Past NEET Entrance Papers

Hint

(a)

\(
I=n \cdot A \cdot e \cdot v_d
\)
The diameter of the first wire is \(d\). The cross-sectional area \(A_1\) is given by:
\(
A_1=\frac{\pi d^2}{4}
\)
\(
I=n \cdot\left(\frac{\pi d^2}{4}\right) \cdot e \cdot v
\)
The diameter of the second wire is \(\frac{d}{2}\). The cross-sectional area \(A_2\) is given by:
\(
A_2=\frac{\pi\left(\frac{d}{2}\right)^2}{4}=\frac{\pi d^2}{16}
\)
For the second wire, the current is given by:
\(
I=n \cdot A_2 \cdot e \cdot v_d^{\prime}
\)
Substituting \(A_2\) :
\(
I=n \cdot\left(\frac{\pi d^2}{16}\right) \cdot e \cdot v_d^{\prime}
\)
Now, we can solve for \(v_d^{\prime}\) :
\(
\frac{d^2}{4} \cdot v=\frac{d^2}{16} \cdot v_d^{\prime}
\)
Multiplying both sides by 16 :
\(
4 d^2 \cdot v=d^2 \cdot v_d^{\prime}
\)
Dividing both sides by \(d^2\) :
\(
4 v=v_d^{\prime}
\)
Thus, the mean drift velocity \(v_d^{\prime}\) in the second wire (with diameter \(\frac{d}{2}\) ) is:
\(
v_d^{\prime}=4 v
\)

Hint

(a)

When the wire is stretched to twice its length, its cross-sectional area decreases by half (since the volume remains constant).
\(
\begin{aligned}
R & =\rho \frac{l}{A} \\
R^{\prime} & =\rho \frac{2 l}{A / 2} \\
R^{\prime} & =4 \rho \frac{l}{A} \\
R^{\prime} & =4 R \\
R^{\prime} & =4 \cdot 10 \Omega \\
R^{\prime} & =40 \Omega
\end{aligned}
\)
When the wire is bent into a circle, the resistance across any diameter is equivalent to two \(20 \Omega\) resistors in parallel (since the circle is divided into two equal halves).
\(
\begin{aligned}
& \frac{1}{R_{e q}}=\frac{1}{20 \Omega}+\frac{1}{20 \Omega} \\
& \frac{1}{R_{e q}}=\frac{2}{20 \Omega} \\
& R_{e q}=10 \Omega
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
&\begin{aligned}
& \frac{8}{x}=\frac{12}{40-x} \\
& \frac{2}{x}=\frac{3}{40-x} \\
& 80-2 x=3 x \\
& 16=x
\end{aligned}\\
&\text { from } B\\
&=40-16=24 \mathrm{~cm}
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
\text { Current in circuit } i & =\frac{10}{4+1}=2 \mathrm{~A} \\
\text { Terminal voltage } & =E-i R \\
& =10-2 \times 1=8 \mathrm{~V}
\end{aligned}
\)

Hint

(b) Given: Original length of the wire, \(l\).
Total resistance of the wire \(R=100 \Omega\).
The wire is divided into 10 equal parts.
Resistance of each part:
Since the wire is divided into 10 equal parts, the length of each part is \(\frac{l}{10}\). Resistance is proportional to length (as long as the cross-sectional area and material of the wire remain constant). Therefore, the resistance of each part, denoted as \(r\), is \(\frac{1}{10}\) th of the total resistance:
\(
r=\frac{R}{10}=\frac{100 \Omega}{10}=10 \Omega
\)
First 5 parts in series:
When resistors are connected in series, the total resistance is the sum of the individual resistances:
\(
R_{\text {series }}=5 \times 10 \Omega=50 \Omega
\)
Next 5 parts in parallel:
When resistors are connected in parallel, the total resistance \(R_{\text {parallel }}\) can be calculated using the reciprocal formula:
\(
\frac{1}{R_{\text {parallel }}}=\frac{1}{10 \Omega}+\frac{1}{10 \Omega}+\frac{1}{10 \Omega}+\frac{1}{10 \Omega}+\frac{1}{10 \Omega}=5 \times \frac{1}{10 \Omega}=\frac{5}{10 \Omega}=\frac{1}{2 \Omega}
\)
Thus,
\(
R_{\text {parallel }}=2 \Omega
\)
Final combination in series:
The total resistance of the combination, where the series and parallel groups are again connected in series, will be:
\(
R_{\text {total }}=R_{\text {series }}+R_{\text {parallel }}=50 \Omega+2 \Omega=52 \Omega
\)

Hint

(b) Find the resistances of the heaters.
\(
\begin{aligned}
& P=\frac{V^2}{R} \\
& R=\frac{V^2}{P} \\
& R_A=\frac{V^2}{1000} \\
& R_B=\frac{V^2}{2000}
\end{aligned}
\)
Calculate the total resistance in series.
\(
\begin{aligned}
& R_s=R_A+R_B \\
& R_s=\frac{V^2}{1000}+\frac{V^2}{2000} \\
& R_s=\frac{3 V^2}{2000}
\end{aligned}
\)
Calculate the power output in series.
\(
\begin{aligned}
P_s & =\frac{V^2}{R_s} \\
P_s & =\frac{V^2}{\frac{3 V^2}{2000}} \\
P_s & =\frac{2000}{3} \mathrm{~W}
\end{aligned}
\)
Calculate the total resistance in parallel.
\(
\begin{aligned}
& \frac{1}{R_p}=\frac{1}{R_A}+\frac{1}{R_B} \\
& \frac{1}{R_p}=\frac{1000}{V^2}+\frac{2000}{V^2} \\
& \frac{1}{R_p}=\frac{3000}{V^2} \\
& R_p=\frac{V^2}{3000}
\end{aligned}
\)
Calculate the power output in parallel.
\(
\begin{aligned}
& P_p=\frac{V^2}{R_p} \\
& P_p=\frac{V^2}{\frac{V^2}{3000}} \\
& P_p=3000 \mathrm{~W}
\end{aligned}
\)
Find the ratio of power outputs.
\(
\frac{P_s}{P_p}=\frac{\frac{2000}{3}}{3000}=\frac{2}{9}
\)

Hint

(b) In option (2),
\(
\frac{10}{15}=\frac{10}{5+R_D}
\)
The diode can conduct and have resistance \(R_D=10 \Omega\) because diode have dynamic resistance. In that case bridge will be balanced.

Hint

(b) \(R+(R/2)=3R/2\)
\(R_{AB}=[(3R/2) \times R]/[(3R/2)+R]=3R/5\)

Hint

(d) \(2 \times R = 4 \times 1\)
\(R = 2 \Omega\)

Hint

(b) When A and B are connected in series:
1. Total Resistance in Series:
The total resistance \(R_s\) when connected in series is:
\(
R_s=R_1+R_2
\)
2. Substituting the Resistance Values:
Using the earlier expressions for \(R_1\) and \(R_2\) :
\(
R_s=\frac{V^2}{H} \cdot t_1+\frac{V^2}{H} \cdot t_2
\)
Simplifying gives:
\(
R_s=\frac{V^2}{H}\left(t_1+t_2\right)
\)
3. Heat Generation for Series Connection:
The heat generated in time \(t_s\) is:
\(
H=\frac{V^2}{R_s} \cdot t_s
\)
4. Final Equation for Time \(t_s\) :
Setting the heat generated equal gives:
\(
\frac{V^2}{R_s} \cdot t_s=H
\)
Substituting \(R_s\) :
\(
t_s=t_1+t_2
\)

Hint

(a) When \(A\) and \(B\) are connected in parallel:
1. Heat Generation Formula: The heat generated by a coil can be expressed using the formula:
\(
H=\frac{V^2}{R} \cdot t
\)
where \(H\) is the heat generated, \(V\) is the voltage, \(R\) is the resistance, and \(t\) is the time.
2. Resistance Calculation:
For coil A:
\(
R_1=\frac{V^2}{H} \cdot t_1
\)
For coil B:
\(
R_2=\frac{V^2}{H} \cdot t_2
\)
3. Equivalent Resistance in Parallel:
The equivalent resistance \(R_p\) for coils in parallel is given by:
\(
\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}
\)
4. Substituting the Resistance Values:
Substitute \(R_1\) and \(R_2\) into the equation:
\(
\frac{1}{R_p}=\frac{H}{V^2 t_1}+\frac{H}{V^2 t_2}
\)
Simplifying gives:
\(
\frac{1}{R_p}=\frac{H}{V^2}\left(\frac{1}{t_1}+\frac{1}{t_2}\right)
\)
5. Heat Generation for Parallel Connection:
The heat generated in time \(t_p\) is:
\(
H=\frac{V^2}{R_p} \cdot t_p
\)
Setting the heat generated equal for both cases:
\(
H=H
\)
Thus:
\(
\frac{V^2}{R_p} \cdot t_p=H
\)
6. Final Equation for Time \(t_p\) :
By substituting \(R_p\) back into the equation, we find:
\(
t_p=\frac{t_1 t_2}{t_1+t_2}
\)

