Saved Bookmarks
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. |
9.8m/s |
|
Answer» 10 m/s |
|
| 2. |
An alternating voltagee = 220 sqrt2 sin (100 t) connected to 4muF capacitor and an ammeter . The ammeter will read : |
|
Answer» 11mA |
|
| 3. |
A circular disc rolls down on an inclined plane without slipping. What fraction of its total energy to translational? |
|
Answer» `2/3` Rotational kinetic energy, `K_(R)=1/2Iomega^(2)` Total kinetic energy, K`=K_(T)+K_(R)` `=1/2Mv^(2)+1/2Iomega^(2)` `=1/2Mv^(2)+1/2(1/2MR^(2))(V^(2))/(R^(2))` (`becausev=Romega)` `=1/2Mv^(2)[1+1/2]` (`becauseI_("circular disc")=1/2MR^(2))` `THEREFORE(K_(T))/K=(1/2Mv^(2))/(1/2Mv^(2)[1+1/2])=1/([1+1/2])=2/3` |
|
| 4. |
Choose the incorrect statement for the polarisation by reflection. |
|
Answer» The reflected light and refracted light is at `90^0` in complete polarization by reflection |
|
| 5. |
The phenomenon utilised in an optical fibre is |
|
Answer» refraction |
|
| 6. |
Rain, pouring down at an angle alpha with the vertical has a speed of 10 m/s. A girl runs against the rain with a speed of 8 m/s and sees that the rain makes an angle beta with the vertical, then relation between alpha and beta is |
|
Answer» `TANBETA=(8+10sinalpha)/(10cosalpha)` |
|
| 7. |
An alpha particle of mass m moves on a circular path of radius r in a plane inside and normal to uniform magnetic field B. The time taken by this particle to complete one revolution is ________ |
|
Answer» `(4pieB)/m` `(MV^(2))/r=Bqv` `(mv)/r=Bq` `(mxromega)/r=B(2e)` `(2pi)/T=(2Be)/m""thereforeT=(2pim)/(2Be)=(pim)/(Be)` |
|
| 8. |
Show that a current carrying solenoid is equivalent to a bar magnet. |
Answer» Solution : The magnitude of magnetic field at a point .P. due to a circular element of THICKNESS .dx. is given by `dB=(mu_0/(4pi)) (2pindxla^2)/(((r-X)^2 + a^2)^(3/2))` The magnitudeof totalmagnetic fieldis given by `B=(mu_0/(4pi)) 2pinIa^2 int_(-1)^1 (dx)/([(r-x)^2 +a^2]^(3/2))` For r > > aand r > > l , `[(r-x)^2 + a^2]^(3//2) ~~ r^3` `thereforeB =(mu_0/(4pi)) (2pinIa^2)/r^3 int_(-l)^l dx "" int_(-l)^ldx = (x)^(-l)^l` i.e.,`B=(mu_0/(4pi)) (2pinIa^2)/r^3 2l` =l + l =2l We note that the productn(2l) I `(pi a^2)` = MAGNETICMOMENT .m. `therefore B=(mu_0/(4pi)) (2m)/r^3` --(1) This expression is similar to the magnetic field at a point on the axis of a short MAGNET. Therefore a bar magnet and a SOLENOID produce similar magnetic fields. |
|
| 9. |
निम्न में से कौन-सी संख्या है ? |
|
Answer» पूर्णांक |
|
| 10. |
A 15.0 muF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current? |
|
Answer» Solution :The CAPACITIVE REACTANCE is `X_(C )=(1)/(2pi v C)=(1)/(2pi (50Hz)(15.0xx10^(-6)F))=212Omega` The rms current is `I=(V)/(X_(C ))=(220V)/(212Omega)=1.04A` The peak current is `i_(m)=sqrt(2)I=(1.41)(1.04A)=1.47A` This current OSCILLATES between +1.47A and –1.47 A, and is ahead of the voltage by `pi//2`. If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled. |
|
| 11. |
A bar magnet of magnetic moment (M) is cut into two parts of equal length. The magnetic moment of each part will be : |
|
Answer» Zero |
|
| 12. |
Two charges q and - 3q are placed fixed on x-axis separated by distance'd'. Where should a third charge 2q be placed such that it will not experience any force ? |
