NEET Practice Questions (MCQs) with Answers & Solutions

current i= $\frac{\mathrm{E}}{\mathrm{R}+\mathrm{r}}$

           2=$\frac{\mathrm{E}}{2+\mathrm{r}}$                  ...(i)  

         0.5=$\frac{\mathrm{E}}{9+\mathrm{r}}$                 ...(ii)

From Eqs. (i) and (ii), We have

               $$\begin{gathered} \frac{2}{0.5}=\frac{9+\mathrm{r}}{2+\mathrm{r}} \\ 4=\frac{9+\mathrm{r}}{2+\mathrm{r}} \\ 3\mathrm{r}=1 \\ \mathrm{r}=\frac{1}{3}\mathrm{\Omega } \end{gathered}$$

Metallic bonding is formed due to attraction of valence (free) electrons with the positive ion cores.Their conductivity decreases with rise of temperature.When visible light falls on a metallic crystal, the electrons of atom absorb visible light, so they are opaque to visible light.However, some orbital state.They then return to their normal state, remitting light of same frequency.

 Kirchhoff's first law follows from the conservation of charge.

 Kirchhoff's second law follow from the conservation of energy.

According to Ohm's law

          $\frac{dV}{dI}=-r$

and $V=\epsilon if I=0 \left[As V+Ir=\epsilon \right]$

$\therefore Slope of the graph=-r and intercept=\epsilon$

Key Idea: In streching, specific resistance remains unchanged.  After streching,  specific resistance$\left(\rho \right)$ will remain same. original resistance of the wire, 

             $R=\frac{pl}{A}$

or   $R\propto \frac{l}{A} or R\propto \frac{l^2}{V} (as V=Al)$

and $R^{\text{'}}\propto \frac{{\left(l+10\%l\right)}^2}{V}$

Therefore,$\frac{R^{\text{'}}}{R}=\frac{{\displaystyle {\left(l+{\displaystyle \frac{10}{100}}l\right)}^2}}{l^2}$

or  $\frac{R^{\text{'}}}{R}=\frac{\left(\frac{{\displaystyle {\left(11l\right)}^2}}{10}\right)}{l^2}$=$\frac{121}{100}$

or $R^{\text{'}}=1.21 R$

Heat evovled due to Joule's effect is used up in boiling water. 

    VIt=ms$∆t$

or    t=$\frac{msVt}{VI}$

putting under given values 

I=4 A, V=220 volt, m= 1 kg 

$$\begin{gathered} ∆t={\left(100-20\right)}^{°}C,s=4200J/k{g}^{°}C \\ \therefore t=\frac{1\times 4200\times 80}{220\times 4}=6.3 min \end{gathered}$$

This problem is based on the application of potentiometer in which we find the internal resistance of a cell. In potentiometer experiment in which we find internal resistance of a cell, let E be the emf of the cell and V the terminal potential difference, then $\frac{E}{V}=\frac{l_1}{l_2}$   

where $l_1 and l_2$ are lengths of potentiometer wire with and without short circuited through a resistance. 

since, $$\begin{gathered} \frac{E}{V}=\frac{R+r}{R}\left[\therefore E=I\left(R+r\right)and V=IR\right] \\ \therefore \frac{R+r}{R}=\frac{I_2}{I_2} \\ or 1+\frac{r}{R}=\frac{110}{100} \\ or \frac{r}{R}=\frac{110}{100}-1 \\ or r=\frac{1}{10}\times 10=1\Omega \end{gathered}$$

${\mathrm{i}}_{\mathrm{g}}=\frac{\mathrm{V}}{\mathrm{G}+\mathrm{R}} \mathrm{or} {10}^{-3}=\frac{2.5}{10+\mathrm{R}} \mathrm{or} \mathrm{R}=2490 \mathrm{\Omega }$

$S=\frac{i_{g}G}{(I-l_g)}=\frac{0.01\times 1}{1-0.01}=\frac{1}{99}\Omega$

The current through the galvanometer producing full-scale deflection is

$I=\frac{V}{R}=\frac{20\times {10}^{-3}}{80}=2.5\times {10}^{-4} A$

To convert the galvanometer into a voltmeter, a high resistance is connected in series with the galvanometer. Therefore, $5 V=(2.5\times {10}^{-4})(R+80) or R=19.92 k\Omega$