CBSE 12th Standard Physics Subject Case Study Questions With Solution 2021 Part - II
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CBSE 12th Standard Physics Subject Case Study Questions With Solution 2021 Part - II
12th Standard CBSE
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Reg.No. :
Physics
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In 1909, Robert Millikan was the first to find the charge of an electron in his now-famous oil-drop experiment. In that experiment, tiny oil drops were sprayed into a uniform electric field between a horizontal pair of oppositely charged plates. The drops were observed with a magnifying eyepiece, and the electric field was adjusted so that the upward force on some negatively charged oil drops was just sufficient to balance the downward force of gravity. That is, when suspended, upward force qE just equaled Mg. Millikan accurately measured the charges on many oil drops and found the values to be whole number multiples of 1.6 x 10-19 C the charge of the electron. For this, he won the Nobel prize.
(i) If a drop of mass 1.08 x 10-14 kg remains stationary in an electric field of 1.68 x 105 N C-I, then the charge of this drop is(a) 6.40 x 10-19 C (b) 3.2 x 10-19 C (c) 1.6 X 10-19 C (d) 4.8 x 10-19 C (ii) Extra electrons on this particular oil drop (given the presently known charge of the electron) are
(a) 4 (b) 3 (c) 5 (d) 8 (iii) A negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field 100 V m-1.If the mass of the drop is 1.6 X 10-3 g, the number of electrons carried by the drop is (g= 10 m s-2)
(a) 1018 (b) 1015 (c) 1012 (d) 109 (iv) The important conclusion given by Millikan's experiment about the charge is
(a) charge is never quantized (b) charge has no definite value (c) charge is quantized (d) charge on oil drop always increases. (v) If in Millikan's oil drop experiment, charges on drops are found to be \(8 \mu \mathrm{C}, 12 \mu \mathrm{C}, 20 \mu \mathrm{C}\) then quanta of charge is
\(\text { (a) } 8 \mu \mathrm{C}\) \(\text { (b) } 20 \mu \mathrm{C}\) \(\text { (c) } 12 \mu \mathrm{C}\) \(\text { (d) } 4 \mu \mathrm{C}\) -
This energy possessed by a system of charges by virtue of their positions. When two like charges lie infinite distance apart, their potential energy is zero because no work has to be done in moving one charge at infinite distance from the other.
In carrying a charge q from point A to point B, work done \(W=q\left(V_{A}-V_{B}\right)\). This work may appear as change in KE/PE of the charge. The potential energy of two charges q1 and q2 at a distance r in air is \(\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} r}\). It is measured in joule. It may be positive, negative or zero depending on the signs of ql and q2.
(i) Calculate work done in separating two electrons form a distance of 1m to 2m in air, where e is electric charge and k is electrostatic force constant.(a) ke2 (b) e2/2 (c) -ke2/2 (d) zero (ii) Four equal charges q each are placed at four corners of a square of side a each. Work done in carrying a charge -q from its centre to infinity is
(a) zero \(\text { (b) } \frac{\sqrt{2} q^{2}}{\pi \varepsilon_{0} a}\) \(\text { (c) } \frac{\sqrt{2} q}{\pi \varepsilon_{0} a}\) \(\text { (d) } \frac{q^{2}}{\pi \varepsilon_{0} a}\) (iii) Two points A and B are located in diametrically opposite directions of a point charge of +2 \(\mu \mathrm{C}\) at distances 2 m and 1 m respectively from it. The potential difference between A and B is
(a) 3 x 103 V (b) 6 x 104 V (c) -9 X 103 V (d) -3 x 103 V (iv) Two point charges A = +3 nC and B = +1 nC are placed 5 ern apart in air. The work done to move charge B towards A by 1 cm is
(a) 2.0 x 10-7 J (b) 1.35 x 10-7 J (c) 2.7 X 10-7 J (d) 12.1 x 10-7 J (v) A charge Q is placed at the origin. The electric potential due to this charge at a given point in space is V. The work done by an external force in bringing another charge q from infinity up to the point is
\(\text { (a) } \frac{V}{q}\) (b) Vq (c) V + q (d) V -
Metals have a large number of free electrons nearly 1028 per cubic metre. In the absence of electric field, average terminal speed of the electrons in random motion at room temperature is of the order of 105 m s-1 When a potential difference V is applied across the two ends of a given conductor, the free electrons in the conductor experiences a force and are accelerated towards the positive end of the conductor. On their way, they suffer frequent collisions with the ions/atoms of the conductor and lose their gained kinetic energy. After each collision, the free electrons are again accelerated due to electric field, towards the positive end of the conductor and lose their gained kinetic energy in the next collision with the ions/atoms of the conductor. The average speed of the free electrons with which they drift towards the positive end of the conductor under the effect of applied electric field is called drift speed of the electrons.
(i) Magnitude of drift velocity per unit electric field is(a) current density (b) current (c) resistivity (d) mobility (ii) The drift speed of the electrons depends on
(a) dimensions of the conductor (b) number density of free electrons in the conductor (c) both (a) and (b) (d) neither (a) nor (b) (iii) We are able to obtain fairly large currents in a conductor because
(a) the electron drift speed is usually very large (b) the number density of free electrons is very high and this can compensate for the low values of the 6 electron drift speed and he very small magnitude of the electron charge (c) the number density of free electrons as well as the electron drift speeds are very large and these compensate for the very small magnitude of the electron charge (d) the very small magnitude of the electron charge has to be divided by the still smaller product of the number density and drift speed to get the electric current (iv) Drift speed of electrons in a conductor is very small i.e., i = 10-4 m s-1. The Electric bulb glows immediately. When the switch is closed because
(a) drift velocity of electron increases when switch is closed (b) electrons are accelerated towards the negative end of the conductor (c) the drifting of electrons takes place at the entire length of the conductor (d) the electrons of conductor move towards the positive end and protons of conductor move towards negative end of the conductor (v) The number density offree electrons in a copper conductor is 8.5 x 1028 m-3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 x 10-6m2 and it is carrying a current of 3.0 A.
