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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
In one part of above network, steady current is flowing. Values of resistances are as shown in the diagram. Find energy stored in capacitor. |
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Answer» SOLUTION :When the capacitor becomes fully charged, there is no current in the branch CONTAINING capacitor. Now, current in the branch AD and DE is 1 + 2 = 3 A. Current in the branch EB is 1 A. Now, `V_(AD)= 3xx 5 = 15 ` V `V_(DE) = 3 xx 1 = 3 V ` ` V_(EB)= 1 xx 2= 2 `V Potential difference across capacitor is, `V = V_(AD) + V_(DE)+ V_(EB)` = 15 + 3 + 2 `therefore V = 20 ` Volt Now, electrostatic potential energy stored in the capacitor is, `U_(E)= (1)/(2) CV^(2)` `therefore U_(E) = (1)/(2) xx 4 xx 10^(-6) xx (20)^(2)` `thereforeU_(E)= 8 xx 10^(-4) `J |
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| 2. |
The conductivity of semiconductor increases with increase in temperature because |
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Answer» both number density of CHARGE carriers and RELAXATION time increase |
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| 3. |
The ratio of areas within the electron orbits for the first excited state to the ground state for hydrogen atom is |
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Answer» `16:1` |
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| 4. |
A jar of height h is filled with a transparent liquid of refractive index mu(See figure). At the centre of the jar on the bottom surface is a dot. Find the minimum diameter of a disc, such that when placed on the top surface symmertically aobut the centre , the dot is invisible |
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Answer» Solution :`rArr` Suppose required minimum DIAMETER of given disc is d. Here for the light rays emanating from point like object O and then becoming incident on water surface (from inside), if `i gt C,` then that object O wil not be see by the observer while observing from outside the water. (Where C = critical angle for water to air) Suppose angle i in the figure is equal to C . Now, according to formula. `sin C = 1/mu` `THEREFORE sin i = 1/mu( because C = i)` From the figure, `tan i=(d/2)/(h)` `therefore d/2 = h tan i ......(1)` Now, sin ` i = 1/mu` `thereforecos i = SQRT(1 sin ^2i) = sqrt(1 - 1/(mu^2)) = (sqrt(mu^2 -1))/(mu)` `therefore tan i = (sin i)/(cos i) = (1/u)/(sqrt(mu^2-1)/(mu))=(1)/(sqrt(mu^2-))` `therefore` From equation (1), `d/2 = h xx (1)/(sqrt(mu^2 - 1))` `therefore d = (2h)/(sqrt(mu^2-1))` Above equation GIVES required result. |
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| 5. |
The persons, who were given Nobel prize twice, are |
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Answer» MADAME CURIE and ALBERT Einstein |
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| 6. |
For a transistor circuit in common emitter configuration, the voltage gain is 100. if the input voltages is 20mV, then the output voltage is |
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Answer» 400m |
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| 7. |
the origins of the two frames coincide at t=t'=0 and the relative speed is 0.980c. Two micrometeorites collide at coordinates x = 100 km and t = 200 mus according to an observer in frame S. What are the (a) spatial and (b) temporal coordinate of the collision according to an observer in frame S'? |
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Answer» |
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| 8. |
(A) : Photoemission of electrons from a photosensitive surface is possible only if the incident radiation has a frequency above threshold frequency (R): Unless the hv > hv_(0) = work function of metals, no photoemission of electrons is possible |
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Answer» A and R are TRUE and R is the correct EXPLANATION of A. |
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| 9. |
Assertion: A charge is lying at the centre of the line joining two identical charges which are fixed. The system will be in equilibrium if that charge is one fourth of the identical charges with opposite sign. Reason:For charge to be in equilibrium, sum of the forces on charge due to rest of the charges must be non zero. |
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Answer» Both Assertion and Reason are TRUE and Reason is the correct explanation of Assertion |
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| 10. |
Describe the working of nuclear reactor with a block diagram. |
