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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.
| 39351. |
In the above question "(A body falls from rest. In the last second of its fall, it covers half of the total distance)" the total height of fall is : |
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Answer» Solution :Here `s=4.9xxn^(2)` or `s=4.9(2+sqrt(2))^(2)` =56 m |
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| 39352. |
प्रकाश एक माध्यम से जिसका अपवर्तनांक n_1 है, दूसरे माध्यम में जिसका अपवर्तनांक n_2 है, जाता है। यदि आपतन का कोण i तथा अपवर्तन का कोण r हो, तो sin i /sinr बराबर होता है |
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Answer» `n_1`के |
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| 39353. |
Emission of electrons bythe absorption of heat energy is callled……………………emission. |
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Answer» PHOTOELECTRIC |
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| 39354. |
Two light waves superposing at the midpoint of the screen are coming from coherent sources of light of phase difference 3 pi radian. Their amplitudes are 1 cm. What will be the resultant amplitude at the given point. |
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Answer» |
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| 39355. |
Electrostatic field is always normal to the surface of a charged conductor. Justify the statement |
| Answer» SOLUTION :DIVERGING LENS | |
| 39356. |
Side rail of length 2L are fixed on a horizontalplane at a distane l from each other These ends aer connected by two identical ideal batteries with emf E by sistancelesss wires (see) On therails is a rod of mass m, which may slide along them The entire system is placed in a uniform vertical magnetic field B Assuming that the resistance of the rodi is R and the resistanceper unit length of each of therails equal to p find the period of small oscillations (in sec) arsing from shifting therod from the equlibrium along the rails Neglect friction internal resistance of batteries and induced emf in the rod [Take : B = pi T, varepsilon = pi volt l = 0.5, L =1m,p = 1Omega//m,R =0.25Omega m = 100 gm] . |
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Answer» |
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| 39357. |
The focal length of a cocave mirror is f and the distane from the object to the principal focus is x. The ratio of the size of the image to the size of the object is : |
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Answer» `(F + x)/(f)` |
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| 39358. |
(a) State Bohr's postulateto define stableorbits in hydrogenatom. Howdoes de Broglie'shypothesis explainthe stability of these orbits ?(b) A hydrogen atom initially in the groundstate absorbs a photon whichexcites itto then n = 4 level . Estimate the frequenyof the photon. |
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Answer» Solution :Bohr. s postulate for stable orbits ,states The electron , in an atom revolves around the mucleus only in those orbits for which its angular MOMENTUM is an intergal multiple of` (h)/( 2h ) (h= `planck. s constant ) [Also accept mvr=`n.( h)/(2PI ) (n=1,2,3......) ` As per de Brogle.s hyothesis ` lambda =(h)/(p) =(h)/( mv) ` For a stable orbit we, must have circumference of the orbit `=n( n=1,2,3..........) ` ` therefore 2pi r =n.mv` or `mvr =(nh)/( 2pi) ` Thus de- Broglie showed that formation of stationary PATTERN for integral .n. gives RISE to stability of the atom . This is nothing but the Bohr.s postulate |
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| 39359. |
A rod of length l is perpendicular to the lines of induction of a uniform magnetic field of induction B. The rod revolves at an angular speed omega about an axis passing through the rod's end parallel to the lines of induction. Find the voltage across the rod's ends. |
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Answer» (a) By integration. We have `triangle Psi=underset(0)overset(1)int B omega x dx=B omega [x^2/2]_(0)^l=1/2 Bwl^2` (b) Graphically. Plot a graph of the strength of the induced field `E^*=triangleepsi //trianglex=B omega x`. Since this is a LINEAR function, its graph is of the form shown in Fig 30.4 b. The voltage is numerically equal to the AREA under the graph : `triangle Psi=(lBwl)/(2)-(Bwl^2)/2` 
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| 39360. |
The internal resistance and internal reactance of an alternating current generator are R_g and X_g respectively. Power from this source is supplied to a load consisting of resisting R_g and reactance X_L. For maximum power to be delivered from the generator to the load. The value of X_L is equal to |
