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.
| 40851. |
Twelve equal wires , each of resistance r ohm are connected so as to form a skeleton cube. Find the equivalent resistance between the diagonally opposite points 1 and 7. |
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Answer» Solution :CONNECT a source between point 1 and 7 The network is SYMMETRICAL about the diagonal 1-7 . Therefore current in resistors are distributed symmetrically about the diagonal . The current distribution is shown in FIG(b) . CHOOSE a close loop 1-2-3-7-9-10-1, we have `-r1/3 ri/6 ri/3 + V=0 ` (or) `V/i = 5/6 r` (or)` R_(17) = V/i = 5/6 r` |
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| 40852. |
Column II show five systems in which two objects are labelled as X and Y. also in such case a point P is shown . Column I gives some statements about X and/or Y . Match these statements to the appropriate system(s) from Column.II |
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| 40853. |
An arc of radius r carries charge. The linear density of charge is lamda and the arc subtends an angle (pi)/(3) at the centre. |
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Answer» `(LAMDA)/(4 epsilon_(0))` |
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| 40854. |
An object is placed in front of a convex mirror at a distance of 50 cm. A plane mirror is introduced covering the lower halfof the convex mirror. If the distance between the object and the plane mirror is 30 cm, it is found that there is no gap between the images formed by the two mirrors. The radius of the convex mirror is |
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Answer» 12.5 cm |
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| 40855. |
What is amplification? |
| Answer» SOLUTION :It is PROCESS of INCREASING the strength of a MESSAGE signal using amplifier. | |
| 40856. |
On Earth, a body suspended on a spring of negligible mass causes extension L and undergoes oscillations along length of the spring with frequency f. On the Moon, the same quantities are L/n and f' respectively. The ratio f'/f is |
| Answer» Solution :Oscillations ALONG spring length are INDEPENDENT of GRAVITATION. | |
| 40857. |
A block slides down a slope of angle thetawith constant velocity. It is then projected up with a velocity of 10ms^(-1) g=10ms^(-2) & theta= 30^(@). The maximum distance it can go up the plane before coming to stop is |
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Answer» 10 m |
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| 40858. |
A ray of light is incident at an angle of 45^@ with a velocity 3xx10^8 m/s on a second medium. The angle of refraction in second medium is 30^@. The velocity of light in second medium is: |
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Answer» `1.5xx10^8 m//s` |
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| 40859. |
Give reasons The Zener diode is fabricated by heavily doping both the p and n sides of the junction. |
| Answer» Solution :HEAVY doping makes the depletion region very thin. This makes the electric field of the junction very high, even for a small REVERSE BIAS voltage. This in turn helps the Zener DIODE to act as a .voltage regulator.. | |
| 40860. |
In a moving coil galvanometer the deflection phi of the coil is related to the electric current I by the relation |
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Answer» `I PROP TAN PHI` |
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| 40861. |
Consider a plane wavefront of electromagnetic fields travelling with a speed c in the right (say+Z) direction, it is given that vecE and vecB are transverse to each other and uniform throughoutthe left of the wavefront and zero on the right of the wavefront. [This is contrived, but not incorrect, configuration chosen for simplicity. In the usual monochromatic plane wave, vecE and vecB very sinusoidally in space and time]. (a) Use Faraday's law to show that E=cB. (b) Use Ampere's law (with displacement current included) to show that c=1//sqrt(mu_0in_0) |
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Answer» Solution :Let `vecE` be in the x-direction and `vecB` in the y- direction. (a) Consider the rectangular loop in the XZ plane with ONE side of LENGTH l prallel to `vecE`. At the instant under consideration, the rectangle is PARTIALLY on left and partially on the RIGHT of the wavefront. Rate of change of magnetic flux =Blc The line integral of `vecE=oint vec E.vec(dl)=El`. From Faraday's laws of electromagnetic induction `:. El=Blcor E=Bc....(i)` (b) Consider a similar rectangle in the YZ plane. Rate of change of electric flux=Elc. The line integral of `vecB` is Bl. from Ampere's law or `Bl=mu_0in_0Elc` or `B=mu_0in_0Ec` or `B=mu_0in_0c(Bc) ` [from EQ(i)] or `c^2=1/(mu_0in_0) or c=1/(sqrt(mu_0in_0))` |
