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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Use Lenz's law to determine the direction of induced current in the situations described by figure. (a) A wire of irregular shape turning into a circular shape, (b) A circular loop being deformed into a narrow straight wire. |
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Answer» Solution :(a) As per the statement when GIVEN loop changes its shape from irregular to circular inward MAGNETIC flux linked with it will INCREASE. (because of increase in area). Hence according to Lenz.s law, induced current should flow in such a direction which can decrease the inward magnetic flux. For this, induced current should flow anticlockwise. i.e. in a direction `a to d to c to b to a to d to c`in the given loop. (b) As the loop CONTRACTS, OUTWARD magnetic flux linked with it will decrease. Hence according to Lenz.s law, induced current should flow in such a direction which can increase outward magnetic flux. For this, induced current should flow clockwise i.e. in a direction `a. to b. to c. to d. to a.`in the given loop. |
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| 2. |
A p-n-p transistor is used in common- emitter mode in an amplifier circuit. A change of 40 muA in the base current brings a change of 2 mA in collector current and 0.04 V in base-emitter voltage. Find the : (i) inputresistance (R_("in")) and (ii) the base cuirent amplification factor (beta). If a load of 6kOmega is used, then also find the voltage gain of the amplifier. |
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Answer» Solution :Given `DeltaI_(B)=40muA=40xx10^(-6)A` `DeltaI_(C)=2mA=2xx10^(-3)A` `DeltaV_(BE)=0.04" volt, "R_(L)=6kOmega=6xx10^(3)Omega` (i) Input Resistance, `R_("in")=(DeltaV_(BE))/(DeltaI_(B))=(0.04)/(40xx10^(-6))=10^(3)Omega=1kOmega` (II) Current amplification FACTOR, `beta=(DeltaI_(C))/(DeltaI_(B))=(2xx10^(-3))/(40xx10^(-6))=50` (iii) Voltage gain in COMMON - emitter configuration. `A_(u)=beta(R_(L))/(R_("lmp"))=50xx(6xx10^(3))/(1xx10^(3))=300` |
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| 3. |
Photoelectron effect support quantum nature of light because |
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Answer» there is a MINIMUM frequency which no photo electrons are emitted. |
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| 4. |
S and is to be piled up on a horizontal ground in the form of a regular cone of a fixed base of radius R. Coefficient of static friction between the sand layers is mu. Maximum volume of the sand can be piled up in the form of cone without slipping on the ground is |
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Answer» `(mu R^(3))/(3 pi)`  `f sin phi = N COS phi` where, f= friction. N = normal reaction and `f= muN` `rArr""mu N sin phi = N cos phi` `rArr" "tan phi =(1)/(mu) =(R)/(h)` (from figure) So, maximum VOLUME of sand cone that can be formed over level ground is `V_("max") =(1)/(3) pi R^(2) h =(1)/(3) pi R^(2) (mu R) =(mu pi R^(3))/(3)` |
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| 5. |
For the hollow thin cylinderical current carrying pipe which statement is correct :- |
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Answer» magnetic field inside the PIPE is not ZERO |
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| 6. |
Define 'electric flux |
| Answer» SOLUTION :The number of electric field LINES CROSSING a given area KEPT normal to the electric field lines is called electric flux. Its unit is `Nm^(2)C^(-1)`. Electric flux is a SCALAR quantity. | |
| 7. |
A solid cylindrical conductor of radius R carries a current along its length.The current density J_(1) however, is not unifrom over the cross section of the conductor but is a function of the radius according to J=br, where b is a constant.the magnetic field B at r_(1) lt R is (mu_(0)b r_(1)^(2))/N.Then find value of N |
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Answer» `B_(1)=(mu_(0)br_(1)^(2))/3 ,:'i_(1)=underset(0)OVERSET(f_(1))intdi==underset(0)overset(f_(1))int(dr)(2pir dr)=(2pibr_(1)^(3))/3` 