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Answer» Solution :Nuclear reactor : Nuclear reactor is a system in which the nuclear fission takes PLACE in a self -sustained controlled manner and the energy produced is used either for research purpose or for power generation. The main parts of a nuclear reactor are fuel, MODERATOR and control rods. In addition to this, there is a cooling system which is connected with power generation set up. Fuel : The fuel is fissionable material usually uranium or plutonium. Naturally occuring uranium contains only `0.7 %` of `""_(92)^(235)"U"` and `99.3%` are only `""_(92)^(235)"U"`. So the `""_(92)^(238)"U"` must be enriched such that it contains at least 2 to 4% of `""_(92)^(235)"U"`. In addition to this, a neutron source is required to initiate the chain reaction for the first time. A mixture of beryllium with plutonium or polonium is used as the neutron source.During fission of `""_(92)^(235)"U"`, only fast neutrons are emitted but the probability of initiating fission by it in another nucleus is very low.Therefore , slow neutrons. Usually the moderators are chosen in such a way that it must be veruy lightnucleus having mass comparable to that of neutrons.Hene , these light nuclei undergo collision with fats neutrons and the speed of the neutron is reduced. Most of the reactors use water , heavy water `(D_(2)O)` and graphite as moderators. The blocks or uranium stacked together with blocks of graphite (the moderator) to form ALARGE pile.  Control rods: The control rods are used to adjust the reaction rate. During each fission, on an average 2.5 neutrons are emitted and in order to have the controlled chainreactions. Only one neutron is allowed to cause another fission and the remaining neutrons are absorbed by the control rods. Usually cadmium or boron acts as control rod material and these rods are inserted intothe uranium blocks. Depending on the insertion depth of control rod into the uranium, the average number of neutrons produced per fission is set to be equal to one or greater than one. If the average number of neutrons producedper fission is equal to one, then reactor is said to be in critical state. In the fact, all the nuclear reactors are maintained in critical state by suitable adjustment of control rods. If it is greater than one , then reactor is said to be in super - critical and it may explode sooner or may cause massive destruction. Shielding : For a protection against harmful radiations, the nuclear reactor is SURROUNDED by a concrete wall of thickness of about 2 to 2.5 m. Cooling system : The cooling system removes the heat generated in the reactor core. Ordinary water,heavy water and liquid sodium are used as coolant since they have very high specific heat capacity and have large boiling point under high pressure. Thsi coolant passes through the fuel block and carries away the heat to the steam generator through heat exchanger. The steam runs the turbines which produces electricity in power reactors. |
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| 11. |
A body is under the action of a force 5 newton moves through 10 m in a straight line. If work done is 25 joule, what is the angle at which force acts with the direction of motion ? |
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Answer» Solution :`W=FS COS THETA implies25xx5xx10 cos theta`. `cos theta=1/2 implies theta=60^(@)` |
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| 12. |
The electromagnetic waves travel with a velocity |
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Answer» EQUAL to velocity of sound = velocity of light |
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| 13. |
A ray of light passes from vacuum into amedium of refractive index mu, the angle f incidence is found to be twice the angle of refractive. Then angle of refractive. Then angle of incidences is: |
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Answer» `cos^(-1) (MU)/(2)` `thereforemu=(SINI)/(sin""(i)/(2))=(2sin""(i)/(2)cos""(i)/(2))/(sin""(i)/(2))=2cos""(i)/(2)` `therefore cos""(i)/(2) = (mu)/(2)` `therefore (i)/(2) = cos^(-1)(mu)/(2)` `therefore I = 2 cos^(-1)(mu)/(2)` |
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| 15. |
An ideal battery passes a current of 5A through a resistor. When it is connected to another resistance of 10Omega in parallel, the current is 6A. Find the resistance of the first resistor. |
Answer» Solution : `"Current through "R_(1)" in the FIRST case "i_(1)=5A` `"Current in the second case "i_(2)=6A` EFFECTIVE resistance in the second case `R=(R_(1)R_(2))/(R_(1)+R_(2)), V=I_(1)R_(1) and V=I_(2)(R_(1)R_(2))/(R_(1)+R_(2))` `I_(1)R_(1)=I_(2)(R_(1)R_(2))/(R_(1)+R_(2))RARR I_(1)=I_(2)(R_(2))/(R_(1)+R_(2))` `5=6xx(10)/(R_(1)+10)rArr 5(R_(1)+10)=60` `5R_(1)+50=60, 5R_(1)=10` `R_(1)=(10)/(5)=2OmegarArr R_(1)=2Omega` |
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| 16. |