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Answer» zero |
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| 39361. |
A ray of light travelling in a rarer medium, strikes a plane boundary between the rarer medium and a denser medium at an angle of incident 'i' such that the reflected and the refracted rays are mutually perpendicular. Another ray of light of same frequency is incident on the same boundary from the side of denser medium. Find the minimum angle of incidence at the denser-rarer boundary so that the second ray is totally reflected. |
| Answer» SOLUTION :`theta_C = SIN^(-1) ( COT i)` | |
| 39362. |
Draw the labelled circuit diagram of a common-emitter transistoramplifie : Explain clearly how the input and output signals are in opposite phase. |
Answer» Solution :The circuit details for using an npn transistor as common emitter amplifier are shown in the fig.  The input (base-emitter) circuit is forward biased and the output (collector-emitter) circuit is reverse biased. When no a.c. signal is applied, the potential difference `V_(c)` between the collector and the emitter , is given by `V_(C)=V_(CE)-I_(C)xxR_(L)` ...(i) Where, `V_(C)` is the voltageof battery `V_(CE)` When an a.c. signal is fed to the input circuit, the forward bias increases during the positive half cycle of the input. This results in an INCREASE in `I_(C)` and a consequent decrease in `V_(C)`, as is clear from (i). THUS during positive half cycle of the input, the collector becomes less positive. During the negative half cycle of theinput, the forward bias is decreased resulting in a decrease in `I_(E)` and hence `I_(C)`. Therefore, from (i) `V_(C)` would increase, MAKING the collector more positive. Hence, in a common-emitter amplifier, the output voltage is `180^(@)` out of phase with the input voltage. |
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| 39363. |
The graph Shows the variation in magnetic flux phi(t) with time through a coil. Which of the statements given below is not correct |
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Answer» There is a change in the direction as well as MAGNITUDE of the induced EMF between B and D |
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| 39364. |
Differentiate between half-life and average life of a radioactive substance. |
Answer» SOLUTION :
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| 39365. |
A TV tower has a height of 100m. What is the maximum distance upto which TV transmission can be received ? (R = 6.4 xx 10^(6) m) |
| Answer» Answer :A | |
| 39366. |
Stationary wave donot transfer |
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Answer» energy |
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| 39367. |
A pendulam of length L and bob of mass M has a speing of force constant K connected a speing of force constsnt K connected horizontally to it at a distance h below is point of suspension. The rod is in equilibrium in vertical podition. The rod of length L. used for vertical suspension s rigid and massless. The frequency of vibration of the system for small values of theta is |
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Answer» `(1)/(2piL)sqrt(gL+(kh)/(m))` Forsmall angle sin `theta ~~ theta` `T=lalpha=-(mgL+kh^(2))theta` `alpha=-(mgL+kh^(2))theta//mL^(2)` `Sof=(1)/(2piL)sqrt(gL+(kh^(2))/(m))` |
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| 39368. |
Electric field given by the vector vec E = x hat i + y hat j is present in the x y plane. A small ring carrying +Q, which can freely slide on a smooth nonconducting rod, is projected along the rod from the point (0, L) such that it can reach the other end of the rod. What minimum velocity should be given to the ring ? (Assume zero gravity). . |
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Answer» `(Q L^2//m)^(1//2)` or `(0 + (1)/(2) mv^2) + q Delta V = 0` `underset(V_(2))overset(V_(1))(int) d V = - [ int E_x d x + int E_y d y]` `Delta V = - [underset(0)overset(L)intx d x + underset(L)overset(0)(int) y dy] = 0` HENCE `v = 0`. |
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| 39369. |
(A): The BE per nucleon of the fused heavier nuclei is more than the B.E per nucleon of the lighter nuclei. (R): The final system is more tightly bound to the initial system. |
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Answer» Both .A. and .R. are TRUE and .R. is the CORRECT EXPLANATION of .A. |
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| 39370. |
The magnifying power of an astronomical telescope for normal adjustment is 10 and the length of the telescope is 110cm..Find the magnifying power of the telescope when the image is formed at the least distance of distinct vision for normal eye. |
