भौतिक विज्ञान (Physics) NEET प्रश्न हिंदी में — उत्तर सहित MCQ अभ्यास

$$\begin{gathered} 2. \mathrm{Molecular} \mathrm{mass} \mathrm{of} \mathrm{He} ; \mathrm{M}=4 \mathrm{gm} \\ \Rightarrow \mathrm{Molar} \mathrm{value} \mathrm{of} {\mathrm{C}}_{\mathrm{V}}={\mathrm{Mc}}_{\mathrm{V}}=4\times 3=12\frac{\mathrm{J}}{\mathrm{mole}-\mathrm{kelvin}} \\ \mathrm{At} \mathrm{constant} \mathrm{volume} \mathrm{P}\propto \mathrm{T} \mathrm{therefore} \mathrm{on} \mathrm{doubling} \mathrm{the} \mathrm{pressure} \mathrm{temperature} \mathrm{also} \mathrm{doubles} \\ \mathrm{i}.\mathrm{e}., {\mathrm{T}}_2=2{\mathrm{T}}_1 \Rightarrow ∆\mathrm{T}={\mathrm{T}}_2-{\mathrm{T}}_1 =273 \mathrm{K} \\ \mathrm{Also} {\left(∆\mathrm{Q}\right)}_{\mathrm{V}}={\mathrm{\mu C}}_{\mathrm{V}}∆\mathrm{T}=\frac{1}{2}\times 12\times 273=1638 \mathrm{J} \end{gathered}$$

$$\begin{gathered} 1. \left(\mathrm{P}+\frac{{\mathrm{aT}}^2}{\mathrm{V}}\right){\mathrm{V}}^{\mathrm{c}}=\left(\mathrm{RT}+\mathrm{b}\right) \\ \Rightarrow \mathrm{P}=\left(\mathrm{RT}+\mathrm{b}\right){\mathrm{V}}^{-\mathrm{c}} -\left({\mathrm{aT}}^2\right){\mathrm{V}}^{-1} \\ \mathrm{Comparing} \mathrm{this} \mathrm{equation} \mathrm{with} \mathrm{P}={\mathrm{AV}}^{\mathrm{m}}-{\mathrm{BV}}^{\mathrm{n}} \\ \mathrm{we} \mathrm{get} \mathrm{m}=-\mathrm{c} \mathrm{and} \mathrm{n}=-1 \end{gathered}$$

2. गैस अणुओं की औसत चाल $\sqrt{\frac{8 \mathrm{kT}}{\mathrm{\pi } \mathrm{m}}}.$ है। यह तापमान तथा अणु द्रव्यमान पर निर्भर करती है। अतः

  ${\mathrm{O}}_2$ की औसत चाल (A) तथा (C) में समान होगी।

$$\begin{gathered} 4. \mathrm{From} \mathrm{Dalton}\text{'}\mathrm{s} \mathrm{law} \mathrm{final} \mathrm{pressure} \mathrm{of} \mathrm{the} \mathrm{mixture} \mathrm{of} \mathrm{nitrogen} \mathrm{and} \mathrm{oxygen} \\ {\mathrm{P}}_{\mathrm{mix}}={\mathrm{P}}_1+{\mathrm{P}}_2=\frac{{\mathrm{\mu }}_1\mathrm{RT}}{\mathrm{V}}+\frac{{\mathrm{\mu }}_2\mathrm{RT}}{\mathrm{V}} \\ =\frac{{\mathrm{m}}_1}{{\mathrm{M}}_1}\frac{\mathrm{RT}}{\mathrm{V}}+\frac{{\mathrm{m}}_2}{{\mathrm{M}}_2}\frac{\mathrm{RT}}{\mathrm{V}} =\frac{8}{32}\frac{\mathrm{RT}}{\mathrm{V}}+\frac{7}{28}\frac{\mathrm{RT}}{\mathrm{V}} =\frac{\mathrm{RT}}{2 \mathrm{V}} \\ \Rightarrow 10=\frac{\mathrm{RT}}{2 \mathrm{V}} ........\left(\mathrm{i}\right) \\ \mathrm{When} \mathrm{oxygen} \mathrm{is} \mathrm{absorbed} \mathrm{then} \mathrm{for} \mathrm{nitrogen} \mathrm{let} \mathrm{pressure} \mathrm{is} \mathrm{P}=\frac{7}{28}\frac{\mathrm{RT}}{\mathrm{V}} \\ \Rightarrow \mathrm{P}=\frac{\mathrm{RT}}{4 \mathrm{V}} ..........\left(\mathrm{ii}\right) \\ \mathrm{From} \mathrm{equation} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right) \mathrm{we} \mathrm{get} \mathrm{pressure} \mathrm{of} \mathrm{the} \mathrm{nitrogen} \mathrm{P}=5 \mathrm{atm}. \end{gathered}$$

