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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.
| 42301. |
(I) Cadmium show photoelectric emission for ultraviolet light. (II) Photosensitive materials eject photo electrons (III) The photocurrent becomes zero at a particular negative potential (IV) The liberation of electrons fromany surface of a substance is electron emission |
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Answer» I,II,III and IV |
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| 42302. |
The average number of neutrons released per fission in the fission of U^235 is |
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Answer» 7 |
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| 42303. |
An armada of spaceships that is 1.00 ly long (as measured in its rest frame) moves with speed 0.850c relative to a ground station is frame S. A messenger travels from the rear of the armada to the front with a speed of 0.950 relative to S. How long does the trip take as measured (a) in the rest frame of the messenger, (b) in the rest frame of the armada. and (c ) by an observer in the ground frame S? |
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Answer» |
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| 42304. |
Length of one metal rod is 1m. 2T is the intensity of the magnetic field. When this rod is rotating with the frequency of 10 Hz perpendicular to the field line by keeping its one end fixed as centre, induced emf is …… |
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Answer» SOLUTION :`EPSILON=(BomegaL^2)/2` `=(Bxx2pifxxL^2)/2` `=(2xx2pixx10xx(1)^2)/2` `=20pi` |
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| 42305. |
State Malus law. |
| Answer» Solution :If at any INSTANT the axis of polarizer subtends an angle `'theta'` with the direction of vibration of incident LIGHT, then the intensity of the EMERGENT `I=I_(0)cos^(2) 0`, where `I_0` is the MAXIMUM intensity corresponding to `theta=0^@`. | |
| 42306. |
A population inversion for two energy levels is often described by assigning a negative Kelvin temperature to the system. What negative temperature would describe a system in which the population of the upper energy level exceeds that of the lower level by 10% and the energy difference between the two levels is 2.32 eV? |
| Answer» SOLUTION :`-2.82xx10^(5)K` | |
| 42307. |
Atomic number of H-like atom with ionization potential 122.4 V for n = 1 is |
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Answer» 1 `THEREFORE z^(2) = (IE xx n^(2))/(13.6) = (122.4 xx (1)^(2))/(13.6) = 9 rArr z=3` |
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| 42308. |
Roforning to previous illustration, what is the time or crossing mo river assuming width of the river is 500 m? |
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Answer» Solution :`v_(m)=sqrt(v_(mv)^(2)+v_(m)^(2)+2V_(mv)V_(OMEGA)costheta)` `v_(m)=sqrt(v^(2)+4v^(2)-2V.2v.(1)/(2))` `v=(v_(m))/(sqrt3)=v_(MW)` `t=(d)/(v_(mw)costheta)=(500)/((5//sqrt3)sqrt(1-1//4))=(200xxsqrt3)/(sqrt3)=100sec` |
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| 42309. |
In a series LCR circuit, resistance R = 10 Omegaand the impedance z = 20 Omega . The phase difference between the current and voltage is |
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Answer» `pi/6` |
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| 42310. |
Give reason : "If net flux assocaited with closec surface is zero, then net charge enclosed b that surface is zero." |
Answer» Solution :The LAW implies that the TOTAL electric flux through a closed surface is ZERO if no charge is enclosed by the surface.  The electric field is uniform and we an considering a closed cylindrical surface with it axis parallel to the uniform field `vecE`. `phi_(1)` and `phi_(2)` represent the flux through the surfaces 1 and 2 of the cylinder and `phi_(3)`is the flux through the curved cylindrical part of the closed surface.  Hence, `phi_(1) =-ES_(1) =-ES (therefore S_(1) =S)` `phi_(2) = ES_(2) = ES (therefore S_(2) =S)` where S is the AREA of circular cross-section. THUS, the total flux is zero, the total charge contained in the closed surface is zero. |
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| 42311. |
