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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.
| 4551. |
Frequency of sound wave is 600Hz. This wave Incidens perpendicular to open door of width 0.75m. At which angle is the first minima obtained ? (speed of light in air = 330 ms^(-1)) |
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Answer» `20.8` `0.75 sin theta_(1)= lambda ....(1)` But `v=lambda F rArr =(v)/(f)` `0.75 sin theta_(1)=(v)/(f)` `:. sin theta_(1)=(v)/(fxx0.75)=(330)/(600x0.75)=0.7333` `:.sintheta_(1)=0.7333` `:.theta_(1)=47^(@)=10.~~47.2^(@)` |
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| 4552. |
Write the relation between path difference and phase difference? |
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Answer» Solution :Phase difference `(phi)`: It is the difference expressed in DEGRESS or RADIANS between two waves having same frequency and referenced to same point in TIME. Path difference `DELTA)`: It is the difference between the LENGHTS of two paths of the two different having same frequency and travelling at same velocity. `delta=(lambda)/(2pi)phi` |
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| 4553. |
What is the pressure inside a small air bubble of 0.1 mm radius situated just below the surface of water ( T= 72 xx 10^ -3N/m) and atmospheric pressure = 1.013 xx 10^5 N//m^2) |
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Answer» `1.157 XX 10^5 N//m` |
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| 4554. |
Poonam's mother is diagnosed cancer. The attending physician told her that she has to undergo radiotherapy. While telling her the side effects of the treatement, the doctor told that her beutiful hair may fall and she may become bald. Poonam's mother refuses to get the treatment. Read the above passage and answer the following questions: (i) What would you do if you were in Poonam's place? (ii) What values are associated with your attitude? |
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Answer» Solution :(i) If I were in POONAM's place, I would TELL may mother that treatment of cancer is a must. Hair fallis a temprary effect end the hair would grow slowly after the therapy.Convincing my mother to get the treatment, using all sorts of reasoning will be my top priority. (III) Concern for my mother's health is of utmost importance. I will not mind taking HELP FORM specialists to convience my mother for getting the treatment. |
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| 4555. |
The most important use of studying nuclear fission and fusion process was to harness nuclear energy for useful purpose of humanity. Nuclear reactor is the machine designed to harness nuclear energy to electricity. The basic principle involved is fission of uranium to krypton and barium by slow moving neutrons as in following ""_(0)n^(1)+""_(92)^(235)Uto""_(32)^(96)Kr+""_(56)^(141)Ba+3_(0)n^(1)+Q(200MeV) reaction: But the difficulty is that one neutron striking with ""_(92)^(235)U produces three(2.6 on an average) neutrons which can cause further fission The neutron intensity available for further fission is controlled by measuring reproduction factor defined as the average number of neutrons from each fission the causes further fission. Maximum value of K 2.6. For optimum operation of nuclear reactor, the value of Kis mentioned in Column-1 for various stages in Columu-Il match the correct columns : {:("Column-I","Column-II"),((A)K=1,(p)"Uncontrolled chain reaction"),((B)Klt1,(q)"Critical"),((C)Kge1,(r)"Sub critical"),((D)K~~25,(s)"Super critical"):} |
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| 4556. |
Figure shows a square loop ABCD with edge length a. The resistance of the wire ABC is r and that of ADC is 2r. Find the magnetic field B at the centre of the loop assuming uniform wires. |
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| 4557. |
An equilateral prism deviates a ray through 45^@ for two angles of incidence differing by 20^@, mu of the prism is: |
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Answer» 1.567 |
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| 4558. |
An electric field converges at the origin whose magnitude is given by the expression E = 100 r N/C, where r is the distance measured from the origin. |
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Answer» TOTAL CHARGE contained in any spherical volume with its centre at origin is negative. |
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| 4559. |
A pendulum suspended from ceiling of a train has a time period T, when the train is at rest. When the train is accelerating with uniform acceleration 'a', the period of oscillation will |
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Answer» Decrease When train accelerates then net acceleration is `a_("net")=sqrt(g^(2)+a^(2))gt g` `:. T.=2pi sqrt((l)/(sqrt(g^(2)+a^(2))))` Sotime PERIOD decreases. Correct choice is (a). |
