Oswal 36 Sample Question Papers CBSE Class 12 Physics Solutions
Section-A
1. (d) Zero
Forces of repulsion on 1 µC charge at O due to 3 µC charge, at A and C are equal and opposite. So they cancel each other. Similarly, forces of attraction of 1 µC charge at O due to −4 µC charges at B and D are also equal and opposite. So, they also cancel each other.
Hence the net force on the charge of 1 µC at O is zero.
2. (c) remains unchanged
The force will still remain
$$\frac{q_1q_2}{4\pi\epsilon_{0}r^{2}}\space\text{according to}\space\\\text{the superposition principle.}$$
3. (d) vd/4
As we know that,
I = nAevd
$$\text{or\qquad}v_{d}∝\frac{1}{\pi r^{2}}\\\text{If r}\xrightarrow{} 2r,\text{then}\\ v_{d'}=\frac{1}{\pi(2r)^{2}}\\v_{d'} =\frac{v_{d}}{4}.$$
4. (b) the line integral of magnetic field along the boundary of the open surface is equal to μ0 times the total current passing through the surface.
According to Ampere’s circuital law,
$$\oint\vec{\text{B}}.\vec{\text{dl}} =\mu_{0}\text{I}$$
$$\textbf{5.\space(d)\space}\frac{\mu_{\textbf{0}}\textbf{NIR}^{\textbf{2}}}{\textbf{2(R}^{\textbf{2}} \textbf{+ x}^{\textbf{2}})^{\frac{\textbf{3}}{\textbf{2}}}}$$
The magnetic field intensity at point P due to current carrying coil having N turns and radius R is given by,
$$\text{B =}\frac{\mu_{0}}{4\pi}\frac{2\pi(\text{NI})\text{R}^{2}}{(\text{R}^{2} +x^{2})^{\frac{3}{2}}}\\\text{or\space}=\frac{\mu_{0}\text{NIR}^{2}}{2(\text{R}^{2} +x^{2})^{\frac{3}{2}}}$$
6. (c) 105 A/m
Given, l = 5 cm = 5 × 10–2 m
A = 2 cm2 = 2 × 10–4 m2
and M = 1 Am–2
∴ Magnetisation of bar magnet is given by,
$$\text{I =}\frac{\text{M}}{\text{V}} =\frac{\text{M}}{\text{A×I}}\\=\frac{1}{2×10^{\normalsize-4}×5×10^{\normalsize-2}}=\frac{1}{10^{\normalsize-5}}$$
= 105 A/m
7. (d) 0.05 mH
From the formula,
$$\text{M =}\frac{\mu_{0}n_{2}n_{1}\text{A}}{l}\\=\\\frac{4\pi×10^{\normalsize-7}×500×200×\frac{4}{10^{4}}}{1}$$
= 160π × 10–7
≃ 0.05 × 10–3H ≃ 0.05 mH
8. (c) 7 A
$$\because\space\text{I}_{rms} =\frac{\text{E}_{\text{rms}}}{2}\\=\frac{200}{40}$$
= 5 A
$$\text{I}_{0} =\text{I}_{\text{rms}}\sqrt{2}$$
= 7.07 A
9. (a) Infrared region
Paschen series of atomic spectrum of hydrogen gas lies in infrared region.
$$\text{As\space}\bar{v} =\text{R}\bigg[\frac{1}{3^{2}} -\frac{1}{n^{2}}\bigg]$$
n = 4, 5, 6 and so on
10. (a) 49°
$$\mu =\frac{1}{\text{sin}\space i_{c}}\\\Rarr\space \text{sin}i_{c} =\frac{1}{\mu} =\frac{3}{4} =0.75\\\Rarr\space i_{c} =\frac{\theta}{2} =\text{sin}^{\normalsize-1}0.75\\ =48.6\degree\\\therefore\space\text{Semi-vertical angle = 48.6°.}$$
11. (d) 0·7 eV
Here, E2 – E1= 10·3eV < 11 eV
∴ Electrons can excite the atoms,
Kmin = (11 – 10·3) = 0·7 eV.
12. (a) 364.6 nm
The shortest wavelength possible for Balmer series is 364.6 nm.
13. (a) Both A and R are true and R is the correct explanation of A.
In semiconductors, the energy gap between conduction band and valence band is small (≈ 1 eV). Due to temperature rise, electron in the valence band gain thermal energy and may jump across the small energy gap (to the conduction band). Thus, conductivity increases and hence resistance decreases.
14. (d) A is false, and R is also false.
Phase difference between any two points on a wavefront is zero. Also two points on a given wavefront are equidistant from source hence independent of distance.