(a) 8.1 x 104 s (b) 2.7 x 104 s (c) 9 x 103 s (d) 3 x 103 S -
Various methods can be used to measure the mass of an atom. One possibility is through the use of a mass spectrometer. The basic feature of a Banbridge mass spectrometer is illustrated in figure. A particle carrying a charge +q is first sent through a velocity selector and comes out with velocity v = E/B.
The applied electric and magnetic fields satisfy the relation E = vB so that the trajectory of the particle is a straight line. Upon entering a region where a second magnetic field \(\vec{B}_{0}\) pointing into the page has been applied, the particle will move in a circular path with radius r and eventually strike the photographic plate.
(i) In mass spectrometer, the ions are sorted out in which of the following ways?(a) By accelerating them through electric field (b) By accelerating them through magnetic field (c) By accelerating them through electric and magnetic field (d) By applying a high voltage (ii) Radius of particle in second magnetic field Bo is
\(\text { (a) } \frac{2 m v}{q E_{0}}\) \(\text { (b) } \frac{m v}{q E_{0}}\) \(\text { (c) } \frac{m v}{q B_{0}}\) \(\text { (d) } \frac{2 m E_{0} v}{q B_{0}}\) (iii) Which of the following will trace a circular trajectory wit largest radius?
(a) Proton (b) -\(\alpha\)particle (c) Electron (d) A particle with charge twice and mass thrice that of electron (iv) Mass of the particle in terms q, Bo, B,r and E is
\(\text { (b) } \frac{q B_{0} B r}{E}\) \(\text { (c) } \frac{q B r}{E B_{0}}\) \(\text { (d) } \frac{q B r E}{B_{0}}\) (v) The particle comes out of velocity selector along a straight line, because
(a) electric force is less than magnetic force (b) electric force is greater than magnetic force (c) electric and magnetic force balance each other (d) can't say. -
When the atomic dipoles are aligned partially or fully, there is a net magnetic moment in the direction of the field in any small volume of the material. The actual magnetic field inside material placed in magnetic field is the sum of the applied magnetic field and the magnetic field due to magnetisation. This field is called magnetic intensity (H).
\(H=\frac{B}{\mu_{0}}-M\)
where M is the magnetisation of the material, llo is the permittivity of vacuum and B is the total magnetic field. The measure that tells us how a magnetic material responds to an external field is given by a dimensionless quantity is appropriately called the magnetic susceptibility: for a certain class of magnetic materials, intensity of magnetisation is directly proportional to the magnetic intensity.
(i) Magnetization of a sample is(a) volume of sample per unit magnetic moment (b) net magnetic moment per unit volume (c) ratio of magnetic moment and pole strength (d) ratio of pole strength to magnetic moment (ii) Identify the wrongly matched quantity and unit pair.
(a) Pole strength Am (b) Magnetic susceptibility dimensionless number (c) Intensity of magnetisation A m-1 (d) Magnetic permeability Henry m (iii) A bar magnet has length- 3 cm, cross-sectional area 2 cm2 and magnetic moment 3 A m2. The intensity of magnetisation of bar magnet is
\(\text { (a) } 2 \times 10^{5} \mathrm{~A} / \mathrm{m}\) \(\text { (b) } 3 \times 10^{5} \mathrm{~A} / \mathrm{m}\) \(\text { (c) } 4 \times 10^{5} \mathrm{~A} / \mathrm{m}\) \(\text { (d) } 5 \times 10^{5} \mathrm{~A} / \mathrm{m}\) (iv) A solenoid has core of a material with relative permeability 500 and its windings carry a current of 1 A. The number of turns of the solenoid is 500 per metre. The magnetization of the material is nearly
\(\text { (a) } 2.5 \times 10^{3} \mathrm{Am}^{-1}\) \(\text { (b) } 2.5 \times 10^{5} \mathrm{~A} \mathrm{~m}^{-1}\) \(\text { (c) } 2.0 \times 10^{3} \mathrm{~A} \mathrm{~m}^{-1}\) \(\text { (d) } 2.0 \times 10^{5} \mathrm{~A} \mathrm{~m}^{-1}\) (v) The relative permeability of iron is 6000. Its magnetic susceptibility is
(a) 5999 (b) 6001 (c) 6000 x 10-7 (d) 6000 x 107 -
In year 1820 Oersted discovered the magnetic effect of current. Faraday gave the thought that reverse of this phenomenon is also possible i.e., current can also be produced by magnetic field. Faraday showed that when we move a magnet towards the coil which is connected by a sensitive galvanometer. The galvanometer gives instantaneous deflection showing that there is an electric current in the loop.
Whenever relative motion between coil and magnet takes place an emf induced in coil. If coil is in closed circuit then current is also induced in the circuit. This phenomenon is called electromagnetic induction.