Solve for x and y, using substitution method: 2x + y = 7, 4x − 3y + 1 = 0 |
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Answer» (2,3) |
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| 17. |
Time period of the satellite very close to the earth’s surface is , |
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Answer» a) 35898 km |
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| 18. |
For a particular transverse standing wave on a long string, one of the antinodes is at r=0 and an adjacent node is at x=0.10 m. The displacement y(t) of the string particle at x =0 is shown in Fig. 16-40, where the scale of the y axis is set by y_(s)=4.0cm. When t = 0.50 s, what is the displacement of the string particle at (a) x= 0.20 m and (b) x= 0.30 m? What is the transverse velocity of the string particle at x =0.20 m at (c) t=0.50s and (d) t=1.0s? (e) Sketch the standing wave at t=0.50s for the range x=0 to x=0.40m. |
Answer» SOLUTION :(a) 0.40m, (B) 0, (C) 0 m/s , (d) 1.0 s (E) 
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| 19. |
An inductor, a capacitor and a resistor are in series with an alternator of frequency 50Hz. The potential difference across them are 50 V, 80 V and 40 V respectively. Find the voltage of the alternator? |
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Answer» Solution :The POTENTIAL difference across the inductor and the capacitor have a PHASE difference of `pi`. The resultant potential difference across the combination of inductor and capacitor is `80-50=30V`. This 30V potential difference and the potential difference across the resistor, namely 40V have phase difference of `pi//2`. By vectroially adding these, the total potential difference, which is same as the VOLTAGE of the alternator is obtained. `V^(2)=30^(2)+40^(2)rArr V=50V` |
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| 20. |
When the electric flux Phi_(e) in a region of space, of relative permittivity (dielectric constant) k = (epsi)/(epsi_(0)), varies with time it gives rise to a displacement current given by |
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Answer» `EPSI(d Phi_(e))/(DT)` |
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| 21. |
y,(x,t) = 0.8//[(4x + 5t)^2 + 5]represents a moving pulse,where x,y are in metre and t in second. then |
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Answer» pulse is moving in +x-direction |
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| 22. |
A beta-particle enters a magnetic field making an angle of 45^@ with the field lines. The path of the particle is |
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Answer» CIRCULAR |
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| 23. |
What is dielectric strength and in what way it helps ? |
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Answer» |
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| 24. |
Two coherent monochromatic light sources are located at two vertices of an equilateral triangle. If the intensity due to each of the source independently is 1 Wm^(-2) at the third vertex. The resultant intensity due to both the source at the third vertex is (in Wm^(-2)). |
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Answer» |
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| 25. |
What is nuclear fission ? Give an example to illustrate it. What is the importance of fission phenomenon ? |
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Answer» Example. When slow NEUTRONS are bombarded on a heavy nucleus of `""_(92)U^(235)` it produces two lighter nuclei of Ba and KR and tremendous amount of energy. The nuclear reaction can be represented as follows : `""_(92)U^(235)+n^(1) rarr ""_(56)Ba^(141)+""_(36)Kr^(92)+3 ""_(0)n^(1)+Delta Q` Importance of nuclear fission 1. Controlled nuclear fission is used in NUMBER of YIELDS e.g. (i) to prepare radio isotopes. (ii) to generate power for propulsion of ships, submarine and aircrafts. (iii) to produce plutonium for explosive PURPOSE. 2. Uncontrolled nuclear fission is used in atom bomb. |
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| 26. |
Which of the following radiations has least wavelength: |
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Answer» `gamma-rays` |
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| 27. |
A certain double plane window consists of two glass sheets each 80 cm xx 80 cm xx 0.30 cm, separated by 0.3 cm stagnant air space between them (see figure-4.49). The indoor surface temperature is 20^(@)C, while the outdoor surface temperature is 0^(@)C. Find: (a) the temperature of the surface of the sheets in contact.with the stagnant air (b) the power transmitted from the inside to the outside. Given that K_(glass)=2.0xx10^(-3) cal s^(-1) cm^(-1)""^(@)C , K_(air)=2.0xx10^(-4) cal s^(-1) cm^(-1)""^(@)C^(-1) |
