Electric Resistance and Ohm's Law — Competition Zone

When 5 V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 × 10^-4 m s^-1. If the electron density in the wire is 8 × 10²⁸ m^-3, the resistivity of the material is close to :

1. 1.6 × 10^-8 Ω m
2. 1.6 × 10^-7 Ω m
3. 1.6 × 10^-6 Ω m
4. 1.6 × 10^-5 Ω m

Answer

1.6 × 10^-5 Ω m

Reason —

Given, V = 5 V, l = 0.1 m, v_d = 2.5 × 10^-4 m s^-1 and n = 8 × 10²⁸ m^-3.

The electric field in the wire is

$\text E = \dfrac{\text V}{l} \\[1em] \text E = \dfrac{5}{0.1} \\[1em] \text E = 50\ \text V\ \text m^{-1}$

The current density is

$j = ne v_d \\[1em] j = (8 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (2.5 \times 10^{-4}) \\[1em] j = 3.2 \times 10^{6}\ \text A\ \text m^{-2}$

By Ohm's law in vector form, $j = \dfrac{\text E}{\rho}$, so

$\rho = \dfrac{\text E}{j} \\[1em] \rho = \dfrac{50}{3.2 \times 10^{6}} \\[1em] \rho = 1.6 \times 10^{-5}\ \Omega\ \text m$

In an aluminium (Al) bar of square cross-section, a square hole is drilled and is filled with iron (Fe) as shown in the figure. The electrical resistivities of Al and Fe are 2.7 × 10^-8 Ω m and 1.0 × 10^-7 Ω m, respectively. The electrical resistance between the two faces P and Q of the composite bar is :

1. $\dfrac{2475}{64}\ \mu\Omega$
2. $\dfrac{1875}{64}\ \mu\Omega$
3. $\dfrac{1875}{49}\ \mu\Omega$
4. $\dfrac{2475}{132}\ \mu\Omega$

Answer

$\dfrac{1875}{64}\ \mu\Omega$

Reason —

Given, side of the Al bar = 7 mm, side of the Fe core = 2 mm, length l = 50 mm = 0.05 m, ρ_Al = 2.7 × 10^-8 Ω m and ρ_Fe = 1.0 × 10^-7 Ω m.

Areas of cross-section :

$\text A_{\text{Fe}} = (2)^2 \\[1em] \text A_{\text{Fe}} = 4\ \text{mm}^2 \\[1em] \text A_{\text{Fe}} = 4 \times 10^{-6}\ \text m^2 \\[1em] \text A_{\text{Al}} = (7)^2 - (2)^2 \\[1em] \text A_{\text{Al}} = 45\ \text{mm}^2 \\[1em] \text A_{\text{Al}} = 45 \times 10^{-6}\ \text m^2$

Resistances of the two parts :

$\text R_{\text{Al}} = \dfrac{(2.7 \times 10^{-8}) \times 0.05}{45 \times 10^{-6}} \\[1em] \text R_{\text{Al}} = 3 \times 10^{-5}\ \Omega \\[1em] \text R_{\text{Al}} = 30\ \mu\Omega \\[1em] \text R_{\text{Fe}} = \dfrac{(1.0 \times 10^{-7}) \times 0.05}{4 \times 10^{-6}} \\[1em] \text R_{\text{Fe}} = 1.25 \times 10^{-3}\ \Omega \\[1em] \text R_{\text{Fe}} = 1250\ \mu\Omega$

The two parts carry the current side by side between the faces P and Q, so they are in parallel,

$\text R = \dfrac{\text R_{\text{Al}}\text R_{\text{Fe}}}{\text R_{\text{Al}} + \text R_{\text{Fe}}} \\[1em] \text R = \dfrac{30 \times 1250}{30 + 1250} \\[1em] \text R = \dfrac{37500}{1280} \\[1em] \text R = \dfrac{1875}{64}\ \mu\Omega$

Across a metallic conductor of non-uniform cross-section a constant potential difference is applied. The quantity which remains constant along the conductor is :

1. current
2. drift velocity
3. electric field
4. current density

Answer

current

Reason — For a steady current the same charge crosses every cross-section of the conductor in a given time, so the current I is the same at every section. The current density $\left(j = \dfrac{\text I}{\text A}\right)$, the drift velocity $\left(v_d = \dfrac{\text I}{ne\text A}\right)$ and the electric field $\left(\text E = \rho j\right)$ all depend on the area of cross-section, which is not uniform. Hence only the current remains constant.

A filament bulb (500 W, 100 V) is to be used in a 230 V main supply. When a resistance R is connected in series, it works perfectly and the bulb consumes 500 W. The value of R is :

1. 13 Ω
2. 230 Ω
3. 46 Ω
4. 26 Ω

Answer

26 Ω

Reason —

Given, the bulb is rated 500 W, 100 V and the main supply is 230 V.

For the bulb to work perfectly it must consume 500 W at 100 V, so the current through it is

$\text I = \dfrac{\text P}{\text V} \\[1em] \text I = \dfrac{500}{100} \\[1em] \text I = 5\ \text A$

The remaining voltage is dropped across the series resistance R,

$\text V_\text R = 230 - 100 \\[1em] \text V_\text R = 130\ \text V \\[1em] \text R = \dfrac{\text V_\text R}{\text I} \\[1em] \text R = \dfrac{130}{5} \\[1em] \text R = 26\ \Omega$

The temperature dependence of resistances of Cu and undoped Si in the temperature range 300-400K, is best described by :

1. Linear decrease for Cu, linear decrease for Si.
2. Linear increase for Cu, linear increase for Si.
3. Linear increase for Cu, exponential increase for Si.
4. Linear increase for Cu, exponential decrease for Si.

Answer

Linear increase for Cu, exponential decrease for Si.

Reason — Copper is a metal. On heating, the relaxation time τ of its free electrons decreases while the free-electron density n remains practically unchanged, so its resistance increases nearly linearly with temperature.

Undoped silicon is a semiconductor. On heating, the number of free electrons n increases greatly, and this increase far outweighs the decrease in τ. Hence the resistance of silicon decreases exponentially with a rise in temperature.

An infinite line charge of uniform electric charge density λ lies along the axis of an electrically conducting infinite cylindrical shell of radius R. At time t = 0, the space inside the cylinder is filled with a material of permittivity ε and electrical conductivity σ. The electrical conduction in the material follows Ohm's law. Which one of the following graphs best describes the subsequent variation of the magnitude of current density j(t) at any point in the material?

Answer

Reason — The line charge produces a radial electric field inside the conducting material, which drives a radial current. This current carries charge away from the axis, so the enclosed charge, and hence the field, goes on decreasing with time.

Let λ be the charge per unit length on the axis at any instant t, and λ₀ its value at t = 0. At a point distant r from the axis, the field in the material and, by Ohm's law, the current density are

$\text E = \dfrac{\lambda}{2\pi\varepsilon r} \\[1em] j = \sigma\text E \\[1em] j = \dfrac{\sigma\lambda}{2\pi\varepsilon r}$

The current flowing out through a coaxial cylindrical surface of radius r and length l reduces the charge λl enclosed by it,

$\dfrac{\text d(\lambda l)}{\text{dt}} = -j(2\pi r l) \\[1em] \dfrac{\text d\lambda}{\text{dt}} = -\dfrac{\sigma\lambda}{2\pi\varepsilon r}(2\pi r) \\[1em] \dfrac{\text d\lambda}{\text{dt}} = -\dfrac{\sigma}{\varepsilon}\lambda$

Separating the variables and integrating from t = 0 to t,

$\int_{\lambda_0}^{\lambda}\dfrac{\text d\lambda}{\lambda} = -\dfrac{\sigma}{\varepsilon}\int_{0}^{t}\text{dt} \\[1em] \ln\dfrac{\lambda}{\lambda_0} = -\dfrac{\sigma}{\varepsilon}t \\[1em] \lambda = \lambda_0 e^{-\sigma t/\varepsilon}$

Since j ∝ λ, the magnitude of the current density falls off exponentially from an initial value and tends to zero,

$j(t) = j_0 e^{-\sigma t/\varepsilon}$

Hence, the graph shown in option (d) is the correct one.