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| 40862. |
Distinguish between resistance, reactance, and inpedance of an ac circuit. |
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Answer» SOLUTION :Resistance: It is the oppoosition offered by an ohomic RESISTOR . Reactance : It is the opposition offered to the current by an inductore or a CAPACITOR . Impedance : It is the effective opposition in an AC circuit containing any two or all three elements - INDUCTOR, capacitor and resistor. |
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| 40863. |
In T.V. broadcasting both picture and sound are transmitted simultaneously. In this |
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Answer» AUDIO signal is frequency MODULATED and video signal is amplitude modulated |
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| 40864. |
We have two radioactive nuclei A and B. Both convert into a stable nucleus C. Nucleus A converts into C after emitting two alpha -particles and three beta -particles. Nucleus B converts into C after emitting one alpha -particle and five beta -particles. A time t = 0, nuclei of A are 4N_0 , and that of B are N_0 . Half-life of A into the conversion of C) is 1min and that of B is . 2 min. Initially number of nuclei of C are zero (a) If mass numbers of A and B are A_1 and A_2respectively. Then find A_1 -A_2 (b) What are number of nuclei of C, when number of nuclei of A and B are equal (c) At what time rate of disintegrations of A and B are equal |
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| 40865. |
In the circuit shown in Fig. 27-47, there is a cube of resistances which are all having the same resistance of R. Find the equivalent resistance of the network between diagonally opposite points as shown in the figure. |
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| 40866. |
The surface tension of water = 72xx10^-3Jm^2. Find the work done in splitting a drop of water of 1mm into 64 small drops. |
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Answer» Solution :`4/3 pi R^3 = 64 xx 4/3 pi r^3` `therefore r^3 = (R^3)/(64)` `therefore r = R/4 = 1/4` mm initial surface energy `E_1 = 4piT (R^2)` FINAL surface energy `E_2 = 64 xx 4 piT (r^2)` `E_2 - E_1 = 4piT (64r^2 - R^2)` work done = `4piT (64r^2 - R^2) = 4piT(4 xx 10^-6 - 10^-6)` `12T xx 10^-6 = 2.7 xx 10^-6 J` |
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| 40867. |
What is the probable reason for children remaining barefoot? |
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Answer» LACK of money |
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| 40868. |
A charged particle moving in a uniform magnetic field penetrates a layer of lead and thereby loses one half of its kinetic energy. How does the radius of curvature of its path change? |
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Answer» (a) Nonrelativistic particles. The momenta of the particle are proportional to the square roots of their kinetic energies, and therefore also proportional to the radii of their tracks. (b) Relativistic particles. In this case the dependence of the MOMENTUM on the kinetic energy of the particle is more complex: `p=1/c sqrt(K(2epis_(0)+K))` Hence we obtain the desired RATIO of the radii of the tracks. |
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| 40869. |
A thin uniform ring of radius R and axis of rotation transverse through its centre. What is its radius of gyration? |
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Answer» 2 R |
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| 40870. |
First move towards left and then towards right For the circuit shown, which of the following statements is true ? |
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Answer» with `S_(1)` closed , `V_(1) = 15 V, V2 = 20V` |
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| 40871. |
Calculate the ratio of the frequencies of the radiation emitted due to transition of the electron in a hydrogen atom from its (i) second permitted energy level to the first level, and (ii) highest permitted energy level to the second permitted level. |
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Answer» Solution :We know that frequency of RADIATION emitted due to transition of an electron in a hydrogen atom from N state to n, stale is given by `v = (13.6 e)/(h) [(1)/(n_(J)^(2))-(1)/(n_(i)^(2))]` For 2-1 transition (i.e., `n_(i) = 2` and `n_(j) = 1`) we , have `V = (13.6e)/(h) [(1)/((1)^(2)) - (1)/((2)^(2))] = (3)/(4) xx (13.6e)/(h)"".......(i)` Forn - 2 transition(i.e., `n_(i) = oo ` and `n_(j) = 2`)we have . `v = (13.6e)/(h) [(1)/((2)^(2)) - (1)/((oo)^(2))] = (1)/(4) xx (13.6 e)/(h)` `rArr""(v)/(v)= (3)/(1)orv = 3 v` |
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| 40872. |