15. (b) Both A and R are true and R is not the correct explanation of A
Light is a transverse, electromagnetic wave that can be seen by the typical human. The wave nature of light was first illustrated through experiments on diffraction and interference. Like all electromagnetic waves, light can travel through a vacuum.
16. (b) Both A and R are true and R is not the correct explanation of A.
As we know that,
Intrinsic (Semiconductor) + Pentavalent(Impurity)
$$\Rarr\space\text{N-type}$$
Section-B
17. (i) Gamma radiations are produced during the radioactive decay of a nucleus.
Frequency range is > 3 × 1017 Hz.
(ii) Welding sparks produces ultraviolet rays.
Frequency range is 7.5 × 1014 Hz – 3 × 1016 Hz.
(iii) T.V. remote control uses infrared radiations.
Frequency range is 3 × 1012 Hz – 4.3 × 1014 Hz.
18. The Lorentz magnetic force is given by the following relation :
$$\vec{F} = q(\vec{v}×\vec{\text{B}})$$
Here q, is the magnitude of the moving charge.
The direction of the magnetic force is perpendicular to the plane containing the velocity vector and
$$\text{the magnetic field vector\space}\vec{\text{B}.}$$
19.
The magnifying power m is the ratio of the angle β subtended at the eye by the final image to the angle α which the object subtends at the lens or the eye. Hence,
$$m =\frac{\beta}{\alpha} =\frac{h}{f_{e}}.\frac{f_{0}}{h} =\frac{f_{0}}{f_{e}}$$
20. Conductors:
(i) In case of conductors, the valence band is completely filled and the conduction band can have two caseseither it is partially filled with an extremely small energy gap between the valence and conduction bands or it is empty, with the two bands overlapping each other.
(ii) Even when a small current is applied, conductors can conduct electricity.
Insulators :
(i) In case of insulators, the energy gap between the conduction and valence bands is very large and the conduction band is practically empty.
(ii) When an electric field is applied to such kind of material, the electrons find hard to receive such a large amount of energy to reach the conduction band. Thus, the conduction band remains empty. That is why no current flows through insulators.
Semiconductors:
(i) In case of semiconductor, the energy band structure of semiconductors is similar to insulators. But in this case, the size of forbidden energy gap is quite smaller than that of the insulators.
(ii) When an electric field is applied to a semiconductor, the electrons in the valence band find it relatively easier to jump to the conduction band. So, the conductivity of semiconductors lies between the conductivity of conductors and insulators.
21. Limitations of Rutherford model:
(i) It is not in accordance with the Maxwell’s theory and could not explain the stability of an atom.
(ii) It did not say anything about the arrangement of electrons in an atom.
OR
In Rutherford experiment, a few a-particles are scattered through large angles. Therefore, massive and positive part of atom should be concentrated in a small portion of the atom. The small positive part having almost the entire mass of the atom was called nucleus. Therefore, the main feature of any atom model should require the presence of the central positive and massive nucleus.
22. Consider a rod PQ of length l, moving in a magnetic field $$\vec{\textbf{B}\space}\textbf{with a constant velocity\space}\vec{\textbf{v}}.$$
The length of the rod is perpendicular to the magnetic field and also the velocity is perpendicular to both the rod and field. The free electrons of the rod also move at this velocity $$\vec{\textbf{v}\space}\textbf{because of which}\\\textbf{it experiences a magnetic force.}$$
$$\vec{\text{F}_{b}} =\vec{\text{qv}}×\vec{\text{B}}$$
This force is towards Q to P.
Thus, the free electrons will move towards P and positive charge will appears at Q. An electrostatic field E is developed within the wire from Q to P. This field exerts a force.
$$\vec{\text{F}_{e}} =\vec{\text{qv}}×\vec{\text{B}} $$
on each free electron, the charge keeps on gathering until
$$\vec{\text{F}_{b}} = \vec{\text{F}_{e}}$$
$$\Rarr\qquad |q\vec{v}×\vec{\text{B}}| =|q\vec{\text{E}}|$$
vB = E
After this, resultant force on the free electrons of the wire PQ becomes zero. The potential difference between the ends Q and P is given by,
V = El = vBl
Thus, the potential difference is maintained by the magnetic force on the moving free electron and hence, produces an emf, e = Bvl
OR
Mutual inductance of two coils is equal to the e.m.f. induced in one coil when rate of change of current through the other coil is unity.