The charge flowing through a resistance R varies with time t as Q = at − bt², where a and b are positive constants. The total heat produced in R is :

1. $\dfrac{a^3 \text R}{6b}$
2. $\dfrac{a^3 \text R}{3b}$
3. $\dfrac{a^3 \text R}{2b}$
4. $\dfrac{a^3 \text R}{b}$

Answer

$\dfrac{a^3 \text R}{6b}$

Reason —

Given, Q = at − bt².

The current is the rate of flow of charge,

$i = \dfrac{\text{dQ}}{\text{dt}} \\[1em] i = \dfrac{\text d}{\text{dt}}(at - bt^2) \\[1em] i = a - 2bt$

The current becomes zero when

$a - 2bt = 0 \\[1em] t = \dfrac{a}{2b}$

so the charge flows only up to this instant. The total heat produced is

$\text H = \int_{0}^{a/2b} i^2\text R\thinspace\text{dt} \\[1em] \text H = \text R\int_{0}^{a/2b}(a - 2bt)^2\thinspace\text{dt}$

Expanding the square and integrating term by term,

$\text H = \text R\int_{0}^{a/2b}\left(a^2 - 4abt + 4b^2t^2\right)\text{dt} \\[1em] \text H = \text R\left[a^2t - \dfrac{4abt^2}{2} + \dfrac{4b^2t^3}{3}\right]_{0}^{a/2b} \\[1em] \text H = \text R\left[a^2t - 2abt^2 + \dfrac{4b^2t^3}{3}\right]_{0}^{a/2b}$

Substituting the limits,

$\text H = \text R\left[a^2\left(\dfrac{a}{2b}\right) - 2ab\left(\dfrac{a}{2b}\right)^2 + \dfrac{4b^2}{3}\left(\dfrac{a}{2b}\right)^3\right] \\[1em] \text H = \text R\left[\dfrac{a^3}{2b} - \dfrac{a^3}{2b} + \dfrac{a^3}{6b}\right] \\[1em] \text H = \dfrac{a^3\text R}{6b}$

The resistance of a wire is 'R' ohm. If it is melted and stretched to 'n' times its original length, its new resistance will be :

1. nR
2. $\dfrac{\text R}{n}$
3. n²R
4. $\dfrac{\text R}{n^2}$

Answer

n²R

Reason — When the wire is melted and stretched, its volume remains constant. If the length becomes n times, the area of cross-section becomes $\dfrac{1}{n}$ times. Hence

$\text R' = \rho\dfrac{nl}{\left(\dfrac{\text A}{n}\right)} \\[1em] \text R' = n^2\left(\rho\dfrac{l}{\text A}\right) \\[1em] \text R' = n^2\text R$

A current of 5 A passes through a copper conductor (resistivity = 1.7 × 10^-8 Ωm) of radius of cross-section 5 mm. Find the mobility of the charges if their drift velocity is 1.1 × 10^-3 m/s.

1. 1.0 m²/Vs
2. 1.8 m²/Vs
3. 1.5 m²/Vs
4. 1.3 m²/Vs

Answer

1.0 m²/Vs

Reason —

Given, I = 5 A, ρ = 1.7 × 10^-8 Ωm, r = 5 mm = 5 × 10^-3 m and v_d = 1.1 × 10^-3 m/s.

Area of cross-section :

$\text A = \pi r^2 \\[1em] \text A = 3.14 \times (5 \times 10^{-3})^2 \\[1em] \text A = 7.85 \times 10^{-5}\ \text m^2$

Current density :

$j = \dfrac{\text I}{\text A} \\[1em] j = \dfrac{5}{7.85 \times 10^{-5}} \\[1em] j = 6.37 \times 10^{4}\ \text A\ \text m^{-2}$

Electric field :

$\text E = \rho j \\[1em] \text E = (1.7 \times 10^{-8}) \times (6.37 \times 10^{4}) \\[1em] \text E = 1.08 \times 10^{-3}\ \text V\ \text m^{-1}$

Mobility :

$\mu = \dfrac{v_d}{\text E} \\[1em] \mu = \dfrac{1.1 \times 10^{-3}}{1.08 \times 10^{-3}} \\[1em] \mu = 1.0\ \text m^2/\text{Vs}$

Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross-section 5 mm² is v. If the electron density in copper is 9 × 10²⁸/m³, the value of v in mm/s is close to (Take, charge of electron to be = 1.6 × 10^-19 C) :

1. 0.02
2. 0.2
3. 2
4. 3

Answer

0.02

Reason —

Given, I = 1.5 A, A = 5 mm² = 5 × 10^-6 m², n = 9 × 10²⁸ m^-3 and e = 1.6 × 10^-19 C.

From I = n e A v_d,

$v = \dfrac{\text I}{ne\text A} \\[1em] v = \dfrac{1.5}{(9 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (5 \times 10^{-6})} \\[1em] v = \dfrac{1.5}{7.2 \times 10^{4}} \\[1em] v = 2.08 \times 10^{-5}\ \text m/\text s$

Converting to mm/s,

$v = 2.08 \times 10^{-5} \times 10^{3} \\[1em] v = 0.0208 \approx 0.02\ \text{mm/s}$

Space between two concentric conducting spheres of radii a and b (b > a) is filled with a medium of resistivity ρ. The resistance between the two spheres will be :

1. $\dfrac{\rho}{2\pi}\left(\dfrac{1}{a} + \dfrac{1}{b}\right)$
2. $\dfrac{\rho}{4\pi}\left(\dfrac{1}{a} + \dfrac{1}{b}\right)$
3. $\dfrac{\rho}{2\pi}\left(\dfrac{1}{a} - \dfrac{1}{b}\right)$
4. $\dfrac{\rho}{4\pi}\left(\dfrac{1}{a} - \dfrac{1}{b}\right)$

Answer

$\dfrac{\rho}{4\pi}\left(\dfrac{1}{a} - \dfrac{1}{b}\right)$

Reason — Consider a thin spherical shell of the medium of radius x and thickness dx lying between the two spheres. The current flows radially, so for this shell the length is dx and the area of cross-section is 4πx². Its resistance is

$\text{dR} = \rho\dfrac{\text{dx}}{4\pi x^2}$

All such shells from x = a to x = b are in series, so the total resistance is

$\text R = \int_{a}^{b}\dfrac{\rho\thinspace\text{dx}}{4\pi x^2} \\[1em] \text R = \dfrac{\rho}{4\pi}\left[-\dfrac{1}{x}\right]_{a}^{b} \\[1em] \text R = \dfrac{\rho}{4\pi}\left(\dfrac{1}{a} - \dfrac{1}{b}\right)$

A metal wire of resistance 3 Ω is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle 60° at the centre, the equivalent resistance between these two points will be :

1. $\dfrac{12}{5}\ \Omega$
2. $\dfrac{5}{3}\ \Omega$
3. $\dfrac{5}{2}\ \Omega$
4. $\dfrac{7}{2}\ \Omega$

Answer

$\dfrac{5}{3}\ \Omega$

Reason —

Given, the original resistance is 3 Ω and the wire is elongated to double its length.

On elongating, the volume remains constant, so $\text R \propto l^2$ and

$\text R' = (2)^2 \times 3 \\[1em] \text R' = 12\ \Omega$

This wire is bent into a circle. Two points making an angle 60° at the centre divide the circle into an arc of 60° and an arc of 300°. The resistance of an arc is proportional to the angle it subtends, so

$\text R_1 = \dfrac{60}{360} \times 12 \\[1em] \text R_1 = 2\ \Omega \\[1em] \text R_2 = \dfrac{300}{360} \times 12 \\[1em] \text R_2 = 10\ \Omega$

The two arcs are in parallel between the two points,

$\text R_{\text{eq}} = \dfrac{2 \times 10}{2 + 10} \\[1em] \text R_{\text{eq}} = \dfrac{20}{12} \\[1em] \text R_{\text{eq}} = \dfrac{5}{3}\ \Omega$

A wire of resistance R is bent to form a square ABCD as shown in the figure. The effective resistance between E and C is : (E is mid-point of arm CD)

1. R
2. $\dfrac{1}{16}\text R$
3. $\dfrac{7}{64}\text R$
4. $\dfrac{3}{4}\text R$

Answer

$\dfrac{7}{64}\text R$

Reason —

Given, a wire of total resistance R is bent to form the square ABCD, and E is the mid-point of the arm CD.

