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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.
| 38801. |
What is the force between two small charged spheres having charges of2xx10^(-7) C and 3xx10^(-7)C placed 30 cm apart in air? |
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Answer» Solution :Here `q_1 =2xx10 ^(-7)C, q_2 =3XX10 ^(-7)C, r = 30 cm = 0.3 ` `THEREFORE ` Force in AIR ` F= (1)/( 4pi in _0).(q_1q_2)/( r^(2)) =(9xx 10 ^(9) xx2 xx 10 ^(-7) )/( (0.3) ^(2)) =6xx 10 ^(-3) N` The force is repulsive in nature. |
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| 38802. |
Two concentric circular coils, one of small radius r_1 and the other of large radius r_2, such that r_1 lt lt r_2, are placed co-axially with centres coinciding. Obtain the mutual inductance of the arrangement. |
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Answer» Solution :Consider two EXTREMELY long co-axial CIRCULAR coils (or solenoids) of equal length L, with smaller one inside the bigger one. Let small and big coils be coil no. 1 and coil no. 2 respectively. Their common area of cross section is suppose `a=pir_1^2`. When `I_1` amount of current is passed through `N_1` turns of small coil, magnetic FLUX linked with big coil is, `Phi_2=N_2B_1 a cos 0^@` (Here, magnetic field is present only inside the small coil and so cross-sectional area of small coil is considered). `therefore Phi_2=N_2((mu_0N_1I_1)/l)a` `therefore Phi_2/I_1=(mu_0N_1N_2a)/l` Here `Phi_2/I_1=M_21`=mutual inductance of second coil towards FIRST coil. `therefore M_21=(mu_0N_1N_2a)/l`....(1) Similarly when `I_2` amount of current is passed through `N_2` no. of turns of big coil, magnetic flux linked with small coil is, `Phi_1=N_1B_2a` `=N_1((mu_0N_2I_2)/l)a` `therefore Phi_1/I_2=(mu_0N_1N_2a)/l` Here `Phi_1/I_2=M_12`=mutual inductance of first coil towards second coil. `therefore M_12 =(mu_0N_1N_2a)/l` ....(2) Equations (1) and (2) prove that `M_12=M_21`...(3) From equations (1),(2) and (3) `M_12=M_21=M=(mu_0N_1N_2a)/l` (where `a=pir_1^2`=cross-sectional area of small coil) |
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| 38803. |
Name the theorem used to find the electric flux through a surface enclosing charge. |
| Answer» SOLUTION :GAUSS THEOREM. | |
| 38804. |
A linear object AB is at a distance of 36cm from an equal convex lens of focal length 30cm. In front of lens there is a plane mirror which is inclined at an angle 45^(@)with the principal axis of the lens at a distance I=1 from the mirror as shownn in figure. A container with water layer d=20cm is placed as shown in the figure. The choose the CORRECT statement(s). Take the refractive index of water is 4/3 |
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Answer» After reflection from the mirror the image of `AB` will be PARALLEL to principal AXIS of the lens |
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| 38805. |
Consider a spongy block of mass m floating on a flowing river. The maximum mass of the block is related to the speed of the river flow v, acceleration due to gravity g and the density of the block rho such that m_(max)=kv^(x)g^(y)rhz^(z) (k is constant). The values of x, y and z should then respectively be (Mass of the spongy block is assumed to vary due to absorption of water by it) |
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Answer» Solution :Since, the maximum mass of the block floating on river depends, speed of flow of the river =v, acceleration due to gravity =g and density of the block =`rho`, `m_(max)=KV^(x)g^(y)rho^(z)` WRITE the dimensional formula of the both side, we get, `[M^(1)L^(0)T^(0)]=[LT^(-1)]^(x)[LT^(-2)]^(y)[ML^(-3)]^(z)` `[M^(1)L^(0)T^(0)]=[M^(Z)L^(x+y-3z)T^(-x-2y)]` Compairing the dimensions of M, L and T on both sides, we get `z=1""....(i)` `x+y-3z=0""....(ii)` `-x-2y=0""......(iii)` `x+y-3xx1=0` [From Eq. (i) and (ii)] `x+y=3""....(iv)` From Eqs. (iii) and (iv), we get `-y=3rArry=-3` From Eq. (iv), we have `x-3=3` `impliesx=6` Hence, the value of x, y and z will be (6,-3,1). |
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| 38806. |
What are ohmic and non ohmic devices ? |
| Answer» Solution :Materials for which the CURRENT against voltage graph is a STRAIGHT line through the origin, are said to obey Ohm.s LAW and their behaviour is said to be OHMIC. Materials or DEVICES that do not follow Ohm.s law are said to be non-ohmic. | |
| 38808. |
