InterviewSolution
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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.
| 1801. |
A neutron having kinetic energy E_(0) collides with singly ionised He atom at rest and move along initial direction. Which of the following statement(s) is/are true for the above mentioned collision. |
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Answer» COLLISION will be perfectly inelastic if `E_(0)=17eV` `v=(v_(0))/5` `1/2mv_(0)^(2)=1/2.(5m).(v_(0)^(2))/25+/_\E` `1/2 mv_(0)^(2)-1/2 . (v^(2))/25m=/_\E` `implies 1/2 mv_(0)^(2) {4/5}=/_\E` `1/2 mv_(0)^(2)=5/4/_\E` `/_\E=13.6` `/_\E=13.6{1- 1/6}=3/4/ (13.6)` `1/2 mv_(0)^(2)=5/4xx13.6 3/5xx13.6` `=17 eV, 8.16 eV` |
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| 1802. |
In the above question, if the initial capacitance of the capacitor was 2muF, the amount of heat produced when the dielectric is inserted. |
| Answer» Answer :C | |
| 1803. |
(a) With the help of a suitable ray diagram, derive the mirror formula for a concave mirror. (b) The near point of a hypermetropic person is 50 cm from the eye. What is the power of the lens required to enable the person to read clearly a book held at 25 cm from the eye ? |
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Answer» Solution :(c ) `f_(0)=1.25cm` `f_(0)=5CM` `u_(0)=-2.5cm` `(1)/(f_(0))=(1)/(v_(0))-(1)/(u_(0))` `(1)/(v_(0))=(1)/(f_(0))+(1)/(u_(0))=(1)/(1.25)+((-1)/(2.5))` `=(1)/(1.25)-(10)/(25)` `v_(0)=2.5cm` As the final image is formed at infinity, it means thatthe object must be placed at the focus of eyepiece. Therefore, the DISTANCE between objective lens and eyepiece `=v_(0)+f_(E)` `=2.5+5` `=7.5cm` |
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| 1804. |
Match list - I with List - II |
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Answer» `a-e, b-f, C-g, d-h` |
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| 1805. |
A wave is represente by y = Asin^2 (kx - omegat + phi). The amplitude and wavelength of wave is given by |
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Answer» `2 A, (2PI)/K` |
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| 1806. |
A silver wire has a resistance of 2.1 Omega at 27.5^(@)C, and a resistance of 2.7 Omega at 100^(@)C. Determine the temperature coefficient of resistivity of silver. |
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Answer» Solution :USE `R_(t)=R_(0)(1_alpha THETA)` `2.1=R_(0)(1_alpha xx27.5)"….(1)"` `2.7=R_(0)(1+alphaxx100)"….(2)"` `"Solve equation (1) and (2) "alpha=0.0039^(@)C^(-1)` |
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| 1807. |
Using the principle of Wheatstone Bridge, describe the method to determine the specific resistance of a wire in the laboratory. Draw the circuit diagram and write the formula used. Write any two important precautions you would observe while performing the experiment. |
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Answer» Solution :Metre Bridge: Special Case of Wheatstone Bridge It is a device based on Wheatstone bridge to determine the unknown resistance of a wire. If ratio of arms resistors in Wheatstone bridge is constant, then no current flows through the galvanometer (or bridge wire). Construction: It consists of a uniform 1 metre long wire AC of constantan or manganin fixed along a scale on a wooden base (fig.) The ends A and C of wire are joined to two L-shaped copper STRIPS carrying connecting screws as shown in the figure. In between these copper strips, there is a third straight copper strip having three connecting screws. The middle screw D is CONNECTED to a sensitive galvanometer. The other terminal of galvanometer is connected to a sliding jockey B. The jockey can be made to move anywhere parallel to wire AC. By pressing the knob of jockey, it begins to touch the wire. Circuit: To find the unknown resistance S, the circuit is complete as shown in fig. The unknown resistance wire of resistance S is connected across the gap between points C and D and a resistance box (R) is connected across the gap between the points A and D. A cell, a rheostat and a key (K) is connected between the points A and C by means of connecting screws. In the experiment when the sliding jockey touches the wire AC at any point, then the wire is divided into two parts. These two parts AB and BC act as the resistances P and Q of the Wheatstone bridge. In this way the resistances of arms AB, BC, AD and DC from the resistances P, Q, R and S of Wheatstone bridge. Thus the circuit of metre bridge is the same as that of Wheatstone bridge. Method : To determine unknown resistance FIRST of all key K is closed and a resistance R is taken in the resistance box in such a way that on pressing jockey B at end points A and C, the deflection in galvanometer is on both the sides. Now jockey is slided on wire at such a position that on pressing the jockey on the wire at that point, there is no deflection in the galvanometer G. In this position the points B and D are at the same potential, therefore the bridge is balanced. The point B is called the null point. The length of both parts AB and BC of the wire are read on the scale. The condition of balance of Wheatstone bridge is )`P/Q=R/S therefore unknown resistance S=(Q/P)R`…(i) To Determine Specific Resistance: If r is the resistance PER cm length of wire AC and l cm is the length of wire AB, then length of wire BC will be (100 – l) cm. P = resistance of wire AB = lr, Q = resistance of wire BC = (100–l)r Substituting these values in equation (i), we get,`S=((100-1)r)/(lr) times R or S=(100-1)/l R`…….