Q: A long straight wire of radius a carries a steady current \$I.\$ The current is uniformly distributed over its cross-section. The ratio o the magnetic fields \$B\$ and \$B',\$ at radial distances \$\dfrac{a}{2}\$ and \$2a\$ respectively, from the axis of the wire is

Consider two amperian loops of radius \$a/2\$ and \$2a\$ as shown in the diagram\$.\$Applying ampere's circuital law for these loops\$,\$ we get\$\oint {B.dL = } {\mu _0}{I_{enclosed}}\$For the smaller loop\$,\$\$ \Rightarrow B \times 2\pi \dfrac{a}{2} = {\mu _0} \times \dfrac{1}{{\pi {a^2}}} \times {\left( {\dfrac{a}{2}} \right)^2}\$\$ = {\mu _0}I \times \dfrac{1}{4} = \dfrac{{{\mu _0}I}}{4}\$\$ \Rightarrow {B_1} = \dfrac{{{\mu _0}I}}{{4\pi a}}\$\$B' \times 2\pi \left( {2a} \right) = {\mu _0}I\$\$\dfrac{B}{{B'}} = \dfrac{{{\mu _0}I}}{{4\pi a}} \times \dfrac{{4\pi a}}{{{\mu _0}I}} = 1\$Hence,option \$(C)\$ is correct answer.

Q: The temperature coefficient of resistance of a wire is \$0.00125\$ per degree celcius. At \$300K\$ its resistance is \$1\$ ohm. The resistance of the wire will be \$2\$ ohm at following temperature :-

\$\textbf{Step 1: Applying Relation between Temperature and Resistance}\$ \$R = R_0(1+\alpha (T_2 - T_1))\$Given Initial resistance at \$300K\$ is \$1\Omega\$.Let temperature be T when final resistance becomes \$2\Omega\$ \$\Rightarrow\ \ 2\Omega = 1\Omega(1+ 0.00125 (T-300 K))\$ \$\Rightarrow \ \ 1 = \cfrac{5}{4} \times 10^{-3}(T-300K)\$ \$\therefore\ \ \ T = 1100K\$Hence correct Option is A

Q: An electric lamp of resistance \$20 \, \Omega\$ and a conductor of resistance \$4\, \Omega\$ are connected to a \$6\, V\$ battery as shown in the circuit. Calculate:(a) The total resistance of the circuit(b) The current through the circuit(c)The potential difference across the (i) Electric lamp and (ii) Conductor, and (d) Power of the lamp.

a). the total resistance of the circuit =\$20\Omega +4 \Omega = 24\Omega \$b). The current through the circuit= current through the bulb= current through the conductor.The current through the circuit= \$\dfrac{voltage\ applied}{Total\ Resistance}.\$the current through the circuit \$=\dfrac{6}{24}amp\$The current through the circuit = \$0.25amp\$c). The potential difference across the lamp and the conductor.(i) The potential difference across the electrical lamp \$=0.25\times 20= 5 volt\$(ii) The potential difference conductor \$=0.25 \times 4= 1 \ volt\$d). Power can be calculated as \$\dfrac{voltage \ across\ the \ lamp}{current\ through\ the\ lamp}\$ power=\$\dfrac{5}{0.25}\rightarrow 20 W\$.

Q: \$V-I\$ graph for two wires \$A\$ and \$B\$ are shown in the figure. If the both wires are of same length and same thickness, which of the two is made of a material of high resistivity? Given justification for your answer.

From the graph, slope of wire \$A\$ is greater slope \$=\dfrac { dy }{ dx } =VIR=VI=\$ slopeHence wire \$A\$ has greater resistance. The heat produced by resistor is given by \$P={ I }^{ 2 }.RT\$. It is directly proportional to square of current paring through it resistance and time for which the current passes through the conductor. Hence \$A\$ wire is made of a material of high resistivity.

Q: A battery of emf \$10 V\$ and internal resistance \$3\Omega\$ is connected to a resistor. The current in the circuit is \$0.5 A\$. The terminal voltage of the battery when the circuit is closed is

\$\textbf{Terminal Voltage across the battery:} \$ We know that the Terminal voltage across the battery(\$V_{AB}\$) is given by: \$V_{AB}=E-Ir \$ \$= 10V-(0.5A)3\Omega = 8.5 Volts\$ Hence, Correct Option is D

Q: A wire of length L and resistance R is stretched so that the length is doubled and area of cross section halved. How will (i) resistance change and (ii) resistivity change.

\$ \Rightarrow\$ Length \$(l)\$ is doubled & Area of cross section (A) is halved.\$l\$ becomes \$2l\$ & A becomes \$\frac{A}{2}\$\$R = P\frac{L}{A}\$now take \$l = 2l,A = \frac{A}{2}\$\$R = \frac{{P2l}}{{A/2}} = 4P\frac{L}{A} \Rightarrow 4R\$Resistance becomes 4 times1) Resistivity does not depend on its length 02 Area.Hence, it does not change.

