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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.
| 151. |
At the first minimum adjacent to the central maximum of a single slit diffraction pattern, the phase difference between the Huygen's theh wavelet from the edge of the slit and the wavelet from the edge of the slit and the wavelet from the edge of the slit and the wavelet from the midpoint of the slit is |
| Answer» Answer :C | |
| 152. |
How does refractive index (mu) of a material vary with respect to wavelenght (lambda). (A and B are constants). |
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Answer» `MU = A + (B)/(lambda^(2))` |
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| 153. |
In a chamber, a uniform magnetic field of 6.5G ( 1G = 10^(-4) T) is . Maintained . An electron is shot into field witha speed of 4.8 xx 10^(6) ms^(-1) normal to the field . Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e = 1.6 xx 10^(-19) C, m = 9.1 xx 10^(-31) kg) |
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Answer» Solution :`B=6.5 XX 10^(-4) T,v= 4.8 xx 10^(6) ms ^(-1)` `BQV = (MV^2)/(r) , r = (mV)/(QB)= (9.1 xx 10^(-31) xx 4.8 xx 10^6)/(1.6 xx 10^(-19) xx 6.5 xx 10^(-4)) = 4.2 xx 10^(-2) m = 4.2 cm ` |
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| 154. |
Draw a graph of number of undecayed nuclei to the time, for a radioactive nuclei. |
Answer» SOLUTION :
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| 155. |
Choose the correct statement for an isolated system. |
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Answer» `DeltaU(C to D)` = NEGATIVE |
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| 156. |
Give an example for isobars. |
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Answer» Solution :EXAMPLE : 1) ` ""_(1)H^3 and ""_(2)He^3` 2) `""_(13)AR^(40)and ""_(20)Ca^(40)` |
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| 157. |
Draw a circuit diagram to show biasing of a solar cell. Draw its characteristic curve and explain it |
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Answer» Solution :A circuit diagram showing biasing of a TYPICAL p-n junction solar cell has been GIVEN in When light (with photon ENERGY hv ` gt E_(g)`) falls at the junction, electron-hole pairs are generated which MOVE in mutually opposite directions due to the junction field. Ifno external load is connected or if the circuit of solar cell is open, electrons and holes are collected at the two sides of the junction giving a photo-voltage `V_(oc)` . When external load R is connected, a photo-current `I_(L)`flows. For R = 0, the current has a maximum value `I_(sc)` which is known as short circuit current. V- I characteristics is shown in . The graph is in the fourth quadrant because a solar cell does not draw current but supplies current to the load. 
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| 158. |
A mixture of two gases is contained in a vessel. The Gas 1 is monoatomic and gas 2 is diatomic and the ratio of root mean square speeds of the molecules of two gases is |
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Answer» 2 |
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| 159. |
In a p-n junction the width of depletion region on either side of junction is ______ or even less. |
| Answer» SOLUTION :`10^(-6)` m | |
| 160. |
Development goals of different sections of our society can be achieved by: |
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Answer» Force |
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| 161. |
If "^imu_g represent the refractive index when a light ray goes from medium 'I' to medium 'g' then the product 2mu_1xx^3mu_^4mu_3 is equal to: |
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Answer» `"^3mu_2` |
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| 162. |
Two point charges +q and -2qare placed at the vertices B and C of an equilateral triangle ABC of side 'a' as shown Obtain the expession for (i) the magnitude ,and (ii) the direction of the resultant electric field at the vertex A due to these two charges . |
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Answer» |
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| 163. |
