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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.
| 1951. |
What is modulation ? Explain its necessity giving example of size of antenna. |
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Answer» Solution :MODULATION is the process by which some characteristics of a relatively high FREQUENCY wave is varied in ACCORDANCE with the INSTANTANEOUS value of a low frequency. Need for modulation For transmission of signal, the height of antenna should be comparable to the wavelength of signal. For audio signal, the height of antenna should be from `15 km` to `15000km`, which is not possible, hence modulation of signal is required. |
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| 1952. |
Tow cells of emf E_(1)and E_(2) ( E_(1)gt E_(2))areconnected individuallyto a potentiometerand theircorresponding balancing lengthare 625 cm and500 cm, then the ratio (E_(1))/(E_(2))is . |
| Answer» ANSWER :A | |
| 1953. |
Using the concept of free electrons in a conductior , derive the expression for the conductivity of a wire in terms of number density and relaxation time . Hence obtain the relation between current density and the applied electric field E. |
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Answer» Solution :Derivationof the expression for conductivity Relation between density and electric field The acceleration ` OVERSET to ( a) = - ( in )/( m ) oversetto E ` The average drift velocity , vd, is given by ` vd = - ( eE) /( m ) ` ` ( TAU `=average time betweencollisions / relaxtion time ) If n is the number of free ELECTRONS per unit volumes , the current I is given by ` I= neA | vd| ` = ( e^(2) A)/( m ) tau n |E| ` But I= |J| A ( j = current density ) we . THEREFORE . get |
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| 1954. |
Derive, making use of an integral, the formula for the moment of inertia of a sphere about its diameter. |
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Answer» `dm=(Mpir^2dz)/(4//3piR^3)=(3M)/(4R^3)(R^3-z^2)dz` Since `r^2=R^2-z^2` . The moment of inertia of such a DISK is `dI=(r^2dm)/2=(3M)/(8R^2)(R^2-z^2)dz` The moment of inertia of the sphere is `I = 2 int_0^R(3M)/(8R^3)(R^4-2R^2z^2+z^4)dz`= `=(3M)/(4R^3)[R^4int_0^Rdz-2R^2int_0^(R)z^2dz+int_0^Rz^4dz]=` `=(3M)/(4R^3)[R^4z-2R^(2)(z^3)/2+z^5/5]_0^R=` `=(3M)/(4R^3)(R^5-(2R^5)/(3)+R^5/5)=2/5MR^2` 
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| 1955. |
Which experiment established the quantum nature of charge ? |
| Answer» SOLUTION :Millikan.s OIL DROP EXPERIMENT. | |
| 1956. |
(a) Deriver and expression for the average power consumed in a series LCR circuit connected to a.c., source in which the phase different between the voltage and the current in the circuit is phi. (b) Define the quality factor in an a.c. circuit. Why should the quality factor have high value in receiving circuits ? Name the factors on which it depends. |
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Answer» Solution :(a) Average power in LCR circuit: Let the alternating e.m.f. applied to an LCR circuit is `E = E_(0) sin omega t`..(i) If alternating current developed lags behid the applied, e.m.f. by a phase angle `phi` then `I = I_(0) sin (omega t - phi)`...(ii) Total work done over a complete cycle is `W = int_(0)^(T) EI DT` `= int_(0)^(T) E_(0) sin ometa t. I_(0) sin (omega t - phi) dt = E_(0) I_(0) int_(0)^(T) sin omega t sin (omega t - phi) dt` `= (E_(0)I_(0))/(2) int_(0)^(T) 2 sin omega t sin (omega t - phi) dt` `= (E_(0)I_(0))/(2) int_(0)^(T) " " [cos (omega t + omega t + phi) - cos (omega t + omega t - phi)] dt` `[ :' 2 sin A sin B]` `= [cos (A -B) - cos (A + B)] = (E_(0)I_(0))/(2) int_(0)^(T) [cos phi - cos (2 omega t - phi)] dt` `= (E_(0)I_(0))/(2) [ t cos phi - (sin (2 omega t - phi))/(2 omega)]_(0)^(T)` `= (E_(0) I_(0))/(2) [T cos phi] " " { :' int_(0)^(T) (sin (2 omega t - phi))/(2 omega) = 0}` `= (E_(0)I_(0))/(2).cos phi.T.` `:.` Average power in LCR circuit over a complete cycle is `P = (W)/(T) = (E_(0)I_(0))/(2) cos phi = (E_(0))/(sqrt2).(I_(0))/(sqrt2) cos phi` `:. P = E_(v) I_(v) cos phi` (b) Quality factor in an A.C. : The Q -factor of a resonant LCR circuit is defined as ratio of the VOLTAGE drop across inductor or capacitor to the applied voltage.`:. Q = ("Voltage across L or C")/("applied voltage")` Since, `V_(L) = IX_(L) and V = IR " " :. Q = (IX_(L))/(IR) = (omega_(0)L)/(R)` Hence, at high FREQUENCIES, the Q-factor `(omega_(0)L)/(R)` is quite large. `:.` The voltage drop across the inductor will be quite large as compared to the applied voltage Also, `Q = (1)/(sqrt(LC)).(L)/(R) " " [ :' omega_(0) = (1)/(sqrt(LC))]` `:. Q = (1)/(R) sqrt((L)/(C))` The Q-factor of LCR-series circuit depends on L., C and R |