Silion and copper are cooled from 300 K to a temperature of 60 K. Then conductivity |
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Answer» for SILION INCREASES and for COPPER decreases |
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| 40873. |
In a conducting hollow sphere of inner and outer radii 5cm and 10cm respectively, a point charge 1 muC is placed at point A, thaUs 3cm from the centre C of the hollow sphere. An external uniform electric field of magnitude 20 N/C is also applied. Net electric force on the this. charge is 15N, away from the away from thecentre of the sphere as shown in figure-1.455. Find magnitude of force exerted by the charge placed at point A on the sphere: centre of the sphere as shown in figure. Find magnitude |
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| 40874. |
Soft iron is used in many parts of electrical machines for |
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Answer» low hysteresis loss and low PERMEABILITY |
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| 40875. |
Distinguish between 'sky waves' and 'space waves' modes of propagation in communication system. (a) Why is sky wave mode propagation restricted to frequencies upto 40 MHz ? (b) Give two examples where space wave mode of propagation is used. |
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Answer» Solution :LONG distance communication between two points on the earth is achieved through reflection of electromagnetic waves by ionosphere. Such waves are called sky waves. Sky wave propagation takes place up to frequency of about 40 MHz. A space wave travels in a straight line from transmitting antenna to the receiving antenna. It is used for line of SIGHT (LOS) communication as well as satellite communication. (a) The ionospheric LAYER acts as a reflector for a certain range of FREQUENCIES (3 to 30 MHz). Electro-magnetic waves of frequencies higher than 30 MHz (upto 40 MHz) penetrate the ionosphere and escape. (b) In television broadcast, microwave links and satellite communication, space wave mode of propagation is used. |
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| 40876. |
Carriers of electric current in superconductors are |
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Answer» electrons |
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| 40877. |
Which of the following helps to slow down fast neutrons in a reactor ? |
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Answer» SAFETY rods |
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| 40878. |
A card sheet divided into squares each of size 1 mm^(2) is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 10 cm) held close to the eye. (a) What is the magnification produced by the lens? How much is the area of each square in the virtual image? (b) What is the angular magnification (magnifying power) of the lens? (c) Is the magnification in (a) equal to the magnifying power in (b)? Explain. |
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Answer» Solution :Area of each piece `= 1mm^2` `therefore u = - 9 cm , f= + 10 cm` (convex LENS) Lens formula, `1/f = 1/v - 1/u` ` therefore 1/v = 1/f + 1/u` `therefore1/v = 1/10 + 1/(-9) = (9-10)/(90) = (-1)/(90)` `thereforev = -90 cm ` `therefore` Value of magnification `m= v/(|u|) = 90/9 = 10 ` `rArr` Area of each piece in virtual IMAGE, `A.= (10)^2 xx 1 = 100 mm^2` `= 1 cm^2` (b) Magnifiying power , `M = (D)/(|u|) = (25)/(9) = 2.8` (c) No, magnification of lesnand ANGULAR magnification of optical instrument both are different. `rArr` Value of magnification `= (|v|)/(|u|)` and angular magnification (magnifying power) ` = (D)/(|u|) = (25)/(|u|)` `rArr` Only when image is obtained at near point, then `|v|` = 25 cm and hence magnification and magnifying power are same. |
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| 40879. |
The area under velocity-time graph gives the displacement whereas the slope of velocity-time graph at any point gives acceleration at that point. Slope is +ve if the line makes an acute angle in the anticlockwise direction with respect to the positive direction along which the independent variable increases. From the calculus, we also know that the slope of the tangent to a y vs. x graph is given by (dy)/(dx) If (dy)/(dx) is +ve then y increases with x, otherwise it decreases. Consider two persons namely Ram, and Rahim, who have gone to picnic at a nice place. Finding a beautiful place they rush towards it simultaneously. Let us take the time when they start simultaneously at t= 0 and at this instant the position of Ram and Rahim be x = 0 and x = + 48, respectively. Ram maintains a uniform velocity of +10 m/s throughout the journey whereas Rahim starts from rest and continuously accelerates with + 1 m//s^(2) QShown is a graph of either position or velocity-time graph Choose the correct alternative. |