Mutual inductance of two coaxial solenoids :
Consider two long co-axial solenoid each of length l with number of turns N1 and N2 wound one over the other. Number of turns per unit length in solenoid,
$$n =\frac{\text{N}_{1}}{1}.\space\text{If I}_{1}\space\text{is the current flowing}$$
in primary solenoid, the magnetic field produced within this solenoid.
$$\text{B}_{1} =\frac{\text{I}_{0}\text{N}_{1}\text{I}_{1}}{l}$$
The flux linked with each turn of inner solenoid coil is Φ2 = B1A2, where A2 is the cross-sectional area of inner solenoid. The total flux linkage with inner coil of N2 turns.
Φ2 = N2Φ2
= N2B1A2
$$= \text{N}_{2}\bigg(\frac{\mu\text{N}_{1}\text{I}_{1}}{l}\bigg)\text{A}_{2}\\=\frac{\mu_{0}\text{N}_{1}\text{I}_{1}\text{A}_{2}}{l}\\\text{Mutual inductance,}\\\text{M}_{21} =\frac{\phi_{2}}{\text{I}_{1}}\\=\frac{\mu_{0}\text{N}_{1}\text{N}_{2}\text{I}_{1}\text{A}_{2}}{l.\text{I}_{1}}\\=\frac{\mu_{0}\text{N}_{1}\text{N}_{2}\text{A}_{2}}{l}$$
If n1 is number of turns per unit length of outer solenoid and r2 is radius of inner solenoid, then
M = μ0n1N2πr22
$$\textbf{23.}\space\vec{u_{i}} =-\bigg(\frac{e}{2m}\bigg)\vec{l}$$
It follows from the definition of μ1 and I.
$$\mu_{i} = \text{iA} =\bigg(\frac{-e}{\text{T}}\bigg)\pi r^{2}\\\text{...(i)}\\\text{l = mvr =m}\bigg(\frac{2\pi r}{\text{T}}\bigg)r\\\text{...(ii)}$$
Where, r is the radius of the circular orbit, which the electron of mass m and charge (– e) completes in time T.
Divide (i) by (ii),
$$\frac{\mu_{i}}{l} =\frac{-e}{\text{T}}\pi r^{2}×\\\frac{\text{T}}{\text{m×2}\pi r^{2}} =\frac{\normalsize-e}{2m}\\\vec{\mu_{i}} =-\bigg(\frac{e}{\text{2m}}\bigg)\vec{l}$$
$$\text{Clearly\space}\mu_{1}\space\text{and}\space\vec{l}\space\text{will be antiparallel}$$
(both being normal to the plane of the orbit).
$$\text{In contrast}\space\frac{\mu_{s}}{s}=\frac{e}{m}.$$
It is obtained on the basis of the quantum mechanics.
24.
| Radio waves | Gamma rays | |
| 1. | They are originated from rapid acceleration and deceleration of electrons. They are generated artificially by transmitter. | They are originated from the settling process of an excited nucleus of a radionuclide after it undergoes radio active decay. |
| 2. | The wavelengths of radio waves is greater than gamma rays . | The wavelength of gamma rays are smaller than radio waves. |
| 3. | Radio waves are used in communication and broadcasting system. | They are used in medicine (radiotherapy), industry (sterilization and disinfection) and the nuclear industry. |
25. Given: The radius of the innermost orbit of a hydrogen atom,
r1 = 5.3 × 10–11 m.
Consider r2 be the radius of the orbit at n = 2. It is related to the radius of the innermost orbit.
r2 = (n)2r1
= 4 × 5.3 × 10–11
= 2.12 × 10–10 m
For n = 3, r3 = (n)2 r1
= 9 × 5.3 × 10–11
= 4.77 × 10–10 m
Thus, the radii of an electron for n = 2 and n = 3 orbits are 2.12 × 10–10 m and 4.77 × 10–10 m respectively.
26. Forward biasing : If the positive terminal of the battery is connected to the p-type semiconductor and negative terminal to the n-type semiconductor then it is said to be forward biased.
Characteristics :
Reverse Biasing : If the positive terminal of the battery is connected to the n-type semiconductor and the negative terminal to the p-type semiconductor then it is said to be reverse biased.
Characteristics :
27. The measure of the ability of a conductor to store charges is known as capacitance
Q ∝ V
Q = CV
if V = 1, Then, C = Q
Capacitance of a conductor is defined as the charge required to raise its potential through one unit.
The S.I. unit of capacitance is farad or coulomb per volt. A farad is very large unit, therefore, pico and microfarads are commonly used units of capacitance.
The capacitance of a capacitor depends on the following factors :
(i) geometry of the plates.
(ii) distance between the capacitor plates and
(iii) nature of dielectric medium separating the plates.
The symbol of capacitor is
28. (a) The important property of photons that is useful in establishing Einstein’s photoelectric equation is their ability to hold on to the electrons of an atom by their forces of attraction.
