Current Electricity
A wire of resistance $160 ~\Omega$ is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be
The current flowing through R$_2$ is :
Two identical heater filaments are connected first in parallel and then in series. At the same applied voltage, the ratio of heat produced in same time for parallel to series will be:
The current sensitivity of moving coil galvanometer is increased by $25 \%$. This increase is achieved only by changing in the number of turns of coils and area of cross section of the wire while keeping the resistance of galvanometer coil constant. The percentage change in the voltage sensitivity will be:
The equivalent resistance of the circuit shown below between points a and b is :
The equivalent resistance between A and B as shown in figure is:
Figure shows a part of an electric circuit. The potentials at points $a, b$ and $c$ are $30 \mathrm{~V}, 12 \mathrm{~V}$ and $2 \mathrm{~V}$ respectively. The current through the $20 ~\Omega$ resistor will be,
A student is provided with a variable voltage source $\mathrm{V}$, a test resistor $R_{T}=10 ~\Omega$, two identical galvanometers $G_{1}$ and $G_{2}$ and two additional resistors, $R_{1}=10 ~M \Omega$ and $R_{2}=0.001 ~\Omega$. For conducting an experiment to verify ohm's law, the most suitable circuit is:
Given below are two statements : One is labelled as Assertion $\mathbf{A}$ and the other is labelled as Reason $\mathbf{R}$.
Assertion A : For measuring the potential difference across a resistance of $600 \Omega$, the voltmeter with resistance $1000 \Omega$ will be preferred over voltmeter with resistance $4000 \Omega$.
Reason R : Voltmeter with higher resistance will draw smaller current than voltmeter with lower resistance.
In the light of the above statements, choose the most appropriate answer from the options given below.
Equivalent resistance between the adjacent corners of a regular n-sided polygon of uniform wire of resistance R would be :
The equivalent resistance between $A$ and $B$ of the network shown in figure;
The drift velocity of electrons for a conductor connected in an electrical circuit is $\mathrm{V}_{\mathrm{d}}$. The conductor in now replaced by another conductor with same material and same length but double the area of cross section. The applied voltage remains same. The new drift velocity of electrons will be
The charge flowing in a conductor changes with time as $\mathrm{Q}(\mathrm{t})=\alpha \mathrm{t}-\beta \mathrm{t}^{2}+\gamma \mathrm{t}^{3}$. Where $\alpha, \beta$ and $\gamma$ are constants. Minimum value of current is :
With the help of potentiometer, we can determine the value of emf of a given cell. The sensitivity of the potentiometer is
(A) directly proportional to the length of the potentiometer wire
(B) directly proportional to the potential gradient of the wire
(C) inversely proportional to the potential gradient of the wire
(D) inversely proportional to the length of the potentiometer wire
Choose the correct option for the above statements :
Ratio of thermal energy released in two resistors R and 3R connected in parallel in an electric circuit is :
The resistance of a wire is 5 $\Omega$. It's new resistance in ohm if stretched to 5 times of it's original length will be :
A uniform metallic wire carries a current 2 A, when 3.4 V battery is connected across it. The mass of uniform metallic wire is 8.92 $\times$ 10$^{-3}$ kg, density is 8.92 $\times$ 10$^{3}$ kg/m$^3$ and resistivity is 1.7 $\times$ 10$^{-8}~\Omega$-$\mathrm{m}$. The length of wire is :
A cell of emf 90 V is connected across series combination of two resistors each of 100$\Omega$ resistance. A voltmeter of resistance 400$\Omega$ is used to measure the potential difference across each resistor. The reading of the voltmeter will be :
As shown in the figure, a network of resistors is connected to a battery of 24V with an internal resistance of 3 $\Omega$. The currents through the resistors R$_4$ and R$_5$ are I$_4$ and I$_5$ respectively. The values of I$_4$ and I$_5$ are :
Explanation:
In the circuit $I=\frac{9}{3}=3 \mathrm{~A}$
$ \mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{A}}=2 \times 1.5=3~~......(I) $
$ \mathrm{V}_{\mathrm{C}}-\mathrm{V}_{\mathrm{B}}=4 \times 1.5=6~~......(II) $
$\mathrm{Eq}^{\mathrm{n}}(\mathrm{II})-\mathrm{Eq}^{\mathrm{n}}(\mathrm{I})$
$ \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=6-3=3 \text { Volt } $
When a resistance of $5 ~\Omega$ is shunted with a moving coil galvanometer, it shows a full scale deflection for a current of $250 \mathrm{~mA}$, however when $1050 ~\Omega$ resistance is connected with it in series, it gives full scale deflection for 25 volt. The resistance of galvanometer is ____________ $\Omega$.