The resistance of each side of the square is $\dfrac{\text R}{4}$, and since E is the mid-point of CD,

$\text R_{\text{EC}} = \text R_{\text{ED}} = \dfrac{\text R}{8}$

Between the points E and C there are two paths in parallel :

(i) the direct path E → C, of resistance $\dfrac{\text R}{8}$

(ii) the path E → D → A → B → C, of resistance

$\dfrac{\text R}{8} + \dfrac{\text R}{4} + \dfrac{\text R}{4} + \dfrac{\text R}{4} = \dfrac{7\text R}{8}$

Hence

$\text R_{\text{eq}} = \dfrac{\left(\dfrac{\text R}{8}\right)\left(\dfrac{7\text R}{8}\right)}{\dfrac{\text R}{8} + \dfrac{7\text R}{8}} \\[1em] \text R_{\text{eq}} = \dfrac{\dfrac{7\text R^2}{64}}{\text R} \\[1em] \text R_{\text{eq}} = \dfrac{7}{64}\text R$

The actual value of resistance R, shown in the figure is 30 Ω. This is measured in an experiment as shown using the standard formula $\text R = \dfrac{\text V}{\text I}$, where V and I are the readings of the voltmeter and ammeter, respectively. If the measured value of R is 5% less, then the internal resistance of the voltmeter is :

1. 600 Ω
2. 570 Ω
3. 350 Ω
4. 35 Ω

Answer

570 Ω

Reason —

Given, the actual value of the resistance is R = 30 Ω and the measured value is 5% less.

$\text R_{\text{measured}} = 30 - (5$

In the circuit shown the voltmeter is connected across R, so the ammeter measures the current through the parallel combination of R and the voltmeter resistance R_V. Hence the measured value is this parallel combination,

$\dfrac{\text R\thinspace\text R_\text V}{\text R + \text R_\text V} = 28.5 \\[1em] \dfrac{30\thinspace\text R_\text V}{30 + \text R_\text V} = 28.5 \\[1em] 30\text R_\text V = 855 + 28.5\text R_\text V \\[1em] 1.5\text R_\text V = 855 \\[1em] \text R_\text V = 570\ \Omega$

In an experiment, the resistance of a material is plotted as a function of temperature (in some range). As shown in the figure, it is a straight line.

One may conclude that :

1. $\text R(\text T) = \text R_0 e^{\text T^2/\text T_0^2}$
2. $\text R(\text T) = \dfrac{\text R_0}{\text T^2}$
3. $\text R(\text T) = \text R_0 e^{-\text T^2/\text T_0^2}$
4. $\text R(\text T) = \text R_0 e^{-\text T_0^2/\text T^2}$

Answer

$\text R(\text T) = \text R_0 e^{-\text T_0^2/\text T^2}$

Reason —

From the figure, the graph of ln R(T) against $\dfrac{1}{\text T^2}$ is a straight line with a negative slope, so

$\ln \text R(\text T) = \ln \text R_0 - \text T_0^2\left(\dfrac{1}{\text T^2}\right)$

where ln R₀ is the intercept and $\text T_0^2$ the magnitude of the slope. Taking antilogarithms,

$\text R(\text T) = \text R_0 e^{-\text T_0^2/\text T^2}$

In the given circuit, an ideal voltmeter connected across the 10 Ω resistance reads 2 V. The internal resistance r, of each cell is :

1. 1 Ω
2. 0.5 Ω
3. 1.5 Ω
4. 0 Ω

Answer

0.5 Ω

Reason —

Given, two cells each of emf 1.5 V and internal resistance r are joined in series, the 15 Ω and 10 Ω resistances are in parallel, and the 2 Ω is in series with them. The voltmeter reads 2 V across the 10 Ω.

Equivalent resistance of the parallel part :

$\text R_\text P = \dfrac{15 \times 10}{15 + 10} \\[1em] \text R_\text P = \dfrac{150}{25} \\[1em] \text R_\text P = 6\ \Omega$

Current in the circuit : The voltmeter is ideal, so the potential difference across the parallel combination is 2 V. Hence

$\text I = \dfrac{2}{6} \\[1em] \text I = \dfrac{1}{3}\ \text A$

Applying the circuit equation : The total emf is 1.5 + 1.5 = 3 V and the total resistance is (6 + 2 + 2r) Ω,

$3 = \dfrac{1}{3}(6 + 2 + 2r) \\[1em] 9 = 8 + 2r \\[1em] r = 0.5\ \Omega$

A 200 Ω resistor has a certain colour code. If one replaces the red colour by green in the code, the new resistance will be :

1. 100 Ω
2. 400 Ω
3. 500 Ω
4. 300 Ω

Answer

500 Ω

Reason — A resistance of 200 Ω is written as 20 × 10¹ Ω, so its first three colour bands are Red (2), Black (0) and Brown (10¹).

If the red colour of the first band is replaced by green, the first significant figure becomes 5, and the code now reads Green (5), Black (0) and Brown (10¹). Hence the new resistance is

$\text R = 50 \times 10^1 \\[1em] \text R = 500\ \Omega$

A 2 W carbon resistor is colour coded with green, black, red and brown respectively. The maximum current which can be passed through this resistor is :

1. 0.4 mA
2. 63 mA
3. 20 mA
4. 100 mA

Answer

20 mA

Reason —

Given, the colour bands are green, black, red and brown, and the power rating of the resistor is 2 W.

Using the colour code, Green corresponds to the figure 5, Black to the figure 0 and Red to the multiplier 10²,

$\text R = 50 \times 10^2 \\[1em] \text R = 5000\ \Omega$

The maximum current which can be passed is obtained from P = I²R,

$\text I = \sqrt{\dfrac{\text P}{\text R}} \\[1em] \text I = \sqrt{\dfrac{2}{5000}} \\[1em] \text I = \sqrt{4 \times 10^{-4}} \\[1em] \text I = 2 \times 10^{-2}\ \text A \\[1em] \text I = 20\ \text{mA}$

In the given circuits, the readings of voltmeters and the ammeters will be :

1. V₁ = V₂ and I₁ = I₂
2. V₂ > V₁ and I₁ > I₂
3. V₂ > V₁ and I₁ = I₂
4. V₁ = V₂ and I₁ > I₂

Answer

V₁ = V₂ and I₁ = I₂

Reason —

In Circuit-1 the voltmeter V₁ is connected across the 10 Ω resistance, and in Circuit-2 the voltmeter V₂ is connected in series with an additional 10 Ω and this arrangement is joined across the first 10 Ω.

An ideal voltmeter has infinite resistance, so it draws no current. Hence in Circuit-2 no current flows through the additional 10 Ω and there is no potential drop across it. Both the voltmeters therefore read the potential difference across the 10 Ω resistance of the main circuit, that is,

$\text V_1 = \text V_2$

Since the voltmeters draw no current, the main circuit is the same in both cases and the ammeters also read the same current,

$\text I_1 = \text I_2$

A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W. Dissipated power when an ideal power supply of 11 V is connected across it is :

1. 11 × 10^-4 W
2. 11 × 10^-5 W
3. 11 × 10⁵ W
4. 11 × 10^-3 W

Answer

11 × 10^-5 W

Reason —

Given, I = 2 mA = 2 × 10^-3 A and the power dissipated P = 4.4 W.