(I) The north pole of a long bar magnet was pushed slowly into a short solenoid connected to a galvanometer. The magnet was held stationary for a few seconds with the north pole in the middle of the solenoid and then withdrawn rapidly. The maximum deflection of the galvanometer was observed when the magnet was(a) moving towards the solenoid (b) moving into the solenoid (c) at rest inside the solenoid (d) moving out of the solenoid. (ii) Two similar circular loops carry equal currents in the same direction. On moving the coils further apart, the electric current will
(a) remain unaltered (b) increases in one and decreases in the second (c) increase in both (d) decrease in both (iii) A closed iron ring is held horizontally and a bar magnet is dropped through the ring with its length along the axis of the ring. The acceleration of the falling magnet is
(a) equal to g (b) less than g (c) more than g (d) depends on the diameter of the ring and length of magnet (iv) Whenever there is a relative motion between a coil and a magnet, the magnitude of induced emf set up in the coil does not depend upon the
(a) relative speed between the coil and magnet (b) magnetic moment of the coil (c) resistance of the coil (d) number of turns in the coil (v) A coil of metal wire is kept stationary in a non-uniform magnetic field
(a) an emf and current both are induced in the coil (b) a current but no emf is induced in the coil (c) an emf but no current is induced in the coil (d) neither emf nor current is induced in the coil -
A transformer is essentially an a.c. device. It cannot work on d.c. It changes alternating voltages or currents. It does not affect the frequency of a.c. It is based on the phenomenon of mutual induction. A transformer essentially consists of two coils of insulated copper wire having different number of turns and wound on the same soft iron core.
The number of turns in the primary and secondary coils of an ideal transformer are 2000 and 50 respectively. The primary coil is connected to a main supply of 120 V and secondary coil is connected to a bulb of resistance \(0.6 \Omega\)
(i) The value of voltage across the secondary coil is(a) 5V (b) 2V (c) 3 V (d) 10 V (ii) The value of current in the bulb is
(a) 7 A (b) 15 A (c) 3 A (d) 5 A (iii) The value of current in primary coil is
(a) 0.125 A (b) 2.52 A (c) 1.51 A (d) 3.52 A (iv) Power in primary coil is
(a) 20W (b) 5W (c) 10 W (d) 15W (v) Power in secondary coil is
(a) 15W (b) 20 W (c) 7W (d) 8 W -
Electrons oscillating in a circuit give rise to radiowaves. A transmitting antenna radiates most effectively the radiowaves of wavelength equal to the size of the antenna. The infrared waves incident on a substance set into oscillation all its electrons, atoms and molecules. This increases the internal energy and hence the temperature of the substance.
(i) If vg, vx and vm are the speeds of gamma rays, X-rays and microwaves respectively in vacuum, then(a) Vg > VX > Vm (b) Vg (c) Vg > VX > Vm (d) Vg= VX= Vm (ii) Which of the following will deflect in electric field?
(a) X-rays (b) \(\Upsilon\)-rays (c) cathode rays (d) ultraviolet rays (iii) \(\Upsilon\)-rays are detected by
(a) point contact diodes (b) thermopiles (c) ionization chamber (d) photocells (iv) The frequency of electromagnetic wave, which best suited to observe a particle of radius 3 x 10-4 cm is the order of
(a) 1015Hz (b) 1014Hz (c) 1013Hz (d) 1012Hz (v) We consider the radiation emitted by the human body. Which one of the following statements is true?
(a) The radiation emitted is in the infrared region. . (b) The radiation is emitted only during the day. (c) The radiation is emitted during the summers and absorbed during the winters (d) The radiation emitted lies in the ultraviolet region and hence it is not visible -
An astronomical telescope is an optical instrument which is used for observing distinct images of heavenly bodies libe stars, planets etc. It consists of two lenses. In normal adjustment of telescope, the final image is formed at infinity. Magnifying power of an astronomical telescope in normal adjustment is defined as the ratio of the angle subtended at the eye by the angle subtended at the eye by the final image to the angle subtended at the eye, by the object directly, when the final image and the object both lie at infinite distance from the eye. It is given by,\(m=\frac{f_{0}}{f_{e}}\) To increase magnifying power of an astronomical telescope in normal adjustment, focal length of objective lens should be large and focal length of eye lens should be small.
(i) An astronomical telescope of magnifying power 7 consists of the two thin lenses 40 cm apart, in normal adjustment. The focal lengths of the lenses are(a) 5cm,35cm (b) 7cm,35cm (c) 17cm,35cm (d) 5cm,30cm (ii) An astronomical telescope has a magnifying power of 10. In normal adjustment, distance between the objective and eye piece is 22 cm. The focal length of objective lens is
(a) 25 cm (b) 10 cm (c) 15 cm (d) 20 cm (iii) In astronomical telescope compare to eye piece, objective lens has
(a) negative focal length (b) zero focal length (c) small focal length (d) large focal length (iv) To see stars, use
(a) simple microscope (b) compound microscope (c) endoscope (d) astronomical telescope (v) For large magnifying power of astronomical telescope
\((a) f_{v}< \((b) f_{v}=f_{\mathrm{e}}\) \((c) f_{o}>>f_{\mathrm{e}}\) (d) none of these -
Consider the situation shown in figure. The two slits S1 and S2 placed symmetrically around the central line are illuminated by monochromatic light of wavelength \(\lambda\). The separation between the slits is d. The light transmitted. by the slits falls on a screen S0 place at a distance D from the slits. The slits S3 is at the central line and the slit S4 is at a distance from S3. Another screen Sc is placed a further distance D away from Sc.