(ii) As the resistance (R) of wire (AB) are known, the resistance S may be calculated.A number of observations are taken for different resistances taken in resistance box and S is calculated each time and the mean value of S is found Specific resistance`p=(SA)/l=(Spir^2)/L ` Knowing resistance S, radius r by screw gauge and length of wire L by metre scale, the value ofmay be calculated. Why very small resistances cannot be measured accurately: In the derivation of formula, the resistances of copper straps and connecting wires are ignored. That is why metre bridge cannot be used for measuring very small resistances. Precautions: (i) In this experiment the resistances of the copper strips and connecting screws have not been taken into account. These resistances are called end-resistances. Therefore very small resistances cannot be found accurately by metre bridge. The resistance S should not be very small. (ii) The current should not flow in the metre bridge wire for a long time, otherwise the wire will become hot and its resistance will be changed. (iii) The resistivity of copper is several times less than the resistivity of the experimental alloy wire. As such area of thick copper strips is more, so copper strips almost offer zero resistance in the circuit. (iv) If any one resistance in Wheatstone bridge is either very small (or very large) in respect of other, then balance point might be very close to terminal A or terminal B. So generally balance point is taken in the middle of the bridge wire 
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| 1808. |
Law of conservation of mass was given by . . |
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Answer» Lavoisier |
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| 1809. |
For a single slit of width a, the first minimum of diffraction pattern of amonochromatic light of wavelength lambda occurs at an angle of lambda//a. At the same angle of lambda//a, we get a maximum for two narrow slits separated by a distance a. Explain. |
| Answer» Solution :In the interference pattern due to narrow slits separated by distance `a`, path differecne`= a sin theta`. For constructive interference, `a sin theta = 1 lambda, sin theta ~= theta = lambda//a`, we get a MAXIMUM. In the diffraction pattern due to a SINGLE SLIT, when path differencefrom secondary waves from the ends of wavefront is `lambda`, then for every pointin one half of the LOWER half for which path diff. between secondary waves is `lambda//2`. Therefore, interference is destructive and we obtain first minimum at the same angle. | |
| 1810. |
Match list - I with List - II |
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Answer» `a-h, B-g,c-e,d-f` |
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| 1811. |
Match list - I with List - II |
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Answer» `a-f, b-g, c-e, d-h` |
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| 1812. |
Match list - I with List - II |
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Answer» `a-g,b-h,c-e, d-f` |
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| 1813. |
Match list - I with List - II |
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Answer» `a-g, b-h, c-e, d-f` |
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| 1814. |
(A):Body projectedvertically up or downfrom the top of a tower with same velocity will reach the ground with same velocity. (R ):Both the bodies projected vertically up and town will have same displacement and acceleration. |
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Answer» |
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| 1815. |
(A): While temperature of a semi conductor is increased its resistance decreases (R): The energy gap between conduction band and valency band is very small and on heating the electrons can be shifted from valence band to conduction band. |
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Answer» Both 'A' and 'R' are true and 'R' is the CORRECT EXPLANATION of 'A' |
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| 1816. |
Define electrical resistivity of material of a conductor. |
| Answer» SOLUTION : Resistivity of the material of a conductor may be defined as numerically equal to the RESISTANCE of UNIT AREA of section and unit length of that conductor. | |
| 1817. |