Q: A storage battery of emf 8.0 V and internal resistance \$0.5\Omega\$ is being charged by a 120 V dc supply using a series resistor of \$15.5\Omega\$. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

\$E=8 V\$, internal resistance \$R_1=0.5\Omega\$\$V=120 V\$, resistance \$R_2=15.5\Omega\$Effective voltage \$V^{'}=V-E=112\ V\$\$I=\cfrac{V^{'}}{R_1+R_2}=7\ A\$Voltage across \$R_1=IR_1=3.5\ V\$Terminal voltage \$=8+3.5=11.5\ V\$Series resistor in the charging circuit limits the current drawn from the external source.

Q: A battery of e.m.f. 15 V and internal resistance 3 ohm is connected to two resistors of resistances 3 ohm and 6 ohm in series. The potential difference between the terminals of the battery is

\$ \textbf {Given data: }\$\$ E = 15\ V, r = 3 \Omega, R_1 = 3 \Omega. R_2 = 6 \Omega \$ \$ \textbf {Step 1: Draw circuit diagram and apply KVL [Refer Fig. 1]} \$ While applying KVL in clockwise direction, increase in voltage is considered as positive and decrease in voltage as negative. \$ E - iR_1 - iR_2 - ir = 0\$ .........\$(1) \$ \$ i = \dfrac {E}{R_1+R_2+r}\$ ............\$(2) \$ \$\textbf{Step 2: Finding terminal Potential Difference of Battery [Refer Fig. 2]}\$While writing potential difference \$V_{AB}\$, we go from B to A, increase in voltage is considered as positive and decrease in voltage as negative. \$ V_{\text{terminal}} = V_{AB} = -ir + E = E - \dfrac{Er}{R_1 + R_2 + r}\$ \$ = 15 - \dfrac{15 \times 3}{3+6+3} = 11.25 V\$ \$ V_{\text{terminal}} = 11.25\ V\$

Question 1

A long straight wire of radius a carries a steady current $I.$ The current is uniformly distributed over its cross-section. The ratio o the magnetic fields $B$ and $B',$ at radial distances $\dfrac{a}{2}$ and $2a$ respectively, from the axis of the wire is

Accepted Answer

Consider two amperian loops of radius $a/2$ and $2a$ as shown in the diagram$.$Applying ampere's circuital law for these loops$,$ we get$\oint {B.dL = } {\mu _0}{I_{enclosed}}$For the smaller loop$,$$ \Rightarrow B \times 2\pi \dfrac{a}{2} = {\mu _0} \times \dfrac{1}{{\pi {a^2}}} \times {\left( {\dfrac{a}{2}} \right)^2}$$ = {\mu _0}I \times \dfrac{1}{4} = \dfrac{{{\mu _0}I}}{4}$$ \Rightarrow {B_1} = \dfrac{{{\mu _0}I}}{{4\pi a}}$$B' \times 2\pi \left( {2a} \right) = {\mu _0}I$$\dfrac{B}{{B'}} = \dfrac{{{\mu _0}I}}{{4\pi a}} \times \dfrac{{4\pi a}}{{{\mu _0}I}} = 1$Hence,option $(C)$ is correct answer.

Question 2

The temperature coefficient of resistance of a wire is $0.00125$ per degree celcius. At $300K$ its resistance is $1$ ohm. The resistance of the wire will be $2$ ohm at following temperature :-

Accepted Answer

$\textbf{Step 1: Applying Relation between Temperature and Resistance}$           $R = R_0(1+\alpha (T_2 - T_1))$Given Initial resistance at $300K$ is $1\Omega$.Let temperature be T when final resistance becomes $2\Omega$          $\Rightarrow\ \ 2\Omega = 1\Omega(1+ 0.00125 (T-300 K))$          $\Rightarrow \ \ 1 = \cfrac{5}{4} \times 10^{-3}(T-300K)$          $\therefore\  \ \ T = 1100K$Hence correct Option is A

Question 3

An electric lamp of resistance $20 \, \Omega$ and a conductor of resistance $4\, \Omega$ are connected to a $6\, V$ battery as shown in the circuit. Calculate:(a) The total resistance of the circuit(b) The current through the circuit(c)The potential difference across the (i) Electric lamp and (ii) Conductor, and (d) Power of the lamp.

Accepted Answer

a). the total resistance of the circuit =$20\Omega +4 \Omega = 24\Omega $b). The current through the circuit= current through the bulb= current through the conductor.The current through the circuit= $\dfrac{voltage\  applied}{Total\ Resistance}.$the current through the circuit $=\dfrac{6}{24}amp$The current through the circuit = $0.25amp$c). The potential difference across the lamp and the conductor.(i) The potential difference across the electrical lamp $=0.25\times 20= 5 volt$(ii) The potential difference conductor $=0.25 \times 4= 1 \ volt$d).  Power can be calculated as $\dfrac{voltage \ across\ the \ lamp}{current\  through\  the\  lamp}$     power=$\dfrac{5}{0.25}\rightarrow 20 W$.

