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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 851. |
The sound waves after being converted into electrical waves are not transmitted as such because : |
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Answer» they travel with the SPEED of sound. |
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| 852. |
A spherical shell of radius 10 cm is carryinga charge q . If the electric potentialat distancesof the sphericalshell is V_(1), V_(2) and V_(3) respectively , then |
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Answer» `V_(1) = V_(2) lt V_(3)` `V= {("CONSTANT" = (1)/(4pi in_(0)) (Q)/(R ): "inside"),("VARIABLE" =(1)/(4pi in_(0)) (Q)/(R ):" outise"):}` `V_(1)= V_(2) gt V_(3)` 
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| 853. |
A radio station has two channels. One is AM at 1020kHz and the other is FM at 89.5MHz. For good results, you will use |
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Answer» longer antenna for the AM channel and shorter for the FM |
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| 854. |
A sphere moves with a relativistic velocity v through a gas whose unit volume contains n slowly moving particles, each of mass m. Find the pressure p exerted by the gas on a spherical surface element perpendicular to the velocity of the sphere, provided that the particles scatter elastically. Show that the pressure is the same both in the reference frame fixed to the sphere and in the reference frame fixed to the gas. |
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Answer» <P> Solution :In the FRAME fixed to the sphere :- The momentum transferred to the eastically scattered particle is`(2mv)/(sqrt(1-v^2/c^2))` The density of the moving element is, from `1.369`, `n(1)/(sqrt(1-v^2/c^2))` and the momentum transferred per unit time per unit area is `p=` the pressure`=(2mv)/(sqrt(1-v^2/c^2))n(1)/(sqrt(1-v^2/c^2))*v=(2mnv^2)/(1-v^2/c^2)` In the frame fixed to the gas :- When the sphere hits a stationary particle, the latter recoils with a velocity `=(v+v)/(1+v^2/c^2)=(2V)/(1+v^2/c^2)` The momentum transferred is `((m*2v)/(1+v^2//c^2))/(sqrt(1-(4v^2//c^2)/((1-v^2//c^2)^2)))=(2mv)/(1-v^2/c^2)` and the pressure is `(2mv)/(1-v^2/c^2)*n*v=(2mnv^2)/(1-v^2/c^2)` |
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| 855. |
An alpha particle of energy 5 MeV is scattered through 180^(@) by a fixed uranium nucleus. The distance of the closest approach is of the order of ........ |
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Answer» `1Å` `:.` kinetic energy = potential energy `:.5MeV=(1)/(4piepsi_(0)).(q_(1)q_(2))/(R)` `:.5xx10^(6)xxe=(9xx10^(9)xx(92e)(2e))/(r)` `:.r=(9xx10^(9)xx92xx2xxe)/(5xx10^(6))=5.3xx10^(-14)m` `:.r=5.3xx10^(-12)cm` |
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| 856. |
A TV tower has a height of 70 m. If the average population density around the tower is 1000 km^(-2), the population Covered by the TV tower |
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Answer» `2.816 xx 10^(6)` |
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| 857. |
In a young's double slit experiment the separation between the slits is halved and the distance between the slits and the screen is doubled. The fringe width is |
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Answer» unchanged |
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| 858. |
A body moves along a circular track of radius R. It starts from one end of a diameter and reaches the other end of diameter. What is the ratio of the distance travelled by the body to its displacement : |
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Answer» `pi/2` 
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| 859. |
A device 'X' is connected to an AC source. The variation of voltage, current and power in one complete cycle is shown in figure. a) Which curve shows power consumption over a full cycle? b) What is the average power consumption over a cycle? c) Identify the device X. |
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Answer» Power = P = VI that is curve of power will be having MAXIMUM amplitude, equals to MULTIPLICATION of amplitudes of voltage (V) and current (I) curve. So, the curve will be represented by A. b) As shown by SHADED area in the diagram, the full cycle of the graph consists of one positive and one negative symmetrical area.  HENCE, average power over a cycle is zero. c) As the average power is zero, hence the device may be inductor (L) or capacitance (C) or the SERIES combination of L and C. |