What kind of Zeeman effect, normal or anomalous, is observed in a weak magnetic field in the case of spectral lines caused by the follwing transitions: (a).^(1)P rarr .^(1)S,(b).^(2)D_(5//2) rarr .^(2)P_(3//2), (c ) .^(3)D_(1) rarr .^(3)P_(0) , (d) .^(5)I_(5) rarr .^(5)H_(4) ? |
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Answer» Solution :(a) The term `.^(1)P_(1)` splits into `3` lines with `M_(Ƶ)` where `g=1+(1xx2+0-1xx2)/(2xx1xx2)=1` The term `.^(1)S_(0)` does not split in weak magnetic field. THUS the transition between `.^(1)P_(1)&.^(1)S_(0)` will result in `3` lines i.e., normal Ƶeeman triplet. (b) The term `.^(2)D_(5//2)` will split ot in`6` terms in accordance with the formula `DeltaE=-gmu_(B)M_(Ƶ)` `M_(Ƶ)= +-(5)/(2), +-(3)/(2),+-(1)/(2)`, and `g=1+(5xx7+1xx3-4xx2xx3)/(2xx5xx7)=(6)/(5)` THER term `.^(2)P_(3//2)` will also split into 4 lines in accordance with the above formula with `M_(Ƶ)= +-(3)/(2)+-(1)/(2)` and `g=1+(3xx5+1xx3-4xx1xx2)/(2xx3xx5)=(4)/(3)=(4)/(3)` It is seen that the `Z` eeman splitting is auomalous as `g` factors are DIFFERENT. (c )`.^(3)D_(1) rarr ^(3)P_(0)` The term `.^(3)D_(1)` splits into `3` levels`(g=5//2)` The term `.^(3)P_(0)` does not split. Thusthe `Z` eeman spectrum is normal. (d) For the `5l_(5)` term `g=1+(5xx6+2xx3-6xx7)/(2xx5xx6)` `=1+(36-42)/(60)=1-(1)/(10)=(9)/(10)` For the `.^(5)H_(4)` term `g=1+(4xx5+2xx3-5xx6)/(2xx4xx5)=1+(26-30)/(40)=(9)/(10)` We see that the splitting in the two levels given by `DeltaE= 0g mu_(B)BM_(Ƶ)` is the same though the number of levels is different (11 and 9). It is then easy to see that only the lines with following energies occur `ħ omega_(0),ħ omega_(0)+- gmu_(B)B`. The `Z` eeman pattern is normal. 
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| 42312. |
If luminous efficiency of a lamp is 2 lumen/watt and its luminous intensity is 42 candela, then power of the lamp is |
| Answer» ANSWER :d | |
| 42313. |
Using Bohr's postulates of the atomic model, derive the expression for radius of th electron orbit. Thus, obtain the expression of Bohr's radius. |
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Answer» Solution :We know that when an electron revolves in a stable orbit, the centripetal force is provided by ELECTROSTATIC force of attraction acting on it due to a proton present in the nucleus. `therefore""(m v_(n)^(2))/(r_(n)) = (1)/(4pi in_(0)) .(e^(2))/(r_(n)^(2))or v_(n) = (NH)/( 2pi m r_(n))"".....(i)` andfromBohr.s quantum CONDITION , we have . `m v_(n) r_(n) =(n h)/(2 pi) or v_(n) =(n h)/( n h)/(4 pi in_(0) m r_(n))""........(ii) ` Squaring(ii) and then equatingit with (i) , we get . `(n^(2)h^(2))/(4 pi^(2) m^(2) r_(n)^(2)) = (e^(2))/(4 pi in_(0) m.r_(n))rArrr_(n) = (n^(2) h^(2))/(4pi^(2) m^(2)) xx (4 pi in_(0) m)/(e^(2)) = ( in_(0)h^(2))/( pi m e^(2)) = n^(2)` In stable orbitof hydrogen atom n = 1and then radiusof 1st ORBITIS called Bohr.sradius`a_(0)` . Obviously ` a_(0)= (in_(0) h^(2))/(pi m e^(2))` |
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| 42314. |
ThreevectorsvecA, vecB, vecCareshownin thefigureanglebetween(i)vecA and vecB(ii)vecBandvecC,(iii)vecAand vecC |
Answer» SOLUTION :To FIND the ANGLE between TWO vectors we CONNECT the tails of the two vectors. We can shift the vectors parallel to themselves such that tails of A,B and C are connected as shown in figure.  Now we observe that angle_between A and B is 60°, B and C is 15° and between A and C is 75° |
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| 42315. |
A parallel plate capacitor is charged by a battery. After sometime the battery is disconnected and a dielectric slab with its thickness equal to the plate separation is inserted between the plates. How will (i) the capacitance of the capacitor, (ii) potential difference between the plates and (iii) the energy stored in the capacitor be affected ? Justify your answer in each case. |
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Answer» Solution : (i) Capacitance, `C = (K epsi_0 A)/(d)`.HENCE capacitance increases K times. (ii) Potential DIFFERENCE, V =` (V_0)/(K)`,hence potential difference decreases by a FACTOR K. (iii) Energy stored , ` E = 1/2 CV^2` ,as capacitance becomes K times & potential difference becomes 1/K times therefore energy stored becomes 1/K times. |
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| 42316. |
Mobility of electric charge mean., ...... per unit electric field. |
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Answer» RESISTANCE `[ RARR mu = (E)/(v_(d)) ]` |