(b) Einstein’s photoelectric equation states that.
Emax = hν – Φ
If Emax > 0 then hν – Φ > 0
or hν – hν0 > 0
Here, ν0 is the threshold frequency.
When v < v0 there will be no emission of electrons
Hence, n must greater than n0.
Again, loss in KE = gain in electromagnetic PE.
At v = v0 and Emax = eVs = 0.
where, Vs is stopping potential.
or Emax = eVs
where Vs represents the stopping potential.
OR
Einstein’s photoelectric equation is given by
$$\text{K}_{\text{max}} =\frac{1}{2}\text{mv}^{2}_{max}\\\text{K}_{max =}hv -\phi_{0}\\\text{or\space}hv = hv_{0} +\frac{1}{2}mv^{2}_{max}$$
where Kmax = Maximum kinetic energy of the photoelectron
vmax = Maximum velocity of the emitted photoelectron
m = Mass of the photoelectron
ν = Frequency of the light radiation
Φ0 = Work function
h = Planck’s constant
If ν0 is the threshold frequency, then the work function can be written as
W = Φ0 = hν0
$$\Rarr\space \text{K}_{max} =\frac{1}{2}\text{mv}^{2}\space_\text{max}\\=\space\text{hv - hv}_{0} =h(v-v_{0})$$
The above equations explains the following results:
(1) If ν < ν0, then the maximum kinetic energy is negative, which is impossible. Hence, photoelectric emission does not take place for the incident radiation below the threshold frequency. Thus, the photoelectric emission can take place if ν > ν0.
(2) The maximum kinetic energy of emitted photoelectrons is directly proportional to the frequency of the incident radiation. This means, that maximum kinetic energy of photoelectron depends only on the frequency of incident light not on the intensity. According to the photoelectric equation,
$$\text{K}_{max} =\frac{1}{2}\text{mv}^{2}_{\text{max}} \\=\text{hv -}\phi_{0}\\\text{K}_{1} =\frac{\text{hc}}{\lambda_{1}}-\phi_{0}\\\text{...(i)}$$
Let the maximum kinetic energy for the wavelength λ2 be K2.
$$\text{K}_{2}= \frac{hc}{\lambda_{2}}-\phi_{0}\\\text{...(ii)}$$
K2 = 2K1 (given)
From (i) and (ii), we have
$$\frac{\text{hc}}{\lambda_{2}}-\phi_{0} = 2\bigg(\frac{hc}{\lambda_{1}}-\phi_{0}\bigg)\\\Rarr\space\phi_{0} =\text{hc}\bigg(\frac{2}{\lambda_{1}}-\frac{1}{\lambda_{2}}\bigg)\\\Rarr\space \text{hv}_{0} =\text{hc}\bigg(\frac{2}{\lambda_{1}} -\frac{1}{\lambda_{2}}\bigg)\\\Rarr\space\frac{c}{\lambda_{0}} =c\bigg(\frac{2}{\lambda_{1}}-\frac{1}{\lambda_{2}}\bigg)\\\Rarr\space\frac{1}{\lambda_{0}} =\bigg(\frac{2}{\lambda_{1}}-\frac{1}{\lambda_{2}}\bigg)\\\Rarr\space\lambda_{0} =\bigg(\frac{\lambda_{1}\lambda_{2}}{2\lambda_{2}-\lambda_{1}}\bigg)$$
Work function is the energy required to eject a photoelectron from the metal.
$$\text{W} =\frac{hc}{\lambda_{0}}\\\therefore\space\\\text{W} =\frac{hc(2\lambda_{2}-\lambda_{1})}{\lambda_{1}\lambda_{2}}$$
Section-D
29. (i) (c) In Young's double slit experiment, the fringe width is
$$\beta =\frac{\text{D}\lambda}{d}\space\text{where D is the distance}\\\text{of the slits from the screen,}\\\text{d is the separation of the slits and λ,}$$
the wavelength. Therefore the fringe width β can be changed either by changing the separation between the sources or the distance of the screen from the sources.
(ii) (c) As the width of one of the slits is increased to 2w, the amplitude due to slit become 2a.
(iii) (d) Δx = (μA – 1)tA – (μB – 1)tB
= μAtA – μBtB – tA + tB = tB – tA
If Δx > 0, then fringe pattern will shift upward.
If Δx < 0, then fringe pattern will shift downwards.
(iv) (b) Contrast between the bright and dark fringes will be reduced.
OR
(a) Since, one of the slit is covered, interference will not occur and fringe pattern will disappear.
30.
(i) (c) In an ideal transformer, there is no power loss. The efficiency of an ideal transformer is h = 1 (i.e. 100%) i.e. input power = output power.