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| 4560. |
Using Bohr's formula for energy quantization determine the T.P. of the ground level of Li^++ atom. |
| Answer» SOLUTION :As `E= -13.6z^2eV/n^2, E_inf - E_1 = 122.4eV`. | |
| 4561. |
How does the dissipated power change, if a transmitted carrier wave is frequency modulated? |
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| 4562. |
(a) Use Gauss'slaw to derive the expression for theelectric filed (vecE) due tostraight uniformaly charges infinite line of charges density lambda C//m. (b) Draw a graphto show the variationof E withperpendicular from thelineof charge. (c) Find the work donein brining a charge q from prependicular distance r_(1) to r_(2) (r_(2) gt r_(1)). |
Answer» Solution :(a)  `cancel(O) = ointvecE .dveca` `cancel(O) = int E.da_(1) cos 90^(@) + intvecE.dveca_(2) cos 0^(@) + int vecE.dveca_(3) cos 90^(@)` `cancel(O) = int Eda_(2)` `cancel(O) = E_(2) pirl_(1)` `cancel(O) = (sumq)/(sum_(0)) = (lambdal_(1))/(sum_(0))` EQUATE them : `(dl_(1))/(sum_(0)) = E_(2)pirl_(1)` `E= (lambda)/(2pirsum_(0))` (b)  (c) WORK done to MOVING from `r_(1)` to `r_(2) dv underset(r_(1))overset(r_(2))int (lambda)/(2pisum_(0)r)dr` `dv = (lambda)/(2pirsum_(0))underset(r_(1))overset(r_(2))int(1)/(r) dr` `cancel(O) = (lambda)/(2pisum_(0))-[log.(r_(2))/(r_(1))]` `w = 9V` `w = 9((lambda)/(2pisum_(0))log((r_(2))/(r_(1))))` |
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| 4563. |
The focal length of the objective and eyepiece of a telescope are respectively 100 cm and 2 cm. The moon subtends angle of 0.5^(@), the angle subtended by the moon's image will be |
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Answer» `10^(@)` |
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| 4564. |
For the given circuit, If internal resistance of cell is 1.5Omega, then |
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Answer» <P>`V_(P)-V_(Q)=0` |
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| 4565. |
The magnetic moment is 5Am^2 and the pole strength is 25 A.m What is the length of the magnet ? |
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| 4566. |
An electric charge produces |
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Answer» ELECTRIC FIELD only |
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| 4567. |
A valance electron is known in a sodium atom is in the state with principle quantum numbern=3, with the total angular momentum being the greatest possible, What is its magnetic moment in that state? |
| Answer» Solution :A valence ELECTRON in a sodium ATOM is in the state with principal quantum number `n=3`, with the TOTAL angular MOMENTUM being the greatest possible, What is its magnetic moment in that state? | |
| 4568. |
Magnetic field lines show the direction (at every point) along which a small magnetised needle aligns (at the point). Do the magnetic field lines also represent the lines of force on a moving charged particle at every point ? |
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Answer» Solution :(a) No. The magnetic force is always normal to B (remember magnette force `=qv xx B).` It is misleading to call magiette fteld ithes as lines of force. (b) If field lines were entirely confined between two ENDS of a straight solenoid, the flux through the cross-section at each end would be non-zero. But the flux of field B through any closed surface must always be zero. For a toroid, this difficulty is absent because it has no .ends.. (c) Gauss.s law of magnetism states that the flux of B through any closed surface is always zero `int_(s) B. triangleS=0` If monopoles existed, the right hand side would be equal to the MONOPOLE (magnetic charge) `q_(m)` enclosed by S. (Analogous to Gauss.s law of electrostaties, `int_(S) B. triangleS=mu_(0)q_(m)" where "q_(m)` is the (monopole) magnetic charge enclosed by S.) (d) No. There is no force or torque on an element due to the field produced by that element itself. But there is a force (or torque) on an element of the same wire. (For the special case of a straight wire, this force is zero.) (E) Yes. THe average of the charge in the system may be zer. Yet, the mean of the magnetic moments due to various current loops may not be zero. We will come across such examples in connection with PARAMAGNETIC material where atoms have net dipole moment thorugh their net charge is zero. |
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| 4569. |
An alternating current is given by i=i_(1)cosomegat+i_(2)sinomegat. The rms current is given by |
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Answer» `(i_(1)+i_(2))/(SQRT2)` |
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| 4570. |
A point changes 60nC is placed at the centre of a a thick , inssulated , metallic spherical shell has radii 10 and 12 cm Find the electric firld at distance 5 11 and 15 cm from the centre what is the force between the point changes and the shell? |