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| 860. |
Atrain approaching a railway platform with a speed of 20 ms^(-1) starts blowing the whistle. Speed of sound in air is 340 ms^(-1) . If the frequency of the emitted sound from the whistle is 640 Hz, the frequency of sound to a person standing on the platform will appear to be |
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Answer» 600 Hz frequency of the source, `v_(0) = 640 Hz` The frequency heard by the person standing on the PLATFORM is, `u.=u_(0) v/(v-v_(s)) = 640 XX [340/(340 -20)] = (640 xx 340)/320 = 680` Hz |
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| 861. |
What are holes ? |
| Answer» Solution :Removal of an electron from the orbit of ATOM leaves a VACANCY behind which is CALLED HOLE. | |
| 862. |
How we state Faraday's 2nd law of induction? |
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Answer» SOLUTION :The induces e.m.f. is equal to the rate of CHANGE of magnetic flux LINKED with the CIRCUIT `THEREFORE E = dphi/dt` |
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| 863. |
Assertion: Electric field intensity inside a solid non-conducting uniformly charged solid sphere increases linearly inside the sphere. Reason: Electric field intensity outside the solid non-conducting uniformly charged solid sphere is same as the electric field generated by a point charge kept at the centre of sphere. |
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Answer» If both assertionand REASON are correct and reason is a correct EXPLANATION of the ASSERTION. |
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| 864. |
How much energy is needed to excite an electron in H-atom from ground state to first excited state? |
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Answer» `-13.6` eV REQUIRED EXCITATION ENERGY = `E_(2) - E_(1) = -3.4 + 13.6= + 10.2 eV` |
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| 865. |
Optical and radiotelescopes are built on the ground, but X - ray astronomy is possible only from satellites orbiting the earth.Why ? |
| Answer» SOLUTION :ATMOSPHERE ABSORBS X-rays, while VISIBLE and RADIOWAVES can pentrate it. | |
| 866. |
In Bohr's model, the atomic radius of the first orbit is r_(0). Then, the radius of third orbit is |
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Answer» `r_(0)/9` `therefore v_(3) = r_(0)(3)^(2) = 9r_(0)` |
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| 867. |
The electric field components in Figure are E_(x)=alpha x^(1/2), E_y=E_(z)=0 in which alpha=800 N//Cm^(1/2). Calculate (a) the flux through the cube, and (b) the charge within the cube. Assume that a=0.1m. |
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Answer» Solution :a. Since the electric field has only an x component, for FACES perpendicular to x DIRECTION, the angle between E and `triangleS" is "+pi/2`. Therefore, the flux `phi=E triangleS` is separately zero for each face of the CUBE except the two shaded ones. Now the magnitude of the electric field at the left face is `E_L= ax^(1/2)=alpha a^(1/2)` (x= a at the left face). The magnitude of electric field at the right face is `E_a=alpha x^((-1)/2)= alpha (2a)^(1/2)` (x=2 a at the right face). The corresponding fluxes are`phi_(L)=E_(L). triangleS=triangleS E_(L). hatn_(L)=E_(L) triangleS cos theta=-E_(L) triangleS," since "theta=180^(@), phi_(R)=E_(R). triangleS=E_(R) triangleS cos theta=E_(R) triangleS,"" sin ce"theta=0^(@)` Net flux through the cube `=phi_(R)+phi_(L)=E_(R)a^(2)-E_(L)a^(2)=a^(2) (E_(R)-E_(L)) =alpha a^(2) [(2a)^(1/2)-a^(1/2)]` `=alpha a^(5/2) (sqrt2-1) =800 (0.1)^(5/2) (sqrt2-1)=1.05 Nm^(2) C^(-1)` b. We can use Gauss. law to find the TOTAL charge q inside the cube. We have `phi=(q)/(epsi_(0)) or q=phi epsi_(0). " Therefore ",q=1.05 xx 8.854 xx 10^(-12) C=9.27 xx 10^(-12)C`. |
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| 868. |
A conductor carries a certain charge. When it is connected to another uncharged conductor of finite capacity, then the energy of the combined system is |
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Answer» more than that of the FIRST conductor |
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| 869. |
How electromagnetic waves are generated ? |
| Answer» SOLUTION :ACCELERATING ELECTRIC CHARGES | |
| 870. |