Explanation:
$\frac{250 \ \text{mA} \times 5}{5 + R_G} = i$
$i = \frac{25}{1050 + R_G}$
Equating the two expressions for current, $i$:
$\frac{250 \ \text{mA} \times 5}{5 + R_G} = \frac{25}{1050 + R_G}$
This equation simplifies to:
$100(5 + R_G) = 1050 \times 5 + R_G \times 5$
Solving for the resistance of the galvanometer, $R_G$:
$95 R_G = 4750$
$R_G = 50 \ \Omega$
So, the resistance of the galvanometer is $50 \ \Omega$.
A potential $\mathrm{V}_{0}$ is applied across a uniform wire of resistance $R$. The power dissipation is $P_{1}$. The wire is then cut into two equal halves and a potential of $V_{0}$ is applied across the length of each half. The total power dissipation across two wires is $P_{2}$. The ratio $P_{2}: \mathrm{P}_{1}$ is $\sqrt{x}: 1$. The value of $x$ is ___________.
Explanation:
The power dissipation, $P_1$, can be calculated using the formula:
$P_1 = \frac{V_0^2}{R}$
Now, let's consider the case where the wire is cut into two equal halves. Each half will have half the original resistance, $\frac{R}{2}$. The potential $V_0$ is applied across the length of each half.
For each half of the wire, the power dissipation, $P'$, can be calculated using the formula:
$P' = \frac{V_0^2}{\frac{R}{2}} = \frac{2V_0^2}{R}$
Since there are two halves of the wire, the total power dissipation across the two wires, $P_2$, is:
$P_2 = 2P' = 2\left(\frac{2V_0^2}{R}\right) = \frac{4V_0^2}{R}$
Now, let's find the ratio $P_2 : P_1$:
$\frac{P_2}{P_1} = \frac{\frac{4V_0^2}{R}}{\frac{V_0^2}{R}} = 4$
Comparing this to the given ratio $\sqrt{x} : 1$, we have:
$\frac{P_2}{P_1} = \sqrt{x}$
So, $\sqrt{x} = 4$.
Squaring both sides, we get:
$x = 16$
The value of $x$ is 16.
The current flowing through a conductor connected across a source is $2 \mathrm{~A}$ and 1.2 $\mathrm{A}$ at $0^{\circ} \mathrm{C}$ and $100^{\circ} \mathrm{C}$ respectively. The current flowing through the conductor at $50^{\circ} \mathrm{C}$ will be ___________ $\times 10^{2} \mathrm{~mA}$.