(i) Find the path difference if \(z=\frac{\lambda D}{2 d}\)\((a) \lambda\) \((b) \lambda / 2\) \((c) 3 / 2 \lambda\) \((d) 2 \lambda\) (ii) Find the ratio of the maximum to minimum intensity observed on \(S_{c} \text { if } z=\frac{\lambda D}{d}\)
(a) 4 (b) 2 (c) \(\infty\) (d) 1 (iii) Two coherent point sources S1 and S2 are separated by a small distance d as shown in figure. The fringes obtained on the screen will be
(a) Concentric Circles (b) points (c) Straight lines (d) semi-circles (iv) In the case of light waves from two coherent sources S1 and S2. there will be constructive interference at an arbitrary point P, if the path difference S1P - S2P is
\(\text { (a) }\left(n+\frac{1}{2}\right) \lambda \) \( \text { (b) } n \lambda\) \( \text { (c) }\left(n-\frac{1}{2}\right) \lambda\) \( \text { (d) } \frac{\lambda}{2}\) (v) Two monochromatic light waves of amplitudes 3A and 2A interfering at a point have a phase difference of 60°. The intensity at that point will be proportional to
(a) 5A2 (b) 13A2 (c) 7A2 (d) 19A2 -
If we allow radiations of a fixed frequency to fall on plate and the accelerating potential difference between the two electrodes is kept fixed, then the photoelectric current is found to increase linearly with the intensity of incident radiation. Here, radiation pressure is P = \(\left(\frac{1+e}{C}\right) I\). As, atmosphere pressure at sea level is 105Pa. If the intensity of light of a given wavelength, is increased, there is an increase in the number of photons incident on a given area in a given time. But the energy of each photon remain the same.
(i) The number of photons hitting the cone second\(\text { (a) } \pi R^{2} I / 2 E\) \(\text { (b) } 2 \pi R^{2} I / E\) \(\text { (c) } \pi R^{2} I / 4 E\) \(\text { (d) } \pi R^{2} I / E\) (ii) A radiation of energy E falls normally on a perfect reflecting surface. The momentum transferred to the surface is
\(\text { (a) } \frac{E}{c}\) \(\text { (b) } \frac{2 E}{c}\) \(\text { (c) } E c\) \(\text { (d) } \frac{E}{c^{2}}\) (iii) Which one is correct?
\(\text { (a) } E^{2}=p^{2} c^{2}\) \(\text { (b) } E^{2}=p^{2} c\) \(\text { (c) } E^{2}=p^{2}\) \(\text { (d) } E^{2}=\frac{p^{2}}{c^{2}}\) (iv) The incident intensity on a horizontal surface at sea level from the Sun is about 1 k W m-2. Assuming that 50% of this intensity is reflected and 50% is absorbed, determine the radiation pressure on this horizontal surface.
(a) 8.2 x 10-2 Pa (b) 5 x 10-6 Pa (c) 3 x 10-5 Pa (d) 6 x 10-5 Pa (v) Find the ratio of radiation pressure to atmospheric pressure P0 about 1 x 105 Pa at sea level.
(a) 5 x 10-11 (b) 4 x 10-8 (c) 6 x 10-12 (d) 8 x 10-11 -
In 1911, Rutherford, along with his assistants, H. Geiger and E. Marsden, performed the alpha particle scattering experiment. H. Geiger and E. Marsden took radioactive source \(\left(\begin{array}{c} 214 \\ 83 \end{array} \mathrm{Bi}\right)\) for a-particles. A collimated beam of a-particles of energy 5.5 MeV was allowed to fall on 2.1 x 10-7 m thick gold foil. The a-particles were observed through a rotatable detector consisting of a Zinc sulphide screen and microscope. It was found that a-particles got scattered. These scattered a-particles produced scintillations on the zinc sulphide screen. Observations of this experiment are as follows.
(I) Most of the a-particles passed through the foil without deflection.
(II) Only about 0.14% of the incident a-particles scattered by more than 1°.
(III) Only about one a-particle in every 8000 a-particles deflected by more than 90°.
These observations led to many arguments and conclusions which laid down the structure of the nuclear model of an atom.
(i) Rutherford's atomic model can be visualised as
(ii) Gold foil used in Geiger-Marsden experiment is about 10-8 m thick. This ensures
(a) gold foil's gravitational pull is small or possible (b) gold foil is deflected when a-particle stream is not incident centrally over it (c) gold foil provides no resistance to passage of a-particles (d) most a-particle will not suffer more than 1° scattering during passage through gold foil (iii) In Geiger-Marsden scattering experiment, the trajectory traced by an a-particle depends on
(a) number of collision (b) number of scattered a- particles (c) impact parameter (d) none of these (iv) In the Geiger-Marsden scatteririg experiment, in case of head-on collision, the impact parameter should be
(a) maximum (b) minimum (c) infinite (d) zero (v) The fact only a small fraction of the number of incident particles rebound back in Rutherford scattering indicates that
(a) number of a-particles undergoing head-on-collision is small (b) mass of the atom is concentrated in a small volume (c) mass of the atom is concentrated in a large volume (d) both (a) and (b). -
Bohr's model explains the spectral lines of hydrogen atomic emission spectrum. While the electron of the atom remains in the ground state, its energy is unchanged. When the atom absorbs one or more quanta of energy, the electrons moves from the ground state orbit to an excited state orbit that is further away.
