Taylor

Optics Synopsis Ray Optics

Laws of reflection:

The incident ray, the reflected ray, and the normal to the reflecting surface are coplanar.

The angle of incidence (i) = angle of reflection (r).

Some important facts about reflection in a plane mirror:

The image and the object are equidistant from the plane mirror.

The image formed by the plane mirror is virtual and of the same size as the object but laterally inverted.

When the plane mirror is turned through an angle θ, the reflected ray is turned through 2θ in the same direction.

The number of images (N) formed due to two plane mirrors inclined at an angle θ is given by

“ray optics class 12 notes “

⇒ \(N=\frac{360^{\circ}}{\theta}-1 \text { when } \frac{360^{\circ}}{\theta} \text { is even }\)

⇒ and \(N=\frac{360^{\circ}}{\theta} \text { when } \frac{360^{\circ}}{\theta} \text { is odd }\)

For θ = 60°, N = 6-1 = 5; and for θ = 40°, N = 9.

The minimum height of the plane mirror required to view the whole image of an object is half the height of the object.

The deviation of a ray is the angle between the directions of the incident and reflected rays.

Reflection from spherical surfaces (concave and convex mirrors):

The image formed by a concave mirror is real for u ≥ f and virtual for u < f.

The image formed by a convex mirror is always virtual and diminished.

⇒ Mirror formula: \(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\)

chapter 9 physics class 12 notes

⇒ Lateral or transverse magnification = \(m_{\mathrm{T}}=-\frac{v}{u}\)

⇒ Longitudinal magnification = \(m_{\mathrm{L}}=-m_{\mathrm{T}}^2=-\left(\frac{v}{u}\right)^2\)

Refraction through a plane surface:

Symmetrical form of Snell’s law:

μ₁ sinθ₁ = μ₂ sinθ₂ = μ₃ sinθ₃ = … .

Here μ is the refractive index of the medium.

Note that this law is true for both plane and curved surfaces.

Refractive index (μ) and speed of light (c):

⇒ \(\mu=\frac{\text { speed of light in vacuum }}{\text { speed of light in the medium }}=\frac{c_0}{c}\)

In symmetrical form, μ₁c₁ = μ₂c₂ =c₀.

During refraction, the speed of light (c) and its wavelength (λ.) change but its frequency (f) remains invariant. Hence,

⇒ \(\mu_1 \lambda_1=\mu_2 \lambda_2=\ldots=\frac{c_0}{f}\)

Apparent depth (h):

1. Refractive index = \(\frac{\text { real depth }}{\text { apparent depth }}=\frac{H}{h}\)
2. Shift of image = \(H-h=\left(1-\frac{1}{\mu}\right) H\)

Critical angle \(\left(\theta_{\mathrm{c}}\right): \sin \theta_{\mathrm{c}}=\frac{\mu_2}{\mu_1}, \text { where } \mu_1>\mu_2\)

Refraction through a spherical surface separating two transparent media:

⇒ \(\frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R}\)

Magnification = \(m=\frac{\mu_1}{\mu_2} \cdot \frac{v}{u}\)

⇒ Lens-maker’s formula: \(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

The power of a lens is the reciprocal of its focal length.

⇒ So, \(P(\text { in dioptres })=\frac{100}{f(\text { in centimetres })}\)

P is positive for a convex lens and negative for concave lens.

“physics class 12 chapter 9 notes “

For two lenses in contact,

⇒ \(\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2} \text { and } P=P_1+P_2\)

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Path of a ray through a prism: The emergent ray bends towards the base of a prism.

A = r + r’ and A + δ = i + i’

Condition for minimum deviation:

i = i’ and r = r’.

⇒ \(\mu=\frac{\sin \frac{A+\delta_m}{2}}{\sin \frac{A}{2}}\)

Deviation through a thin prism (with A small) = δ = (μ-1)A.

Dispersion (Cauchy’s relation): \(\mu=A+\frac{B}{\lambda^2}\)

Mean deviation = δ = (μ_y-1)A.

Angular dispersion = θ = δ_V– δ_R =(μ_V – μ_R.)A.

⇒ Dispersive power = \(\omega=\frac{\text { angular dispersion }}{\text { mean deviation }}=\frac{\theta}{\delta_{\mathrm{y}}}=\frac{\mu_{\mathrm{V}}-\mu_{\mathrm{R}}}{\mu_{\mathrm{y}}-1}\)

Dispersion without deviation (direct-vision spectroscope):

1. Condition: \(\frac{A_1}{A_2}=-\frac{\mu_{2 y}-1}{\mu_{1 y}-1} \)
2. Net angular dispersion = θ = A₁(μ_1y – 1)(ω₁ – ω₂).

Deviation without dispersion (achromatic combination of prisms):

Condition: \(\frac{A_1}{A_2}=-\frac{\omega_2\left(\mu_{2 y}-1\right)}{\omega_1\left(\mu_{1 y}-1\right)}\)

Achromatic combination of lenses: \(\frac{\omega_1}{f_1}+\frac{\omega_2}{2}=0\)

Optical instruments: These are devices to increase the visual angle.

Magnification (m) of a simple microscope:

1. \(m=1+\frac{D}{f}\) when the final image is at a distance D.
2. \(m=\frac{D}{f}\) for normal adjustment, i.e., when the eye is most relaxed (v = ∞).

Magnification (m) of a compound microscope:

1. \(m=\frac{-v_{\mathrm{o}}}{u_{\mathrm{o}}}\left(1+\frac{D}{f_{\mathrm{e}}}\right) \approx-\frac{L}{f_{\mathrm{o}}}\left(1+\frac{D}{f_{\mathrm{e}}}\right)\)
2. For normal adjustment, m = \(-\frac{L}{f_0} \frac{D}{f_e}\)

Astronomical telescope:

1. Magnification = m = \(-\frac{f_0}{f_{\mathrm{e}}}\)
2. Tube length = L = f₀ +f_e.

Wave Optics

Light (or any other electromagnetic wave) is a transverse wave motion.

A wavefront is an imaginary surface containing all the points vibrating in the same phase.

“class 12th physics ray optics “

Coherent sources: Sources are said to be coherent if the phase difference between them remains invariant with time.

Path difference is

Δx = S₂P-S₁P ≈ S₂N = d sin θ,

where d = separation between the coherent sources S₁ and S₂.

Phase difference = \(\phi=\frac{2 \pi}{\lambda}(\Delta x)\)

⇒ General relation: \(\frac{\Delta x}{\lambda}=\frac{\phi}{2 \pi}=\frac{\Delta t}{T}\), where Δt = time delay.

Optical path = p(path length in the medium).

Intensity ∝ (amplitude)².

Resultant intensity (I):

A₁ = A1₂ + A₂² + 2A₁A₂ Cos Φ

or I = I₁ +1₂ + 2√I₁I₂cos Φ.

Maxima and minima:

For maxima, cos Φ = +1 ⇒ Φ = 2nπ for n = 0, 1, 2, … .

For minima, cos Φ = -1 ⇒ Φ = (2n +1)π for n = 0, 1, 2, … .

The following table shows the conditions for maxima and minima.

Fringe width = \(\beta=\frac{\lambda D}{d}\), where λ = wavelength, D = distance between the screen and the plane of the slits, and d = slit-separation.

Angular width of a fringe = θ = \(\frac{\beta}{D}=\frac{\lambda}{d}\)

⇒ Maximum number of observable fringes = \(N_{\max }=\frac{d}{\lambda}\)

Shift of a fringe system: When a thin film of thickness t and refractive index μ is pasted over one slit, the optical path introduced is Δx = (μ-1)t.

number of fringes shifted = N = \(=\frac{(\mu-1) t}{\lambda}\)

“ray optics in physics “

The angular position of minima in diffraction:

For a single slit of width a:

⇒ a sin θ_n = nλ ⇒ \(\theta_n \approx n\left(\frac{\lambda}{a}\right)\)

For a circular aperture of diameter b:

⇒ b sin θ =1.22λ ⇒ \(\theta \approx 1.22\left(\frac{\lambda}{b}\right)\)

Airy disc: It is the central bright circular region of the diffraction pattern formed due to a circular aperture (objective lens, pupil, etc.).

1. Angular width of the principal maximum due to a single slit = \(2 \theta_1=2\left(\frac{\lambda}{a}\right)\)
2. Angular radius of the Airy disc = θ = \(1.22\left(\frac{\lambda}{b}\right)\)

Rayleigh criterion: Two-point sources cannot be resolved if their separation is less than the radius of the Airy disc.

For resolution, the angular separation is \(\theta \geq \ 1.22\left(\frac{\lambda}{b}\right)\).

Resolving power (RP):

1. For a telescope, RP = \(\frac{1}{\Delta \theta}=\frac{b}{1.22 \lambda}\) where b = diameter of the objective.
2. For a microscope,r RP = \(\frac{1}{d_{\min }}=\frac{2 \mu \sin \beta}{1.22 \lambda}=\frac{2 \mathrm{NA}}{1.22 \lambda}\), where NA = numerical aperture = μsinβ

Polarization:

Brewster’s law: μ = tanθ_p.

Intensity of the polarized light = I = \(\frac{1}{2} I_0\), where I₀ = intensity of the unpolarized light.

Malus’s law: I = I₀cos²θ, where I₀ = initial intensity of the incident polarized light, I = intensity after its emergence and θ = angle between the pass axes.

Current Electricity

Question 1. A wire of resistance 4 is stretched to twice its original length. The resistance of the stretched wire would be

1. 4
2. 6
3. 8
4. 16

Answer: 4. 16

Resistance = R = \(\rho \frac{l}{A}\).

When l is doubled, A is reduced to \(\frac{A}{2}\), so that volume remains constant.

∴ \(R^{\prime}=\rho \frac{l^{\prime}}{A^{\prime}}=\rho \frac{2 l}{A / 2}=4\left(\rho \frac{l}{A}\right)=4 R=4(4 \Omega)=16 \Omega\)

Question 2. A wire of resistance 12 SI m”1 is bent to form a complete circle of radius 10 cm. The resistance between its two diametrically opposite points A and B as shown in the figure is

electricity quiz

1. 0.6
2. 3
3. 6
4. 6

Answer: 1. 0.6

The total resistance of the wire,

⇒ \(R=\left(12 \Omega \mathrm{m}^{-1}\right) \text { (total length) }=\left(12 \Omega \mathrm{m}^{-1}\right)(2 \pi r)\)

⇒ \(\left(12 \Omega \mathrm{m}^{-1}\right)\left(2 \pi \times 10 \times 10^{-2} \mathrm{~m}\right)=2.4 \pi \Omega\)

Resistance of each half across AB is \(\frac{R}{2}\) = 1.2π Ω

∴ equivalent resistance across AB will be 0.6π Ω.

Read And Learn Also NEET Physics Multiple Choice Question and Answers

Question 3. A wire of a certain material is stretched slowly by 10 percent. What will be its resistance and resistivity respectively?

1. 1.2 times and 1.1 times
2. 1.2 times and remains unchanged
3. 1.1 times and 1.2 times
4. None of the above

Answer: 2. 1.2 times and remains unchanged

\(R=\rho \frac{l}{A}=\rho \frac{l}{V n}=\frac{\rho}{V} l^2\) (where volume V is constant).

Fractional change in the resistance,

\(\frac{d R}{R}=2 \frac{d l}{l}\) = 2(10%) = 0.2.

Increase in resistance, dR = 0.2R.

Hence, resistance becomes R + 0.2R = 1.2R, but resistivity remains unchanged.

current electricity pyq

Question 4. When a uniform wire of resistance R is stretched so that its radius r becomes r/2 then its resistance becomes

1. R
2. 4R
3. 12R
4. 16R

Answer: 4. 16R

Since volume V = Al remains constant,

⇒ \(A_1 l_1=A_2 l_2 \Rightarrow \frac{l_1}{l_2}=\frac{A_2}{A_1}=\frac{\pi(r / 2)^2}{\pi r^2}=\frac{1}{4}\)

⇒ Now, \(\frac{R_1}{R_2}=\frac{\rho l_1 / A_1}{\rho l_2 / A_2}=\left(\frac{l_1}{l_2}\right)\left(\frac{A_2}{A_1}\right)=\left(\frac{1}{4}\right)\left(\frac{1}{4}\right)=\frac{1}{16}\)

∴ \(R_2=16 R_1=16 R\)

Question 5. Three conductors have their conductances 2 S, 4 S, and 6 S. When they are joined in parallel, their equivalent conductance Will be

1. 12 S
2. \(\frac{1}{12} \mathrm{~S}\)
3. \(\frac{12}{11} \mathrm{~s}\)
4. \(\frac{11}{12} \mathrm{~S}\)

Answer: 1. 12 S

Conductance is reciprocal of the resistance and measured in Siemens ormhos. When connected in parallel,

\(\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=(2+4+6) \mathrm{S}\).

∴ Equivalent conductance = \(G=\frac{1}{R}=12 \mathrm{~S}\).

Question 6. A and B are two points on a uniform ring of resistance R. The ∠ACB = θ, where C is the center of the ring. The equivalent resistance between A and B is

1. \(\frac{R \theta(2 \pi-\theta)}{4 \pi^2}\)
2. \(R\left(1-\frac{\theta}{2 \pi}\right)\)
3. \(\frac{R \theta}{2 \pi}\)
4. \(\frac{R(2 \pi-\theta)}{4 \pi}\)

Answer: 1. \(\frac{R \theta(2 \pi-\theta)}{4 \pi^2}\)

Let σ be the resistance per unit length of the wire. Hence, the resistance of the upper part of APB is

R₁ = σ (length APB)

= σ (θr), where r is the radius of the circle.

Similarly, the resistance of the lower part of AQB is

R₂ = σr(2π-θ).

These segments are joined in parallel, so their equivalent resistance is

⇒ \(R_{\text {eq }}=\frac{R_1 R_2}{R_1+R_2}=\frac{\sigma(\theta r) \sigma r(2 \pi-\theta)}{\sigma \theta r+\sigma r(2 \pi-\theta)}=\frac{\sigma r \theta(2 \pi-\theta)}{2 \pi} .\)

Substituting σ \(\sigma=\frac{R}{2 \pi r}\), we get

∴ \(R_{\text {eq }}=\left(\frac{R}{2 \pi r}\right)\left(\frac{r \theta}{2 \pi}\right)(2 \pi-\theta)=\frac{R \theta(2 \pi-\theta)}{4 \pi^2} .\)

Question 7. The masses of three wires of copper are in the ratio of 1 : 3: 5 and their lengths are in the ratio of 5 : 3: 1. The ratio of their resistances is

1. 1:3:5
2. 5:3:1
3. 1:25:125
4. 125:15:1

Answer: 4. 125:15:1

The Ratio of the masses = m₁: m₂: m₃ = 1:3:5

m₁=m, m₂ : 3m, m₃ = 5m.

Ratio of the lengths = l₁ : l₂ : l₃ = 5:3:1

l₁=5l, l₂=3l, l₃=l.

Now, resistance \(R=\rho \frac{l}{A}=\rho \frac{l^2}{V}=\rho D \frac{l^2}{m}\)

where V= volume and D = density.