What is the unit of magnetic permeability? |
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Answer» `wb^-1Am^(-1)` |
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| 164. |
Answer the following questions : (a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances ? Explain. (b) A virtual image, we always say, cannot be caught on a screen. Yet when we 'see' a virtual image, we are obviously bringing it on to the 'screen' (i.e., the retina) of our eye. Is there a contradiction ? (c) A diver under water, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter to the diver than what he actually is ? (d) Does the apparent depth of a tank of water change if viewed obliquely ? If so, does the apparent depth increase or decrease ? (e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter ? |
Answer» Solution :(a) We know that both plane and convex mirrors produce virtual images of real objects. They may produce real image of virtual objects i.e., when the incident light beam is highly convergent, converging towards a point O behind the mirror as shown in Fig. 9.05(a) for a plane mirror and Fig. 9.05(b) for a convex mirror respectively. (b) A virtual image MEANS that the reflected/ refracted light rays form a divergent beam and HENCE it cannot be converged on a screen. However, when we .see. a virtual image, this virtual image acts as an object and the convex lens of eye converges it on the screen (the retina) of our eye. Therefore, there is no CONTRADICTION of any sort. (c) As shown in Fig. 9.06, when an object OO. situated in air is viewed from inside water its virtual, erect and magnified image II. is formed. It is on this account that to a diver under water a fisherman standing on the bank of lake appears to be taller than what he actually is.  (d) The apparent depth decreases when a tank filled with water is viewed obliquely compared to the depth when SEEN near normally. (e) Refractive index of diamond is much greater than that of glass. In fact, it is 2.42 as compared to 1.5 of glass. Consequently, CRITICAL angle for diamond air interface is quite small (about 24°). Thus, a skilled diamond cutter has a large option to cut the diamond faces for any value of angle of incidence varying from 24° to 90°. It makes his work easy. |
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| 165. |
Find the velocity of the excited hydrogen atoms if the first line of the Lyman series is displaced by Delta lambda= 0.20 nm when their radiation is observed at an angle theta=45^(@) to their motion direction. |
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Answer» Solution :Since `DELTA lambda(= 0.20nm)lt lt lambda(= 121nm)` of the first LYMAN line of `H` atom, we need not worry about `V^(2)//c^(2)` effects. Then `omega'=(omega)/(1-betacos theta),beta=(v)/(c )` Hence `1- beta COS theta=(omega)/(omega')=(lambda')/(lambda)` or `beta cos theta=1-(lambda')/(lambda)=(Delta lambda)/(lambda)` But `omega=(3)/(4)R so lambda=(2pic)/(omega)=(8pic)/(3R)` Hence `v=cbeta=(3RDeltalambda)/(8pi cos theta)` Substitution gives `( cos theta=(1)/(sqrt(2)))` `v= 7.0xx10^(5)m//s` |
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| 166. |
What is the magnetic induction due to a magnet of pole strength 20A-m and length 20 cm at a distance of 0.5m from its centre on the axial line? |
| Answer» SOLUTION :`69.44 XX 10^(-7)T ` | |
| 167. |
A uniform magnetic field vecB is set up along the positive x-axis. A particle of charge q and mass m moving with a velocity vecv enters the field at the origin in X - Y plane such that it has velocity components both along and perpendicular to the magnetic field vecB. Trace, giving reasons, the trajectory followed by the particle. Find out the expression for the distance moved by the particle along the magnetic field in one rotation. |