The applied a.c power to a full-wave rectifier is 400W. The d.c power output obtained is 200W. Rectifier efficiency is nearly |
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Answer» 0.25 |
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| 871. |
If an electron revolves around a nucleus in a circular orbit of radius R with frequency n, then the magnetic field produced at the centre of the nucleus will be |
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Answer» `(mu_(0)EN)/(2R)` |
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| 872. |
Answer the following questions : (a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances ? Explain. (b) A virtual images, we always say, cannot be caught on a screen, Yet when we see a virtual image, we bring it to the screen i.e, retina of our eye. Is there a contradiction ? (c ) A diver under water looks obliquelyat a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter than what he actually is ? (d) Does the apparent depth of a tank of water change if viewed obliquely ? If so, does the apparent depth increase or decrease ? (e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of same use to diamond cutter ? |
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Answer» Solution :(a) Yes, when rays incident on a plane mirror or a convex mirror are tending to converge to a point behind the mirror, they are relflected to a point on a screen in front of the mirror. Hence a real image is formed (when the object is virtual ). (b) No, there is no contradiction. Eye lens is convergent. It forms a real image of the virtual object (i.e., the virtual image being seen ) on the retina. ( c) As fisherman is ni air, LIGHT travels from rarer to denser medium. It bends towards the normal, appearingto come from a larger distance. Therefore, to the diver under water. fisherman looks taller. (d) Yes, the apparent depth DECREASES further, when water TANK is viewed obliquely compared to the depth when seen near normally. (e) As `MU = (1)/(sin C):. sin C = (1)/(mu)` As refractive index of diamond is much greater than that of ordinary glass, critical angle `C` for diamond is much smaller `(~~ 24^(@))` as compared to that for glass `(~~ 42^(@))`. A skilled diamond cutter exploits the large range of angles of INCIDENCE of light `(24^(@)` to `90^(@))` to ensure that light entering the diamond suffers multiple total internal reflection within the diamond. This produces sparking effect in the diamond. |
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| 873. |
Define an inductor. Derive equation of energy U=1/2LI^2 stored in inductor. |
Answer» Solution :In a CIRCUIT, a component which has SELF inductance is called an inductor. Symbol for an inductor in a circuit is the one shown in figure.  This self-induced emf is also called the back emf as it opposes any change in the current in a circuit. So, work needs to be done against the back emf (e) in establishing the current. This work done is stored as magnetic potential energy. For the current I at an instant in a circuit, the rate of work done is, `(dW)/(dt)=|epsilon|I=LI(dI)/(dt)` `therefore` dW=LIdI Total amount of work done in establishing the current I is, `W=int_0^1 dW=int_0^1 LIdI=1/2LI^2` Thus, the electrical energy required to build up the current I is, `U=1/2LI^2` This expression reminds us of `mv^2//2` for the (MECHANICAL) kinetic energy of a particle of mass m, and SHOWS that L is analogous to m (i.e., L is electrical INERTIA and opposes growth and decay of current in the circuit). |
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| 874. |
A conducting ring of circular cross - section with inner and outer radii a and b is made out of a material of resistivity rho. The thickness of the ring is h. It is placed coaxially in a vertical cylindrical region of a magnetic field B=krt. Where k is a positive constant, r is the distance from the axis and t is the time. If the current through the ring is I=((kh)/(alphap))[b^(3)-a^(3)] , then what is the value of alpha? |
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| 875. |
A projectile is thrown with angle of projection tan^(-1)(4/3) The ratio of its horizontal range to the greatest height reaches is : |
| Answer» ANSWER :B | |
| 876. |
Is the de Broglie wavelength of a photons of an em variation equal to the wavelength of electron variation ? |