The given figure shows an energy level diagram of the hydrogen atom. Several transitions are marked as I, II, III and so on. The diagram is only indicative and not to scale.
(i) In which transition is a Balmer series photon absorbed?(a) II (b) III (c) IV (d) VI (ii) The wavelength of the radiation involved in transition II is
(a) 291 nm (b) 364 nm (c) 487 nm (d) 652 nm (iii) Which transition will occur when a hydrogen atom is irradiated with radiation of wavelength 103 nm?
(a) I (b) II (c) IV (d) V (iv) The electron in a hydrogen atom makes a transition from n = n1 to n = n2 state. The time period of the electron in the initial state is eight times that in the final state. The possible values of n1 and n2 are
(a) n1 = 4, n2 = 2 (b) n1 = 8, n2 = 2 (c) n1 = 8, n2 = 3 (d) n1 = 6, n2 = 2 (v) The Balmer series for the H-atom can be observed
(a) if we measure the frequencies of light emitted when an excited atom falls to the ground state (b) if we measure the frequencies of light emitted due to transitions between excited states and the first excited state. (c) in any transition in a H-atom (d) none of these. -
Neutrons and protons are identical particle in the sense that their masses are nearly the same and the force, called nuclear force, does into distinguish them. Nuclear force is the strongest force. Stability of nucleus is determined by the neutron proton ratio or mass defect or packing fraction. Shape of nucleus is calculated by quadrupole moment and spin of nucleus depends on even or odd mass number. Volume of nucleus depends on the mass number. Whole mass of the atom (nearly 99%) is centred at the nucleus.
(i) The correct statements about the nuclear force is/are(a) change independent (b) short range force (c) non-conservative force (d) all of these. (ii) The range of nuclear force is the order of
(a) 2 x 10-10 m (b) 1.5 x 10-20 m (c) 1.2 x 10-4 m (d) 1.4 x 10-15 m (iii) A force between two protons is same as the force between proton and neutron. The nature of the force is
(a) electrical force (b) weak nuclear force (c) gravitational force (d) strong nuclear force. (iv) Two protons are kept at a separation of 40 \(\dot A\). Fn is the nuclear force and Fe is the electrostatic force between them. Then
(a) Fn << Fe (b) Fn = Fe (c) Fn >> Fe (d) Fn \(\approx \) Fe (v) All the nucleons in an atom are held by
(a) nuclear forces (b) van der Waal's forces (c) tensor forces (d) coulomb forces -
Rectifier is a device which is used for converting alternating current or voltage into direct current or voltage. Its working is based on the fact that the resistance of p-n junction becomes low when forward biased and becomes high when reverse biased. A half-wave rectifier uses only a single diode while a full wave rectifier uses two diodes as shown in figures (a) and (b) .
(i) If the rms value of sinusoidal input to a full wave rectifier is \(\frac{V_{0}}{\sqrt{2}}\) then the rms value of the rectifier's output is\(\text { (a) } \frac{V_{0}}{\sqrt{2}}\) \(\text { (b) } \frac{V_{0}^{2}}{\sqrt{2}}\) \(\text { (c) } \frac{V_{0}^{2}}{2}\) \(\text { (d) } \sqrt{2} V_{0}^{2}\) (ii) In the-diagram, the input ac is actoss the terminals A and C. The output across Band D is
(a) same as the input (b) half wave rectified (c) zero (d) full wave rectified (iii) A bridge rectifier is shown in figure. Alternating input is given across A and C. If output is taken across BD, then it is
(a) zero (b) same as input (c) half wave rectified (d) full wave rectified (iv) A p-n junction (D) shown in the figure can act as a rectifier. An alternating current source (V) is connected in the circuit. The current (I) in the resistor(R) can be shown by
(v) With an ac input from 50 Hz power line, the ripple frequency is
(a) 50 Hz in the dc output of half wave as well as full wave rectifier (b) 100 Hz in the de output of half wave as well as full wave rectifier (c) 50 Hz in the de output of half wave and 100 Hz in dc output offull wave rectifier (d) 100 Hz in the dc output of half wave and 50 Hz in the de output of full wave rectifier
Part A
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CBSE 12th Standard Physics Subject Case Study Questions With Solution 2021 Part - II Answer Keys
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(i) (a): As, \(q E=m g \Rightarrow q=\frac{1.08 \times 10^{-14} \times 9.8}{1.68 \times 10^{5}}\)
\(=6.4 \times 10^{-19} \mathrm{C}\)
(ii) (a): \(q=n e \text { or } \Rightarrow n=\frac{6.4 \times 10^{-19}}{1.6 \times 10^{-19}}=4\)
(iii) (c) : For the drop to be stationary,
Force on the drop due to electric field = Weight of the drop
qE=mg
\(q=\frac{m g}{E}=\frac{1.6 \times 10^{-6} \times 10}{100}=1.6 \times 10^{-7} \mathrm{C}\)
