NEET Practice Q & A

Hint

(c) During the charging of battery terminal, potential difference is always greater than emf of circuit.
\(
V=E+i r
\)

Hint

(b) Specific resistance of silver, copper and aluminium are \(1.6 \times 10^{-8} \Omega-\mathrm{m}, 1.7 \times 10^{-8} \Omega-\mathrm{m}\) and \(2.7 \times 10^{-8} \Omega-\mathrm{m}\), respectively.
Since, conductivity \((\sigma)=\frac{1}{\text { resistivity }(\rho)}\)
Hence, \(\sigma_{\mathrm{Al}}<\sigma_{\mathrm{Cu}}<\sigma_{\mathrm{Ag}}\)
Thus, correct sequence is \(\mathrm{Al}, \mathrm{Cu}, \mathrm{Ag}\).

Hint

(a) Voltmeter has high resistance. So, most of the main current will flow through ammeter which is in parallel. So, it will burn out. No damage will occur to voltmeter.

Hint

\(
\text { (c) } V=E-i r=E-\left(\frac{E}{R+r}\right) r=E\left(\frac{R}{R+r}\right)=\frac{E}{1+r / R}
\)

Hint

(b) \(\text { In parallel combination of cell } i=\frac{E}{R+\frac{r}{n}}\)

Hint

(b) The minimum resistance can be achieved when we connect all resistances in parallel.
So, equivalent resistance of combination \(=\frac{r}{10}\).

Hint

(b) When wire is bent in the form of a circle, then between two points in any diameter, it is equivalent to two resistances in parallel, \(R_{\mathrm{eq}}=\frac{12}{2}=6 \Omega\)

Hint

(a) is correct.

Explanation:
When a steady current flows through a conductor, it means that the number of positive charges flowing in one direction is equal to the number of negative charges flowing in the opposite direction. This results in a net charge of zero in any segment of the conductor.

Hint

(c) Out of \(n\) cells, two cells will cancel out each other’s emf.
So, net emf \(=(n-2) E\).
Total resistance \(=R+n r\)
Current, \(i=\frac{(n-2) E}{n r+R}\)

Hint

(d) Current, \(i=\frac{E}{r+R}\)
Power, \(P=i^2 R \Rightarrow P=\frac{E^2 R}{(r+R)^2}\)
Power will be maximum, when \(r=R\).
\(
P_{\max }=\frac{E^2}{4 r}
\)

Hint

(b) Slope of the \(V-i\) curve at any point is equal to the reciprocal of the resistance at that point.
From the curve, slope for \(T_1>\) slope for \(T_2\)
\(
R_{T_1}<R_{T_2}
\)
\(\Rightarrow\) Also at higher temperature resistance will be higher, so
\(
T_2>T_1
\)

Hint

(c) Current in the circuit will increase because another resistance is connected in parallel to the circuit and hence potential drop across the ammeter will decrease. So, the potential difference over voltmeter will increase because total potential difference over ammeter and voltmeter is equal to emf (constant).

Hint

(a) Resistance of wire, \(R=\rho L / A\)
On stretching the wire, the volume of the wire remains same.
\(
\Rightarrow \quad \frac{A^{\prime}}{A}=\frac{l}{l^{\prime}}
\)
So, new resistance, \(R^{\prime}=\rho l^{\prime} / A^{\prime}\)
\(
\begin{array}{ll}
\Rightarrow & \frac{R^{\prime}}{R}=\left(\frac{l}{l}\right)\left(\frac{A}{A^{\prime}}\right)=\frac{l^{\prime}}{l} \cdot \frac{l^{\prime}}{l}=\left(\frac{l}{l}\right)^2=\left(\frac{1.1 l}{l}\right)^2=1.21 \\
\therefore & R^{\prime}=1.21 R=1.21 \times 10=12.1 \Omega=12 \Omega
\end{array}
\)

Hint

(d) So, equivalent resistance of circuit, \(R_{\text {eq }}=5 \Omega\)
\(\therefore\) Current in the circuit \(=\frac{20}{5}=4 \mathrm{~A}\)
As in parallel, current is divided according to resistance, so current flowing through each resistance \(=2 \mathrm{~A}\).

Hint

\(
\begin{aligned}
&\text { (a) } \text { Potential difference }=10 \mathrm{~V}\\
&\text { So, } i=\frac{10}{5}=2 \mathrm{~A}
\end{aligned}
\)

Hint

\(
\text { (a) Internal resistance, } r=R\left(\frac{E}{V}-1\right)=(10)\left(\frac{1.5}{1.25}-1\right)=2 \Omega
\)

Hint

(b) For twisted wire, there are two halves each of resistance \(2 \Omega\) in parallel.
So, \(R_{\mathrm{eq}}=\frac{2 \times 2}{2+2}=1 \Omega\)

Hint

(a) When wire is divided in 10 equal parts, then each part will have a resistance \(=R / 10=r\).
Let equivalent resistance be \(r_R\), then
\(
\begin{array}{llrl}
& & \frac{1}{r_R} & =\frac{1}{r}+\frac{1}{r}+\frac{1}{r}+\ldots 10 \text { times } \\
& \therefore & \frac{1}{r_R} & =\frac{10}{r}=\frac{10}{(R / 10)}=\frac{100}{R} \\
& \therefore & & r_R=\frac{R}{100}=0.01 R
\end{array}
\)

Hint

(b) Given, \(\frac{R_1 R_2}{R_1+R_2}=\frac{6}{8} \dots(i)\)
When one resistance say \(R_2\) is broken, then
\(
R_1=2 \Omega \dots(ii)
\)
From Eqs. (i) and (ii), we get
\(
R_2=\frac{6}{5} \Omega
\)

Hint

\(
\text { (a) We have, } \quad R=\rho \frac{l}{A}
\)
\(
\begin{gathered}
R \propto l \\
\frac{R_1}{R_2}=\frac{l_1}{l_2}=\frac{1}{2} \dots(i) \\
R_2=2 R_1
\end{gathered}
\)
\(
\because \quad R_1+R_2=9 \dots(ii)
\)
\(
\begin{aligned}
&\text { From Eqs. (i) and (ii), we get }\\
&\begin{aligned}
& & R_1+2 R_1 & =9 \\
\Rightarrow & & R_1 & =3 \Omega \\
\therefore & & R_2 & =2 R_1=2 \times 3=6 \Omega
\end{aligned}
\end{aligned}
\)
\(
\text { Net resistance, } R_{\text {net }}=\frac{R_1} R_2}{R_{\mathrm{l}}+R_2}=\frac{3 \times 6}{3+6}=2 \Omega
\)

Hint

(b) Let resistance, \(R=100 \Omega\)
\(
\therefore \quad R^{\prime}=100+100 \times \frac{10}{100}=110 \Omega
\)
\(\therefore\) Required shunt, \(\Delta R=R^{\prime}-R=110-99=11 \Omega\)

Hint

(d) Resistance,
\(
\begin{aligned}
& R_t=R_0(1+\alpha t) \\
& 20=R_0(1+20 \alpha) \text { and } 60=R_0(1+500 \alpha)
\end{aligned}
\)
From these equations, we can find
\(
\begin{aligned}
R_0 & =18.33 \Omega \\
\alpha & =4.54 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}
\end{aligned}
\)
\(
\begin{array}{ll}
& R_t=25 \Omega \\
\text { Again, } & R_t=R_0(1+\alpha t) \\
& 25=18.33\left(1+4.54 \times 10^{-3} t\right) \\
\text { We find, } & t=80^{\circ} \mathrm{C}
\end{array}
\)

Hint

(c) The current \(i\) crossing area of cross-section \(A\), can be expressed in terms of drift velocity \(v_d\) and the moving charges as
\(
i=n e v_d A
\)
where, \(n\) is number of charge carriers per unit volume and \(e\) the charge on the carrier.
\(
\begin{aligned}
\therefore \quad v_d=\frac{i}{n e A} & =\frac{24 \times 10^{-3}}{\left(3 \times 10^{23}\right)\left(1.6 \times 10^{-19}\right)\left(10^{-4}\right)} \\
& =5 \times 10^{-3} \mathrm{~ms}^{-1}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (a) }\\
&\begin{aligned}
\text { Potential gradient } & =\frac{V}{L}=\frac{i R}{L}=\frac{i \rho L}{L A}=\frac{i \rho}{A} \\
& =\frac{0.2 \times 40 \times 10^{-8}}{8 \times 10^{-6}}=10^{-2} \mathrm{~V} / \mathrm{m}
\end{aligned}
\end{aligned}
\)

Hint

(b) In order to balance the bridge, the effective value of \(S\) must be equal to \(2 \Omega\).
So,
\(
\begin{aligned}
2 & =\frac{S \times 6}{S+6} \\
2 S+12 & =6 S \Rightarrow 4 S=12 \Rightarrow S=3 \Omega
\end{aligned}
\)

Hint

\(
\text { (b) Current, } \begin{aligned}
i=q f=q\left(\frac{v}{2 \pi r}\right) & =\frac{1.6 \times 10^{-19}\left(2.2 \times 10^6\right)}{(2 \pi)\left(5 \times 10^{-11}\right)} \\
& =1.12 \times 10^{-3} \mathrm{~A}
\end{aligned}
\)

Note: Current is defined as charge per unit time: \(I=\frac{q}{T}\).
The time period \(T\) for one revolution is related to the speed \(v\) and radius \(r\) by \(T=\frac{2 \pi r}{v}\).