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| 42317. |
The instantaneous current and voltage of an a.c. circuit are given by I = 10 sin 314 t A and V = 50 sin (314t + pi/2)V.What is the power dissipation in the circuit ? |
| Answer» Solution :As VOLTAGE is leading the current HENCE power FACTOR `cos phi=0`and power DISSIPATION in the circuit is zero. | |
| 42318. |
Which of the statements given in Exerciseis true for p-type semiconductors. |
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Answer» ELECTRONS are majority CARRIERS and TRIVALENT ATOMS are the dopants. A p-type SEMICONDUCTOR is obtained when trivalent atoms, such as aluminium are doped in silicon atoms. In a p-type semiconductor, the holes are the majority carriers, while the electrons are the minority carriers. |
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| 42319. |
A wire loop confined in a plane is rotated in its oen plane with some angular velocity. A uniform magnetic field exists in the region. Fid the emf induced in the loop. |
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Answer» SOLUTION :`i=V(Bxxl)` `=vvBl. Cos theta` `theta is the angle between normal to plane and B which is 90^(@)` `HENCE emf INDUCED = 90^(@).` `so, i=vBl cos 90^(@)=0. |
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| 42320. |
A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor? |
| Answer» SOLUTION :`1.5 XX 10^(-8)` J | |
| 42321. |
The energy of the ground state of hydrogen is -13.6eV. The energy of the first excited state is |
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Answer» `-27.2` eV `therefore E_(2) = (E_(1))/(E_(2)) = (-13.6 eV)/(4) = -3.4 eV` |
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| 42322. |
An inductor of self inductance 12 H carries a steady current of 2A. How can a 60V self-induced e.m.f. be made to appear in the inductor? |
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Answer» SOLUTION :`L=12H,""I=2A,VAREPSILON=60V` Let `(dI)/(DT)` be the rate of variation of current required, so that, `(dI)/(dt)=(varepsilon)/(L)=(60)/(12)=5A//s""therefore (dI)/(dt)xxt=2andt=(2)/((dI)/(dt))=(2)/(5)=0.4s` The current should be uniformly reduced from 2 to zero in 0.4 seconds. |
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| 42323. |
A sound wave propagates in a medium of Bulk modulus B by means of compressions and rare fractions. If P_( c) and P_(e )are the pressures at compression and rarefaction respectively, .a. be the wave amplitude and k be the angular wave number then, |
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Answer» `P_c` is maximum and `P_r` is MINIMUM. |
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| 42324. |
In the Bohr 's hydrogen atom model,the radius of the stationary orbit is directly porportional to the principle quantum number n as |
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Answer» `N^-2` |
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| 42325. |
Column-I shows four systems each of same length L for producing standing waves. The lowest possible natural frequency of a system is called it's fundamental frequency whose wavelength is denoted as lambda_(1) Match each system with statements given is Column-II describing the nature and wave length of the standing waves. |
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Answer» |
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| 42326. |
Figure shows a circuit that contains three identical resistors with resistance R=9.0Omega each, two identical inductors with inductance L=2.0mH each, and an ideal battery with emf e=18V. The current i through the battery just after the switch closed is,.....: |
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Answer» `0.2A` `i_(1)-=(e)/(R )=(18)/(9)=2A` :.CURRENT through the BATTERY is `i=2i_(1)=2xx2=4A` 
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| 42327. |
In Fig. a cockroach with mass m rides on a disk of mass 6.00 m and radius R. The disk rotates like a merry-go-round around its central axis at angular speed omega_(i)=1.50rad//s. The cockroach is initially at radius r = 0.800R, but then it crawls out to the rim of the disk. Treat the cockroach as a particle. What then is the angular speed? |