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| 1957. |
A convex lens made up of glass of refractive index 1.5 is dipped, in turn, in (i) a medium of refractive in dex 1.65, (ii) a medium of refractive index 1.33. (a) Will it behave as a converging or a diverging lens in the two cases ? (b) How will its focal length change in the two media ? |
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Answer» SOLUTION :`(1)/(f) = ( mu -1) ( ( 1)/(R_(1)) - ( 1)/( R_(2)))` (a) (i) DIVERGING LENS or concave lens (ii) Converging lens or convex lens (b) (i) Focal length will BECOME negative nad its magnitude would increase (ii) Focal length increases. |
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| 1958. |
A block is on a horizontal surface (a shake table) that is moving back and forth horizontally with simple harmonic motion of frequency 2.0 Hz. The coefficient of static friction between block and surface is 0.45. How great can the amplitude of the SHM be if the block is not to slip along the surface? |
| Answer» Solution :0.028m, A LARGER amplitude REQUIRES a larger force at the END points of the motion. The surface cannot SUPPLY the larger force and the BLOCK slips. | |
| 1959. |
When was Evelyn's deafness noticed? |
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Answer» when she was 18 |
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| 1960. |
Giving 2V to LED passes a 10 mA current. If you want to connect this diode to a 6V battery, then the value of resistance in series kept…… |
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Answer» `400 Omega` Suppose initially the resistance of LEDis `R_(1)` `therefore R_(1) = (V)/(I)=(2)/(10xx10^(-3))=200Omega` Now, if there is resistance `R_(2)` by connecting the LED to the 6V battery, In V = IR, I is CONSTANT `thereforeV prop R` `therefore (V_(2))/(V_(1))=(R_(2))/(R_(1))` `(6)/(2)=(R_(2))/(200)` `therefore R_(2)=600Omega` `therefore` Resistance joining in SERIES `=R_(2)-R_(1)` `= 600-200=400Omega` |
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| 1961. |
An organ pipe P_1closed at one end and vibrating in its first overtone, and another pipe P_2 open at both ends and vibrating in its third overtone, are in resonance with a given tuning fork. The ratio of the length of P_1 to that of P_2 is |
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Answer» `8/3` |
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| 1962. |
A dog while barking delivers about 1 mW of power. If this power is uniformly distributed over a hemispherical area, what is the sound level at a distance of 5 m? What would the sound level be if instead of 1 dog, 5 dogs start barking at the same time each delivering 1 mW of power |
| Answer» Answer :A | |
| 1963. |
A point charge q=2muC is at the origin. It has velocity 2 hatj m//s.Find the magnetic field at the following point in vector from (at the moment when the charged particle passes through the origin) : (i) (2,0,0) , (ii) (0,2,0) , (iii) (0,0,2) , (iv) (2,1,2) (v)Is the magnitude of the magnetic field on the circumference of the circle (in yz plane) y^(2)+z^(2)=c^(2) where c^(2) is a contant is same every where.Is it same in direction also. (vI) Answer the above (vii) for the circle of same equation but in a plane x=a where a is a contant. |
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| 1964. |
A metal has a work function of 4.3eV What does it mean ? |
| Answer» Solution :It means that if the electrons in the metal are provided with an energy of 4.3 eV from some external source, they will cross over the surface barrier to leave the metal. The lower the woik.FUNCTION of-an emitter, the SMALLER is the temperature required to CAUSE ELECTRON emission. For this REASON, emitters should have low work function. | |
| 1965. |
The following flow diagram represents the manufacturing of sodium carbonate ? 2NH_(3)+H_(2)O+CO_(2)rarr (NH_(4))_(2)CO_(3)overset((a))rarrNH_(4)HCO_(3)overset((b))rarr underset(Na_(2)CO_(3)+CO_(2)+H_(2)O)underset(darr (d))(NaHCO_(3))+(c) Which of the following option describes the underlined reagents, products and reaction conditions ? |
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Answer» `{:("OPTION",(a),(b),(c ),(d)),(,"CARBON dioxide",NaCl,NH_(4)Cl,"Heat"):}` `NH_(4)HCO_(3)+NaClrarrNaHCO_(3)+NH_(4)Cl`. `2NaHCO_(3)UNDERSET(1373K)overset("Heat")rarrNa_(2)CO_(3)+CO_(2)+H_(2)O` |
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| 1966. |
A parallel beam of monochromatic light falls normally on a single narrow slit, how does the angular width of the principal maximum in the resulting diffraction pattern depend on width of the slit ? |