OR
(d) Transformer is used to obtain desired ac voltage and current.
(ii) (c) For a transformer,
$$\frac{\text{V}_{s}}{\text{V}_{P}} =\frac{\text{N}_{s}}{\text{N}_{P}}$$
where N denotes number of turns and V = voltage.
$$\therefore\space\frac{\text{V}_{s}}{\text{V}_{p}} =\frac{\text{N}_{s}}{\text{N}_{p}}$$
where N denotes number of turns and V = voltage.
$$\therefore\space\frac{\text{V}_{s}}{220} =\frac{10}{20}\\\therefore\space\text{V}_{s} =110\text{V}$$
(iii) (a) In a transformer the primary and secondary currents are related by
$$\text{I}_{s} =\bigg(\frac{\text{N}_{p}}{\text{N}_{s}}\bigg)\text{I}_{p}$$
and the voltages are related by
$$\text{V}_{s} =\bigg(\frac{\text{N}_{s}}{\text{N}_{p}}\bigg)\text{V}_{p}$$
where subscripts p and s refer to the primary and secondary of the transformer.
$$\text{here,\space}\text{V}_{p} =\text{V}_{s}\frac{\text{N}_{p}}{\text{N}_{s}} = 4\\\therefore\space\text{I}_{s} =4\text{I}_{p}\\\text{and}\space\text{V}_{s}=\bigg(\frac{1}{4}\bigg)\text{V} =\frac{\text{V}}{4}$$
(iv) (c) The efficiency of the transformer is
$$\eta =\frac{\text{Output power}(\text{P}_{\text{out}})}{\text{Input power}(\text{P}_{\text{in}})}×100$$
Here, Pout = 100 W,
Pin = (220 V) (0.5 A) = 110 W
$$\therefore\space\eta =\frac{100\space\text{W}}{110\space\text{W}}×100 = 90\%$$
Section-E
31. (a) Wave front: A wave front is the locus of all the points in space that reach a particular distance by a propagating wave in same phase at any instant.
Huygen’s principle: It is based on two assumptions:
(i) Each point of the wavefront behaves like a source of secondary disturbances and secondary
wavelets from there points spread out in all directions with the same speed as that of the original wave front.
(ii) When we draw an envelope in the forward direction of the secondary disturbances at any instant, And this envelope tells the new position of the wavefront at that instant.
(b) (i) Both the reflection and refraction takes place due to the interaction of light with the atoms at the surface of the separation. Light incident on these atoms, force them to vibrate with the frequency of light. But, the light emitted by these charged atoms is equal to their own frequency of oscillation. So, both the reflected and refracted lights have same frequency No, Hence, frequency remains changed.
(ii) For a given frequency the energy carried by a wave depends on the amplitude of the wave. It does not depend on the speed of the wave propagation. Hence the energy of the wave remains same and does not decrease.
(iii) The intensity of light is determined by the number of photons incident per unit area around the point at which intensity is to be determined.
OR
(a) In Δ ABC,
$$\text{sin\space i}=\frac{\text{BC}}{\text{AC}} =\frac{v_{1}t}{\text{AC}}\\\text{sin r} =\frac{\text{AE}}{\text{AC}} =\frac{v_{2}t}{\text{AC}}\\\frac{\text{sin i}}{\text{sin r}} =\frac{\text{BC}}{\text{AE}}\\=\frac{v_{1}t}{v_{2}t}=\frac{v_{1}}{v_{2}}\\\mu_{1} =\frac{c}{v_{1}}\\\mu_{2} =\frac{c}{v_{2}}\\\therefore\space\frac{\mu_{2}}{\mu_{1}} =\frac{v_{1}}{v_{2}}\\\therefore\space \frac{\text{sin i}}{\text{sin r}} =\frac{\mu_{2}}{\mu_{1}}$$
or μ2 sin r = μ1 sin i
It is Snell’s law.
(b) The refractive index of a medium is defined as the ratio of speed of light in vacuum to that of the speed of the light in medium.
$$\text\space {i.e\space} \space n=\frac{c}{v}$$
Where, n = refractive index of medium when light ray passes from v acuum into a medium.
c = velocity of light in vacuum
v = velocity of light in the medium
Proof of Snell’s law of refraction
When a wavefront travels from one medium to other, it deviates from its path. In travelling from one medium to other, the frequency of wave remains same and speed and wavelength changes. Let, XY be a surface separating two media ‘1’ and ‘2’. Let the speed of waves of v1 and v2.