Hence,

⇒ \(\rho D=\text { constant, so } R \propto \frac{l^2}{m} \text {. }\)

⇒ \(R_1: R_2: R_3=\frac{l_1^2}{m_1}: \frac{l_2^2}{m_2}: \frac{l_3^2}{m_3}=\frac{25 l^2}{m}: \frac{9 l^2}{3 m}: \frac{l^2}{5 m}\)

∴ \(25: 3: \frac{1}{5}=125: 15: 1\)

Question 8. N equal resistors are first connected in series and then connected in parallel. What is the ratio of the maximum to the minimum resistance?

1. N
2. \(\frac{1}{N^2}\)
3. \(N^2\)
4. \(\frac{1}{N}\)

Answer: 3. \(N^2\)

Equivalent resistance is maximum when connected in series, so

⇒  \(R_{\max }=(R+R+\ldots)_{N \text { times }}=N R\).

It will be minimal when connected in parallel, so

⇒ \(\frac{1}{R_{\min }}=\left(\frac{1}{R}+\frac{1}{R}+\ldots\right)_{N \text { times }}=\frac{N}{R} \Rightarrow R_{\min }=\frac{R}{N}\)

∴ Hence, the ratio \(\frac{R_{\max }}{R_{\min }}=\frac{N R}{R / N}=N^2\).

Question 9. Two wires of the same metal have the same length, but their cross sections are in the ratio 3:1. They are joined in series. The resistance of the thicker wire is 10 Ω. The total resistance of the combination will be

1. 10
2. 20
3. 40
4. 100

Answer: 3. 40

Given that \(l_1=l_2 \text { and } A_1: A_2=3: 1\)

Ratio of resistances, \(\frac{R_1}{R_2}=\frac{l_1}{l_2} \frac{A_2}{A_1}=\frac{1}{3}\)

Resistance of the thicker wire, \(R_1=10 \Omega\)

Hence, \(R_2=3 R_1=30 \Omega\)

Equivalent resistance when connected in series is \(R=R_1+R_2=10 \Omega+30 \Omega=40 \Omega\).

Question 10. A wire of resistance R is melted and recast to half of its length. The resistance of the new wire becomes

1. \(\frac{R}{4}\)
2. \(\frac{R}{2}\)
3. R
4. 2R

Answer: 1. \(\frac{R}{4}\)

When the length is halved, the area of the cross-section will be doubled so that volume remains constant

⇒ \(\frac{R^{\prime}}{R}=\frac{\rho l^{\prime} / A^{\prime}}{\rho l / A}=\left(\frac{l^{\prime}}{l}\right)\left(\frac{A}{A^{\prime}}\right)=\left(\frac{l / 2}{l}\right)\left(\frac{A}{2 A}\right)=\frac{1}{4}\)

∴ \(\text { Hence, } R^{\prime}=\frac{R}{4}\)

Question 11. Three resistances, each of 4 Q, are connected to form a triangle. The resistance between any two terminals will be

1. 12
2. 2
3. 6
4. \(\frac{8}{3} \Omega\)

Answer: 4. \(\frac{8}{3} \Omega\)

The combination across any two comers (say B and C) consists of 8 Ω and 4 Ω in parallel, for which the equivalent resistance is

∴ \(R=\frac{(8 \Omega)(4 \Omega)}{(8 \Omega)+(4 \Omega)}=\frac{8}{3} \Omega\)

Question 12. When a wire of uniform cross-section a, length L, and resistance R is bent into a circle, the resistance between two of its diametrically opposite points will be

1. 4R
2. 2R
3. \(\frac{R}{4}\)
4. \(\frac{R}{2}\)

Answer: 3. \(\frac{R}{4}\)

Total resistance = R. This is equally divided into two parts across the diameter, each equal to \(\frac{R}{2}\).

Hence, for parallel combination, \(\frac{1}{R_{\mathrm{eq}}}=\frac{1}{R / 2}+\frac{1}{R / 2}=\frac{4}{R} \Rightarrow R_{\mathrm{eq}}=\frac{R}{4}\)

Question 13. N resistors, each of rΩ, when connected in parallel give an equivalent resistance of R Ω. If these resistors were connected in series, the combination would have a resistance in ohms equal to

1. \(N^2 R\)
2. \(\frac{R}{N^2}\)
3. \(\frac{R}{N}\)
4. NR

Answer: 1. \(N^2 R\)

Equivalent resistance of N resistances, each of resistance r, connected in parallel is \(R=\frac{r}{N} \Omega\).

When connected in series, the equivalent resistance is \(R^{\prime}=N r \Omega=N(N R) \Omega\)

\(N^2 R \Omega\).  [∵\(R=\frac{r}{N}\)]

Question 14. A circuit consisting of five resistors, each of resistance R, forms a Wheatstone bridge. What is the equivalent resistance of the circuit?

1. 2R
2. R
3. \(\frac{R}{2}\)
4. \(\frac{2 R}{3}\)

Answer: 2. R

Equivalent resistance means resistance across the terminals of the cell which is a combination of two resistance of 2R in parallel. This is equal to R.

Question 15. Two metal wires of identical dimensions are connected in series. If \(\sigma_1 \text { and } \sigma_2\) are the conductivities of the metal wires respectively, the effective conductivity of the combination is

1. \(\frac{\sigma_1+\sigma_2}{\sigma_1 \sigma_2}\)
2. \(\frac{\sigma_1 \sigma_2}{\sigma_1+\sigma_2}\)
3. \(\frac{\sigma_1+\sigma_2}{2 \sigma_1 \sigma_2}\)
4. \(\frac{2 \sigma_1 \sigma_2}{\sigma_1+\sigma_2}\)

Answer: 4. \(\frac{2 \sigma_1 \sigma_2}{\sigma_1+\sigma_2}\)

Resistances of the component resistors are \(R_1=\rho_1 \frac{l}{A}\)

and \(R_2=\rho_2 \frac{l}{A}\)

∴ \(R_{\mathrm{eq}}=R_1+R_2=\left(\rho_1+\rho_2\right) \frac{l}{A}\)

If this is replaced by a single wire of equivalent resistivity p then \(\rho \frac{2 l}{A}=\left(\rho_1+\rho_2\right) \frac{l}{A}\)

\(2 p=\rho_1+\rho_2\).

Since conductivity \(\sigma=\frac{1}{\rho}\), we have

⇒ \(\frac{2}{\sigma}=\frac{1}{\sigma_1}+\frac{1}{\sigma_2}=\frac{\sigma_1+\sigma_2}{\sigma_1 \sigma_2}\)

∴ \(\sigma=\frac{2 \sigma_1 \sigma_2}{\sigma_1+\sigma_2}\)

Question 16. A 12-cm-long wire is given the shape of a right-angled triangle ABC having sides 3 cm, 4 cm, and 5 cm as shown in the figure. The resistance between two ends (AB, BC, CA) of the respective sides are measured one by one by a multimeter, the resistances will be in file ratio

1. 9:16:25
2. 3:4:5
3. 27:32:35
4. 21:24:25

Answer: 3. 27:32:35

Let r be the resistance per unit length of the wire, so the three sides have the resistances 3r, 4r, and 5r respectively.

Resistances across AB, \(R_1=\frac{9 \times 3}{12} r\);

across BC, \(R_2=\frac{8 \times 4}{12} r \text { and across AC, } R_3=\frac{7 \times 5}{12} r\).

Hence, the ratio \(R_1: R_2: R_3=27: 32: 35\).

Question 17. A ring is made of a wire having a resistance of 12 Q. Find the points A and B, as shown in the figure at which a current-carrying conductor should be connected so that the resistance R of the sub-circuit between these two points is equal to (8/3)

1. \(\frac{l_1}{l_2}=\frac{5}{8}\)
2. \(\frac{l_1}{l_2}=\frac{1}{2}\)
3. \(\frac{l_1}{l_2}=\frac{1}{3}\)
4. \(\frac{l_1}{l_2}=\frac{3}{8}\)

Answer: 2. \(\frac{l_1}{l_2}=\frac{1}{2}\)

Let r be the resistance per unit length.

∴ Resistance of length\(l_1 \text { is } R_1=r l_1 \text { and } R_2=r l_2\).

Total resistance = \(\left(R_1+R_2\right)=r\left(l_1+l_2\right)=12 S\).

Equivalent resistance across AB is \(\frac{R_1 R_2}{R_1+R_2}=\frac{\left(r l_1\right)\left(r l_2\right)}{r\left(l_1+l_2\right)}=\frac{8}{3} \Omega\)

⇒ \(R_1 R_2=\frac{8}{3}\left(R_1+R_2\right)=\frac{8}{3} \times 12 \Omega=32 \Omega\)

Hence, \(R_1=8 \Omega, R_2=4 \Omega \text { and } \frac{R_1}{R_2}=\frac{l_1}{l_2}=\frac{8}{4}=\frac{2}{1}\).

The ratio is 2: l or 1: 2.

“electricity physics questions “

Question 18. Three resistors P, Q, and R each of 2Ω, and an unknown resistance S form the four arms of a Wheatstone bridge circuit When a resistor of 6 Ω is connected parallel to S, the bridge gets balanced. What is the value of S?

1. 1
2. 3
3. 6
4. 2

Answer: 2. 3

For the bridge to be balanced, the fourth arm must have equivalent resistance = \(2 \Omega\)

⇒ \(\frac{(6 \Omega) S}{6 \Omega+S}=2 \Omega\)

⇒ \(12 \Omega+2 S=6 S\)

⇒ \(4 S=12 \Omega \Rightarrow S=3 \Omega\)

Question 19. In a Wheatstone bridge, all four arms have equal resistance R. If the resistance of the galvanometer arm is also R, the equivalent resistance of the combination as seen by the battery is

1. \(\frac{R}{4}\)
2. \(\frac{R}{2}\)
3. R
4. 2R

Answer: 3. R

The equivalent resistance as seen by the battery is equivalent to the effective resistance across the battery, which is \(\frac{(2 R)(2 R)}{(2 R)+(2 R)}=R\). The galvanometer resistance R will be ineffective as current through the galvanometer, \(I_g=0\).

Question 20. A fuse wire is a wire of

1. High resistance and high melting point
2. High resistance and low melting point
3. Low resistance and low melting point
4. Low resistance and high melting point

Answer: 2. High resistance and low melting point

A fuse wire is used to reduce the damage of electrical appliances when a high current passes through the wire. A fuse wire should have high resistance and a low melting point so that it may melt easily if the current suddenly becomes high.

Question 21. The resistance of each arm of a Wheatstone bridge is 10 Ω. If a resistance of Ω is connected in series with the galvanometer, the equivalent resistance across the battery will be

1. 10
2. 15
3. 20
4. 40

Answer: 1. 10

When the die Wheatstone bridge is balanced, no current flows through the galvanometer, so galvanometer resistance remains ineffective. Any addition of resistance \((10 \Omega)\) connected in series or across the galvanometer will not change the effective resistance across the battery. In this case,

⇒ \(R_{\mathrm{eq}}=\frac{(20 \Omega)(20 \Omega)}{40 \Omega}=10 \Omega\)

Question 22. What is the equivalent resistance between A and B in the given figure?

1. 40
2. 10
3. 20
4. 50

Answer: 3. 20

The given network of resistors represents a balanced Wheatstone bridge whose each branch has the same resistance of \(20 \Omega\). Hence, the equivalent resistance is \(R=\frac{(40 \Omega)(40 \Omega)}{40 \Omega+40 \Omega}=\frac{1600}{80} \Omega=20 \Omega\).

Question 23. The resistance across the terminal points A and B of the given infinitely long circuit will be

1. \((\sqrt{3}-1) \Omega\)
2. \((2-\sqrt{3}) \Omega\)
3. \((\sqrt{3}+1) \Omega\)
4. \((2+\sqrt{3}) \Omega\)

Answer: 3. \((\sqrt{3}+1) \Omega\)

Let the equivalent resistance of the given infinitely long circuit be x. Hence, the given circuit can be replaced by the circuit shown in the adjoining figure. The equivalent resistance across AB will also be x.

Hence, \(x=1 \Omega+\frac{(1 \Omega) x}{1 \Omega+x}+1 \Omega\)

⇒ \(x^2-2 x-2=0\)

⇒ \(x=(\sqrt{3}+1) \Omega\).

Question 24. The equivalent resistance between points X and Y in the given figure will be

1. 10
2. 7
3. 5
4. 3

Answer: 3. 5

The given network of resistances can be redrawn in a simple form as shown in the adjoining diagram. 3 Ω and 7 Ω in series is equal to 10 Ω which in parallel with 10 Ω is equivalent to 5 Ω. Next, this 5 Ω x and 5 Ω in series is 10 Ω which finally combines in parallel with 10 Ω on XY and gives R = 5 Ω.

Question 25. Five identical resistors, each of resistance r, are connected as shown in the figure. A battery of V volt is connected between A and B. The current flowing through the branch AFCEB will be

1. \(\frac{3 V}{r}\)
2. \(\frac{V}{r}\)
3. \(\frac{V}{2 r}\)
4. \(\frac{2 V}{r}\)

Answer: 3. \(\frac{V}{2 r}\)

The given network of resistors can be redrawn as shown in this figure. This constitutes a balanced Wheatstone bridge for which the equivalent resistance across AB is \(\frac{(2 r)(2 r)}{4 r}=r\).

Current delivered by the cell is I = \(\frac{V}{r}\)

This current is divided equally into the upper and lower branches. Hence, current through the branch AFCEB is \(\frac{I}{2}=\frac{V}{2 r}\).

Question 26. The net resistance of a circuit between A and B is

1. \(\frac{8}{3} \Omega\)
2. \(\frac{14}{3} S\)
3. \(\frac{16}{3} \Omega\)
4. \(\frac{22}{3} \Omega\)

Answer: 2. \(\frac{14}{3} S\)

The given circuit represents a balanced Wheatstone bridge, since \(\frac{P}{Q}=\frac{3}{4}=\frac{R}{S}=\frac{6}{8}\).

Hence, the cross-connected resistance of 7 Ω is ineffective.

Resistance of upperbranch = 7 Ω

and that oflowerbranch =14 Ω.

∴ equivalent resistance = \(R=\frac{(7 \Omega)(14 \Omega)}{(7+14) \Omega}=\frac{14}{3} \Omega\)

Question 27. In the given figure, each resistor has its resistance r =10. What will be the equivalent resistance between the terminals A and D?

1. 10
2. 20
3. 30
4. 40

Answer: 3. 30

In order to find the equivalent resistance across AD, we imagine a cell whose terminals are connected to A and D respectively. The equivalent circuit is shown in the figure. The equivalent resistance across AD is \(R=r+\frac{(2 r)(2 r)}{2 r+2 r}+r=3 r=3(10 \Omega)=30 \Omega\).