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Answer» Solution :(1) The cockroach.s crawl changes the mass DISTRIBUTION of the cockroach-disk system. (2) The angular momentum of the system does not change because there is no EXTERNAL torque to change it. (3) The magnitude of the angular momentum of a rigid body or a particle is given by Eq. `(L=Iomega)`. Calculations: We want to find the FINAL angular speed. Our key is to equate the final angular momentum `L_(f)` to the initial angular momentum `L_(i)`, because both involve angular speed. They also involve rotational inertia I. So, let.s start by finding the rotational inertia of the system of cockroach and disk before and after the crawl. The rotational inertia of a disk rotating about its central axis is given by `1//2MR^(2)`. Substituting 6.00 m for the mass M, our disk here has rotational ineria `I_(d)=3.00mR^(2)`  From Eq., we know that the rotational inertia of the cockroach is equal to `mr^(2)`. Substituting the cockroach.s initial RADIUS (r = 0.800R) and final radius (r = R), we find that its initial rotational inertia about the rotation axis is `I_(ci)=0.64mR^(2)` and its final rotational inertia about the rotation axis is `I_(cf)=mR^(2)` So, the cockroach-disk system initially has the rotational inertia `I_(i)=I_(d)+I_(ci)=3.64mR^(2)`, and finally has the rotational inertia `I_(f)=I_(d)+I_(cf)=4.00mR^(2)` Next, `(L=Iomega)` to write the fact that the system.s final angular momentum `L_(f)` is equal to the system.s initial angular momentum `L_(i)`: `I_(f)omega_(f)=I_(i)omega_(i)` or `4.00mR^(2)omega_(f)=3.64mR^(2)(1.50rad//s)`. After canceling the unknowns m and R, we COME to `omega_(f)=1.37rad//s`. |
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| 42328. |
"Know your face beauty through complexion meter" was one of the stall on a science exhibition. A student interestedto know his/her face beauty was made to stand on a platform and light from a lamp was made to fall on his/her face. The reading of complexion meter indicated the face beauty of the student which might be very fair, fair, semifair, semidark and dark, etc. Read abve passage and answer the following questions: (i) What is the basic concept used in the working of complexion meter? (ii) How is the face beauty recorded by face complexion meter? (iii) What basic values do you learn from the above study? |
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Answer» Solution :(i) The basic concept used in the working of complexion meter is PHOTOELECTRIC effects. (ii) The reading of complexion meter DEPENDS on the current generated from the reflected from the FACE. Fairer the COULOR more is the LIGHT reflected and vice versa. (iii) Complexion meters judge your face beauty only. The inner beauty of a person is judged by the virtues he or she possesess. In may opinion, inner beauty is much more valuable than the face beauty. |
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| 42329. |
Two insulated spheres of radii R_1 and R_2 having changes Q_1and Q_2 respectively are connected to each other. There is: |
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Answer» no change in the ENERGY of the SYSTEM |
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| 42330. |
Draw backs of Amplitude modulation |
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Answer» During transmission extraneous noise creeps in. |
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| 42331. |
When an excess charge is placed on an isolated conductor, it will distribute itself on the surface of that conductor so that all points of conductor whether on the surface or inside the conductor becomes to same potential. This is true even if there is a cavity inside the conductor. If an isolated conductor is placed in an external electric field, all points of conductor still come to a single potential regardless of whether the conductor has an excess charge. The free electrons distribute themselves on the surface in such a way that the electric field they produce at interior point cancels the external field that would otherwise be there. Choose the incorrect charge distribution |
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Answer»
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| 42332. |