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Answer» Solution :For diffraction at a single-slit the ANGULAR WIDTH of the principal MAXIMUM is inversely PROPORTION to the width of the slit i.e., angular width `thetaprop(1)/(a)`. |
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| 1967. |
A man stading in a swimming pool looks at a stone lying at the bottom. The depth of the swimming pool is h, At what distance from the surface of water is the image of the formed (Line of vision is normal , Refractive index of water is n) |
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Answer» H/n |
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| 1968. |
An a.c. circuit consists of an inductor of inductance 0.5 H and a capacitor of capacitance 8 uF in series. The current in the circuit is maximum when the angular frequency of a.c. source is |
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Answer» `500 RAD s^(-1)` `omega_(0) =1/sqrt(LC) = 1/sqrt(0.5 xx 8 xx 10^(-6)) = 500 rad s^(-1)` |
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| 1969. |
A gun mounted on the top of a moving truck is aimed in the backward direction at an angle of 30^(@) to the vertical. If the muzzle velocity of the bullet is 4 ms^(-1) the value of speed of the truck that will make the bullet come of out vertically is |
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Answer» `1 MS^(-1)` |
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| 1970. |
Two clocks are being tested against a standard clock located in a national laboratory .At 12:00:00 noon by the standard clock, the readings of the two clocks are {:(,"Day 1","Day 2","Day 3","Day 4" , "Day 5"),("Clock 1",12:00:04,12:02:19,12:01:50,11:59:04,11:59:08),("Clock 2",11:15:24,11:15:01,11:14:59,11:15:00,11:14:58):} Which of the two clocks will you prefer to measure the time intervals during an experiment ? |
| Answer» SOLUTION :Zero error is not as SIGNIFICANT as the variation in the time INTERVAL or Clock 2 | |
| 1971. |
Three infinite straight wires A, B and C carry currents as shown in figure. The resultant force on wire B is directed |
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Answer» towards A 
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| 1972. |
For a circular coil of radius R and N carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by B = (mu_0 I R^2 N)/(2(x^2 + R^2)^(3//2)) (a) Show that this reduces to the familiar result for field at the centre of the coil. (b) Consider two parallel co-axial circular coil of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by , B = 0.72 (mu_0 N I)/(R) , approximately. [Such an arrangment to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.] |
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Answer» Solution :(a) for the centre of the coil putting x = 0 in the given relation, we have `B = (mu_0 I R^2 N)/(2 R^3) = mu_0 (NI)/(2R)` (b) In a small region length 2d about the mid-point between the coils, the total magnetic field due to both current carrying coils has a value `B = (mu_0 I R^2 N)/(2) xx [{(R/2 + d)^2 + R^2}^(-3//2) + {(R/2 - d)^(2) + R^2}^(-3//2)]` As d is very small, neglecting the term CONTAINING `d^2`, we have `B = (mu_0 I R^2 N)/(2) xx ((5R^2)/(4))^(-3//2) xx [(1 + (4d)/(5R))^(-3//2) + (1 - (4d)/(5R))^(-3//2)]` On EXPANDING by binominal theorem and neglecting higher powers of `((4d)/(5R))`, we get `B = (mu_0 I R^2 N)/(2 R^3) xx (4/5)^(+3//2) xx [1 - (6d)/(5R) + 1+ (6d)/(5R)]` `implies B = (4/5)^(3//2) (mu_0 I N)/(R) = 0.72 (mu_0 I N)/(R)`. |
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| 1973. |
Both the ring and the conducting sphere are given the same charge Q. Determine the potential of the sphere. Assume that the centre of the sphere lies on the axis of th ring. Is it necessary that the charge on the ring be uniformly distributed to answer the above question ? |
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| 1974. |
Absorptive power of a body depend upon |
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Answer» SURFACE AREA |
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| 1975. |
An a.c. circuit consists of ohmic resistance only. If the frequency of a.c. source increases then current in the circuit also increases. |
| Answer» SOLUTION : False - In an ohmic RESISTIVE CIRCUIT the current remains constant and does not change with FREQUENCY of a.c. | |
| 1976. |
For Rayleigh scattering of light, value of a is ...... |
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Answer» `ALPHA LT lt λ` |
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| 1977. |
In Division Germer Experiment accelerating potential is kept constant at 54 volt. As detector is rotated, the first intensity maximum is obtained at an angle of : |
| Answer» ANSWER :A | |