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| 8. |
A body moving under the action of a force. Suddenly force is increased to such an extent that it's kinetic energy is increased by 100%. The momentum increases by |
| Answer» ANSWER :D | |
| 9. |
Three identical particals, each of charge Q=0.5 mu C and mass m=7.5xx10^(-10) kg are placed at the verticle of an equilateral triangle of side r_0=1m and released.The particle repel each other by coulomb forces to larger equilateral triangle. Find the relative velocity between anytwo of them when the triangle has grown to a side length of r=2m. |
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Answer» |
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| 10. |
Velocity of light = …… |
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Answer» `SQRT(in_(0)mu_(0))` |
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| 11. |
An electron is travelling with a constant velocity vecv=v_(0)hatj in the presence of a uniform electric field vecE=E_(0)hatk, where E_(0)=(mg)/e (Here, m denotes the mass of the electron and e denotes its charge g denotes the acceleration due to gravity). Gravity acts in the -Z direction. At t=0, when the electron is at the origin, a uniform magnetic fiedl vecB=B_(x)hati+B_(y)hatj is switched on. Here B_(x) and B_(y) are fixed positive constants and B_(0)=sqrt(B_(x)^(2)+B_(y)^(2)). The coordinates of the point whre the electron intersects the first time after t=0 are given by: |
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Answer» `((2pimv_(0))/e((B_(x)^(2))/(B_(0)^(3))),(2pimv_(0))/e((B_(x)B_(y))/(B_(0)^(3))))` Let the angle made by the magnetic field vector with the Y-axis be `theta`. So `(B_(x))/(B_(y))=tan theta` The magnetic field is given by `vecB=(B_(0)sin theta)hati+(B_(0)cos theta)hatj` The electron will follow a helical PATH. The component of velocity of the electron parallel to the field will REMAIN unchanged, and the components of its velocity in a plane perpendicular to the field will oscillate. At `t=0` component of velocity parallel to field `v_(y)=v_(0)cos theta` And, component of velocity perpendicular to field, `v_(_|__)=v_(0)sin theta` So radius of the helical path `r=(mv_(_|_))/(eB_(0))` And time perido `T=(2pim)/(eB_(0))` The electron will return to the X-Y plane when it has completed one full rotation, i.e. at `t=T=(2pim)/(eB_(0))` During this time, the displacement of the electron (in a DIRECTION parallel to the field), `L=v_(y)((2pim)/(eB_(0)))=(2pimv_(0)cos theta)/(eB_(0))` , coordinates of the point where the electron intersects the X-Y plane are ` (L sin theta, L cos theta)` i.e. `((2pi mv_(0)sin theta cos theta)/(eB_(0)), (2pi m v_(0)cos^(2) theta)/(eB_(0)))`, i.e. `(((2pi m v_(0))/e)((B_(x)B_(y))/(B_(0)^(3))),((2pimv_(0))/e)((B_(y)^(2))/(B_(0)^(3))))` |
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| 12. |
A power transmission line feeds input power at 2400 V to a step down ideal transformer having 4000 turns in its primary. What should be number of turns in its secondary to get power output at 240V? |
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Answer» <P> Solution :In ideal TRANSFORMER `P_("in") = P_(0)``V_(p) I_p = V_s I_s` `(V_s)/(V_p) = (I_P)/(I_S) = (N_S)/(N_P) N_S ((V_S)/(V_P)) N_P = (240)/(2400) xx 4000 = 400` |
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| 13. |
White coheral light (400 nm700 nm ) is sent through the slit of a Young.s double slit experiment. The separation between the slitsis 0.5 mm and the screen is 50 away from the slits . There is a Which wavelengths (s) will be absent in the light coming from the hole ? |
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Answer» 400 NM ,667 nm |
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| 14. |
Two free point charges +q and +4q are placed apart a distance x. A third charge is so placed that all the three charges are in equilibrium. Then |