$$\begin{gathered} 1. \mathrm{When} \mathrm{inter} \mathrm{molecular} \mathrm{forces} \mathrm{vanish} \mathrm{away} \mathrm{the} \mathrm{water} \mathrm{will} \mathrm{act} \mathrm{as} \mathrm{an} \mathrm{ideal} \mathrm{gas}. \\ \mathrm{The} \mathrm{volume} \mathrm{of} \mathrm{one} \mathrm{mole} \mathrm{of} \mathrm{water} \mathrm{at} \mathrm{S}.\mathrm{T}.\mathrm{P}. =22.4 \mathrm{litre} \\ \mathrm{We} \mathrm{know} \mathrm{that} 1 \mathrm{mole} \mathrm{of} \mathrm{water}=18 \mathrm{gm} \\ \mathrm{So} \mathrm{volume} \mathrm{of} 4.5 \mathrm{kg} \mathrm{of} \mathrm{water} =\frac{22.4\times 4.5\times {10}^3}{18}=5.6\times {10}^3 \mathrm{litre} =5.6 {\mathrm{m}}^3 \end{gathered}$$

$$\begin{gathered} 3. \mathrm{According} \mathrm{to} \mathrm{Boyle}\text{'}\mathrm{s} \mathrm{law} {\left({\mathrm{P}}_1{\mathrm{V}}_1\right)}_{\mathrm{At} \mathrm{top} \mathrm{of} \mathrm{the} \mathrm{lake}}={\left({\mathrm{P}}_2{\mathrm{V}}_2\right)}_{\mathrm{At} \mathrm{the} \mathrm{bottom} \mathrm{of} \mathrm{the} \mathrm{lake}} \\ \Rightarrow {\mathrm{P}}_1{\mathrm{V}}_1=\left({\mathrm{P}}_1+\mathrm{h}\right){\mathrm{V}}_2 \\ \Rightarrow 10\times \frac{4}{3}\mathrm{\pi }{\left(\frac{5\mathrm{\pi }}{4}\right)}^3 =\left(10+\mathrm{h}\right)\times \frac{4}{3}{\mathrm{\pi r}}^3 \\ \Rightarrow \mathrm{h}=\frac{610}{64} =9.53 \mathrm{m} \end{gathered}$$

$$\begin{gathered} 4. \mathrm{Escape} \mathrm{velocity} \mathrm{from} \mathrm{the} \mathrm{earth}\text{'}\mathrm{s} \mathrm{surface} \mathrm{is} 11.2 \mathrm{km}/\sec . \\ \mathrm{So}, {\mathrm{v}}_{\mathrm{rms}}={\mathrm{v}}_{\mathrm{escape}}=\sqrt{\frac{3\mathrm{RT}}{\mathrm{M}}} \\ \Rightarrow \mathrm{T}=\frac{{\left({\mathrm{V}}_{\mathrm{escape}}\right)}^2\times \mathrm{M}}{3\mathrm{R}} \\ =\frac{{\left(11.2\times {10}^3\right)}^2\times \left(2\times {10}^{-3}\right)}{3\times 8.31} =10063 \mathrm{K} \end{gathered}$$