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Answer» Solution :YES , DE BROGLIE wavelength `lambda` = h/p, for a photon and for an electron `lambda = (hv)/c` . HENCE `lambda` = h/p = c/v . Hence , two wavelength are same. |
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| 877. |
The velocity of sound in air at 9 atmosphere pressure and that at 1 atm. Pressure would be |
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Answer» `1: 1` At constant temperature the ratio `(P)/(rho)` REMAINS same. HENCE speed of sound is in DEPENDENT of change of pressure. Hence correct choice is (a). |
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| 878. |
A chain reaction dies out sometimes, why? |
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Answer» Solution :A CHAIN REACTION may die out due toany of the following reasons: (i) Size of FISSIONABLE material may be less than the critical size. (ii) Mass of fissionable material may be less than the critical mass. (iii) Neutron ABSORBING material (arrestor) might absorb neutrons at a faster rate then the rate at which they are being PRODUCED . |
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| 879. |
An engine standing at the platform blows a whistle of frequency 305 vib/sec. if the velocity of sound be 1220 km/h, the frequency of the whistle as heard by a man towards the engine with a speed of 20 km/h is : |
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Answer» 300 vib/s Hence the correct choicer is (c). |
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| 880. |
What is transducer ? How is it used ? |
| Answer» Solution :It is a device which converts one form of energy into another. TRANSDUCER is used in electronoic communication SYSTEM. For example, a microphone is a transducer which converts SOUND energy into electric energy and hence acts as a transducer. | |
| 881. |
A fish is at a depth of 12 cm in water is viewed by an observer on the bank of a lake. To what height the image of the fish is raised in cm ? |
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| 882. |
In Young's double slit experiment the light emitted from source has lambda = 6.5 xx 10^(-7) m and the distance between the two slits is 1 mm. Distance between the screen and slit is 1 metre. Distance between the screen and slit is 1 metre. Distance between third dark and fifth bright fringes will be : |
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Answer» `3.2` mm `x_(3) = (2n - 1) (lambda)/(2) (D)/(d) = 16.25 xx 10^(-4) m` `x_(5) - x_(3) = 1.63 mm`. |
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| 883. |
At a certain place, the vertical component of the earth's magnetic field is 0.4 xx 10^(-4) T and horizontal component is 0.3 xx 10^(-4) T. What will be the total intensity of magnetic field of the earth ? |
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Answer» `0.5 xx 10^(-4)` T `= sqrt((0.4 xx 10^(-4) )^(2) + (0.3 xx 10^(-4) )^(2) )` `= sqrt((0.16 + 0.09 ) xx 10^(-8))` `therefore B= 0.5 xx 10^(-4)` T |
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| 884. |
Give any two consequences of refraction of light. |
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Answer» Solution :(i) The phenomena of lateral shift and NORMAL shift are due to REFRACTION of light. (II) The early SUNRISE and LATE sunset are due to atmospheric refraction of light. |
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| 885. |
Huygen.s principle is used |
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Answer» to determine the velocity of light |
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| 886. |
The equation of a transverse wave travelling on a rope is given by y = 10sin pi (0.01x-2.000t) where y and x are in cm and t in seconds. The maximum transverse speed of a particle in the rope is about |
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Answer» 62.8 cm/s |
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| 887. |
A spring is attached with a block of mass m and a fixed horizontal roof as shown. The block is lying on a smooth horizontal table and initially the spring is vertical and unstretched. Natural length of spring is 3l_(0). A constant horizontal force F is applied on the block so that block moves inthe directionof force. When length of the spring becomes 5l_(0), block leaves contact with the table. Find the constant force F, if initial and final velocity of block is zero. |
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| 888. |