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Answer» Unknown charge at `(-4Q)/(9)` |
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| 15. |
A circular coil of radius 10 cm, 500 turns and resistance 2Omega is placed with its plane, perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through 180^@ in 0.25 s. The induced e.m.f. in the coil is ..... (Take H_E=3.0xx10^(-5) T) |
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Answer» `6.6xx10^(-4)` V =NAB(-1)-NAB =-2NAB `epsilon=-(Deltaphi)/(Deltat)` `=(2NAB)/(Deltat)` `=(2xx500xx3.14xx(10xx10^(-2))^2 xx 3xx10^(-5))/0.25` `=(1000xx3.14xx10^(-2)xx3xx10^(-5))/(25xx10^(-2))` `=376.8xx10^(-5) APPROX 3.8xx10^(-3)` V |
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| 16. |
If vecu = 2hati - 2hatj + 3hatkand the final velocity is vecv = 2hati - 4hatj + 5hatkand it is covered in a time of 10 sec, find the acceleration vector. |
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Answer» `(3bari-2barj+2bark)/(10)` |
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| 17. |
You are given the two circuits as shown in figure. Show that circuit (a) acts as OR gate while the circuit (b) acts as AND gate. |
Answer» Solution :(a) The output of the NOR gate is given in the INPUT of the NOT gate so its truth table can be WRITTEN as follows:  From the truth table it can be said that Y = Y.., which shows that the circuit behaves as an OR gate. (b) Here the output of the two NOT gate is given in the input of the NOR gate so its truth table can be obtained as follows:  From the truth table it can be said that the given circuit ACTS acts as AND. |
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| 18. |
Draw the circuit diagram of transistor as an oscillator and explain its working. |
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Answer» Solution :An L-C circuit can be used to produce oscillations of desired frequency. It consists of a tank circuit consisting of inductor L and CAPACITOR C connected in parallel. The frequency of the tank circuit is given by `v=(1)/(2pisqrt(LC))` Due to the resistance of inductive coil, there occurs a small but CONSTANT energy loss and oscillations thus produced are damped. To transmit speech or MUSIC, we require undamped electromagnetic waves called carrier waves.  To do so an L-C circuit is coupled with transistor in such a way that there is a proper feedback to the L-C circuit at the proper timings so that the energy of L-C circuit remains the same throughout oscillaitons. When key K is pressed, the collector ATTAINS positive potential due to which a weak collector current will start rising in `L_(1)`. The increasing magnetic flux is linked with `L_(1)` and HENCE with L. |
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| 19. |
An object is placed on the principal axis of a convex mirror. Distance of onject from the mirror is 40cm. A plane mirror is placed between the object and the convex mirror, covering lower half below principal axis of the mirror. Distance between the object and the plane mirror is 30cm. If there is no parallax between the two images formed by plane mirror and convex mirror, the focal length of the convex mirror is |
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Answer» SOLUTION :`u=-40cm`, `v=+20cm` `(1)/(F)=(1)/(v)+(1)/(u)impliesf=40cm` 
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| 20. |
A regular hexagon of side 10 cm has a charge 5 muC at each of its vertices. Calculate the potential at the centre of the hexagon. |
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Answer» Solution :Here CHARGE on each vertex of hexagon q = 5`MU C = 5 XX 10^(-6) C` SIDE of hexagon r = 10 cm=0.1 m It is clear that distance of hexagon centre O from any of the vertices of hexagon = r = 0.1 m `:.` Potential at centre of hexagon `V = 6 xx (1/(4piepsi_0).q/r)` ` = (6 xx 9 xx 10^9 xx 5 xx 10^(-6))/0.1 =2.7 xx 10^6V` |
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| 21. |