Explanation:
First, you establish a relationship between the currents and resistances at $0^{\circ} \mathrm{C}$ and $100^{\circ} \mathrm{C}$:
$i_0 R_0 = i_{100} R_{100}$
Plugging in the given values for $i_0$ and $i_{100}$:
$2 R_0 = 1.2 R_0 (1 + 100\alpha) ~\cdots (1)$
From this equation, you find the value of $\alpha$:
$1 + 100\alpha = \frac{5}{3} \Rightarrow 100\alpha = \frac{2}{3} \Rightarrow 50\alpha = \frac{1}{3}$
Now, you need to find the current $i_{50}$ at $50^{\circ} \mathrm{C}$. To do this, you calculate the resistance $R_{50}$ using the found value of $\alpha$:
$R_{50} = R_0 (1 + 50\alpha) = R_0 (1 + \frac{1}{3})$
Using the fact that the voltage across the conductor remains constant, you can find the current $i_{50}$:
$i_{50} = \frac{i_0 R_0}{R_{50}} = \frac{2 \times R_0}{R_0 (1 + \frac{1}{3})} = \frac{2}{1 + \frac{1}{3}} = 1.5 \mathrm{~A}$
Thus, the current flowing through the conductor at $50^{\circ} \mathrm{C}$ is $15 \times 10^2 \mathrm{~mA}$
Two identical cells each of emf $1.5 \mathrm{~V}$ are connected in series across a $10 ~\Omega$ resistance. An ideal voltmeter connected across $10 ~\Omega$ resistance reads $1.5 \mathrm{~V}$. The internal resistance of each cell is __________ $\Omega$.
Explanation:
Since the two cells are connected in series, their internal resistances add up, and the total internal resistance of the series combination is $2r$. The total EMF of the series combination of cells is $1.5\,\text{V} + 1.5\,\text{V} = 3\,\text{V}$.
Let's use Kirchhoff's Voltage Law (KVL) for the closed loop in the circuit:
$\text{EMF}_{total} - I(R + 2r) = 0$
We are given that the voltage across the $10\,\Omega$ resistor, as measured by the ideal voltmeter, is $1.5\,\text{V}$. According to Ohm's law, the current in the circuit can be determined as:
$I = \frac{V}{R} = \frac{1.5\,\text{V}}{10\,\Omega} = 0.15\,\text{A}$
Now, substitute the given values into the KVL equation:
$3\,\text{V} - 0.15\,\text{A}(10\,\Omega + 2r) = 0$
Solve for $2r$:
$3\,\text{V} - 1.5\,\text{V} = 0.15\,\text{A} \cdot 2r$
$1.5\,\text{V} = 0.3\,\text{A} \cdot r$
Now, solve for the internal resistance $r$:
$r = \frac{1.5\,\text{V}}{0.3\,\text{A}} = 5\,\Omega$
So, the internal resistance of each cell is $5\,\Omega$.
In the circuit diagram shown in figure given below, the current flowing through resistance $3 ~\Omega$ is $\frac{x}{3} A$.
The value of $x$ is ___________
Explanation:
$ \mathrm{E}_2-\mathrm{E}_1=8-4=4 \mathrm{~V} $
$ \begin{aligned} & \frac{1}{3}+\frac{1}{6}=\frac{1}{2}=\frac{1}{R} \\\\ & R=2 \Omega \end{aligned} $
$ I=\frac{4}{8}=0.5 \mathrm{~A} $
$ \begin{aligned} & \mathrm{I}_1=\left(\frac{6}{3+6}\right) \times 0.5 \\\\ & \mathrm{I}_1=\frac{2}{3} \times 0.5=\frac{1}{3} \mathrm{~A} \end{aligned} $
$ I_1=\frac{x}{3}=\frac{1}{3} \therefore x=1 $
A rectangular parallelopiped is measured as $1 \mathrm{~cm} \times 1 \mathrm{~cm} \times 100 \mathrm{~cm}$. If its specific resistance is $3 \times 10^{-7} ~\Omega \mathrm{m}$, then the resistance between its two opposite rectangular faces will be ___________ $\times 10^{-7} ~\Omega$.
Explanation:
The resistance of a material can be calculated using the formula:
$ R = \rho \frac{L}{A} $
where
- $R$ is the resistance,
- $\rho$ (rho) is the resistivity or specific resistance of the material,
- $L$ is the length (or distance over which the resistance is being measured), and
- $A$ is the cross-sectional area through which the current flows.
In the context of a rectangular parallelepiped, the "length" and "cross-sectional area" can vary depending on which faces of the shape are considered.