Number of electrons carried by the drop is
\(n=\frac{q}{e}=\frac{1.6 \times 10^{-7} \mathrm{C}}{1.6 \times 10^{-19} \mathrm{C}}=10^{12}\)
(iv) (c)
(v) (d): Millikan's experiment confirmed that the charges are quantized, i.e., charges are small integer multiples of the base value which is charge on electron. The charges on the drops are found to be multiple of 4. Hence, the quanta of charge is 4 \(\mu \)C. -
(i) (c): \(W=(\text { P.E. })_{\text {final }}-(\text { P.E. })_{\text {initial }}\)
\(=\frac{k e^{2}}{2}-\frac{k e^{2}}{1}=\frac{-k e^{2}}{2}\)
(ii) (b) : Potential at the centre of the square due to four equal charges q at four corners
\(V=\frac{4 q}{4 \pi \varepsilon_{0}(a \sqrt{2}) / 2}=\frac{\sqrt{2} q}{\pi \varepsilon_{0} a}\)
\(W_{0 \rightarrow \infty}=-W_{\infty \rightarrow 0}=-(-q) V=\frac{\sqrt{2} q^{2}}{\pi \varepsilon_{0} a}\)
(iii) (c): Here, \(q=2 \mu \mathrm{C}=2 \times 10^{-6} \mathrm{C}, r_{A}=2 \mathrm{~m}, r_{B}=1 \mathrm{~m}\)
\(\therefore \ V_{A}-V_{B}=\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{1}{r_{A}}-\frac{1}{r_{B}}\right]\)
\(=2 \times 10^{-6} \times 9 \times 10^{9}\left[\frac{1}{2}-\frac{1}{1}\right] \mathrm{V}=-9 \times 10^{3} \mathrm{~V}\)
(iv)(b) : Required work done = Change in potential energy of the system
\(W=U_{f}-U_{i}=k \frac{q_{1} q_{2}}{r_{f}}-k \frac{q_{1} q_{2}}{r_{i}}=k q_{1} q_{2}\left[\frac{1}{r_{f}}-\frac{1}{r_{i}}\right]\)
\(\therefore \ W=\left(9 \times 10^{9}\right)\left(3 \times 10^{-9} \times 1 \times 10^{-9}\right)\) \(\times\left[\frac{1}{4 \times 10^{-2}}-\frac{1}{5 \times 10^{-2}}\right]\)
\(=27 \times 10^{-7} \times(0.05)=1.35 \times 10^{-7} \mathrm{~J}\)
(v) (b) -
(i) (d): Mobility is defined as the magnitude of drift velocity per unit electric field
Mobility, \(\mu=\frac{\left|v_{d}\right|}{E}\)
(ii) (c): Drift velocity \(v_{d}=\frac{I}{n e A}\)
where the symbols have their usual meanings
(iii) (b): I = neAvd
vd is of order offew m S-I, e = 1.6 x 10-19 C,
A is of the order of mm2, so a large I is due to a large value of n in conductors.
(iv) (c): When we close the circuit, an electric field is established instantly with the speed of electromagnetic wave which causes electrons to drift at every portion of the circuit, due to which the current is set up in the entire circuit instantly. The current which is set up does not wait for electrons to flow from one end of the conductor to another. Thus, the electric bulb glows immediately when switch is closed.
(v) (b): Here,
Number density of free electrons, n = 8.5 x 1028 m-3
Area of cross-section of a wire, A = 2.0 x 10-6 m2
Length of the wire, 1= 3.0 m
Current, I = 3.0 A
The drift velocity of an electron is \(v_{d}=\frac{I}{n e A}\) ...(i)
The time taken by the electron to drift from one end to other end of the wire is
\(t=\frac{l}{v_{d}}=\frac{\ln e A}{I}\)
\(=\frac{(3.0 \mathrm{~m})\left(8.5 \times 10^{28} \mathrm{~m}^{-3}\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)\left(2.0 \times 10^{-6} \mathrm{~m}^{2}\right)}{(3.0 \mathrm{~A})}\)
= 2.7 x 104 s -
(i) (c): In mass spectrometer, the ions are sorted out by accelerating them through electric and magnetic field.
(ii) (c): As \(\frac{m v^{2}}{r}=q v B_{0} \therefore r=\frac{m v}{q B_{0}}\)
(iii) (b): As radius \(r \propto \frac{m}{q}\)
\(\therefore\) r will be maximum for \(\alpha\) - particle.
(iv) (b) : Here, \(r=\frac{m v}{q B_{0}} \text { or } m=\frac{r q B_{0}}{v}\)
As \(v=\frac{E}{B}, \therefore m=\frac{q B_{0} B r}{E}\)
(v) (c): From the relation v = E/B, it is clear electric and magnetic force balance each other. -
(i) (b)
(ii) (d): Magnetic permeability - Henry m-1
(iii) (d): Given, L= 3 cm, A = 2 cm2, M = 3 A m2
.Intensity of magnetisation \(=\frac{M}{l A}=\frac{3}{3 \times 10^{-2} \times 2 \times 10^{-4}}\)
\(=\frac{1}{2 \times 10^{-6}}=0.5 \times 10^{6}=5 \times 10^{5} \mathrm{~A} / \mathrm{m}\)
(iv) (b): Here, n = 500 turns/m
\(I=1 \mathrm{~A}, \mu_{-}=500\)
Magnetic intensity \(H=n I=500 \mathrm{~m}^{-1} \times 1 \mathrm{~A}=500 \mathrm{~A} \mathrm{~m}^{-1}\)
As \(\mu_{r}=1+\chi \quad \text { or } \chi=\left(\mu_{r}-1\right)\)
Magnetisation, M = XH
\(=\left(\mu_{r}-1\right) H=(500-1) \times 500 \mathrm{~A} \mathrm{~m}^{-1}\)
\(=2.495 \times 10^{5} \mathrm{~A} \mathrm{~m}^{-1} \approx 2.5 \times 10^{5} \mathrm{~A} \mathrm{~m}^{-1}\)
(v) (a): Relative permeability of iron \(\mu_{r}=6000\)
Magnetic susceptibility \(\chi_{m}=\mu_{r}-1=5999\) -
(i) (d) : More rapid is the movement of bar magnet, more is the deflection observed in the galvanometer
(ii) (c) : Two circular loops carrying current in the same direction will attract each other. If they are now separated, induced currents will try to keep status quo, by increasing the current in both the coils.