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Answer» 14 |
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| 39371. |
Nuclear radius of ""_(82)^(205)Pb is 7 fermi. Calculate the nuclear radius of ""_(8)^(16)O. |
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| 39372. |
A galovameter has resistance 500 ohm. It is shunted so that its sensitivity decreases by 100 times. Find the shunt resistance. |
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Answer» Solution :`"Sensitivity" PROP (1)/("range") "" therefore n = 100` `S = (G)/((n-1)) = (500)/((100-1)) = (500)/(99) Omega = 5.05 Omega` |
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| 39373. |
The wavelength of a monochromatic light in vacuum is gamma. It travels from vacuum to a medium of absolute refractive index mu. The ratio of wavelength of the incident and refracted wave is |
| Answer» ANSWER :C | |
| 39374. |
Four capacitors, each of 1 muF, are joined as shown in Fig.The equivalent capacitance between the points A and B is |
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Answer» `1 mu F` |
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| 39375. |
Two capacitors are in parallel and when connected to a source of 3000 V, store 250 J of energy. When they are connected in series to the same source, the energy stored decreases by 190 J for the same potential. Their capacities are in the ratio |
| Answer» ANSWER :A | |
| 39376. |
In the above problem the value of E in the space outside the sheets is |
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Answer» `sigma//epsilon_(0)` |
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| 39377. |
Consider two concentric spherical metal shells of radii r_(1) and r_(2) (r_(2)gt r_(1)). Find the charge on the inner shell and charge distribution. |
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Answer» Solution :Let charge on inner shell be `q`. The potential at `P` `V_(P) = (1)/(4 pi in_(0)) ((q)/(r_(1)) + (Q)/(r_(2))) = 0` `q = -(Qr_(1))/(r_(2))` Charge on inner shell `= -Qr_(1)//r_(2)` and it will be on OUTER surface. `(Qr_(1)//r_(2)` has been transferred to earth leaving negative charge on inner shell.) Charge DISTRIBUTION : The facing will surfaces willhave EQUAL and opposite charges and total charge on outer shell `Q`.  
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| 39378. |
If the Fresnel's biprism experiment is performed in water instead of air, then the distance (d) between the coherent sources becomes nearly : |
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Answer» 1.3 times |
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| 39379. |
How much positive and negative charge is there in a cup of water? |
| Answer» Solution :Let us ASSUME that the mass of one cup of WATER is 250 g the MOLECULAR mass of water is 18 g THUS one moleof water is 18 g therefore the number ofmolecules in onecup of wateris `(250//18)xx6.02 xx10^(23)`each molecule of water containstwo hydrogen atoms and one oxygentotal negative charge has the same magnitude it is equal to `(250//18)xx6.02 xx10xx1.6 xx10^(-10^(-19) C=1.34 xx10^(7)C` | |
| 39380. |
In a parabola spectrograph, the velocities of four positive ions P,Q,R and S are v_(1),v_(2),v_(3) and v_(4) respectively |
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Answer» `v_(1) gtv_(2)gtv_(3)gtv_(4)` |
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| 39381. |
Which of the following graphs shows the variation of magnetic induction B with distance 'r' fromalong wire carrying current ? |
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Answer»
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| 39383. |
In Young.s double-slit experiment using monochromatic light of wavelength lambda, the intensity of light at a point on the screen where path difference is lambda, is K units. What is the intensity of light at a point where path difference is lambda"/"3? |
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Answer» Solution :Phase difference corresponding to `lambda` is `2PI`. Phase difference corresponding to `(lambda)/(3)" is "(2pi)/(3)` `I= I_(1)+I_(2)+2 sqrt(I_(1)I_(2)) cos phi""` ASSUME that `I_(1)= I_(2)= I_(0)` In the first case, `K= I_(0)+I_(0)+2I_(0) cos 2pi = 4I_(0)` In the second case, `K= I_(@)+I_(0)+2I_(0) cos ""(2pi)/(3)` `I_(@)+I_(0)-2I_(0)(1/2)= I_(0)` Now `(K.)/(K)= (I_0)/(4I_0)= (1)/(4)" or "K.= (K)/(4)`. |
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| 39384. |
Below Figure represents a painter in a crate which hangs alongside a building. When the painter of mass 100 kg pulls the rope, the force exerted by him on the floor of the crate is 450N. If the crate weighs 25 kg, find the acceleration and tension in the rope, (g = 10 m//s^(2)) |