Hint

(c) In general, resistance of voltmeter is large. For ideal voltmeter, resistance should be infinite.
So, voltmeter of range 100 V , resistance will be less than the ideal voltmeter.
In general, resistance of ammeter is small. For smaller range, shunt resistance should be larger to facilitate larger current through the galvanometer. For larger range, shunt resistance should be smaller to facilitate smaller current through the galvanometer. So, the resistance of ammeter with larger range will be smaller than the resistance of ammeter with smaller range.
\(
\mathrm{D}>\mathrm{A}>\mathrm{B}>\mathrm{C}
\)

Hint

(b) Use Ohm’s Law to calculate the resistance.
\(
\begin{aligned}
R & =\frac{V}{I} \\
R & =\frac{200 \mathrm{~V}}{20 \mathrm{~A}} \\
R & =10 \Omega
\end{aligned}
\)
Calculate the uncertainty in the voltage.
\(
\frac{\Delta V}{V}=\frac{4}{200}=0.02
\)
Calculate the uncertainty in the current.
\(
\frac{\Delta I}{I}=\frac{0.2}{20}=0.01
\)
Calculate the uncertainty in the resistance.
\(
\begin{aligned}
& \frac{\Delta R}{R}=\frac{\Delta V}{V}+\frac{\Delta I}{I} \\
& \frac{\Delta R}{R}=0.02+0.01=0.03
\end{aligned}
\)
Calculate the absolute uncertainty in the resistance.
\(
\begin{aligned}
& \Delta R=0.03 \cdot R \\
& \Delta R=0.03 \cdot 10 \Omega \\
& \Delta R=0.3 \Omega
\end{aligned}
\)
The value of the resistance is \((10 \pm 0.3) \Omega\).

Hint

\(
\begin{aligned}
&\text { Substitute the given values into the equation for the charge amplitude. }\\
&\begin{aligned}
& Q=Q_0 e^{-R t / 2 L} \\
& 0.50 Q_0=Q_0 e^{-(1.5 \Omega) t / 2(12 \mathrm{mH})}
\end{aligned}
\end{aligned}
\)
\(
0.50=e^{-62.5 t}
\)
Take the natural logarithm of both sides of the equation.
\(
\begin{aligned}
& \ln (0.50)=\ln \left(e^{-62.5 t}\right) \\
& \ln (0.50)=-62.5 t
\end{aligned}
\)
\(
\begin{aligned}
&\text { Solve for } \boldsymbol{t} \text {. }\\
&\begin{aligned}
& t=\frac{\ln (0.50)}{-62.5} \\
& t=\frac{-\ln (2)}{-62.5} \\
& t=\frac{0.693}{62.5} \\
& t=0.01109 \mathrm{~s} \\
& t \approx 11.09 \mathrm{~ms}
\end{aligned}
\end{aligned}
\)

Hint

(c) In the steady state Capacitor will be fully charged and behave like open ciruit.
\(
I=\frac{V}{R}=\frac{10}{5}=2 A
\)

Hint

(b)

\(
\mathrm{R}=\frac{\rho \mathrm{L}}{\mathrm{~A}}
\)
If diameter becomes thrice then cross section area will become 9 times so
\(
\mathrm{R} \propto \frac{1}{\mathrm{~A}}
\)
Resistance will become \(\frac{1}{9}\) times
\(
\mathrm{R}^{\prime}=\frac{81 \Omega}{9}=9 \Omega
\)

Hint

(b)

\(
\begin{aligned}
& \mathrm{I}=\mathrm{neAV}_{\mathrm{d}} \\
& \mathrm{~V}_{\mathrm{d}}=\frac{\mathrm{I}}{\mathrm{neA}}=\frac{10}{10^{22} \times 1.6 \times 10^{-19} \times \pi \times 10^{-6}} \\
& \mathrm{~V}_{\mathrm{d}}=\frac{6.25}{\pi} \times 10^3 \mathrm{~m} / \mathrm{sec}
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
& \mathrm{r}=\left(\frac{\ell_{\mathrm{o}}-\ell_{\mathrm{c}}}{\ell_{\mathrm{c}}}\right) \mathrm{R} \\
& 1=\left(\frac{330-\ell_{\mathrm{c}}}{\ell_{\mathrm{c}}}\right) \times 2 \\
& 3 \ell_{\mathrm{c}}=660 \\
& \ell_{\mathrm{c}}=220 \mathrm{~cm}
\end{aligned}
\)

Hint

(a)

\(
\mathrm{i}=\frac{10-5}{10}=\frac{5}{10} \mathrm{~A}
\)
\(
=0.5 \mathrm{~A}
\)
from \(A\) to \(B\) through \(E\).

Hint

(d) For no reading galvanometer. Potential across it is same.
\(
\begin{aligned}
& \mathrm{i}_{400 \Omega} \Rightarrow \frac{10-2}{400}=\frac{8}{400}=\frac{1}{50}=\mathrm{i}_{\mathrm{R}} \\
& \mathrm{i}_{\mathrm{R}} \Rightarrow \frac{\mathrm{~V}_{\mathrm{R}}}{\mathrm{R}} \Rightarrow \frac{2}{\mathrm{R}}=\frac{1}{50} \Rightarrow \mathrm{R}=100 \Omega
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
& \mathrm{I}_{\mathrm{S}}=\frac{E}{10 R} \ldots . .(1) \\
& \mathrm{I}_{\mathrm{P}}=\frac{\mathrm{E}}{\mathrm{R} / 10}=\frac{10 \mathrm{E}}{\mathrm{R}} \ldots .(2) \\
& \mathrm{n}=\frac{\mathrm{I}_{\mathrm{P}}}{\mathrm{I}_{\mathrm{S}}}=100 \Rightarrow \mathrm{n}=100
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
& \mathrm{R}_{\mathrm{T}}=\mathrm{R}_0\left[1+\alpha\left(\mathrm{T}-\mathrm{T}_0\right)\right] \\
& 6.8=2[1+\alpha(80-\alpha)] \\
& \alpha=\frac{2.4}{80}=0.03 /{ }^{\circ} \mathrm{C}=3 \times 10^{-2} /{ }^{\circ} \mathrm{C}
\end{aligned}
\)

Hint

(b) Silver has the smallest resistivity.
Explanation: Resistivity is a measure of how much a material resists the flow of electric current, and a lower resistivity indicates better conductivity. Silver is considered the best electrical conductor among common metals, meaning it has the lowest resistivity.

Why other options are incorrect:
Germanium: While a semiconductor, it has a much higher resistivity than silver.
Glass: Glass is an insulator, meaning it has extremely high resistivity.
Silicon: Like germanium, silicon is a semiconductor with higher resistivity than silver.

Hint

(b)

\(
\begin{aligned}
&\text { Write the resistance formula for wire } \boldsymbol{A} \text {. }\\
&R_A=\frac{\rho L_A}{A_A}
\end{aligned}
\)
\(
\begin{aligned}
&\text { Write the resistance formula for wire } B \text {. }\\
&R_B=\frac{\rho L_B}{A_B}
\end{aligned}
\)
Since the wires are made of the same material and have the same length, we can simplify the equations.
\(
\begin{aligned}
R_A & =\frac{\rho L}{A_A} \\
R_B & =\frac{\rho L}{A_B}
\end{aligned}
\)
\(
\begin{aligned}
&\text { Write the cross-sectional area of wire } A \text { in terms of its radius } \boldsymbol{r}_{\boldsymbol{A}} \text {. }\\
&A_A=\pi r_A^2
\end{aligned}
\)
\(
\begin{aligned}
&\text { Write the cross-sectional area of wire } B \text { in terms of its radius } r_{\boldsymbol{B}} \text {. }\\
&A_B=\pi r_B^2
\end{aligned}
\)
Since the diameter of wire \(B\) is thrice the diameter of wire \(A\), the radius of wire \(B\) is also thrice the radius of wire \(\boldsymbol{A}\).
\(
r_B=3 r_A
\)
\(
\begin{aligned}
&\text { Substitute } r_{\boldsymbol{B}}=3 r_A \text { into the equation for } \boldsymbol{A}_{\boldsymbol{B}} \text {. }\\
&\begin{aligned}
& A_B=\pi\left(3 r_A\right)^2 \\
& A_B=9 \pi r_A^2
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&\text { Substitute } A_B=9 \pi r_A^2 \text { into the equation for } R_B \text {. }\\
&R_B=\frac{\rho L}{9 \pi r_A^2}
\end{aligned}
\)
\(
\begin{aligned}
&\text { Substitute } A_A=\pi r_A^2 \text { into the equation for } \boldsymbol{R}_{\boldsymbol{A}} \text {. }\\
&R_A=\frac{\rho L}{\pi r_A^2}
\end{aligned}
\)
\(
\begin{aligned}
&\text { Compare the equations for } R_A \text { and } R_B \text {. }\\
&R_B=\frac{R_A}{9}
\end{aligned}
\)
\(
\begin{aligned}
&\text { Substitute } \boldsymbol{R}_{\boldsymbol{A}}=81 \boldsymbol{\Omega} \text { into the equation for } \boldsymbol{R}_{\boldsymbol{B}} \text { : }\\
&\begin{aligned}
& R_B=\frac{81 \Omega}{9} \\
& R_B=9 \Omega
\end{aligned}
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
&R=\frac{1}{G}\\
&\text { Thus reciprocal of resistance (R) is conductance (G) }
\end{aligned}
\)