Hint

(c) Total cells \(=m \times n=24 \Rightarrow m n=24 \dots(i)\)
For maximum current in the circuit,
\(
R=\frac{m r}{n} \Rightarrow 3=\frac{m}{n} \times(0.5) \Rightarrow m=6 n \dots(ii)
\)
On solving Eqs. (i) and (ii), we get \(m=12\) and \(n=2\)

Hint

\(
\begin{aligned}
& \text { (c) Current, } I_V=\frac{V_V}{R_V}=\frac{5}{20 \times 10^3}=0.25 \times 10^{-3} \mathrm{~A} \\
& \therefore \quad V=I_V(R+r) \\
& \qquad \begin{array}{rr}
110 & =0.25 \times 10^{-3}\left(R+20 \times 10^3\right) \\
R  =420 \mathrm{k} \Omega
\end{array}
\end{aligned}
\)

Hint

(c) We have, \(i_1=\frac{E}{r+R_1}\) and \(i_2=\frac{E}{r+R_2}\) From these two equations, we get
\(
r=\frac{i_2 R_2-i_1 R_1}{i_1-i_2}
\)

Hint

(b)
\(
\begin{aligned}
\frac{1}{R_P} & =\frac{1}{R}+\frac{1}{R}+\frac{1}{R}=\frac{3}{R} \\
R_P & =\frac{R}{3} \Omega
\end{aligned}
\)
\(
\begin{aligned}
\text { and } R_S & =R+R=2 R \Omega \\
R_{\text {net }} & =R_P+R_S \\
R_{\text {net }} & =2 R+\frac{R}{3}=\frac{7 R}{3} \Omega
\end{aligned}
\)

Hint

(c) \(i=\frac{2+2}{1+1.9+0.9}=\frac{4}{3.8} \mathrm{~A}\)
For cell \(A, E=V+i r\)
\(
\begin{aligned}
& V=E-i r=2-\frac{4}{3.8} \times 1.9 \\
& V=0 \text { (zero) }
\end{aligned}
\)

Hint

(c) In circuit, resistance between \(P Q, Q R\) and \(R S\) are in parallel. Now circuit reduces to

\(
R_{A B}=1+1+\frac{1}{3}=\frac{7}{3} \Omega
\)

Hint

(a) Resistance, \(R=\rho \cdot \frac{l}{A}\)
For same material and same length, \(\frac{R_2}{R_1}=\frac{A_1}{A_2}=\frac{3}{1}\)
\(
\therefore \quad R_2=3 R_1
\)
Resistance of thick wire, \(R_1=10 \Omega\) (given)
\(\therefore\) Resistance of thin wire, \(R_2=3 \times 10=30 \Omega\)
Total resistance of series combination \(=10+30=40 \Omega\)

Hint

(c) Total potential drop across the given wire
\(
=\left(1 \times 10^{-3}\right)\left(10^2\right)=0.1 \mathrm{~V}
\)

Therefore, the potential difference across \(R[latex] should be 1.9 V.
[latex]
\text { Now, } \quad \frac{1.9}{0.1}=\frac{R}{3} \text { or } R=57 \Omega
\)

Hint

(a) The resistance of a wire is given by \(R=\rho \frac{l}{A}\), where \(\rho\) is the resistivity, \(l\) is the length, and \(A\) is the cross-sectional area.
The area of a circle is \(A=\pi r^2\).
Density is given by density \(=\frac{\text { mass }}{\text { volume }}\).
Volume of a cylinder is \(V=A \cdot l\).
For resistors in parallel, the equivalent resistance is given by \(\frac{1}{R_{e q}}=\frac{1}{R_1}+\frac{1}{R_2}\).
Since the wires have the same mass \(m\) and are made of the same material, their densities \(\rho\) are equal.
The volume of wire \(A\) is \(V_A=\pi r_A^2 l_A\).
The volume of wire \(B\) is \(V_B=\pi r_B^2 l_B\).
Since mass \(=\) density \(\times\) volume, and the masses are equal:
\(m=\rho V_A=\rho V_B\).
Therefore, \(V_A=V_B\).
\(
\pi r_A^2 l_A=\pi r_B^2 l_B
\)
Substituting \(r_A=2 r_B\) :
\(\pi\left(2 r_B\right)^2 l_A=\pi r_B^2 l_B\).
\(4 \pi r_B^2 l_A=\pi r_B^2 l_B\).
\(4 l_A=l_B\).
The resistance of wire \(A\) is \(R_A=\rho \frac{l_A}{\pi r_A^2}\).
The resistance of wire \(B\) is \(R_B=\rho \frac{l_B}{\pi r_B^2}\).
Substituting \(r_A=2 r_B\) and \(l_B=4 l_A\) :
\(
\begin{aligned}
R_B & =\rho \frac{4 l_A}{\pi r_B^2} \\
R_A & =\rho \frac{l_A}{\pi\left(2 r_B\right)^2}=\rho \frac{l_A}{4 \pi r_B^2}
\end{aligned}
\)
Therefore, \(R_B=16 R_A\).
When connected in parallel, the equivalent resistance \(R_{e q}\) is given by:
\(\frac{1}{R_{e q}}=\frac{1}{R_A}+\frac{1}{R_B}\).
Substituting \(R_B=16 R_A\) :
\(\frac{1}{R_{e q}}=\frac{1}{R_A}+\frac{1}{16 R_A}\).
\(\frac{1}{R_{e q}}=\frac{16+1}{16 R_A}=\frac{17}{16 R_A}\).
\(R_{e q}=\frac{16 R_A}{17}=\frac{R_B}{17}\).
\(
\text { If } R_{\text {eq }}=4 \text {, then } R_A=4.25 \Omega \text { or } R_B=68 \Omega \text {. }
\)

Hint

(d) Balanced condition for Wheatstone’s bridge,
\(
\frac{R_1}{R_2}=\frac{R_3}{R_4}
\)
We have, \(\frac{24+X}{84+Y}=\frac{10}{30}=\frac{1}{3} \dots(i)\)
\(
\text { and } \frac{3}{1}=\frac{24+84+X+Y}{10+30} \quad[\because V=I R]
\)
\(
=\frac{108+X+Y}{40} \dots(ii)
\)
\(
\begin{aligned}
&\text { Solving Eqs. (i) and (ii), we get }\\
&X=Y=6 \Omega
\end{aligned}
\)

Hint

(b) Current, \(i=\frac{E}{R+r}\). When \(R\) decreases to \(0, i=\frac{E}{r}\) and \(V=I R=0\).

Hint

(c) Initially, \(\frac{P}{Q}=\frac{l}{100-l}=\frac{20}{100-20}=\frac{20}{80}=\frac{1}{4}\)
When a resistance of \(15 \Omega\) is connected in series with the smaller of the two, i.e. with \(P\), then
\(
\begin{aligned}
& \qquad \begin{aligned}
\frac{P+15}{Q} & =\frac{40}{60} \Rightarrow \frac{P}{Q}+\frac{15}{Q}=\frac{2}{3} \\
\frac{1}{4}+\frac{15}{Q} & =\frac{2}{3} \Rightarrow \frac{15}{Q}=\frac{5}{12} \\
\Rightarrow \quad Q & =36 \Omega \\
\text { Resistance, } P & =\frac{1}{4} \times 36=9 \Omega
\end{aligned}
\end{aligned}
\)

Hint

(c) When a resistance of \(100 \Omega\) is connected in series, then the current flowing will be
\(
i=\frac{V}{100+R} \dots(i)
\)
When a resistance of \(1000 \Omega\) is connected in series, the range of galvanometer gets doubled.
Current, \(i=\frac{2 V}{1100+R} \dots(ii)\)
From Eqs. (i) and (ii), we get
\(
\frac{V}{100+R}=\frac{2 V}{1100+R} \Rightarrow R=900 \Omega
\)