The equation of a transverse wave is given by y = 0.05 sin pi (2t - 0.02 x) , where x,y are in metre and t is in second. The minimum distance of separation between two particles which are in phase and the wave velocity are respectively |
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Answer» 50m 50 `ma^(-1)` y = 0.05 sin `(2pi t - 0.02 pi x)` COMPARING it with standard equation y = r sin `(omega t - KX) ` We have `omega= 2pi and K = 0.02 pi` and k = `(2pi)/(lambda) RARR -.01 pi= (2pi)/(lambda) ` `lambda = (2pi)/(0.02 pi) = 100 ` m so CORRECT choice is (b). |
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| 22. |
In a thermodynamic process, a system absorbs 2 kilo calorie of heat and at the same time does 500 J of work. What is the change in internal energy of the system: |
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Answer» `-500 J` `therefore dU =dQ-dW` Here `dQ =2000" CAL. "=2000xx4 CDOT 2=8400 J and dW=500 J` `therefore dU =8400 -500=7900 J` THUS, correct choice is (c ). |
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| 23. |
(I) Amplitude modulation is used in radio and TV brodcasting. (II) Phase of the carrier signal remain constant in amptitude modulation. (III) Noise level is low in AM. Baseband signal carries information |
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Answer» I,II and III only |
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| 24. |
A thin fixed ring of radius 'R' and positive charge 'Q' is placed in a vertical plane. A particle of mass 'm' and negative charge 'q' is placed at the centre of ring. If the particle is given a small horizontal displacement, show that it executes SHM. Also find the time period of small oscillations of this particle, about the centre of ring. (Ignore gravity) |
| Answer» SOLUTION :`2pisqrt((4piepsilon_(0)MR^(3))/(QQ))` | |
| 25. |
Explain about compound mircoscope and obtain the equation for magnification. Compound microscope: |
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Answer» Solution :The LENS NEAR the object, called the objective, forms a real, magnified image of the object. This serves as the object for the second lens which is the eyepiece. Eyepiece serves as a SIMPLE microscope that produce finally an enlarged and virtual image. The first inverted image FORMED by the objective is to be adjusted close to, but within the focal plane of the eyepiece so that the final image is formed nearly at infinity or at the near point. The final image is inverted with respect to the original. We can obtain the magnificaiton for a compound microscope. Magnificationof compound microscope From the ray diagram, the linear magnification DUE to the objective is, `m_(0) = (h.)/(v)` From the figure, `tan beta = (h)/(f_(0)) = (h.)/(L.)` then `(h.)/(h)=(L)/(f_(0)),m_(0)=(L)/(f_(0))` Here, the distance L is between the first focal point of the eyepiece to the second focal point of the objective. This is called the tube lenght Lof the microscope as `f_(0) and f_(e)` are comparatively smaller than L. If the final image is formed at p (nearpoint focusing), the magnification `m_(e)` of the eyepiece is, `m_(e) = 1 + (D)/(f_(e))` The total magnification m in nearpoint focusing is, `m=m_(o)m_(e)=((L)/(f_(o)))(1+(D)/(f_(e)))` If the image is formed at infinty (normal focusing), the magnificaiton `m_(e)` of the eyepiece is, `m_(e) = (D)/(f_(e))` The total magnification m in normal focusing is, `m=m_(o)m_(e)=((L)/(f_(o)))((D)/(f_(e)))` 
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| 26. |
What is the de Broglie wavelength associated with (a) an electron moving with a speed of 5.4xx10^(6) m//s, and (b) a ball of mass 150 g travelling at 30.0 m//s? |
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Answer» Solution :(a) For the electron : Mass `m=9.11xx10^(-31)kg`. speed `V=5.4xx10^(6)m//s`. Then. Momentum `p=m v=9.11xx10^(-31)(kg)xx5.4xx10^(9)(m//s)` `p=4.92xx10^(-24)kgm//s` de BROGLIE wavelength, `lambda=h//p` `=(6.63xx10^(-34)Js)/(4.92xx10^(-24)kgm//s)` `lambda=0.135nm` (b) For the ball : Mass `m.=0.150 kg," speed v. "=30.0m//s` Then momentum `..=m.v.=0.150(kg)xx30.0(m//s)` `p.=4.50kg m//s` de Broglie wavelength `lambda.=h//p.` `=(6.63xx10^(-34)Js)/(4.50xxkgm//s)` `lambda.=1.47xx10^(-34)m` The de Broglie wavelength of electron is comparable with X-ray WAVELENGTHS. However, for the ball it is about `10^(-19)` times the size of the proton, quite beyond experimental measurement. |
