Saved Bookmarks
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.
| 41451. |
Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius r lt lt h. Now a high voltage source (HV) is connected across the conducting plates such that the bottom plate is at +V_0 and the top plate at -V_0. Due to their conducting surface, the balls will get charged, will become equipotential with the plate and are repelled by it. The balls will eventually collide with the top plate, where the coefficient of restitution can be taken to be zero due to the soft nature of the material of the balls. The electric field in the chamber can be considered to be that of a parallel plate capacitor. Assume that there are no collisions between the balls and the interaction between them is negligible. (Ignore gravity) Which one of the following statements is correct? |
|
Answer» (a) The balls will stick to the top plate and remain there |
|
| 41452. |
A drop of water of mass m and density rho is placed between two weill cleaned glass plates, the distance between which is d. What is the force of attraction between the plates? (T= surface tension) |
|
Answer» `(2Tm)/(pid^(2)RHO)` |
|
| 41453. |
The wavelength of K_(alpha) line of an X-ray spectrum of an element of atomic no. Z= 11 is alpha.. The wavelength of same line of element of atomic no. Z' is 4lamda, then Z' is : |
|
Answer» 4 `V=^(2)(z-B)^(2)` `(v_(1))/(v_(2))=((z_(1)-b)^(2))/((z_(2)-b)^(2))` But `v=(c)/(lambda)` `(lambda_(2))/(lambda_(1))=((z_(1)-1)^(2))/((z_(2)-1)^(2))` `(4LAMBDA)/(lambda)=((11-1)^(2))/((z.-1)^(2))=4` Then `z=6` |
|
| 41454. |
An electric circuit requires a total capacitance of 2mu F across a potential of 1000V. Large number of 1muF capacitances are available each of which would breakdown if the potential is more than 350 V. How many capacitances are required to mae the circuit ? |
|
Answer» 24 Voltage rating of each capacitor = 350 V Supply voltage = 1000 V TOTAL capacitance `= 2mu F` Let n capacitors of `1mu F` each be connected in series in a row and m such rows be connected in parallel as SHOWN in the figure.  `because` Each capacitor can withstand 350 V. `:. n = (1000)/(350) = 2.8` As n cannot be fraction, therefore n = 3 Capacitance of each row of 3 capacitors of `1mu F` each in series is `C_(s) = (1)/(3) mu F` Total capacitance of m such rows in parallel `= (m)/(3) mu F` `:. (m)/(3) mu F = 2mu F` or m = 6 Total NUMBER of capacitors `= n xx m = 3 xx 6 = 18` |
|
| 41455. |
When the source moves with a velocity v, the ratio of wavelengths received by A and B will be (c= speed of Sound): |
|
Answer» `(C+V)/(c+v)` |
|
| 41456. |
The efective resistance in series combination of two equal resistance is .s.. When they are joined in parallel the total resistance is p. If s = np then the minimum possible value of .n. is |
|
Answer» 4 |
|
| 41457. |
Assertion :- The mutual inductance M between two coils of self inductances L_(1) and L_(2) is given by M gt sqrt(L_(1)L_(2)) Reason :- The energy density (i.e. energy per unit volume) at an poit, where the magnetic field B is given by (B^(2))/(2 mu_(0)) |
|
Answer» If the ASSERTION & REASON are True& the Reason is a correct explanation of the Assertion . |
|
| 41458. |
The number of electrons to be put on a spherical conductor of radius 0.1m to produce an electric field of 0.036N/C just above its surface is |
|
Answer» `2.7 XX 10^(5)` |
|
| 41459. |
If distance between source and screen increases by 2%, then intensity obtained at screen will be ....... |
|
Answer» <P> INCREASED by 4% `:. I prop (I)/(r^(2))` `:. (I_(2))/(I_(1))=((r_(1))/(r_(2)))^(2)` but `r_(1)=r` `2% "of" r_(2)=r_(1)+r_(1)` `=((r)/(1.02r))^(2) "" =r+rxx0.02` `=1.02r ` `=((1)/(1.02))^(2)` `=0.96` `:. I_(2)=0.96 I_(1) ""I_(1) gt I_(2)` `(I_(1)-I_(2))/(I_(1))xx100=(1-096I_(1))/(I_(1))xx100%` `=0.04xx100%=` decreasesed by 4 |