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Answer» |
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| 28. |
Demostrate that the transformation forumulas (3.6h) fllowfrom theformulas (3.6i)at v lt lt c. |
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Answer» SOLUTION :In the non relativistic limit, we neglect `v^(2)//c^(2)` and WRITE, `{:(VEC(E')_(||)=vec(E)_(||)),(vec(E)_(bot)~=vec(E)_(bot)+vec(v)xxvec(B)):}}{:(vecB'_(||)=vec(B)_(||)),(vec(B')_(bot)~=vec(B)_(bot)-vec(v)xxvec(E)//c^(2)):}` These two equcationscan becombined to give, `vec(E') = vec(E) + vec(v) xx vec(B), vec(B') = vec(B) - vec(v) xx vec(E)//c^(2)` |
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| 29. |
What are Neutrons? |
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Answer» SOLUTION :A neutrons is a NEUTRAL particle i.e.,has no CHARGE It.s mass is alomost identical to that of PROTON charge on neutron=zero mass of neutrons =`m_n`=`1.6750xx10^(-27)`= 1.008665a.m.u. |
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| 30. |
A straight wire is first bent into a circle of radius *r' and then into a square of side 'x' each of one turn. If currents flowing through them are in the ratio 4:5, the ratio of their effective magnetic moments is |
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Answer» `(PI)/(8)` |
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| 31. |
A square ground of side a = 10/sqrt(2) m has a circular running track of radius a/2 with its centre coinciding the centre of the ground. A man is running on the track with an angular velocity ft) = 22 rad/s while a car is moving on a road adjacent to ground as shown in the figure. The car moves in such away that the car, the man and the centre of the ground always lie on the same straight line. If a source of sound of frequency v = 300 Hz is being placed at the centre of the ground find the minimum frequency received by the man in the car. Assume velocity of sound in air is v - 330m/s. |
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Answer» 200 Hz |
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| 32. |
Find the Brewstar angle for air-water surface for yellow light. Refractive index of water for yellow light =1.33 |
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Answer» |
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| 33. |
A heavy but uniform rope of length L is suspended from a ceiling, A particle is dropped from the ceiling at the instant the bottom end is given the jerk. Where will the particle meet the pulse? |
| Answer» SOLUTION :At a DISTANCE L/3 from the BOTTOM | |
| 34. |
An inductor of 3H is connected to a battery of emf 6V through a resistance of 100 Omega. Calculate the time constant. What will be the maximum value of current in the circuit ? |
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Answer» SOLUTION :Given that L = 3H.E=6V.R=100 `OMEGA` Time constant `tau_(L)=(L)/(R)=(3)/(100)=0.03` sec Maximum Current `I_(0)=(E)/(R)=(6)/(100) "AMP"=0.06`amp |
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| 35. |
Draw a schematic diagram of a reflecting telescope (Cassegrain). Write its two advantages |
| Answer» Solution :A LABELLED ray diagram of a reflecting type telescope has been shown in Fig. 9.51.Reflecting type telescopes are (i) generally FREE from chromatic and spherical ABERRATIONS, (n) easier to install and operate and (in) can have wider APERTURE and consequently higher resolving POWER as well as range. | |
| 36. |
A motorist drives north for 35.0 minutes at 85.0 km/h and then stops then stops for 15.0minutes .He next continues north ,travelling 130 km in 2.00 hours . (a)What is his total displacement ? b)What is his average velocity ? |
| Answer» SOLUTION :(a)76.6km, (B)63.4km/hr. | |
| 37. |
A circular disc of mass 10 kg and radius 0.2 m is set into rotation about an axis passing through its centre and perpendicular to its plane by applying torque of 10 Nm. The angular velocity of the disc at end of 6sec,from start is |
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Answer» 200 rad/sec |
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| 38. |
What is p-type semiconductor ? |
| Answer» SOLUTION :When the SEMICONDUCTOR is doped with an ACCEPTOR IMPURITY. is CALLED p-type. | |
| 39. |
At a point A on the earth's surface the angle of dip delta= +25^(@). At a point B on the earth's surface the angle of dip, delta = -25^(@). We can interpret that : |
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Answer» A and B are both located in the southern HEMISPHERE. |
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| 40. |
Speed of a revolvingelectron aroundthe nucleusvaries with principal quantum numbern as : |