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Answer» Solution :CONSIDER a charged particle of mass m and charge q entering a uniform magnetic field `vecB` with a VELOCITY `vecv` along an angle `theta` with the direction of magnetic field in the plane of peper . Velocity `vecv` may be resolved into two components(i) `v cos theta` along the magnetic field, and (ii) `v sin theta` perpendicular to the magnetic field as shown in fig. Obviously due to normal COMPONENT `v sin theta` the charged particle will ecperience a force `F = q(v sin theta) B` along a direction perpendicular to `vecB` as well as `(v sin theta)` i.e, force F is along the z-axis. Under its effect, the charged particle describe a CIRCULAR path of radius r, such that `B q v sin theta = (m (v sin theta)^2)/(r)` `implies r = (m v sin theta)/(B q) "" ......(i)` The time period of revolution will be given by `T = (2 pi r)/((v sin theta)) = (2 pi m)/(B q) "" .......(ii)` or Frequency of revolution `v = 1/T = (Bq)/(2 pi m) ""........(iii)` Due to component `v cos theta` the charged particle does not experience force due to magnetic field and tends to move linearly with a constant speed. Thus, under the combined effect of both the velocity componetns we can say that the charged particlewill describe a helical path having its axis parallel to magnetic field. In one COMPLETE revolution of its circular path the charged particle covers a linear distance along `vecB`, which is called as "pitch" of helix and is denoted by p. Obviously `p = (v cos theta) T = (2 pi m v cos theta)/(Bq)` 
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| 168. |
What ideas led Maxwell to think about e.m. wave? |
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Answer» Solution :Faraday from his exprimental study on electromagnetic induction, conclude that a magnetic field changing with time in a region produces an electric field there. Maxwell thought that there is a GREAT SYMMETRY in nature. He from his theortical study concluded that an electric field changing with time in a region produces a magnetic field there. It means a change in either field (electric or magnetic) with time produces the other field. This IDEAL led Maxwell to conclude that thevariation in electric and magnetic field VECTORS perpendicular to each other leads to the production of e.m. wave, which can travel in SPACE. |
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| 169. |
A short-circuited coil is placed in a time varying magnetic field. Electrical power is dissipated due to the current induced in the coil. If the number of turns were to be quadrupled and the radius of the wire is to be halved, then find the electrical power dissipated |
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Answer» SOLUTION : Current is induced in the short-circuited coil due to the imposed time - varying magnetic field. Power P = `(e^2)/(R )` Here e ` = - (dphi)/(dt)` where `PHI = NBA` and `R = (RHO l)/(pi r^2) ` where l and r are LENGTH and radius of the wire ` therefore P = (pi r^2)/(rho l) [ (d)/(dl) (NBA) ]^2 " or " P = (pi r^2)/(rho l) N^2 A^2 ((dB)/(dt))^2` or P = (constant) `(N^2r^2)/(l)` When `r_2= (r_1)/(2)` then `l_2 = 4l_1` ` P_2/P_1 = ((4N)^2)/(N^2) xx ((r)/(2r))^2 xx ((l)/(4l))` ` therefore P_2/P_1 = (16N^2 xx r^2 xx l)/(N^2 xx 4r^2 xx 4l) " or " P_2/P_1 = 1/1 ` `therefore ` Power dissipated is the same |
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| 170. |
A paricle starts from the origin at t = 0s with a velocity of 10.0hatjm//s and moves in the xyplane with a constant acceleration of (8.0hati+2.0hatj)ms^(-2). What time is the xcoordinate of the particle 16m? |
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Answer» t = 2s |
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| 171. |
A ball is thrown upwards from the ground . It is at a height of 100m in upwards and downward journeys at times t_1 and t_2 respectively , If g = 10m/s^2 , then t_1t_2 is equal to |
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Answer» 10 |
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| 172. |
A radioactive sample decays with a mean life of 20ms. A capacitor of capacitance 100muF is charged to some potential and then the plates are connected to a resistance .R.. If the ratio of the charge on the capacitance to the activity of radioactive sample remains constant in time is R=100(x)Omega, then x= |
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Answer» 2 |
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| 174. |