|
| 41460. |
{:("List-I","List-2"),((a)"Electron volt",(e)746 W),((b) "kWH", (f) 10^(15)m),((c)"Horse power" ,(g) 36xx10^(5)J),((d) "Fermi ,(h) 1.6xx10^(-19)J):} |
|
Answer» a - H`""` b- G `""`c -e `""` d-F |
|
| 41461. |
Calculate the electric and magnetic fields produced by the radiation coming from a 100 W bulb at a distance of 3 m. Assume that the efficiency of the bulb is 2.5% and it is a point source. |
|
Answer» Solution :The bulb, as a point source, radiates light in all directions UNIFORMLY. At a distance of 3 m, the surface area of the surrounding sphere is `A=4 pi r^(2) =4 pi (3)^(2)=113 m^(2)` The intensity I at this distance is `I=("POWER")/("Area")=(100 Wxx2.5%)/(113 m^(2))` `=0.022 W//m^(2)` Half of this intensity is provided by the electric FIELD and half by the magnetic field. `(1)/(2) I =(1)/(2) (epsi_(0) E_("rms")^(2) c)` `=(1)/(2) (0.022 W//m^(2))` `E_("rms") = sqrt((0.022)/((8.85xx10^(12))(3xx10^(8))) V//m` `=2.9 V//m` The value of E found above is the root mean SQUARE value of the electric field. Since the electric field in a light beam is sinusoidal, the peak electric field, `E_(0)` is `E_(0)=sqrt(2) E_("rms") =sqrt(2) xx2.9 V//m` `=4.07 V//m` Thus, you see that the electric field strength of the light that you use for reading is fairly large. Compare it with electric field strength of TV or FM WAVES, which is of the order of a few microvolts per metre. Now, let us calculate the strength of the magnetic field. It is `B_("rms") =(E_("rms"))/(C ) =(2.9 V m^(-1))/(3xx10^(8) ms^(-1))=9.6xx10^(-9) T` Again, since the field in the light beam is sinusoidal, the peak magnetic field is `B_(0) =sqrt(2) B_("rms") =1.4xx10^(-8) T.` Note that although the energy in the magnetic field is equal to the energy in the electric field, the magnetic field strength is evidently very weak. |
|
| 41462. |
When the number of turns and the length of a solenoid are doubled keeping the area of cross-section same, the inductance: |
|
Answer» is halved |
|
| 41463. |
A force between the two stationary charges separated by certain distance (a) obeys Newton.s third law (b) is a central force ( c) is non conservative force (d) is a scalar |
|
Answer» a is correct |
|
| 41464. |
Domain of a function is |
|
Answer» SET A |
|
| 41465. |
A capacitor having capacitance 2 muF is charged to a potential difference of 50 V. it is then diconnected from battery and connected to an inductor of inductance 5 mH. Peak current that flows through the inductor is |
|
Answer» 1A `(1)/(2) Li_(0)^(2) = (1)/(2)CV^(2)` `implies 5xx10^(-3)xxi_(0)^(2)=2xx10^(-6)xx2500` `implies i_(0)^(2)= (2xx10^(-6)xx2500)/(5xx10^(-3))=1` `implies i_(0)=1 A ` |
|
| 41466. |
Light with an energy flux of 18 w/cm^2 falls on a non-reflecting surface at naormal incidence. The pressure exerted on the surface is: |
|
Answer» `2N/m^2` |
|
| 41467. |
you have learnt in the text how Huygens principle leads to the laws of relfection and refraction. Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equla to the distance of the object from the mirror. |
|
Answer» Solution :In Fig. P is a POINT object PLACED at a distance `r` from a PLANE mirror `M_(1), M_(2)`. With `P` as centre and `PO = r` as radius, draw a spherical arc , `AB`. This is the spherical wavefront from the object, incident on `M_(1),M_(2)`. If mirror were not present, the position of wave front `AB` would appear as`P"B"`, according to huyghen's construction. As is CLEAR from the figure, `A'B'` and `A"B"` are two spherical arcs located symmetricallyon either side of `M_(1) M_(2)`. Therefore, `A'PB'` can be treated as reflected image of `A"PB"`. From simple geometry, we find `OP = OP'`. which was to be proved. 