(iii) (b): Acceleration of the magnet will not be equal to g. It will be less than g. This is because, as the magnet falls, amount of magnetic flux linked with the ring changes.
An induced emf is developed in the ring which opposes the downward motion of the magnet.
(iv) (c) : The magnitude of induced emf set up in the coil does not depend upon the resistance of the coil whereas induced current set up in the coil depend upon the resistance of the coil.
(v) (d) : As long as a coil of metal is kept stationary in a magnetic field, even if it is non-uniform, unless it is changing with respect to time, there will be no induced emf or current in the coil. -
(i) (c) : As \(\frac{E_{s}}{E_{p}}=\frac{n_{s}}{n_{p}} \Rightarrow E_{s}=E_{p} \cdot \frac{n_{s}}{n_{p}}\)
\(=\frac{120 \times 50}{2000}=3 \mathrm{~V}\)
(ii) (d) : \(I_{s}=\frac{E_{s}}{R} \Rightarrow I_{s}=\frac{3}{0.6}=5 \mathrm{~A}\)
(iii) (a) : As \(\frac{I_{p}}{I_{s}}=\frac{E_{s}}{E_{p}}\)
\(\Rightarrow I_{p}=\frac{E_{s}}{E_{p}} \times I_{s}=\frac{\ 3}{120} \times 5=0.125 \mathrm{~A}\)
(iv) (d) : Power in primary \(P_{p}=E_{p} \times I_{p}=120 \times 0.125\)
= I5W
(v) (a) : Power in secondary coil \(P_{s}=E_{s} \times I_{s}=3 \times 5\)
= 15W -
(i) (d): All electromagnetic waves travel in vacuum with the same speed.
(ii) (c): Cathode rays (beam of electrons) get deflected in an electric field.
(iii) (c) : \(\Upsilon\)-rays are detected by ionization chamber.
(iv) (b): Size of particle \(=\lambda=\frac{c}{v}\)
\(v=\frac{c}{\lambda}=\frac{3 \times 10^{10} \mathrm{~cm} \mathrm{~s}^{-1}}{3 \times 10^{-4} \mathrm{~cm}}=3 \times 10^{14} \mathrm{~Hz}\)
(v) (a): Every body at a temperature T > 0 K emits radiation in the infrared region. -
(i) (a): \(m=\frac{f_{o}}{f_{e}}=7\)
\(f_{o}=7 f_{e}\)
In normal adjustment, distance between the lenses
\(f_{o}+f_{e}=40 \)
\(7 f_{0}+f_{e}=40 \Rightarrow f_{e}=\frac{40}{8}=5 \mathrm{~cm} \)
\(f_{o}=7 f_{e}=7 \times 5=35 \mathrm{~cm}\)
(ii) (d): \(m=-10 ; L=22 \mathrm{~cm}\)
\(\text { As } m=\frac{-f_{o}}{f_{e}} \Rightarrow-10=-\frac{f_{o}}{f_{e}}\)
\(f_{o}=10 f_{\mathrm{e}} \)
\(\text { As } L=f_{o}+f_{e} \)
\(22=10 f_{e}+f_{e}=11 f_{e} \)
\(\text { or } f_{e}=\frac{22}{11}=2 \mathrm{~cm}\)
\(f_{o}=10 f_{e}=20 \mathrm{~cm}\)
(iii) (d): Objective lens has larger focal length than eye-piece.
(iv) (d): Astronomial telescope is used to see stars, sun etc.
(v) (c) :f0>>fe -
(i) (b): \(\text { As } z=\frac{\lambda D}{2 d}\)
\(\text { At } S_{4}: \frac{\Delta x}{d}=\frac{z}{D}\)
\(\Rightarrow \Delta x=\frac{\lambda D}{2 d} \frac{d}{d}=\frac{\lambda}{2}\)
(ii) (c): \(z=\frac{\lambda D}{d}\)
\(\Delta x \text { at } S_{4}: \Delta x=\frac{\lambda D}{d} \frac{d}{d}=\lambda\)
Hence, maxima at S4 as well as S3'
Resultant intensity at S4 I = 4I0
\(\therefore \quad \frac{I_{\max }}{I_{\min }}=\frac{\left[\left(4 I_{0}\right)^{1 / 2}+4\left(4 I_{0}\right)^{1 / 2}\right]^{2}}{\left[\left(4 I_{0}\right)^{1 / 2}-\left(4 I_{0}\right)^{1 / 2}\right]^{2}}=\infty\)
(iii) (a): When the screen is placed perpendicular to the line joining the 'sources, the fringes will be concentric circles.
(iv) (b): Constructive interference occurs when the path difference (S1P - S2P) is an integral multiple of \(\lambda\).or S1P - S2P = n\(\lambda\), where n = 0,1,2,3, .....