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Answer» It is position-time GRAPH of RAM as OBSERVED by Rahim |
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| 40880. |
In Young.s double slit experiment, blue-green light of wavelength 500nm is used. The slits are 1.20mm apart, and the viewing screen is 5.40 m away from the slits. What is the fringe width. |
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Answer» `6.2mm` |
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| 40881. |
A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direetion east to west. The earth's magnetic held at the place is 0.39 G, and the angle of dip 35^@. The magnetic declination is nearly zero. What are the resultant magnetic felds at points 4.0 cm below the cable? |
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Answer» Solution :Magnetic field at a point due to 4 wires , `= 4 ((mu_(o))/(4pi)) ((2i)/(a))` =` (4xx10^(-7)xx2xx1)/ (4xx10^(-2)) = 0.2 XX 10^(-4)` T = 0.2 G Below the CABLE , `B_(H)^(1) = underset(("field pointing inward"))0.39cos35^(@)-underset(("field pointing outward"))0.2 = 0.12G` `B_(V)^(1) = 0.39 sin 35^(@) - 0.22 G ` `:. ` Resultant magnetic field =`sqrt((B_(H)^(1))^(2)+(B_(V)^(1))^(2))` = ` sqrt((0.12)^(2)+(0.22)^(2))` = 0.25 G Resultant field direction `theta = tan ^(-1) (B_(v)^(1))/(B_(H)^(1))` `= tan ^(-1) ((0.22)/(0.11))` `= tan ^(-1) (2)` `theta = 62^(@)` Above the cable : `B_(v)^(1) = 0.39 sin 35^(@)` +0.2 = 0.52 G `B_(v)^(1)` = 0.22 G `:. ` Resultant field = `sqrt((0.52)^(2)+(0.22)^(2))` =0.57 G `theta ^(1) = tan ^(-1) ((0.22)/(0.52))` = `tan^(-1) (0.4231)` = `23^(@)` |
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| 40882. |
Two concentric rings are kept in the same plane. Number of turns in both the rings is 20. Their radii are 40 cm and 80 cm and they carry electric currents of 0.4 A and 0.6 A respectively, in mutually opposite directions. The magnitude of the magnetic field produced at their centre is ______ T. |
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Answer» `4mu_(0)` `r_(1)=40cm=0.4m` `r_(2)=80cm=0.8m` N = 20 `I_(1)=0.4A` `I_(2)=0.6A` Magnetic field due to ELECTRIC CURRENT `I_(1)`, `B_(1)=(mu_(0)NI_(1))/(2r_(1))=(mu_(0)(20)(0.4))/(2xx(0.4))=10mu_(0)T` Magnetic field due to electric current `I_(2)`, `B_(2)=(mu_(0)NI_(2))/(2r_(2))=(mu_(0)(20)(0.6))/(2xx(0.8))=13/4mu_(0)T` Both the magnetic fields are in opposite direction, so RESULTING magnetic field, `B=B_(1)-B_(2)=10mu_(0)-30/4mu_(0)` `thereforeB=10/4mu_(0)T` |
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| 40883. |
A solid sphere has a uniformly distributed mass of 1.0 xx 10^4 g and a radius of 1.0 m. What is the magnitude of the gravitational force due to the sphere on a particle of mass m = 2.0 kg when the particle is located at a distance of (a) 1.5 m and (b) 0.50 m from the center of the sphere? (c) Write a general expression for the magnitude of the gravitational force on the particle at a distance r le 1.0 , from the center of the sphere. |
| Answer» Solution :(a) `5.9 XX 10^(-7)N,` (b) `6.7 xx 10^(-7)N,` (C ) `(1.3xx 10^(-6)N//m)r` | |
| 40884. |
Light of wavelength 5000 Å falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray? |
| Answer» SOLUTION :`5000Å, 6XX10^(14)HZ, 45^(@)` | |
| 40885. |
In the circuit diagram shown in figure, initially switch S is opened and the circuit is in steady state. At time t=0, the switch S is closed and the new steady state is reached after some time. Choose the correct option(s) |
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Answer» Current in the INDCUTOR when the circuit reaches the new steady state is 4A. |
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| 40886. |
The bulk modulus of metal is 8 xx 10^ 9 N/m^2. What amount of pressure is required to reduce the volume of a spherical ball by 5%? |
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Answer» SOLUTION : `K = (vdp)/(dv) therefore dp = k (dv)/V` `= 8 XX 10^9 xx 5/100 = 4 xx 10^8 N/m^2` |
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| 40887. |
Who predicted the existence of electromagnetic waves? Give the wavelength range of electromagnetic spectrum. |
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Answer» Solution :MAXWELL. PREDICTED the existence of electromagnetic waves. The wavelength range of electromagnetic SPECTRUM is of the ORDER of `10^(-4) m-10^(7)m` |
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| 40888. |
During each cycle, a heat engine absorbs 400J of heat from its high-temperature source and discards 300J of heat into its low-temperature sink. What is the efficiency of this engine? |
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Answer» `1/7` |
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| 40889. |