Hint

(d) Drift velocity, \(v_d=\frac{i}{n A e}=\frac{i \times 4}{n \pi D^2 e}\)
i.e. \(v_d \propto \frac{1}{D^2}\)
\(
\therefore \quad \frac{v_{d_1}}{v_{d_2}}=\frac{D_2^2}{D_1^2}=\left(\frac{1}{2}\right)^2=\frac{1}{4}
\)

Hint

(c) Since the power consumed by both bulbs is the same, we have:
\(P_1=P_2\)
Using the formula \(P=\frac{V^2}{R}\), we get:
\(\frac{V_1^2}{R_1}=\frac{V_2^2}{R_2}\)
Rearranging to find the ratio of resistances:
\(\frac{R_1}{R_2}=\frac{V_1^2}{V_2^2}=\frac{200^2}{300^2}=\frac{40000}{90000}=\frac{4}{9}\)
Find the ratio of potential differences:
In a series circuit, the voltage drop across each resistor is proportional to its resistance:
\(\frac{V_{R 1}}{V_{R 2}}=\frac{R_1}{R_2}\)
Substituting the ratio of resistances:
\(\frac{V_{R 1}}{V_{R 2}}=\frac{4}{9}\)
In a series circuit, the power consumed is given by \(P=I^2 R\). Since the current is the same through both bulbs:
\(\frac{P_1}{P_2}=\frac{I^2 R_1}{I^2 R_2}=\frac{R_1}{R_2}\)
Substituting the ratio of resistances:
\(\frac{P_1}{P_2}=\frac{4}{9}\)

Hint

(b) Current required by each bulb,
\(
i=\frac{P}{V}=\frac{100}{220} \mathrm{~A}
\)
If \(n\) bulbs are joined in parallel, then \(n i=i_{\text {fuse }}\) or \(\quad n \times \frac{100}{220}=10 \quad\) or \(n=22\)

Hint

(d) The electric field required to give an electron an average energy of 2 eV over its mean free path.
Mean free path of electrons: \(\lambda=4 \times 10^{-8} \mathrm{~m}\)
Average energy gained by an electron: \(E=2 \mathrm{eV}\)
\(1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}\)
The energy gained by a charge \(q\) in an electric field \(E\) over a distance \(d\) is given by \(E=q E d\).
How to solve?
Calculate the electric field using the formula relating energy, charge, electric field, and distance.
Step 1: Convert the energy from eV to Joules
\(
E=2 \mathrm{eV}=2 \times 1.602 \times 10^{-19} \mathrm{~J}=3.204 \times 10^{-19} \mathrm{~J}
\)
Step 2: Calculate the electric field
\(E=q E d\), where \(q[latex] is the charge of an electron [latex]\left(1.602 \times 10^{-19} \mathrm{C}\right), E\) is the electric field, and \(d\) is the mean free path \(\lambda\).
Rearrange the formula to solve for \(E: E=\frac{E}{q \lambda}\)
Substitute the values: \(E=\frac{3.204 \times 10^{-19} \mathrm{~J}}{\left(1.602 \times 10^{-19} \mathrm{C}\right)\left(4 \times 10^{-8} \mathrm{~m}\right)}\)
\(
\begin{aligned}
& E=\frac{3.204 \times 10^{-19}}{6.408 \times 10^{-27}} \mathrm{~V} / \mathrm{m} \\
& E=0.5 \times 10^8 \mathrm{~V} / \mathrm{m} \\
& E=5 \times 10^7 \mathrm{~V} / \mathrm{m}
\end{aligned}
\)

Hint

(a) Initially, \(R_1=50+2950=3000 \Omega\)
\(
E=3 V \Rightarrow I_1=\frac{3}{3000}=1 \times 10^{-3} \mathrm{~A}
\)
To reduce deflection to 20 ,
\(
I_2=\frac{2}{3} \times 1 \times 10^{-3} \mathrm{~A}
\)
\(
\begin{aligned}
&\begin{array}{ll}
\therefore & R \times \frac{2}{3} \times 10^{-3}=3000 \times 1 \times 10^{-3} \\
\Rightarrow & R=\frac{3000 \times 3}{2}=4500 \Omega
\end{array}\\
&\text { So, resistance to be added }=4500-50=4450 \Omega
\end{aligned}
\)

Hint

(b) We have, shunt, \(S=\frac{G}{(n-1)} \dots(i)\)
where, \(G\) is resistance of galvanometer.
Again, \(\quad S^{\prime}=\frac{G}{\left(n^{\prime}-1\right)} \dots(ii)\)
\(
\begin{array}{lr}
\therefore & \frac{S}{S^{\prime}}=\frac{\left(n^{\prime}-1\right)}{(n-1)} \\
\Rightarrow & \frac{S(n-1)}{S^{\prime}}=\left(n^{\prime}-1\right)
\end{array}
\)
\(
n^{\prime}=\frac{S(n-1)+S^{\prime}}{S^{\prime}}
\)

Hint

(d) Power is related to voltage and resistance by \(P=\frac{V^2}{R}\).
Resistance is related to resistivity, length, and cross-sectional area by \(R=\rho \frac{L}{A}\).
The cross-sectional area of a wire is \(A=\pi r^2\), where \(r\) is the radius.
Since the voltage is the same for both bulbs, the power is inversely proportional to the resistance:
\(P=\frac{V^2}{R}\)
\(R=\frac{V^2}{P}\)
The resistance of a wire is given by:
\(R=\rho \frac{L}{A}\)
Since the length \(L\) and resistivity \(\rho\) (material is the same) are the same for both filaments, resistance is inversely proportional to the cross-sectional area \(A\) :
\(A=\frac{\rho L}{R}\)
Combine the equations
Substitute the expression for \(R\) from Step 1 into the equation for \(A\) from Step 2:
\(A=\frac{\rho L}{\frac{V^2}{P}}\)
\(A=\frac{\rho L P}{V^2}\)
Compare the cross-sectional areas
Since \(\rho, L\), and \(V\) are constant, the cross-sectional area \(A\) is directly proportional to the power \(P\) :
\(A \propto P\)
Therefore, the bulb with higher power will have a larger cross-sectional area.
Since \(P_2=60 \mathrm{~W}>P_1=25 \mathrm{~W}\), the 60 W bulb has a thicker filament.
The 60 W bulb has a thicker filament.

Hint

(b) Three resistors are in parallel.
The equivalent resistance is \(1 \Omega\).
Two resistors are in the ratio \(1: 2\).
No resistance value is fractional.
For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances: \(\frac{1}{R_{e q}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\).
How to solve?
Set up an equation using the parallel resistance formula and the given ratio, then solve for the unknown resistances.
Step 1: Define the resistances
Let the three resistances be \(\boldsymbol{R}_1, \boldsymbol{R}_2[latex], and [latex]\boldsymbol{R}_3\).
Let \(R_1=x[latex] and [latex]R_2=2 x\).
Step 2: Set up the equation for parallel resistance
The equivalent resistance is \(1 \Omega\), so:
\(
\frac{1}{1}=\frac{1}{x}+\frac{1}{2 x}+\frac{1}{R_3}
\)
Step 3: Simplify the equation
\(
\begin{aligned}
& 1=\frac{2}{2 x}+\frac{1}{2 x}+\frac{1}{R_3} \\
& 1=\frac{3}{2 x}+\frac{1}{R_3}
\end{aligned}
\)
Step 4: Solve for \(\boldsymbol{R}_3\)
\(
\begin{aligned}
& \frac{1}{R_3}=1-\frac{3}{2 x} \\
& \frac{1}{R_3}=\frac{2 x-3}{2 x} \\
& R_3=\frac{2 x}{2 x-3}
\end{aligned}
\)
Step 5: Find possible integer values for x and \(R_3\)
Since no resistance value is fractional, \(2 x-3\) must divide \(2 x\).
Test integer values for \(x\) starting from 2 (since \(x\) must be greater than \(\frac{3}{2}\) ).
If \(x=2, R_3=\frac{2(2)}{2(2)-3}=\frac{4}{1}=4\).
If \(x=3, R_3=\frac{2(3)}{2(3)-3}=\frac{6}{3}=2\).
If \(x>3, R_3\) will not be an integer.
Step 6: Determine the largest resistance
For \(x=2\), the resistances are \(2 \Omega, 4 \Omega\), and \(4 \Omega\) (not unequal).
For \(x=3\), the resistances are \(3 \Omega, 6 \Omega\), and \(2 \Omega\).
The largest resistance is \(6 \Omega\).
The largest of the three resistances is \(6 \Omega\).