A light bulb and an open coil inductor are connected to an ac source through a key as shown in Fig. The switch is closed and after sometime, an iron rod is inserted into the interior of the inductor. The glow of the light bulb (a) increases, (b) decreases, (c) is unchanged, as the iron rod is inserted. Give your answer with reasons. |
| Answer» Solution :As the iron rod is INSERTED, the magnetic field inside the coil MAGNETIZES the iron INCREASING the magnetic field inside it. Hence, the inductance of the coil increases. CONSEQUENTLY, the inductive reactance of the coil increases. As a result, a larger fraction of the applied ac voltage appears across the inductor, leaving less voltage across the bulb. Therefore, the glow of the light bulb DECREASES. | |
| 38809. |
A 10 kg satellite circles earth once every 2h in an orbit having a radius of 8000 km. Assuming that Bohr's angular momentum postulate applies to satellites just as it does to an electron in the hydrogen atom, find the quantum number of the orbit of the satellite. |
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Answer» Solution :`mv_(n) r_(n) `= nh /`2pi` . Here m = 10 kg and `r_(n) 8 xx 10^(6)`m . we have the time period T of the circling satellite as 2h. That is T = 7200 s. THUS the velocity `v_(n) = 2pi r_(n)`/T The quantum number of the orbit of satellite n = `(2pi xx 8 xx 10^(6) m)^(2) xx 10 `/(7200s `xx 6.64 xx 10^(-34)`JS) = 5.3 `xx 10^(45)` Note that the quantum number for the satellite motion is extremely large! In FACT for such large quantum numbers the results of quantization condition tend to those of CLASSICAL PHYSICS. |
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| 38810. |
A point source of light is placed 60 cm away from screen. Intensity detected at point P is I . Now a diverging lens of focal length 20cm is placed 20cm away from S between S and P. The lens transmits 75%of light incident on it. Find the new value intensity at P. |
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Answer» SOLUTION :`u=-20,f=-20rArrv=-10` Let P=Power of source `I=(P)/(4pi(60)^(2))` ENERGY receibed by lens, `E_(2)=(P)/(4pi(20)^2)A_1` `:. I_(2)=(0.75E_(2))/(A_(2))` From SIMILAR TRIANGLES `(A_(2))/(A_(1))=25` `:. I_(2)=0.27I` 
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| 38811. |
Two identical positive charges are placed at the diagonally opposite corners of a square and two more identical but negative charges are placed at the remaining two corners. What is the work done in putting all charges together at the centre if q is the magnitude of each charge and 2a is the length of each side of the square? |
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Answer» |
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| 38812. |
One conducting metal sphere contains 10^23 atoms. If from 0.1 % atoms one electrons are removed, then electric charge deposited on sphere is ....... |
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Answer» 16 C `N = 0.1%` of `10^(23) = 10^(23) xx 1/1000 = 10^(20)` `therefore` Charge on the sphere Q = Ne `therefore Q = 10^(20) xx 1.6 xx 10^(-19) = 16 C` |
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| 38813. |
A step up transformer operates on a 230 V line and a load current of 2 ampere. The ratio of the primary and secondary windings is 1:25. What is the current in the primary ? |
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Answer» Solution :USING the relation `(N_(P))/(N_(S)) = (I_(S))/(I_(P)), I_(P) = (N_(S)I_(S))/(N_(P))` Here `N_(P)//N_(S) = 1//25` or `N_(S)//N_(P) = 25//1 = 25` and `I_(S) = 2A` Current in primary, `I_(P) = 25 xx 2 = 50 A` |
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| 38814. |
In the three parts of a transistor, 'Emitter' is of |
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Answer» MODERATE size heavily doped |
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| 38815. |
A susceptibility of a certain magnetic material is 400. What is the class of the magnetic material ? |
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Answer» DIAMAGNETIC |
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| 38816. |
Whichbasic mode of communcationsis used in satellite communication ? What typeof wave propagation is used this mode ? Write, giving reason, the frequency range used inthis mode of propagation. |