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| 27. |
In a Wheatstone's bridge three resistances P,Q,R connected in three arms and the fourth arm is formed by two resistances S_(1),S_(2) connected in parallel. The condition for bridge to be balanced will be |
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Answer» `(P)/(Q)=(R)/(S_(1)+S_(2))` |
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| 28. |
A plane harmonic electromagnetic wave with plane polarization propagates in vacuum. The elctric component of the wave has a strength amplitude E_(m)= 50 m V//m, the frequency is v=100MHz. Find : (a) the efficient value of the displacement current density , (b) the mean energyflow density averaged ove an oscillation period. |
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Answer» Solution :`E = E_(m)cos (2pi vt - KX)` (a) `j_(dis) = (delD)/(delt) = - 2pi epsilon_(0)vE_(m)sin (OMEGAT - kx)` Thus `(j_(dis))_(rms) = lt j_(dis)^(2)gt^(1//2)` `= sqrt(2) pi epsilon_(0)vE_(m) = 0.20 mA//m^(2)`. (B) `lt S_(x) gt = (1)/(2) sqrt((epsilon_(0))/(mu_(0))) E_(m)^(2)` as in. Thus `lt S_(x) gt = 3.3 muW//m^(2)` |
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| 29. |
Assertion: Light waves can be polarised but sound waves cannot be polarised. Reason: Sound waves in air are longitudinal in nature. |
| Answer» Solution :Being transverse in nature LIGHT WAVES can be polarised but being longitudinal in nature SOUND waves in air cannot be polarised. | |
| 30. |
Assertion: Fuse wire must have high resistance and low melting point. Reason: Fuse is used for small current flow only |
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Answer» If both assertion and REASON are true and the reason in the correct EXPLANATION of the assertion. |
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| 31. |
Critical angle of glass-air is 42^@, then speed of light in glass is ..... |
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Answer» `3 xx 10^8` m`s^(-1)` Speed in AIR c=`3xx10^8 ms^(-1)` Refractive INDEX n=`c/v` but n=`(1)/(sinC)` `THEREFORE(1)/(sinC)=(3xx10^8)/(v)` `therefore v=3xx10^8xxsinC` `=3xx10^8xxsin42^@``=3xx10^8xx0.6691` `=2.0073xx10^8` `therefore v~~2xx10^8ms^(-1)` |
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| 32. |
In put characterisitcs are shown. For CE configuration of n-p-n transistor for different ouput voltages. Here |
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Answer» `V_(CE_(1))gtV_(CE_(2))` |
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| 33. |
An object moves towards a stationary plane mirror at a speed of 4 ms^(-1) with what speed will his image move towards him? |
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Answer» `2 ms^(-1)` |
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| 34. |
What is the magnetic induction at the mid-point of the straight line joining the two polesof ta horse shoe magnet , separated by a distance'd' ? (The pole strength of each pole is 'm'. ) |
| Answer» Solution :Magnetic INDUCTION `B = (mu_0 2m)/(4PI d^(2))` , towardsthe S-pole and away FORM N-pole . | |
| 35. |
Derive the condition of balance for Wheatstone bridge. |
Answer» SOLUTION : Circuit arrangement for a Wheatstone bridge is being shown in the adjoining figure. In balance condition of bridge no current flows through galvanometer and it gives no deflection. Currents in various branches of network as per Kirchhoff.s first law are, thus, REPRESENTED in figure.  Applying Kirchhoff.s second law to mesh ABDA,we have `-P*I_(1)+R*(I-I_(1))=0` or `PI_(1)=R(I-I_(1)) "" ...(i)` Again for mesh BCDB, we have `-Q*I_(1)+S*(I-I_(1))=0 or QI_(1)=S(I-I_(1)) "" ...(ii)` DIVIDING (i) by (ii), we GET `(P)/(Q)=(R )/(S)` which is the balance condition of Wheatstone.s bridge. |