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Answer» ` v prop m ` |
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| 41. |
What refraction means ? |
| Answer» Solution :A BEAM of light from ONE transparetn medium passes OBLIQUELY to another medium,the BENDING of ray is called REFRACTION. | |
| 42. |
Assertion:The illumination of earth's surface from sun is more at noon than in the morning. Reason: Luminance of a surface refers to brightness of the surface. |
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Answer» If both ASSERTION and REASON are true and the reason is the CORRECT explanation of the assertion |
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| 43. |
Figure shows planar loops of different shapes moving out of or into a region of a magnetic field which is directed normal to the plane of the loop away from the reader. Determine the direction of induced current in each loop using Lenz's law. |
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Answer» SOLUTION :i. The magnetic flux through the RECTANGULAR loop abcd increases, due to the MOTION of the loop into the region of magnetic field. The induced current must flow ALONG the path bcdab so that it opposes the increasing flux. ii. Due to the outward motion, magnetic flux through the triangular loop abc decreases due to which the induced current flows along bacb, so as to oppose the change in flux. iii. As the magnetic flux decreases due to motion of the irregular shaped loop abcd out of the region of magnetic field, the induced current flows along cdabc, so as to oppose change in flux. [Note that there are no induced current as long as the loops are COMPLETELY inside or outside the region of the magnetic field.] |
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| 44. |
Two point charges A = +3 nC and B = +1 nC are placed 5 cm apart in air. The work done to move charge B towards A by 1 cm is |
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Answer» `1.35xx10^(-7)J` `B=+1 nC = 1xx10^(-9)C` Distance, `r_(1)=5 cm = 0.05 m = 5XX10^(-2)m` WORKDONE `(W)=U_(B)-U_(A)` `= (kq_(1)q_(2))/(r_(2))-(kq_(1)q_(2))/(r_(1))` Here, `r_(2)=r_(1)-1` `r_(2)=5-1=4 cm = 0.04 m = 4xx10^(-2)m` `= kq_(1)q_(2)[(1)/(r_(2))-(1)/(r_(1))]` `= 9xx10^(9)xx3xx10^(-9)xx1xx10^(-9)` `= [(1)/(4xx10^(-2))-(1)/(5xx10^(-2))]` `=(27xx10^(-9)xx1)/(5xx4xx10^(-2))` `= (27)/(20)xx10^(-7)` `= 1.35xx10^(-7)J` |
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| 46. |
The B.E of ""_(10)Ne^(20) is 160.6 MeV. Find its atomic mass. The rest mass of proton =1.007825 a.m.u. and that of neutron =1.008665 a.m.u : |
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Answer» 9.6243 a.m.u. `=(7.27 xx 1.6 xx 10^(-19) xx 10^(6))/(3) xx n` `=4 xx 10^(21) xx 10^(6)("given")` `n=1.03163 xx 10^(40)` mass of helium atoms burnt/sec `=1.03163 xx 10^(40) xx 6.9 xx 10^(-27)kg` `=7.1 xx 10^(13)kg` |
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| 47. |
The gyromagnetic ratio of an electron in an H-atom, according to Bohr model, is |
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Answer» independent of which orbit it is in. Gyromagnetic ratio = `("Magnetic moment of e")/("ANGULAR MOMENTUM of e")` `mu_(l)=M_(e)/L_(e)` = `(-evr)/(2m_(e)VR)" "(becauseM_(e)=(-evr)/2" "L_(e)=m_(e)vr)` `mu_(l)=(-e)/(2m_(e))` Magnetic moment doesn.t depend on velocity of electron. But only depends on charged particle e and it is negative. So, option (A) and (B) are TURE. |
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| 48. |
STATEMENT-1 The minimum slit separation d for interference to produce at least one maximum other than central maximum in YDSE is 3. STATEMENT-2 For a maximum ,path difference=n .The maximum value of path difference =d,slit separation. |
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Answer» Statement-1 is TRUE Statemetnt-2 is True,Statement -2 is a correct EXPLANATION for statement -1. For `n=1,d=lambda `and we will have three maximum. |
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| 49. |
What is the minimum voltage applied to an X-ray tube with a vanadium cathode for which the lines of the K_(alpha)-series appear? |
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Answer» Hence `varphigt(hc)/(elamda_(alpha)),orvarphigt(3(Z-1)^(2)HCR)/(4e)`. |
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| 50. |
In a flexible balloon 2, moles of SO_2 having an initial volume of 1 kL at a temperature of 27.^@C is filled . The gas is first expanded to thrice its initial volume isobarically and then further expanded adiabatically so as to attain its initial temperature . Assuming the gas to be ideal , the work done by the gas in the whole process is [gammaSO_2=4/3,R=25/3J mol ^(-1) K^(-1)] |
| Answer» ANSWER :C | |