Hint

(b) 

From Kirchhoff’s loop law :
\(
\begin{aligned}
& V-I r-I R=0 \\
& \Rightarrow I=\left(\frac{V}{r+R}\right)
\end{aligned}
\)
Terminal potential difference across cell,
\(
\begin{aligned}
& V_{A B}=I R \\
& V_{A B}=\frac{R V}{r+R} \\
& =\frac{7.5 \times 4}{0.5+7.5}=3.75 \mathrm{~V}
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
&\text { As both resistors are in parallel combination so potential drop }(V) \text { across both are same. }\\
&\begin{aligned}
& P=\frac{V^2}{R} \Rightarrow P \propto \frac{1}{R} \\
& \frac{P_1}{P_2}=\frac{R_2}{R_1}=\frac{200}{100}=\frac{2}{1} \\
& =2: 1
\end{aligned}
\end{aligned}
\)

Hint

(a) The radius \(r\) of the wire is given as:
\(
r=\frac{10^{-2}}{\sqrt{\pi}} \mathrm{~m}
\)
The cross-sectional area \(A\) of the wire can be calculated using the formula for the area of a circle:
\(
A=\pi r^2
\)
Substituting the value of \(r\) :
\(
A=\pi\left(\frac{10^{-2}}{\sqrt{\pi}}\right)^2=\pi\left(\frac{10^{-4}}{\pi}\right)=10^{-4} \mathrm{~m}^2
\)
The electric field \(E\) is given as \(10 \mathrm{~V} / \mathrm{m}\) and the length \(L\) of the wire is 10 m . The potential difference \(V\) across the wire can be calculated using the formula:
\(
V=E \cdot L
\)
Substituting the values:
\(
V=10 \mathrm{~V} / \mathrm{m} \times 10 \mathrm{~m}=100 \mathrm{~V}
\)
\(
\begin{aligned}
&I=\frac{V}{R}\\
&\text { Substituting the values: }\\
&I=\frac{100 \mathrm{~V}}{10 \Omega}=10 \mathrm{~A}
\end{aligned}
\)
Current density \(J\) is defined as the current per unit area:
\(
J=\frac{I}{A}
\)
Substituting the values of current \(I\) and area \(A\) :
\(
J=\frac{10 \mathrm{~A}}{10^{-4} \mathrm{~m}^2}=10 \times 10^4 \mathrm{~A} / \mathrm{m}^2=10^5 \mathrm{~A} / \mathrm{m}^2
\)

Hint

(b)

\(
\begin{aligned}
&\text { The resistances in wheat stone bridge follows: }\\
&\mathrm{P} / \mathrm{X}=\mathrm{Q} / \mathrm{Y}
\end{aligned}
\)
As \(P / X=Q / Y\) so we can conclude that Resistance of \(P\) and \(Q\) should be roughly identical since it reduces experiment error.

Hint

(a) The plot of current (I) versus the applied voltage (V) across a metallic conductor is generally a straight line, obeying Ohm’s law. Ohm’s law states that the current flowing through a conductor is directly proportional to the voltage applied across its ends, at a constant temperature:
\(
\mathrm{I}=\mathrm{V} / \mathrm{R}
\)
Where,
\(I\) is the current in amperes (A),
\(V\) is the voltage in volts \((\mathrm{V})\),
R is the resistance of the conductor in ohms \((\Omega)\).

Hint

(a) \(\frac{4}{3}+\frac{4}{3}+2+1=\frac{17}{3}\)

Hint

(c) 

\(
\text { The formula for the meter bridge is } \frac{R}{S}=\frac{l}{100-l}
\)
Let \(R\) be the resistance in the left gap and \(Y\) be the resistance in the right gap.
Use the meter bridge formula to find the value of \(\frac{R}{Y}\).
\(
\begin{aligned}
\frac{R}{Y} & =\frac{l}{100-l} \\
\frac{R}{Y} & =\frac{30}{100-30} \\
\frac{R}{Y} & =\frac{30}{70} \\
\frac{R}{Y} & =\frac{3}{7}
\end{aligned}
\)
When a resistance of \(16 \Omega\) is connected in parallel with \(Y\), the equivalent resistance becomes \(\frac{16 Y}{16+Y}\).
Use the meter bridge formula again to find the value of \(\frac{R(16+Y)}{16 Y}\).
Substitute \(\frac{R}{Y}=\frac{3}{7}\) into the equation.
The value of the resistance \(Y\) is \(\frac{64}{3} \Omega\).

Hint

(c) Let net resistance of the given infinite network be ‘ \(R\) ‘
The above infinite network can be considered as

Now the total resistnace is: \(R=1+1+\frac{R}{R+1}\)
\(
\begin{aligned}
& \therefore R=2+\frac{R}{R+1} \\
& R^2+R=2 R+2+R \\
& R^2-2 R-2=0 \\
& R=\left(\frac{2 \pm \sqrt{4+8}}{2}\right) \\
& R=(1+\sqrt{3}) \Omega
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
&\begin{aligned}
& \frac{R}{4}=.25 \\
& \mathrm{R}=1
\end{aligned}\\
&\text { Now these four resistances are arranged in series }\\
&\begin{aligned}
& \mathrm{R}_{\mathrm{S}}=\mathrm{R}+\mathrm{R}+\mathrm{R}+\mathrm{R}=4 \mathrm{R} \\
& =4 \times 1=4 \Omega
\end{aligned}
\end{aligned}
\)

Hint

(a) Current density, \(\mathrm{J}=\frac{I}{A}=\) \(nev_{d}\)
Drift velocity, \({V}_{d}=\frac{e E}{m} \tau\);
Electrical resistivity, \(\rho=\frac{m}{n e^2 \tau}\) or \(\rho=\frac{E}{J}\)
Relaxation period, \(\tau=\frac{m}{n e^2 \rho}\)
\(A \rightarrow R\)
\(B \rightarrow S\)
\(D \rightarrow Q\)
\(C \rightarrow P\)

Hint

(c)

\(
\begin{aligned}
&I_3=\frac{I_1 r_2}{r_1+r_3}\\
&\frac{I_3}{I_1}=\frac{I_1 r_2}{\left(r_2+r_3\right) I_1}=\frac{r_2}{r_2+r_3}
\end{aligned}
\)

Hint

(b) For some metals like copper, resistivity is nearly proportional to temperature although a non linear region always exists at very low temperature (\(\text { lower than } 0^{\circ} \mathrm{C}\))

Hint

(a) The relation between the mobility and drift velocity is,
Mobility, \(\mu=\frac{v_d}{E}\)
\(
\begin{aligned}
& =\frac{7.5 \times 10^{-4}}{3 \times 10^{-10}} \\
& =2.5 \times 10^6 \mathrm{~m}^2 V^{-1} \mathrm{~s}^{-1}
\end{aligned}
\)

Hint

(a) Initially, \(\frac{P}{10}=\frac{l_1}{l_2}=\frac{3}{2}\)

\(
\Rightarrow P=\frac{30}{2}=15 \Omega
\)
Now Resistance, \(R=\frac{\rho l}{A}\)
\(
\begin{aligned}
& \frac{R_1}{R_2}=\frac{l_1}{l_2} \\
& \Rightarrow \frac{15}{1}=\frac{1.5}{l_2} \\
& l_2=0.1 m \\
& =1.0 \times 10^{-1} \mathrm{~m}
\end{aligned}
\)

Hint

(b)

\(
I=\frac{2+4}{4+1+1}=\frac{6}{6}=1 A m p
\)

Hint

(b)

\(
R=\frac{\rho L}{A}
\)
Since both conductors are made of the same material, have the same length, and the same resistance, we can write:
\(
R_1=R_2
\)
This implies:
\(
\frac{\rho L}{A_1}=\frac{\rho L}{A_2}
\)
Since \(\rho\) (resistivity) and \(L\) (length) are the same for both conductors, we can cancel them out from both sides of the equation:
\(
\frac{1}{A_1}=\frac{1}{A_2}
\)
Thus, the ratio of the areas is:
\(
\frac{A_1}{A_2}=1
\)

Hint

(c) The temperature coefficient of resistance is negative for semiconductor and insulators, and it is positive for metals.