The Fig. shows experimental set up of a metre bridge. Whenthe two unknown resistances X and Yare inserted, the null point Dis obtained 40 cm from the end A. When a resistance of 10Omegais connected in series with X, the null point shifts by 10 cm. Find the position of the null point when the 10Omegaresistance is instead | А connected in series with resistance ‘Y. Determine the values of the resistances X and Y. |
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Answer» Solution :When two unknown resistances X and Y are inserted as shown in the figure, then null POINT D is obtained at 40 CM from end A i.e., l = 40 cm. ` therefore X/Y = (40)/((100 - 40)) = 40/60` ` rArr Y = 3/2 X`....(i) When a resistance of 10`OMEGA`is connected in series with X, the null point should shift away from point A. As null point SHIFTS by 10 cm, hence `l_1`= 40 + 10 = 50 cm. ` therefore (X+10)/(Y) = (50)/(100 - 50) = 50/50` `rArr X+ 10 = Y`..(II) Solving (i) and (ii), we get X = 20 `Omega`and Y = 30 `Omega`. If a resistance of 10`Omega`is connected in series with resistance Y then the null point will be obtained at a distance `l_2`from point A, such that `(X)/(Y+ 10) = (l_2)/(100 - l_2) " or" (20)/(30 + 10) = 20/40 = (l_2)/(100 - l_2)` ` rArr l_2 = 100/3 cm= 33.3 cm ` |
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| 1818. |
If the total charge enclosed by a surface is zero, does it imply that the electric field every where on the surface is zero ? |
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Answer» Solution :When q = 0, then by Gauss.s THEORAM `oint_s OVERSET to E. overset to (dS)=q/epsilon_@ = 0` Then electrostatic field MAY be normal to the SURFACE, it is not necessary to be zero EVERY where on the surface. |
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| 1819. |
r.m.s. value of current i=3+4sin (omegat+pi//3) is |
| Answer» Answer :B | |
| 1820. |
An electron is trapped in a one-dimensional infinite potential well that is 100 pm wide, the electron is in its ground state. What is the probability that you can detect the electron in an interval of width Deltar 5.0 pm centered at x = (a) 25 pm, (b) 50 pm, and (c) 90 pm? (Hint: The interval Deltar is so narrow that you can take the probability density to be constant within it.) |
| Answer» SOLUTION :(a) 0.050, (B) 010, (C) 0.0095 | |
| 1821. |
बहुकेन्द्रकीय अवस्था उपस्थित होती है : |
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Answer» शान्त क्षेत्र (QUIESCENT CENTRE ) |
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| 1822. |
Microwaves which travel with the speed of light are reflected from a distant aeroplane approaching the waye source radar. It is found that when the reflected waves are beat against• the wave sradiated from the source, the beat frequency is 990 Hz. If the micro waves are 0.1 m in wavelength, what is the approaching speed of the aeroplane. |
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Answer» |
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| 1823. |
A small body of mass m is tried to a string and revolved in a vertical circle of radius r. If the tension in the string at the highest point is mg, what is its speed there? |
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Answer» Solution :If `v_1` is the speed and `T_1` the TENSION in the string at the highest POINT, `T_1=mg=(mv_1^2)/(r)-mg` `therefore v_1^2=2rg "" therefore v_1=sqrt(2rg)` |
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| 1824. |
The displacement of a particle after time is given by x = k/b^2 (1-e^(-bt)) where 'b' is a constant. The acceleration of the particle is |
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Answer» `ke^(-bt)` |
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| 1825. |
Derive an expression for potential energy of a system of three charges in the absence of external electric field. |
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Answer» Solution :Consider pointcharges `q_(1),q_(2),q_(3)` let `r_(12),r_(23),r_(31)` be the DISTANCE between `q_(1),q_(2),q_(3),q_(1)` RESPECTIVELY the potential energy between `q_(1)` & `q_(2)`will be similary `U_(2)=(1)/(4 pi epsilon_(0)).(q_(2)q_(3))/(r_(23))` between`q_(2)` and `q_(3)` totoalenergy of the system as a resultof assembiling the charge `U=U_(1)+U_(2)+U_(3)` `U=(1)/(4pi epsilon_(0))(q_(1)q_(2))/(r_(12))+(q_(2)q_(3))/(r_(23))+(q_(3)q_(1))/(r_(31))` |
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| 1826. |
In a series LCR circuit , and alternating emf (v) and current (i) are given by the equationv=v_(0) sin omegat, i=i_(0)sin (omegat+(pi)/(3)). The average power dissipated in the circuit over a cycle of AC is |
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Answer» <P>`(v_(0)i_(0))/(2)` `P_(ave)=V_(rms)I_(rms)cosphi` `=((V_(0))/(sqrt(2)))((I_(0))/(sqrt(2)))COS((pi)/(3))=(V_(0)I_(0))/(2)xx(1)/(2)=(V_(0)I_(0))/(4)` |
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| 1827. |
Half life of certain radioactive nuclei is 3 days and its activity is 8 times the safe limit. After how much time will the activity of the radioactive sample reach the safe limit ? |