Suppose, a plane wavefront AB in first medium is incident obliquely on the boundary surface XY and its end touches the surface at A at time t = 0, while the other end B reaches the surface at point B’ after time-interval ‘t’ Clearly, BB’ = v1t.
In the same time, wavelets starts from A and reaches A’ in time ‘t’ with velocity v2.
Therefore, AA’ = v2 t
According to Huygen’s principle, A’B’ is the new position of the wavefront in second medium. A’B’ is a refracted wavefront.
Let, the incident wavefront (AB) and refracted wavefront (A’B’) makes angle i and r with surface XY.
In ΔAB’B,
∠ABB’ = 90°
$$\therefore\space\text{sin t} =\frac{\text{BB'}}{\text{AB'}} =\frac{v_{1}t}{\text{AB'}}\\\text{...(i)}$$
In ΔAA’B’,
∠AA’B’ = 90º
$$\therefore\space\text{sin r} =\frac{\text{AA'}}{\text{AB'}} =\frac{v_{2}t}{\text{AB'}}\\\text{...(ii)}$$
Dividing equation (i) by (ii), we get
$$\frac{\text{sin i}}{\text{sin r}} =\frac{v_{1}}{v_{2}} =\text{constant}$$
Hence, ratio of sine of angle of incidence and the sine of angle of refraction for a given pair of media is constant. This is Snell’s law of refraction.
32. (a) (i) Drift velocity is defined as the average velocity with which the free electrons are drifted towards the positive terminal under the effect of applied electric field. Thermal velocities are randomly distributed and average thermal velocity is zero.
$$\frac{\vec{u}_{1} +\vec{u}_{2}+.... +\vec{u_{\text{N}}}}{\text{N}} = 0\\v_{d} =-\frac{e\text{ET}}{m}$$
(ii) We know that the current flowing through the conductor is :
I = nAevd
$$\therefore\space\text{I = neA}\bigg(-\frac{e\text{Et}}{m}\bigg)\\\text{Using\space}\text{E} =-\frac{\text{V}}{\text{t}}\\\text{I = neA}\bigg(\frac{\text{eV}}{\text{ml}}\bigg)\tau\\=\bigg(\frac{ne^{2}\text{A}\tau}{\text{ml}}\bigg)\text{V} =\frac{1}{\text{R}}\text{V}$$
I ∝ V → Which is Ohm’s law.
$$\text{Where R =}\frac{ml}{\text{nAe}^{2}\tau}\space\text{is a constant for}$$
a particular conductor at a particular temperature and is called the resistance of the conductor.
$$\text{R} =\bigg(\frac{m}{ne^{2}\tau}\bigg)\frac{l}{\text{A}} =\frac{pl}{\text{A}}\\\rho =\bigg(\frac{m}{ne^{2}\tau}\bigg)$$
Where, ρ is the specific resistance or resistivity of the material of the wire. It depends on number of free electron per unit volume and temperature.
(b) The equivalent resistance between A and C,
$$\text{R' =}\frac{\text{R}\frac{\text{R}_{0}}{2}}{\text{R + }\frac{\text{R}_{0}}{2}} =\frac{\text{RR}_{0}}{\text{2R + R}_{0}}$$
Equivalent resistance between A and B,
$$\text{R}_{\text{eq}} = \frac{\text{R}\text{R}_{0}}{\text{2R +R}_{0}} +\frac{\text{R}_{0}}{2}\\\text{R}_{\text{eq}} =\frac{\text{R}_{0}(\text{4R + R}_{0})}{2(\text{2R +R}_{0})}$$
Current in the circuit,
$$\text{I =}\frac{\text{V}}{\text{R}_{\text{eq}}}\\=\frac{\text{V}.2(\text{2R +R}_{0})}{\text{R}_{0}(\text{4R} + \text{R}_{0})}$$
∴ Voltage across R,
VR = I.R′
$$\text{V}_{\text{R}} =\frac{\text{V}.2(\text{2R + R}_{0})}{\text{R}_{0}(\text{4R +R}_{0})}×\\\frac{\text{RR}_{0}}{\text{2R + R}_{0}}\\\text{V}_{\text{R}} =\frac{\text{2\text{VR}}}{\text{4R} +\text{R}_{0}}$$
OR
(a) (i) Conductivity of a metallic wire is defined as its ability to allow electric charges or heat to pass through it.
Numerically, conductivity is reciprocal of resistivity.
SI unit : ohm– 1 m–1 or mho m–1 or Sm–1.
(ii) Consider a potential difference V be applied across a conductor of length l and cross-section A.
Electric field inside the conductor,
$$\text{E} =\frac{\text{V}}{\text{l}}.$$
Due to the external field the free electrons inside the conductor drift with velocity vd.