A particle having a charge of 100 μc and mass of 10mg is projected in a uniform magnetic field of 25 Tesla, with a speed of 10ms^(- 1). The velocity is perpendicular to the field. The radius of path and period of revolution are |
| Answer» Answer :C | |
| 889. |
The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B_(0) = 510 nT. What is the amplitude of the electric field part ofthe wave ? |
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Answer» Solution :Here, `B_(0)=510xx10^(-9) T` `E_(0)=CB_(0)=3XX10^(8)xx510xx10^(-9)=153 NC^(-1)`. |
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| 890. |
In Q.No. 69, work done in complete cycle is : |
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Answer» `3xx10^(-3)J` `=(1)/(2)BC xxAC` `=(1)/(2)3xx10^(-3)xx2` `=3xx10^(-3)J`. Thus, correct choice is (a). |
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| 891. |
a. Name the experiment which establishes the wave nature of moving electrons. b. In that experiment, electrons are accelerated through a potential of 54 V and is made to fall normally on the nickel crystal of interatomic separation 0.91 A. Draw the polar graph showing the variationof intensity of the scattered electrons electrons and latitudeangle at this potential. c. Calculate the d-Broglie wavelength of the electron. |
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Answer» Solution :a. Davisson-Germer experiment. b.  c. Intensity DECREASES with LATITUDE angle and then increases to maximum at `50^(@)` d. `lambda=(h)/(p)=(h)/(sqrt(2ME))=(6.63xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx5.4xx1.6xx10^(-19)))=1.67Å` |
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| 892. |
A thin disc of mass M and radius R has mass per unit area sigma( r) = kr^(2), where r is the distnace from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is xMR^(2), where x is _________ |
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Answer» M.I. of ELEMENT ring of radius r and infinitesimally small thickness dr about the axis is `dI=(DM).r^(2) ={(kr^(2))2pi rdr}r^(2)`  `rArr I=2pi kR^(6) //6` ...........(1) Also, MASS of ring, `M = int dm = int_(r=0)^(R) (kr^(2))(2pi r dr)` Or, `M=2pi k R^(4)/4`........(2) From (2) and (2) `I= 4/6 MR^(2) = 2/3 MR^(2)` |
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| 893. |
Solve,x + 2y + 1 = 0, 2x - 3y - 12 = 0 |
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Answer» X = -3,y = -2 |
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| 894. |
Electromagnetic radiation of intensity I is made incident normally on a perfect nonreflecting surface (i.e perfect absorber). Then pressure developed on this surface would be …… . |
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Answer» IC `P_(S)=(F)/(A)=(((p)/(t)))/(A)=(p)/(At)=(((U)/(C )))/(At)=(U)/(ACt)=(I)/(C )` (where p = MOMENTUM, U = radiant ENERGY I = INTENSITY of radiation `= (U)/(At)`) |
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| 895. |
Which of following waves cannot propagate in vacuum ? |
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Answer» X - rays |
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| 897. |
Potential near the junction in p-n junction is zero then potential at …….. is ……… |
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Answer» n-side, NEGATIVE |
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| 898. |
Two photons of energies twice and thrice the work function of a metal are incident on the metal surface .Then, the ratio of maximum velocities of the photoelectrons emitted in the two cases respectively ,is |
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Answer» `SQRT2:1` |
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| 899. |
I is the intensity due to a source of light at any point P on the screen.Now the light reaches at P via two different paths (i) direct (ii) after reflection from a plane mirror.If the path difference between two paths is 3lambda//2, the intensity at P is : |
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Answer» zero `THEREFORE` Total path difference `= (3 lambda)/(2) + (lambda)/(2) = 2 lambda` This SATISFIES the condition of maxima. Hence the total intensity `infty (a + a)^(2)` `infty 4a^(2) = 4I`. |
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| 900. |
A load is supported using three wires of same cross section area as shown. Then fox |
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Answer» <P>`{:(P,Q,R,S),(2,3,1,4):}` |
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