In this particular calculation, the resistance is being measured between the two smaller faces of the parallelepiped, which are squares of side length 1 cm:
The cross-sectional area $A$ should be:$A = 100 \, \text{cm} \times 1 \, \text{cm} = 100 \, \text{cm}^2$
Converting this to meters gives:
$A = 100 \, \text{cm}^2 = 1 \, \text{m} \times 0.01 \, \text{m} = 0.01 \, \text{m}^2$
So, if we use these values for the length $L$ and cross-sectional area $A$ in the resistance formula, we get:
$ R = \rho \frac{L}{A} = 3 \times 10^{-7} \Omega \, m \times \frac{0.01 \, m}{0.01 \, \text{m}^2} = 3 \times 10^{-7} \Omega $
Therefore, the resistance between the two smaller faces of the rectangular parallelepiped is $3 \times 10^{-7} \Omega$.
10 resistors each of resistance 10 $\Omega$ can be connected in such as to get maximum and minimum equivalent resistance. The ratio of maximum and minimum equivalent resistance will be ___________.
Explanation:
When resistors are connected in series, the equivalent resistance is the sum of individual resistances. Therefore, the maximum equivalent resistance will be when all 10 resistors are connected in series, giving a total resistance of 100 $\Omega$.
When resistors are connected in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances. Therefore, the minimum equivalent resistance will be when all 10 resistors are connected in parallel, giving a total resistance of 1 $\Omega$.
The ratio of maximum to minimum equivalent resistance is thus $\frac{100}{1}=100$.
The number density of free electrons in copper is nearly $8 \times 10^{28} \mathrm{~m}^{-3}$. A copper wire has its area of cross section $=2 \times 10^{-6} \mathrm{~m}^{2}$ and is carrying a current of $3.2 \mathrm{~A}$. The drift speed of the electrons is ___________ $\times 10^{-6} \mathrm{ms}^{-1}$
Explanation:
Using the formula:
$ I = n e A v $
where $I$ is the current, $n$ is the number density of free electrons, $e$ is the charge of an electron, $A$ is the cross-sectional area of the wire, and $v$ is the drift speed of the electrons.
We can isolate $v$ to find:
$ v = \frac{I}{n e A}$
Substituting the given values:
$ v = \frac{3.2 \, \text{A}}{8 \times 10^{28} \, \text{m}^{-3} \times 1.6 \times 10^{-19} \, \text{C} \times 2 \times 10^{-6} \, \text{m}^2}$
This simplifies to:
$ v = \frac{3.2}{16 \times 1.6 \times 10^3} \, \text{ms}^{-1} = 125 \times 10^{-6} \, \text{ms}^{-1} $
So, the drift speed of the electrons is indeed $125 \times 10^{-6} \, \text{ms}^{-1}$.
A current of $2 \mathrm{~A}$ flows through a wire of cross-sectional area $25.0 \mathrm{~mm}^{2}$. The number of free electrons in a cubic meter are $2.0 \times 10^{28}$. The drift velocity of the electrons is __________ $\times 10^{-6} \mathrm{~ms}^{-1}$ (given, charge on electron $=1.6 \times 10^{-19} \mathrm{C}$ ).
Explanation:
The drift velocity $v_d$ can be found using the formula for current $I$ in a conductor:
$I = nqAv_d$,
where:
- $n$ is the number density of free electrons (number of free electrons per unit volume),
- $q$ is the charge of an electron,
- $A$ is the cross-sectional area of the conductor, and
- $v_d$ is the drift velocity of the electrons.
We can rearrange the above formula to solve for $v_d$:
$v_d = \frac{I}{nqA}$.
Given that $I = 2 \, \text{A}$, $n = 2.0 \times 10^{28} \, \text{m}^{-3}$, $q = 1.6 \times 10^{-19} \, \text{C}$, and $A = 25.0 \, \text{mm}^{2} = 25.0 \times 10^{-6} \, \text{m}^{2}$, we can substitute these values into the formula to find $v_d$:
$v_d = \frac{2}{(2.0 \times 10^{28})(1.6 \times 10^{-19})(25.0 \times 10^{-6})}$
$v_d = 25 \times 10^{-6} \, \text{ms}^{-1}$.