Hint

(d) 

\(
\begin{aligned}
&\text { (d) Applying Kirchhoff’s law at point } D \text {, we get }\\
&\begin{aligned}
i_1 & =i_2+i_3 \\
\frac{V_A-V_D}{10} & =\frac{V_D-0}{20}+\frac{V_D-V_C}{30} \\
70-V_D & =\frac{V_D}{2}+\frac{V_D-10}{3} \\
\Rightarrow \quad V_D & =40 \mathrm{~V} \Rightarrow i_1=\frac{70-40}{10}=3 \mathrm{~A} \\
i_2 & =\frac{40-0}{20}=2 \mathrm{~A}
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&\text { and }\\
&i_3=\frac{40-10}{30}=1 \mathrm{~A}
\end{aligned}
\)

Hint

(b) \(E_{\text {net }}=\frac{E_1 / r_1-E_2 / r_2}{1 / r_1+1 / r_2}\)
Current through \(R\) will be zero, if \(E_{\text {net }}=0\)
\(
\therefore \quad \frac{E_1}{r_1}=\frac{E_2}{r_2}
\)

Hint

\(
\begin{aligned}
&\text { (b) No current will flow through the grounded wire as there is no return path. }\\
&\therefore \quad I=\frac{V}{2 R}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (b) We have, } I_G=0\\
&\begin{aligned}
& \therefore \quad V_X=E_2=2 \mathrm{~V} \\
& \frac{V_{500 \Omega}}{V_X}=\frac{(12-2)}{2}=\frac{500}{X} \\
& \therefore \quad X=100 \Omega
\end{aligned}
\end{aligned}
\)

Hint

\(
\text { (d) The circuit can be reduced as, }
\)

\(
\begin{aligned}
&\text { Now, equivalent resistance between } A \text { and } B \text { is }\\
&R_{\mathrm{eq}}=2+4+2=8 \Omega
\end{aligned}
\)

Hint

(b) Potential difference across \(100 \Omega\) resistance is 5 V. As voltmeter and \(100 \Omega\) resistance are in parallel. It means equivalent resistance of voltmeter and \(100 \Omega\) should be \(50 \Omega\).
So, resistance of voltmeter must be \(100 \Omega\).

Hint

\(
\begin{aligned}
&\text { (d) Net resistance of circuit, } R=1+6+4=11 \Omega\\
&\begin{aligned}
I & =\frac{6}{11} \mathrm{~A} \text { (Ammeter reading) } \\
\therefore \quad V_1 & =\frac{6}{11} \times 1=\frac{6}{11} \mathrm{~V}
\end{aligned}
\end{aligned}
\)
Voltage across voltameter is \(= – \frac{6}{11}+6=\frac{60}{11}\mathrm{~V}\)

Hint

(a) The given circuit forms a balanced Wheatstone bridge between points \(A\) and \(B\).

\(
\begin{aligned}
R_{A B} & =(R+R) \|(R+R) \\
& =2 R \| 2 R \\
& =\frac{2 R \times 2 R}{2 R+2 R}=R
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
R_{\mathrm{net}} & =\frac{10 \times(8+24 / 7)}{10+(8+24 / 7)} \\
& =5.3 \Omega \simeq 5 \Omega
\end{aligned}
\)

Hint

(a) The given circuit consists of a balanced Wheatstone bridge.

\(
R_{\text {net }}=R+\frac{2}{3} R=\frac{5}{3} R
\)

Hint

\(
\text { (c) } q=\int_0^5 i d t=\int_0^5(1.2 t+3) d t=30 \mathrm{C}
\)

Hint

(a) Resistor in OD and resistor in OB are obsolete as it is part of balanced wheatstone bridge. We can remove them.

\(
\begin{aligned}
& \frac{1}{R_{\mathrm{eq}}}=\frac{1}{2 R}+\frac{1}{2 R}+\frac{1}{2 R} \\
& R_{\mathrm{eq}}=\frac{2}{3} R
\end{aligned}
\)

Hint

(d) The circuit can be drawn as,

\(
\text { Resistance between points } P \text { and } S=4 \Omega
\)

Hint

(d) Power of the heater: \(\boldsymbol{P}=1.08 \mathrm{~kW}=1080 \mathrm{~W}\)
Mass of steam: \(m=100 \mathrm{~g}=0.1 \mathrm{~kg}\)
Temperature of water: \(T=100^{\circ} \mathrm{C}\)
Latent heat of vaporization of water: \(L_v=2.26 \times 10^6 \mathrm{~J} / \mathrm{kg}\)
Power is the rate of energy transfer: \(P=\frac{E}{t}\)

How to solve?
Calculate the energy required to vaporize the water and then use the power formula to find the time.

Step 1: Calculate the energy required to vaporize the water
The energy required to vaporize a mass \(m\) of water at \(100^{\circ} \mathrm{C}\) is given by:
\(E=m L_v\)
\(E=(0.1 \mathrm{~kg})\left(2.26 \times 10^6 \mathrm{~J} / \mathrm{kg}\right)\)
\(E=2.26 \times 10^5 \mathrm{~J}\)

Step 2: Calculate the time required
The time \(t\) required to supply energy \(E\) at a power \(P\) is given by:
\(t=\frac{E}{P}\)
\(t=\frac{2.26 \times 10^5 \mathrm{~J}}{1080 \mathrm{~W}}\)
\(t \approx 209.26 \mathrm{~s}\)
The time required to produce 100 g of steam is approximately 210 s.

Hint

(b) When shunt resistance is \(4 r\), the ammeter reads 0.03 A.
When shunt resistance is \(r\), the ammeter reads 0.06 A.
When a shunt is connected to a galvanometer, the potential difference across the galvanometer and the shunt are equal.

How to solve?
Set up two equations based on the given information using the shunt formula and solve for the galvanometer resistance and maximum current.

Step 1: Set up the equation for the first scenario.
Let \(I_g\) be the maximum current through the galvanometer.
Let \(G\) be the resistance of the galvanometer.
The current through the shunt is \(0.03-I_g\).
The potential difference across the galvanometer is \(I_g G\).
The potential difference across the shunt is \(\left(0.03-I_g\right) 4 r\).
Equating the potential differences:
\(I_g G=\left(0.03-I_g\right) 4 r\)

Step 2: Set up the equation for the second scenario.
The current through the shunt is \(0.06-I_g\).
The potential difference across the shunt is \(\left(0.06-I_g\right) r\).
Equating the potential differences:
\(I_g G=\left(0.06-I_g\right) r\)

Step 3: Solve for \(I_g\).
Divide the equation from Step 1 by the equation from Step 2:
\(\frac{I_g G}{I_g G}=\frac{\left(0.03-I_g\right) 4 r}{\left(0.06-I_g\right) r}\)
Simplify:
\(1=\frac{4\left(0.03-I_g\right)}{0.06-I_g}\)
Solve for \(\boldsymbol{I}_{\boldsymbol{g}}\) :
\(0.06-I_g=0.12-4 I_g\)
\(3 I_g=0.06\)
\(I_g=\frac{0.06}{3}\)
\(I_g=0.02\)
The maximum current that can be sent through the galvanometer is 0.02 A.

Hint

(c)

When key K is opened in the circuit, bulb B1 will decrease in brightness and bulb B2 will increase in brightness.

Explanation:
Bulb B1: When key K is closed, the voltage across B1 is split with B2 and B3. When K is opened, the voltage drop across B1 increases because it is now the only path for the current. This increase in voltage, however, is balanced by a decrease in current, as B3 are no longer part of the circuit. The power consumed by B1 (and thus its brightness) decreases.

Bulb B2: With key K open, B3 will draw no current from the source. This results in the terminal voltage of the source increasing, which leads to an increase in the voltage across B2. The increase in voltage will cause the bulb B2 to increase in brightness.