Hint

(b)

By \(K V L\)
\(
-I_2 R_2-E_2+E_3+I_3 R_1=0
\)
or
\(
I_2 R_2+E_2-E_3-I_3 R_1=0
\)

Hint

(b)

\(
\begin{aligned}
& R_{e q}=\left(4+8+\frac{12 \times 6}{12+6}\right) \Omega \\
& =(12+4) \Omega=16 \Omega
\end{aligned}
\)

Hint

(c) Resistance for ideal voltmeter \(=\infty\)
Resistance for ideal ammeter \(=0\)
For \(1^{\text {st }}\) circuit,
\(
\mathrm{V}_1=\mathrm{i}_1 \times 10=\frac{10}{10} \times 10=10 \mathrm{volt}
\)
For \(2^{\text {nd }}\) circuit,
\(10 \Omega\) is in series with ideal voltmeter. Therefore it will not affect the circuit.
So, \(\mathrm{V}_2=\mathrm{i}_2 \times 10=\frac{10}{10} \times 10=10 \mathrm{volt}\)
\(
\therefore \mathrm{v}_1=\mathrm{V}_2
\)
Hence, \(\mathrm{i}_1=\mathrm{i}_2=\frac{10 \mathrm{~V}}{10 \Omega}=1 \mathrm{~A}\)

Hint

(a) Fuse is used as circuit protector

Hint

(b) When all bulbs are glowing

\(
\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R}}{3}+\frac{\mathrm{R}}{3}=\frac{2 \mathrm{R}}{3}
\)
\(\operatorname{Power}\left(P_i\right)=\frac{E^2}{R_{e q}}=\frac{3 E^2}{2 R} \dots(i)\)
When two from section \(A\) and one from section \(B\) are glowing, then

\(
\mathrm{R}_{\mathrm{eq}}=\frac{\mathrm{R}}{2}+\mathrm{R}=\frac{3 \mathrm{R}}{2}
\)
\(
\operatorname{Power}\left(\mathrm{P}_{\mathrm{f}}\right)=\frac{2 \mathrm{E}^2}{3 R} \dots(ii)
\)
Dividing equation (i) by (ii) we get
\(
\frac{P_i}{P_f}=\frac{3 E^2 3 R}{2 R 2 E^2}=9: 4
\)

Hint

(a) Current in first branch \(=\frac{2}{50} \mathrm{~A}\)
Current in second branch \(=\frac{2}{50} \mathrm{~A}\) \(\Delta \mathrm{V}\) from A to P
\(
\Delta \mathrm{V}_1=2-\frac{2}{50} \times 20
\)

\(
\begin{aligned}
& \Delta \mathrm{V} \text { from } \mathrm{A} \text { to } \mathrm{Q} \\
& \Delta \mathrm{~V}_2=2-\frac{2}{50} \times 30 \\
\Rightarrow & \text { Voltage difference }=0.4 \mathrm{~V}
\end{aligned}
\)

Hint

(d) Yes, the bridge will work. For a balanced condition, the current drawn from the battery will be zero
Also, \(p \propto l_1\) and \(Q \propto l_2\) Therefore, the condition \(\frac{P}{Q}=\frac{l_1}{l_2}\) will remain same after interchanging the cell and galvanometer.

Hint

(a) In series grouping equivalent resistance
\(
\mathrm{R}_{\text {series }}=\mathrm{nR}
\)
In parallel grouping equivalent resistance
\(
\begin{aligned}
& \mathrm{R}_{\text {parallel }}=\frac{R}{n} \\
& I=\frac{E}{n R+R} \dots(i) \\
& 10 \mathrm{I}=\frac{E}{\frac{R}{n}+R} \dots(ii)
\end{aligned}
\)
Dividing eq. (ii) by (i),
\(
10=\frac{(n+1) R}{\left(\frac{1}{n}+1\right) R}
\)
Solving we get, \(\mathrm{n}=10\)

Hint

(a) \(I=\frac{n E}{n r}=\frac{E}{r}\)

From the above equation, it is clear that the total current in the circuit is independent of \({n}\), the number of cells. 

Hint

(b) We know that, \(\mathrm{R}=\frac{\rho^{\ell}}{\mathrm{A}}\)
or \(\mathrm{R}=\frac{\rho \ell^2}{\text { Volume }} \Rightarrow \mathrm{R} \propto \ell^2\)
\(
\begin{aligned}
&\text { According to question } \ell_2=\mathrm{n} \ell_1\\
&\begin{aligned}
& \frac{R_2}{R_1}=\frac{\mathrm{n}^2 l_1^2}{l_1^2} \\
& \text { or, } \frac{\mathrm{R}_2}{\mathrm{R}_1}=\mathrm{n}^2 \\
& \Rightarrow \quad \mathrm{R}_2=\mathrm{n}^2 \mathrm{R}_1
\end{aligned}
\end{aligned}
\)

Hint

(b) A potentiometer is an accurate and versatile device to make electrical measurements of E.M.F. because the method involves zero deflection without any current in galvanometer.

Note: Reading of potentiometer is accurate because during taking reading it does not draw any current from the circuit.

Hint

(c) At time, \(\mathrm{t}=0\) i.e., when switch is closed, inductor in the circuit provides very high resistance (open circuit) while capacitor starts charging with maximum current (low resistance). Equivalent circuit of the given circuit Current drawn from battery,

\(
i=\frac{\varepsilon}{(R / 2)}=\frac{2 \varepsilon}{R}=\frac{2 \times 18}{9}=4 \mathrm{~A}
\)

Hint

(a)

Equivalent resistance across point AC
\(
R_{A C}=\frac{\frac{R_0}{4} \times R}{\frac{R_0}{4}+R}=\frac{R R_0}{R_0+4 R}
\)
From voltage divider rule
\(
\begin{aligned}
& V_{A C}=\frac{R_{A C}}{R_{A C}+R_{C B}} V_0=\frac{\frac{R R_0}{R_0+4 R} V_0}{\frac{R R_0}{R_0+4 R}+\frac{3 R_0}{4}} \\
& =\frac{4 R R_0 V_0}{4 R R_0+3 R_0^2+12 R R_0}=\frac{4 R V_0}{3 R_0+16 R}
\end{aligned}
\)

Hint

(c) Use the Proportionality Relation:
The relationship between the EMFs and the lengths is given by:
\(
\frac{V 1}{V 2}=\frac{L 1}{L 2}
\)
Rearrange the Equation to Find L2
Rearranging the equation gives:
\(
L 2=\frac{L 1 \times V 2}{V 1}
\)
Substitute the Known Values:
Substitute the known values into the equation:
\(
L 2=\frac{36 \mathrm{~cm} \times 2.5 \mathrm{~V}}{1.5 \mathrm{~V}}
\)
Calculate L2:
\(
\begin{aligned}
& L 2=\frac{36 \times 2.5}{1.5} \\
& L 2=\frac{90}{1.5} \\
& L 2=60 \mathrm{~cm}
\end{aligned}
\)

Hint

(d) \(V_A-V_B=(2 A \times 2 \Omega)+3 V+(2 A \times 1 \Omega)=4 V+3 V+2 V=9 V\)

Hint

(c) Resistance of bulb, \(R_B=\frac{V^2}{P}=\frac{(100)^2}{500}=20 \Omega\)
Power of the bulb in the circuit?

\(
\begin{aligned}
P & =V I \\
I & =\frac{P}{V_B} \\
& =\frac{500}{100}=5 \mathrm{~A}
\end{aligned}
\)
\(
\begin{aligned}
& V_R=I R \Leftrightarrow(230-100)=5 \times R \\
\therefore \quad & R=26 \Omega
\end{aligned}
\)

Hint

(c) Suppose two cells have emfs \(\varepsilon_1\) and \(\varepsilon_2\) (also \(\varepsilon_1>\varepsilon_2\) ).
Potential difference per unit length of the potentiometer wire \(=k\) (say)
When \(\varepsilon_1\) and \(\varepsilon_2\) are in series and support each other then
\(
\varepsilon_1+\varepsilon_2=50 \times k \dots(i)
\)
When \(\varepsilon_1\) and \(\varepsilon_2\) are in opposite direction
\(
\varepsilon_1-\varepsilon_2=10 \times k \dots(ii)
\)
On adding eqn. (i) and eqn. (ii)
\(
\begin{aligned}
2 \varepsilon_1 & =60 k \Rightarrow \varepsilon_1=30 k \text { and } \varepsilon_2=50 k-30 k=20 k \\
\therefore \quad \frac{\varepsilon_1}{\varepsilon_2} & =\frac{30 k}{20 k}=\frac{3}{2}
\end{aligned}
\)

Hint

(a) Given, \(Q=a t-b t^2\)
\(
\therefore \quad I=\frac{d Q}{d t}=a-2 b t
\)
At \(t=0, Q=0 \Rightarrow I=0\)
Also, \(I=0\) at \(t=a / 2 b\)
\(\therefore \quad\) Total heat produced in resistance \(R\),
\(
H=\int_0^{a / 2 b} I^2 R d t=R \int_0^{a / 2 b}(a-2 b t)^2 d t
\)
\(
=\frac{a^3 R}{b}\left[\frac{1}{2}+\frac{1}{6}-\frac{1}{2}\right]=\frac{a^3 R}{6 b}
\)

Hint

(c)

As both metal wires are of identical dimensions, so their length and area of cross-section will be same. Let them be \(l\) and \(A\) respectively. Then The resistance of the first wire is
\(
R_1=\frac{l}{\sigma_1 A} \dots(i)
\)
and that of the second wire is
\(
R_2=\frac{l}{\sigma_2 A} \dots(ii)
\)
As they are connected in series, so their effective resistance is
\(
\begin{aligned}
R_s & =R_1+R_2 \\
& =\frac{l}{\sigma_1 A}+\frac{l}{\sigma_2 A} \quad \text { (using (i) and (ii)) } \\
& =\frac{l}{A}\left(\frac{1}{\sigma_1}+\frac{1}{\sigma_2}\right) \dots(iii)
\end{aligned}
\)
If \(\sigma_{\text {eff }}\) is the effective conductivity of the combination, then
\(
R_s=\frac{2 l}{\sigma_{\mathrm{eff}} A} \dots(iv)
\)
\(
\begin{aligned}
&\text { Equating eqns. (iii) and (iv), we get }\\
&\begin{aligned}
& \frac{2 l}{\sigma_{\mathrm{eff}} A}=\frac{l}{A}\left(\frac{1}{\sigma_1}+\frac{1}{\sigma_2}\right) \\
& \frac{2}{\sigma_{\mathrm{eff}}}=\frac{\sigma_2+\sigma_1}{\sigma_1 \sigma_2} \text { or } \quad \sigma_{\mathrm{eff}}=\frac{2 \sigma_1 \sigma_2}{\sigma_1+\sigma_2}
\end{aligned}
\end{aligned}
\)