Is photoelectric emission possible at all frequencies ? Give reason for your answer . |
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Answer» SOLUTION :Kinetic energy of photoelectrons is GIVEN by Einstein.s PHOTOELECTRIC equation `K_max = 1/2 mv_max^2 = hv - omega_0` = `(hv)/lembda - omega_0` `therefore v_max^2 prop 1/lambda` `v_max prop 1/sqrtlembda`. As the wavelength of incident LIGHT decreases, the velocity of photoelectrons increases. |
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| 175. |
From the two e.m.f. equations e_1 = E_0 sin (100 pit) and e_2 = E_0 sin (100 pit + pi/3), we find that |
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Answer» `e_1` leads `e_2` by `60^@` |
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| 176. |
A 900 pF capacitor is charged by 50 V battery. Find the electrostatic potential energy of this capacitor. (2) Now this capacitor is disconnected from the battery and connect to another equivalent capacitor, then what is the total electrostatic energy stored by the system respectively ? |
| Answer» SOLUTION :`1.125 xx10^(-6) J, 0.5625 xx10^(-6)`J | |
| 177. |
The mutual inductance of a pair of coils if a current of 3 ampere in one coil causes the flux in the second coil of 2000 turns to change by 6xx10^(-4) weber per turn of the secondary coil is |
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Answer» `6xx10^(-4)` henry |
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| 178. |
The north pole of a magnet is brought from a coil, then the direction of induced current will be __ |
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Answer» in the CLOCKWISE DIRECTION |
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| 179. |
The value of horizontal component of the earth's field, at a place where the dip is 45^(@) and total field 3.4 xx 10^(-5) tesla is |
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Answer» `2.404 XX 10^(-5) T` |
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| 180. |
A pellet with charge +Q is centered inside a spherical cavity in a conducting ball. The conducting ball has a charge -2q on it. The cavity in the ball is off-center, as shown in Fig. (a) How much charge resides on the inner surface of the cavity ? (b) Is the charge on the cavity surface distributed uniformly or nonuniformly ? Why ? (c ) How much charge resides on the outer surface of the conducting ball ? (d) Is the charge on the ball's outer surface distribution uniformly or nonuniformly ? |
| Answer» SOLUTION :(a) -Q, (B) UNIFORMALY, (C ) -q, (d) uniformaly | |
| 181. |
How does a lubricant help in decreasing friction ? |
| Answer» Solution :Lubricant gets FILLED in the irregularities and THUS FORMS a layer in between the two surfaces. Under such circumstances firm interlocking is not POSSIBLE. THEREFORE, there is a decrease in force of faction. | |
| 183. |
The earth (mass=6xx10^24 kg), revolves around the sun with angular velocity of 2xx10^(-7) rad/s in a circular orbit of radius 1.5xx10^8 km. What is the force exerted by the sun on the earth ? |
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Answer» Solution :The GRAVITATIONAL force exerted by the sun of the earth is EQUAL to the CP force `(mv^2)/r =mrw^2` `therefore F=mrw^2` `6xx10^24xx1.5xx10^8xx10^3xx(2xx10^(-7))^2` `3.6xx10^32` N. |
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| 184. |
While doing an experiment with potentiometer (see figure) it was found that the deflection is one sided and (i) the deflection decreased while moving from one and A of the wire, to the end R, (ii) the deflection increased, while the jockey was moved towards the end D. (a)Which terminal positive and negative of the cell E_1 is connected at X in case (i) and how is E_1 related to E ? (b)Which terminal of the cell E_1 is connected at X in case (i in a) ? |
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| 185. |
A ball is suspended a cord from the ceiling of a motor car. What will be the effect on the position of the ball if the car is moving with constant velocity |
| Answer» SOLUTION :The BALL will REMAINS SUSPENDED VERTICALLY | |
| 186. |