When an excess charge is placed on an isolated conductor, it will distribute itself on the surface of that conductor so that all points of conductor whether on the surface or inside the conductor becomes to same potential. This is true even if there is a cavity inside the conductor. If an isolated conductor is placed in an external electric field, all points of conductor still come to a single potential regardless of whether the conductor has an excess charge. The free electrons distribute themselves on the surface in such a way that the electric field they produce at interior point cancels the external field that would otherwise be there. In electrical field, equipotential surfaces must |
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Answer» be PLANE surfaces |
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| 42333. |
An electric field 'E' whose direction is radially outward varies as distance from origin 'r' as shown in the graph E is taken as positive if its direction is away from the origin. Then the work done by electric field on a 2 C charge if it is taken from (1,1,0) to (3,0,0) is : |
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Answer» `20 (3 + SQRT(2))J` |
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| 42335. |
A student uses a simple pendulum of exactly 1 m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 s for this and records 40 s for 20 oscillations. For this observation, which of the following statements is true? |
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Answer» Error `DeltaT` in measuring T, the time PERIOD, is0.02 SECONDS. `(Deltat)/t=(1s) /(40S)=1/40` Time period , `T=(40 s)/20=2s` Error in measurement of time period `DeltaT=T xx (Deltat)/t= 2 s xx 1/40`= 0.05 s The time period of simple PENDULUM is `T=2pisqrt(l/g)` or `T^2=(4pi^2l)/g` or `g=(4pi^2l)/T^2` `therefore (Deltag)/g=(2DELTAT)/T=2xx1/40 =1/20 (because (DeltaT)/T = (Deltat)/t)` Percentage error in determination of g is `(Deltag)/g xx 100=1/20 xx 100`=5% |
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| 42336. |
इनमें से कौन नदी प्रदूषण के लिए उत्तरदाई है? |
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Answer» फैक्ट्रियों से निकलने वाले गंदगी का नदी में मिलना। |
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| 42337. |
सतलुज नदी में से किसकी है सहायक नदी है? |
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Answer» गंगा |
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| 42338. |
Solar cell converts_____energy to _____energy. |
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Answer» |
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| 42339. |
An observer can see through a pin-hole at the top end of a thin rid of height h, placed as shown in figure. The beaker height is 3h and its radius is h. When the beaker is filled with a liquid upto a height 2h, he can see the lower end of the rod. Find the refractive index of the liquid. |
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Answer» SOLUTION :`PQ =QR =2h` ` therefore(2525)i=45^@` `thereforeST =RT=h=KM=MN` so` KS= sqrt(h^2 +(2h)^2)=hsqrt(5)` ` thereforesinr = (h)/(h sqrt(5))=(1)/(sqrt(5))` ` thereforemu = (sin i)/( sin r ) = ( sin 45^@ )/( 1//sqrt(5))=sqrt((5)/(2))` |
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| 42340. |
A ray of light is passed from an optically denser medium to a rearer medium. Critical angle for the pair of media is C. The maximum angle of deviation of the ray will be |
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Answer» `pi - C` |
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| 42341. |
Three identical metal balls, each of the radius r are placed touching each other on a horizontal surface such that an equilateral triangle is formed when centres of three balls are joined. The centre ofthe mass of the system is located at |
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Answer» LINE joining centres of any TWO balls  Centre of mass of each ball lies on the centre. `rArr` Centre of mass of combined body will be at the CENTROID of equilateral TRIANGLE. |
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| 42342. |
A charged particle of mass m and charge q is projected on a rough horizontally xy- plane surface with z- axis in the vertically upward direction. Both electric and magnetic fields are acting in the region and given by vec(E)=-E_(0)hat(k) and vec(B)=-B_(0)hat(k) respectively. The particle enters into the field at (a,0,0) with velocity vec(v)=upsilon_(0)hat(j). The particle starts moving into a circular path on the plane. If the coefficient of friction between the particle and the plane is mu. Then calculate tha : (a) time when the particle will come to rest (b) time when the particle will hit the centre. ( c) distance travelled by the particle when it comes to rest. |