Resistance of the unknown resistor :

$\text R = \dfrac{\text P}{\text I^2} \\[1em] \text R = \dfrac{4.4}{(2 \times 10^{-3})^2} \\[1em] \text R = \dfrac{4.4}{4 \times 10^{-6}} \\[1em] \text R = 1.1 \times 10^{6}\ \Omega$

Power dissipated at 11 V :

$\text P' = \dfrac{\text V^2}{\text R} \\[1em] \text P' = \dfrac{(11)^2}{1.1 \times 10^{6}} \\[1em] \text P' = \dfrac{121}{1.1 \times 10^{6}} \\[1em] \text P' = 1.1 \times 10^{-4} \\[1em] \text P' = 11 \times 10^{-5}\ \text W$

Two electric bulbs rated at 25 W, 220 V and 100 W, 220 V are connected in series across a 220 V voltage source. If the 25 W and 100 W bulbs draw powers P₁ and P₂ respectively, then :

1. P₁ = 16 W, P₂ = 4 W
2. P₁ = 4 W, P₂ = 16 W
3. P₁ = 9 W, P₂ = 16 W
4. P₁ = 16 W, P₂ = 9 W

Answer

P₁ = 16 W, P₂ = 4 W

Reason —

Given, the bulbs are rated 25 W, 220 V and 100 W, 220 V, joined in series across 220 V.

Resistances of the bulbs :

$\text R_1 = \dfrac{(220)^2}{25} \\[1em] \text R_1 = 1936\ \Omega \\[1em] \text R_2 = \dfrac{(220)^2}{100} \\[1em] \text R_2 = 484\ \Omega$

Current in the series circuit :

$\text I = \dfrac{220}{1936 + 484} \\[1em] \text I = \dfrac{220}{2420} \\[1em] \text I = \dfrac{1}{11}\ \text A$

Powers drawn :

$\text P_1 = \text I^2\text R_1 \\[1em] \text P_1 = \dfrac{1}{121} \times 1936 \\[1em] \text P_1 = 16\ \text W \\[1em] \text P_2 = \text I^2\text R_2 \\[1em] \text P_2 = \dfrac{1}{121} \times 484 \\[1em] \text P_2 = 4\ \text W$

Six similar bulbs are connected as shown in the figure with a DC source of emf E and zero internal resistance.

The ratio of power consumption by the bulbs when (i) all are glowing and (ii) in the situation when two from section A and one from section B are glowing, will be :

1. 1 : 2
2. 2 : 1
3. 4 : 9
4. 9 : 4

Answer

9 : 4

Reason —

Let the resistance of each bulb be R. Section A has three bulbs in parallel and section B has three bulbs in parallel, and the two sections are joined in series across the source of emf E.

(i) When all the six are glowing :

$\text R_\text A = \dfrac{\text R}{3}, \qquad \text R_\text B = \dfrac{\text R}{3} \\[1em] \text R_1 = \dfrac{\text R}{3} + \dfrac{\text R}{3} \\[1em] \text R_1 = \dfrac{2\text R}{3} \\[1em] \text P_1 = \dfrac{\text E^2}{\text R_1} \\[1em] \text P_1 = \dfrac{3\text E^2}{2\text R}$

(ii) When two from A and one from B are glowing :

$\text R_\text A' = \dfrac{\text R}{2}, \qquad \text R_\text B' = \text R \\[1em] \text R_2 = \dfrac{\text R}{2} + \text R \\[1em] \text R_2 = \dfrac{3\text R}{2} \\[1em] \text P_2 = \dfrac{\text E^2}{\text R_2} \\[1em] \text P_2 = \dfrac{2\text E^2}{3\text R}$

Ratio :

$\dfrac{\text P_1}{\text P_2} = \dfrac{\dfrac{3\text E^2}{2\text R}}{\dfrac{2\text E^2}{3\text R}} \\[1em] \dfrac{\text P_1}{\text P_2} = \dfrac{3}{2} \times \dfrac{3}{2} \\[1em] \dfrac{\text P_1}{\text P_2} = \dfrac{9}{4}$

Two solid conductors are made up of same material, have same length and same resistance. One of them has a circular cross section of area A₁ and the other one has a square cross section of area A₂. The ratio A₁/A₂ is :

1. 2
2. 1.5
3. 1
4. 0.8

Answer

1

Reason — The two conductors are of the same material, so they have the same resistivity ρ, and they have the same length l and the same resistance R. From

$\text R = \rho\dfrac{l}{\text A} \\[1em] \text A = \dfrac{\rho l}{\text R}$

Since ρ, l and R are the same for both, their areas of cross-section must also be equal, whatever be the shape of the cross-section. Hence

$\dfrac{\text A_1}{\text A_2} = 1$

Consider four conducting materials copper, tungsten, mercury and aluminium with resistivity ρ_C, ρ_T, ρ_M and ρ_A respectively :

1. ρ_M > ρ_A > ρ_C
2. ρ_C > ρ_A > ρ_T
3. ρ_A > ρ_M > ρ_C
4. ρ_A > ρ_T > ρ_C

Answer

ρ_M > ρ_A > ρ_C

Reason — Of the four materials, mercury has by far the largest resistivity (about 98 × 10^-8 Ω m), followed by tungsten, aluminium and copper. Among aluminium and copper, copper has the lower resistivity, which is why copper wires are used as connecting wires. Hence the correct order among these three materials is

$\rho_\text M \gt \rho_\text A \gt \rho_\text C$

An electrical power line, having a total resistance of 2 Ω, delivers 1 kW at 220 V. The efficiency of the transmission line is approximately :

1. 85%
2. 96%
3. 72%
4. 91%

Answer

96%

Reason —

Given, the power delivered P = 1 kW = 1000 W at V = 220 V, and the total resistance of the line R = 2 Ω.

Current in the line :

$\text I = \dfrac{\text P}{\text V} \\[1em] \text I = \dfrac{1000}{220} \\[1em] \text I = 4.545\ \text A$

Power lost in the line :

$\text P_{\text{loss}} = \text I^2\text R \\[1em] \text P_{\text{loss}} = (4.545)^2 \times 2 \\[1em] \text P_{\text{loss}} = 41.3\ \text W$

Efficiency of transmission :

$\eta = \dfrac{\text{power delivered}}{\text{power delivered} + \text{power lost}} \times 100 \\[1em] \eta = \dfrac{1000}{1041.3} \times 100 \\[1em] \eta = 96$

Two resistors 400 Ω and 800 Ω are connected in series across a 6 V battery. The potential difference measured by a voltmeter of 10 kΩ across 400 Ω resistor is close to:

1. 2.05 V
2. 1.8 V
3. 2 V
4. 1.95 V

Answer

1.95 V

Reason —

Given, R₁ = 400 Ω and R₂ = 800 Ω in series across a 6 V battery, and a voltmeter of resistance R_V = 10 kΩ = 10000 Ω is connected across the 400 Ω.

Resistance of the 400 Ω with the voltmeter across it :

$\text R' = \dfrac{400 \times 10000}{400 + 10000} \\[1em] \text R' = \dfrac{4 \times 10^{6}}{10400} \\[1em] \text R' = 384.6\ \Omega$

Current drawn from the battery :

$\text I = \dfrac{6}{384.6 + 800} \\[1em] \text I = \dfrac{6}{1184.6} \\[1em] \text I = 5.06 \times 10^{-3}\ \text A$

Potential difference measured :

$\text V = \text I\text R' \\[1em] \text V = (5.06 \times 10^{-3}) \times 384.6 \\[1em] \text V = 1.95\ \text V$

A circuit to verify Ohm's law uses ammeter and voltmeter in series or parallel connected correctly to the resistor. In the circuit :

1. Ammeter is always connected in series and voltmeter in parallel
2. Both ammeter and voltmeter must be connected in parallel
3. Ammeter is always used in parallel and voltmeter is series
4. Both ammeter and voltmeter must be connected in series

Answer

Ammeter is always connected in series and voltmeter in parallel

Reason — An ammeter measures the current through the resistor, so the whole current must pass through it and it has to be joined in series. It is given a very low resistance so that it does not appreciably alter the current in the circuit.