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Answer» |
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| 39385. |
A parallel-plate capacitor of plate area 40cm^(2) and separation between the plates 0.10 mm is connected to a battery of emf 2.0 V throgh a 16Omega resistor. Find the electric field in the capacitor 10 ns after the connections are made. |
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Answer» `51muV//m` |
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| 39386. |
Light with wavelength lambda falls on a system of crossed polarizer P and analyzer A between which a Babinet compesator C is inserted (Fig.). The compensator consists of two q2uartz wedges with the optical axis of one of them being parallel to the edge, and of the other, perpendicular to it. The principal directions of the polarizer and the analyser from an angle of 45^(@) with the optical axes of the compensator. The refracting angle of the wedges is equal to Theta(Theta lt lt 1) qand the difference of refractive indices of quartz is n_(e) - n_(0). The insertion of investigated birefringent sample S, with the optical axis oriented as shown in the figure, results in displacement of the fringe upward by deltax mm. Find: the width of the fringe Deltax, (b) the magnitude and the sign of the optical path difference of ordinary and extraordianry rays, which appears due to the sample S. |
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Answer» Solution :The light from `P` is plane polarized with its electric vector vibrating at `45^(@)` with the plane of the paper. At first the sample `S` is absent Light from `P` can be resolved into components vibrating in and perpendicuar to the plane of the paper. The former is the `E` ray in the left half of the Babinet components and the latter is the `O` ray. In the right half the nomenclature is the opposite. In the compenstor the two components acquire a phase difference which depends on the relative position of the ray. If the ray is incident at a distance `x` above the central line through the compensator then the `E` ray acquires a phase `(2pi)/(lambda) (n_(E) (l - x) + n_(0) (l +x))tan Theta` while the `O` ray acquires `(2pi)/(lambda) (n_(0)(l - x) +n_(E) (l + x)) tan Theta` so the phase difference between the two reays is `(2pi)/(lambda) (n_(0) - n_(E)) 2 x tan Theta = delta` we get dark FRINGES when ever `delta = 2k pi` because then the emergent light is the same as that coming form the polarizer and is quenched by the analyser. {If `delta = (2k +1) pi`, we get BRIGHT fringes beacuse in this case, the plane of polarization of th emergent hight has rotated by `90^(@)` and is therefore fully transmitted by the analyser.} If follows that the fringe width `Deltax` is given by `Delta x = (lambda)/(2|n_(0) - n_(E)|tan Theta)` (b) If the finges are displaced upwards by `delta x`, then the path difference intoduced by the sample between the `O` and the `E` rays must be such as to be exactly CANCELLED by the compensator. Thus `(2pi)/(lambda) [d (N'_(0) - n'_(E)) + (n_(E) - n_(0)) 2 delta x tan Theta] = 0` or `d(n'_(0) - n'_(E)) =- 2(n_(E) - n_(0)) delta x Theta` usng `tan Theta_(~=) Theta`.  
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| 39387. |
a. What is the purpose of Cu_(2)? b. Which plate becomes positive? c. Which plate becomes negative? |
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Answer» SOLUTION :a. Cuprous oxide is a photosensitive semiconductor. b. Cu c. AG |
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| 39388. |
Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel (a) What is the total capacitance of the combination? (b) Determine the charge on each capacitor if the combination is connected to a 100 V supply |
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Answer» Solution :For all parallel combination of the capacitors equivalent capacitance, `C_(p) = C_(1) +C_(2)+C_(3)` where `C_(1)=2` pF `C_(2)= 3 pF` and `C_(3) = 4 pF ` =2+3+4 =9 pF `= 9xx10^(-12)` F (B)  SUPPOSE, the `q_(1),q_(2)` and `q_(3)` are the charges on capacitor `C_(1), C_(2)` and `C_(3)` respectively and supply voltage is 100 V. `:. q_(1)=C_(1)V=2xx10^(-12) xx100=2xx10^(-10)C` `q_(2)=C_(2)V= 3XX10^(-12)xx10^(-12)xx100=3xx10^(10)C ` `q_(3)= C_(3)V= 4xx10^(-12)xx100= 4xx10^(10)C` |
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| 39389. |
Broadcasting antennas are generally |
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Answer» OMNIDIRECTIONAL TYPE |
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| 39390. |
The total energy of an electron in the firstexcited state of hydrogen atom is about –3.4eV. Its kinetic energy in this state is |