Question 28. Six resistors, each of 3 Q., are connected along the sides of a hexagon and three resistors of 6Q each are connected along AC, AD, and AE as shown in the figure. The equivalent resistance F between A and B is equal to

1. 6
2. 2
3. 3
4. 9

Answer: 2. 2

Resistors AF and FE, each of 3 Ω in series, give 6 £2. This 6 Ω with AE (= 6 Ω) in parallel gives 3 Ω. This, 3 £2 with ED (=3 Ω) gives 6 Ω which in parallel with AD (=6 Ω) gives 3 Ω. Proceeding in the same way, the equivalent resistance across AC is found to be Ω. Finally, we are left with 6 Ω and 3 Ω across AB in parallel which gives \(R_{\mathrm{AB}}=\frac{(6 \Omega)(3 \Omega)}{6 \Omega+3 \Omega}=2 \Omega\)

Question 29. In the given circuit, if the current through the 3 Q resistor is 0.8 A then the potential drop across the 4Q resistor is

1. 9.6 V
2. 1.2V
3. 4.8 V
4. 2.6 V

Answer: 3. 4.8 V

Potential difference (p.d.) across the 3-Ω resistor,

V-IR = (0.8 A)(3 ft) = 2.4 V.

p.d. is the same across the 6-Ω resistor, J'(6 Q) = 2.4 V.

Hence, current through the 6-Ω resistor,\(I^{\prime}=\frac{2.4 \mathrm{~V}}{6 \Omega}=0.4 \mathrm{~A}\).

Total current through the circuit = \(\left(I+I^{\prime}\right)=(0.8 \mathrm{~A})+(0.4 \mathrm{~A})=1.2 \mathrm{~A}\)

∴ p.d. across 4 Q resistor = (1.2 A)(4 £2) = 4.8 V

Question 30. In the given network, each resistance r =1 £1 The effective resistance between A and B is

1. \(\frac{4}{3} \Omega\)
2. \(\frac{8}{7} \Omega\)
3. \(7 \Omega\)
4. \(\frac{3}{2} \Omega\)

Answer: 2. \(\frac{8}{7} \Omega\)

By symmetry, the current (I₁ + I₂), which enters at A must leave at B. So, does not get divided at O. Hence, point O can be detached from the arm. This circuit can be redrawn equivalently as shown in figure (2), for which the upper and lower arms have the resistances \(r+\frac{2 r}{3}+r=\frac{8}{3} r \text { and } 2 r\)

equivalent resistance across AB \(R_{\mathrm{AB}}=\frac{\left(\frac{8}{3} r\right)(2 r)}{\frac{8}{2} r+2 r}=\frac{8}{7} r=\frac{8}{7}(1 \Omega)=\frac{8}{7} \Omega\)

Question 31. In the network shown in the figure, each resistance is equal to r. The equivalent resistance between terminals A and b is

1. 3r
2. 6r
3. \(\frac{3 r}{2}\)
4. \(\frac{2 \pi}{3}\)

Answer: 4. \(\frac{2 \pi}{3}\)

When the current enters the network through a, it gets equally divided through branches. Points c, d, and e will be at the same potential. Hence, no current flows through cd and de, and these two resistors are ineffective. The circuit is thus equivalent to three equal resistors in parallel, each equal to 2r.

∴ \(\frac{1}{R}=\frac{1}{2 r}+\frac{1}{2 r}+\frac{1}{2 r}=\frac{3}{2 r} \Rightarrow R=\frac{2}{3} r\)

“physics questions on electricity “

Question 32. In the given figure, let the equivalent resistance between terminals A and B be (when A°- switch S is open) and Rc (when switch S is closed). The ratio R₀:R_c will be

1. 9: 8
2. 8:9
3. 4:3
4. 3:4

Answer: 1. 9:8

The two adjoining figures have been drawn with switch S open and with switch S closed.

So, \(R_{\mathrm{o}}=\frac{(18 \Omega)(18 \Omega)}{(18 \Omega)+(18 \Omega)}=9 \Omega\)

⇒ and \(R_c=\frac{(12 \Omega)(6 \Omega)}{(12 \Omega)+(6 \Omega)}+\frac{(6 \Omega)(12 \Omega)}{(6 \Omega)+(12 \Omega)}=8 \Omega\)

∴ \(\frac{R_0}{R_c}=\frac{9}{8}\)

Question 33. The equivalent resistance between terminals A and B in the given figure is

1. 2
2. 5
3. 4
4. 6

Answer: 2. 5

Resistance of 3 Ω each in DE and EF in series gives 6 Ω which joined in

parallel with 3 Ω in branch DG gives \(\frac{6 \times 3}{9} \Omega=2 \Omega\).

This 2 Ω with 4 Ω in series with branch CD gives 6 ft which joined in parallel with 3 Ω in branch CH gives 2 ft.

⇒ Finally, \(R_{\mathrm{AB}}=3 \Omega+2 \Omega=5 \Omega\).

Question 34. In the given circuit, what should be the value of R so that no current flows through it?

1. 9
2. 12
3. 18
4. Any value

Answer: 4. Any value

The given network of resistors constitutes a balanced Wheatstone bridge since the ratio \(\frac{3 \Omega}{6 \Omega}=\frac{4 \Omega}{8 \Omega}=\frac{1}{2}\).

Hence, no current will flow through R for any value of R.

Alternative method

The voltage of 5 V of the source is divided in the upper and lower branches in the same proportion; so potential difference, \(V_c-V_d=0\). Hence, current through R is zero for any value of R.

Question 35. The equivalent resistance for the given network of resistors between terminals a and b are R1 and R2 respectively. The ratio R1/R2 is

1. \(\frac{4}{3}\)
2. \(\frac{3}{16}\)
3. \(\frac{2}{3}\)
4. \(\frac{1}{2}\)

Answer: 2. \(\frac{3}{16}\)

In the first case, one end of all resistors is connected at a common point a, and the other ends to a common point b, so they are in parallel,

Hence, \(R_1=\frac{r}{4}\).

In the second case, the network can be redrawn as shown in the given figure.

Here,\(R_2=r+\frac{r}{3}=\frac{4}{3} r\).

Hence, \(\frac{R_1}{R_2}=\frac{r / 4}{4 r / 3}=\frac{3}{16}\)

Question 36. In the given circuit, the potential difference between A and B is

1. zero
2. 5 V
3. 10 V
4. 15 V

Answer: 3. 10 V

The p-n junction is forward biased, hence its resistance is negligible and the circuit can be redrawn as shown.

Equivalent resistance, \(R=10 \mathrm{k} \Omega+5 \mathrm{k} \Omega=15 \mathrm{k} \Omega\).

Maincurrent, \(I=\frac{30 \mathrm{~V}}{15 \mathrm{k} \Omega}\).

This current is equally divided atA, so p.d. across AB is

∴ \(V_{\mathrm{AB}}=\left(\frac{I}{2}\right)(10 \mathrm{k} \Omega)=\frac{(30 \mathrm{~V})(10 \mathrm{k} \Omega)}{2(15 \mathrm{k} \Omega)}=10 \mathrm{~V}\).

Question 37. For the given circuit, the value of the current I is

1. 10 A
2. 5 A
3. 2.5 A
4. 20 A

Answer: 1. 10 A

The four branches of the resistors7 network have equal resistance (=5 Ω), so they satisfy the balance condition of the Wheatstone bridge. The cross-connected 5-Ω resistor is ineffective, so the equivalent circuit can be redrawn as shown.

For a parallel combination of resistors,

⇒ \(\frac{1}{R}=\frac{1}{10}+\frac{1}{10}+\frac{1}{5}=\frac{2}{5}\)

⇒ \(R=\frac{5}{2} \Omega\)

Hence, \(I=\frac{25 \mathrm{~V}}{5 / 2 \Omega}=10 \mathrm{~A}\).

Question 38. Across a metallic conductor of a nonuniform cross-section, a constant potential difference is applied. The quantity which remains constant along the conductor is

1. Current density
2. Electric field
3. Travelocity
4. Current

Answer: 4. Current density

In a current-carrying metallic conductor, there is no accumulation of charge anywhere. So, the rate of flow of charge (ΔQ/Δt), or the current, remains constant Drift speed, current density, and electric field vary with change in cross-sectional area.

“electricity test “

Question 39. A, B, and C are voltmeters of resistances R, 1.5R, and . 3R respectively as shown in the figure. x When some potential difference is maintained between X and Y, the voltmeter readings are VA, VB, and Vc respectively. Then,

1. \(V_{\mathrm{A}}=V_{\mathrm{B}}=V_{\mathrm{C}}\)
2. \(V_{\mathrm{A}} \neq V_{\mathrm{B}}=V_{\mathrm{C}}\)
3. \(V_{\mathrm{A}}=V_{\mathrm{B}} \neq V_{\mathrm{C}}\)
4. \(V_{\mathrm{A}} \neq V_{\mathrm{B}} \neq V_{\mathrm{C}}\)

Answer: 1. \(V_{\mathrm{A}}=V_{\mathrm{B}}=V_{\mathrm{C}}\)

The equivalent resistance of B and C is

⇒ \(R_{e q}=\frac{(1.5 R)(3 R)}{(1.5 R)+(3 R)}=\frac{(1.5)(3 R)}{(4.5)}=R \Omega\)

If l is the current through A then

⇒ \(V_{\mathrm{X}}-V_{\mathrm{a}}=V_{\mathrm{A}}=I R, V_{\mathrm{a}}-V_{\mathrm{b}}=V_{\mathrm{B}}=V_{\mathrm{C}}=I R_{\mathrm{eq}}=I R\)

∴ Thus, \(V_A=V_B=V_C\)

Question 40. The reading of the voltmeter in the given circuit is

1. 2.25 V
2. 4.25 V
3. 2.75 V
4. 6.25 V

Answer: 1. 2.25 V

Equivalent resistance in the given circuit is

⇒ \(R+R_{\text {eq }}=40 \Omega+\frac{(60 \Omega)(40 \Omega)}{60 \Omega+40 \Omega}=40 \Omega+24 \Omega=64 \Omega\)

Current through the cell is

⇒ \(I=\frac{6 \mathrm{~V}}{64 \Omega}=\frac{3}{32} \mathrm{~A}\)

Hence, the reading of the voltmeter is

⇒ \(V=I\left(R_{\text {eq }}\right)=\left(\frac{3}{32} \mathrm{~A}\right)(24 \Omega)=2.25 \mathrm{~V}\)

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Question 41. The internal resistance of a cell of emf 2 V is 0.1 Ω. It is connected to a resistance of 3.9 Ω. The voltage across the cell (terminal voltage) will be

1. 1.90 V
2. 1.95 V
3. 0.5 V
4. 2 V

Answer: 2. 1.95 V

Given that \(\varepsilon=2 \mathrm{~V}, r=0.1 \Omega, R=3.9 \Omega\)

Current through the circuit is \(I=\frac{\varepsilon}{R+r}=\frac{2 \mathrm{~V}}{4 \Omega}=0.5 \mathrm{~A}\).

∴ The terminal voltage across the cell is

⇒ \(V=\varepsilon-I r=2 \mathrm{~V}-(0.5 \mathrm{~A})(0.1 \Omega)\)

= 2 V-0.05 V =1.95 V.

Question 42. In the given circuit, the current flowing through the 25 V cell is

1. 7.2 A
2. 14.2 A
3. 10 A
4. 12 A

Answer: 4. 12 A

The current through each branch is shown in the figure. Applying Kirchhoff’s voltage loop rule, we have for each closed loop formed with a 25 V cell,

-30+11 I_i -25 = 0;

20+5 I₂ -25 = 0;

-5 +10 I₃ -25 = 0;

10 + 5 I₄ -25 = 0.

⇒ \(I_1=\frac{55}{11} \mathrm{~A}=5 \mathrm{~A}, I_2=1 \mathrm{~A}, I_3=3 \mathrm{~A}, I_4=3 \mathrm{~A}\)

Hence, current through the 25-V cell is

⇒ \(I=I_1+I_2+I_3+I_4=12 \mathrm{~A}\)

Question 43. A meter bridge is set up to determine resistance X using a standard 10 Ω resistor. The galvanometer shows a null point when the tapping key is at the 52 cm mark. The end-corrections are 1 cm and A 2 cm respectively for the ends A and B. The determined value of X is

1. 10.2
2. 10.8
3. 10.6
4. 11.1

Answer: 3. 10.6

From the balance condition,

⇒ \(\frac{X}{10 \Omega}=\frac{(52+1) \mathrm{cm}}{(48+2) \mathrm{cm}}=\frac{53}{50}\)

Unknown resistance is

⇒ \(X=\frac{53}{50} \times 10 \Omega=10.6 \Omega\).

Question 44. When a current is passed through a conductor, a 5°C rise in temperature is observed. If the strength of the current is made three times, the rise in temperature will be

1. 5°C
2. 15 °C
3. 45 °C
4. 20 °C

Answer: 3. 45 °C

Amount of heat produced, \(H=I^2 R t\)

⇒ \(\frac{H_1}{H_2}=\frac{I_1^2}{I_2^2}\)

⇒ \(\frac{H_1}{H_2}=\frac{m c \theta_1}{m c \theta_2}=\frac{I_1^2}{9 I_1^2}=\frac{1}{9} \quad\left(\text { given } I_2=3 I_1\right)\)

Hence, \(\theta_2=9 \theta_1=9\left(5^{\circ} \mathrm{C}\right)=45^{\circ} \mathrm{C}\)

Question 45. The amount of heat developed by a 100-W bulb in 1 minute will be

1. 100 J
2. 1000 J
3. 600 J
4. 6000 J

Answer: 4. 6000 J

Heat energy developed is

H = power x time = (100 W)(l m)

⇒ \(\left(100 \mathrm{~J} \mathrm{~s}^{-1}\right)(60 \mathrm{~s})=6000 \mathrm{~J}\)

“electricity test “

Question 46. A bulb of 25 W, 200 V’ and another bulb of 100 W, 200 V’ are connected in series with a supply line of 220 V. Then

1. Both the bulbs will glow with the same brightness
2. Both the bulbs will get fused
3. The 25-W bulb will glow more brightly
4. The 100-W bulb will glow more brightly

Answer: 3. The 25-W bulb will glow more brightly

Since watt = volt x ampere, so

⇒ \(\text { watt }=V(I)=V\left(\frac{V}{R}\right)=\frac{V^2}{R}\)

⇒ \(R=\frac{V^2}{\text { watt }}\)

Hence, the resistance of the 25-watt bulb will be more than that of the 100-watt bulb. Since the bulbs are in series, the same current flows through them.

Hence, heat \(\left(H=I^2 R t\right)\) will be more in the 25-W bulb and it will glow more brightly.

Question 47. A uniform wire consumes power P when connected across a given potential difference V. If it is cut into two equal halves and joined in parallel with the same potential difference V, it will consume power

1. P
2. 2P
3. 4P
4. \(\frac{P}{4}\)

Answer: 3. 4P

Power (in watt) \(P=V I=\frac{V^2}{R}\)

When the wire is cut and joined the equivalent resistance is \(R^{\prime}=\frac{R}{4}\)

∴ \(P^{\prime}=\frac{V^2}{R^{\prime}}=\frac{V^2}{R / 4}=4\left(\frac{V^2}{R}\right)=4 P\).