A voltmeter measures the potential difference across the resistor, so it has to be joined in parallel with it. It is given a very high resistance so that it draws practically no current from the circuit.

A battery of 3.0 V is connected to a resistor dissipating 0.5 W of power. If the terminal voltage of the battery is 2.5 V, the power dissipated within the internal resistance is :

1. 0.10 W
2. 0.072 W
3. 0.50 W
4. 0.125 W

Answer

0.10 W

Reason —

Given, emf of the battery E = 3.0 V, power dissipated in the external resistor P = 0.5 W and terminal voltage V = 2.5 V.

Current in the circuit : The power dissipated in the resistor is P = VI,

$\text I = \dfrac{\text P}{\text V} \\[1em] \text I = \dfrac{0.5}{2.5} \\[1em] \text I = 0.2\ \text A$

Potential drop across the internal resistance :

$\text V_r = \text E - \text V \\[1em] \text V_r = 3.0 - 2.5 \\[1em] \text V_r = 0.5\ \text V$

Power dissipated within the internal resistance :

$\text P_r = \text V_r\text I \\[1em] \text P_r = 0.5 \times 0.2 \\[1em] \text P_r = 0.10\ \text W$

Column-I gives certain physical terms associated with flow of current through a metallic conductor.

Column-II gives some mathematical relations involving electrical quantities. Match Column-I and Column-II with appropriate relations.

1. (A) - (R), (B) - (Q), (C) - (S), (D) - (P)
2. (A) - (R), (B) - (S), (C) - (P), (D) - (Q)
3. (A) - (R), (B) - (S), (C) - (Q), (D) - (P)
4. (A) - (R), (B) - (P), (C) - (S), (D) - (Q)

Answer

(A) - (R), (B) - (S), (C) - (P), (D) - (Q)

Reason — Taking the terms of Column-I one by one :

(A) Drift Velocity : The drift velocity acquired by a free electron in a field E is $v_d = \dfrac{e\text E}{m}\tau$, which is (R).

(B) Electrical Resistivity : From Ohm's law in vector form, $\text J = \sigma\text E = \dfrac{\text E}{\rho}$, so $\rho = \dfrac{\text E}{\text J}$, which is (S).

(C) Relaxation Period : Since $\rho = \dfrac{m}{ne^2\tau}$, the relaxation time is $\tau = \dfrac{m}{ne^2\rho}$, which is (P).

(D) Current Density : The current density is $\text J = ne v_d$, which is (Q).

The effective resistance of a parallel connection that consists of four wires of equal length, equal area of cross-section and same material is 0.25 Ω. What will be the effective resistance if they are connected in series?

1. 4 Ω
2. 0.25 Ω
3. 0.5 Ω
4. 1 Ω

Answer

4 Ω

Reason — Let the resistance of each wire be R'. The four wires are of equal length, equal area of cross-section and the same material, so they are of equal resistance.

In parallel :

$\text R_\text P = \dfrac{\text R'}{4} \\[1em] \text R_\text P = 0.25 \\[1em] \text R' = 1\ \Omega$

In series :

$\text R_\text S = 4\text R' \\[1em] \text R_\text S = 4 \times 1 \\[1em] \text R_\text S = 4\ \Omega$

Three resistors having resistances r₁, r₂ and r₃ are connected as shown in the given circuit. The ratio $\dfrac{\text I_3}{\text I_1}$ of currents in terms of resistances used in the circuit is :

1. $\dfrac{r_2}{r_1 + r_3}$
2. $\dfrac{r_1}{r_2 + r_3}$
3. $\dfrac{r_2}{r_2 + r_3}$
4. $\dfrac{r_1}{r_1 + r_2}$

Answer

$\dfrac{r_2}{r_2 + r_3}$

Reason —

From the figure, the current I₁ enters at A through r₁ and then divides into I₂ through r₂ and I₃ through r₃, since r₂ and r₃ are in parallel. Hence

$\text I_1 = \text I_2 + \text I_3$

The potential difference across the parallel combination is the same for both branches,

$\text I_2 r_2 = \text I_3 r_3 \\[1em] \text I_2 = \dfrac{\text I_3 r_3}{r_2}$

Substituting in the first equation,

$\text I_1 = \dfrac{\text I_3 r_3}{r_2} + \text I_3 \\[1em] \text I_1 = \text I_3\left(\dfrac{r_3 + r_2}{r_2}\right) \\[1em] \dfrac{\text I_3}{\text I_1} = \dfrac{r_2}{r_2 + r_3}$

What equal length of an iron wire and a copper-nickel alloy wire, each of 2 mm diameter connected parallel to give an equivalent resistance of 3 Ω? (Given : resistivities of iron and copper-nickel alloy wire are 12 µΩ cm and 51 µΩ cm respectively)

1. 97 m
2. 110 m
3. 90 m
4. 82 m

Answer

97 m

Reason —

Given, the two wires are of equal length l and each of diameter 2 mm, so the radius is 1 mm = 10^-3 m. Their resistivities are

$\rho_1 = 12\ \mu\Omega\ \text{cm} \\[1em] \rho_1 = 12 \times 10^{-8}\ \Omega\ \text m \\[1em] \rho_2 = 51\ \mu\Omega\ \text{cm} \\[1em] \rho_2 = 51 \times 10^{-8}\ \Omega\ \text m$

Area of cross-section :

$\text A = \pi r^2 \\[1em] \text A = 3.14 \times (10^{-3})^2 \\[1em] \text A = 3.14 \times 10^{-6}\ \text m^2$

Resistances of the two wires :

$\text R_1 = \dfrac{(12 \times 10^{-8})l}{3.14 \times 10^{-6}} \\[1em] \text R_1 = 3.82 \times 10^{-2}\thinspace l \\[1em] \text R_2 = \dfrac{(51 \times 10^{-8})l}{3.14 \times 10^{-6}} \\[1em] \text R_2 = 16.24 \times 10^{-2}\thinspace l$

In parallel, the equivalent resistance is 3 Ω :

$\dfrac{\text R_1\text R_2}{\text R_1 + \text R_2} = 3 \\[1em] \dfrac{(3.82 \times 10^{-2})(16.24 \times 10^{-2})l^2}{(3.82 + 16.24) \times 10^{-2}\thinspace l} = 3 \\[1em] \dfrac{(62.0 \times 10^{-4})l}{20.06 \times 10^{-2}} = 3 \\[1em] (3.09 \times 10^{-2})l = 3 \\[1em] l = 97\ \text m$

Consider a galvanometer shunted with 5 Ω resistance and 2% of current passes through it. What is the resistance of the given galvanometer?

1. 245 Ω
2. 344 Ω
3. 300 Ω
4. 226 Ω

Answer

245 Ω

Reason —

Given, the shunt resistance S = 5 Ω and 2% of the total current passes through the galvanometer.

Let the total current be I. Then the current through the galvanometer is 0.02 I and that through the shunt is 0.98 I.

The galvanometer and the shunt are in parallel, so the potential difference across them is the same,

$\text I_\text g\text R_\text g = \text I_\text s\text S \\[1em] (0.02\thinspace\text I)\text R_\text g = (0.98\thinspace\text I) \times 5 \\[1em] \text R_\text g = \dfrac{0.98 \times 5}{0.02} \\[1em] \text R_\text g = 245\ \Omega$

The colour coding on a carbon resistor is shown in the given figure. The resistance value of the given resistor is :

1. (7500 ± 375) Ω
2. (5700 ± 285) Ω
3. (7500 ± 755) Ω
4. (5700 ± 375) Ω

Answer

(7500 ± 375) Ω

Reason —

From the figure the colour bands in order are Violet, Green, Red and Gold.

Using the colour code, Violet corresponds to the figure 7, Green to the figure 5, Red to the multiplier 10², and Gold to a tolerance of 5%. Hence

$\text R = 75 \times 10^2 \\[1em] \text R = 7500\ \Omega \\[1em] \text{Tolerance} = 5$

Hence the resistance is (7500 ± 375) Ω.