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Answer» Solution :`(a) N=mg+qE_(0)……………….(1)` `qB_(0)upsilon=(m upsilon^(2))/( R )………………(2)` and `-m(d upsilon)/(DT) = mu N …………………(2)` From equation `(2), ""R=( m upsilon)/(B_(0)q)……………..(4)` From equations `(1)` and `(3)` `-m.(d upsilon)/(dt)=mu ( mg +qE_(0))( o r)` `-m int_(v_(0))^(0)d upsilon=mu(mg+qE_(0))int_(0)^(t)dt` Thus `t=(m upsilon_(0))/(mu(mg+qE_(0)))` `(b)`From equation `(4)` `dR=(m)/(B_(0)q)d upsilon=-(mu(mg+qE_(0))dt)/(qB_(0))` `int_(R_(1))^(0)dR=(-mu(mg+qE_(0)))/(qB_(0))INT _(0)^(t)dt` or or`t=(qB_(0)R_(i))/(mu(mg+qE_(0)))` Here `R_(i)=(m upsilon_(0))/(B_(0)q)` Thus`t=(m upsilon_(0))/(mu(mg+qE_(0)))` `(c )``-m upsilon.(d upsilon)/(dl)=mu (mg+qE_(0))` or, `-m int_(upsilon_(0))^(0)upsilond upsilon=mu(mg+qE_(0))int_(0)^(t)dt ` or `l=(m upsilon_(0)^(2))/(2mu(mg+qE_(0)))` |
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| 42343. |
A ray is incident at an angle 38^@ with a mirror. The angle between normal and reflected ray is ........... . |
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Answer» `52^@` i+`38^@=90^@` `therefore i=52^@`  From the law of REFLECTION, angle of reflection = angle of incident `therefore R=i` `therefore` r=`52^@` |
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| 42344. |
A student performed the experiment to measure the speed of sound in air using resonance air - column method. Two resonance in the air - column were obtained by lowering thewater level. The resonance with the shorter air column is the first resonance and that with the longer air column is the second resonance. Then : |
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Answer» the intensity of the SOUND heard at the first resonance was more than that at the second resonance. |
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| 42345. |
As the mass number increases, the binding energy per nucleon in a nucleus |
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Answer» increases |
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| 42346. |
In a Hall effect measurements in magnetic field with induction B= 5.0kG the transverse electric field strength in an impurity-free germanium turned out be eta=10 times less than the longitudinal electric field strength. Find the difference in the mobilities of conduction electrons and holes in the given semiconductor. |
Answer» SOLUTION : If an electric field `E_(x)` is present in a sample containing equal AMOUNTS of both electrons and holes, the two drift in opposite directions. In the presence of a magnetic field `B_(Ƶ)=B` they SET up Halll voltage in opposite directions. The net Hall electric field is GIVEN by `E_(y)=(v_(x)^(+)-v_(x)^(-))B` `=(v_(x)^(+)-v_(x)^(-))B` `=(u_(0)^(+)u_(0)^(-))E_(x)B` But `(E_(y))/(E_(x))=(1)/(ETA)` Hence `|u_(0)^(+)-u_(0)^(-)|=(1)/(etaB)` Substitution gives `|u_(0)^(+)-u_(0)^(-)|= 0.2m^(2)//"volt-sec"` |
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| 42347. |
For a ball of mass 0.12 kg and moving with speed of 20 m/s .Calculate de-Broglie wave length. |
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Answer» <P> SOLUTION :p=mV`=0.12xx20` |
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| 42348. |
Self inductance of a solenoid is directly proportional to ___ |
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Answer» current passing through SOLENOID. Hence we have the relation `L PROP A` |
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| 42349. |
Which of the following is diamagnetic substance ? |
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Answer» Nickel |
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| 42350. |
A capacitor having a capacitance of 100muFis charged to a potential difference of 24 V. The charging battery is disconnected and the capacitor is connected to another battery of emf 12 V with the positive plate of the capacitor joined with the positive terminal of the battery.Find the heat developed during the flow of charge after reconnection. |
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Answer» 6.2 mJ |
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