Question 48. An electric bulb marked ’40 W, 200 V’ is connected to a circuit of supply voltage 100 V. The power consumed will be

1. 100 W
2. 20 W
3. 10 W
4. 40 W

Answer: 3. 10 W

Resistance of the bulb, \(R=\frac{V^2}{P}\)

Here, V = 200 V, P = 40 W, hence \(R=\frac{(200 \mathrm{~V})^2}{(40 \mathrm{~W})}=1000 \Omega\)

When connected across a 100-V supply, the power consumed is

⇒  \(P=\frac{V^2}{R}=\frac{(100 \mathrm{~V})^2}{(1000 \Omega)}=10 \mathrm{~W}\)

Question 49. The potential difference \(\left(V_{\mathrm{A}}-V_{\mathrm{B}}\right)\) between the points A and B as shown in the figure is

image

1. -3 V
2. +3 V
3. +6 V
4. +9 V

Answer: 4. +9 V

Consider two points a and b across the cell.

Hence, \(V_{\mathrm{A}}-V_{\mathrm{B}}=\left(V_{\mathrm{A}}-V_{\mathrm{a}}\right)+\left(V_{\mathrm{a}}-V_{\mathrm{b}}\right)+\left(V_{\mathrm{b}}-V_{\mathrm{B}}\right)\)

⇒ \((2 A)(2 \Omega)+(3 V)+(2 A)(1 \Omega)=4 V+3 V+2 V=9 V\)

Question 50. Two cells, one of emf 18 V and internal resistance 2 and the other of emf 12 V and internal resistance 1 ft, are connected as shown. The reading of the voltmeter will be

1. 15 V
2. 30 V
3. 14 V
4. 18 V

Answer: 3. 14 V

The given circuit is redrawn as shown in the figure. Let I be the current in the closed loop consisting of two cells only.

Applying the loop rule,

12 V +1(1 Ω)+(2 Ω)1-18 V =0

⇒ 1 = 2 A.

Voltmeter reading will be the p.d. across a and b, which is

⇒ \(V_{\mathrm{a}}-V_{\mathrm{b}}=(18 \mathrm{~V})-(2 \Omega)(2 \mathrm{~A})=18 \mathrm{~V}-4 \mathrm{~V}=14 \mathrm{~V}\)

Alternative method

The equivalent emf of the cells is

⇒ \(\varepsilon=\frac{e_1 r_2+e_2 r_1}{r_1+r_2}=\frac{(18 \mathrm{~V})(1 \Omega)+(12 \mathrm{~V})(2 \Omega)}{2 \Omega+1 \Omega}\)

⇒ \(\frac{(18+24) V \Omega}{3 \Omega}=\frac{42}{3} V=14 \mathrm{~V}\).

Question 51. The current through the cell in the circuit is

1. 1 A
2. \(\frac{2}{3} \mathrm{~A}\)
3. \(\frac{2}{9} \mathrm{~A}\)
4. \(\frac{1}{8} \mathrm{~A}\)

Answer: 1. 1 A

The circuit is redrawn as shown. Equivalent resistance across the cell is R, where \(\frac{1}{R}=\frac{1}{3}+\frac{1}{6}=\frac{2+1}{6}=\frac{1}{2} \Rightarrow R=2\)

Current through the cell is

⇒ \(I=\frac{\varepsilon}{R}=\frac{2 \mathrm{~V}}{2 \Omega}=1 \mathrm{~A}\).

Question 52. From the graph of current I and voltage V as shown here, identify the region corresponding to negative resistance.

1. DE
2. CD
3. BC
4. AB

Answer: 2. CD

Resistance is. measured by \(\frac{\Delta V}{\Delta I}\) which is the slope of the V-I graph.

When the slope is negative, resistance is negative, hence the region where R is negative is CD.

Question 53. In a Wheatstone bridge, the resistance of each of the four arms is 10 Q. If the resistance of the galvanometer is also 10 £2 then the effective resistance of the bridge will be

1. 10
2. 5
3. 20
4. 40

Answer: 1. 10

When each of the four arms has the same resistance of 10 Ω, the bridge is balanced, and no current flows through the galvanometer. Hence, the effective resistance is

⇒ \(R=\frac{(20 \Omega) \times(20 \Omega)}{20 \Omega+20 \Omega}=10 \Omega\)

Question 54. A cell has an emf of 1.5 V. When connected across an external resistance of 2 Ω, the terminal voltage falls to 1.0 V. The internal resistance of the cell is

1. 2
2. 1.5
3. 1.0
4. 0.5

Answer: 3. 1.0

Given that ε =1.5 V, R = 2 Ω, V =1.0 V.

Terminal voltage, V = ε -Ir. →(1)

But V = IR => \(I=\frac{\dot{V}}{R}=\frac{1 \mathrm{~V}}{2 \Omega}=0.5 \mathrm{~A}\).

SubstitutingI = 0.5 A in (1),

⇒ \(1.0 \mathrm{~V}=1.5 \mathrm{~V}-0.5 r \Rightarrow r=\frac{0.5 \mathrm{~V}}{0.5 \mathrm{~A}}=1.0 \Omega\).

Question 55. Two batteries of emf 4 V and 8 V with internal resistances 1 and 2 are connected in a circuit with an external resistance R = 9 as shown in the figure. The current and potential differences between points P and Q are

1. \(\frac{1}{3} \mathrm{~A} \text { and } 3 \mathrm{~V}\)
2. \(\frac{1}{12} \mathrm{~A} \text { and } 12 \mathrm{~V}\)
3. \(\frac{1}{6} \mathrm{~A} \text { and } 4 \mathrm{~V}\)
4. \(\frac{1}{9} \mathrm{~A} \text { and } 9 \mathrm{~V}\)

Answer: 1. \(\frac{1}{3} \mathrm{~A} \text { and } 3 \mathrm{~V}\)

Net resistance, R = l Ω + 2 Ω + 9 Ω = 12 Ω.

Net voltage, V =8 V- 4 V = 4 V.

Current in the direction Q to P is

⇒ \(I=\frac{V}{R}=\frac{4 \mathrm{~V}}{12 \Omega}=\frac{1}{3} \mathrm{~A}\)

Now, the potential difference between P and Q is

⇒ \(V_{\mathrm{P}}-V_{\mathrm{Q}}=\left(V_{\mathrm{P}}-V_{\mathrm{a}}\right)+\left(V_{\mathrm{a}}-V_{\mathrm{b}}\right)+\left(V_{\mathrm{b}}-V_{\mathrm{c}}\right)+\left(V_{\mathrm{c}}-V_{\mathrm{Q}}\right)\)

⇒ \(-\left(\frac{1}{3} A\right)(1 \Omega)+(-4 V)+\left(-\frac{1}{3} A\right)(2 \Omega)+8 V\)

⇒ \(-\frac{1}{3} V-4 V-\frac{2}{3} V+8 V=3 V\).

Question 56. The resistivity of the material of a potentiometer wire is 10^-7m and its area of cross section is 10^-6 m². When a current I = 0.1 A flows through the wire, its potential gradient is

1. \(10^{-2} \mathrm{Vm}^{-1}\)
2. \(10^{-4} \mathrm{~V} \mathrm{~m}^{-1}\)
3. \(10 \mathrm{~V} \mathrm{~m}^{-1}\)
4. \(0.1 \mathrm{~V} \mathrm{~m}^{-1}\)

Answer: 1. \(10^{-2} \mathrm{Vm}^{-1}\)

Resistance, \(R=\rho \frac{l}{A}\)

Hence, the potential difference across the ends of the wire is

⇒ \(V=I R=I\left(\rho \frac{l}{A}\right)\)

Potential gradient is

⇒ \(\frac{V}{l}=\frac{I \rho}{A}=\frac{(0.1 \mathrm{~A})\left(10^{-7} \Omega \mathrm{m}\right)}{10^{-6} \mathrm{~m}^2}=10^{-2} \mathrm{~V} \mathrm{~m}^{-1}\)

Question 57. The resistances of the four arms P, Q, R, and S of a Wheatstone bridge are 10 Ω, 30 Ω, 30, and 90 respectively. The emf and internal resistance of the cell are 7 V and 5respectively. If the galvanometer resistance is 50 Ω, the current drÿwn from the cell will be

1. 1.0 A
2. 0.2 A
3. 0.1 A
4. 2.0 A

Answer: 2. 0.2 A

With the given values of the four arms of the bridge, the balance condition \(\frac{P}{Q}=\frac{R}{S}\) is. satisfied.

So, the current through the galvanometer is zero for all values of G.

The equivalent resistance of the bridge is

⇒ \(R=\frac{(40 \Omega)(120 \Omega)}{40 \Omega+120 \Omega}=30 \Omega\)

Net resistance in the circuit is

= R + r = 30 Ω + 5 Ω = 35 Ω.

Hence, the current drawn from the cell is

⇒ \(I=\frac{\varepsilon}{R+r}=\frac{7 \mathrm{~V}}{35 \Omega}=0.2 \mathrm{~A}\).

Question 58. The resistance for balance in the two arms of a meter bridge is 5 and R respectively. When the resistance R is shunted with an equal resistance, the balance point is at 1.6 l₁. The resistance R is

1. 10
2. 15
3. 25
4. 20

Answer: 2. 15

In the first case of balance,

⇒ \(\frac{5 \Omega}{R}=\frac{l_1}{100 \mathrm{~cm}-l_1}\) → (1)

When R is shunted by an equal resistance R, its equivalent resistance will be

⇒  \(\frac{R}{2}\).

For this new condition of balance,

⇒ \(\frac{5 \Omega}{R / 2}=\frac{1.6 l_1}{100 \mathrm{~cm}-1.6 l_1}\)  →(2)

Taking the ratio \(\frac{l}{2}\), we get,

⇒ \(\frac{1}{2}=\frac{l_1}{100 \mathrm{~cm}-l_1} \times \frac{100 \mathrm{~cm}-1.6 l_1}{1.6 l_1}\)

⇒ \(\frac{1.6}{2}=\frac{100 \mathrm{~cm}-1.6 l_1}{100 \mathrm{~cm}-l_1} \Rightarrow l_1=25 \mathrm{~cm}\)

∴ From (1) \(R=5 \Omega \frac{(100 \mathrm{~cm}-25 \mathrm{~cm})}{25 \mathrm{~cm}}=15 \Omega\)

Question 59. Three resistors P, Q, and R, each of 2 Ω, and an unknown resistance S form the four arms of a Wheatstone bridge circuit. When a resistance of 6 is connected in parallel to S, the bridge gets balanced. The value of S is

1. 2
2. 3
3. 1
4. 6

Answer: 2. 3

For balance, the fourth arm must be of 2.

Hence, \(\frac{1}{S}+\frac{1}{6 \Omega}=\frac{1}{2 \Omega}\)

⇒ S = 3Ω.

Question 60. In the circuit shown, if a conducting wire is connected between the points A and B, the current in this wire will

1. Flow from A to B
2. Flow from B to A
3. Be zero
4. Flow in the direction which will be decided by the value of V

Answer: 2. Flow from B to A

Let the current through the upper and lower branches be I₁ and I2 respectively.

Hence,

VP – VQ = I₁ (4+4) =8 I₁ ,

VP – VQ = I₂ (1 + 3) = 4 I₂.

So, 4I2 = 8 I₁ ⇒ I₂  = 2 I_{1  →} (1)

Now, VP- VA = 4 I₁ ⇒ VA = Vp- 4 I₁. → (2)

And Vp-VB = 1 ( I₂ ) = 2 I₁ ⇒ VB = VP – 2 I₁. → (3)

From (2) and (3), VB>VA.

Hence, on joining A and B by a conducting wire, current will flow from B to A

Question 61. A cell can be balanced against 110 cm and 100 cm of a potentiometer wire respectively, with and without being short-circuited through a resistance of 10 Ω. Its internal resistance is

1. 1.0
2. 0.5
3. 2.0
4. 0.2

Answer: 1. 1.0

With the switch open, balance l =110 cm is obtained across emf £ of
the cell. Thus, ε=kl.

When S is closed, the cell is shunted by a 10-Ω resistance so that the terminal voltage V is balanced at l’=100 cm.

Thus, V = kl’.

So, \(\frac{\varepsilon}{V}=\frac{l}{l^{\prime}}=\frac{110 \mathrm{~cm}}{100 \mathrm{~cm}}\)  → (1)

But \(\varepsilon=I(R+r) \text { and } V=I R\)

∴ \(\frac{\varepsilon}{V}=\frac{R+r}{R}=\frac{10 \Omega+r}{10 \Omega}\)  → (2)

Equating (1) and (2),

⇒ \(\frac{10 \Omega+r}{10 \Omega}=\frac{110}{100} \Rightarrow r=1 \Omega\)

Question 62. A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery used across the potentiometer wire has an emf of 2.0 V and negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance R, connected across the given cell, has values of

• infinity and
• 9.5 ft, the balancing lengths on the potentiometer wire are found to be 3 m and 2.85 m respectively. The value of the internal resistance of the cell is

1. 0.25
2. 0.95
3. 0.5
4. 0.75

Answer: 3. 0.5

When the cell is without any parallel resistance, the balance length is at l =3 m and when shunted with R = 9.5 Ω, the balance length l’ = 2.85 m.

We know that internal resistance is

⇒ \(r=\left(\frac{l}{l^{\prime}}-1\right) R=\left(\frac{3 \mathrm{~m}}{2.85 \mathrm{~m}}-1\right) 9.5 \Omega=\frac{0.15 \mathrm{~m}}{2.85 \mathrm{~m}} \times 9.5 \Omega=0.5 \Omega\)

Question 63. A potentiometer wire of length l and resistance r is connected in series with a battery of emf80 and a resistance rx. A cell of unknown emf8 is balanced against the length of the potentiometer wire. The emf if will be given by

1. \(\frac{L \varepsilon_0 r}{l r_1}\)
2. \(\frac{\varepsilon_0 r}{\left(r+r_1\right)} \frac{l}{L}\)
3. \(\frac{\varepsilon_0 l}{L}\)
4. \(\frac{L \mathcal{E}_0 r}{\left(r+r_1\right) l}\)

Answer: 2. \(\frac{\varepsilon_0 r}{\left(r+r_1\right)} \frac{l}{L}\)

Current through the potentiometer wire by the driving cell ε₀ is

⇒ \(I=\frac{\varepsilon_0}{r+r_1}\)

The potential drop across the ends of the potentiometer wire is

⇒ \(V=I r=\frac{\varepsilon_0 r}{r+r_1}\)

Hence, potential gradient = \(\frac{V}{L}=\frac{\mathcal{E}_0 r}{\left(r+r_1\right) L}\)

The balance point is obtained at distance Z, so the emf of the cell is

⇒ \(\varepsilon=\frac{V l}{L}=\frac{\varepsilon_0 r l}{\left(r+r_1\right) L}\)

Question 64. A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite directions. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of emf is

1. 5:4
2. 3:2
3. 3:4
4. 5 :1

Answer: 2. 3:2

When the cells are in supporting order,

⇒ \(\left(\varepsilon_1+\varepsilon_2\right)=k l_1=k(50 \mathrm{~cm})\)

When the cells are in reverse order

⇒ \(\begin{aligned}
\left(\varepsilon_1-\varepsilon_2\right)=k l_2=k(10 \mathrm{~cm})\end{aligned} \)

⇒ \(\begin{aligned}\frac{\varepsilon_1+\varepsilon_2}{\varepsilon_1-\varepsilon_2}=5, \text { hence } \frac{\varepsilon_1}{\varepsilon_2}=\frac{3}{2}
\end{aligned}\).