If you are provided a set of resistances 2 Ω, 4Ω, 6Ω and 8 Ω, connect these resistances so as to obtain an equivalent resistance of $\dfrac{46}{3}\ \Omega$.

1. 2Ω and 6Ω are in parallel with 4Ω and 8Ω in series.
2. 4Ω and 6Ω are in parallel with 2Ω and 8Ω in series.
3. 2Ω are 4Ω are in parallel with 6Ω and 8Ω in series.
4. 6Ω and 8Ω are in parallel with 2Ω and 4Ω in series.

Answer

2Ω are 4Ω are in parallel with 6Ω and 8Ω in series.

Reason — Taking the 2 Ω and the 4 Ω in parallel,

$\text R_\text P = \dfrac{2 \times 4}{2 + 4} \\[1em] \text R_\text P = \dfrac{8}{6} \\[1em] \text R_\text P = \dfrac{4}{3}\ \Omega$

Joining the 6 Ω and the 8 Ω in series with this combination,

$\text R = \dfrac{4}{3} + 6 + 8 \\[1em] \text R = \dfrac{4 + 18 + 24}{3} \\[1em] \text R = \dfrac{46}{3}\ \Omega$

which is the required equivalent resistance.

The equivalent resistance of the given circuit between the terminals A and B is :

1. 0Ω
2. $\dfrac{9}{2}\ \Omega$
3. 3Ω
4. 1Ω

Answer

1Ω

Reason —

Given,

• Upper wire, from the left-hand corner towards A : 2 Ω up to the middle junction M, then 2 Ω from M
• Vertical arms between the upper and the lower wires, from left to right : 5 Ω, 2 Ω (at M), 3 Ω and 3 Ω
• A plain connecting wire joins the upper and the lower wires at the left-hand end, across the 5 Ω
• A is at the right-hand end of the upper wire and B at the right-hand end of the lower wire

The left-hand vertical connecting wire joins the two ends of the 5 Ω resistance, so the 5 Ω is short-circuited and carries no current. Hence the left-hand upper corner is at the same potential as B.

The two 3 Ω arms are joined directly between A and B, so

$\text R_1 = \dfrac{3 \times 3}{3 + 3} \\[1em] \text R_1 = 1.5\ \Omega$

The 2 Ω vertical arm at the middle junction M and the 2 Ω which reaches B through the left-hand corner are both connected between M and B, that is, they are in parallel,

$\text R_2 = \dfrac{2 \times 2}{2 + 2} \\[1em] \text R_2 = 1\ \Omega$

Adding the 2 Ω between A and M in series,

$\text R_3 = 2 + 1 \\[1em] \text R_3 = 3\ \Omega$

Finally R₃ and R₁ are in parallel between A and B,

$\text R = \dfrac{3 \times 1.5}{3 + 1.5} \\[1em] \text R = \dfrac{4.5}{4.5} \\[1em] \text R = 1\ \Omega$

A current of 5A is passing through a non-linear magnesium wire of cross-section 0.04 m². At every point the direction of current density is at an angle of 60° with the unitvector area of cross-section. The magnitude of electric field at every point of the conductor is (Resistivity of magnesium ρ = 44 × 10^-8 Ωm) :

1. 11 × 10^-5 V m^-1
2. 11 × 10^-7 Vm^-1
3. 11 × 10^-3 Vm^-1
4. 11 × 10^-2 V m^-1

Answer

11 × 10^-5 V m^-1

Reason —

Given, I = 5 A, A = 0.04 m², θ = 60° and ρ = 44 × 10^-8 Ωm.

The current is the flux of the current density through the area of cross-section,

$\text I = \vec{j}\cdot\vec{\text A} \\[1em] \text I = j\text A\cos\theta \\[1em] j = \dfrac{\text I}{\text A\cos\theta} \\[1em] j = \dfrac{5}{0.04 \times \cos 60\degree} \\[1em] j = \dfrac{5}{0.04 \times 0.5} \\[1em] j = \dfrac{5}{0.02} \\[1em] j = 250\ \text A\ \text m^{-2}$

By Ohm's law in vector form, E = ρ j,

$\text E = (44 \times 10^{-8}) \times 250 \\[1em] \text E = 11000 \times 10^{-8} \\[1em] \text E = 1.1 \times 10^{-4} \\[1em] \text E = 11 \times 10^{-5}\ \text V\ \text m^{-1}$

A copper (Cu) rod of length 25 cm and cross-sectional area 3 mm² is joined with a similar aluminium (Al) rod as shown in figure. Find the resistance of the combination between the ends A and B. (Take resistivity of copper = 1.7 × 10^-8 Ωm, Resistivity of aluminium = 2.6 × 10^-8 Ω m.)

1. 2.170 mΩ
2. 0.858 mΩ
3. 0.0858 mΩ
4. 1.420 mΩ

Answer

0.858 mΩ

Reason —

Given, l = 25 cm = 0.25 m, A = 3 mm² = 3 × 10^-6 m², ρ_Cu = 1.7 × 10^-8 Ωm and ρ_Al = 2.6 × 10^-8 Ω m.

Resistance of the copper rod :

$\text R_{\text{Cu}} = \dfrac{(1.7 \times 10^{-8}) \times 0.25}{3 \times 10^{-6}} \\[1em] \text R_{\text{Cu}} = 1.417 \times 10^{-3}\ \Omega$

Resistance of the aluminium rod :

$\text R_{\text{Al}} = \dfrac{(2.6 \times 10^{-8}) \times 0.25}{3 \times 10^{-6}} \\[1em] \text R_{\text{Al}} = 2.167 \times 10^{-3}\ \Omega$

From the figure the two rods are joined in parallel between A and B,

$\text R = \dfrac{\text R_{\text{Cu}}\text R_{\text{Al}}}{\text R_{\text{Cu}} + \text R_{\text{Al}}} \\[1em] \text R = \dfrac{(1.417 \times 2.167) \times 10^{-6}}{(1.417 + 2.167) \times 10^{-3}} \\[1em] \text R = \dfrac{3.070 \times 10^{-6}}{3.584 \times 10^{-3}} \\[1em] \text R = 0.857 \times 10^{-3}\ \Omega \approx 0.858\ \text m\Omega$

In the given figure, a battery of emf E is connected across a conductor PQ of length l and different area of cross-section having radii r₁ and r₂ (r₂ < r₁).

Choose the correct option as one moves from P to Q.

1. Electric field decreases.
2. Drift velocity of electron increases.
3. Electric current decreases.
4. All of the above.

Answer

Drift velocity of electron increases.

Reason —

For a steady current the same current I flows through every cross-section of the conductor PQ, so the current does not decrease. Since r₂ < r₁, the area of cross-section goes on decreasing as one moves from P to Q.

The drift velocity is

$v_d = \dfrac{\text I}{ne\text A}$

so $v_d \propto \dfrac{1}{\text A}$, and the drift velocity increases from P to Q.

The electric field is E = ρ j, and the current density $j = \dfrac{\text I}{\text A}$ also increases, so the electric field increases as well. Hence only the second statement is correct.

The resistance of a conductor at 15°C is 16Ω and at 100°C is 20Ω. What will be the temperature coefficient of resistance of the conductor?

1. 0.042°C^-1
2. 0.003°C^-1
3. 0.033°C^-1
4. 0.010°C^-1

Answer

0.003°C^-1

Reason —

Given, R₁ = 16 Ω at t₁ = 15°C and R₂ = 20 Ω at t₂ = 100°C.