Let, number of electrons per unit volume = n, charge on an electron = e
∴Total electrons in length l = nAl
and Total charge q = neAl
Time taken by electrons to enter and leave the conductor.
$$t = \frac{1}{v_{d}}\\\text{Current, I} =\frac{q}{t} =\frac{\text{neAI}}{\frac{\text{I}}{v_{d}}}$$
= neAvd
$$\text{Current density, J} =\frac{\text{I}}{\text{A}}\\\text{= nev}_{d}$$
We know,
$$v_{d} =\frac{e\text{Et}}{m} =\frac{e\text{V}\tau}{ml}\\\begin{bmatrix}\because\space \text{E}\tau =\frac{\text{Vt}}{l}\end{bmatrix}\\\therefore\space \text{I = neAv}_{d} =\frac{\text{neAVe}\tau}{\text{ml}}\\\Rarr\space\frac{\text{V}}{\text{I}} = \frac{\text{ml}}{\text{ne}^{2}\tau \text{A}}\\\Rarr\space\text{R} =\frac{\text{ml}}{\text{ne}^{2}\tau \text{A}}\\\begin{bmatrix}\because\space\frac{\text{V}}{\text{I}} =\text{R};\space \text{Ohm's law}\end{bmatrix}\\\text{Resistivity;}\space\rho =\frac{\text{RA}}{l}\\=\frac{mlA}{lne^{2}\tau \text{A}}\\\text{or\space}\rho =\frac{m}{ne^{2}\tau}$$
$$\text{Since conductivity,}\sigma =\frac{\text{I}}{\rho}\\\therefore\space \sigma =\frac{ne^{2}\tau}{m}\\\text{…(ii)}$$
Relation between current density and field :
For an electron, charge q = – e.
and current density,
$$\text{J =}\frac{\text{I}}{\text{A}} =-\text{nev}_{d}\\\text{[from (i)]}\\\text{J = (- ne)}\bigg(\frac{-\text{eE}\tau}{m}\bigg) \\=\bigg(\frac{ne^{2}\tau}{m}\bigg)\text{E}$$
$$\Rarr\space\text{J =}\sigma\text{E}$$
[from (ii)]
which is the required relation.
(b) All free electrons suffer collisions with the heavy fixed ions inside the conductor. After collisions, these electrons again moves with the same speed, but in random directions. So, at given time, net velocity of the electrons is zero i.e.,
$$\vec{u_{avg}} =\frac{\vec{u}_{1} +\vec{u}_{2} +......\vec{u}_{n}}{n} =0$$
If the electric field established inside the conductor, electrons get accelerated, so
$$a =-\frac{e\text{E}}{m}$$
Now, the average velocity of all electrons is given by
$$\vec{u}_{d} = \vec{u}_{avg}-\frac{\epsilon\text{E}}{m}\tau\\\vec{u_{d}} = 0 -\frac{\epsilon\text{E}}{m}\tau$$
where τ is the average relaxation time.
$$\text{But\space E} =\frac{\text{V}}{\text{l}}\\\vec{u_{d}} = -\frac{\epsilon\text{V}\tau}{\text{ml}}\\u_{d}∝\frac{1}{t}$$
As the drift velocity is inversely proportional to its length, so the drift velocity would be reduced by onethird if the length of the conductor is tripled.
33. (a) Initially when there is vacuum between the two plates, the capacitance of the two parallel plates is
$$\text{C} =\frac{\epsilon_{0}\text{A}}{d}$$
Here, A is area of plates and d is separation between the plates.
Let capacitor is connected with battery. Thus electric field inside the capacitor will be produced and it is given by E.
As soon as dielectric slab of thickness
$$t =\frac{d}{2}\space\text{inserted the electric field}$$ inside slab (t) will be E0 and in (d – t) will be E0, (E0 = kE).