Question 65. A circuit contains an ammeter, a battery of 30 V, and a resistance of 40.8 all connected in series. If the ammeter has a coil of resistance 480 and a shunt of 20 then the reading in the ammeter will be

1. 0.5 A
2. 1 A
3. 2 A
4. 0.25 A

Answer: 1. 0.5 A

Equivalent resistance across ab is

⇒ \(R_{\mathrm{eq}}=\frac{G S}{G+S}=\frac{480 \times 20}{500} \Omega=19.2 \Omega\)

Total resistance in the circuit,

R = 40.8 Ω +19.2 Ω = 60 Ω.

Main currents

∴ \(I=\frac{30 \mathrm{~V}}{60 \Omega}=0.5 \mathrm{~A}\)

The current measured by the ammeter is the current flowing through the circuit, so the ammeter reading will be 0.5 A.

Question 66. A millivoltmeter of 25 mV range is to be converted into an ammeter of 25 A range. The value (in ohm) of the required shunt will be

1. 0.001
2. 0.01
3. 1
4. 0.05

Answer: 1. 0.001

For full-scale deflection, the current through the meter,

⇒ \(I_g=\frac{V}{G}=\left(\frac{25 \mathrm{mV}}{G}\right) A\) where G = resistance of the meter.

In order to increase the range to 25 A when used as an ammeter, the value of the required shunt is

⇒ \(S=\frac{I_{\mathrm{g}} \mathrm{G}}{I-I_{\mathrm{g}}} \approx \frac{I_{\mathrm{g}} \mathrm{G}}{I}\) [∵ Ig<<I]

⇒ \(\frac{25 \times 10^{-3} \mathrm{~V}}{25 \mathrm{~A}}=0.001 \Omega\).

Question 67. A galvanometer which has a coil of resistance 100 Ω, gives a full-scale deflection for 30 mA current. If it is to work as a voltmeter in the 30-V range, the resistance required to be added will be

1. 900
2. 1800
3. 500
4. 1000

Answer: 1. 900

Given that G =100 Ω, Ig = 30 mA.

The voltage across the galvanometer for full-scale deflection is

⇒ \(V=I_{\mathrm{g}} \cdot G=(30 \mathrm{~mA})(100 \Omega)=3 \mathrm{~V}\).

The factor n by which its measuring range as a voltmeter is increased is

⇒ \(n=\frac{30 \mathrm{~V}}{3 \mathrm{~V}}=10\)

The resistance required to be added in series with the meter will be

⇒ \(R=(n-1) G=(10-1)(100 \Omega)=900 \Omega\).

Question 68. The resistance of an ammeter is 13 Ω and its scale is graduated for a current up to 100 A. After an additional shunt has been connected to this ammeter, it becomes possible to measure currents up to 750 A by this meter. The value of the resistance of this shunt is

1. 20
2. 2
3. 0.2
4. 2k

Answer: 2. 2

Given that G =13 Ω, Ig =100 A.

p.d. across the ammeter for full-scale deflection is

V = Ig x G = (100 A)(13 Ω) =1300 V

When the measuring range is increased (I = 750 A), let the shunt resistance be S. Equivalent resistance of the parallel combination of G and S is \(\frac{G S}{G+S}\) and potential difference across this = \(I\left(\frac{G S}{G+S}\right)\)

For full-scale deflection, the p.d. across this combination must be V =1300 V.

⇒ \(1300 \mathrm{~V}=(750 \mathrm{~A}) \frac{13 \Omega \times S}{13 \Omega+S}\).

Hence, shunt=S = 2Ω.

Question 69. A battery is charged at a potential of 15 V for 8 h when the current flowing is 10 A. The battery during discharge supplies a current of 5 A for 15 h. The mean terminal voltage during discharge is 14 V. The watt-hour efficiency of the battery is

1. 82.5%
2. 87.5%
3. 80%
4. 90%

Answer: 2. 87.5%

During the charging process, the input energy is

E₁ = coulomb x volt = (It)V J

= (10 A)(8 h)(15 V) =1200 W h.

During discharge, the energy output is

E₀ = (5 A)(15 h)(14 V) =1050 W h

⇒ Efficiency = \(\eta=\frac{\text { output }}{\text { input }} \times 100 \%=\frac{1050}{1200} \times 100 \%=87.5 \%\)

Question 70. Two cells having the same emf are connected in series through an external resistance R. The cells have internal resistances r₁ and r₂ (r₁ > r₂) respectively. When the circuit is closed the potential difference across the first cell is zero. The value of R is

1. \(r_1-r_2\)
2. \(\frac{r_1-r_2}{2}\)
3. \(\frac{r_1+r_2}{2}\)
4. \(r_1+r_2\)

Answer: 1. \(r_1-r_2\)

Current through the circuit is

⇒ \(I=\frac{2 e}{R+r_1+r_2}\)

The potential difference across the first cell is

⇒ \(V_{\mathrm{a}}-V_{\mathrm{b}}=e-I r_1=0 \text { (given) } \Rightarrow e=I r_1\)

⇒ \(e=\frac{2 e r_1}{R+r_1+r_2}\)

⇒ \(R+r_1+r_2=2 r_1 \Rightarrow R=r_1-r_2\)

Question 71. Refer to the electrical circuit shown in the figure. Which of the following equations is the correct representation for the given circuit?

1. \(\varepsilon_1-\left(I_1+I_2\right) R-I_1 r_1=0\)
2. \(\varepsilon_2-I_2 r_2-\varepsilon_1-I_1 r_1=0\)
3. \(-\mathcal{E}_2-\left(I_1+I_2\right) R+I_2 r_2=0\)
4. \(\varepsilon_1-\left(I_1+I_2\right) R+I_1 r_1=0\)

Answer: 1. \(\varepsilon_1-\left(I_1+I_2\right) R-I_1 r_1=0\)

According to Kirchhoff’s junction rule, the current through the resistance R is ( I₁ + I₂ ). Applying Kirchhoff’s loop rule for the upper closed loop,

⇒ \(\begin{aligned}
R\left(I_1+I_2\right)+I_1 r_1-\varepsilon_1=0\end{aligned}\)

⇒ \(\begin{aligned}\varepsilon_1-\left(I_1+I_2\right) R-I_1 r_1=0
\end{aligned}\)

Question 72. In the given circuit, the cells A and B have negligible resistances. For VA =12 V, Rx = 500 Q, and R =100 Q, the galvanometer G shows null deflection. The value of VB is

1. 4 V
2. 2 V
3. 12 V
4. 6 V

Answer: 2. 2 V

For null deflection in the galvanometer, a and b must be at the same potential, and

⇒ \(\begin{aligned}
V_{\mathrm{B}}  =V_{\mathrm{a}}-V_{\mathrm{c}}=I(100 \Omega)\end{aligned} \)

⇒ \(\begin{aligned}\frac{12 \mathrm{~V} \times 100 \Omega}{600 \Omega}=2 \mathrm{~V}
\end{aligned}\).

Question 73. Six identical bulbs are connected as shown in the figure with a DC source of emf E and zero internal resistance. The ratio of power consumed by the bulbs when

1. All are glowing and
2. Two from section A and one from section B are glowing and will be

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

Answer: 2. 9:4

Equivalent watt in series is given by \(\frac{1}{W_S}=\frac{1}{W_1}+\frac{1}{W_2}+\frac{1}{W_3}+\ldots\),and in parallel \(W_P=W_1+W_2+W_3+\ldots\)

If the voltage of each bulb is W then in case (1)

⇒ \(W_1=\frac{W}{2}+\frac{W}{2}+\frac{W}{2}=\frac{3}{2} W\)

and in case (2),

⇒ \(\frac{1}{W_2}=\frac{1}{2 W}+\frac{1}{W}=\frac{3}{2 W} \Rightarrow W_2=\frac{2 W}{3}\)

∴ Required ratio, \(\frac{W_1}{W_2}=\frac{3 W / 2}{2 W / 3}=\frac{9}{4}\).

Question 74. Which of the following acts as a circuit-protecting device?

1. Conductor
2. Inductor
3. Switch
4. Fuse

Answer: 4. Fuse

A fuse is an electrical safety device that operates to provide over-current protection to an electrical circuit. It has high resistance and low melting point so that when a high current flows through the circuit, the fuse wire melts and the current stops flowing.

Question 75. In the circuit shown below, the readings of voltmeters and ammeters will be

1. \(V_2>V_1 \text { and } I_1=I_2\)
2. \(V_1=V_2 \text { and } I_1>I_2\)
3. \(V_1=V_2 \text { and } I_1=I_2\)
4. \(V_2>V_1 \text { and } I_1>I_2\)

Answer: 3. \(V_1=V_2 \text { and } I_1=I_2\)

An ideal voltmeter has infinite resistance whereas an ideal ammeter has effectively zero resistance. Under this ideal condition,

⇒ \(I_1=\frac{10 \mathrm{~V}}{10 \Omega}=1 \mathrm{~A}, V_1=(1 \mathrm{~A})(10 \Omega)=10 \mathrm{~V}\),

⇒ and \(I_2=\frac{10 \mathrm{~V}}{10 \Omega}=1 \mathrm{~A}, V_2=(1 \mathrm{~A})(10 \Omega)=10 \mathrm{~V}\).

Thus, V₁ = V₂ and I₁ = I₂

Question 76. The resistive network shown is connected to a DC source of 16 V. The power consumed by the network is 4 W. The value of R is

1. l
2. 16
3. 8
4. 4

Answer: 3. 8

The equivalent resistance of the given combination of resistors is

R_net = 2R+ 2R + 4R = SR.

Power consumed = \(4 \mathrm{~W}=\frac{V^2}{R_{\text {net }}}\)

∴ \(4=\frac{(16)^2}{8 R}, \text { hence } R=\frac{16 \times 16}{4 \times 8} \Omega=8 \Omega\)

Question 77. In the given circuit, an ideal voltmeter connected across the 10-Q resistor reads 2 V. The internal resistance r of each cell is

1. 0.5
2. 0
3. 1.5
4. 1

Answer: 1. 0.5

The potential difference across the 10-Ω and 15-Ω resistors is 2 V, so the current is

⇒ \(i_1=\frac{V}{R_1}=\frac{2}{10} \mathrm{~A}\)

Similarly, \(i_2=\frac{V}{R_2}=\frac{2}{15} \mathrm{~A}\).

∴ total current \(I=i_1+i_2=\left(\frac{2}{10}+\frac{2}{15}\right) \mathrm{A}=\frac{1}{3} \mathrm{~A}\)

Applying Kirchhoff’s loop rule,

(10 Ω)i₁ + (2 Ω)I + (2r)I = 3 V

⇒ \(\left(10 \times \frac{2}{10}\right) V+\left(2 \times \frac{1}{3}\right) V+(2 r)\left(\frac{1}{3} A\right)=3 V\)

⇒ \(\frac{2}{3} r=\left(3-2-\frac{2}{3}\right) \Omega\)

∴ internal resistance = r = 0.5 Ω.

Question 78. An ideal battery of 4 V and a resistance R are connected in series in the primary circuit of a potentiometer of length 1 m and resistance 5 Ω. The value of R to give a potential difference of 5 mV across 10 cm of the potentiometer wire is

1. 480
2. 490
3. 495
4. 395

Answer: 4. 395

The total resistance in the main circuit is (R +5 Ω) and current through the l primary circuit is

⇒ \(I=\frac{4 V}{R+5 \Omega}\)

Potential gradient is

⇒ \(\frac{V}{L}=\frac{I \times 5 \Omega}{100} \mathrm{~V} \mathrm{~cm}^{-1}\)

Hence, p.d. across the 10-cm wire will be

⇒ \(\left(\frac{I}{20} \times 10\right) \mathrm{V}=5 \mathrm{mV}=5 \times 10^{-3} \mathrm{~V}\)

From (1),

⇒ \(\frac{1}{2}\left(\frac{4 V}{R+5 \Omega}\right)=5 \times 10^{-3} V\)

⇒ \(\frac{2}{R+5 \Omega}=\frac{5}{1000} \Rightarrow R=395 \Omega\)

Question 79. For the circuit shown with R₁=1 Ω, R₂ =2 Ω, ε₁=2 V, and ε₂ = 4 V, the potential difference between points a and b is

1. 2.7 V
2. 3.3 V
3. 2.3 V
4. 3.7 V

Answer: 1. 2.7 V

From the Kirchhoff loop rule for the closed path ABCDA,

1(x)+2 +(1)x-y-2-y = 0

⇒ 2x-2y = 0 ⇒ x = y  → (1)

For the closed-loop AbaDA,

x-4+2(x + y)+x+2 = 0

⇒ 4x + 2y = 2

⇒ 2x+y =1 → . (2)

From (1) and (2),

⇒ \(x=\frac{1}{3}=y\)

⇒ \(V_{\mathrm{a}}-V_{\mathrm{b}}=-2(x+y)+4=4 \mathrm{~V}-2\left(\frac{2}{3}\right) \mathrm{V}=\frac{8}{3} \mathrm{~V}=2.67 \mathrm{~V} \approx 2.7 \mathrm{~V}\)

Question 80. A current of 2 mA was passed through an unknown resistor which dissipated 4.4 W of power. When a supply of 11 V is connected across this resistor, the dissipated power will be

1. \(11 \times 10^{-4} \mathrm{~W}\)
2. \(11 \times 10^{-5} \mathrm{~W}\)
3. \(11 \times 10^{-6} \mathrm{~W}\)
4. \(11 \times 10^{-3} \mathrm{~W}\)

Answer: 2. \(11 \times 10^{-5} \mathrm{~W}\)

Power dissipation = 4.4 W = I2R

⇒ \(R=\frac{4.4 \mathrm{~W}}{\left(2 \times 10^{-3} \mathrm{~A}\right)^2} \Omega=11 \times 10^5 \Omega\)

With a supply voltage of 11 V, the power dissipated is

⇒ \(\frac{V^2}{R}=\frac{(11 \mathrm{~V})^2}{11 \times 10^5 \Omega}=11 \times 10^{-5} \mathrm{~W}\).

Question 81. The drift speed of electrons, when A of electric current flows through a copper wire of cross section 5 mm², is v. If the number density of electrons in copper is 9 x l0²⁸ m^-3, the value of v in mm s^-1 is close to

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

Answer: 1. 0.02

Current 1 through a conductor is given by I = Avdne, where I =1.5 A, rt = number density = 9 x1028 m-3, and A- cross-sectional area =5 mm2.