(v) (d): Here, A1 = 3A, A2 = 2A and \(\varphi\)= 60° The resultant amplitude at a point is
\(R =\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos \phi} \)
\(=\sqrt{(3 A)^{2}+(2 A)^{2}+2 \times 3 A \times 2 A \times \cos 60^{\circ}} \)
\(=\sqrt{9 A^{2}+4 A^{2}+6 A^{2}}=A \sqrt{19}\)
As, Intensity \(\infty\) (Amplituder)2 Therefore, intensity at the same point is
\(I \propto 19 A^{2}\) -
(i) (d): Power of light received by the cone \(=I\left(\pi R^{2}\right)\)
Let number of photons hitting the cone per second is n.
Then, \(n E=I \pi R^{2} \Rightarrow n=\pi R^{2} I / E\)
(ii) (b): Initial moment \(p_{i}=\frac{E}{c}\)
For a perfectly selecting surface
Final momentum \(p_{f}=\frac{-E}{c}\)
\(\Delta p=p_{f}-p_{i}=\frac{-E}{c}-\frac{E}{c}=\frac{-2 E}{c}\)
Hence a momentum \(\frac{2 E}{c}\) is transferred to the reflecting surface.
(iii) (a): According to the theory of relatively,
\(E=m c^{2}=m c . c=p c\)
\(\text { or } \quad E^{2}=p^{2} c^{2}\)
where p is the momentum of a photon.
(iv) (b): I = 1 kWm-2
As, 50% of light is reflected, thus e = 0.5
Radiation pressure \(P=\frac{(1+e) I}{c}\)
\(\therefore \quad P=\frac{(1+0.5) \times 1000}{3 \times 10^{8}}=5 \times 10^{-6} \mathrm{~Pa}\)
(v) (a) : \(\frac{P_{\mathrm{rad}}}{P_{0}}=\frac{5 \times 10^{-6}}{1 \times 10^{5}}=5 \times 10^{-11}\) -
(i) (d) : Rutherford's atom had a positively charged centre and electrons were revolving outside it.It is also called the planetary model of the atom as in option (d).
(ii) (d): As the gold foil is very thin, it can be assumed that a-particles will suffer not more than one scattering during their passage through it. Therefore, computation of the trajectory of an a-particle scattered by a single nucleus is enough.
(iii) (c) : Trajectory of a-particles depends on impact parameter which is the perpendicular distance of the initial velocity vector of the a particles from the centre of the nucleus. For small impact parameter a particle close to the nucleus suffers larger scattering.
(iv) (b): At minimum impact parameter, a particles rebound back (\(\theta\) = \(\pi\)) and suffers large scattering.
(v) (d): In case of head-on-collision, the impact parameter is minimum and the a-particle rebounds back. So, the fact that only a small fraction of the number of incident particles rebound back indicates that the number of a-particles undergoing head-on collision is small. This in turn implies that the mass of the atom is concentrated in a small volume. Hence, option (a) and (b) are correct. -
(i) (d): For Balmer series, n1 = 2; n2 = 3, 4, ...
(lower) (higher)
Therefore, in transition (VI), photon of Balmer series is absorbed.
(ii) (c): In transition II
\(E_{2}=-3.4 \mathrm{eV}, E_{4}=-0.85 \mathrm{eV}\)
\(\Delta E=2.55 \mathrm{eV} \Rightarrow \Delta E=\frac{h c}{\lambda} \Rightarrow \lambda=\frac{h c}{\Delta E}=487 \mathrm{nm}\)
(iii) (d): Wavelength of radiation = 1030 \(\dot A\)
\(\Delta E=\frac{12400}{1030 \dot A}=12.0 \mathrm{eV}\)
So, difference of energy should be 12.0 eV (approx.) Hence for n1 = 1 to n2 = 3.
\(E_{n_{3}}-E_{n_{1}}=-1.51 \mathrm{eV}-(-13.6 \mathrm{eV}) \approx 12 \mathrm{eV}\)
Therefore, transition V will occur.
(iv) (a): \(T^{2} \propto r^{3} \text { and } r \propto n^{2} \Rightarrow T^{2} \propto n^{6} \Rightarrow T \propto n^{3}\)
\(\frac{T_{1}}{T_{2}}=\left(\frac{n_{1}}{n_{2}}\right)^{3} \Rightarrow 8=\left(\frac{n_{1}}{n_{2}}\right)^{3} \text { or } \frac{n_{1}}{n_{2}}=2\)
(v) (b) -
(i) (d) : All options are basic properties of nuclear forces. So, all options are correct.
(ii) (d) : The nuclear force is of short range and the range of nuclear force is the order of 1.4 x 10-15 m
Now, volume \(\propto R^{3} \propto A\)
(iii) (d)
(iv) (a) : Nuclear force is much stronger than the electrostatic force inside the nucleus i.e., at distances of the order of fermi. At 40 \(\dot A\), nuclear force is ineffective and only electrostatic force of repulsion is present.This is very high at this distance because nuclear force is not acting now and the gravitational force is very feeble.Fnuclear<< Felectrostatiinc this case.
(v) (a). -
(i) (a) :The rms value of the output voltage at the load resistance \(V_{\mathrm{rms}}=\frac{V_{0}}{\sqrt{2}}\)
(ii) (d)
(iii) (a)
(iv) (c): The given circuit works as a half wave rectifier. In this circuit, we will get current through R when p-n junction is forward biased and no current when p-n junction is reverse biased. Thus the current (I) through resistor (R) will be shown in option (c).
(v) (c)
Part A