Therefore, the drift velocity of the electrons is $25 \times 10^{-6} \, \text{ms}^{-1}$.
As shown in the figure, the voltmeter reads $2 \mathrm{~V}$ across $5 ~\Omega$ resistor. The resistance of the voltmeter is _________ $\Omega$.
Explanation:
$ i=\frac{1 V}{2 \Omega}=\frac{1}{2} A $
$\therefore$ Current through voltmeter $=i-i_1$
$ =\frac{1}{2}-\frac{2}{5}=\frac{5-4}{10}=\frac{1}{10} \mathrm{~A} $
$\therefore$ For voltmeter
$ 2=\left(\frac{1}{10}\right) R \Rightarrow R=20 \Omega $
The length of a metallic wire is increased by $20 \%$ and its area of cross section is reduced by $4 \%$. The percentage change in resistance of the metallic wire is __________.
Explanation:
The resistance ($R$) of a wire can be calculated by the formula:
$ R = \rho \frac{L}{A}, $
where
- $\rho$ is the resistivity (a property of the material),
- $L$ is the length of the wire, and
- $A$ is the cross-sectional area of the wire.
If the length ($L$) is increased by 20%, $L$ becomes $1.2L$.
If the cross-sectional area ($A$) is reduced by 4%, $A$ becomes $0.96A$.
The new resistance $R'$ is then:
$ R' = \rho \frac{1.2L}{0.96A} = 1.25R, $
so the resistance has increased by 25%.
Therefore, the percentage change in the resistance of the metallic wire is 25%.
In the given circuit, the value of $\left| {{{{\mathrm{I_1}} + {\mathrm{I_3}}} \over {{\mathrm{I_2}}}}} \right|$ is _____________
Explanation:
$\begin{aligned} & \mathrm{I}_1=\mathrm{I}_2=\frac{20-10}{10}=1 \mathrm{~A} \\\\ & \mathrm{I}_3=1 \mathrm{~A} \\\\ & \left|\frac{\mathrm{I}_1+\mathrm{I}_3}{\mathrm{I}_2}\right|=2\end{aligned}$
In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf $1.5 \mathrm{~V}$ is found to be $60 \mathrm{~cm}$. If this cell is replaced by another cell of emf E, the length-of null point increases by $40 \mathrm{~cm}$. The value of $E$ is $\frac{x}{10} V$. The value of $x$ is ____________.
Explanation:
$E \propto l$
$ \begin{aligned} & \frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}} \\\\ & \frac{1.5}{E}=\frac{60}{100} \\\\ & E=\frac{150}{60}=\frac{5}{2}=\frac{25}{10} \\\\ & \text { so } x=25 \end{aligned} $
Two identical cells, when connected either in parallel or in series gives same current in an external resistance $5 ~\Omega$. The internal resistance of each cell will be ___________ $\Omega$.
Explanation:
$\varepsilon_{\text {series }}=\varepsilon_{1}+\varepsilon_{2}=2 \varepsilon$
$r_{\text {series }}=r_{1}+r_{2}=2 r$
$i=\frac{2 \varepsilon}{5+2 r}$ ......(1)
$\varepsilon_{\text {parallel }}=\frac{\frac{\varepsilon_{1}}{r_{1}}+\frac{\varepsilon_{2}}{r_{2}}}{\frac{1}{r_{1}}+\frac{1}{r_{2}}}=\varepsilon$
$ r_{\text {parallel }}=\frac{r}{2}$
$ i=\frac{\varepsilon}{\frac{r}{2}+5} $ .......(2)
Equating (1) and (2), we get
$\Rightarrow \frac{2 \varepsilon}{2 r+5}=\frac{\varepsilon}{\frac{r}{2}+5}$
$\Rightarrow r+10=2 r+5 $
$\Rightarrow r=5 \Omega$
If the potential difference between $\mathrm{B}$ and $\mathrm{D}$ is zero, the value of $x$ is $\frac{1}{n} \Omega$. The value of $n$ is __________.