Using R_t = R₀(1 + αt) for the two temperatures,

$\text R_1 = \text R_0(1 + \alpha t_1), \qquad \text R_2 = \text R_0(1 + \alpha t_2)$

Eliminating R₀,

$\alpha = \dfrac{\text R_2 - \text R_1}{\text R_1 t_2 - \text R_2 t_1}$

Substituting the values,

$\alpha = \dfrac{20 - 16}{(16 \times 100) - (20 \times 15)} \\[1em] \alpha = \dfrac{4}{1600 - 300} \\[1em] \alpha = \dfrac{4}{1300} \\[1em] \alpha = 3.08 \times 10^{-3} \approx 0.003\ \degree\text C^{-1}$

A wire of 1 Ω has a length of 1 m. It is stretched till its length increases by 25%. The percentage change in resistance to the nearest integer is :

1. 56%
2. 76%
3. 12.5%
4. 25%

Answer

56%

Reason — On stretching, the volume of the wire remains constant, so

$\text R \propto l^2$

The length increases by 25%, so l' = 1.25 l. Hence

$\dfrac{\text R'}{\text R} = (1.25)^2 \\[1em] \dfrac{\text R'}{\text R} = 1.5625 \\[1em] \dfrac{\text R' - \text R}{\text R} \times 100 = (1.5625 - 1) \times 100 \\[1em] \dfrac{\text R' - \text R}{\text R} \times 100 = 56.25$

To the nearest integer, the percentage change in resistance is 56%.

A conducting wire of length 'l', area of cross-section A and electric resistivity ρ is connected between the terminals of a battery. A potential difference V is developed between its ends, causing an electric current. If the length of the wire of the same material is doubled and the area of cross-section is halved, the resultant current would be :

1. $\dfrac{1}{4}\dfrac{\text{VA}}{\rho l}$
2. $4\dfrac{\text{VA}}{\rho l}$
3. $\dfrac{1}{4}\dfrac{\rho l}{\text{VA}}$
4. $\dfrac{3}{4}\dfrac{\text{VA}}{\rho l}$

Answer

$\dfrac{1}{4}\dfrac{\text{VA}}{\rho l}$

Reason — The resistance of the wire is $\text R = \dfrac{\rho l}{\text A}$, so the current is

$\text I = \dfrac{\text V}{\text R} \\[1em] \text I = \dfrac{\text{VA}}{\rho l}$

When the length is doubled and the area of cross-section is halved,

$\text R' = \dfrac{\rho(2l)}{\left(\dfrac{\text A}{2}\right)} \\[1em] \text R' = \dfrac{4\rho l}{\text A} \\[1em] \text R' = 4\text R$

Hence the resultant current is

$\text I' = \dfrac{\text V}{4\text R} \\[1em] \text I' = \dfrac{1}{4}\dfrac{\text{VA}}{\rho l}$

A current through a wire depends on time as i = α₀t + βt^2t where α₀ = 20As^-1 and β = 8As^-2. Find the charge crossed through a section of the wire in 15s.

1. 2250 C
2. 11250 C
3. 21000 C
4. 260 C

Answer

11250 C

Reason —

Given, i = α₀t + βt² with α₀ = 20 As^-1 and β = 8 As^-2, and t = 15 s.

The current is the rate of flow of charge, so

$\text q = \int_{0}^{15} i\thinspace\text{dt} \\[1em] \text q = \int_{0}^{15}(20t + 8t^2)\thinspace\text{dt} \\[1em] \text q = \left[\dfrac{20t^2}{2} + \dfrac{8t^3}{3}\right]_{0}^{15} \\[1em] \text q = \left[10t^2 + \dfrac{8t^3}{3}\right]_{0}^{15} \\[1em] \text q = \left[10 \times (15)^2\right] + \left[\dfrac{8 \times (15)^3}{3}\right] \\[1em] \text q = 2250 + \dfrac{8 \times 3375}{3} \\[1em] \text q = 2250 + 9000 \\[1em] \text q = 11250\ \text C$

Note: The current in the question is printed as i = α₀t + βt^2t. The exponent 2t is a misprint; the units given for β (As^-2) show that the term must be βt², and only then does the integration give the answer in the options.

A current of 10 A exists in a wire of cross sectional area of 5 mm² with a drift velocity of 2 × 10^-3 ms^-1. The number of free electrons in each cubic metre of the wire is ............... .

1. 2 × 10⁶
2. 625 × 10²⁵
3. 2 × 10²⁵
4. 1 × 10²³

Answer

625 × 10²⁵

Reason —

Given, I = 10 A, A = 5 mm² = 5 × 10^-6 m², v_d = 2 × 10^-3 ms^-1 and e = 1.6 × 10^-19 C.

From I = n e A v_d,

$n = \dfrac{\text I}{e\text A v_d} \\[1em] n = \dfrac{10}{(1.6 \times 10^{-19}) \times (5 \times 10^{-6}) \times (2 \times 10^{-3})} \\[1em] n = \dfrac{10}{1.6 \times 10^{-27}} \\[1em] n = 6.25 \times 10^{27} \\[1em] n = 625 \times 10^{25}\ \text m^{-3}$

(A) The drift velocity of electrons decreases with the increase in the temperature of conductor.

(B) The drift velocity is inversely proportional to the area of cross-section of given conductor.

(C) The drift velocity does not depend on the applied potential difference to the conductor.

(D) The drift velocity of electron is inversely proportional to the length of the conductor.

(E) The drift velocity increases with the increase in the temperature of conductor.

Choose the correct answer from the options given below.

1. (A) and (B)
2. (A) and (D)
3. (B) and (E)
4. (B) and (C)

Answer

(A) and (D)

Reason — The drift velocity in terms of the potential difference applied across a wire of length l is

$v_d = \dfrac{e\text V}{ml}\tau$

(A) is correct : With a rise in temperature the metal ions vibrate more vigorously, so the relaxation time τ decreases and hence the drift velocity decreases.

(D) is correct : From the above relation $v_d \propto \dfrac{1}{l}$, so the drift velocity is inversely proportional to the length of the conductor.

(B) is wrong, because the expression contains no area of cross-section. (C) is wrong, because v_d is directly proportional to V. (E) is wrong, being the opposite of (A).

Two metallic wires of identical dimensions are connected in series. If σ₁ and σ₂ are the conductivities of these wires respectively, the effective conductivity of the combination is :

1. $\dfrac{\sigma_1 \sigma_2}{\sigma_1 + \sigma_2}$
2. $\dfrac{2\sigma_1 \sigma_2}{\sigma_1 + \sigma_2}$
3. $\dfrac{\sigma_1 + \sigma_2}{2\sigma_1 \sigma_2}$
4. $\dfrac{\sigma_1 + \sigma_2}{\sigma_1 \sigma_2}$

Answer

$\dfrac{2\sigma_1 \sigma_2}{\sigma_1 + \sigma_2}$

Reason — Let each wire be of length l and area of cross-section A. Since $\rho = \dfrac{1}{\sigma}$, the resistances are

$\text R_1 = \dfrac{l}{\sigma_1\text A}, \qquad \text R_2 = \dfrac{l}{\sigma_2\text A}$

In series,

$\text R = \text R_1 + \text R_2 \\[1em] \text R = \dfrac{l}{\text A}\left(\dfrac{1}{\sigma_1} + \dfrac{1}{\sigma_2}\right)$

The combination is a single conductor of length 2l and the same area A. If σ is its effective conductivity,

$\dfrac{2l}{\sigma\text A} = \dfrac{l}{\text A}\left(\dfrac{\sigma_1 + \sigma_2}{\sigma_1\sigma_2}\right) \\[1em] \dfrac{2}{\sigma} = \dfrac{\sigma_1 + \sigma_2}{\sigma_1\sigma_2} \\[1em] \sigma = \dfrac{2\sigma_1\sigma_2}{\sigma_1 + \sigma_2}$

Given below are two statements :

Statement I : A uniform wire of resistance 80Ω is cut into four equal parts. These parts are now connected in parallel. The equivalent resistance of the combination will be 5Ω.

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 :

1. Both statement I and statement II are correct.
2. Both statement I and statement II are incorrect.
3. Statement I is correct but statement II is incorrect.
4. Statement I is incorrect but statement II is correct.

Answer

Statement I is correct but statement II is incorrect.

Reason —

Statement I : When a wire of resistance 80 Ω is cut into four equal parts, the resistance of each part is $\dfrac{80}{4} = 20\ \Omega$. Joining these four in parallel,

$\text R = \dfrac{20}{4} \\[1em] \text R = 5\ \Omega$

Hence Statement I is correct.