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| 4571. |
State the working of a.c. generator with the help of a labelled diagram. The coil of an a.c. generator having N turns, each of area A, is rotated with a constant angular velocity omega. Deduce the expression for the alternating emf generated in the coil. What is the source of energy generation in this device? |
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Answer» Solution :For labelled diagram of an ac generator, see Fig. 6.55. Construction : An a.c. generator consists of the following four parts : (i) Armature: It is a rectangular coil consisting of a LARGE number of turns of insulated copper wire wound over a laminated soft iron core. The coil can be rotated about its central axis. (ii) Field magnet: Nand S are the pole pieces of a strong electromagnet, between which the armature coil is rotated. (iii) Slip rings : `R_(1) and R_(2)` are two hollow metallic rings, to which two ends of armature coil are connected. The rings rotate with the rotation of the coil. (iv) Brushes : `B_(1) and B_(2)` are two flexible metal plates, which are fixed and are kept in light contact with slip rings. Principle : See Short Answer Question Number 53. Theory and Working: As the armature coil ABCD is rotated in the magnetic field, angle between the field and normal to the coil changes continuously. Therefore, the magnetic FLUX linked with the coil changes and an emf is induced in the coil. Let initially, the armature coil be moving as shown in Fig. 6.59(a) so that AB moves OUTWARDS (towards US) and CD inwards (into the paper). The amount of magnetic flux linked with the coil is changing and in accordance with Fleming.s right hand rule, induced current in AB is from A to B and in CD it is from C to D. In the external circuit current flows from `B_(2) to B_(1)`. After half the rotation of the coil, AB moves inwards and CD outwards as shown in Fig. 6.59 (b). Now applying Fleming.s right hand rule, we find that the induced current in AB is from B to A and in CD  it is from D to C. In the external circuit, current flows from B to B . Thus, it is clear that the induced current in the external circuit changes its direction after every half rotation of the coil. Hence, the current induced is alternating in nature. Whenever plane of armature coil is perpendicular to the magnetic field, magnetic flux linked with the coil is maximum but change in magnetic flux is zero and hence magnitude of induced emf/current is zero. However, when plane of armature coil is parallel to the magnetic field, magnetic flux linked with the coil is zero and change in magnetic flux is maximum. Hence the magnitude of induced emf is maximum. In this manner, magnitude of induced emf/current goes on changing continuously For expression of induced emf, see Short Answer Question Number 53. |
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| 4572. |
What is the two important phenomena that govern the nano particles ? |
| Answer» Solution :QUANTUM confinement effects and surface effects are the two IMPORTANT PHENOMENA that govern NANO particles. | |
| 4574. |
Two coherent sources of intensity ratio 81:64 interfere. Deduce the ratio of intensities beween the maxima and minimain the interference pattern. |
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| 4575. |
The de Broglie wavelength of an electron moving with a velocity c/2 (c=velocity of light in vacuum) is equal to the wavelength of a photon. The ratio of the kinetic energies of electron and photon is |
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Answer» `1:4` `lambda_e = (h)/(pe) = (h)/(m_e (c/2))=(2h)/(m_ec)`.....(i) where `m_e` is the MASS of an electron Kinetic energy of an electron, `K_e =1/2 m_e (C/2)^2 = 1/8 m_e C^2` Kinetic energy of a photon, `K_(ph)=(HC)/(lambdaph)` `because lambda_e = lambda_(ph) `(Given) `THEREFORE K_(ph)= (hc)/(lambda_e)= (hc)/((2h)/(m_eC))=(m_ec^2)/(2)` (Using(i)) `therefore (K_e)/(K_(ph))= 1/8 m_eC^2 xx (2)/(m_ec^2)=1/4` |
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| 4576. |
A parallel beam of light of wavelength 6000 Å gets diffracted by a single slit of width 0.3 mm. The angular position of the first minima of differacted light is : |
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Answer» `2 xx 10^(3)` rad `0.3 xx 10^(-3) xx sin theta = 6000 xx 10^(-10)` `sin theta = 2 xx 10^(-3)` or `theta = 2 xx 10^(-3)` rad. |
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| 4577. |
A : At the first glance, the top surf Morpho butterfly.s wing appears a beautifu blue green. If the wing moves the colour changes. R : Different pigments in the wing reflect light at different angles. |
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Answer» Both A and R are true and R is the correct explanation of A |
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| 4578. |