Explanation:
The circuit is a Wheatstone bridge, so
${{{{6 \times 3} \over {6 + 3}}} \over {{{x \times 1} \over {x + 1}}}} = {{1 + 2} \over x}$
$ \Rightarrow {{2(x + 1)} \over x} = {3 \over x}$
$ \Rightarrow x = {1 \over 2}$
So $n = 2$
In the following circuit, the magnitude of current I1, is ___________ A.
Explanation:
The indicated diagram shows current flow diagram loops for writing Kirchhoff's law are also indicated, writing the equation
$2{I_3} + {I_1} + {I_3} + {I_2} = 5$
or ${I_1} + {I_2} + 3{I_3} = 5$ ..... (1)
${I_2} - 5 = 2({I_3} - {I_2}) + ({I_1} + {I_3} - {I_2})$
or ${I_1} - 4{I_2} + 3{I_3} = - 5$ ...... (2)
$({I_1} + {I_3}) + ({I_1} + {I_3} - {I_2}) = 2$
or $2{I_1} - {I_2} + 2{I_3} = 2$ ...... (3)
on solving ${I_1} = {3 \over 2}A,{I_2} = 2,{I_3} = {1 \over 2}A$
$ = 1.50$
A null point is found at 200 cm in potentiometer when cell in secondary circuit is shunted by 5$\Omega$. When a resistance of 15$\Omega$ is used for shunting, null point moves to 300 cm. The internal resistance of the cell is ___________$\Omega$.
Explanation:
Let the emf is E and internal resistance is r of this secondary cell so
${{RE} \over {r + R}} \propto l$
so ${{{R_1}E} \over {r + {R_1}}} \propto {l_1}$
& ${{{R_2}E} \over {r + {R_2}}} \propto {l_2}$
$ \Rightarrow {{{R_1}(r + {R_2})} \over {{R_2}(r + {R_1})}} = {{{l_1}} \over {{l_2}}}$
or ${{5(r + 15)} \over {15(r + 5)}} = {{200} \over {300}}$
$ \Rightarrow r = 5\,\Omega $
When two resistance $\mathrm{R_1}$ and $\mathrm{R_2}$ connected in series and introduced into the left gap of a meter bridge and a resistance of 10 $\Omega$ is introduced into the right gap, a null point is found at 60 cm from left side. When $\mathrm{R_1}$ and $\mathrm{R_2}$ are connected in parallel and introduced into the left gap, a resistance of 3 $\Omega$ is introduced into the right gap to get null point at 40 cm from left end. The product of $\mathrm{R_1}$ $\mathrm{R_2}$ is ____________$\Omega^2$
Explanation:
As per given information
${{{R_1} + {R_2}} \over {10}} = {{0.6} \over {0.4}}$ ...... (1)
& ${{{{{R_1}{R_2}} \over {{R_1} + {R_2}}}} \over 3} = {{0.4} \over {0.6}}$ ..... (2)
$ \Rightarrow \left. \matrix{ {R_1} + {R_2} = 15 \hfill \cr \& \,{R_1}{R_2} = 30 \hfill \cr} \right] \Rightarrow {R_1}{R_2} = 30\,{\Omega ^2}$
In a metre bridge experiment the balance point is obtained if the gaps are closed by 2$\Omega$ and 3$\Omega$. A shunt of X $\Omega$ is added to 3$\Omega$ resistor to shift the balancing point by 22.5 cm. The value of X is ___________.
Explanation:
$\Rightarrow I=40 \mathrm{~cm}$
as $3 \Omega$ is shunted the balance point will shift towards $3 \Omega$. So, new length $l^{\prime}=22.5+I=62.5$
So, $\frac{62.5}{37.5}=\frac{2}{3 x}(3+x)$
$\Rightarrow x=2 \Omega$
Two cells are connected between points A and B as shown. Cell 1 has emf of 12 V and internal resistance of 3$\Omega$. Cell 2 has emf of 6V and internal resistance of 6$\Omega$. An external resistor R of 4$\Omega$ is connected across A and B. The current flowing through R will be ____________ A.