Hint

(d)

The current through the potentiometer wire is
\(
I=\frac{E_0}{\left(r+r_1\right)}
\)
and the potential difference across the wire is
\(
V=I r=\frac{E_0 r}{\left(r+r_1\right)}
\)
The potential gradient along the potentiometer wire is
\(
k=\frac{V}{L}=\frac{E_0 r}{\left(r+r_1\right) L}
\)
As the unknown e.m.f. \(E\) is balanced against length \(l\) of the potentiometer wire,
\(
\therefore \quad E=k l=\frac{E_0 r}{\left(r+r_1\right)} \frac{l}{L}
\)

Hint

(c)

Resistance of the ammeter is
\(
R_A=\frac{(480 \Omega)(20 \Omega)}{(480 \Omega+20 \Omega)}=19.2 \Omega
\)
(As \(480 \Omega\) and \(20 \Omega\) are in parallel)
As ammeter is in series with \(40.8 \Omega\),
\(\therefore\) Total resistance of the circuit is
\(
R=40.8 \Omega+R_A=40.8 \Omega+19.2 \Omega=60 \Omega
\)
By Ohm’s law,
Current in the circuit is
\(
I=\frac{V}{R}=\frac{30 \mathrm{~V}}{60 \Omega}=\frac{1}{2} \mathrm{~A}=0.5 \mathrm{~A}
\)

Hint

(c)

\(
\begin{aligned}
& \text { Required potential gradient }=1 \mathrm{mV} \mathrm{~cm} \\
& =\frac{1}{10} \mathrm{Vm}^{-1}
\end{aligned}
\)
Length of potentiometer wire, \(l=4 \mathrm{~m}\)
So potential difference across potentiometer wire
\(
=\frac{1}{10} \times 4=0.4 \mathrm{~V} \dots(i)
\)
In the circuit, potential difference across \(8 \Omega\)
\(
=I \times 8=\frac{2}{8+R} \times 8 \dots(ii)
\)
\(
\begin{aligned}
&\text { Using equation (i) and (ii), we get, } 0.4=\frac{2}{8+R} \times 8\\
&\begin{aligned}
& \frac{4}{10}=\frac{16}{8+R}, 8+R=40 \\
\therefore \quad & R=32 \Omega
\end{aligned}
\end{aligned}
\)

Hint

(c)

Effective resistance of B and C
\(
=\frac{\mathrm{R}_{\mathrm{B}} \cdot \mathrm{R}_{\mathrm{C}}}{\mathrm{R}_{\mathrm{B}}+\mathrm{R}_{\mathrm{C}}}=\frac{1.5 \mathrm{R} \times 3 \mathrm{R}}{1.5 \mathrm{R}+3 \mathrm{R}}=\frac{4.5 \mathrm{R}^2}{4.5 \mathrm{R}}=\mathrm{R}
\)
i.e., equal to resistance of voltmeter A .
In parallel potential difference is same so, \(\mathrm{V}_{\mathrm{B}}=\mathrm{V}_{\mathrm{C}}\) and in series current is same So, \(\mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{B}}=\mathrm{V}_{\mathrm{C}}\)

Hint

(d) The area of cross section of conductor is non uniform so current density will be different but the flow of electrons will be uniform so current will be constant.

Hint

(b) Distance between twe cities \(=150 \mathrm{~km}\)
Resistance of the wire, \(R=\left(0.5 \Omega \mathrm{~km}^{-1}\right)(150 \mathrm{~km})=75 \Omega\)
Voltage drop across the wire,
\(
V=\left(8 \mathrm{~V} \mathrm{km}^{-1}\right)(150 \mathrm{~km})=1200 \mathrm{~V}
\)
Power loss in the wire is
\(
P=\frac{V^2}{R}=\frac{(1200 \mathrm{~V})^2}{75 \Omega}=19200 \mathrm{~W}=19.2 \mathrm{~kW}
\)

Hint

(b) In the first case: This is a balanced wheatstone bridge condition,
\(
\frac{5}{R}=\frac{\ell_1}{100-\ell_1} \dots(i)
\)
In the second case:
At balance point
\(
\frac{5}{(R / 2)}=\frac{1.6 l_1}{100-1.6 l_1} \dots(ii)
\)
Divide eqn. (i) by eqn. (ii), we get
\(
\begin{aligned}
& \frac{1}{2}=\frac{100-1.6 l_1}{1.6\left(100-l_1\right)} \\
& 160-1.6 l_1=200-3.2 l_1 \\
& 1.6 l_1=40 \text { or } l_1=\frac{40}{1.6}=25 \mathrm{~cm}
\end{aligned}
\)
Substituting this value in eqn. (i), we get
\(
\begin{aligned}
& \frac{5}{R}=\frac{25}{75} \\
& R=\frac{375}{25} \Omega=15 \Omega
\end{aligned}
\)

Hint

(c) The internal resistance of the cell is
\(
r=\left(\frac{l_1}{l_2}-1\right) R
\)
Here, \(l_1=3 \mathrm{~m}, l_2=2.85 \mathrm{~m}, R=9.5 \Omega\)
\(
\therefore \quad r=\left(\frac{3}{2.85}-1\right)(9.5 \Omega)=\frac{0.15}{2.85} \times 9.5 \Omega=0.5 \Omega
\)

Hint

(d) Resistance of a wire, \(R=\rho \frac{l}{A}=4 \Omega\)
When wire is stretched twice, its new length be \(l^{\prime}\). Then
\(
l^{\prime}=2 l
\)
On stretching volume of the wire remains constant.
\(\therefore l A=l^{\prime} A^{\prime}\) where \(A^{\prime}\) is the new cross-sectional area or \(\quad A^{\prime}=\frac{l}{l^{\prime}} A=\frac{l}{2 l} A=\frac{A}{2}\)
\(\therefore \quad\) Resistance of the stretched wire is
\(
\begin{aligned}
R^{\prime} & =\rho \frac{l^{\prime}}{A^{\prime}}=\rho \frac{2 l}{(A / 2)}=4 \rho \frac{l}{A} \\
& =4(4 \Omega)=16 \Omega \text { (Using (i)) }
\end{aligned}
\)

Hint

(a)

\(
\begin{aligned}
&\begin{aligned}
& I=\frac{\varepsilon}{R+r} \\
& \Rightarrow I R+I r=\varepsilon
\end{aligned}\\
&\text { Here, }\\
&\begin{aligned}
& R=10 \Omega, r=?, \varepsilon=2.1 V, I=0.2 A \\
& \therefore 0.2 \times 10+0.2 \times r=2.1 \\
& 2+0.2 r=2.1 \\
& 0.2 r=0.1 \Rightarrow r=\frac{1}{2}=0.5 \Omega
\end{aligned}
\end{aligned}
\)

Hint

(a) For a balanced Wheatstone’s bridge
\(
\begin{aligned}
& \frac{P}{Q}=\frac{R}{S} \\
\therefore & \frac{10 \Omega}{30 \Omega}=\frac{30 \Omega}{90 \Omega} \text { or } \frac{1}{3}=\frac{1}{3}
\end{aligned}
\)
It is a balanced Wheatstone’s bridge. Hence no current flows in the galvanometerarm. Hence, resistance 50 \(\Omega\) becomes ineffective.
\(\therefore\) The equisalent resistance of the circuit is
\(
\begin{aligned}
\mathrm{R}_{\mathrm{qg}} & =5 \Omega+\frac{(10 \Omega+30 \Omega)(30 \Omega+90 \Omega)}{(10 \Omega+30 \Omega)+(30 \Omega+90 \Omega)} \\
& =5 \Omega+\frac{(40 \Omega)(120 \Omega)}{40 \Omega+120 \Omega}=5 \Omega+30 \Omega=35 \Omega
\end{aligned}
\)
Current drawn from the cell is
\(
I=\frac{7 \mathrm{~V}}{35 \Omega}=\frac{1}{5} \mathrm{~A}=0.2 \mathrm{~A}
\)

Hint

(a) Let \(\rho\) be the resistivity of the material of the wire and r be the radius of the wire.
Therefore, the resistance of 1 m wire is
\(
\begin{aligned}
& R=\frac{\rho(1)}{\pi r^2}=\frac{\rho}{\pi r^2} \\
& \left(\because R=\frac{\rho l}{A}\right)
\end{aligned}
\)
Let \(\varepsilon\) be emf of each cell.
In the first case, 10 cells each of emf \(\varepsilon\) are connected in series to heat the wire of length 1 m by \(\Delta\) \(T\left(=10^{\circ} \mathrm{C}\right)\) in time t.

