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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 2151. |
Briefly explain the action of p-n-p transistor. |
Answer» Solution :Action of p-n-p transistor. In a p-n-p transistor, when the emitter is forward biased, the holes in the emitter and the electrons in the BASE begin to move towards the junction, holes being attracted by the negative TERMINAL and the electrons by the positive terminal of the battery. On reaching the base-emitter junction a small fraction of total number of holes with electrons to get neutralized. As the base layer is very thin the collector is kept at high negative potential , almost all the holes are attracted by the collector, producing a hole-current between the emitter and collector.  The emitter current `I_(E)` is the sum of base current and the collector current. i.e. `I_(e)=I_(b)+I_(c)` So `I_(c) lt I_(e)` Working of p-n-p In working of p-n-p, as each hole reaches the collector, an ELECTRON is emitted from the negative terminal of battery and the hole is neutralized. Simultaneously, a covalent bond is broken near the emitter, the electron so produced ENTER the positive terminal of the forward bias battery and the hole starts its journey towards the emitter base junction. Holes are the current carriers in p-n-p transistor, however the current in the outer circuit is always due to flow of electrons. |
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| 2152. |
Mass of photon in motion is ……… |
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Answer» `(C )/(HF)` `mc^(2)=hf [E=mc^(2)]` `THEREFORE m=(hf)/(c^(2))` |
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| 2153. |
For which positions of the object does a concave mirror produce an inverted, magnified and real image ? |
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Answer» At CENTER of Curvature |
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| 2154. |
a.c. power is transmitted over long distances at voltages so that power loss during transmission is ……… |
| Answer» SOLUTION :HIGH, LEAST | |
| 2155. |
What is the ratio of electric and magnetic field of the wave ? |
| Answer» SOLUTION :VELOCITY C=E/B | |
| 2156. |
Where on the earth's surface is the magnetic dip zero? |
| Answer» SOLUTION :MAGNETIC DIP is ZERO at the magnetic EQUATOR. | |
| 2157. |
A U-tube contains a non-viscous liquid up to a height of 20 cm in each of the column. It is pressed in one of the columns and then released. The liquid executes a S.H.M. of period : |
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Answer» 0.89 s `implies T=2pi sqrt((20)/(980))` `T=0.89" s".` So CORRECT CHOICE is (a). |
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| 2158. |
A metalwireof mass m sideswithout friction magneticfields of inductionB. A battery ofconstantemf epsilon is connectedto the rails . What is the terminalspeed of the slider ? |
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| 2159. |
Draw logic diagrams for the Boolean expressions given below. A * barB+ barA * B=Y |
Answer» Solution :The required logic DIAGRAM for the given Boolean EXPRESSION is given in figure. Here the input B before applying to first AND gate and input A before applying to second AND gate have been inverted. The output of these gales are, therefore, `A barB` and `barA `B respectively. These OUTPUTS are FED to OR gate which gives `Y = A barB+ barA B `as shown in figure. 
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| 2160. |
(a) Three point charges q,-4q and 2q are placed at the vertices of an equilateral triangle ABC of side 'l' as shown in the figure. Obtain the expression for the magnitude of the resultant electric force acting on the charge q. (b) Find out the amount of the work done to separate the charges at infinite distance. |
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Answer» Solution :FORCE on charge Q due to the charge -4q ` F_1 =(1)/( 4pi in _0alpha ) ((4q^(2))/( l^(2)) ` along AB Force on the charge q. due to the charge 2q ` F_2 =(1)/( 4pin in _0) ((4q^(2))/( l^(2)) `along CA the forces `F_1 and F_2 ` are inclined to each other at an angle of `120^(@) ` Hence , resultant electric force on charge q ` F= sqrt( F_1^(2) +F_2^(2) +2F_1F_2cos THETA ) ` ` =sqrt( F_1^(2) +F_2^(2) + 2F_1F_2cos theta 120^(@) ) ` ` = sqrt( F_1^(2) +F_2^(2) -F_1F_2 ) ` ` =((1)/( 4pi in _0) (q^(2) )/( l^(2))) sqrt( 16+4-8) ` ` =(1)/( 4pi in _0) ((2sqrt3q^(2))/( l^(2)))` 
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| 2161. |
Among the following four spectral regions, the photons has the highest energy in |
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Answer» Infrared |
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| 2162. |
Find gamma for the mixture of 11 gm CO_(2) and 14 gm N_(2)? |
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Answer» `gamma_(mix)=(7)/(5)` |