In an P.N.P transistor circuit, the collector current is 10 mA. If 90% of the electrons emitted reach the collector. |
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Answer» Solution :Here, `I_E=10mA` As 90% of the holes REACH the COLLECTOR, so the collector, so the collector currents. `I_C=90%of I_E` `I_C=(90)/(100)I_E` `I_E=(100)/(90)I_C=(100)/(90)xx10` `I_E~=11mA` BASE current, `I_B=I_E-I_C=11-10` `I_B=1 mA` |
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| 4579. |
The carrier freqeuncy of a station is 40 MHz. A resistor of 10k Omega and capacitor of 12 pF are available in the detector circuit. The possible value of C will be |
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Answer» 12 `C=12pF=12xx10^(-12)F` `v_(c)=40MHz=40xx10^(6)Hz=4xx10^(7)Hz`. Time PERIOD of carrier frequency `T_(c)=(1)/(v_(c))-(1)/(4xx10^(7))=2.5xx10^(8)s` time CONSTANT of C-R circuit, As `T_(c)lt tau`, therefore, circuit is good enough for detection All the values of C given in the options are POSSIBLE. |
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| 4580. |
A gas undergoes a change in which its pressure P and volume V are related as pVn=constant, where n is a constant. If the specific heat of the gas in this change is zero, then the value of n is |
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Answer» `n=gamma-1` |
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| 4581. |
The period of a geostationarysatellite is : |
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Answer» Solution :The time period of geostationary satellite is 24 HOURS. So correct CHOICE is (a). |
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| 4582. |
A network of four 10 µF capacitors is connected to a 500 V supply, as shown in Fig. 2.29. Determine (a) the equivalent capacitance of the network and (b) the charge on each capacitor. (Note, the charge on a capacitor is the charge on the plate with higher potential, equal and opposite to the charge on the plate with lower potential.) |
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Answer» Solution :(a) In the given network, `C_(1),C_(2)` and `C_(3)` are connected in series. The effective capacitance C. of these three capacitors is given by `1/C. = 1/C_(1) + 1/C_(2) + 1/C_(3)` For `C_(1) = C_(2) = C_(3) = 10 muF, C. = (10//3) muF`.The network has C′ and `C_(4)`connected in parallel. THUS, the equivalent capacitance C of the network is `C = C. + C_(4) =(10/3 + 10) muF = 13.3 muF` (B) Clearly, from the figure, the charge on each of the capacitors, `C_(1), C_(2)` and `C_(3)` is the same, say Q, Let the charge on `C_(4)` be Q.. Now, since the potential difference across AB is `Q//C_(1)`. across BC is `Q//C_(2)`, across CD is `Q//C_(3)`,we have `Q/C_(1) + Q/C_(2) + Q/C_(3) = 500 V` Also, `Q.//C_(4) = 500 V` This gives for the given VALUE of the capacitances. `Q = 500 V xx 10/3 muF = 1.7 xx 10^(-3)` C and `Q. = 500 V xx 10 muF = 5.0 xx 10^(-3)` C |
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| 4583. |
The number density of free electrons in a copper conductor as estimated is 8.5xx 10^28 m^(-3) . Howlong does an electron take to drift from one end of a wire 3.0 m long to its other end ? The areaof cross-section of the wire is 2.0 xx 10^(-6) m^2and it is carrying a current of 3.0 A. |
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Answer» Solution : Here number density of free ELECTRONS `N = 8.5 xx 10^28 m^(-3)`, length of wire l = 3.0 m, cross-section area`A = 2.0 xx 10^(-6) m^2` and I = 3.0 A From the relation I=ne `A v_d`, we have `v_d = (I)/("neA") = l/t` where t is the time taken by free electrons to DRIFT from one end of conductor to the other end `t = ("neAl")/(I) = (8.5 xx 10^28 xx 1.6 xx 10^(-19) xx 2.0 xx 10^(-6) xx 3.0)/(3.0) = 2.72 xx 10^4 s` ` = (2.72 xx 10^4)/(60 xx 60) h = 7.5 h (appx)` |
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| 4584. |
The angle of minimum deviation for a prism is 40^(@) and the angle of the prism is 60^(@). The angle of incidence in this position will be |
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Answer» `30^(@)` `rArr""2anglei=100^(2) or anglei=50^(@)` |
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| 4585. |
27 small drops of water having same chargt and same radius are combined to form one big drop. The ratio of capacitance of one big drop to small drop is ....... |
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Answer» `2:1` `:. C_(B) : C_(S)= (27)^((1)/(3)):1` `:. C_(B) : C_(3)=3:1` |
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| 4586. |
Two radio transmitters radiating in phase are located at point A and B, 250 m apart. The ratio wave have frequency of 3MHz. A radio receiver is moved out from point B along a line BC (perpendicular to AB). The distance from B beyond which the detector does not detect any minima si : |
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Answer» 200 m |
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| 4587. |
