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

$I-C,II-F,III-A,IV-B$

सूची -i के विकल्पों की जाँच

I. $$\begin{gathered} Potential = \frac{Work}{Charge} \\ Work =Potential X Charge \\ Joule =Coulomb \times Volt \end{gathered}$$

$$\begin{gathered} II. Power ={\left( Current\right)}^2 \times Resis\tan ce \\ Watt = Amper{e}^2 \times Ohm \end{gathered}$$

$$\begin{gathered} III. Voltage \left(V\right)= \frac{Li}{t} \\ where L : Inductor (Unit : Henry) \\ i: Current \\ Putting units, \\ Volt =\frac{Henry \times Ampere}{Second} \end{gathered}$$

IV. $$\begin{gathered} For a capacitor , \\ Q =C\times V \\ Coulomb = Farad X Volt \end{gathered}$$

आवेग = बल × समय $=(kg\text{-}m\text{}/\text{}{s}^2)\times s=kg\text{-}m\text{}/\text{}s$

$$\begin{gathered} V=\frac{Li}{t} \\ L= \frac{Vt}{i} \\ R=\frac{V}{i} \\ Then, \\ \frac{L}{R} =\frac{\frac{Vt}{i}}{\frac{V}{i}} =t \end{gathered}$$

$\frac{L}{R}$ = समय नियतांक

$RC=T$

∵ $\left[R\right]=\left[M{L}^2{T}^{-3}{I}^{-2}\right]$ तथा $\left[C\right]=\left[M^{-1}{L}^{-2}{T}^4{I}^2\right]$

$$\begin{gathered} \mathrm{V}=\mathrm{k}.{\mathrm{g}}^{\mathrm{x}}.{\mathrm{\lambda }}^{\mathrm{y}}.{\mathrm{\rho }}^{\mathrm{z}} \\ \left[{\mathrm{M}}^0{\mathrm{LT}}^{-1}\right]={\left[{\mathrm{LT}}^{-2}\right]}^{\mathrm{x}}.{\left[\mathrm{L}\right]}^{\mathrm{y}}{\left[{\mathrm{ML}}^{-8}\right]}^{\mathrm{z}} \\ \left[{\mathrm{M}}^0{\mathrm{LT}}^{-1}\right]=\left[{\mathrm{M}}^{\mathrm{z}}{\mathrm{L}}^{\mathrm{x}+\mathrm{y}-3\mathrm{z}} {\mathrm{T}}^{-2\mathrm{y}}\right] \\ \mathrm{On} \mathrm{comparing} - \\ \mathrm{z}=0 ......\left(\mathrm{i}\right) \\ \mathrm{x}+\mathrm{y}-3\mathrm{z}=1 \Rightarrow \mathrm{x}+\mathrm{y}=1 ........\left(\mathrm{ii}\right) \\ -2\mathrm{y}=-1 \\ \mathrm{y}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \Rightarrow \mathrm{x}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ \mathrm{V}=\mathrm{k}.\sqrt{\mathrm{g\lambda }} \Rightarrow {\mathrm{V}}^2=\mathrm{k}.\mathrm{g\lambda } \end{gathered}$$

$\rho =\frac{RA}{l}$ अर्थात् प्रतिरोधकता की विमा $\left[M\text{}{L}^3{T}^{-1}{Q}^{-2}\right]$ है  ........(i)

ओम के नियम का उपयोग करते हुए,

V=IR

$$\begin{gathered} R=\frac{V}{I}=\frac{{\displaystyle \frac{Work }{Charge}}}{Current}=\frac{{\displaystyle \frac{M{L}^2{T}^{-2}}{Q}}}{{\displaystyle \frac{Q}{T}}}=\frac{M{L}^2{T}^{-1}}{Q^2} \\ Putting this value of R in \left(i\right) \\ \rho =\frac{M{L}^2{T}^{-1}}{Q^2}\times \frac{L^2}{L} \\ =\frac{M{L}^3{T}^{-1}}{Q^2} \end{gathered}$$

$v\propto g^p{h}^q$ (दिया है)

प्रत्येक राशि की विमा रखकर तथा दोनों ओर की घातों की तुलना करके हमें $\left[L{T}^{-1}\right]={\left[L{T}^{-2}\right]}^p{\left[L\right]}^q$ प्राप्त होता है

$\Rightarrow p+q=1,-2p=-1,\therefore p=\frac{1}{2},q=\frac{1}{2}$

$\frac{R}{L}=\frac{V/I}{V\times T/I}=\frac{1}{T}=$आवृत्ति  

प्रतिरोधकता $\left[\rho \right]=\frac{\left[R\right].\left[A\right]}{\left[l\right]}\text{where }\left[R\right]=\left[M{L}^2{T}^{-1}{Q}^{-2}\right]$

$\therefore \left[\rho \right]=\left[M{L}^3{T}^{-1}{Q}^{-2}\right]$  

$$\begin{gathered} \left[\mathrm{C}\right]=\left[{\mathrm{M}}^0{\mathrm{LT}}^{-1}\right] \\ \left[\mathrm{G}\right]=\left[{\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}\right] \\ \left[\mathrm{h}\right]=\frac{\left[\mathrm{Ed}\right]}{\left[\mathrm{C}\right]}=\left[{\mathrm{ML}}^2{\mathrm{T}}^{-1}\right] \\ \mathrm{Let} \mathrm{m}\propto {\mathrm{C}}^{\mathrm{x}}{\mathrm{G}}^{\mathrm{y}}{\mathrm{h}}^{\mathrm{z}} \\ \mathrm{m}={\mathrm{KC}}^{\mathrm{x}}{\mathrm{G}}^{\mathrm{y}}{\mathrm{h}}^{\mathrm{z}} \\ \left[{\mathrm{ML}}^0{\mathrm{T}}^0\right]=\mathrm{K}{\left[{\mathrm{LT}}^{-1}\right]}^{\mathrm{x}}\times {\left[{\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}\right]}^{\mathrm{y}}\times {\left[{\mathrm{ML}}^2{\mathrm{T}}^{-1}\right]}^{\mathrm{z}} \\ \left[{\mathrm{ML}}^0{\mathrm{T}}^0\right]=\left[{\mathrm{M}}^{-\mathrm{y}+\mathrm{z}}·{\mathrm{L}}^{\mathrm{x}+3\mathrm{y}+2\mathrm{z}}·{\mathrm{T}}^{-\mathrm{x}-2\mathrm{y}-\mathrm{z}}\right] \\ \mathrm{Comparing} \mathrm{the} \mathrm{dimensions}:- \\ -\mathrm{y}+\mathrm{z}=1 ......\left(\mathrm{I}\right) \\ \mathrm{x}+3\mathrm{y}+2\mathrm{z}=0 .......\left(\mathrm{II}\right) \\ -\mathrm{x}-2\mathrm{y}-\mathrm{z}=0 \\ \Rightarrow \mathrm{x}+2\mathrm{y}+\mathrm{z}=0 .......\left(\mathrm{III}\right) \\ \mathrm{From} \mathrm{eq}.\left(\mathrm{II}\right)- \\ \left(\mathrm{x}+2\mathrm{y}+\mathrm{z}\right)+\mathrm{y}+\mathrm{z}=0 \\ 0+\mathrm{y}+\mathrm{z}=0 \\ \Rightarrow \mathrm{Y}=-\mathrm{z} \\ \mathrm{in} \mathrm{eq}.\left(\mathrm{I}\right)- \\ \mathrm{z}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ \mathrm{y}=\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ \mathrm{x}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \end{gathered}$$