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Answer» SOLUTION :`((A)/(8A))=(1/2)^(t//T_(1//2))` or `(1/2)^3=(1/2)^(t//3)` or `3=t/3` `RARR t=9` DAYS. |
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| 1828. |
Binding energy per nucleon of ._1H^2 and ._2He^4 are 1.1 MeV and 7.0 MeV respectively. Energy released in the process ._1H^2 + ._1H^2 = ._2He^4 is - |
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Answer» 20.8 MeC |
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| 1829. |
The figure shown blocks A and B are mass 2 kg and 8 kg and they are connected through strings to a spring connected to ground. The blocks are in equilbrium. (g=10m//s^(2)) Now the force on A is suddenly removed. The acceleration of block B becomes :- |
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Answer» `1.0n//s` |
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| 1830. |
Given a point source of light, which of the following can produce a parallel beam of light? |
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Answer» Convex mirror |
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| 1831. |
A charged ball from a silk thread, which makes an angle theta with a large charged conducting sheet. The surface charge density sigma of the sheet is proportional to….. |
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Answer» `tan theta`  As SHOWN in the figure If T is the TENSION produced in the string then if `F_e` is electric force then, (i) For horizontal equilibrium of bob, `T sin theta- F_q= qE=q(sigma/(2in_0))`…..(1) (ii) For VERTICAL equilibrium of bob `T cos theta= mg`.......(2) Taking ratio of EQUATION (1) to equation (2) `T cos theta= mg and T sin theta= (q sigma)/(2 in_0)` `THEREFORE (T sin theta)/(T cos theta)= sigma/(2in_0) times q/(mg)` `therefore tan theta= (q/(2 in_0 mg)). sigma` `therefore tan theta infty sigma`[ `because` Remaining terms are constant] |
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| 1832. |
In Young's double slit experiment the widht of one slit isdouble that of the other. The ratio of intensity of bright band to that ofa dark band in the interference pattern prodcued by them, is 9:1 |
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Answer» `9:1` `(a_(1)^(2))/(a_(2)^(2))=(W_(1))/(W_(2))` `(a_(1)^(2))/(a_(2)^(2))=(2)/(1) :. (a_(1))/(a_(2))=(SQRT(2))/(1)` `a_(1)=sqrt(2)a_(2)` `(I_("max"))/(I_("min"))=((a_(1)+a_(2))^(2))/((a_(1)-a_(2))^(2))=((2a_(1)+a_(2))^(2))/((2a_(1)-a_(2))^(2))` `=(9)/(1)` |
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| 1833. |
The flat bottom of cylinder tank is silvered and water (mu = 4/3)is filled in the tank upto a height h. Asmall bird is hovering at a height 3h from the bottom of the tank. When a small hole is opened near the bottom of the tank, the water level falls at the rate of 1 cm/s. The bird will perceive that his velocity of image is 1/x cm/sec (in downward directions) where x is |
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Answer» |
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| 1834. |
Electrons give up energy at the rate of Rl^(2) per second to the thermal energy. What time scale would number associate with energy in problem (a) ? N = no of electron/ volume = 10^(29)//m^(3) , length of circuit = 10 cm , cross-section = A = (1 mm )^(2) |
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Answer» SOLUTION :Power dissipated, P = `I^(2) ` R `(E)/(l) = (1)^(2) xx6 ` `therefore (E)/(t) = 6 W` `therefore t = (E)/(6) = (2XX 10^(-17))/(6) "" [ because "K.E." = E]` `therefore t = (1)/(3)xx 10^(-17)` `therefore tapprox 0.33 xx 10^(-17) ` s `therefore t approx 3.3 xx 10^(-18) ` S |
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| 1835. |
Which of the following is a dimensional constant ? |
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Answer» Refractive INDEX |
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| 1836. |
If the earth shrinks in its radius by 6% mass remaining constant, the value of 'g' on its surfacewill: |
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Answer» decrease `therefore` with decrease in RADIUS the value of g INCREASES. Hence correct CHOICE is (b). |
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| 1837. |
In a simple pendulum experiment, length is measured as 31.4 cm with an accuracy of Imm. The time for 100 oscillations of pendulum is 112.0s with an accuracy of 0.1s. The percentage accuracy in g is |
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Answer» 1 |
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| 1838. |
What is MODEM ? |
| Answer» Solution :MODEM performs the functions of both modulator and DEMODULATOR. While TRANSMITTING SIGNAL, modem acts as a modulator while in RECEIVING mode, modem acts as a demodulator. | |
| 1839. |
A machine gun of mass 10kg, fires 10gm bullets at the rate of two every second. Each bullet comes out with a velocity of 200 m/s. The velocity of recoil of the gun at the end of the fourth second, after firing starts is |