Drift speed is

⇒ \(v_{\mathrm{d}}=\frac{I}{A n e}\)

⇒ \(\frac{1.5 \mathrm{~A}}{\left(5 \times 10^{-6} \mathrm{~m}^2\right)\left(9 \times 10^{28} \mathrm{~m}^{-3}\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)}\)

⇒ \(\frac{1.5 \mathrm{~A}}{72 \times 10^3 \mathrm{C} \mathrm{m}^{-1}}=0.02 \times 10^{-3} \mathrm{~m} \mathrm{~s}^{-1}=0.02 \mathrm{~mm} \mathrm{~s}^{-1}\)

Question 82. A copper wire is stretched to make it 0.5% longer. The percentage change in its electrical resistance is

1. 2.0%
2. 1.0%
3. 0.5%
4. 2.5%

Answer: 2. 1.0%

Resistance = \(R=\rho \frac{l}{A}=\rho \frac{l^2}{V}\), where Z = length and V = volume.

⇒ \(\frac{\Delta R}{R}=2 \frac{\Delta l}{l}=2(0.5 \%)=1.0 \%\).

Question 83. When the switch S in the circuit shown is closed, the value of the current I will be

1. 4 A
2. 3 A
3. 2 A
4. 5 A

Answer: 4. 5 A

Let V be the potential at junction C.

We have, \(i_1=\frac{20-V}{2}, i_2=\frac{10-V}{4}\)

∴ i₁ + i₂ = i = \(\frac{V}{2}\)

⇒ \(\frac{20-V}{2}+\frac{10-V}{4}=\frac{V}{2}\) ⇒ V = 10 volts.

∴ Current = i \(\frac{V}{2 \Omega}=\frac{10 \mathrm{~V}}{2 \Omega}\) = 5 A.

Question 84. Two equal resistances when connected in series to a battery consume 60 W of electric power. If these resistances are connected in parallel combination to the same battery, the electric power consumed will be

1. 240 W
2. 120 W
3. 30 W
4. 60 W

Answer: 1. 240 W

In a series combination, equivalent wattage is

⇒ \(W_S=\frac{W_1 W_2}{W_1+W_2} \Rightarrow 60 \mathrm{~W}=\frac{W_1 \times W_2}{2 W_1}\)

The wattage of each resistor is

W₁ = 120 W

Hence, the consumption of power when connected in parallel is

WP = W₁ + W₂ = 2 W₁ = 2(120 W) = 240 W.

Question 85. The resistance of a galvanometer is 50and the maximum current of 0.002 A gives full-scale deflection. What resistance must be connected to it so as to convert it into an ammeter of range 0-0.5 A?

1. 0.5
2. 0.02
3. 0.2
4. 0.002

Answer: 3. 0.2

The initial measuring range is

I = 0.002 A and the final range is nl = 0.5 A.

∴ \(n=\frac{0.5 \mathrm{~A}}{0.002 \mathrm{~A}}=250\)

Required shunt is

⇒ \(S=\frac{G}{n-1}=\frac{50 \Omega}{249} \approx 0.2 \Omega\).

Question 86. A metal wire of resistance 3 is elongated to make it a uniform wire of double its previous length. This new wire is now bent and its ends are joined to make a circle. If two points P and Q on this circle make an angle of 60° at the center, the equivalent resistance across these two points will be

1. \(\frac{5}{2} \Omega\)
2. \(\frac{5}{3} \Omega\)
3. \(\frac{12}{5} \Omega\)
4. \(\frac{7}{2} \Omega\)

Answer: 2. \(\frac{5}{3} \Omega\)

Initial resistance, R =3 Ω.

When the length is doubled, its new resistance will become 4R =12 Ω.

Resistance of part \(P Q=\frac{1}{6}(12 \Omega)=2 \Omega\)

and that oflower part=12Ω- 2 Ω =10 Ω.

Equivalent resistance across PQ

∴ \(R_{\mathrm{eq}}=\frac{r_1 r_2}{r_1+r_2}=\frac{(2 \Omega)(10 \Omega)}{12 \Omega}=\frac{5}{3} \Omega\).

Question 87. A potentiometer wire AB having length L and resistance 12r is joined to a cell D of emf £ and internal resistance r. A cell C of emf \(\) and internal resistance 3r is connected to A as shown. The length AJ at which the galvanometer shows null deflection is

1. \(\frac{5 L}{12}\)
2. \(\frac{11 L}{12}\)
3. \(\frac{13 L}{24}\)
4. \(\frac{11 L}{24}\)

Answer: 3. \(\frac{13 L}{24}\)

The total resistance in the main circuit = r +12r =13r.

∴ current through the potentiometer wire = \(I=\frac{\varepsilon}{13 r} .\)

The potential drop across length L of the wire AB is

⇒ \(V=I(12 r)=\frac{\varepsilon}{13 r}(12 r)=\frac{12}{13} \varepsilon\)

Potential gradient = \(\frac{V}{L}=\frac{12}{13} \frac{\varepsilon}{L} \mathrm{~V} \mathrm{~m}^{-1}\).

When the galvanometer shows null deflection at length AJ = x, we have

⇒  \(\frac{\varepsilon}{2}=\frac{12}{13} \frac{\varepsilon}{L} x \Rightarrow x=\frac{13 L}{24}\)

Question 88. In the. given circuit, the cells have. zero internal resistance. The current (in ampere) passing through resistances JRj and R2 respectively are

1. 0.5,0
2. 1.0,2
3. 2, 2
4. 0,1.5

Answer: 1. 0.5,0

In the closed loop A,

-10 +(x + y) 20 = 0

⇒ \(x+y=\frac{1}{2}\)  → (1)

In the closed loop B,

(x + y)20 + 20y-10 = 0

⇒ 2(x+y)+2y =l

⇒ 2x+4y =l

⇒ \(x+2 y=\frac{1}{2}\) → (2)

Solving (1) and (2),

⇒ \(x=\frac{1}{2}\) A and y = 0.

∴ Current in R₁ is \((x+y)=\frac{1}{2}\) A and in R₂ is y = 0.

Question 89. The given circuit contains two ideal diodes, each with a forward resistance of 50 ft. If the battery voltage is 6 V, the current through the 100-12 resistance will be

1. 0.027 A
2. 0.036 A
3. 0.020 A
4. 0.030 A

Answer: 3. 0.020 A

In the given circuit, diode D2 is reverse-biased. Hence, nonconducting. D1 will be conducted with forward resistance of 50.

Thus, the total resistance is

R = 50 Ω+150 Ω +100 Ω = 300 Ω.

⇒ So, \(I=\frac{6 \mathrm{~V}}{300 \Omega}=0.02 \mathrm{~A}\)

Question 90. In the given circuit, the currents I₁=-03 A, I₄ = 0.8 A, and I₅ = 0.4 A are flowing as shown. The currents I₂, I₃ and I₆ are respectively

1. 1.1 A, 0.4 A, 0.4 A
2. 1.1 A, -0.4 A, 0.4 A
3. 0.4 A, 1.1 A, 0.4 A
4. 0.4 A, 0.4 A, 1.1 A.

Answer: 1. 1.1 A, 0.4 A, 0.4 A

Applying junction Me:

At R, I₂ +(-0.3 A) = 0.8 A

⇒ I₂ = 1.1 A.

At S, I₃ + 0.4 A = 0.8 A

⇒ I₃ = 0.4 A.

At P, I₆ = 0.4 A.

Thus, the values are 1.1 A, 0.4 A, and 0.4 A.

Question 91. In the figure shown, what is the current (in ampere) drawn by the battery?

1. \(\frac{7}{38}\)
2. \(\frac{20}{3}\)
3. \(\frac{9}{32}\)
4. \(\frac{13}{24}\)

Answer: 3. \(\frac{9}{32}\)

The given circuit can be redrawn in a simplified form as shown. The equivalent resistance of the parallel combination is

⇒ \(R_1=\frac{50 \times 10}{60} \Omega=\frac{25}{3} \Omega .\)

Net resistance in the circuit is

⇒ \(R=\frac{25}{3} \Omega+15 \Omega+30 \Omega=\frac{160}{3} \Omega\)

current delivered by the battery is

⇒ \(I=\frac{\varepsilon}{R}=\frac{15 \mathrm{~V}}{\frac{160}{3} \Omega}=\frac{9}{32} \mathrm{~A}\)

Question 92. The potential difference across A and B in the given circuit is

1. 1 V
2. 3 V
3. 2 V
4. 6 V

Answer: 3. 2 V

The current distribution in the different branches is shown in the diagram.

No current in 5 Ω and 10 Ω. Thus, V_A-V_B = V_a-V_b.

For the upper loop, x+y+1-2+y = 0 => x + 2y = 1A.

For the lower loop, -y+2-3+x=0 => x-y =1 A.

Solving, we get x=1 A, y =0.

Now for any branch across ab,

V_a-V_b = -(1 Ω)x +3V =-1V +3V = 2V.

Question 93. The Wheatstone bridge shown in the figure gets balanced when the carbon resistor has the color code orange, red, and brown. The resistors R₂ and R₄ are 80 Ω and 40 Ω respectively. Assuming that the color code for the carbon resistor gives accurate values, the color code for R₃ would be

1. Brown, blue, black
2. Brown, blue, brown
3. Grey, black, brown
4. Red, green, brown

Answer: 2. Brown, blue, brown

R₁ has the color code

Orange – 3

Red – 2

Brown – 1

⇒ \(R_1=32 \times 10^1=320 \Omega\) .

⇒ For balance \(\frac{R_1}{R_2}=\frac{R_3}{R_4}\)

⇒ \(\quad R_3=\frac{R_4}{R_2} R_1=\frac{40 \Omega}{80 \Omega} \times 320 \Omega=160 \Omega \text {. }\)

The corresponding color code is

Brown – 1

Blue – 6

Brown -10¹

⇒ 16 x 10¹ Ω = 160 Ω

Question 94. A carbon resistor has the color code resistor indicated in the figure. What is the value of the resistance?

1. 5.3 M ± 5%
2. 64 k±10%
3. 6.4 M ±5%
4. 64 k±10%

Answer: 4. 64 k±10%

From the given color code

Green – 5

Orange – 6

Yellow – 4

Golden – 5%

⇒ R = 53 x 10⁴ ± 5% = 530 kΩ ± 5% .

Question 95. A resistor is shown in the figure. Its value and tolerance are given respectively as

1. 270Ω and 5%
2. 27kΩ and 20%
3. 27kΩ and 10%
4. 270kΩ and 10%

Answer: 3. 27kΩ and 10%

The color rings are

Red – 2

Violet – 7

Orange – 3

Silver – 10%

∴ Resistance = R = 27 x 10³ Ω,10% = 27 kΩ,10%.

Question 96. A cell of internal resistance r drives current through an external resistance R. The power delivered by the cell to the external resistance will be maximum when

1. R = O.Olr
2. R = 2r
3. R = r
4. R =10r

Answer: 3. R = r

According to the maximum power output theorem/ the external resistance R must be equal to the internal resistance r of the voltage source.

Question 97. A moving-coil galvanometer has a resistance of 50 Ω and it indicates full deflection at 4-mA current. A voltmeter is made using this galvanometer and a 5-kΩ resistance. The maximum voltage that can be measured using this voltmeter will be close to

1. 10 V
2. 15 V
3. 40 V
4. 20 V

Answer: 4. 20 V

Given that G = 50 Ω, I_g = 4 mA.

While using this galvanometer ‘ as a voltmeter (with measuring range V), a resistance of 5 kΩ is connected in series so as to get the same current I_g (=4 x10^-3A) for full-scale deflection

∴ Hence, \(I_{\mathrm{g}}=\frac{V}{G+R}\)

⇒ V = (G + R)I_g = (50 Ω + 5000 Ω)(4 x10-3 A) = 20.2 V ≈ 20 V.

Question 98. A 200 resistor has a certain color code. If one replaces the red color with green in the code then the new resistance will be

1. 100
2. 200
3. 400
4. 500

Answer: 4. 500

For the 200-Ω resistor, the color code is red, black, and black. Red corresponds to 2 which is replaced by green (≡ 5) so the new resistance will be 500 Ω.

Question 99. A 2 W-carbon resistor is color-coded with green, black, red, and brown respectively. The maximum current that can be passed through the resistor is

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

Answer: 3. 20 mA

The color codes are

Green – 5

Black – 0

Red – 2

Brown – 1

R_min = 50 x 10²

⇒ \(I_{\max }=\sqrt{\frac{P}{R_{\min }}}=\sqrt{\frac{2 \mathrm{~W}}{50 \times 10^2 \Omega}}=2 \times 10^{-2} \mathrm{~A}=20 \mathrm{~mA}\)

Question 100. The given figure shows a cell of emf S =s 3 V and internal resistance r connected across an external resistance R. If power dissipation in R is 0.5 W and the terminal voltage across the cell is 2.5 V then the power dissipation in the internal resistance r is

1. 0.5 W
2. 0.1W
3. 0.3 W
4. 1.0 W

Answer: 2. 0.1W

Power dissipation in external resistance is

W = V x I ⇒ 0.5 W = (2.5V) I.

∴ \(\text { current }=I=\frac{0.5 \mathrm{~W}}{2.5 \mathrm{~V}}=\frac{1}{5} \mathrm{~A}\)

The potential drop across internal resistance is

V_r = 3 V- 2.5 V = 0.5 V.

Power dissipation in the internal resistance r is

⇒ \(W_r=V_r \cdot I=(3 \mathrm{~V}-2.5 \mathrm{~V})\left(\frac{1}{5} \mathrm{~A}\right)=0.5 \times \frac{1}{5} \mathrm{~W}=0.1 \mathrm{~W}\)

101. The resistance of the voltmeter in the given circuit is 10 k£l. Find the voltmeter reading.

1. 4 V
2. 3.25 V
3. 1.25 V
4. 1.95 V

Answer: 4. 1.95 V

Let V_A-V_B = V = voltmeter reading.

The main current is

⇒ \(I=I_1+I_2=\frac{V}{10000}+\frac{V}{400}\)

V + 7(800 Ω) = 6 V

⇒ \(V+V\left(\frac{1}{10000}+\frac{1}{400}\right)(800 \Omega)=6 \mathrm{~V}\)

⇒ \(V+\frac{8 V}{100}+2 V=6 \mathrm{~V} \Rightarrow V=1.95 \mathrm{~V}\).

Question 102. In the given circuit, 1.02 V is balanced at 51 cm from end A of a 1.0-m-long potentiometer wire. Find the potential gradient along AB.

1. \(0.1 \mathrm{~V} \mathrm{~cm}^{-1}\)
2. \(0.02 \mathrm{~V} \mathrm{~cm}^{-1}\)
3. \(0.3 \mathrm{~V} \mathrm{~cm}^{-1}\)
4. \(0.4 \mathrm{~V} \mathrm{~cm}^{-1}\)

Answer: 2. \(0.02 \mathrm{~V} \mathrm{~cm}^{-1}\)

For null deflection, potential drop = ΔV =1.02 V for length l = 51 cm of the potentiometer wire.

Hence, the potential gradient is

⇒ \(\frac{\Delta V}{\Delta l}=\frac{1.02 \mathrm{~V}}{51 \mathrm{~cm}}=0.02 \mathrm{~V} \mathrm{~cm}^{-1}\)

Question 103. A cylindrical shell of length l with inner and outer radii Rx and R2 is made of a material of resistivity p. Find its resistance if current flows radially outward.