\(
\therefore \frac{(10 \varepsilon)}{R} r=m s \Delta T \quad \ldots \text { (i) }
\)
In second case, Resistance of same wire of length 2 m is

\(
R^{\prime}=\frac{\rho(2)}{\pi r^2}=\frac{2 \rho}{\pi r^2}=2 R
\)
Let n cells each of emf \(\varepsilon\) are connected in series to heat the same wire of length 2 m , by the same temperature \(\Delta \mathrm{T}\left(=10^{\circ} \mathrm{C}\right)\) in the same time t.
\(
\therefore \frac{(n \varepsilon)^2 t}{2 R}=(2 m) s \Delta T
\)
Divide (ii) by (i), we get
\(
\begin{aligned}
& \frac{n^2}{200}=2 \Rightarrow n^2=400 \\
& \therefore n=20
\end{aligned}
\)

Hint

(d) Here,
Length of each rod, \(l=1 \mathrm{~m}\)
Area of cross-section of each rod,
\(
A=0.01 \mathrm{~cm}^2=0.01 \times 10^{-4} \mathrm{~m}^2
\)
Resistivity of copper rod,
\(
\begin{aligned}
\rho_{\mathrm{cu}} & =1.7 \times 10^{-6} \Omega \mathrm{~cm} \\
& =1.7 \times 10^{-6} \times 10^{-2} \Omega \mathrm{~m}=1.7 \times 10^{-8} \Omega \mathrm{~m}
\end{aligned}
\)
Resistivity of iron rod,
\(
\begin{aligned}
\rho_{\mathrm{Fe}} & =10^{-5} \Omega \mathrm{~cm} \\
& =10^{-5} \times 10^{-2} \Omega \mathrm{~m}=10^{-7} \Omega \mathrm{~m}
\end{aligned}
\)
\(\therefore\) Resistance of copper rod,
\(
R_{\mathrm{Cu}}=\rho_{\mathrm{Cu}} \frac{l}{A}
\)
and resistance of iron rod, \(R_{\mathrm{Fe}}=\rho_{\mathrm{Fe}} \frac{l}{A}\)
As copper and iron rods are connected in series, therefore equivalent resistance is
\(
R=R_{\mathrm{Cu}}+R_{\mathrm{Fe}}=\rho_{\mathrm{Cu}} \frac{l}{A}+\rho_{\mathrm{Fe}} \frac{l}{A}=\left(\rho_{\mathrm{Cu}}+\rho_{\mathrm{Fe}}\right) \frac{l}{A}
\)
\(
\begin{aligned}
&\text { Voltage required to produce } 1 \text { A current in the tods is }\\
&\begin{aligned}
V & =I R=(1)\left(R_{\mathrm{Cu}}+R_{\mathrm{Fe}}\right) \\
& =\left(\rho_{\mathrm{Cu}}+\rho_{\mathrm{Fe}}\right)\left(\frac{l}{A}\right) \\
& =\left(1.7 \times 10^{-8}+10^{-7}\right)\left(\frac{1}{0.01 \times 10^{-4}}\right) \mathrm{V} \\
& =10^{-7}(0.17+1)\left(10^6\right) \mathrm{V}=17 \times 10^{-1} \mathrm{~V}=0.117 \mathrm{~V}
\end{aligned}
\end{aligned}
\)

Hint

(c) Resistance is directly proportional to length,
\(
\begin{aligned}
& \frac{1}{R_{A B}}=\frac{1}{3}+\frac{1}{4+5}=\frac{(4+5)+3}{(3)(4+5)} \\
& R_{A B}=\frac{3 \times(4+5)}{3+(4+5)}=\frac{27}{12}
\end{aligned}
\)
Similarly,
\(
\begin{aligned}
& R_{B C}=\frac{4 \times(3+5)}{4+(3+5)}=\frac{32}{12} \\
& R_{A C}=\frac{5 \times(3+4)}{5+(3+4)}=\frac{35}{12} \\
& \therefore R_{A B}: R_{B C}: R_{A C}=27: 32: 35
\end{aligned}
\)

Hint

(b) Since the galvanometer shows no deflection so current will flow as shown in the figure.

\(
\begin{aligned}
&\text { Current, }\\
&\begin{aligned}
& I=\frac{V_A}{R_1+R}=\frac{12 V}{(500+100) \Omega}=\frac{12}{600} A \\
& V_B=I R=\left(\frac{12}{600} A\right)(100 \Omega)=2 V
\end{aligned}
\end{aligned}
\)

Hint

(c) Power, \(P=\frac{V^2}{R}\)
As the resistance of the bulb is constant
\(
\begin{aligned}
& \therefore \quad \frac{\Delta P}{P}=\frac{2 \Delta V}{V} \\
& \% \text { decrease in power }=\frac{\Delta P}{P} \times 100=\frac{2 \Delta V}{V} \times 100 \\
& =2 \times 2.5 \%=5 \%
\end{aligned}
\)

Hint

(d) Let \(x\) is the resistance per unit length then

\(
\text { equivalent resistance } R=\frac{R_1 R_2}{R_1+R_2} \frac{\left(x l_1\right)\left(x l_2\right)}{x l_1+x l_2}
\)
\(
\begin{aligned}
& \Rightarrow \quad \frac{8}{3}=x \frac{l_1 l_2}{l_1+l_2} \\
& \frac{8}{3}=x \frac{l_1}{\frac{l_1}{l_2}+1} \dots(i)
\end{aligned}
\)
\(
\text { also } \begin{aligned}
R_0 & =x l_1+x l_2 \\
12 & =x\left(l_1+l_2\right) \\
12 & =x l_2\left(\frac{l_1}{l_2}+1\right) \dots(ii)
\end{aligned}
\)
Divide (i) by (ii), we get
\(
\frac{\frac{8}{3}}{\frac{12}{1}}=\frac{\frac{x l_1}{\left(\frac{l_1}{l_2}+1\right)}}{x l_2\left(\frac{l_1}{l_2}+1\right)}=\frac{l_1}{l_2\left(\frac{l_1}{l_2}+1\right)^2}
\)
\(
\left(\frac{l_1}{l_2}+1\right)^2 \times \frac{8}{36}=\frac{l_1}{l_2}
\)
\(
\begin{aligned}
& \left(y^2+1+2 y\right) \times \frac{8}{36}=y\left(\text { where } y=\frac{l_1}{l_2}\right) \\
& \quad 8 y^2+8+16 y=36 y \\
& \Rightarrow 8 y^2-20 y+8=0 \\
& \Rightarrow 2 y^2-5 y+2=0 \\
& \Rightarrow 2 y^2-4 y-y+2=0 \\
& \Rightarrow 2 y(y-2)-1(y-2)=0 \\
& \Rightarrow \\
& (2 y-1)(y-2)=0 \\
& \Rightarrow \quad y=\frac{l_1}{l_2}=\frac{1}{2} \text { or } 2
\end{aligned}
\)

Hint

(b) The equivalent resistance of the given circuit is
\(
R_{\mathrm{eq}}=\frac{5 R}{5+R}
\)
Power dissipated in the given circuit is
\(
\begin{aligned}
& P=\frac{V^2}{R_{\mathrm{eq}}} \text { or } 30=\frac{(10)^2}{\left(\frac{5 R}{5+R}\right)} \\
& 150 R=100(5+R) \\
& 150 R=500+100 R \text { or } 50 R=500 \\
& R=\frac{500}{50}=10 \Omega
\end{aligned}
\)

Hint

(b)

Current in the circuit, \(I=\frac{\varepsilon}{R+r}\)
Potential difference across R,
\(
\begin{aligned}
& V=I R=\left(\frac{\varepsilon}{R+r}\right) R \\
& V=\frac{\varepsilon}{1+\frac{r}{R}}
\end{aligned}
\)
When \(\mathrm{R}=0, \mathrm{~V}=0\) and when \(\mathrm{R}=\infty, \mathrm{V}=\varepsilon\)
Thus \(V\) increases as \(R\) increases upto certain limit, but it does not increase fürther.

Hint

(b) Current flows through the \(9 \Omega\) resistor is
\(
\begin{aligned}
& I_1^2=\frac{36}{9}=4 \text { (As } P=I^2 R \text { ) }\\
& I_1=2 \mathrm{~A}
\end{aligned}
\)
As the resistors \(9 \Omega\) and \(6 \Omega\) are connected in parallel, therefore potential difference across them is same.
\(
\begin{aligned}
\therefore \quad 9 I_1= & 6 I_2 \\
& I_2=\frac{9 \times 2}{6}=3 \mathrm{~A}
\end{aligned}
\)
Current drawn from the battery is
\(
I=I_1+I_2=(2+3) \mathrm{A}=5 \mathrm{~A}
\)
The potential difference across the \(2 \Omega\) resistor is
\(
=(5 \mathrm{~A})(2 \Omega)=10 \mathrm{~V}
\)

Hint

(b) Let \(\varepsilon\) be the emf and \(r\) be internal resistance of the battery.