If the coefficient of mutual induction of the primary and secondary coils of an induction coil is 6 H and a current of 5A is cut off in 1/5000 second, calculate the emp induced in the secondary coil. |
| Answer» SOLUTION :`|e| = M (DI)/(DT) , e = 6 XX (5)/(1//5000) V = 15 xx 10^4V` | |
| 4588. |
The speed of light in air is 3 xx 10^(8) ms^(-1) . If refractive index of glass is 1.5 , find the time taken by light to travel a distance of 10 cm in glass . |
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| 4589. |
A 100 Omega resistor is connected to a 220 V. 50Hz as supply. (a) What is the rms value of current in the circuit? (b) What is the net power consumed over a full cycle? |
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Answer» SOLUTION :Here, `V_(rms) = 220 V` and `R = 100 OMEGA` (a) `therefore I_(rms) = (V_(rms))/R = 220/110 = 2.2 A` (b) Power consumed over a full CYCLE `P = V_(rms) I_(rms) = 220 xx 2.2 = 484 W`. |
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| 4590. |
Time taken by a ray of light to travel through a glass slab of thickness 2.0 cm and R.I. 1.5 for normal incidence will be: (Velocity of light in air = 3xx10^8 m/s) |
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Answer» `0.66xx10(-8)s` |
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| 4591. |
The electric potential (V) in a certain region of space depends only on x-coordinate of point as V=-alpha x^(3)+beta (alpha and beta constants). Find the volume charge density (rho) of this region of space |
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| 4592. |
In an oscillator 0^(@) the electrostatic energy associated with the inductor is converted into ……………. . |
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Answer» ELECTROMAGNETIC ENERGY |
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| 4593. |
A battery of 6 volts is connected to the terminals of 3 m long wire of uniform thickness and resistance Fig. of the order of 60 Omega.The difference of potential between two points separated by 50 cm on the wire will be |
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Answer» 1V |
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| 4594. |
The average value of the alternating current over a complete cycle is |
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Answer» ZERO |
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| 4595. |
Give the expression for limit of resolution of a telescope along with the explanation of the symbols used. |
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Answer» SOLUTION :`d THETA=(1.22lambda)/(D)` where `'lambda'` WAVELENGTH of light emitted by the distance source/object D - DIAMETER of the aperture. |
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| 4596. |
A battery is used to charge a parallel plate capacitor till the potential difference between the plates become equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work clone by the battery will be ....... |
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Answer» 1 `= CV xx V = CV^(2)` The energy stored in capacitor `U =(1)/(2) CV^(2)` `:. " Ratio " (U)/(W)= ((1)/(2)CV^(2))/(CV^(2))=(1)/(2)` |
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| 4597. |
In some algae, the two gametes are so similar in appearance that is not possible to categorize them into male and female gametes. These gametes are called |
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Answer» Isogametes |
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| 4598. |
A particle performing S.H.M. has a velocity of 10 m/s, when it crosses the mean position. If the amplitude of oscillation is 2 m, then what is the velocity at midway between mean position and extreme position ? |
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Answer» SOLUTION :Velocity at MEAN position is maximum `v_max=aomega` `THEREFORE 10=2xxomega` ` therefore omega=5` v=omegasqrt(a^2-x^2) =5sqrt((2)^2-(1)^2)` `=5xxsqrt3 m//s.` |
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| 4599. |
Rectangle ABCD has area 200 sq. unit.An ellipse with area 200pi sq.unit passes through Aand C has focii at B and D. Then perimeter of the rectangle ABCD is |
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Answer» 80 UNIT  Area of ellipse `piab=200pi` `rArr` AB=200 …(1) Given AB x AD =200 …(2) AB+AD=2A Squaring bothsides `AB^2+AD^2 +2AB xx AD =4A^2` `BD^2 + 2 xx 200 = 4a^2` `4a^2e^2 + 400 =4a^2` `100=a^2 - a^2e^2` `100=b^2` `rArr` b=100 `therefore` ab=200 a=20 Perimeter of rectangle ABCD=2 (AB+AD) =2 x 2a =4a = 4 x 20 =80 |
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| 4600. |
A stone is projected vertically up to reach maximum height h. The ratio of its kinetic energy to its potential energy at a height (4h)/5 will be |
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Answer» A. `5:4` |
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