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| 2163. |
A photon beam of energy 12.1 eV is incident on a hydrogen atom. The orbit to which electron of H-atom be excited is |
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Answer» 2nd orbit. |
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| 2164. |
The true statements regarding refractive index of a material are (A) it depends on the nature of the material (B) it increases with temperature (C ) it decreases with increases in wavelength (D) it is inversely proportional to velocity of light in the medium |
| Answer» Answer :A | |
| 2165. |
The breaking stress for a wire of radius 'r' of given material is F N/m^2 . The breaking stress for the wire of same material of radius 2 r is |
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Answer» a)F |
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| 2166. |
A point charge q is placed at a distance d from the centre of a circular disc of radius R. Find electric flux flowing through the disc due to that charge |
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Answer» Solution :We know that total flux originated from a point CHARGE Q in all directions is `(q)/(in_(0))`. This flux is originated in a solid ANGLE `4pi`. In the given case solid angle subtended by the cone subtended by the disc at the point charge is `Omega = 2PI (1- cos theta)` So, the flux of q which is passing through the surface of the disc is `phi = (q)/(in_(0)) (Omega)/(4pi) = (q)/(2in_(0))(1- cos theta)` From the FIGURE, `cos theta = (d)/(sqrt(d^(2) + R^(2)))`, so `phi = (q)/(2 in_(0)) {1-(d)/(sqrt(d^(2) + R^(2)))}` |
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| 2167. |
A.M.C.G. is based on |
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Answer» heating EFFECT of CURRENT |
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| 2168. |
A mass of 1kg carrying a charge of 2C is accelerated through a potential of 1V. The velocity acquired by it is |
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Answer» `SQRT(2) ms^(-1)` |
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| 2169. |
When an electron jumps from 2nd stationary orbit to 1st stationary orbit of hydrogen atom, the emitted energy is______ ev. |
| Answer» SOLUTION : `10.2 EV [E_(2)- E_(1) = (-13.6)/((2)^(2)) -[-(13.6)/((1)^(2))] = +3.4eV]` | |
| 2170. |
lambda_(1) and lambda_(2) are the wavelength of the first numbers of the Lyman and Paschen series respectively, then find lambda_(1): lambda_(2) |
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Answer» Solution : Wavelength of first LINE in Lyman series `lambda_(1)` is, `:.(1)/(lambda_(1))=R[(1)/(1^(2))-(1)/(n_(i)^(2))] ` [ `:.` for first line `n_(1)=2]` `:.(1)/(lambda_(1))=R[(1)/(1^(2))-(1)/(r^(2))]=R[1-(1)/(4)]` `:. (1)/(lambda_(1))=(3R)/(4)....(1)` `rArr` Wavelength of first line in PASCHEN series `lambda_(2)`, `:.(1)/(lambda_(1))=R[(1)/(3^(2))-(1)/(4^(2))] [ :.` for first line `n_(i)=4]` `:.(1)/(lambda_(2)) [(1)/(9)-(1)/(16)]=R[(7)/(14)]` `:.(1)/(lambda_(2))=(7R)/(144) ...(2)` `rArr` Taking ratio of equation (2) and equation (1), `(lambda_(1))/(lambda_(2))=(7R)/(144)xx(4)/(3R)` `:. (lambda_(1))/(lambda_(2))=(7)/(108)` |
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| 2171. |
Quark composition of proton is...... |
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Answer» uuu |
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| 2172. |
If the reflected image formed is magnified and virtual, then the mirror system is |
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Answer» CONCAVE only |
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| 2173. |
A car moves along a horizontal circular road of radius r with velocity v. The coefficient of friction between the wheels and the road is mu. Which of the following statements is not true ? |
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Answer» The car will slip if `v gt SQRT(mu rg)` |
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| 2174. |
How inductive reactance depends on the frequency of AC ? |
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Answer» SOLUTION :It is directly PROPORTIONAL to the frequency of AC `X_(L) - OMEGA L` |
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| 2175. |
Two discs of moments of inertia l_(1) and l_(2) and angular speeds omega_(1) and omega_(2) are rotating along collinear axes pass through their centre of mass and perpendicular to their plane. If the two are made to rotate jointly along the same axis, the rotational K.E. of system will be |
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Answer» `((l_(1)omega_(1)+l_(2)omega_(2))^(2))/(2(l_(1)+l_(2)))` `I_(1)omega_(1)+I_(2)omega_(2)=(I_(1)+I_(2))OMEGA` Angular VELOCITY of the system `omega=(I_(1)omega_(1)+I_(2)omega_(2))/(I_(1)+I_(2))` Rotational kinetic energy `=1/2(I_(1)+I_(2))omega^(2)` `=1/2(I_(1)+I_(2)).((I_(1)omega_(1)+I_(2)omega_(2))/(I_(1)+I_(2)))^(2)=((I_(1)omega_(1)+I_(2)omega_(2))^(2))/(2(I_(1)+I_(2)))` |