| 1978. |
What is the formula of angle of dip ? |
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Answer» (A) `TAN^(-1) (B_h)/( B_v)` `rArr phi = tan^(-1) (B_v)/( B_h)` |
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| 1979. |
For the small damping oscillator, the mass of the block is 500 g and value of spring constant is k = 50 N/m and damping constant is 10 g s^(-1) The time period of oscillation is (approx.) |
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Answer» `2PI s` |
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| 1980. |
(A): The drift velocity of electrons in a metallic wire will decrease, if the temperature of the wire is increased. (R): On increasing temperature, conductivity of metallic wire decreases. |
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Answer» Both 'A' and 'R' are true and 'R' is the correct EXPLANATION of 'A' |
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| 1981. |
Assertion: Ampere’s circuital law is independent of Biot-Savart’s law. Reason: Ampere’s circuital law can be derived from the Biot-savart’s law. |
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Answer» Assertionis correct, REASON is correct, reason is a correct explanation for ASSERTION. |
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| 1982. |
A molecule of a substance has a permanent electric dipole moment of magnitude 10^(-29) Cm. A mole of this substance is polarised (at low temperature) by applying a strong electrostatic field of magnitude 10^6 V m^(-1). The direction of the field is suddenly changed by an angle of 60^@. Estimate the heat released by the substance in aligning its dipoles along the new direction of the field. Assume 100% polarisation of the sample. |
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Answer» Solution :Here, dipole moment of each molecules `= 10^(-29)Cm` As 1 mole of the substance contains `6 xx 10^(23)` molecules, total dipole moment of all the molecules, `p= 6 xx 10^(23) xx 10^(-29) Cm = 6 xx 10^(6)Cm` Initial POTENTIAL energy, `U= - pE cos THETA =-6 xx 10^(-6) xx 10^(6) cos 0^@ =- 6 J` Final potential energy `("when "theta= 60^@), U_f= - 6 xx 10^(-6) xx 10^6 cos 60^@= -3 J` Change in potential energy = -3J -(-6J) =3J So, there is loss in potential energy. This must be the energy released by the substance in the form of heat in aligning its dipoles. |
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| 1983. |
The India born and USA based Nobel Laureate Prof. Chandrasekhara is known for his work on |
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Answer» STUDY of cosmic rays |
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| 1984. |
A cell placed in hypotonic solution bursts up. It is |
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Answer» ANIMAL cell |
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| 1985. |
A galvanometer has a current range of 15mA and voltage range of 750mv. To convert this galvanometer into an ammeter of range 25A, the sunt resistance required is nearly |
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Answer» `0.2Omega` |
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| 1986. |
A person is driving a vehicle at uniform speedof 5m s^(-1) on a level curved track of radius5m. The coefficient of static friction between tyres and road is 0.1. Will the person slip while taking the turn with the same speed ? Take g=10 ms^(-2). Choose the correct statement. |
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Answer» A person will SLIP if `V^(2) = 5MS^(-1)` `v_("max") = sqrt(0.1 xx 5 xx 10) = sqrt(5)` or `v_("max")^(2) = 5m^(2)//s^(2)` `therefore` PERSONS or vehicle will slip if the velocity is more than `sqrt(5)m//s` |
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| 1987. |
What is a diode. |
| Answer» Solution :A device which ALLOWS flow of CURRENT in ONE direction and not opposite. | |
| 1988. |
Two blocks of masses m_(1) = 2kg and m_(2) = 4kg are attached to two ends of a light ideal spring of force constantk = 1000 N//m. The system is kept on a smooth inclined plane inclined at 30^(@) with horizontal. The block m_(2) is attached to a light string whose other end is connected to a mass m_(3) = 1 kg. A force F = 15N is applied on m_(1) and the system is released from rest. Assume initially the spring is at relaxed position and the system is released from rest. Find the (a) maximum extension of the spring. ( |
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| 1989. |
A screen is placed 2 m away from a single narrow slitwhich is illuminated by the light of wavelength 6xx10^(-7)m. If the first minimum lies 4 mm on either side of the central maximum, find the width of the slit. |