|
|
| 41468. |
A ray of light incident at an angle theta on a refracting face of a prism emerges from the other face normally. If the angle of prism is 5^(@) and the prism is made of a material of refractive index 1.5, the angle of incidence is |
|
Answer» `7.5^(@)` `(sin theta)/(sin r_(1))=n or (sin theta)/(sin 5^(@))=1.5` As angle are small hence, we have `(theta)/(5^(@))=1.5 or theta=5xx1.5-7.5^(@)`. |
|
| 41469. |
A Man rowing a boat upstream at rest with respect to shore |
| Answer» SOLUTION :The man is doing work because he is applying a force. This is due to the fact that there is a RELATIVE motion between the BOAT and the stream. | |
| 41470. |
For the circuit shown in the figure. Find the expressions for the impedance of the circuit and phase of current |
|
Answer» Solution :Let ac SOURCE` V = V_(@) sin omegat` `I_(R)=(V_(@))/(R)SINOMEGAT,I_(L)=(V_(@))/(X_(L))sin(omegat-(pi)/(2))andI_(C)=(V_(@))/(X_(c))sin(omegat+(pi)/(2))` And as all these are in parallel, `I=I_(R)+I_(L)+I_(c)` `=[(V_(@))/(R)sinomegat+V_(@)(omegaC-(1)/(OMEGAL))cosomegat]` `(V_(@))/(R)=I_(0)sinphiV_(o)(omegaC-(1)/(omegaL)=I_(o)COsomegat` `I=I_(0)sin(omegat-phi)` `tanphi=(omegaC-(1)/(omegaL))//(1)(R)=R(omegaC-(1)/(omegaL))` `I_(0)=(V_(0))/(Z)V_(0)=[(10)/(R^(2))+(omegaC(1)/(omegaL))^(2)]^(t//2)` `z=(1)/(R^(2))+[(1)/(R^(2))+((1)/(X^(c))-(1)/(X_(L))^(2)]^(t//2)` |
|
| 41471. |
A current carrying loop of irregular shape is located in an external magnetic field. If the wire is flexible , why does it become circular? |
| Answer» SOLUTION : For a given PERIMETER, a CIRCLE encloses greatest area than any other shape. Thus in ORDER to involve the maximum number of field lines, the coil assumes circular shape with its plane normal to the field lines. | |
| 41472. |
(A): The working of dynamo is based on the principle of self induction. (R) : Self induction of a coil is numerically equal to the magnetic flux linked with the coil, when a unit current flow through it. |
|
Answer» Both .A. and .R. are TRUE and .R. is the CORRECT EXPLANATION of .A. |
|
| 41473. |
A p-n junction offers a high resistance in forward bias and a low resistance is reverse bias. |
| Answer» Solution :FALSE - A pn junction offers SMALL resistance in FORWARD bias and a high resistance in REVERSE bias. | |
| 41474. |
1 Mole glucose contain.... mole oxygen |
|
Answer» 4 |
|
| 41475. |
A horizontal stream of charged particles is accelerated to velocity 3 xx10^(7) m s^(-1) immediately before being allowed into an electric field between teo horizontal plates separated by 2 cm and maintained at a p.d. of 100 V. The stream is deflected by 5 mm. Calculate e/m of the charged particles of the stream. (Length of plates =10 cm). [Hint: Electric field between plates = rate of change of potential] |
|
Answer» |
|
| 41476. |
The flux associated with surface shown in figure is...... |
|
Answer» `(4e)/epsilon_(0)` Charge of down quark `=-e/3` `THEREFORE` TOTAL charge enclosed `q=(2e)/3-e/3-e/3` `therefore q=0` `therefore` Flux associated, `phi=q/epsilon_(0)=0` |
|
| 41477. |
The greatest height to which a man can throw a ball is ‘A’, the greatest distance to which he can throw it will be : |
|
Answer» H Then `H_(max)=(u^(2))/g` `R_(max)=u^(2)/g` Clearly `R_(max)=2H_(max)=2h` |
|
| 41478. |
Answer the following questions: (c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures? |
| Answer» SOLUTION :At low pressures, IONS have a chance to reach their respective electrodes and constitute a CURRENT. At ordinary pressures, ions have no chance to do so because of COLLISIONS with gas molecules and RECOMBINATION. | |
| 41479. |