Statement II : In a parallel combination the same potential difference V acts across both the resistances, so the thermal energy developed is

$\text H = \dfrac{\text V^2 t}{\text R} \\[1em] \text H \propto \dfrac{1}{\text R} \\[1em] \dfrac{\text H_{3\text R}}{\text H_{2\text R}} = \dfrac{2\text R}{3\text R} \\[1em] \dfrac{\text H_{3\text R}}{\text H_{2\text R}} = \dfrac{2}{3}$

The ratio is 2 : 3 and not 3 : 2, so Statement II is incorrect.

A 1 m long wire is broken into two unequal parts X and Y. The X part of the wire is stretched into another wire W. Length of W is twice the length of X and the resistance of W is twice that of Y. Find the ratio of length of X and Y.

1. 1 : 4
2. 1 : 2
3. 4 : 1
4. 2 : 1

Answer

1 : 2

Reason — Let the resistance per unit length of the original wire be k, and let the lengths of the parts X and Y be x and y. Then

$\text R_\text X = kx, \qquad \text R_\text Y = ky$

The part X is stretched into the wire W whose length is twice that of X. On stretching, the volume remains constant, so $\text R \propto l^2$ and

$\text R_\text W = (2)^2\text R_\text X \\[1em] \text R_\text W = 4kx$

It is given that R_W = 2R_Y,

$4kx = 2ky \\[1em] y = 2x \\[1em] \dfrac{x}{y} = \dfrac{1}{2}$

Following 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.

1. Both A and R are true and R is the correct explanation of A.
2. Both A and R are true but R is NOT the correct explanation of A.
3. A is true but R is false.
4. A is false but R is true.

Answer

Both A and R are true and R is the correct explanation of A.

Reason —

Assertion A is true: A standard resistance coil must keep its marked value even when it warms up on carrying a current, and alloys such as constantan and manganin are used in making such coils.

Reason R is true: Constantan and manganin have a very small temperature coefficient of resistance, so their resistance is hardly affected by a change of temperature.

The Reason is the very property that makes these alloys suitable for standard resistance coils, so R is the correct explanation of A.

Two identical cells each of emf 1.5 V are connected in parallel across a parallel combination of two resistors each of resistance 20Ω. A voltmeter connected in the circuit measures 1.2 V. The internal resistance of each cell is :

1. 2.5Ω
2. 4Ω
3. 5 Ω
4. 10Ω

Answer

5 Ω

Reason —

Given, two identical cells each of emf 1.5 V and internal resistance r joined in parallel, and two resistors of 20 Ω each joined in parallel. The voltmeter reads V = 1.2 V.

External resistance :

$\text R = \dfrac{20 \times 20}{20 + 20} \\[1em] \text R = 10\ \Omega$

Equivalent emf and internal resistance : For two identical cells in parallel the emf remains 1.5 V and the internal resistance becomes $\dfrac{r}{2}$.

Terminal voltage :

$\text V = \dfrac{\text{ER}}{\text R + \dfrac{r}{2}} \\[1em] 1.2 = \dfrac{1.5 \times 10}{10 + \dfrac{r}{2}} \\[1em] 10 + \dfrac{r}{2} = \dfrac{15}{1.2} \\[1em] 10 + \dfrac{r}{2} = 12.5 \\[1em] \dfrac{r}{2} = 2.5 \\[1em] r = 5\ \Omega$

What will be the most suitable combination of three resistors A = 2Ω, B = 4 Ω, C = 6Ω, so that $\left(\dfrac{22}{3}\right)\Omega$ is equivalent resistance of combination?

1. Parallel combination of A and C connected in series with B.
2. Parallel combination of A and B connected in series with C.
3. Series combination of A and C connected in parallel with B.
4. Series combination of B and C connected in parallel with A.

Answer

Parallel combination of A and B connected in series with C.

Reason — Joining A = 2 Ω and B = 4 Ω in parallel,

$\text R_\text P = \dfrac{2 \times 4}{2 + 4} \\[1em] \text R_\text P = \dfrac{8}{6} \\[1em] \text R_\text P = \dfrac{4}{3}\ \Omega$

Joining C = 6 Ω in series with this combination,

$\text R = \dfrac{4}{3} + 6 \\[1em] \text R = \dfrac{4 + 18}{3} \\[1em] \text R = \dfrac{22}{3}\ \Omega$

which is the required equivalent resistance.

An aluminium wire is stretched to make its length, 0.4% larger. The percentage change in resistance is :

1. 0.4%
2. 0.2%
3. 0.8%
4. 0.6%

Answer

0.8%

Reason — On stretching, the volume of the wire remains constant, so

$\text R \propto l^2$

Taking the fractional change,

$\dfrac{\Delta\text R}{\text R} \times 100 = 2 \times \dfrac{\Delta l}{l} \times 100 \\[1em] \dfrac{\Delta\text R}{\text R} \times 100 = 2 \times 0.4$

Two resistors of resistance, 100 Ω and 200Ω are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in 100 Ω to that in 200 Ω in a given time is :

1. 4 : 1
2. 1 : 2
3. 2 : 1
4. 1 : 4

Answer

2 : 1

Reason — In a parallel combination the same potential difference V acts across both the resistors, so the thermal energy developed in a given time is

$\text H = \dfrac{\text V^2 t}{\text R} \\[1em] \text H \propto \dfrac{1}{\text R} \\[1em] \dfrac{\text H_{100}}{\text H_{200}} = \dfrac{\text R_{200}}{\text R_{100}} \\[1em] \dfrac{\text H_{100}}{\text H_{200}} = \dfrac{200}{100} \\[1em] \dfrac{\text H_{100}}{\text H_{200}} = \dfrac{2}{1}$

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 × 10^-3 kg, density is 8.92 × 10³ kg/m³ and resistivity is 1.7 × 10^-8 Ω m. The length of wire is :

1. l = 10 m
2. l = 100 m
3. l = 5 m
4. l = 6.8 m

Answer

l = 10 m

Reason —

Given, I = 2 A, V = 3.4 V, mass m = 8.92 × 10^-3 kg, density d = 8.92 × 10³ kg/m³ and ρ = 1.7 × 10^-8 Ω m.

Resistance of the wire :

$\text R = \dfrac{\text V}{\text I} \\[1em] \text R = \dfrac{3.4}{2} \\[1em] \text R = 1.7\ \Omega$

Volume of the wire :

$\text{Vol} = \dfrac{m}{d} \\[1em] \text{Vol} = \dfrac{8.92 \times 10^{-3}}{8.92 \times 10^{3}} \\[1em] \text{Vol} = 10^{-6}\ \text m^3$

Length of the wire : Since $\text A = \dfrac{\text{Vol}}{l}$,

$\text R = \rho\dfrac{l}{\text A} \\[1em] \text R = \dfrac{\rho l^2}{\text{Vol}} \\[1em] 1.7 = \dfrac{(1.7 \times 10^{-8})l^2}{10^{-6}} \\[1em] 1.7 = (1.7 \times 10^{-2})l^2 \\[1em] l^2 = 100 \\[1em] l = 10\ \text m$

The charge flowing in a conductor changes with time as Q(t) = αt − βt² + γt³, where α, β and γ are constants. Minimum value of current is

1. $\alpha - \dfrac{\gamma^2}{3\beta}$
2. $\beta - \dfrac{\alpha^2}{3\gamma}$
3. $\alpha - \dfrac{\beta^2}{3\gamma}$
4. $\alpha - \dfrac{3\beta^2}{\gamma}$

Answer

$\alpha - \dfrac{\beta^2}{3\gamma}$

Reason —

Given, Q(t) = αt − βt² + γt³.

The current is the rate of flow of charge,

$i = \dfrac{\text{dQ}}{\text{dt}} \\[1em] i = \alpha - 2\beta t + 3\gamma t^2$

For the current to be minimum,

$\dfrac{\text{di}}{\text{dt}} = 0 \\[1em] -2\beta + 6\gamma t = 0 \\[1em] t = \dfrac{\beta}{3\gamma}$