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| 1990. |
In a single slit diffraction experiment first minimum for,lamda_1 = 660nmcoincides with first maxima for wavelength lamda_2. Calculate lamda_2 . |
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Answer» SOLUTION :Position of minima in DIFFRACTION pattern is given by, ` d sin THETA =n lamda` For first minima of `lamda_1` , we have `d sin theta_1 = (1) lamda_1 " or " sin theta_1 =(lamda_1)/(d)`.....(i) The first maxima approximately lies between first and second minima. For wavelength `lamda_2`its position will be, `d sin theta_2 = 3/2 lamda_2 therefore sin theta_2 = (3 lamda_2)/(2d)`...(II) The two will coincide if , `theta_1 = theta_2 " or " sin theta_1 = sin theta_2 therefore (lamda_1)/(d) =(3lamda_2)/(2d)` `lamda_2 = 2/3 lamda_1= 2/3 x 660 NM = 440 nm` |
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| 1991. |
A 100m long antenna is mounted on a 500m tall building. The complex can become a transmission tower of waves with lambda |
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Answer» Solution :Length of antenna, I=100m, As `l=(LAMBDA)/(4)` or `lambda=4l=4xx100=400m`. |
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| 1992. |
From a point on the ground at a distance 2m from the foot of a vertical wall, a ball is thrown at an angle of 45^@ of which just clears the top, of the wall and afterward strikes the ground at a distance 4m on the other side. The height of the wall is |
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Answer» `2//3 m` |
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| 1993. |
The primary coil of a transformer has 200 turns and the secondary has 1000 turns. If the power output from the secondary at 1000 V is 9 kW, calculate (i) the primary voltage, and(ii) the heat loss in the primary coil if the resistance of primary is 0.2 Omegaand the efficiency of the transformer is 90%. |
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Answer» SOLUTION :Here, `N_(P) = 200, N_(s) = 1000 V, V_(s) = 1000 V`, output power = 9 kW = 9000 W, and EFFICIENCY `eta = 90% = 0.9` (i) `therefore V_(s)/V_(p) = N_(s)/N_(p)`, hence, `V_(p) = (V_(s).N_(p))/(N_(s)) = (1000 XX 200)/1000 = 200 V` (ii) `therefore` Efficiency `eta = ("output power")/("input power") = ("Output")/(V_(p).I_(p))` `therefore I_(p) = (9000)/(200 xx 0.9) = 50 A` As resistance of primary coil `R_(p) = 0.2 Omega` `therefore` Power IONS in primary coil as heat `=I_(p)^(2) . R_(p) = (50)^(2) xx 0.2 = 500 W` |
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| 1994. |
After changing a parallel plate capacitor by a battery, the battery is removed. Now a dielectric slab is inserted between the plates. What will be the effect on charge |
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| 1995. |
After changing a parallel plate capacitor by a battery, the battery is removed. Now a dielectric slab is inserted between the plates. What will be the effect on capacitance, |
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| 1996. |
After changing a parallel plate capacitor by a battery, the battery is removed. Now a dielectric slab is inserted between the plates. What will be the effect on P.D. across is doubled. |
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| 1997. |
An observer can see through a pin-hole the top end of a thin of height h, places as shown in figure. The beaker height is 3h and its radius h. When the beaker is filled with a liquid up to a height 2h, he can see the lower end of the rod. Then the refractive index of the liquidis |
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Answer» `(5)/(2)` Since `tanr=(2h)/(2h)=1` `rArrr=45^(@)` `RARR sini=(h)/(hsqrt(5))` `rArr sini=(1)/(sqrt(5))` `:. (1)/(n)=(1//sqrt(5))/(1//sqrt(2))rArrn=sqrt((5)/(2))` 
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| 1998. |
What is magnetic susceptibility? For which material is it low and positive? |
| Answer» Solution :Magnetic susceptibility is DEFINED as the ratio of magnetization and the magnetic intensity i,e `X = M/H`The magnetic susceptibility is low and positive for paramagnetic materials LIKE : SODIUM CHLORIDE, SILICON, bismuth etc. | |
| 1999. |
Ground state energy of hydrogen atom is E_(0) .Match the following two columns. |
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Answer» `:. U=2E,K=-E` |
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| 2000. |
Tow capacitons which have the same dielectric material (epsi_(r)=2) are connected in series and the combnationis putacross a steady p.d. of 220 V. what will be eheir p.d.s if the dielctric of the smaller capacitor is replaced by a dielectric of relative permittivity 5? |
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