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| 2176. |
Focal length of a convex lens will be maximum for |
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Answer» BLUE light |
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| 2177. |
A body of mass 10 gm and having charge 2C is attached to a spring which is suspended from the ceiling. It vibrates with a time period 1 sec. If an electric field of intensity 100 N/C is now applied in the downward direction, find the time period. |
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| 2178. |
The cathode rays have particle nature because of the fact that |
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Answer» They can propagate in vacuum |
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| 2179. |
What are ground waves ? |
| Answer» Solution :Ground waves are the RADIOWAVES which are travelling directly from TRANSMITTING antenna to RECEIVING antenna FOLLOWING the SURFACE of earth. | |
| 2180. |
What is the principle of optical fibre ? |
| Answer» SOLUTION :It is BASED on the principle of total internal reflection of LIGHT. | |
| 2181. |
Figure (a) shows an arrangement of three charged particles separted by distance d. Particles A and C are fixed on the x-axis, but particle B can be moved along a circle centered on particle A. During the movement, a radial line between A and B makes an angle theta relative to the positive direction of the x-axis (Fig. b). The curves in Fig. c give, for two situations, the magnitude F_("net") of the net electrostatic force on particle A due to the other particles. That net force if given as a function of angle theta and as a multiple of a basic amount F_(0). For example on curve 1, at theta=180^(@) we see that F_("net")=2F_(0). (a) For the situation corresponding to curve 1, what is the ratio of the charge of particle C to that of particle B (including sign)? (b) For the situation corresponding tocurve 2, what is that ratio? |
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Answer» |
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| 2182. |
(A) :Communication in UHF/VHF regions can be established by space wave or tropospheric wave (R) : Communication in UHF/VHF regions is limited to line of sight distance |
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Answer» Both 'A' and 'R' are true and 'R' is the CORRECT EXPLANATION of 'A'. |
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| 2183. |
What amount of charge will be in 1600 electrons? |
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Answer» SOLUTION :Here n=1600 So q=ne=1600`XX(-1.6xx10)^(-19)` `=2590xx10^(-19)C`. |
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| 2184. |
A wire of resistance 18 ohm is bent to form an equilateral triangle. The resistance across any two vertices of the triangle is (in ohm) |
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Answer» 12 |
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| 2185. |
A ball of radius 11 cm and mass 8 kg rolls from rest down a ramp of length 2m. The ramp is inclined at 35° to the horizontal. When the ball reaches the bottom, its velocity is (sin 35° = 0.57)? |
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Answer» 2 m/s |
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| 2186. |
(A) Inductance is called the inertia of an electric circuit. (R): An inductor tends to keep the flux constant |
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Answer» Both A and R are true and R is the CORRECT EXPLANATION |
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| 2187. |
A metallic wire of 10pi cm and of magnetic moment M is bent into a semi circle. Calculate the new magnetic moment. |
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Answer» |
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| 2188. |
""_(72)^(180)A overset(a)(rarr)B overset(beta)(rarr)C overset(alpha)(rarr)D overset(gamma)(rarr)E a. What are dfferent types of emission taking place here ? b. Name the product after each emission. c. Justify your answer. |
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Answer» SOLUTION :`alpha`-emission, `beta` - emission and `gamma` - emission. b. `""_(72)^(180)A RARR ""_(70)^(176)B rarr""_(71)^(176)C rarr ""_(69)^(172)D rarr ""_(69)^(172)E`. When `alpha`-emission takes place, atomic number (Z) decrease by 2 and mass number (A) decreases by 4 and we get `""_(70)^(176)B`. When a `beta`-emission takes place, Z increase by 1 and A remains the same, THUS we get `""_(71)^(176)C`. Again an `alpha`- emission takes place and we get `""_(69)^(172)E`. |
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| 2189. |