$$\begin{gathered} 4. \mathrm{Total} \mathrm{internal} \mathrm{energy} \mathrm{of} \mathrm{system} ={\mathrm{U}}_{\mathrm{oxygen}}+{\mathrm{U}}_{\mathrm{argon}}={\mathrm{\mu }}_1\frac{{\mathrm{f}}_1}{2}\mathrm{RT} + {\mathrm{\mu }}_2\frac{{\mathrm{f}}_2}{2}\mathrm{RT} \\ = 2\frac{5}{2}\mathrm{RT} + 4\frac{3}{2}\mathrm{RT} \\ = 5 \mathrm{RT} + 6 \mathrm{RT} \\ = 11 \mathrm{RT} \left[\mathrm{As} {\mathrm{f}}_1=5 \left(\mathrm{for} \mathrm{oxygen}\right) \mathrm{and} {\mathrm{f}}_2=3\left(\mathrm{for} \mathrm{argon}\right)\right] \end{gathered}$$

$$\begin{gathered} 3. \mathrm{At} \mathrm{T} \mathrm{K}, \mathrm{pressure} \mathrm{of} \mathrm{gas} \left(\mathrm{P}\right) \mathrm{in} \mathrm{the} \mathrm{jar} \mathrm{pressure} = \mathrm{Total} \mathrm{pressure} - \mathrm{saturated} \mathrm{vapour} \mathrm{pressure} \\ \Rightarrow \mathrm{P}=\left(830-30\right)=800 \mathrm{mm} \mathrm{of} \mathrm{Hg} \\ \mathrm{New} \mathrm{temperature} \mathrm{T}\text{'}=\left(\mathrm{T}-\frac{\mathrm{T}}{100}\right)=\frac{99 \mathrm{T}}{100} \\ \mathrm{Using} \mathrm{Charle}\text{'}\mathrm{s} \mathrm{law} \frac{\mathrm{P}}{\mathrm{T}}=\frac{\mathrm{P}\text{'}}{\mathrm{T}} \Rightarrow \mathrm{P}\text{'}=\frac{\mathrm{PT}\text{'}}{\mathrm{T}} \\ =\frac{800\times 99 \mathrm{T}}{100 \mathrm{T}}= 792 \mathrm{mm} \mathrm{of} \mathrm{Hg} \\ \mathrm{Saturated} \mathrm{vapour} \mathrm{pressure} \mathrm{at} \mathrm{T}\text{'}=25 \mathrm{mm} \mathrm{of} \mathrm{Hg} \\ \therefore \mathrm{Total} \mathrm{pressure} \mathrm{in} \mathrm{the} \mathrm{jar} = \mathrm{Actual} \mathrm{pressure} \mathrm{of} \mathrm{gas} + \mathrm{Saturated} \mathrm{vapour} \mathrm{pressure} \\ = 792 + 25 = 817 \mathrm{mm} \mathrm{of} \mathrm{Hg}. \end{gathered}$$

$$\begin{gathered} 3. \mathrm{Given} \mathrm{R} = 8.3\frac{\mathrm{Joule}}{\mathrm{mol} \mathrm{K}}=\frac{8.3}{4.2}\frac{\mathrm{cal}}{\mathrm{mol} \mathrm{K}}=2\frac{\mathrm{cal}}{\mathrm{mol} \mathrm{K}} \\ {\mathrm{C}}_{\mathrm{V}}={\mathrm{C}}_{\mathrm{P}}-\mathrm{R}=\left(7.2-2\right) \approx 5 \mathrm{cal}/\mathrm{mol} \mathrm{l}°\mathrm{C} \\ \mathrm{At} \mathrm{constant} \mathrm{volume} \mathrm{required} \mathrm{heat} {\left(∆\mathrm{Q}\right)}_{\mathrm{V}}={\mathrm{\mu C}}_{\mathrm{V}}∆\mathrm{T}=5\times 5\times 10=250 \mathrm{cal} \end{gathered}$$