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Answer» 1.6 m/s |
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| 1840. |
A battery of emf V and internal resistance r is connected to N identical bulbs, all in parallel. Resistance of each bulb is R. It is observed that maximum cumulative power is dissipated in the bulbs if N = 10. Can we say that r=(R)/(10) |
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Answer» |
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| 1841. |
A point charge + q is placed at the mid-point of a cube of side l.The electric flux emerging from the cube is : |
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Answer» ZERO |
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| 1842. |
A free floating magnetic needle at North pole is __________ to the surface of earth. |
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Answer» |
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| 1843. |
The internal resistance of a cell of emf 2V is 0.1Omega. It is connected to a resistance of 3.9Omega. The voltage across the cell is |
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Answer» 0.5 V |
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| 1844. |
Given R_(1)=5.0+-0.2Omega, R_(2)=10*0+-0.1Omega. What is total resistance in parallel with possible % error.? |
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Answer» `15Omega+-2%` `=(50)/(15)=3*3Omega` ALSO `(DeltaR_(p))/(R_(p))XX100=(DeltaR_(1))/(R_(1))xx100+(DeltaR^(2))/(R^(2))xx100` `+(DELTA(R_(1)+R_(2)))/(R_(1)+R_(2))xx100` `=(0*2)/(5*0)xx100+(0*1)/(10*0)xx100+(0*3)/(15)xx100=7%` `:.R_(p)=3*3Omega+-7%` Hence correct answer is `(b)`. |
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| 1845. |
A 100W sodium lamp radiates energyuniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. (a) what is the energy per photon associated with the sodium light? (b) At what rate are the photons delivered to the sphere ? |
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Answer» Solution :Here `LAMDA=589nm=5.89xx10^(-7)m` and power of LAMP `P=100W` (a) Enerrgy per photon of SODIUM light `E=(hc)/(lamda)=(6.63xx10^(-34)xx3xx10^(8))/(5.89xx10^(-7))J=3.38xx10^(-19)J=2.11eV` (b) Number of photons delivered per second `n=(P)/(E)=(100W)/(3.38xx10^(-19)J)=2.95xx10^(20)s^(-1)=3xx10^(20)s^(-1)`. |
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| 1846. |
A gyroscope, a uniform disc of radius R=5.0cm at the end of a rod of length l=10cm (figure), is mounted on the floor of an elevator car going up with a constant acceleration w=2.0m//s^2. The other end of the rod is hinged at the point O. The gyroscope precesses with an angular velocity n=0.5 rps. Neglecting the friction and the mass of the rod, find the proper angular velocity of the disc. |
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Answer» Solution :The moment of inertia of the disc about its symmetry axis is `1/mR^2`. If the angular velocity of the disc is `omega` then the angular MOMENTUM is `1/2mR^2omega`. The precession frequency being `2pin`, we have `|(dvecM)/(DT)|=1/2mR^2omegaxx2pin` This MUST equal `m(g+w)l`, the effective gravitational torques (g being REPLACED by `g+w` in the elevator). Thus, `omega=((g+w)l)/(piR^2n)=300rad//s` |
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| 1847. |
A radioactive substance decay to 1/16thof initial activity in 40 days. The half life of the radioactive substance is |
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Answer» 2.5 |
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| 1848. |
The number of turns in primary and secondary of a transformer are 1000 and 3000 respectively. If 80 V A.C. applied to the primary, the potential difference per turn across secondary would be …… |
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Answer» 0.08V `epsilon_2/epsilon_1=N_2/N_1` `THEREFORE epsilon_2=epsilon_1N_2/N_1` `=(80xx3000)/1000` =240 V The voltage PER turn `=epsilon_2/N_2 =240/3000`=0.08 V |
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| 1849. |
A rough inclined plane inclined at angle 37° with horizontal, has a metallic wire of length 20cm with its length bot to length, of inclined plane ( μ = 0.1) When a current of i is passing through the wire and a magnetic field is applied normal to the plane upwards, the wire starts moving up with uniform velocity for B = 0.5T. Then find the magnitude of current i, (mass of the wire=50g) |
Answer» Solution : When the WIRE is in EQUILIBRIUM `Bil= MG (sin theta + mucostheta)` Bil= `mg(sintheta+μcostheta)` `=5xx10^(-2)xx10(3/5+0.1xx4/5)i=(10^(-1)xx3.4)/(10^(-1))=3.4A` |
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| 1850. |
Assertion :If the ice on the polar caps of the earth melts, then length of day will increase. Reason:Moment of inertia of the earth increases, as ice on polar caps melts. |
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Answer» If both the Asseration and REASON are true and reason EXPLAINS the ASSERTION : |
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