1. \(\frac{\rho}{2 \pi l} \ln \frac{R_2}{R_1}\)
2. \(\frac{\rho}{\pi l} \ln \frac{R_2}{R_1}\)
3. \(\frac{\rho}{4 \pi l} \ln \frac{R_2}{R_1}\)
4. \(\frac{\rho}{3 \pi l} \ln \frac{R_2}{R_1}\)

Answer: 1. \(\frac{\rho}{2 \pi l} \ln \frac{R_2}{R_1}\)

Resistance of a coaxial cylindrical shell of radius r and thickness dr is

⇒ \(d R=\rho \frac{d r}{A}=\rho \frac{d r}{2 \pi r l}\)

total resistance is

⇒ \(R=\int d R=\frac{\rho}{2 \pi l} \int_{R_1}^{R_2} \frac{d r}{r}=\frac{\rho}{2 \pi l} \ln \frac{R_2}{R_1}\)

Question 104. Which of the following graphs represents the variation of resistivity (p) with temperature (T) for copper?

Answer: 2.

With the increase in temperature of a conductor, thermal agitation increases and the collisions become more frequent. Thus, copper being a good conductor, its conductivity decreases and resistivity(ρ)increases nonlinearly with temperature. Hence, graph (b).

105. The color code of a resistor is given in the figure. The value of its resistance with tolerance is

1. 47 kΩ, 10%
2. 4.7 kΩ, 5%
3. 470 Ω, 5%
4. 470 kΩ, 5%

Answer: 3. 470Ω, 5%

The sequence of colors is

Yellow – 4

Violet – 7

Brown – 10

Gold – 5%

Hence, R =(47×10) Ω ± 5% = 470 Ω, 5%.

Question 106. The resistance wire connected in the left gap of a meter bridge balances a 10-Ω resistance in the right gap at a point that divides the bridge wire in the ratio 3: 2. If the length of the resistance wire is 1.5 m then the length of 1 Ω of the resistance wire is

1. \(1.0 \times 10^{-1} \mathrm{~m}\)
2. \(1.5 \times 10^{-1} \mathrm{~m}\)
3. \(1.5 \times 10^{-2} \mathrm{~m}\)
4. \(1.0 \times 10^{-2} \mathrm{~m}\)

Answer: 1. \(1.0 \times 10^{-1} \mathrm{~m}\)

For balance: \(\frac{R}{10 \Omega}=\frac{3}{2}\) R =15Ω.

⇒ Since R<*l, hence \(\frac{R^{\prime}}{R}=\frac{l^{\prime}}{l} \Rightarrow \frac{R^{\prime}}{15 \Omega}=\frac{l^{\prime}}{1.5 \mathrm{~m}}\)

⇒ \(l^{\prime}=\frac{(1.5 \mathrm{~m})}{(15 \Omega)} \times(1 \Omega)=\frac{1}{10} \mathrm{~m}=1.0 \times 10^{-1} \mathrm{~m}\).

Question 107. A charged particle having a drift velocity of \(7.5 \times 10^{-4} \mathrm{~m} \mathrm{~s}^{-1}\) in an electric field of3 x10-10 V m-1 has mobility (in m² V^-1 s^-1) of

1. \(2.5 \times 10^6\)
2. \(2.5 \times 10^{-6}\)
3. \(2.25 \times 10^{-15}\)
4. \(2.25 \times 10^{15}\)

Answer: 1. \(2.5 \times 10^6\)

Mobility is defined as drift velocity per unit electric field \(\left(=\frac{v}{E}\right)\)

Given that

⇒ \(v=7.5 \times 10^{-4} \mathrm{~m} \mathrm{~s}^{-1}, E=3 \times 10^{-10} \mathrm{Vm}^{-1}\).

∴ Hence, mobility = \(\frac{7.5 \times 10^{-4}}{3 \times 10^{-10}} \text { units }=2.5 \times 10^6 \text { units. }\)

Question 108. In the given circuit, the cell has negligible internal resistance. The potential difference across BD is

1. 4 V
2. 3 V
3. 2 V
4. 1 V

Answer: 3. 2 V

Current in the upper branch ABC is

⇒ \(I_1=\frac{40 \mathrm{~V}}{(40+60) \Omega}=\frac{40}{100} \mathrm{~A}=0.4 \mathrm{~A}\)

Similarly, for the lower branch,

⇒ \(I_2=\frac{40 \mathrm{~V}}{90 \Omega+110 \Omega}=\frac{40}{200} \mathrm{~A}=0.2 \mathrm{~A}\)

Now, V_A-V_B = l₁ (40 Ω) = 0.4 x 40 V =16 V;

V_A-V_D = l₂ (90Ω) = 0.2 x 90 V =18 V.

∴ (V_A-V_D)-(V_A-V_B) = V_B-V_D = 2V.

Question 109. For the given circuit, find the potential at B with respect to A.

1. +2V
2. -2 V.
3. +1 V
4. -1 V

Answer: 3. +1 V

From the junction rule at D, the current in branch DC is 1 A.

Now,

V_B-V_A=(V_B-V_D)+(V_D-V_c)-(V_c-V_A)

= (-2V)+(2 V) + (1V) = +1 V.

Question 110. The value of the current l₁ flowing from A to C in the circuit diagram is

1. 1 AB
2. 5 A
3. 4 A
4. 2 A

Answer: 1. 1 AB

Potential difference = V_A-V_c=8 V

∴ \(I_1=\frac{V}{R}=\frac{8 \mathrm{~V}}{8 \Omega}=1 \mathrm{~A}\).

Electricity And Magnetism Notes

An electric current is the rate of flow of electric charges.

In units, \(1 A=\frac{1 C}{1 s}\)

• • Average current, \(I_{\mathrm{av}}=\frac{\Delta Q}{\Delta t}\)
  • Instantaneous current, \(i=\frac{d Q}{d t}\)
• The current density is the current per unit cross-sectional area.

In SI units, \(1 \mathrm{Am}^{-2}=\frac{1 \mathrm{~A}}{1 \mathrm{~m}^2}\)

• • Average current density, \(\vec{j}=\frac{\Delta I}{\Delta s}\)
    • So,\(\Delta I=\vec{j} \cdot \overrightarrow{\Delta s}\)
  • Current density at some specific point, \(\vec{j}=\frac{d I}{d s}\)
    • So, \(I=\int \vec{j} \cdot \overrightarrow{d s}\)

Note that current is not a vector and is additive, but current density is a vector quantity.

“electricity and magnetism “

Drift speed, \(v_d=\frac{l}{n A e}\), where n = number density, e = charge of an electron and A = cross-sectional area.

• Ohm’s law: V = IR (in file scalar form).
  • \(\vec{j}=\sigma \vec{E}\)(in the vector form).
  • Resistance, \((\rho)=\frac{1}{\text { conductivity }(\sigma)}\)
• Resistance, \(R=\rho \frac{l}{A}\).
• Temperature-dependence of resistance: R_θ = R₀ (1+αθ), where a = temperature coefficient of resistance.
• Resistance in series: Rs = R1 + R2 +… where Rs is greater than the greatest resistance.
• Resistance in parallel: \(\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\ldots\)+…, where R is less than the least resistance.

The electromotive force (emf) ε of an electric cell is the potential difference between the terminals of the cell (or terminal voltage) in an open circuit.

• The terminal voltage of a cell of emf ε and internal resistance r:
  • When the cell delivers a current I,
    V = V_A-V_B = (V_A-V_C) + (V_C-V_B)
    = ε-(V_B-V_c)= ε-Ir  [∵V_B-V_c]

• • When the cell is being charged by a steady current I,
    V = V_A-V_B
    = (V_A-V_C) + (V_C-V_B)
    =ε+Ir

Magnetism and Matter NEET Notes

• The current through a shunt and a galvanometer are respectively

⇒ \(I_{\mathrm{s}}=\left(\frac{G}{G+S}\right) I \text { and } I_g=\left(\frac{S}{G+S}\right) I \text {, }\)

physics magnetism and electricity

where S = shunt resistance

and G = Galvanometer

• Galvanometer as an ammeter.
  • I_g G = (I-I_g)S.
  • \(S=\frac{G}{n-1}\)
  • \(n=\frac{I}{I_g}\)

• Galvanometer as a voltmeter:
  • V_A-V_B = I_g (R+G).
  • R = (n-1)G.
  • \(n=\left(\frac{V}{V_{\mathrm{g}}}\right)=\frac{V}{I_{\mathrm{g}} G}\)

Magnetism and Matter NEET Notes

Kirchhoff’s laws:

• • Junction rule: ∑I_i= 0 (= charge conservation),
  • Loop rule: ∑V_i = 0 (= energy conservation).

Grouping of cells:

• • In series: \(I=\frac{N e}{R+N r}.\)
  • In parallel: \(I=\frac{N e}{N R+r}\)
• Equivalent emf of two cells in parallel:

\(\mathcal{E}=\frac{e_1 r_2+e_2 r_1}{r_1+r_2}\), While the main current is

⇒ \(I=\frac{e_1 r_2+e_2 r_1}{R\left(r_1+r_2\right)+r_1 r_2}\).

“physics electricity and magnetism “

• Wheatstone bridge: The bridge shown in the figure is said to be balanced if \(\frac{P}{Q}=\frac{R}{S}\).

• Equivalent resistance in some special cases across A and B:

Magnetic Effect of Current formulae for NEET

\(R_{\mathrm{AB}}=\left(\frac{3 n+1}{n+3}\right) r\)

R_AB = r₁ + r₂

The CD is a conductor.

• Electric Power:

P (in Watts) = V (in Watts) x I (im amperes)= \(\frac{V^2}{R}=I^2 R\)

• Equivalent Power:
  • In series: \(\frac{1}{W_s}=\frac{1}{W_1}+\frac{1}{W_2}+\ldots\)
  • In parallel: WP = W1 + W2 +…
  • A Potentiometer measures the emf of a cell.
  • Potential gradient, \(k=\frac{V_{\mathrm{A}}-V_{\mathrm{B}}}{A B}\)
  • This internal resistance of the cell, \(r=\left(\frac{l}{l}-1\right) S\), where balancing lengths are l and l with key (K) open and closed respectively.

• Growth of change in an RC circuit:
  • Instantaneous charge, \(q=Q_0\left(1-e^{-t / R C}\right)\)
  • Then, instantaneous current, \(i=\frac{d q}{d t}=\frac{Q_0}{R C} e^{-t / R C}=\frac{\varepsilon}{R} e^{-t / R C}\)
• Decay of charge in an RC circuit:
  • Instantaneous charge, \(q=Q_0 e^{-t / R C}=\varepsilon C e^{-t / R C}\).
  • Then, instantaneous current, \(i=\frac{d q}{d t}=-\frac{\varepsilon}{R} e^{-t / R C}\).
  • The time constant of an RC circuit = RC (measured in seconds).

“magnetism notes “

Magnetic effect of current: Moving charges, (collectively equivalent to an an electric current) produce a magnetic field\((\vec{B})\) around themselves.

• Force on a charge q moving with a velocity \(\vec{v}\).
  • In an electric field \(\vec{E}: \quad \vec{F}_{\text {elec }}=q \vec{E}\).
  • In a magnetic field \(\vec{B}: \vec{F}_{\mathrm{mag}}=q(\vec{v} \times \vec{B})\).
• Lorentz force: \(\vec{F}=\vec{F}_{\text {elec }}+\vec{F}_{\text {mag }}=q(\vec{E}+\vec{v} \times \vec{B})\).
• Path of a charged particle moving in a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}\):

The force \(\vec{F}=q(\vec{v} \times \vec{B})\) acting perpendicular to both \(\vec{v}\)and \(\vec{B}\) provides a centripetal force for its circular path of radius r, where \(F=q v B=\frac{m v^2}{r}, v=\frac{q B r}{m}\) (i.e v r)and time of revolution \(T=\frac{2 \pi m}{q B}\)(independent of radius r).

Magnetic Effect of Current formulae for NEET

• Force on a current element \((I \overrightarrow{d l})\)in a magnetic field \(\overrightarrow{\boldsymbol{B}}\):

\(\vec{F}=q \vec{v} \times \vec{B}=I(d t \vec{v}) \times \vec{B}=I(\overrightarrow{d l} \times \vec{B})\)

A magnetic monopole (or an isolated magnetic pole) has no existence. There always exists a magnetic dipole with a current loop.

• Magnetic moment of a current loop:

\(\left.\vec{m}=I \vec{A} \text { (SI unit: } \mathrm{A} \mathrm{m}^2\right)\)

• Torque on a magnetic dipole placed in a uniform magnetic field:

\(\vec{\tau}=\vec{m} \times \vec{B}, \text { where } \vec{m}=I N \vec{A}\) and N = number of turns.

• Work done in deflecting a current loop (= a magnetic dipole) in a uniform magnetic field \(\overrightarrow{\mathrm{B}}\):

\(W=m B(1-\cos \theta)\), where 9 is the angle between B and m.

The potential energy of a magnetic dipole in \(\vec{B}\): \(U=-\vec{m} \cdot \vec{B}=-m B \cos \theta\)

• • When \(\theta=0^{\circ}, U_{\min }=-m B\), and the dipole is stable.
  • When \(\theta=180^{\circ}, U_{\max }=m B\), and the dipole is unstable.

“magnetism and electromagnetism “

Biot-Savart law:

The magnetic field at P due to a current I is \(\overrightarrow{d B}=\frac{\mu_0}{4 \pi}\left(\frac{I \vec{d} \times \hat{r}}{r^2}\right)\) (expressed in teslas)

or \(d B=\frac{\mu_0}{4 \pi}\left(\frac{I d l \sin \theta}{r^2}\right)\), directed into the paper plane.

• Magnetic field due to a circular current loop:
  • At the centre, \(B=\frac{\mu_0 I}{2 R}\).
  • At a point on its axis, \(\vec{B}=\frac{\mu_0 I R^2 \hat{x}}{2\left(R^2+x^2\right)^{3 / 2}}\)
  • At a distance x » R,

⇒ \(B=\frac{\mu_0 I R^2}{2 x^3}=\frac{2 \mu_0}{4 \pi}\left(\frac{\pi R^2 I}{x^3}\right)=\frac{\mu_0}{4 \pi} \cdot 2\left(\frac{A I}{x^3}\right)=\frac{\mu_0}{4 \pi}\left(\frac{2 m}{x^3}\right)\).

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Magnetic Effect of Current NEET Notes for Physics

• • At the centre of an arc, \(B=\frac{\mu_0 I \theta}{4 \pi R}\).

• Magnetic field due to a straight current:
  • Of infinite length: \(B=\frac{\mu_0 I}{2 \pi d}\).
  • Of finite length: \(B=\frac{\mu_0 I}{4 \pi d}\left(\sin \theta_1+\sin \theta_2\right)\).
  • At the centre O, the magnetic field has a zero magnitude and is independent of the angle 0.

• Force on a Current element \(\overrightarrow{I l}\) in magnetic field \(\vec{B}\):

⇒ \(\vec{F}=I \vec{l} \times \vec{B}\).