A ball is suspended by a cord from the ceiling of a motor car. What will be the effect on the position of the ball if the car is moving with accelerated motion |
| Answer» SOLUTION :The BALL will MOVE BACKWARD | |
| 187. |
A ball is suspended by a cord from the ceiling of a motor car. What will be the effect on the position of the ball if the car is turning towards right. |
| Answer» SOLUTION :The BALL will MOVE TOWARDS the LEFT. | |
| 188. |
In Question 3 and 4, what is the net power absorbed by each circuit over a complete cycle ? Explain your answer. |
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Answer» Solution :`P = E_(V) I_(v) cos phi` In a pure inductor, phase difference between alternating VOLTAGE and current is `phi = 90^(@)` `:. P = E_(v) I_(v) cos 90^(@) =` Zero. In Q.4 `P = E_(v) I_(v) cos phi` In a pure capacitor, phase difference between alternating voltage and current is `phi = 90^(@)` `:. P = E_(v) I_(v) cos 90^(@) =` Zero Infact, in both the cases, power spent in one HALF CYCLE is retrieved in HTE other half cycle. |
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| 189. |
In a full wave rectifier circuit operating from50 Hz mains frequency, the fundamental frequency in the ripple would be ……… |
Answer» SOLUTION :100 Hz 
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| 190. |
Two block-spring mass system are moving on smooth horizontal surface as shown in the figure as t=0. find out the minimum time t when the spring is maximum elongated l_(0) is the natural length of spring. |
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Answer» Solution :`T=2pisqrt((4)/(9pi^(2)))=(2xx2)/(3)=(4)/(3)` `t=(3T)/(4)=(3)/(4)xx((4)/(3))=1` |
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| 191. |
A solid disc is rotating at an angular speed of 20 rad//s. It is decelerated at a constant rate of 2 rad // s^2 . Through what angle disc will turn before coming to rest ? |
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Answer» SOLUTION :`omega^2-omega_0^2=2 alpha THETA` `THEREFORE 0-400=2xx(-2)xxtheta therefore theta`=100 RADIAN |
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| 192. |
(A) : In Ampere Maxwell law, displacement current has same physical effect as conduction current. (R) : The units and dimensions of displacement current and conduction current are same. |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the correct EXPLANATION of 'A'. |
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| 193. |
The following objects are placed one after each other in given order onto a central axis with a separation of 40 cm each. A point source of light O. a diverging lens of focal length 40 cm. a converging lens of focal length 40 cm and converging mirror of focal length 80 cm. The aperture diameter of lenses and mirror is d=20 cm. If a point source of light is placed at a perpendicular distance of x from central axis then. (You have to consider single optical event at any optical element) Mark the CORRECT statement(s):- |
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Answer» Final image is formed in the plane of converging lens `i_(1)=(O_(1)f_(1))/(O_(1)-f_(1))=(-4.4)/(4-(-4))dm=-2dm`(dmrarrdecimeter) Since the linear magnification is `m_(1)=(H_(i1))/(H_(O1))=(i_(1))/(O_(1))=(1)/(2)` the height of the first image is `H_(i1)=(x)/(2)=H_(O2)`. This virtual image is seen bythe diverging lens as an object with an object distance of `O_(2)=f_(2)+|i_(2)|=6dm`, THEREFORE the image formed by second lens has an image distance of `i_(2)=(O_(2)f_(2))/(O_(2)-f_(2))=(6.4)/(6-4)dm=12dm` The linear magnification of the second lens is `m_(2)=(i_(2))/(O_(2))=(12)/(6)=2`  Hence the height of the second image is `H_(i2)=m_(2)H_(O2)=m_(2)H_(i1)=xc=H_(O3.)` The second image is seen by the concave mirror as a virtual object bering at a distance of `O_(3)=f_(3)-i_(2)=-8dm`, therefore the image is formed at `i_(3)=(O_(3)f_(3))/(O_(3)-f_(3))=(-8.8)/(-8-8)=(-64)/(-16)=1dm` This is a real image and since the linear magnification of the mirror is: `m_(3)=(i_(3))/(O_(3))=(1)/(2)` the height of the final image is `H_(i3)=m_(3)H_(O3)=(x)/(2)` Note that final image is formed in the plane of the converging lens, therefore the image can only be captured on a screen if its height is greater than the radius of the lens. This happens when x is greater than d, therefore the list condition for perpendicular is d lt x. If x is increased further than`1.5 d`,the rayswillall pass under the concave mirror and there will be no image formation. Therefore the solution of the PROBLEM is : `d lt x lt 1.5 d`.creased further than `1.5 d`, the rays will all pass under the concave mirror and there will be no images formation. Therefore the solution of the problem is: `d lt x lt 1.5 d`. |