In the first case,

\(
2=\frac{\varepsilon}{2+r}
\)
In the second case,

\(
\begin{aligned}
&0.5=\frac{\varepsilon}{9+r}\\
&\text { Divide (i) by (ii), we get }\\
&\begin{aligned}
& \frac{2}{0.5}=\frac{9+r}{2+r} \Rightarrow 4+2 r=4.5+0.5 r \\
& 1.5 r=0.5 \Rightarrow r=\frac{0.5}{1.5}=\frac{1}{3} \Omega
\end{aligned}
\end{aligned}
\)

Hint

(a) We have, \(\mathrm{e}=\mathrm{at}+\mathrm{bt}{ }^2\)
\(
\Rightarrow \frac{\mathrm{de}}{\mathrm{dt}}=\mathrm{a}+2 \mathrm{bt}
\)
At neutral temperature,
\(
\begin{aligned}
& \mathrm{t}=-\frac{\mathrm{a}}{2 \mathrm{~b}} \\
& \therefore \frac{\mathrm{de}}{\mathrm{dt}}=0
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (d) Current from } \mathrm{D} \text { to } \mathrm{C}=1 \mathrm{~A}\\
&\begin{aligned}
& \therefore V_D-V_C=2 \times 1=2 V \\
& \\
& V_A=0 \quad \therefore V_C=1 V, \therefore V_D-V_C=2 \\
& \therefore V_D-1=2 \quad \therefore V_D=3 V \\
&
\end{aligned}
\end{aligned}
\)
\(
\therefore V_D-V_B=2 \quad \therefore 3-V_B=2 \therefore V_B=1 V
\)

Hint

(a)

\(
\text { For minimum deflection of } 1 \text { division required current }=1 \mu \mathrm{~A}
\)
\(
\begin{aligned}
& \Rightarrow \text { Voltage required }=\mathrm{IR}=(1 \mu \mathrm{~A})(10)=10 \mu \mathrm{~V} \\
& \therefore 40 \mu \mathrm{~V} \equiv 1^{\circ} \mathrm{C} \\
& \Rightarrow 10 \mu \mathrm{~V} \equiv \frac{1}{4}^{\circ} \mathrm{C}=0.25^{\circ} \mathrm{C}
\end{aligned}
\)

Hint

(b) (i) When key between the terminals 1 and 2 is plugged in,
P.D. across \(R=I R=k l_1\)
\(
\Rightarrow R=k l_1 \text { as } I=1 A
\)
(ii) When key between terminals 1 and 3 is plugged in.
P.D. \(\operatorname{across}(X+R)=I(X+R)=k l_2\)
\(
\begin{array}{ll}
\Rightarrow & \mathrm{X}+\mathrm{R}=k l_2 \\
\therefore & X=k\left(l_2-l_1\right) \\
\therefore & R=k l_1 \text { and } X=k\left(l_2-l_1\right)
\end{array}
\)

Hint

(d) Kirchhoff’s junction law or Kirchhoff’s first law is based on the conservation of charge. Kirchhoff’s loop law or Kirchhoff’s second law is based on the conservation of energy. Hence both statements (A) and (B) are correct.

Hint

(b) \(E=30 \theta-\frac{\theta^2}{15}\)
For neutral temperature, \(\frac{d E}{d \theta}=0\)
\(
\begin{aligned}
& 0=30-\frac{2}{15} \theta \\
& \therefore \theta=15 \times 15 \\
& =225^{\circ} \mathrm{C}
\end{aligned}
\)
Hence, neutral temperature is \(225^{\circ} \mathrm{C}\).

Hint

(d)

\(
\text { Length of wire }=2 \pi r=2 \pi(0.1)=0.2 \pi \mathrm{~m}
\)

Total resistance of wire \(=12 \Omega \times 2 \pi \times 10^{-1}=2.4 \pi\)
Resistance of each half \(=\frac{2.4 \pi}{2}=1.2 \pi\)
and as about diameter both parts are in parallel
\(
R_{AB}=\frac{1.2 \pi}{2}=0.6 \pi \Omega
\)

Hint

(b) The terminal potential difference of a cell is given by \(\mathrm{V}+\mathrm{Ir}=\varepsilon\)
\(
\begin{aligned}
& \mathrm{V}=\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}} \\
& \Rightarrow \mathrm{~V}=\varepsilon-\mathrm{Ir}
\end{aligned}
\)
\(\Rightarrow \frac{d V}{d I}=-r\), Also for, \(\mathrm{i}=0\) then \(\mathrm{V}=\varepsilon\)
\(\therefore\) slope \(=-r\), intercept \(=\varepsilon\)

Hint

(d)

Applying Kirchhoff’s rule in loop abefa \(\varepsilon_1-\left(\mathrm{i}_1+\mathrm{i}_2\right) \mathrm{R}-\mathrm{i}_1 \mathrm{r}_1=0\).

Hint

\(
\begin{aligned}
&\text { (c) Energy }=2 \mathrm{eV}=e E \lambda\\
&\therefore E=\frac{2 e V}{e \lambda}=\frac{2}{4 \times 10^{-8}}=5 \times 10^7 \mathrm{~V} / \mathrm{m}
\end{aligned}
\)

Hint

(d)

Clearly, \(2 \Omega, 4 \Omega\) and \((1+5) \Omega\) resistors are in parallel. Hence, potential difference is same across each of them.
\(
\therefore I_1 \times 2=I_2 \times 4=I_3 \times 6
\)
Given \(\mathrm{I}_1=3 \mathrm{~A} \quad \therefore \mathrm{I}_1 \times 2=\mathrm{I}_3 \times 6\)
Given \(I_1=3 \mathrm{~A}\).
\(
\begin{aligned}
& \therefore I_1 \times 2=I_3 \times 6 \text { provides } \\
& I_3=\frac{I_1 \times 2}{6}=\frac{3 \times 2}{6}=1 \mathrm{~A}
\end{aligned}
\)
Total P.D. \(=5 \times 1+1 \times 1=6 \mathrm{~V}\)
Now, the potential across the \(5 \Omega\) resistor is 5V.
\(\therefore\) the power dissipated in the \(5 \Omega\) resistor
\(
\mathrm{P}=\frac{\mathrm{V}^2}{\mathrm{R}}=\frac{5^2}{5}=5 \mathrm{watt} .
\)

Hint

(b)

\(
\frac{\Delta l}{l}=0.1 \therefore l=1.1
\)
but the area also decreases by 0.1.
\(
\text { mass }=\rho l A=V \rho, \ln l+\ln A=\ln \text { mass. }
\)
\(
\therefore \frac{\Delta l}{l}+\frac{\Delta A}{A}=0 \Rightarrow \frac{\Delta l}{l}=\frac{-\Delta A}{A}
\)
Length increases by 0.1 , resistance increases, area decreases by 0.1 , then also resistance will increase. Total increase in resistance is approximately 1.2 times, due to increase in length and decrease in area. But specific resistance does not change.

Hint

(c) Power \(=220 \mathrm{~V} \times 4 \mathrm{~A}=880\) watts. \(=880 \mathrm{~J} / \mathrm{s}\). Heat needed to raise the temperature of 1 kg water through \(80^{\circ} \mathrm{C}\)
\(
\begin{aligned}
& =m s . \Delta T \times 4.2 \mathrm{~J} / \mathrm{cal} \\
& =1000 \mathrm{~g} \times 1 \mathrm{cal} / \mathrm{g} \times 80 \times 4.2 \mathrm{~J} / \mathrm{cal} .
\end{aligned}
\)
\(
\begin{aligned}
& \therefore \text { Time taken }=\frac{1000 \times 1 \times 80 \times 4.2}{880}=\frac{336 \times 10^3}{880} \\
& =382 \mathrm{~s}=6.3 \mathrm{~min} .
\end{aligned}
\)

Hint

(a) [In the question, the length \(110 \mathrm{~cm} \& 100\) cm are interchanged ] as \(\varepsilon>\frac{\varepsilon R}{R+r}\) Without being short circuited through \(R\), only the battery \(\varepsilon\) is balanced.
\(
\varepsilon=\frac{V}{L} \times l_1=\frac{V}{L} \times 110 \mathrm{~cm} \dots(i)
\)
\(
\begin{aligned}
&\text { When } R \text { is connected across } \varepsilon \text {, }\\
&R i=R \cdot\left(\frac{\varepsilon}{R+r}\right)=\frac{V}{L} \times I_2 \Rightarrow \frac{R \varepsilon}{R+r}=\frac{V}{L} \times 100 \dots(ii)
\end{aligned}
\)
\(
\begin{aligned}
&\text { Dividing eqn. (i) and (ii), } \frac{(R+r)}{R}=\frac{110}{100}\\
&\begin{aligned}
& \Rightarrow 1+\frac{r}{R}=\frac{110}{100} \Rightarrow \frac{r}{R}=\frac{110}{100}-\frac{100}{100} \\
& \Rightarrow r=R \cdot \frac{10}{100}=\frac{R}{10} . \quad \text { As } R=10 \Omega ; r=1 \Omega
\end{aligned}
\end{aligned}
\)

Hint

(d) As the P.D. between \(4 \Omega\) and \(3 \Omega\) (in parallel), are the same,
\(
4 \times 1 \mathrm{amp}=3 \times i_1 \Rightarrow i_1=\frac{4}{3} \mathrm{~A}
\)
Total resistance of \(4 \Omega\) and \(3 \Omega=12 / 7 \Omega\).
Current in \(M Q P(\) upper one \()=1+\frac{4}{3}=\frac{7}{3} \mathrm{~A}\)
\(
\therefore \text { P.D. }=\frac{12}{7} \times \frac{7}{3}=4 \mathrm{~V}
\)
Current in \(M N P=\frac{4}{1.25}=\frac{4 \times 4}{5}=\frac{16}{5} \mathrm{~A}\)
\(\therefore\) P.D. across \(1 \Omega=\frac{16}{5} \mathrm{~A} \times 1 \Omega=\frac{16}{5}\) volt
\(\Rightarrow\) PD. across \(1 \Omega=3.2\) volt.

