathrm{C}
\end{aligned}
\)

Hint

(b)
Vapour pressure of a substance is independent of amount of substance. It depends only on temperature. So they have ratio of \(1: 1\).

Hint

\(
\begin{array}{ll}
v_{\mathrm{ms}}=\sqrt{\frac{3 R T}{m}} & T_1=273-50 \\
v_{\mathrm{ms}} \propto \sqrt{T} & =223 \mathrm{~K} \\
V_{\mathrm{rms}} \text { is increased by } 3 \text { times } & T_2=?
\end{array}
\)
So, final rms speed \(=v+3 v=4 v\)
\(
\begin{aligned}
& \frac{v_1}{v_2}=\sqrt{\frac{T_1}{T_2}} \\
& \frac{v}{4 V}=\sqrt{\frac{223}{T_2}} \Rightarrow \frac{1}{16}=\frac{223}{T_2} \\
& T_2=3568 \mathrm{~K} \\
& T_2=3568-273=3295^{\circ} \mathrm{C}
\end{aligned}
\)