| Answer» Solution :SATELLITE communications usesbroadcastmode in which SPACE wave MODE of propagation is used. For FREQUENCY range `30MHz- 1000MHz`, `lambda`falls in the range `0.3m` to `10M`. In thisrange, diffractionof wave is highand alsoloosesdirectional property so range of frequency is `5.925 - 6.425 GHz`is satellitecommunication. | |
| 38817. |
Derive the expression for energy stored in a charged capacitor. |
Answer» Solution : CONSIDER a capacitor of CAPACITANCE C. At any INTERMEDIATE stage of charge, Let`Q^(1)`and `-Q^(1)` be the charge on positive and negative plate respectively and V. be the potential difference across the plates. The work done dW to move small charge `dQ^(1)` from -ve plate to +ve plate is `dW=V^(1)dQ^(1)` But `C=(Q)/(V^(1))impliesV^(1)=(Q)/(C)` `therefore dW=(Q)/(C)dQ^(1)` The total work done to transfer Q charge from -ve to +ve plate `W=int_(0)^(0)dQ=int_(0)^(0)(Q)/(C)dQ[int QdQ=(Q^(2))/(2)]` `W=(Q^(2))/(2C)=(1)/(2)(Q^(2))/(C)` The work don is stored as energy U in charged capacitor. Therefore ,`U=(1)/(2)(Q^(2))/(C)=(1)/(2)QV=(1)/(2)CV^(2)` |
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| 38818. |
Consider a region inside which there are various types of charged but the total charge is zero. At points outside the region |
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Answer» the ELECTRIC field is necessarily zero |
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| 38819. |
A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece? |
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Answer» Solution :Here, `f_(0) = 144 CM` and `f_(e) = 6.0 cm` `therefore` Magnifying power `=-f_(0)/f_(e) =(-144)/6 = -24` and separation between the two lenses `=f_(0) + f_(e) = 144 + 6 = 150 cm` or `1.5 m` |
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| 38820. |
The process of changing some characteristics of a carrier wave in accordance with the incoming signal is called: |
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Answer» Amplification |
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| 38821. |
A uniform disc of mass m and diameter 2R moves forward towards another uniform disc of mass 2m & diameter 2R on a frictionless surface as shown in figure When the first disc contacts the second, they stick to each other and move as a single object. |
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Answer» the velocity of combined disc after the COLLISION is `(V_(0))/3` `MV_(0)=2MV` `V=(V_(0))/3` Angular momentum conservation about centre of mass `1/2MR^(2).omega_(0)-4/3MV_(0)R=[(MR^(2))/2+M(4/3R)^(2)+1/2(2M)R^(2)+2M(2/3R^(2))]OMEGA` `omega=3/25 omega_(0)-8/25(V_(0))/R` For `omega=0` `omega_(0)=(8V_(0))/(3R)` The energy loss, `/_\E=E_(i)-E_(f)` `=1/2MV_(0)^(2)+1/2((MR^(2))/2).omega^(2)-1/2 (3M)V^(2)=19/9MV_(0)^(2)` |
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| 38822. |
The central cell after triple fusion becomes the |
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Answer» PEC (primary ENDOSPERM CELL) |
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| 38823. |
Define the term 'stopping potential ' in relation to photoelectric effect. |
| Answer» Solution :THEMINIMUM negative POTENTIAL GIVENTO the anodeof photo- cellfor whichthe photoelectriccurrentbecomes zero is calledstopping potential. | |
| 38824. |
Mutual inductance of system of two coils is 0.3 H. If the current in the one coil is changed from 10A to 40A in 0.01 sec, the average induced emf in the other coil is ...... volt. |
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Answer» 9 `epsilon_2=M_21 (dI_1)/(DT)` `=0.3xx(40-10)/0.01` `=0.3xx30/0.01` `THEREFORE epsilon_2`=900 V |
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| 38825. |
A circular coil of a wire containing 100 turns each of radius 2cm carries a current of 0.20A. The magnetic field at the centre of the coil is….. |
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Answer» `2pi XX 10^(-4)` |
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| 38826. |
A capacitor of capacity 6muF and initial charge 160muC is connected with key S and resistance as shown in figure. Point M is earthed. If key is closed at t= 0, then the current (I_(1) through resistance R (=1"Omega") at t = 16"mu" s is |
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Answer» `10//3e A` `C = 6muF` , <br>`q = 160e^(-t//8/3xx6) = 160 xx e^(-t//16) or I = (dq)/(dt)` <br> `I = (160)/(16) e^(-t//16) = 10e^(-t//16) or 10e^(-1)= 10/e`<br> `I_1 xx 4 = (10/e -I_1) xx 8` <br> or `12I_1 = 80/e or I_1 = 80/(12e) = 20/(3e) A` .)
