Current Electricity MCQs for NEET — Physics Questions with Answers

(c) When a cell is balanced against potential drop across a certain length of potentiometer wire, no current flows through the cell 

$\therefore$emf of cell=potential drop across balance length of potentiometer wire. So, a potentiometer is a more accurate device for measuring emf of a cell or no current flows through the cell during the measurement of emf.

(d) Given, charge $Q= at-b{t}^2,$      ...(i) 

$\because$ We know that current, $I=\frac{dQ}{dt}$

So, eq(i) can be written as

              $I=\frac{d}{dt}\left(at-b{t}^2\right)$

             $I=a-2bt$       ...(ii)

For maximum value of t, till the current exist is given by

$\Rightarrow a-2bt=0$ 

            $t=\frac{a}{2b}$            ...(iii)

$\because$ The total heat produced (H) can be given as

$$\begin{gathered} H={\int }_0^t I^{2}R dt \\ ={\int }_0^{a/2b}{\left(a-2bt\right)}^{2}R.dt \left\{\because t=\frac{a}{2b}\right\} \\ ={\int }_0^{a/2b}\left(a^2+4b^2{t}^2-4abt\right)Rdt \\ H={\left[a^{2}t+4b^2\frac{t^3}{3}-\frac{4ab{t}^2}{2}\right]}_0^{a/2b}R \end{gathered}$$

solving above equation we get,

    $\Rightarrow H=\frac{a^{3}R}{6b}$

$$\begin{gathered} Potential drop in the potentiometer wire=400\times 1\times {10}^{-3}=0.4V \\ Potential drop in the resis\tan ce to be connecteed in series=1.6V \\ Current in the potentiometer wire=\frac{0.4}{8}A \\ Resis\tan ce=\frac{1.6}{0.4}\times 8=32 ohm \end{gathered}$$

In a potentiometer circuit, the unknown EMF E is given by the ratio of the potential drop across the length l of the potentiometer wire to the total potential drop across the entire wire. This is given by the formula: E = (E_o * r * l) / ((r + r_1) * L), where E_o is the EMF of the battery.

Potential difference (drop) between two cities=150x8=1200V

Average resistance of total wire=0.5x150=75Ω  

Power loss=P=V²/R

=1200x1200/75

=19200 W

=19.2 kW

Given, e=2V and l=4m

Potential drop per unit length

φ=e/l=2/4=0.5V/m

$$\begin{gathered} For the first case, \\ e\text{'}=\phi l_1 ...\left(i\right) \\ For the second case, \\ V=\phi l_2 ...\left(ii\right) \\ From Eqs. \left(i\right) and \left(ii\right), \\ e\text{'}/V=l_1/l_2 \\ e\text{'}=l(r+R) and V=IR for the second case \\ So, r=R(l_1/l_2-1)=9.5(3/2.85-1)=9.5(1.05-1)=9.5x0.05=0.475=0.5\Omega \end{gathered}$$

(d) As R'=n²R=2² x 4=4 x 4=16Ω 

(b) As I=E/R+r or E=I(R+r)             

2.1=0.2(10+r) 

10+r=2.1/2 x 10

∴  r=10.5-10=0.5Ω

Power $P=\frac{V^2}{R}$

For small variation

$\frac{∆P}{P}\times 100\%=\frac{2\times ∆V}{V}\times 100\%$

                   $=2\times 2.5=5\%$

Therefore, power would decrease by 5%.

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

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