Hint

(a) Maximum current possible in galvanometer,
So,
\(
\begin{aligned}
I_{\max } & =25 \times 4 \times 1 \\
10^{-2} & =\frac{2.5}{100+R} \\
100+R & =250 \\
\text { Resistance, } R & =150 \Omega
\end{aligned}
\)

Hint

(a) Let, the bulb 400 W is having resistance value of \(R\). For 200 W , necessary value of resistance will be \(2 R\).
Total value of resistance in the circuit will be \(R+R=2 R\)
If \(I\) is the maximum current in the circuit, then \(I^2 R=400 \mathrm{~W}\)
\(
\begin{aligned}
\text { Power of circuit as a whole } & =I^2 \times 2 R=2 \times I^2 R=2 \times 400 \\
& =800 \mathrm{~W}
\end{aligned}
\)

Hint

(b) Maximum current possible in bulb \(=\frac{50}{100}=0.5 \mathrm{~A}\)
Resistance of each bulb \(=\frac{V^2}{P}=\frac{100 \times 100}{50}=200 \Omega\)
If \(n\) be the number of bulbs possible, then total resistance of circuit \(=\frac{200}{n}+10\)
Maximum current in the circuit \(=0.5 \times n\)
So, \(\frac{120}{\frac{200}{n}+10}=0.5 n \Rightarrow n=4\)

Hint

\(
\text { (c) Let } R_{A B}=x \text {. Then, }
\)

\(
\begin{aligned}
R_{A B} & =1+\frac{x}{1+x} \quad \text { or } \quad x=1+\frac{x}{1+x} \\
x+x^2 & =1+x+x \quad \text { or } \quad x^2-x-1=0 \\
x & =\frac{1+\sqrt{1+4}}{2}=\frac{1+\sqrt{5}}{2} \Omega
\end{aligned}
\)

Hint

(a) 

Applying node analysis \(i=i_1+i_2\)
\(i=\frac{20-16}{2}+\frac{20+16}{4}=2+9=11 A\)

Hint

(b) According to the question, \(R=\left(20+\frac{i}{2}\right) \Omega\) Now, current, \(i=\frac{250}{R}=\frac{250}{20+(i / 2)}\)
\(
\Rightarrow \quad i^2+40 i-500=0
\)
Solving, we get
\(
i=10 \mathrm{~A}
\)

Hint

\(
\begin{aligned}
&\text { (d) According to question, }\\
&\begin{aligned}
Q & =a t-b t^2 \Rightarrow i=\frac{d Q}{d t}=a-2 b t \\
i & =0 \text { at } t=\frac{a}{2 b} \\
\frac{d i}{d t} & =-2 b
\end{aligned}
\end{aligned}
\)

Hint

\(
\text { (b) } \begin{aligned}
\text { Voltage sensitivity } & =\frac{\text { Current sensitivity }}{\text { Resistance of galvanometer } G} \\
G & =\frac{\text { Current sensitivity }}{\text { Voltage sensitivity }}=\frac{10}{2}=5 \Omega
\end{aligned}
\)
Full scale deflection current, \(i_g=\frac{150}{10}=15 \mathrm{~mA}\)
Voltage to be measured, \(V=150 \times 1=150 \mathrm{~V}\)
Hence, \(R=\frac{V}{i_g}-G=\frac{150}{15 \times 10^{-3}}-5=9995 \Omega\)

Hint

\(
\text { (b) We have, } \frac{V^2}{R} \times 16=H \quad(\because H=P \times t)
\)
\(
\begin{aligned}
&\frac{V^2}{0.9 R} \times t=H \Rightarrow \frac{V^2}{R} \times 16=\frac{V^2}{0.9 R} \times t\\
&\therefore \text { Time, } t=16 \times 0.9=14.4 \mathrm{~min}
\end{aligned}
\)

Hint

(b) Due to balanced Wheatstone bridge, resistance between \(A\) and \(B\) can be removed.

\(
\begin{aligned}
&\text { The current between } D \text { and } E \text {, }\\
&I_{D E}=\frac{10}{R_{D E}+R_{H G}}=\frac{10}{2+2}=2.5 \mathrm{~A}
\end{aligned}
\)

Hint

(d) When \(K_1\) is closed, \(R_1\) is short-circuited.
When \(K_2\) is open, \(I_0=\frac{E}{r+R_2}=\frac{E}{r+100} \dots(i)\)
\(
\text { When } K_2 \text { is closed, } I_0=\frac{1}{2}\left[\frac{E}{r+50}\right] \dots(ii)
\)
From these two equations, we get \(r=0\)
When \(K_1\) is open and \(K_2\) is closed, \(\frac{I_0}{2}=\frac{E}{2\left(R_1+50\right)} \dots(iii)\)
\(
\begin{aligned}
&\text { From Eqs. (i) and (iii), we have }\\
&R_1=50 \Omega
\end{aligned}
\)

Hint

(c) Let current through \(X Y\) is \(i_3\). Applying Kirchhoff’s law to loops (1) and (2),

\(
\begin{aligned}
i_1+0 \times i_3-3 i_2 & =0 \\
i_1 & =3 i_2 \dots(i)
\end{aligned}
\)
and \(-2\left(i_1-i_3\right)+4\left(i_2+i_3\right)=0\)
So,\(2 i_1-4 i_2=6 i_3 \dots(ii)\)
\(
\begin{aligned}
&\begin{aligned}
& \text { Also, } & 50 & =i_1+2\left(i_1-i_3\right) \\
& \therefore & 3 i_1-2 i_3 & =50 \dots(iii)
\end{aligned}\\
&\text { From Eqs. (i), (ii) and (iii), we get }\\
&i_3=2 \mathrm{~A}
\end{aligned}
\)

Hint

(a) Given, for a wire, \(\frac{R}{l}=\frac{1}{2}\) Length of wire, \(l=5 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}\)
\(
\therefore \quad R=\frac{l}{2}=2.5 \times 10^{-2} \Omega
\)
Potential difference, \(V=1 \mathrm{~V}\) or \(I R=1\)
\(
I=\frac{1}{R}=\frac{1}{2.5 \times 10^{-2}}=\frac{100}{2.5}=40 \mathrm{~A}
\)

Hint

(c) Given, current, \(I=10 \mathrm{~A}\)
Area of cross-section, \(A=4 \times 10^{-6} \mathrm{~m}^2\)
Density of conductors, \(\rho=2.7 \mathrm{~g} / \mathrm{cc}=2.7 \times 10^3 \mathrm{~kg} / \mathrm{m}^3\)
Molecular weight of aluminium, \(M_w=27 \mathrm{~g}=27 \times 10^{-3} \mathrm{~kg}\)
If \(n\) be the total number of electrons in the conductor per unit volume, then
\(
\begin{aligned}
&\begin{aligned}
n & =\frac{\text { Total number of electrons }}{\text { Volume of conductor }(V)} \\
& =\frac{\text { Number of atoms per mole } \times \text { Number of moles }}{V} \\
& =\frac{\text { Avogadro number }}{V} \times\left(\frac{M}{M_w}\right) \\
& =6 \times 10^{23} \times \frac{\rho}{M_w}=6 \times 10^{23} \times \frac{2.7 \times 10^3}{27 \times 10^{-3}} \\
\therefore \quad n & =6 \times 10^{28}
\end{aligned}\\
&\text { We know that, drift velocity, }\\
&\begin{aligned}
v_d & =\frac{I}{n e A}=\frac{10}{6 \times 10^{28} \times 1.6 \times 10^{-19} \times 4 \times 10^{-6}} \\
& =2.6 \times 10^{-4} \mathrm{~m} / \mathrm{s}
\end{aligned}
\end{aligned}
\)

Hint

(b) The situation is shown in the circuit diagram.

Current flowing through the circuit, \(I=\frac{V_i}{R_1+R_2}\)
Voltage across \(R_2, V_2=I R_2\)
\(
V_2=\frac{V_i R_2}{R_1+R_2} \Rightarrow \frac{V_i}{V_2}=\frac{R_1+R_2}{R_2}
\)

Hint

(a) Given, resistivity of copper, \(\rho=1.72 \times 10^{-8} \Omega-\mathrm{m}\)
Electrons density, \(n=8.5 \times 10^{28} / \mathrm{m}^3\)
\(
\begin{aligned}
\therefore \quad \text { Mobility }(\mu)=\frac{1}{\rho n e} & =\frac{1}{1.72 \times 10^{-8} \times 8.5 \times 10^{28} \times 1.6 \times 10^{-19}} \\
& \simeq 4.25 \times 10^{-3} \mathrm{~m}^2 / \mathrm{C} \Omega
\end{aligned}
\)

Hint

(a) Power rating of heater, \(P=1000 \mathrm{~W}\)
Voltage rating of heater, \(V=100 \mathrm{~V}\)
\(\therefore\) Resistance of heater,
\(
R_1=\frac{V^2}{P}=\frac{(100)^2}{1000}=10 \Omega
\)
According to question, power dissipated in heater,
\(
P^{\prime}=62.5 \mathrm{~W}
\)
\(\therefore\) Voltage \(\left(V^{\prime}\right)\) across heater can be calculated as
\(
P^{\prime}=\frac{\left(V^{\prime}\right)^2}{R_1}
\)
\(
\begin{array}{lr}
\Rightarrow & \left(V^{\prime}\right)^2=P^{\prime} \times R_1=62.5 \times 10 \\
\Rightarrow & V^{\prime}=25 \mathrm{~V} \text { (across heater) }
\end{array}
\)
\(\therefore\) Voltage across \(10 \Omega\) resistor,
\(
V^{\prime \prime}=100-25=75 \mathrm{~V}
\)
Current in \(10 \Omega\) resistor
\(
=\frac{V^{\prime \prime}}{10}=\frac{75}{10}=7.5 \mathrm{~A}
\)
Current in heater resistor
\(
=\frac{V^{\prime}}{10}=\frac{25}{10}=2.5 \mathrm{~A}
\)
So, current in \(R, I=7.5-2.5=5 \mathrm{~A}\)
Now, \(V=I R\)
\(
\Rightarrow \quad R=V / I=\frac{25}{5}=5 \Omega
\)

Hint

(c) During charging, \(E=V-ir\) (due to reversed current). In case of charging, emf of a cell is less than its terminal voltage while in case of discharging emf is greater than terminal voltage.
Therefore, Assertion is correct but Reason is incorrect.

