Thermal Properties of Matter MCQs for NEET — Physics Questions with Answers

(c) Heat transferred in one minute is utilised in melting the ice so, $\frac{\mathrm{KA}\left({\mathrm{\theta }}_1-{\mathrm{\theta }}_2\right)\mathrm{t}}{\mathrm{l}}=\mathrm{m}\times \mathrm{L}$

$\Rightarrow \mathrm{m}=\frac{{10}^{-3}\times 92\times \left(100-0\right)\times 60}{1\times 8\times {10}^4}=6.9\times {10}^{-3} \mathrm{kg}$

(d) $\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{KA}}{\mathrm{l}}\mathrm{d\theta }=\frac{0.01\times 1}{0.05}\times 30=6 \mathrm{J}/\sec$

Heat transferred in one day (86400 sec)

$\mathrm{\theta }=6\times 86400=518400 \mathrm{J}$

Now Q = mL$\Rightarrow \mathrm{m}=\frac{\mathrm{Q}}{\mathrm{L}}=\frac{518400}{334\times {10}^3}=1.552 \mathrm{kg}=1552 \mathrm{g}$

(d) Rate of cooling ${\mathrm{R}}_{\mathrm{C}}=\frac{\mathrm{d\theta }}{\mathrm{dt}}=\frac{\mathrm{A\epsilon \sigma }\left({\mathrm{T}}^4-{\mathrm{T}}_0^4\right)}{\mathrm{mc}}$

$\Rightarrow \frac{\mathrm{d\theta }}{\mathrm{dt}}\propto \frac{\mathrm{A}}{\mathrm{V}}\propto \frac{{\mathrm{r}}^2}{{\mathrm{r}}^3}\Rightarrow \frac{\mathrm{d\theta }}{\mathrm{dt}}\propto \frac{1}{\mathrm{r}}$

(b) $\frac{\mathrm{dT}}{\mathrm{dt}}=\frac{\mathrm{\sigma A}}{\mathrm{mcJ}}\left({\mathrm{T}}^4-{\mathrm{T}}_0^4\right)$ [In the given problem fall in temperature of body dT=(200-100)=100 K, temp. of surrounding ${\mathrm{T}}_0$ = 0K, Initial temperature of body T = 200 K]

$$\begin{gathered} \frac{100}{\mathrm{dt}}=\frac{\mathrm{\sigma }4{\mathrm{\pi r}}^2}{{\displaystyle \frac{4}{3}{\mathrm{\pi r}}^3\mathrm{\rho cJ}}}\left({200}^4-0^4\right) \\ \Rightarrow \mathrm{dt}=\frac{\mathrm{r\rho cJ}}{48\mathrm{\sigma }}\times {10}^{-6}\mathrm{s}=\frac{\mathrm{r\rho c}}{\mathrm{\sigma }}.\frac{4.2}{48}\times {10}^{-6} \\ = \frac{7}{80}\frac{\mathrm{r\rho c}}{\mathrm{\sigma }}\mathrm{\mu s}≃\frac{7}{72}\frac{\mathrm{r\rho c}}{\mathrm{\sigma }}\mathrm{\mu s} \left[\mathrm{As} \mathrm{J}=4.2\right] \end{gathered}$$

(c) $\mathrm{Q}=\mathrm{\sigma At}\left({\mathrm{T}}^4-{\mathrm{T}}_0^4\right)$

If T, ${\mathrm{T}}_0$, $\mathrm{\sigma }$ and t are same for both bodies then

$\frac{{\mathrm{Q}}_{\mathrm{sphere}}}{{\mathrm{Q}}_{\mathrm{cube}}}=\frac{{\mathrm{A}}_{\mathrm{sphere}}}{{\mathrm{A}}_{\mathrm{cube}}}=\frac{4{\mathrm{\pi r}}^2}{6{\mathrm{a}}^2}$              ....(i) 

But according to problem, volume of sphere = Volume of cube$\Rightarrow \frac{4}{3}{\mathrm{\pi r}}^3={\mathrm{a}}^3\Rightarrow \mathrm{a}={\left(\frac{4}{3}\mathrm{\pi }\right)}^{1/3}\mathrm{r}$

Substituting the value of a in equation (i) we get 

$$\begin{gathered} \frac{{\mathrm{Q}}_{\mathrm{sphere}}}{{\mathrm{Q}}_{\mathrm{cube}}}=\frac{4{\mathrm{\pi r}}^2}{6{\mathrm{a}}^2}=\frac{4{\mathrm{\pi r}}^2}{6{\left\{{\left({\displaystyle \frac{4}{3}}\mathrm{\pi }\right)}^{1/3}\mathrm{r}\right\}}^2} \\ =\frac{4{\mathrm{\pi r}}^2}{6{\left({\displaystyle \frac{4}{3}}\mathrm{\pi }\right)}^{2/3}{\mathrm{r}}^2}={\left(\frac{\mathrm{\pi }}{6}\right)}^{1/3}:1 \end{gathered}$$

the gas (preferably hydrogen or helium) is enclosed in a glass or fused-quartz bulb connected to a mercury manometer.

According to the KINETIC_THEORY chapter, 'A gas that satisfies Eq. (12.3) [PV = $\mu$RT] exactly at all pressures and temperatures is defined to be an ideal gas.'

The KINETIC_THEORY chapter states, 'Notice that all curves approach the ideal gas behaviour for low pressures and high temperatures.' This is because molecules are far apart and molecular interactions are negligible under these conditions.

The KINETIC_THEORY chapter mentions, 'If we fix $\mu$ and T in Eq. (12.3) [PV = $\mu$RT], we get PV = constant... i.e., keeping temperature constant, pressure of a given mass of gas varies inversely with volume. This is the famous Boyle’s law.'

The KINETIC_THEORY chapter states, 'if you fix P, Eq. (12.1) shows that V $\propto$ T i.e., for a fixed pressure, the volume of a gas is proportional to its absolute temperature T (Charles’ law).'