A: The speed of a planetis maximumat perihelion. R : Theangularmomentumof a planet aboutcentre of sun is conserved . |
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Answer» If both Assertion & Reasonare true . Andthe reasonis the correct explanationof theassertion , then mark (1) |
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| 40890. |
A transformer has turn ratio 100 : 1. If the secondary coil has 4 amp current, then current in primary coil is |
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Answer» 4A `rArr (I_(P))/(4)=(1)/(100)rArr I_(P)=(4)/(100)=0.04 A` |
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| 40891. |
Give one application ofX-rays. |
| Answer» SOLUTION :For the INVESTIGATION of STRUCTURE of CRYSTALS. | |
| 40892. |
Two capacitors 2C and 4C initially charged to potential difference of V_(0) and 3V_(0) with the potential as show are connected to an inductor of inductance L. Initial curren tin the inductor is zer. Now the swtich 'S' is closed. The maxmimum current in the circuit is |
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Answer» `(V_(0))/(8)SQRT((C)/(3L))` |
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| 40893. |
In which of the following the carriers of electric current are electrons only ? |
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Answer» a SUPER conductor |
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| 40894. |
Phot diodes are preferable used in reverse bias conditions for measuring light intensity because |
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Answer» Current in reverse biase is less which is easier to measure compared to higher currents in forward bias SINCE `Deltan=Deltap` {equal hole electron pair generation in BOND breaking] `(Deltan)/(n)lt lt (Deltap)/(p)` `therefore` it is difficult to measure small `(Deltan)/(n)` Compared to `Deltap//p` `therefore` Photodiode is USED in Reverse bios mode since it is easier to measure the higher fractional change of minority carriers. |
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| 40895. |
A bar magnet has a pole strength of 15 A m and magnetic length 20 cm. What is the magnetic induction produced by it at a point at a distance of 30 cm from either pole? (mu_(0)/(4pi) = 10^(-7)Wb//Am) |
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Answer» `1.11xx10^(-5) Wb//m^(2)` The point is at a distance of 30 cm from EITHER pole. Hence it must be on the EQUATOR. `B_(eq) = (mu_(0))/(4pi)M/((d^(2)+l^(2))^(3//2))` But `(d^(2) + l^(2))^(1//2) = 30cm = 0.3m ` `therefore B_(eq ) =10^(-7) xx 3/(0.3)^(3)=(3xx10^(-7))/(27xx10^(-3))=(10^(-4))/9` `IMPLIES B = 1.11 xx10^(-5) Wb//m^(2)` |
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| 40896. |
A student performs an experiment for determintion of g = (4 pi^2 L)/(T^2) , L ~~ 1mand the commits an error of DeltaL for the tajes the time of n oscillations with the stop watch least count DeltaT. For which of the following data the measurement of g will be more accurate? |
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Answer» `DeltaL=0.5,DELTAT = 0.1, N = 20` |
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| 40897. |
A spherical capacitor is made of two conducting sphericalshells of radii a and b = 3a . The space between the shells is filled with a dielectric of dielectric constant K = 3upto a radius c = 2a as shown. If the capacitance of given arrangement is n timesthe capacitance of an isolated of an isolated sphericalconducting shell of radius a. Then find value of n . |
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| 40898. |
Two charges 5mu C and 4mu C are separated by a distanc 20 cm in air. Work to be done to decrease the distance to 10 cm is |
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Answer» 1.8 j |
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| 40899. |
What is SI unit of power of lens. |
| Answer» SOLUTION :Dioputre | |
| 40900. |
Usinf the Hund rules, write the spectral symbol of the basic term of the atom whose only partially filld subshell (a)is filled by 1//3 and S=1, (b)is filled by 70%, and S=3//2. |
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Answer» Solution :(a)If `S=1` is the maximum spin then three must be two electrons (if there are two holes then the shell will be more than half full). This MEANS that there are `6` electrons in the full shell so it is a `p` shell. By PAUL's PRINCIPLE the only antisymmertric combination of two electrons has `L=1` Also `J=L-S` as the shell is less than half full. THUS the term is `.^(3)P_(0)` (b) `S=(3)/(2)` means either `3` electrons or `3` holes. As the shell is more than half full the former possibility is ruled out. Thus we must have seven `d` electrons. Then as in problem 6.120 we get the term `.^(4)F_(9//2)` |
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