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| 36. |
Name the p-n junction diode which emit spontaneous radiation when forward biased. How do we choose the semiconductor, to be used in those diodes, if the emitted radiation is to be in the visible region? |
| Answer» SOLUTION :LIGHT EMITTING DIODE (LED) | |
| 37. |
A glass prism of refractive index 1.5 is immersed in water (refractive index 4/3). A light beam incident normally on the face AB is totally reflected to reach on the face BC if |
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Answer» `SIN theta ge 8/9` At POINT P, glass is denser and water is rarer . `""^gmu_w=(""^amu_w)/(""^amu_g)=(4//3)/(3//2)=8/9` `THEREFORE` From the theory of total internal reflection , sin `i_C=8/9` where `i_C`= critical angle `therefore theta = i_C` or sin `theta`=8/9 for critical angle. 
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| 38. |
The position of the neutral points is on the equatorial line of a bar magnet when the north pole of the magnet is pointing |
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Answer» west |
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| 39. |
Two small similar balls of mass m are hung by silk threads of length l from the same point , a small angle apart and carry similar charges q. They lose charge and draw closer quasistatically .Calculate the rate of loss of charge if the relative velocity varies as v=k/sqrtxwhere k is a constant. |
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| 40. |
The angle of a prism is 60^(@). When light is incident at an angle of 60^(@)on the prism, the angle of emergence is 40^(@). The angle of incidence i for which the light ray will deviate the least is such that |
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Answer» `I LT 40^(@)` |
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| 41. |
The V-I characteristic of a silicon diode is shown in the figure . Calculate the resistance of the diode at (a) I_(D) = 15 mA and (b) V_(D) = -10 V |
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Answer» Solution :Considering the diode characteristics as a straight line between `I = 10 mA` to I = 20 mA PASSING through the origin , we can calculate the resistance USING Ohm.s law . a. From the curve , at I = 20 mA , V = 0.8 V , I = 10 mA , V = 0.7 V `r_(fb) = (DELTAV)/(DELTAI) = (0.1)/(10) = 10 Omega` b. From the curve at `V = -10V , I = -1 mu A` `therefore r_(fb) = (10)/(1 mu) = 1.0 xx 10^(7) Omega` |
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| 42. |
If R_1 and R_2 are two resistors connected in series,what is their effective resistance? |
| Answer» SOLUTION :R=`R_1+R_2` | |
| 43. |
In galvanometer of resistance 20Omega gives a deflection of one division when a potential difference of 4mV is applied across its terminals. Calculate the resistance of the shut if the current of 10A is to be measured by it. The galvanometer has 25 divisions. |
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Answer» |
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| 44. |
A resistor of R = 6Omega, an inductor of L=1 H and C = 17.36 muF are connected in series with an A.C. source. Find the Q-factor. |
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Answer» 2.37 `=1/6sqrt(1/(17.36xx10^(-6)))=1/6sqrt(10^8/1736)` `therefore Q=1/6xx10^4/41.66` = 40.006 `therefore Q approx 40` |
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| 45. |
(a) Describe briefly, with the help of suitable diagram, how the transverse nature of light can be demonstrated by the phenomenon of polarisation. (b) When unpolarised light passes from air to a transparent medium, under what condition does the reflected light is plane polarised ? |