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Answer» `3.4` EV |
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| 39391. |
In a double slit experiment ,the separation between the slits is d=0.25cm and the distance of the screen D=100cm from the slits .if the wavelength of light used in lambda=6000Åand I_(0)is the intensity of the central bright fringe.the intensity at a distance x=4xx10^(-5)in form the central maximum is- |
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Answer» |
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| 39392. |
It is desired to measure a maximum current of 30 A with an ammeter of range 3.0 Ampere and resistance 0.09 Ohm . How will you do this ? |
| Answer» SOLUTION :0.01 `OMEGA` to be CONNECTED in PARALLEL with the AMMETER ] | |
| 39393. |
A ray of light is incident on liquid surface such that reflected and refracted ray are perpendicular to each other. If refractive index is 1.4, find angle of incident. |
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Answer» `54.5^@`  From FIGURE, `theta_1+90^@+theta_2=180^@` `THEREFORE theta_2=90^@-theta_1` Now,`n_1sin theta_1=n_2sin theta_2=n^2sin(90^@-theta_1)` `n_1=1,n_2=1.4,theta_1=?` `therefore (1)sintheta_1=1.4cos theta_1` `therefore (sin theta_1)/(COS theta_1)=tantheta_1=1.4` `therefore theta_1=tan^(-1)(1.4)=54.5^@` |
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| 39394. |
The current in the windings on a toroid is 2.0 A. There are 400 turns and the mean circumferential length is 40 cm. If the inside magnetic field be 1.0 T , the relative permeability of the core of torroid is approximately |
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Answer» 100 |
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| 39395. |
When alpha , betaand gammaradiation pass through a gas, their ionizing powers, in decreasing order, are |
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Answer» `GAMMA, ALPHA , BETA` |
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| 39396. |
The potential energy of a particle of mass 2 kg moving in a plane is given by U = (-6x -8y)J. The position coordinates x and y are measured in meter. If the particle is initially at rest at position (6, 4)m, then |
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Answer» Its ACCELERATION is of magnitude 5 `m/s^2` |
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| 39397. |
A copper disc of radius Im is rotated about its natural axis with an angular velocity 2 rad/sec in a uniform magnetic field of 5 tesla with its plane perpendicular to the field. Find the emf induced between the centre of the disc and its rim. |
| Answer» SOLUTION :`e=(1)/(2)B OMEGA R^(2), e=(1)/(2) xx 5xx2xx1xx1=5` volt | |
| 39398. |
In the YDSE using a monochromatic light of wavelength lambda, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is |
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Answer» `(2n+1) LAMBDA"/"2` |
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| 39399. |
An electron dipole of moment vecp is placed in a uniform electric field vecE . Writethe expression for the torque vectau experienced by the dipole.Identify two pairs of perpendicular vectors in the expression . Shown diagrammatically the orientation of the dipole in the field for which the torque is (i)the maximum , (ii)Half the maximum value , (iii)Zero. |
Answer» Solution :Let a DIPOLE AB is situated in the uniform electric field `vecE`, The torque on dipole is , `vectau=vecpxxvecE`,where `vecp`is dipole moment from the line charge . Also E and unit vector n normal to curved suface are in the same direction , So `theta=0^(@)`.Contribution of curved suface of cylinder towards electric flux will be zero (null) Two pairs of PERPENDICULAR vectors are `vectau andvecpandvectauandvecE`in the expression . (i)The torque is the maximum , when `theta=90^(@)`which is shown in following DIAGRAM :  (ii)The maximum value of torque =pE The HALF of maximum value `=(pE)/(2)` So , we have `pEsintheta=(pE)/(2)` or`sintheta=(1)/(2)` or `theta=30^(@)` The arrangement is as given below or shown as below :  (iii) For zero value of torque , `sintheta=0ortheta=0^(@)` , whichis shown as below : 
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| 39400. |
Two coherent point sources S_(1) and S_(2) vibrating in phase emit light of wavelength lambda. The separation between the sources is 2 lambda. Consider a line passing through S_(2) and perpendicular to the line S_(1)S_(2). What is the smallest distance from S_(2) where a minimum of intensity occurs ? |
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Answer» `(5lambda)/(2)` |
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