Hint

(a) Switching results in high decay/growth rate of current which results in a high current when bulb is turned ON or OFF (due to back emf). So, a bulb is most likely to get fused when it is just turned ON or OFF.

Hint

(c) \(\because\) Current, \(I=n A e v_d\)
or \(\quad v_d \propto \frac{1}{A}\)
If diameter of wire is \(d / 4\), then area will be \(A / 16\), so new drift velocity \(=16 v_{d^*}\)

Hint

(a) It is clear that the two cells oppose each other, hence the effective emf in closed circuit is \(18-12=6 \mathrm{~V}\) and net resistance is \(1+2=3 \Omega\) (because in the closed circuit the internal resistance of two cells are in series).
The current in circuit will be in direction of arrow shown in figure
\(
I=\frac{\text { effective emf }}{\text { total resistance }}=6 / 3=2 \mathrm{~A}
\)
The potential difference across \(V\) will be same as the terminal voltage of either cell.

\(
\begin{aligned}
&\text { Since, current is drawn from the cell of } 18 \mathrm{~V} \text {, hence }\\
&\begin{aligned}
V_1 & =E_1-I r_1=18-(2 \times 2) \\
& =18-4=14 \mathrm{~V}
\end{aligned}
\end{aligned}
\)

Hint

(c) Let \(R\) be the resistance of the wire, then
(i) The heat generated is \(H_1=\frac{V^2 t}{R}\).
(ii) Resistance of each part will be \(R / 2\). When they are connected in parallel, the resistance will be \(R / 4\).
Hence, \(H_2=4 V^2 t / R\).
(iii) In case of four wires connected in parallel, the resistance will be \(R / 8\).
\(\therefore \quad H_3=\frac{8 V^2 t}{R}\)
(iv) \(H_4=\frac{V^2 t}{R / 2}=\frac{2 V^2 \cdot t}{R}\)

Hint

(b) The current in case of voltmeter of range 10 V and resistance \(50 \Omega\) is
\(
I=\frac{10}{50} \mathrm{~A} \dots(i)
\)
Let \(X\) be the resistance connected in order to make a voltmeter with range 15 V having current,
\(
I=\frac{15}{X} \mathrm{~A} \dots(ii)
\)
Equating Eqs. (i) and (ii), we get
\(
\frac{15}{X}=\frac{10}{50}=\frac{(15) \times(50)}{(10)}=75 \Omega
\)
Then, \(R=75-50=25 \Omega\) must be connected in series, because \(V \propto R\) when current is constant.

Hint

(a)  “Charge is a vector quantity” because charge is actually a scalar quantity, meaning it only has magnitude and not direction; so option (a) is incorrect.

Explanation:
Vector quantities: Vectors have both magnitude and direction, like force or velocity.
Scalar quantities: Scalars only have magnitude, like time or temperature.

Why the other options are correct:
(b) Current is a scalar quantity:

Current is defined as the rate of flow of charge, and it only has magnitude.
(c) Charge can be quantized:

This means charge can only exist in discrete units, like individual electrons, not in continuous values.
(d) Charge is additive in nature:

Total charge is the sum of individual charges present in a system.
(e) Charge is conserved:

In a closed system, the total charge remains constant; charge can neither be created nor destroyed. 

Hint

(c) Current remains constant throughout the metallic conductor. Current density \(J=\frac{I}{A}\) is not constant because cross-sectional area is a variable parameter. Drift velocity \(v_d=\frac{I}{n e A}\) is not constant. Since, \(v_d \propto \frac{1}{A}\).

Hint

\(
\begin{aligned}
&\begin{array}{ll}
\text { (c) Slope }=\frac{I}{V} & \\
\because & V=I R \\
\therefore & \frac{I}{V}=\frac{1}{R}
\end{array}\\
&\text { Hence, slope is reciprocal of resistance. }
\end{aligned}
\)

Hint

(a) The resistance of two wires are
\(
R_1=\rho_1 \frac{l}{A} \text { and } R_2=\rho_2 \frac{l}{A}
\)
Now, equivalent resistance of series connection of wire
\(
R_{\mathrm{eq}}=R_1+R_2=\rho_1 \frac{l}{A}+\rho_2 \frac{l}{A}=\frac{l}{A}\left(\rho_1+\rho_2\right)
\)
The equivalent resistance \(R_{\text {eq }}\) can be given by \(R_{\text {eq }}=\rho_{\text {eq }} \frac{2 l}{A}\)
\(
\begin{aligned}
& \Rightarrow \quad \rho_{\mathrm{eq}} \frac{2 l}{A}=\frac{l}{A}\left(\rho_1+\rho_2\right) \\
& \text { Hence, } \rho_{\mathrm{eq}}=\frac{\rho_1+\rho_2}{2}
\end{aligned}
\)

Hint

(a) Total resistance \(=50+3950=4000 \Omega\)
For this circuit, deflection of resistance \(R\) be
\(
\Rightarrow \quad \frac{4000}{R}=\frac{30}{15} \Rightarrow R=\frac{4000}{2}=2000 \Omega
\)
Then, resistance in series should be \(=2000-50=1950 \Omega\)

Hint

(a) Kirchhoff’s first law states that, algebraic sum of currents meeting at a point in a circuit is zero. It is based on conservation of charge.
Kirchhoff’s second law states that, the algebraic sum of potential differences around a closed loop is zero. It is based on the law of conservation of energy.

Hint

(b) Using the formula, \(m=Z I\), where \(m=\) mass of silver deposited \(=0.05 \%\) of \(750 \mathrm{~g}=0.375 \mathrm{~g}=3.75 \times 10^{-4} \mathrm{~kg}\)
Current passing through it, \(I=0.8 \mathrm{~A}\)
\(
Z=\text { ECE of silver }=11.8 \times 10^{-7} \mathrm{kgC}^{-1}
\)
\(\therefore\) The time needed for depositing silver is given by
\(
t=\frac{m}{Z I}=\frac{3.75 \times 10^{-4}}{11.8 \times 10^{-7} \times 0.8}=397 \mathrm{~s}=6 \mathrm{~min} 37 \mathrm{~s}
\)

Hint

\(
\begin{aligned}
&\text { (c) The shunt resistance required is given by }\\
&\begin{aligned}
& S=\left(\frac{I_{D C}}{I-I_{D C}}\right) R_{D C}=\left(\frac{100}{500-100}\right) \times 0.1=\frac{100}{400} \times 0.1 \\
\therefore & S=0.025 \Omega
\end{aligned}
\end{aligned}
\)
If this value of shunt resistance is connected in parallel with the DC ammeter, then the range will be extended to \(0-500 \mathrm{~A}\).

Hint

(a) The circuit diagram can be redrawn as the potential between \(A\) and \(B\) is

\(
\begin{aligned}
& \text { For the loop } A B C D A,+2-2 I^{\prime}+2-2 I=0 \\
& \qquad 2=I+I^{\prime} \dots(i)
\end{aligned}
\)
\(
\begin{aligned}
&\text { For the loop } A B F E A \text {, }\\
&\begin{aligned}
2-2 I+2-2\left(I-I^{\prime}\right) & =0 \\
4-4 I+2 I^{\prime} & =0 \\
2 & =2 I-I^{\prime} \dots(ii)
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&\text { From Eqs. (i) and (ii), we get }\\
&4=3 I \Rightarrow I=4 / 3=1.33 \mathrm{~A}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (c) As we know that, } R_1=\rho \frac{l}{a}=\rho \frac{l^2}{V} \\
& \text { where, } l=\text { length of wire, } \\
& \quad a=\text { area of cross-section of the wire } \\
& \text { and } \quad V=\text { volume of the wire. } \\
& \text { As, } \quad R_1 \propto l^2 \\
& \Rightarrow \quad \frac{R_1}{R_2}=\left(\frac{l_1}{l_2}\right)^2=\left(\frac{1}{2}\right)^2 \\
& \Rightarrow \quad R_2: R_1=4: 1
\end{aligned}
\)

Hint

(d) We know that, potential drop across a resistance,
i.e.
\(
\begin{gathered}
V=I R \Rightarrow V=\text { constant } \\
I \propto \frac{1}{R} \\
I_1: I_2: I_3=\frac{1}{R_1}: \frac{1}{R_2}: \frac{1}{R_3}=\frac{1}{2}: \frac{1}{3}: \frac{1}{4}=6: 4: 3
\end{gathered}
\)

Hint

\(
\begin{aligned}
&\text { (a) Total internal resistance does not change }\\
&\begin{array}{ll}
\therefore & R^{\prime}=4 r \\
\text { Net emf, } & E^{\prime}=E(n-2 m) \\
\text { Here, } & n=\text { total number of cells }=4 \\
& m=\text { wrong connection }=1 \\
& E^{\prime}=E[4-2], \text { i.e. } E^{\prime}=2 E
\end{array}
\end{aligned}
\)

Hint

(a) Given, \(I=3 \mathrm{~A}\) and \(V=6 \mathrm{~V}\)
We know that, \(V=I R\)
\(
R=\frac{V}{I} \Rightarrow R=\frac{6}{3}=2 \Omega
\)
If the ammeter and voltmeter have resistance, then \(R<2 \Omega\).