Hint

\(
\begin{aligned}
&\text { (c) Power dissipiated }=P\\
&=\frac{V^2}{R}=\frac{(18)^2}{6}=54 \mathrm{~W}
\end{aligned}
\)

Hint

(b) A balanced wheatstone bridge simply requires
\(
\frac{P}{Q}=\frac{R}{S} \Rightarrow \frac{2}{2}=\frac{2}{S}
\)
Therefore, S should be \(2 \Omega\).
A resistance of \(6 \Omega\) is connected in parallel.
In parallel combination,
\(
\begin{aligned}
& \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2} \\
& \frac{1}{2}=\frac{1}{6}+\frac{1}{S} \\
& \Rightarrow S=3 \Omega
\end{aligned}
\)

Hint

(c) Current will flow from \(B\) to \(A\)

Potential drop over the resistance \(C A\) will be more due to higher value of resistance. So potential at \(A\) will be less as compared with at \(B\). Hence, current will flow from \(B\) to \(A\).

Hint

(a) 

Current in the circuit

\(
\begin{aligned}
& =\frac{E+E}{r_1+r_2+R}=\frac{2 E}{r_1+r_2+R} \\
& \text { P.D. across first cell }=E-i r_1 \\
& =E-\frac{2 E \times r_1}{\left(r_1+r_2\right)+R} \\
& \text { Now, } E=\frac{2 E r_1}{\left(r_1+r_2\right)+R}=0 \\
& \Rightarrow E=\frac{2 E r_1}{r_1+r_2+R} \Rightarrow 2 r_1=r_1+r_2+R \\
& R=r_1-r_2
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
&\text { Power }=\mathrm{V} . I=I^2 \mathrm{R}\\
&\begin{aligned}
& i_2=\sqrt{\frac{\text { Power }}{R}}=\sqrt{\frac{2}{8}} \\
& =\sqrt{\frac{1}{4}}=\frac{1}{2} A
\end{aligned}
\end{aligned}
\)
Potential over \(8 \Omega=R i_2=8 \times \frac{1}{2}=4 V\)
This is the potential over parallel branch. So,
\(
i_1=\frac{4}{4}=1 A
\)
Power of \(3 \Omega=i_1{ }^2 R=1 \times 1 \times 3=3 W\)

Hint

(a) Kirchhoff ‘ s first law deals with conservation of electrical charge and the second law deals with conservation of electrical energy.

Hint

(d) It is a balanced Wheatstone bridge. Hence resistance \(4 \Omega\) can be eliminated.
\(
\begin{aligned}
& \therefore R_{\mathrm{eq}}=\frac{6 \times 9}{6+9}=\frac{18}{5} \\
& \therefore i=\frac{V}{R_{\mathrm{eq}}}=\frac{5 V}{18}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (b) } P=i^2 R \quad \text { or } \quad 1=25 \times R \\
& R=\frac{1}{25}=0.04 \Omega
\end{aligned}
\)

Hint

(b)

\(
\mathrm{R}_{\mathrm{eq}}=\frac{R_1 R_2}{R_1+R_2}=\frac{\left(\frac{R}{2} \cdot \frac{R}{2}\right)}{\frac{R}{2}+\frac{R}{2}} \quad \therefore R_{\mathrm{eq}}=\frac{R}{4}
\)

Hint

\(
\text { (c) } \mathrm{V}=\frac{\frac{E_1}{r_1}+\frac{E_2}{r_2}}{\frac{1}{r_1}+\frac{1}{r_2}}=\frac{\frac{18}{2}+\frac{12}{1}}{\frac{1}{2}+\frac{1}{1}}=14 \mathrm{~V}
\)  (Since the cells are in parallel)

Hint

(c) \(R=\frac{\rho \ell_1}{A_1}\), now \(\ell_2=2 \ell_1\)
\(
\begin{aligned}
& A_2=\pi\left(r_2\right)^2=\pi\left(2 r_1\right)^2=4 \pi r_1^2=4 A_1 \\
& \therefore \quad R_2=\frac{\rho\left(2 \ell_1\right)}{4 A_1}=\frac{\rho \ell_1}{2 A_1}=\frac{R}{2}
\end{aligned}
\)
\(\therefore\) Resistance is halved, but specific resistance remains the same.

Hint

(b) \(\mathrm{R} \propto \ell\)
For \(300 \mathrm{~cm}, \mathrm{R}=100 \Omega\)
For \(50 \mathrm{~cm}, R^{\prime}=\frac{100}{300} \times 50=\frac{50}{3} \Omega\)
\(
\begin{aligned}
& \therefore \mathrm{IR}=6 \\
& \Rightarrow I R^{\prime}=\frac{6}{R} \times R^{\prime}=\frac{6}{100} \times \frac{50}{3}=1 \mathrm{volt}
\end{aligned}
\)

Hint

(b) The given circuit can be redrawn as shown.

From circuit,
\(
\frac{F C}{C E}=\frac{F D}{D E}=1
\)
Thus, it is balanced Wheatstone’s bridge, so resistance in arm \(C D\) is indffective and so, current flows in arm.
Net resistance of the circuit is
\(
\begin{aligned}
& \frac{1}{R^{\prime}}=\frac{1}{(R+R)}+\frac{1}{(R+R)} \\
& =\frac{1}{2 R}+\frac{1}{2 R}=\frac{2}{2 R}=\frac{1}{R} \\
& R^{\prime}=R
\end{aligned}
\)
So, net current drawn from the battery
\(
i^{\prime}=\frac{V}{R^{\prime}}=\frac{V}{R}
\)
\(
\text { Current through } \mathrm{AFCEB}=\mathrm{V} / 2 \mathrm{R}
\)

Hint

\(
\begin{aligned}
&\text { (c) Efficiency is given by } \eta=\frac{\text { output }}{\text { input }}\\
&=\frac{5 \times 15 \times 14}{10 \times 8 \times 15}=0.875 \text { or } 87.5 \%
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
& P=\frac{V^2}{R} \\
& \Rightarrow R=\frac{V^2}{P}=\frac{(220)^2}{60}=\frac{4 \times(110)^2}{60} \\
& R^{\prime}=\frac{(110)^2}{60}=\frac{R}{4}
\end{aligned}
\)

Hint

(d) When \(n\) resistance of \(r\) ohm connected in parallel then their equivalent resistance is
\(
\begin{aligned}
& \Rightarrow \frac{1}{R}=\frac{1}{r}+\frac{1}{r}+\frac{1}{r}+\ldots \ldots . . . . . n \text { times } \\
& \therefore \quad \frac{1}{R}=\frac{n}{r} \Rightarrow R=\frac{r}{n} \Rightarrow r=n R
\end{aligned}
\)
When these resistance connected in series
\(
\begin{aligned}
R_s & =r+r+\ldots \ldots \ldots . . . . . n \text { times } \\
& =n r=n \times n R=n^2 R
\end{aligned}
\)

Hint

(c) The resistance of each bulb
\(
=\frac{V^2}{P}=\frac{(200)^2}{60} \Omega
\)
When three bulbs are connected in series their resultant resistance
\(
=\frac{3 \times(200)^2}{60}
\)
Thus power drawn by bulb when connected across 200 V supply
\(
P=\frac{V^2}{R_{r e}}=\frac{(200)^2}{3 \times(200)^2 / 60}=20 \mathrm{~W}
\)

Hint

(b) Fuse wire should have high resistance and low melting point.

Hint

(c) In balance Wheatstone bridge, the galvanometer arm can be neglected so equivalent resistance \(={R}\).

Hint

(a) Let \(R_1\) and \(R_2\) be the resistance of the two coils and \(V\) be the voltage supplied.
Effective resistance of two coils in parallel \(=\frac{R_1 R_2}{R_1+R_2}\)
Let \(H\) be the heat required to begin boiling in kettle.
Then \(H=\) Power \(\times\) time \(=\frac{V^2 t_1}{R_1}=\frac{V^2 t_2}{R_2}\)
For parallel combination, \(H=\frac{V^2\left(R_1+R_2\right) t_p}{R_1 R_2}\)
\(
\begin{aligned}
& \Rightarrow \frac{1}{t_p}=\left(\frac{t_2+t_1}{t_2 t_1}\right) \\
& \therefore t_p=\frac{t_1 t_2}{t_1+t_2}=\frac{10 \times 40}{10+40}=8 \text { minute. }
\end{aligned}
\)