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| 38827. |
The incorrect statement regarding the lines of force of the magnetic field B is |
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Answer» magnetic intensity is a MEASURE of linesof forcepassingthrough UNIT area held NORMAL |
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| 38828. |
At which angle of incidence light reflected fron glass becomes completely plane polarised i At this angle of incidence angle of refraction is 33.6^(@) |
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Answer» <P>`90^(@)` `=90^(@)-33.6^(@)` `=56.4^(@)` |
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| 38829. |
An optical fibre (mu = 1.72) is surrounded by a glass coating (mu =1.50 ). Find the critical angle for total internal reflection at the fibre-glass interface. |
| Answer» SOLUTION :`theta_C = SIN^(-1) (75 )/( 86 )` | |
| 38830. |
How rainbow is formed ? |
| Answer» SOLUTION :DISPERSION of SUNLIGHT. | |
| 38831. |
Tritium has half life of 12.5 years undergoing beta decay. What fraction of sample of tritium will remain undecayed after 50 years? |
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Answer» `(1)/(8)` |
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| 38832. |
(a) PM wave is similar to AM wave. (b) PM generally uses a smaller bandwidth than FM (c ) FM signal produced from PM signal is very stable. (d) FM and PM waves are completely different forsquare wave modulating signal. |
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Answer» |
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| 38833. |
A duster weighs 0.5N. It is pressed against a vertical board with a horizontal force of 11N, If the co-efficient of friction is 0.5 the minimum force that must be applied on the duster parallel to the board to move it upwards is |
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Answer» 0.4 N |
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| 38834. |
An under water swimmer is at a depth of 12 m below the surface of water. A bird is at a height of 18 m from the surface of water, directly above his eyes. For the swimmer the bird appears to be a disance from the surface of water equal to ("Refractive Index of water is"(4)/(3)) |
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Answer» 24m |
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| 38835. |
In question 55, which charge has the largest magnitude? |
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Answer» A |
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| 38836. |
Hooke's law is valid for the elastic extension (or compression) of a rod. This law may be written down, by analogy with the formula of the previous problem, by substituting Young's modulus E for the bulk modulus K. Write this formula and express the rigidity of the rod in terms of its dimensions. |
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| 38837. |
Photo electric current depends on the intensity of incident lightWork function of caesium and platinum are 2.14eV and 5.65eV respectively. Which one of the metal has higher threshold wavelength? Why? |
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Answer» Solution :Threshold frequency of CAESIUM `V_0=(omega_0/h) =((2.14xx1.6xx10^-19 )/(6.6xx10^-34))=0.52xx10^15` HZ Threshold wavelength of caesium `lambda_0=C/V_0)=((3xx10^8)` /(0.52xx10^15))=5.77xx10^-7m` |
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| 38838. |
The power in AC circuit is given by P = E_("rms")I_("rms") cos phi. The value of cos phi in series LCR circuit at resonance is : |
| Answer» ANSWER :B | |
| 38839. |
A wire of length 'L' meters carrying a current 'I' amperes is bent in the form of a circle. The magnitude of its magnetic moment is |
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Answer» `(iL)/(4pi)` |
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| 38840. |
Which tribe did the author belong? |
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Answer» FAST thinkers |
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| 38841. |
A point light source lies on the principal axis of a concave spherical mirror with radius of curvature 160 cm. Its image appears to be back of the mirror at a distance of 70 cm from the mirror. Determine the location of the light source. |
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Answer» Solution :`(1)/(U ) +(1)/(v ) = (2)/(R )` Herev=70cm, `R=- 160cm, (1)/(u )= (2)/(R ) -(1)/(V)` ` therefore(1)/(u )= (2)/(-160 CM) - (1)/( 70cm) =- (15)/( 560cm)` ` thereforeu=- (560 )/( 15 ) cm=- 37cm` The force is at a distance of 37 cm in front of the mirror. |
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| 38842. |