Unit of surface charge density (sigma) is……. |
|
Answer» `(C )/(m^(2))` |
|
| 41480. |
In the question number 18, the potential at a point 20 cm from the mid-point of the line joining the two charges in a plance normal to the line and passing through the mid-point is |
|
Answer» `1.4xx10^(5)V` `=1/(4piepsilon_0(PA))(q_1+q_2)` `(because PA=PB=sqrt((0.2)^2+(0.2)^2)=0.28 m) =(9xx10^9(1.8+2.8)xx10^(-6))/0.28=1.4xx10^5` V 
|
|
| 41481. |
The wavelength of X-rays lies between: |
|
Answer» MAXIMUM to FINITE limits |
|
| 41482. |
The circuit shown in Fig. contains two diodes each with a forward resistance of 50 ohm and with infinite reverse resistance. If the battery voltage is 6V, find the current through the 100 ohm resistance. |
Answer» Solution :As per given circuit, diode `D_(1)` is FORWARD BIASED and offers a resistance of 50 ohm. Diode `D_(2)` is REVERSE biased and as its corresponding resistance circuit is as SHOWN in Fig.As all the three resistances are in SERIES, the current through them is  `I=(6V)/((50+150+100)W)=(6)/(300)A=0.02A` |
|
| 41483. |
Red light however bright it is,cannot produce the emission of electrons from a clean zinc surface , but even weak ultraviolet radiation can do so, why ? |
|
Answer» Solution :(i) The photoemission of electron does not depend on the intensity but it depends on the frequency and HENCE on the energy of photon of incident light. (ii) If the energy of photon is GREATER than the WORK function, the photoemission of electrons results however weak the incident radiation MAY be. (iii) The energy of photon of red light is less than the work functionof zinc, so red light cannot emit photoelectrons. (iv) The energy of photon of ultraviolet light is greater than the work function of zinc, so ultraviolet light can emit photoelectrons. |
|
| 41484. |
A clock fixed on a wall shows time 04 : 25:37. What time will its image in a plane mirror hanging on an opposite vertical wall show? |
|
Answer» `07:43:32` |
|
| 41485. |
In Fig. the input is across the terminals A and C and the output is across B and D. Then the output is |
|
Answer» half wave RECTIFIED |
|
| 41486. |
(A) :· The range of given voltmeter can be both increased and decreased. (R) : By adjusting the value of resistance in series with galvanometer the range of votlmeter can be adjusted. |
|
Answer» Both 'A' and 'R' are TRUE and 'R' is the CORRECT EXPLANATION of 'A'. |
|
| 41487. |
E तीव्रता वाले विद्युत - क्षेत्र में q आवेश रखने पर उस पर लगने वाला होगा |
|
Answer» `F=E/q` |
|
| 41488. |
To a person travelling due east with a velocity V, the wind appears to blow from an angle theta North of east. When he starts travelling due North with a velocity 2V, the wind appears to blow from an Angle betaNorth east find the time direction of wind. |
| Answer» Solution :` alpha ` westofsouthandtan ` alpha=(1 + 2cottheta)/( 2 + tan THETA)` | |
| 41489. |
Current of 50/pi Hz frequency is passing through an A.C. circuit having series combination of resistance R = 100Omega and L = 1 H, then phase difference between voltage and current is ………. |
|
Answer» Solution :`f=50/pi` Hz `THEREFORE omega=2pif =2pixx50/pi`=100 rad/s `X_L=omegaL=100xx1=100Omega` and in R-L series A.C. circuit, the phase difference between voltage and current `delta=TAN^(-1)(X_L/R)=tan^(-1) (100/100)=tan^(-1) (1)` `therefore delta=pi/4` or `45^@` |
|
| 41490. |
The ratio of volume of atom to that of nucleus |
|
Answer» `10^15` |
|
| 41491. |
There is a straight wire suspended in magnetic field and there is a ring made out from the wire of same length also, suspended in same magnetic field. When both are made to oscillate like pendulum inside a magnetic field then which one will stop earlier? |