Hint

(b) Given, \(R_1=100 \Omega\)
\(
\begin{aligned}
& R_2=200 \Omega \Rightarrow T_1=100^{\circ} \mathrm{C} \\
& \alpha=0.005 \text { per }{ }^{\circ} \mathrm{C}
\end{aligned}
\)
We know that, new resistance of the bulb filament,
\(
\begin{aligned}
R_2 & =R_1\left[1+\alpha\left(T_2-T_1\right)\right] \\
200 & =100\left[1+0.005\left(T_2-100\right)\right]
\end{aligned}
\)
\(
\begin{array}{r}
2=1+\left[0.005\left(T_2-100\right)\right] \\
0.005\left(T_2-100\right)=1
\end{array}
\)
\(
\begin{array}{rlrl}
& T_2-100 =\frac{1}{0.005} \\
\Rightarrow & T_2-100 =\frac{1000}{5} \\
T_2-100 =200 \\
\Rightarrow & T_2 =200+100=300^{\circ} \mathrm{C}
\end{array}
\)

Hint

(d) Consider the ring as two parts. As two resistances are joined in parallel between two points \(A\) and \(B\), then two resistances would be
\(
\begin{aligned}
& R_1=\frac{R}{2 \pi r} \cdot r \theta=\frac{R}{2 \pi} \theta \\
& R_2=\frac{R}{2 \pi r} r(2 \pi-\theta)=\frac{R}{2 \pi}(2 \pi-\theta)
\end{aligned}
\)
Now, equivalent or effective resistance between \(A\) and \(B\),
\(
\begin{aligned}
R_{\text {eq }} & =\frac{R_1 \times R_2}{R_1+R_2} \\
R_{\text {eq }} & =\frac{\frac{R}{2 \pi} \theta \times \frac{R}{2 \pi}(2 \pi-\theta)}{\frac{R}{2 \pi}[\theta+2 \pi-\theta]}=\left[\frac{\frac{R^2 \theta(2 \pi-\theta)}{4 \pi^2}}{\frac{2 \pi R}{2 \pi}}\right] \\
& =\frac{R^2 \theta(2 \pi-\theta)}{4 \pi^2} \times \frac{2 \pi}{2 \pi R}=\frac{R \theta(2 \pi-\theta)}{4 \pi^2}
\end{aligned}
\)

Hint

(b) Case I For balanced point of meter bridge,
\(
\frac{5}{R}=\frac{l_1}{100-l_1} \dots(i)
\)
Case II When \(R\) is shunted with equal resistance, i.e. \(R\)
\(
\begin{array}{ll}
& \frac{1}{R^{\prime}}=\frac{1}{R}+\frac{1}{R} \Rightarrow R^{\prime}=R / 2 \\
\therefore & \frac{5}{R / 2}=\frac{1.6 l_1}{100-1.6 l_1} \dots(ii)
\end{array}
\)
From Eqs. (i) and (ii), we get
\(
2\left(\frac{l_1}{100-l_1}\right)=\frac{1.6 l_1}{100-1.6 l_1} \Rightarrow l_1=25 \mathrm{~cm}
\)
From Eq. (i), \(\frac{5}{R}=\frac{25}{75} \Rightarrow R=15 \Omega\)

Hint

\(
\text { (a) As, current } I_G=\frac{2}{1000} I
\)

Also, \(I_r=\left(\frac{G}{G+r}\right) I\)
We know that, potential across \(G\) and shunt \(r\) are same.
\(
\begin{aligned}
& \therefore \quad V_G=V_r \Rightarrow I_G(G)=I_r r \\
& \Rightarrow \quad \frac{2 I G}{1000}=\frac{G I}{G+r}(r) \Rightarrow 2(G+r)=1000 r \\
& \Rightarrow \quad G+r=500 r \Rightarrow \frac{G}{r}+1=500 \\
& \Rightarrow \quad G / r=499 \Rightarrow r=\frac{1}{499} G
\end{aligned}
\)

Hint

\(
\text { (c) The required resistance }=\frac{\frac{R}{2} \times \frac{R}{2}}{\frac{R}{2}+\frac{R}{2}}=\frac{R}{4}=\frac{9}{4} \Omega
\)

Hint

(c) Let the values of resistances be \(R_1\) and \(R_2\), respectively.
When \(R_1\) and \(R_2\) resistances are in series, then
\(
R_1+R_2=6 \text { (according to question) } \dots(i)
\)
When \(R_1\) and \(R_2\) resistances are parallel, then
\(
\frac{R_1 R_2}{R_1+R_2}=\frac{4}{3} \dots(ii)
\)
From Eq. (i), we get
\(
\frac{R_1 R_2}{6}=\frac{4}{3} \Rightarrow R_1 R_2=4 \times 2 \Rightarrow R_1 R_2=8 \dots(iii)
\)
We know that,
\(
\begin{array}{ll}
& R_1-R_2=\sqrt{\left(R_1+R_2\right)^2-4 R_1 R_2}=\sqrt{36-4 \times 8} \\
\Rightarrow \quad & R_1-R_2=\sqrt{4} \\
\Rightarrow \quad & R_1-R_2=2 \dots(iv)
\end{array}
\)
From Eqs. (i) and (iv), we get
\(
R_1=4 \Omega, R_2=2 \Omega
\)

Hint

(b) As, resistances \(1 \Omega, 2 \Omega\) and \(1 \Omega\) are in parallel.
So, the required internal resistance,
\(
\frac{1}{r}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=\frac{1}{1}+\frac{1}{2}+\frac{1}{1}=\frac{2+1+2}{2} \Rightarrow r=\frac{2}{5} \Omega
\)
\(
\begin{aligned}
&\text { The potential difference between points } P \text { and } Q \text {, }\\
&\begin{aligned}
E & =\frac{\frac{E_1}{r_1}+\frac{E_2}{r_2}+\frac{E_3}{r_3}}{1 / r}=\frac{\frac{1}{1}+\frac{2}{2}+\frac{3}{1}}{5 / 2} \\
& =\frac{\frac{2+2+6}{2}}{5 / 2}=\frac{10 / 2}{5 / 2}=\frac{10}{5} \times \frac{2}{2}=2 \mathrm{~V}
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (b) As, mobility }=\frac{\text { drift velocity }}{\text { electric field }} \\
& \Rightarrow \mu=\frac{v_d}{E}=\frac{\text { metre }^2}{\text { volt-second }}=\frac{[\mathrm{L}]^2}{(\text { joule/coulomb)-second }}
\end{aligned}
\)
\(
=\frac{[\mathrm{L}]^2}{\frac{\left[\mathrm{~kg}-\text { metre }^2-\text { second }^{-2}\right]-\text { second }}{\text { ampere-second }}}=\frac{\left[\mathrm{L}^2\right]}{\left[\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]}=\left[\mathrm{M}^{-1} \mathrm{~T}^2 \mathrm{~A}\right]
\)

Hint

(a) small and positive.
Explanation:
Positive coefficient:
Most alloys, including those used for resistors, experience an increase in resistance with increasing temperature. This means the temperature coefficient is positive.

Small value:
While the resistance change is positive, for alloys used in resistors, the change is typically very small compared to other materials like pure metals. This makes the temperature coefficient “small and positive”.