Question 4

A student carries out an experiment and plots the V-I graph of three samples of nichrome wirewith resistances R1, R2 and R3 respectively (Figure.12.5). Which of the following is true?

Accepted Answer

Hint: Resistance can be obtained from the slope of the V-I graph. Step 1: Comparing the slopes of the three samples. Slope for $R_1$ is maximum and slope for $R_3$ is the minimum. $(Slope)_{R_1} > (Slope)_{R_2} > (Slope)_{R_3}$Step 2: Comparison of the resistances of the samples from the slopes.We know that,  Resistance of the wire $= \dfrac{1}{Slope \ of \ the \ I-V graph}$Therefore, ${R_3} > {R_2} > {R_1}$Final Step : The correct order of resistance is ${R_3} > {R_2} > {R_1}$

Question 5

$V-I$ graph for two wires $A$ and $B$ are shown in the figure. If the both wires are of same length and same thickness, which of the two is made of a material of high resistivity? Given justification for your answer.

Accepted Answer

From the graph, slope of wire $A$ is greater slope $=\dfrac { dy }{ dx } =VIR=VI=$ slopeHence wire $A$ has greater resistance. The heat produced by resistor is given by $P={ I }^{ 2 }.RT$. It is directly proportional to square of current paring through it resistance and time for which the current passes through the conductor. Hence $A$ wire is made of a material of high resistivity.

Question 6

A battery of emf $10 V$ and internal resistance $3\Omega$ is connected to a resistor. The current in the circuit is $0.5 A$. The terminal voltage of the battery when the circuit is closed is

Accepted Answer

$\textbf{Terminal Voltage across the battery:} $  We know that the Terminal voltage across the battery($V_{AB}$) is given by:                  $V_{AB}=E-Ir $                            $= 10V-(0.5A)3\Omega = 8.5 Volts$ Hence, Correct Option is D

Question 7

A wire of length L and resistance R is stretched so that the length is doubled and area of cross section halved. How will (i) resistance change and (ii) resistivity change.

Accepted Answer

$ \Rightarrow$ Length $(l)$ is doubled & Area of cross section (A) is halved.$l$ becomes $2l$ & A becomes $\frac{A}{2}$$R = P\frac{L}{A}$now take $l = 2l,A = \frac{A}{2}$$R = \frac{{P2l}}{{A/2}} = 4P\frac{L}{A} \Rightarrow 4R$Resistance becomes 4 times1) Resistivity does not depend on its length 02 Area.Hence, it does not change.

Question 8

A storage battery of emf 8.0 V and internal resistance $0.5\Omega$ is being charged by a 120 V dc supply using a series resistor of $15.5\Omega$. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit?

Accepted Answer

$E=8 V$, internal resistance $R_1=0.5\Omega$$V=120 V$,  resistance $R_2=15.5\Omega$Effective voltage $V^{'}=V-E=112\ V$$I=\cfrac{V^{'}}{R_1+R_2}=7\ A$Voltage across $R_1=IR_1=3.5\ V$Terminal voltage $=8+3.5=11.5\ V$Series resistor in the charging circuit limits the current drawn from the external source.

Question 9

Two unequal resistances are connected in parallel. Which of the following statement is true

Accepted Answer

In parallel connection, all the elements are connected between same two points hence the voltage drop across any number of resistors or other elements connected in parallel is same. Hence C is correct.

Question 10

A battery of e.m.f. 15 V and internal resistance 3 ohm is connected to two resistors of resistances 3 ohm and 6 ohm in series. The potential difference between the terminals of the battery is

Accepted Answer

$ \textbf {Given data: }$$ E = 15\ V, r = 3 \Omega, R_1 = 3 \Omega. R_2 = 6 \Omega $ $ \textbf {Step 1: Draw circuit diagram and apply KVL [Refer Fig. 1]} $ While applying KVL in clockwise direction, increase in voltage is considered as positive and decrease in voltage as negative.                     $ E - iR_1 - iR_2 - ir = 0$     .........$(1) $                     $ i = \dfrac {E}{R_1+R_2+r}$              ............$(2) $ $\textbf{Step 2: Finding terminal Potential Difference of Battery [Refer Fig. 2]}$While writing potential difference $V_{AB}$, we go from B to A, increase in voltage is considered as positive and decrease in voltage as negative.                  $ V_{\text{terminal}} = V_{AB} = -ir + E = E - \dfrac{Er}{R_1 + R_2 + r}$                                                                $ = 15 - \dfrac{15 \times 3}{3+6+3} = 11.25 V$                   $ V_{\text{terminal}} = 11.25\ V$