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Answer» Solution :(a) Light waves are transverse in nature. This can be demonstrated by the phenomenon of polarisation, by using two polaroids. When ordinary light from a monochromatic light source S passes through a POLAROID sheet `P_(1)`, known as polariser, the light becomes polarised in which electric vector is oscillating only along the pass axis of polaroid `P_(1)`. if the light is allowed to pass through a second polaroid `P_(2)`, known as analyser, whose pass axis is exactly parallel to that of `P_(1)`, then the light passes through it also and we see maximum light. HOWEVER, if `P_(2)` is rotated about its axis, there is decrease in INTENSITY of transmitted light. when pass axis of polaroid `P_(2)` is rotated about its axis, there is decrease in intensity of transmitted light. when pass axis of polaroid `P_(2)` is making an angle of `90^(@)` from that of `P_(1)`, no light passes through `P_(2)` and there is complete darkness on other side of it. however, on making pass axis of `P_(2)` parallel to `P_(1)` there is again maximum light passing through it. the experiment proves that light passing through polaroid `P_(1)` gets polarised about its pass axis. if the pass axis of `P_(2)` makes an angle `theta` with the pass axis of `P_(1)`, then only a component of electric vector passes through `P_(2)` and intensity of light decreases. when pass axis of `P_(2)` is making an angle of `90^@` with `P_(1)`, no vibrations of electric vectors can be pass through `P_(2)` and intensity of emergent light is ZERO.  (b) Plane polarised light can be obtained by reflection from a transparent surface (say a plane glass plate) of refractive index n when the light is incident on the surface at the polarising angle `i_(p)`. Value of polarising angle is given by Brewster.s law, according to which `tani_(p)=n or i_(p)=tan^(-1)(n)`. |
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| 46. |
A non-conducting sphere ofradius R has a positive charge which is distributed over its volume with densityrho= rho_(0)(1- (x)/(R)) per unit volume, wherex is distance from the centre. If dielectric constant of material of the sphere is k= I, calculate energy stored in surrounding space and total self energy of the sphere: |
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| 47. |
What is the source of energy associated with propagating em waves? |
| Answer» SOLUTION :Oscillating/accelerated CHARGE | |
| 48. |
A convergent lens of power 16D is used as a simple microscope. The magnification produced by the lens, when the final image is formed at least distance of distinct vision is |
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Answer» 6 |
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| 49. |
In a nuclear reactor the function of the Moderator is to decrease |
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Answer» NUMBER of neutrons |
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| 50. |
Two identical parallel plate capacitors A and B are connected to a battery of V volts with the switch S closed . The switch is now opened and the free space between the plates of the capacitors is filled with dielectric of dielectric constant K . Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric . |
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Answer» Solution :Here initially `C_(A) = C_(B) = C` and `V_(A) = V_(B) = V` `THEREFORE` Total initial electrostatic energy STORED in two capacitors `U= (1)/(2) C_A V_(A)^(2) + (1)/(2) C_(B) V_(B)^(2) = (1)/(2) CV^(2) + (1)/(2) CV^(2) = CV^(2) "" ….. (i)` When the space between the plates of the two capacitors is filled with a dielectric of dielectric constant K , capacitance of two capacitors changes to `C_(A) = C_(B) = KC` Now capacitor A is still connected to battery , hence `V._(A) = V_(A) = V` . However , capacitor B is disconnected from the battery , hence `V._(B) = (Q)/(C._(B)) = (CV)/(KC) = (V)/(K)` `therefore` Total final electrostatic energy stored in two capacitors `U. = 1/2 C._A V._A^2 + (1)/(2) C._B V._B^2 = (1)/(2) (KC) V^(2) + (1)/(2) (KC) ((V)/(K))^(2) = (1)/(2) KCV^(2) + (1)/(2) (CV^(2))/(K) = (1)/(2) CV^(2) (K + (1)/(K))"" .... (ii)` `IMPLIES (U)/(U.) = (CV^(2))/((1)/(2) CV^(2) (K + (1)/(K))) = (2)/(K + (1)/(K)) = (2K)/(K^(2) + 1)` |
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