Hint

\(
\begin{aligned}
& \text { (b) Resistance of wire, } R=\rho \frac{l}{A} \\
& \text { Given, } R_1=4 \Omega, l_1=l, A_1=A \\
& \qquad \begin{array}{l}
l_2=2 l, A_2=A / 2, \rho=\text { constant } \\
\therefore \quad \frac{R_1}{R_2}=\frac{l_1 / A_1}{l_2 / A_2}=\frac{l_1 \times A_2}{l_2 \times A_1} \\
\Rightarrow \quad \frac{4}{R_2}=\frac{l \times A / 2}{2 l \times A} \Rightarrow \frac{4}{R_2}=\frac{1}{4} \Rightarrow R_2=16 \Omega
\end{array}
\end{aligned}
\)

Hint

(a) Balanced condition for Wheatstone bridge,
\(
\frac{P}{Q}=\frac{R}{s} \dots(i)
\)
where, \(\frac{1}{s}=\frac{1}{s_1}+\frac{1}{s_2} \Rightarrow s=\frac{s_1 s_2}{s_1+s_2} \dots(ii)\)
From Eqs. (i) and (ii), we get
\(
\frac{P}{Q}=\frac{R\left(s_1+s_2\right)}{s_1 s_2}
\)

Hint

(b) Resistance, \(R=\frac{V}{i_g}-G=\frac{120}{0.01}-10=12000-10\)
\(
R=11990 \Omega
\)
To convert a galvanometer into a voltmeter, high resistance of value \(11990 \Omega\) should be connected in parallel.

Hint

(b) Let, \(R_x\) resistance be connected in series to convert galvanometer to voltmeter.

According to the question, \(R_g=100 \Omega I_g=30 \mathrm{~mA}\)
(Current corresponding to full scale deflection)
Now, we can write, \(V=I_g \times R_g+I_g R_x=I_g\left(R_g+R_x\right) \dots(i)\)
Given,
\(
V=30 \mathrm{~V}
\)
From Eq. (i), we get
\(
\begin{aligned}
& 30=\left(30 \times 10^{-3}\right)\left(100+R_x\right) \\
\Rightarrow & \frac{30}{30 \times 10^{-3}}=100+R_x \\
\Rightarrow & 10^3=100+R_x \Rightarrow R_x=1000-100=900 \Omega
\end{aligned}
\)

Hint

(b) Current, \(I=q f\)
Given, \(q=1.6 \times 10^{-19} \mathrm{C}\) and \(f=10^{19}\)
\(
\therefore \quad I=1.6 \mathrm{~A}
\)

Hint

(c) \(\quad v_d=\frac{I}{n A e}\)
Given, \(\quad I=2 \mathrm{~A}, n=5.86 \times 10^{28} \mathrm{~m}^{-3}\)
\(
A=\pi r^2=\pi\left(0.1 \times 10^{-2}\right)^2, e=1.6 \times 10^{-19} \mathrm{C}
\)
\(\therefore\) Drift velocity, \(v_d=0.68 \times 10^{-4} \mathrm{~ms}^{-1}\)

Hint

\(
\text { (a) As, } R \propto l / d^2 \Rightarrow R_2=R_1\left(\frac{l_2 d_1^2}{l_1 d_2^2}\right)=\frac{1.5(0.31)^2}{(0.155)^2} \times 4.2=25.2 \Omega
\)

Hint

(d) Suppose, \(m\) rows are connected in parallel and each row contains \(n\) identical cells (each cell having \(E=1.5 \mathrm{~V}\) and \(r=1 \Omega\) ) For maximum current in the external resistance \(R\), the necessary condition
\(
\begin{aligned}
R & =\frac{n r}{m} \\
1.5 & =\frac{n \times 1}{m} \\
1.5 m & =n \dots(i)
\end{aligned}
\)
Total cells \(=24=n \times m \dots(ii)\)
On solving Eqs. (i) and (ii), we get
\(
\begin{gathered}
m=4 \\
n=6
\end{gathered}
\)
Therefore, 6 cells in each row are connected in series and 4 such rows are connected in parallel.

Hint

(d)
\(
\begin{aligned}
R_1 & =R_0(1+\alpha t) \\
1 & =R_0(1+0.00125 \times 27) \dots(i)\\
2 & =R_0(1+0.00125 \times t) \dots(ii)
\end{aligned}
\)
From Eqs. (i) and (ii), we get
\(
t=854^{\circ} \mathrm{C}=1127 \mathrm{~K}
\)

Hint

(b) Given, temperature coefficient of carbon,
\(
\alpha_1=4 \times 10^{-3} /{ }^{\circ} \mathrm{C}
\)
Temperature coefficient of copper,
\(
\alpha_2=-0.5 \times 10^{-3} /{ }^{\circ} \mathrm{C}
\)
Hence, \(R_\rho \alpha_1=-R_2 \alpha_2\)
\(
\begin{aligned}
\frac{R_2}{R_1} & =\frac{-4 \times 10^{-3}}{0.5 \times 10^{-3}} \\
\Rightarrow \quad & \frac{R_2}{R_1}=\frac{8}{1}
\end{aligned}
\)

Hint

(a) Let the three conductors having resistances \(R_1, R_2\) and \(R_3\) respectively, and the current drawn by them are 1A, 2A and 3 A respectively, when connected in turn across a battery.
\(
\begin{array}{ll}
\therefore & V=R_1 \quad V=2 R_2 \text { and } V=3 R_3 \\
\text { So, } & R_1=V, R_2=\frac{V}{2} \text { and } R_3=\frac{V}{3}
\end{array}
\)
When the conductors are connected in series, across the same battery, then
\(
\begin{aligned}
& & V & =I\left[R_1+R_2+R_3\right] \\
\Rightarrow & & V & =I\left[V+\frac{V}{2}+\frac{V}{3}\right] \\
\Rightarrow & & 1 & =I\left[\frac{6+3+2}{6}\right] \text { or } I=\frac{6}{11} \mathrm{~A}
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
R \propto l^2 / m & \Rightarrow R_1: R_2: R_3=\frac{l_1^2}{m_1}: \frac{l_2^2}{m_2}: \frac{l_3^2}{m_3} \\
& =\frac{25}{1}: \frac{9}{3}: \frac{1}{5}=125: 15: 1
\end{aligned}
\)

Hint

(b) We have, \(I_1=\frac{I \times 2 R}{3 R}=\frac{2 I}{3}\)
\(
\therefore \quad H_1=I_1^2 R=\frac{4 I^2}{9} \times R \dots(i)
\)
Also, \(\quad I_2=\frac{I \times R}{3 R}=\frac{I}{3}\)
\(
\begin{array}{ll}
\therefore & H_2=I_2^2(2 R)=\frac{I^2}{9} \times 2 R \dots(ii) \\
\text { and } & H_3=I^2(1.5 R) \dots(iii)
\end{array}
\)
From Eqs. (i), (ii) and (iii), we get
\(
\begin{aligned}
H_1: H_2: H_3 & =\frac{4 I^2}{9} \times R: \frac{I^2}{9} \times 2 R: I^2 \times 1.5 R \\
& =\frac{4}{9}: \frac{2}{9}: 1.5=4: 2: 13.5 \\
& =8: 4: 27
\end{aligned}
\)

Hint

(a) A milliammeter will have greater resistance than ammeter because,
1. Number of turns in the coil in milliammeter is increased to compensate low value of current to create large deflection.
2. Milliammeter does not have usually a shunt in parallel because main current is very small. Ammeter usually have a shunt in parallel which makes overall resistance low.

Hint

(c) We know that, the resistance of any wire,
\(
R=\frac{\rho L}{A}
\)
In given case, \(L_1=L, L_2=2 L, \rho\) and \(A\) are constants.
Hence,
\(
\frac{R_1}{R_2}=\frac{L_1}{L_2} \Rightarrow \frac{R_1}{R_2}=\frac{L}{2 L} \Rightarrow \frac{R_1}{R_2}=\frac{1}{2}
\)

Hint

(d) Power when connected in series,
\(
\frac{1}{P}=\frac{1}{P_1}+\frac{1}{P_2}
\)
Given, \(P_1=60 \mathrm{~W}\) and \(P_2=100 \mathrm{~W}\)
Hence, \(P=\frac{60 \times 100}{60+100}\)
\(
\Rightarrow \quad P=\frac{6000}{160} \Rightarrow P=37.5 \mathrm{~W}
\)

Hint

(b) We know that, drift velocity, \(v_d=\frac{e E \tau}{m}=\frac{e \tau}{m} \frac{V}{l}\)
Here, \(\quad v_d \propto v\)
Given, condition \(\quad V_1=V[latex] and [latex]V_2=2 \mathrm{~V}\)
Hence,
\(
\frac{v_d}{v_d^{\prime}}=\frac{V}{2 V} \Rightarrow v_d^{\prime}=2 v_d=2 v
\)

Hint

(d) Specific resistance, or resistivity ( \(\rho\) ), is a fundamental property of the material itself.
It describes how strongly a material opposes the flow of electric current.
Consider the effect of stretching
When a wire is stretched, its length and cross-sectional area change.
However, the material composition of the wire remains the same.
Since specific resistance is a material property, it does not change with the physical dimensions of the wire.
Therefore, stretching the wire will not alter its specific resistance.
Solution
The specific resistance of the wire will remain the same.