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| 194. |
What are drawbacks of the capillary rise method? |
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Answer» Solution :1 It is difficult to measure mean radius .r. over a RANGE of the entire height. 2 The cleaning of such narrow tube is not easy. 3 A correction for the curvature of the surface of liquid in the CONTAINER is necessary but difficult to DETERMINE |
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| 195. |
Calculate the relation between the angle of incidence and angle of deviation of a prism? |
| Answer» SOLUTION :M = 1+D/f. | |
| 196. |
In an L-C-R A.C. series circuit L = 5H, omega = "100 rads"^(-1), R= 100 Omega and power factor is 0.5. Calculate the value of capacitance of the capacitor. |
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Answer» SOLUTION :FIRST Method : Power factor `cosdelta=R/sqrt(R^2+(omegaL-1/(omegaC))^2)` Square on both SIDES `cos^2delta=R^2/(R^2+(omegaL-1/(omegaC))^2)` but , cos`delta=0.5=1/2` `therefore 1/4=R^2/(R^2+(omegaL-1/(omegaC))^2)` `therefore R^2+(omegaL-1/(omegaC))^2` `therefore (omegaL-1/(omegaC))^2=3R^2` `therefore omegaL-1/(omegaC)=sqrt3R` `therefore omegaL-sqrt3R=1/(omegaC)` `therefore C=1/omega(1/(omegaL-sqrt3R))` `=1/100(1/(100xx5-sqrt3xx100))` `=10^(-2)/(500-173.2)=10^(-2)/326.8=306xx10^(-7)` `=30.6xx10^(-6)` F =30.6 `muF` Second Method: Power factor `cosdelta=0.5` `therefore delta=pi/3` rad Now tan`delta=(omegaL-1/(omegaC))/R` `therefore "tan"pi/3 xx R=omegaL-1/(omegaC)` `therefore sqrt3xx100=100xx5-1/(100xxC)` `therefore 1.732xx100=500-1/(100C)` `therefore 1/(100C) =500-173.2` `therefore 1/(100C) =326.8` `therefore C=1/(326.8xx100)`=0.00003059 `approx 30.6xx10^(-6)F=30.6 muF` |
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| 197. |
The capacitance of a conductor ____________ when an earthed conductor is brought near it. |
| Answer» SOLUTION :Increases. On BRINGING an EARTHED conductor CLOSE by, the potential decreases and HENCE capacitance increases. | |
| 198. |
While doing an experiment with potcntiometc as shown in figure in was found that th deflection is one sided and(i) the detlectio, decreased while moving from one end A of the wire to the end B,(ii) the deflection increase while the jockey was moved towards the end (i) Which terminal + or - ve Of the cell E_(1)connected at X in case (i) and how is related to E ? (ii) Which terminal of the cellE_(1) is connected at X in case (ii) ? |
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Answer» SOLUTION :(i) Deflection of galvanometer is on one SIDE when jokey key is moved from A to B if deflection decreases, then it can be SAID that current through primary cell (LOWER circuit) decrease. and p.d. between point A and jokey key increase. This is possible only if positive of `E_(1) ` cell is connected with terminal X, thus `E_(1) gt ` E (ii) When jokey key is moved from A to B if deflection increases then it can be said that current through primary cell (lower circuit) increases and p.d. between jokey key and A decreases. This is possible only if positive of `E_(1) ` cell is connected with X. |
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| 199. |
As shown in the figure, the planes of two concentric coils are mutually perpendicular. The mutual inductance of this system is ……. |
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Answer» `mu_0N^2A` Now, `phi_2=M_21 I_1` in `phi_2=0` so `M_21=0`. 
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| 200. |
The following truth table corresponds to the logical gate |
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Answer» NAND |
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