• Force per unit length between two parallel currents:

⇒ \(F=\frac{\mu_0}{2 \pi}\left(\frac{I_1 I_2}{d}\right)\left(\text { in } \mathrm{N} \mathrm{m}^{-1}\right)\)

• Ampere’s circuital law: \(\oint \vec{B} \cdot \overrightarrow{d l}=\mu_0 I\)
• Magnetic field inside a solenoid: \(B=\frac{\mu_0 N I}{l}=\mu_0 n I\).
• The magnetic field inside a toroid (an endless solenoid):

⇒ \(B=\frac{\mu_0 N I}{2 \pi R}=\mu_0 n I\)

• Velocity selector: A charged particle moving with a velocity \(\vec{v}=v \hat{i}\) and passing through crossed electric and magnetic fields emerges undefeated when

⇒ \(\vec{E}=E(-\hat{j}) \text { and } \vec{B}=B(-\hat{k})\).

The electric force \(\vec{F}_{\text {elec }}=-q(E \hat{j})\) is balanced by the magnetic force \(\vec{F}_{\text {mag }}=q(\vec{v} \times \vec{B})=q v B \hat{i} \times(-\hat{k})=q v B \hat{j}\).

Thus, \(q v B=q E \text { or } v=\frac{E}{B}\).

• Magnetic moment \((\vec{m})\) of a bar magnet:

It is given by \((\vec{m})\) = pole strength x magnetic length = p_m x 2l.

Here \((\vec{m})\) is a vector directed from the SP to the NP (SI unit: A m² ) and pm is positive for the NP and negative for the SP (SI unit: A m).

Electrical And Magnetic Properties NEET

• Magnetic field \((\vec{B})\) due to a bar magnet:
  • At an axial point, \(\vec{B}=\frac{\mu_0}{4 \pi}\left[\frac{2 \vec{m} d}{\left(d^2-l^2\right)^2}\right] \approx \frac{\mu_0}{4 \pi}\left(\frac{2 \vec{m}}{d^3}\right)\).
  • At an equatorial point, \(\vec{B}=\frac{\mu_0}{4 \pi}\left[\frac{-\vec{m}}{\left(d^2+l^2\right)^{3 / 2}}\right] \approx \frac{\mu_0}{4 \pi}\left(\frac{-\vec{m}}{d^3}\right)\).
  • At any point \(\mathrm{P}(r, \theta), B=\frac{\mu_0}{4 \pi}\left(\frac{m}{d^3} \sqrt{1+3 \cos ^2 \theta}\right)\)

“magnetism definition physics “

• Torque on a bar magnet in a uniform magnetic field \((\vec{B})\) :

⇒ \(\vec{\tau}=\vec{m} \times \vec{B} \Rightarrow \tau=m B \sin \theta\).

• • Potential energy, \(U=-(\vec{m} \cdot \vec{B})=-m B \cos \theta\)
• The time period of oscillations of a bar magnet in a uniform magnetic field \(\vec{B}\):

⇒ \(T=2 \pi \sqrt{\frac{I}{m B}}\),

where t= moment of inertia about the rotational axis.

Hence, \(B=\frac{4 \pi^2 I}{m}\left(\frac{1}{T}\right)^2=k f^2\) where f = frequency of oscillations.

• Magnetic meridian: It is a vertical plane at a place on the earth’s surface containing a resultant magnetic field at that place.
• Geographical meridian: A vertical plane at a place on the earth’s surface passing through the line joining the geographical north and south is called the geographical meridian of that place.
• Elements of terrestrial magnetism: Three elements are required to completely describe the magnetic field at a place on the earth (both in magnitude and direction). These are listed and explained below.

• • Declination: The angle between the magnetic meridian and the geographical meridian at a place is called fire declination at that place. It is expressed as 0°E or 0W.
  • Dip or inclination (δ): The angle which the earth’s magnetic field \(\vec{B}\) makes with the horizontal line in the magnetic meridian at a place is called the dip (δ) at that place. At the magnetic north and south poles, the dip is 90°, and on the magnetic equator, δ = 0°.
  • Horizontal component (B_H): It is the component of the resultant field B in the horizontal direction in the magnetic meridian at a place.

From the given figure, \(B_{\mathrm{H}}=B \cos \delta \text { and } B_{\mathrm{V}}=B \sin \delta\).

Electrical And Magnetic Properties NEET

∴ \(\tan \delta=\frac{B_{\mathrm{V}}}{B_{\mathrm{H}}} \Rightarrow B=\sqrt{B_{\mathrm{H}}^2+B_{\mathrm{V}}^2} .\).

• Magnetic properties of matter: The magnetism in solids has its origin in the orbital motions and spin rotations of the orbital electrons of the atoms. These motions cause magnetic moments and add up to produce magnetization in solids in the presence of an external magnetic field.
• Intensity of magnetization (I):

⇒ \(I=\frac{\text { magnetic moment }}{\text { volume }}=\frac{\text { (pole strength) }(\text { length })}{(\text { cross-sectional area })(\text { length })}\)

\(=\frac{\text { pole strength }}{\text { cross-sectional area }}\left(\text { SI unit: } \frac{\mathrm{Am}}{\mathrm{m}^2}=\mathrm{Am}^{-1}\right)\)

• Magnetic intensity (H): When a magnetic substance is placed in an external magnetic field B0, it gets magnetized due to the alignments of its atomic dipoles, and the net field inside the material is given by

\(B=B_0+\mu_0 I=\mu_0(H+I), \text { where } B=B_0=\mu_0 H \text {, }\) in free space (I = 0).

Thus, H = \(H=\frac{B}{\mu_0}-I .\). The SI unit of H is the same as that of \(I\left(\mathrm{~A} \mathrm{~m}^{-1}\right)\).

Different expressions for B and H are as follows.

• • At the centre of a circular coil: \(B=\frac{\mu_0 I}{2 R} \Rightarrow H=\frac{B}{\mu_0}=\frac{I}{2 R}.\).
  • Inside a solenoid: \(B=\mu_0 n I \Rightarrow H=n I\).
  • Biot-Savart law: \(\overrightarrow{d B}=\frac{\mu_0}{4 \pi}\left(\frac{I \vec{d} \times \hat{r}}{r^2}\right) \Rightarrow \vec{H}=\frac{\overrightarrow{d B}}{\mu_0}\).
• Magnetic susceptibility \((\chi)\): Magnetic susceptibility indicates the ability of a material to get magnetized when placed in an external magnetizing field. Thus, the intensity of magnetization (l) is proportional to the magnetic intensity (H).

So, \(I \propto H \Rightarrow I=\chi H .\)

Since both I and H are expressed in amperes per metre, \((\chi)\) is dimensionless.

• • For vacuum, I = 0 and \((\chi)\) = 0.
  • For paramagnetic materials, \((\chi)\) is positive.
  • For diamagnetic materials, \((\chi)\) is negative.
• Magnetic permeability, \(\mu=\frac{B}{H}\)

Relation between relative permeability pr and susceptibility \((\chi)\):

The magnetic field inside a material is

⇒ \(B=\mu_0(H+I)=\mu_0(H+\chi H)=\mu_0 H(1+\chi)\).

But \(B=\mu H\).

Electrical And Magnetic Properties NEET

∴ \(\mu=\mu_0(1+\chi)\)

⇒ \(\frac{\mu}{\mu_0}=\mu_{\mathrm{r}}=1+\chi\).

• Curie’s law: With an increase in temperature, the alignment of the elementary dipoles of a magnetic material is reduced, which decreases its magnetization. According to Curie’s law,

⇒ \(\chi \propto \frac{1}{T} \Rightarrow \chi=\frac{C}{T}\) where C is Curie constant.

• Curie temperature (T_c): It is the temperature at which a ferromagnetic material converts into a paramagnetic one. Thus,

⇒ \(\chi=\frac{C^{\prime}}{T-T_C}\) where Tc is Curie point and C is a constant.

• Electromagnetic induction: An electromotive force (emf) is induced whenever a magnetic flux (O) linked with a coil changes with time. This phenomenon is known as electromagnetic induction.
• Magnetic \(\Phi=N A B \cos \theta=N \vec{A} \cdot \vec{B}\) where N = number of turns, A = area within the coil, B = strength of the magnetic field (in tesla), and 0 = angle between the magnetic field and the normal to the area.

“magnetism definition physics “

Magnetic flux is expressed in Webers (symbol: Wb)

1 Wb =l T m².

• Faraday’s law: The induced emf in a coil is proportional to the rate of change of the magnetic flux linked with the coil. Thus,

⇒ \(\mathcal{E}=-\frac{d \Phi}{d t}=-N \frac{d \Phi_0}{d t}\),

where N = number of turns and O0 = magnetic flux linked with each turn.

• Motional emf: When a wire moves through a magnetic field so as to cut the field lines, an emf is induced in the wire, and it is called the motional emf.

It has the magnitude ε =\(B l v \cos \theta\), where \(\vec{v}\) = velocity of the wire, l = length of the wire, \(\vec{B}\) = magnetic field and θ = angle between \(\vec{v}\) and \(\hat{n}\) (the unit vector perpendicular to the length of the wire).

Self-inductance (L): Magnetic flux = \(\Phi=L I, \text { induced emf }=\varepsilon=-\frac{d \Phi}{d t}\) = \(-L \frac{d I}{d t}\) and magnetic energy linked with the inductor = \(U=\frac{1}{2} L I^2\) where L is the self-inductance of the inductor. The self-inductance
of a solenoid is \(L=\mu_0 n^2 A l=\frac{\mu_0 N^2 A}{l}\), where N = total number of turns, l = length of the solenoid and A = area of each turn.

The SI unit of self-inductance is the Henry (symbol: H).

• The energy density (u) in a magnetic field \(\vec{B}\):

⇒ \(\left.u=\frac{\text { total energy }(U)}{\text { volume }(V)}=\frac{B^2}{2 \mu_0} \quad \text { (SI unit: } \mathrm{J} \mathrm{m}^{-3}\right)\)

• Mutual inductance (M):

\(\Phi_2=M I_1 \text { and } \varepsilon_2=-M \frac{d I_1}{d t}\), where M is the mutual inductance of two inductors.

The mutual inductance of two solenoids is given by \(M=\frac{\mu_0 N_1 N_2 A}{l}\).

Magnetic Effect of Current formulae for NEET

The SI unit of mutual inductance is the Henry (symbol: H).

• Growth of current in an LR circuit:

⇒ \(I=I_0\left[1-e^{-(R / L) t}\right]\).

• Decay of current in an LR circuit:

⇒ \(I=I_0 e^{-(R / L) t}\).

• Time constant of an LR circuit, \(\tau=\frac{L}{R}\).
• Instantaneous magnetic flux in a coil rotating in a magnetic field:

⇒ \(\Phi=N A B \cos \omega t\)

• Induced emf in a coil:

⇒ \(\mathcal{E}=-\frac{d \Phi}{d t}=N A B \omega \sin \omega t=\varepsilon_0 \sin \omega t\),

where \(\varepsilon_0=\text { peak emf }=N A B \omega\).

• Instantaneous current in an AC circuit:

⇒ \(I=I_0 \sin \left(\omega t+\Phi_0\right)\),

where \(I_0=\text { peak current, } \omega=\text { angular frequency }=2 \pi f \text { and } \Phi_0\) = initial phase.

• Average values of an alternating current (AC):
  • In one complete cycle, \(I_{\mathrm{av}}=\frac{1}{T} \int_0^T I d t=0\).
  • In a half cycle, \(I_{\mathrm{av}}=\frac{1}{T / 2} \int_0^{T / 2} I d t=\frac{2 I_0}{\pi}\).
  • Root-mean-square(or virtual) value,\(I_{\text {rms }}=\frac{I_0}{\sqrt{2}}=\frac{\text { peak value }}{\sqrt{2}}\).
• Reactance (X):
  • Reactance of an inductor: \(X_{\mathrm{L}}=\omega L=2 \pi f L\).
  • Reactance of a capacitor: \(X_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C}\).
• Impedance (Z):
  • Impedance of an LR circuit: \(Z=\sqrt{R^2+\omega^2 L^2}\)
  • Impedance of a CR circuit: \(Z=\sqrt{R^2+\frac{1}{\omega^2 C^2}}\)
  • Impedance of an LCR circuit: \(Z=\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}\)
  • Impedance of an LC circuit: \(Z=\left|X_L-X_C\right|=\left|\omega L-\frac{1}{\omega C}\right|\).
• Phase difference \((\phi)\) between the current and the voltage in an AC circuit:
  • With R only: = 0, i.e., the current (I) and the voltage (V) are in the same phase.
  • With L only: \(\phi=\frac{\pi}{2}\), i.e., the current lags behind the voltage by \(\).
  • With C only: \(\phi=\frac{\pi}{2}\), i.e., the current leads the voltage by \(\frac{\pi}{2}\).
  • With L and R in series: \(\tan \varphi=\frac{\omega L}{R} \text {, i.e., } I \text { lags } V \text { by } \tan ^{-1}\left(\frac{X_L}{R}\right)\)
  • With C and R in series: \(\tan \varphi=\frac{X_C}{R}=\frac{1}{\omega C R}\)i.e., I leads V by \(\tan ^{-1}\left(\frac{\mathrm{X}_{\mathrm{C}}}{R}\right)\)
  • With L and C in series: \(\tan \varphi=\frac{\left|X_{\mathrm{L}}-X_{\mathrm{C}}\right|}{0}=\infty \Rightarrow \varphi=90^{\circ}\),
  • i.e., the current leads by \(\frac{\pi}{2}\) for Xc > XL and the current lags by \(\frac{\pi}{2}\) for XL>Xc.
  • In a series LCR circuit: \(\tan \varphi=\frac{X}{R}=\frac{\left|X_{\mathrm{L}}-X_{\mathrm{C}}\right|}{R}\)
• Power in an AC circuit:
• • True average power = \(\frac{I_0}{\sqrt{2}} \cdot \frac{V_0}{\sqrt{2}} \cos \varphi=(\text { rms power }) \cos \varphi\) (rms power)cos <p.
  • \(\text { Power factor }=\frac{\text { true average power }}{\text { rms power }}=\cos \varphi=\frac{R}{Z}\).
• Electrical resonance: A series RLC circuit is said to be at resonance when the current amplitude \(I_0=\frac{V_0}{Z}\) becomes maximum at a specific frequency called the resonant frequency \(\left(f_{\mathrm{r}}=\frac{1}{2 \pi \sqrt{L C}}\right)\).

“magnetism definition physics “

At resonance,

• • The circuit is purely resistive,
  • Reactance = X = X_L-X_c = 0 ,
  • 1 and 5 are in phase, i.e., \(\varphi=0\),
  • power factor =1 (maximum).
• Q (quality) factor: \(Q=\frac{1}{R} \sqrt{\frac{L}{C}}\)
• LC oscillations: \(\frac{1}{2} L I^2+\frac{Q^2}{2 C}\) = constant, \(\omega=2 \pi f=\frac{1}{\sqrt{L C}} \text { and }\) and Q = Qo cos cof.
• Transformer’s turns ratio, \(\frac{N_{\mathrm{s}}}{N_{\mathrm{p}}}=\frac{V_{\mathrm{s}}}{V_{\mathrm{p}}}=\frac{I_{\mathrm{p}}}{I_{\mathrm{s}}}\).