Now potential difference between the plates will be as follows :
V = Et + E0 (d – t)
$$\begin{bmatrix}\text{as t} =\frac{d}{2}\end{bmatrix}\\ =\frac{\text{Ed}}{2} + \frac{\text{E}_{0}d}{2}=\frac{\text{d}}{2}(\text{E + E}_{0})\\=\frac{d}{2}\bigg(\frac{\text{E}_{0}}{k} +\text{E}_{0}\bigg)\\\begin{bmatrix}\because\space\frac{\text{E}_{0}}{\text{E}} =k\end{bmatrix}\\=\frac{\text{E}_{0}d}{2}\frac{(k+1)}{k}$$
$$\text{Now,\space}\text{E}_{0}=\frac{\sigma}{\epsilon_{0}} =\frac{\sigma}{\epsilon_{0}\text{A}}\\\text{V} =\frac{d}{2k}\frac{q}{\epsilon_{0}\text{A}}(\text{k+1})\\\text{As}\space\text{C} =\frac{q}{\text{V}}\\\text{Thus,\space C =}\frac{2k\epsilon_{0}\text{A}}{\text{d}(\text{k+1})}$$
(b) (i) Initial voltage, V1 = V volts and charge stored
Q1 = 360 mC
Q1 = CV1 …(i)
Changed potential,
V2 = V – 120
Q2 = 120 mC
Q2 = CV2 …(ii)
By applying (i) divided by (ii), we get
$$\frac{\text{Q}_{1}}{\text{Q}_{2}} =\frac{\text{CV}_{1}}{\text{CV}_{2}}\\\Rarr\space \frac{360}{120} =\frac{\text{V}}{\text{V - 120}}\\\Rarr\space \text{V = 180 V}\\\text{Hence, C =}\frac{\text{Q}_{1}}{\text{V}_{1}}\\=\frac{360×10^{\normalsize-6}}{180}$$
= 2 × 10–6 F
= 2 μF
(ii) If the voltage applied had increased by 120 V, then
V3 = 180 + 120 = 300 V
Hence charge stored in capacitor,
Q3 = CV3
= 2 × 10–6 × 300 = 600 μC
OR
(a) Let ‘q’ be the charge on the charged capacitor.
Energy stored in it is,
$$\text{U =}\frac{q^{2}}{\text{2C}}$$
When another similar uncharged capacitor is connected, the net capacitance of the system is,
C' = 2C
The charge on the system is constant. So, the energy stored in the system now is,
$$\text{U'} =\frac{q^{2}}{\text{2(C')}^{2}}\\\Rarr\space\text{U' =}\frac{q^{2}}{\text{2(2C)}}\\\Rarr\space \text{U'} =\frac{q^{2}}{\text{4C}}$$
Thus, the required ratio is,
$$\frac{\text{U'}}{\text{U}} =\frac{\frac{q^{2}}{\text{4C}}}{\frac{q^{2}}{\text{2C}}}\\\Rarr\space \frac{\text{U'}}{\text{U}} =\frac{1}{2}$$
(b) When the capacitors are connected in parallel.
Equivalent capacitance, Cp = C1 + C2
The energy stored in the combination of the capacitors,
$$\text{E}_{p} =\frac{1}{2}\text{C}_{p}\text{V}^{2}\\\Rarr\space\text{E}_{p} =\frac{1}{2}(\text{C}_{1} +\text{C}_{2})(100)^{2} \\= 0.25\text{J}\\\Rarr\space\text{(C}_{1} + \text{C}_{2}) = 5×10^{\normalsize-5}\\\text{...(i)}$$
When the capacitors are connected in series.
Equivalent capacitance,
$$\text{C}_{s} =\frac{\text{C}_{1}\text{C}_{2}}{\text{C}_{1} +\text{C}_{2}}$$
The energy stored in the combination of the capacitors,
$$\text{E}_{s} =\frac{1}{2}\text{CV}^{2}\\\Rarr\space \text{E}_{s} =\frac{1}{2}\frac{\text{C}_{1}\text{C}_{2}}{\text{C}_{1} +\text{C}_{2}}(100)^{2}\\ =0.045\space\text{J}\\\Rarr\space\frac{1}{2}×\frac{\text{C}_{1}\text{C}_{2}}{5×10^{\normalsize-5}}×(100)^{2}\\ =0.045\space\text{J}\\\Rarr\space \text{C}_{1}\text{C}_{2} = 0.045×10^{\normalsize-4}×10^{\normalsize-4}$$
= 4.5 × 10–10
(C1 – C2)2 = (C1 + C2)2 – 4C1C2
$$\Rarr\space (\text{C}_{1} -\text{C}_{2}) =\sqrt{ 7×10^{\normalsize-10}}\\=2.64×10^{\normalsize-5}$$
C1 – C2 = 2.64 × 10–5 …(ii)
Solving (i) and (ii), we get
C1 = 38.2 μF and C2 = 11.8 μF
When the capacitors are connected in parallel, the charge on each of them can be obtained as follows :
Q1 = C1V = 38.2 × 10– 6 × 100
= 38.2 × 10–4 C
Q2 = C2V = 11.8 × 10– 6 × 100
= 11.8 × 10–4 C
CBSE 36 Sample Question Papers Science Stream (PCM)
All Subjects Combined for Class 12 Exam 2024
CBSE 36 Sample Question Papers Science Stream (PCB)
All Subjects Combined for Class 12 Exam 2024
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