| Answer» SOLUTION :In case of ring induced CURRENT can flow and cause loss of energy DUE to electric heating, so ring will stop oscillating earlier than rod. In case of rod emf MAY induce but due to absence of closed loop there would not be any current and so it will stop only when the rod loses complete MECHANICAL energy against air resistance, so it will take more time to stop. | |
| 41492. |
What is conductivity ? Its value depend on which factors ? Write its unit and dimension. |
|
Answer» Solution :Reciprocal of resistivity is called conductivity. It is DENOTED as `sigma`. `therefore sigma = (1)/(rho)= (l)/(RA)` UNIT of conductivity is `(1)/(Omega m)` or `Omega m^(-1)`. `Omega` is reciprocal of resistance `(Omega)`. `RARR` Dimensional formula of conductivity is `M^(-1) L^(-3) T^(3)A^(2)`. `rArr` VALUE of condurtivlty depend on type of material, temperature and PRESSURE. `Value of conductivity do not depend on dimension . |
|
| 41493. |
In viewing through a magnifying glass, one usually positions one's eyes very close to the lens, does angular magnification change if the eye is moved back? |
|
Answer» Solution :Yes, it DECREASES a little because the angle subtended at the eye is then slightly less than the angle subtended at the lens. The effect is negligilble if the IMAGE is at a very large distance away. [ Note: When the eye is SEPARATED from the lens , the angles subtended at the eye by the first object and its image are not EQUAL . ] |
|
| 41494. |
The name of the gate obtained by the combination as shown is |
|
Answer» NAND |
|
| 41495. |
The temperaturecoefficientof resistanceof tungsten is 4.5 xx10^(-3) ""^(@)C^(-1)and thatof germaniumis- 5 xx 10^(-2)""^(@)C^(-1) . A tungstenwireof resistance100 Omega isconnectedin series with a germaniumwireofresistanceR. The valueof Rfor which the resistanceof combinationdoes notchange withtemperatureis . |
|
Answer» `9 OMEGA` |
|
| 41496. |
The ratio of velocity of light in emty space to the velocity of light in the medium we call ? |
| Answer» SOLUTION :REFRACTIVE INDEX | |
| 41497. |
A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300" rad s"^(-1) . (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates. |
|
Answer» Solution :(a) `I_("RMS") =V_("rms") omega C=6.9 mu A` (b) Yes. The derivation in Exercise 8.1(b) is true EVEN if i is oscillating in TIME. (c ) The FORMULA `B =(mu_(0) r)/(2 pi R^(2)) i_(d)` goes through even if `i_(d)` (and therefore B) oscillates in time. The formula shows they oscillate in phase. Since `i_(d)= i`, we have `B_(0)=(mu_(0) r)/(2 pi R^(2)) i_(0)`," where "B_(0) and i_(0)` are the amplitudes of the oscillating magnetic field and current, RESPECTIVELY. `i_(0) =sqrt(2) I_("rms")=9.76 mu A.` For `r=3 cm, R=6 cm, B_(0)=1.63xx10^(-11) T.` |
|
| 41498. |
A wooden box lying at rest on an inclined surface of a wet wood is held at static equilibrium by a constant force applied perpendicular to the incline. If the mass of the box is 1kg, the angle of inclination is 30^(@) and the coefficient of static friction between the box and the inclined plane is 0.2, the minimum magnitude of 7 is (Use g= 10m//s^2 ) |
|
Answer» 0 N , as `30^(@)` is less than angle of repose |
|
| 41499. |
An open truck is moving with a uniform velocity of 10 ms^(-1) If rain adds water at the rate of 5 kgs with zero velocity, then the additional force applied by the engine to maintain the same velocity is : |
|
Answer» `0.5N` (C) is the choice |
|
| 41500. |
The charge on 500 cc of water due to protons will be |
|
Answer» `6.1 xx 10^(27) C ` |
|