A ray of light travelling in glass having refractive index _(a)mu_(g)=3//2 is incident at a critical angle C on the glass air interface. If a thin layer of water is poured on glass air interface, then what will be the angle of emergence of this ray in air when it emerges from water air inteface? |
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Answer» `180^(@)` |
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| 2190. |
The limit of resolution of eye is approximately |
| Answer» Answer :B | |
| 2191. |
फलन ={(4,2),(9,1), (6,1), (10,3)} का परिसर है। |
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Answer» (4,9,6,10} |
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| 2192. |
In a given process on an ideal gas dW = 0 and dQ angle 0. Then for the gas : |
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Answer» the volume will increase `& dQ =dU+dW` CHANGE in internal energy `dU lt 0` `rArr` Internal energy decreases and hence temperature also decreases. Thus, CORRECT choice is (a). |
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| 2193. |
Derive the equation for acceptanc angleand numerical aperture, of optical fiber. Acceptance angle in optical fibre: |
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Answer» Solution :To ensure the cirtical angle INCIDENCE in the core - cladding boundary INSIDE the optical fibre, the light should be incident at a certain angle at the END of the optical while entering in to it. This angleis called acceptance angle. It dependson the refractiveincidens of the core`n_(1),` cladding `n_(2)` and the outer medium `n_(3)`, Assume the light is incident at an angle called acceptance angle `i_(a)` at the outer medium and core bounadary at A. The Snell.s law in the product form, equation for this refraction at the POINT A.   `n_(3) sin i_(a) = n_(1) sin r_(a)` To have the total interal reflection inside optical, fibre, the angle of incidence at the core-cadding INTERFACE at B should be atleastcritical angle `i_(c)`,. Snell.s law in the product form, equaiton for the refractionat point B is, `n_(1) sin i_(c) = n_(2) 90^(@)` `n_(1) sin i_(c) n_(2) "" therefore sin 90^(@) = 1` `therefore sin i_(c) n (n_(2))/(n_(1))` From the right angle triangle `DeltaABC`, `i_(c) = 90^(@) - r_(a)` Now, equation (3) becomes, `sin (90^(@) - r_(a)) = (n_(2))/(n_(1))` Using trigonometry, `cos r_(a) = (n_(2))/(n_(1))` `sin r_(a) = sqrt(1 - cos^(2) r_(a))` Substituting for `cos r_(a)` `sinr_(a)=sqrt(1-((n_(2))/(n_(1)))^(2))=sqrt((n_(1)^(2)-n_(2)^(2))/(n_(1)^(2)))` Substituting this in equation (1) `n_(3)sini_(a)=n_(1)=n_(1)sqrt((n_(1)^(2)-n_(2)^(2))/(n_(1)^(2)))=sqrt(n_(1)^(2)-n_(2)^(2))` On further simoplificaiton, `sini_(a)=sqrt((n_(1)^(2)-n_(2)^(2))/(n_(3)))(or)sini_(a)=sqrt((n_(1)^(2)-n_(2)^(2))/(n_(3)^(2))` `i_(a)=sini^(-1)(sqrt((n_(1)^(2)-n_(2)^(2))/(n_(3)^(2))))` If outer medium is air, then `n_(3) = 1`. The acceptance angle `i_(a)` becomes, `i_(a)=sini^(-1)(sqrt(n_(1)^(2)-n_(2)^(2)))` Light can haveany angle of incidence from o to `i_(a)` with the normal at the end of the optical fibre forming a conical shape called acceptancecone. In the equation (6), the term `(n_(3)sini_(a))` `NA=n_(3)sini_(a)sqrt(n_(1)^(2)-n_(2)^(2))` |
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| 2195. |
A particle is vibrating in S.H.M. with a period of 6 sec and amplitude of 3m. Its maximum speed is |
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Answer» `pi/2 m//sec` |
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| 2196. |
Disturbances of two waves are shown as a function of time in the following figure. The ratio of their intensities will be - |
| Answer» ANSWER :D | |
| 2197. |
A 220V ac is more dangerous than 220V dc. Why ? |
| Answer» Solution :Here rms VALUE of ac `V_rms=220V`. But the peak value of the GIVEN ac supply, `V_0=220sqrt2=311V`, In the case of 220V dc, the peak value is the same 220V. That is why the ac is more dangerous than dc of the same VOLTAGE. | |
| 2198. |
Which of the following is not expressed correctly with respect to units ? |
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Answer» Specific heat `RARR` J `kg^(-1) K^(-1)` |
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| 2199. |
Fusion processes, like combining two deutrons ot form a He nucleus are impossible at ordinary temperatures and pressure. The resons for this can be traced to the fact: |
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Answer» nuclear forces have short RANGE |
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| 2200. |
Two similar balls of mass .m. are hungby a silk thread of length L and carry similar charges Q as in Asssuming the separation to be small. The separation between the halls (denoted by x) is equal to |
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Answer» `[(Q^(2).2L)/(4pi epsilon.mg)]^((1)/(3))` |
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