A solid conducting sphere of radius 'a' is surrounded by a thin unchanged concentric conducting shell of radius 2a. A point charge q is plaed at a distance 4a from common centre of conducting sphere and shell. The inner sphere is then grounded The potential of outer shell is |
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Answer» `(q)/(32 PI epsilon_(0)a)` |
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| 38843. |
A solid conducting sphere of radius 'a' is surrounded by a thin unchanged concentric conducting shell of radius 2a. A point charge q is plaed at a distance 4a from common centre of conducting sphere and shell. The inner sphere is then grounded The charge on solid sphere is : |
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Answer» `-(q)/(2)` |
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| 38844. |
A solid conducting sphere of radius 'a' is surrounded by a thin unchanged concentric conducting shell of radius 2a. A point charge q is placed at a distance 4a from common centre of conducting sphere and shell. The inner sphere is then grounded Pick up the correct statement |
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Answer» Charge on surface of INNER sphere is non-uniformly distributed |
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| 38845. |
Assertion : V_("rms") and V_("mean") of gaseous molecules is nearly of the order of velocity of sound. Reason : The sound travels in air because of vibrational molecular motion. |
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Answer» If both assertion and REASON are TRUE and reason is the correct EXPLANATION of assertion. |
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| 38846. |
Must every magnetic field configuration have a north pole and a south pole ? What about the field due to a toroid ? |
| Answer» Solution :It is not NECESSARY that every magnetic field configuration have a north POLE and a south pole. POLES are formed only if the source of field has a net non-zero magnetic moment. In CASE of a toroid and for an infinitely long conductor CARRYING current net magnetic moment is zero and hence there is no pole formation. | |
| 38847. |
A big drops of a liquid of radius 2 cm splits into 1000 small droplets. Then radius of each droplet is, |
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Answer» `2 XX 10^-3 CM` |
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| 38848. |
A hollow metallic sphere of radius 0.1 m is charged to a potential of 5 V. The potential at its centre is |
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Answer» inifinite |
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| 38849. |
A ball is dropped vertically from a height above the ground. It hits the ground and bounces up vertically to a height d/2. Neglecting air resistance its velocity varies with the heighth above the ground as: |
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Answer»
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| 38850. |
Distinguish between Cohernt and Incoherent addition of waves. Develop the theory of constructive interference. |
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Answer» Solution :Coherent sources: The two sources which maintain zero (or) any constant phase relation between themselves are KNOWN as Coherent sources: Incoherent sources: If the phase difference changes with TIME, the two sources are known as incoherent sources. THEORY of constructive and destructive interference: Let the waves of two coherent sources be `y_(1)=a sin omegat.....(1)` `y_(2)=a sin (omegat+phi) .....(2)` where is a amplitude and `phi` is the phase difference between two displacements. According to superposition principle `y=y_(1)+y_(2)` `y=a sin omegat+a sin (omegat+phi)+a sin omegat cos phi+a cos sin phi` `y=a sin omegat[1+cos phi]+cos omegat[a sin phi]....(3)` Let `A cos phi=a(1+cos phi]....(4)` `A sin phi=a sin phi ....(5)` Substituting equations (4) and (5) in equation (3) `y=A sin omegat, cos phi+A cos omegat sin THETA` `y=A sin (omegat+phi).....(6)` Where A is reulstant amplitude. Squaring equations (4) and (5). Then adding `A^(2)[cos^(2)phi+sin^(2)phi]=a^(2)[1+cos^(2)phi+2cos phi+sin^(2)phi]` `A^(2)[1]=a^(2)[1+1+2 +cos phi]""(therefore I=A^(2))` `I=2a^2 xx 2cos^(2) ""(phi)/(2)` `I=4a^(2) cos^(2) ""(phi)/(2)` `I=4I_(0)cos^(2)""(phi)/(2)......(7)""(therefore I_(0)=a^(2))` Case (i) For constructive interference: Intensity should be MAXIMUM. `cos ""(phi)/(2)=1 Rightarrow phi=2npi` Where `n=0,1,2,3,......Rightarrow phi=0, 2pi, 4pi,6x ......I_("max")=4I_(0)` Case (ii) For destructive interference: Intensity should be minimum i.e, `cos phi=0 Rightarrow phi =(2n+1)pi, " where n "=0,1,2,3,......,phi=pi,3pi,5pi Rightarrow I_(min)=0` 
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