Explanation:
$\mathrm{KCL}$ at $A$ gives
$\frac{6-V_{A}}{4}+\frac{0-V_{A}}{6}+\frac{18-V_{A}}{3}=0$
$V_{A}=10$
So current through $4 \Omega=\frac{10-6}{4}=1 \mathrm{~A}$
In the given circuit, the equivalent resistance between the terminal A and B is __________ $\Omega$.
Explanation:
Remove the resistors that have no current. Now the equivalent circuit becomes -
$ \begin{aligned} & \mathrm{R}_{\mathrm{eq}}=3+(2 \| 2)+6 \\\\ & \mathrm{R}_{\mathrm{eq}}=3+1+6 \\\\ & \mathrm{R}_{\mathrm{eq}}=10 \Omega \end{aligned} $
If a copper wire is stretched to increase its length by 20%. The percentage increase in resistance of the wire is __________%.
Explanation:
Considering volume to be conserved
$ \begin{aligned} & \text { Vol. }=\ell_{0} A_{0}=\left(1.2 \ell_{0}\right) \mathrm{A} \\\\ & A_{\text {final }}=\frac{A_{0}}{1.2} \\\\ & R_{\text {in }}=\frac{\rho \ell_{0}}{A_{0}} \\\\ & R_{\text {final }}=\frac{\rho 1.2 \ell_{0}}{\frac{A_{0}}{1.2}}=\frac{\rho \ell_{0}}{A_{0}}(1.2)^{2} \end{aligned} $
$=\mathrm{R}_{\text {in }}(1.44)$
Hence increase $=44 \%$
A hollow cylindrical conductor has length of 3.14 m, while its inner and outer diameters are 4 mm and 8 mm respectively. The resistance of the conductor is $n\times10^{-3}\Omega$. If the resistivity of the material is $\mathrm{2.4\times10^{-8}\Omega m}$. The value of $n$ is ___________.
Explanation:
$R=\rho \frac{l}{\pi\left(r_2^2-r_1^2\right)}$
where $r_2=$ outer radius
$ \begin{gathered} r_1=\text { inner radius } \\\\ \rho=\text { resistivity, } l=\text { length } \\\\ \therefore \rho=\frac{2.4 \times 10^{-8} \times 3.14}{\pi\left(\frac{8^2}{4}-\frac{4^2}{4}\right) \times 10^{-6}} \\\\ =\frac{2.4 \times 10^{-8} \times 4}{48 \times 10^{-6}}=2 \times 10^{-3} \Omega \end{gathered} $
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Alloys such as constantan and manganin are used in making standard resistance coils.
Reason R: Constantan and manganin have very small value of temperature coefficient of resistance.
In the light of the above statements, choose the correct answer from the options given below.
A $1 \mathrm{~m}$ long wire is broken into two unequal parts $\mathrm{X}$ and $\mathrm{Y}$. The $\mathrm{X}$ part of the wire is streched into another wire W. Length of $W$ is twice the length of $X$ and the resistance of $\mathrm{W}$ is twice that of $\mathrm{Y}$. Find the ratio of length of $\mathrm{X}$ and $\mathrm{Y}$.
$
\frac{\mathrm{R}_{\mathrm{X}}}{\mathrm{R}_{\mathrm{Y}}}=\frac{\ell_{\mathrm{X}}}{\ell_{\mathrm{Y}}}
$
Two metallic wires of identical dimensions are connected in series. If $\sigma_{1}$ and $\sigma_{2}$ are the conductivities of the these wires respectively, the effective conductivity of the combination is :
Given below are two statements :
Statement I : A uniform wire of resistance $80 \,\Omega$ is cut into four equal parts. These parts are now connected in parallel. The equivalent resistance of the combination will be $5 \,\Omega$.
Statement II: Two resistances 2R and 3R are connected in parallel in a electric circuit. The value of thermal energy developed in 3R and 2R will be in the ratio $3: 2$.
In the light of the above statements, choose the most appropriate answer from the option given below