|
Answer» Solution :Suppose distance AC = x and AB = d Repulsive force on 2q by q, `E_(q) = (K(q)(2q))/x^(2)`……(1) Attractive force on 2q by `-3q` `E_(-3q)= (k(2q)(3q))/(x+d)^(2)`………..(2) Resultant force on 2q. `F = FQ + F_(-3q)` (`therefore` No force on 2q) `therefore Fq =-F_(-3q)` `k(2q^(2))/x^(2) = +k(6Q^(2))/(x+d)^(2)` `therefore 1/x^(2) =3/(x+d)^(2)` By TAKING square root on both sides, `1/x = sqrt(3)/(x|d) = 1.732/(x+d)` `therefore x+d = 1.732 x` `therefore x +d = 1.732 x` `therefore x = 1.366 d` `=d/2(1+sqrt(3))` |
|
| 13. |
The electric field part of an electromagnetic wave in a medium is represented by E_(x) = 0,E _(y)= 2.5 (N)/( C) cos [(2 pi xx 10^(6) rad/s ]t - ( pi xx 10^(-2) (rad)/m) x]E_(z) = 0 The wave is |
|
Answer» moving along the + xdireaction with FREQUENCY ` 10^(6)` Hz and wavelength 100 m . ` E_(y) = 2.5 N/C cos [(2PI xx 10^(6) (rad)/s) t - ( pi xx 10^(-2) (rad)/m)x]` This SHOWS that the wave is propagating along + direction. Comparing the givenequation with ` E_(y) = E_(0)cos(omegat - kx),`we get ` omega =2 pi xx 10^(6) or v = 10^(6) Hz ` ` and k = pi xx 10^(6) or 2 pi v= 10^(-2) m ` `or lamda= ( 2pi)/( pi xx 10^(-2)) = 200 m` |
|
| 14. |
(A): In metrebridge experiment, a high resis tance is connected in series with the galvanometer. (R): As resistance increases, current through the circuit increases |
|
Answer» Both 'A' and 'R' are TRUE and 'R' is the correct EXPLANATION of 'A' |
|
| 15. |
एक प्रारूपिक आवृतबीजी परागकोष लघुबीजाणुधानियो की संख्या होती है |
|
Answer» 1 |
|
| 16. |
A bus is moving with a velocity 10 mson a straight road. A scooterist wishes to overtake the bus in 100 second. If the bus is at a distance 1 km from the scooter, with what velocity should the scooterist chase the bus? |
| Answer» Answer :A | |
| 17. |
If a bar magnet is placed with its north pole pointingtowards north of earththe neutral points are loacted |
|
Answer» on the axial line |
|
| 18. |
Two bodies of masses m and M are attached to the two ends of a light string passing over a fixed ideal pulley (Mgtgtm). When the bodies are in motion, the tension in the string is approximately |
| Answer» ANSWER :C | |
| 19. |
If two positive integers a and b are written as a =p^3q^2and b= pq^3, p, q are prime numbers, then HCF (a, b) is: |
|
Answer» PA |
|
| 20. |
What are the users of spectometer? |
| Answer» Solution :The SPECTROMETER is an optical instrument used to STUDY the spectra of DIFFERENT sources of light and to measure the REFRACTIVE indices of materials. | |
| 21. |
A 15 muF capacitor is connected to a 220 V, 50 Hz a.c. source. Value of capacitive reactance is ….. Omega . |
|
Answer» 106 v=50 Hz CAPACITIVE REACTANCE, `X_C=1/(omegaC)` `X_C=1/(2pivC)` `THEREFORE X_C=1/(2xx3.14xx50xx15xx10^(-6))` `therefore X_C= 212 Omega` |
|
| 22. |
An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46 cm. What will be length of the air column above mercury in the tube now ? (Atmospheric pressure = 76 cm of Hg) |
Answer» Solution :(76) (8) = (54 - x) (76 - x)  x=38 cm LENGTH of air column= 54 - 38 = 16 cm So correct answer is (c). |
|
| 23. |
Ray of light is incident on surface of glass plate of absolute refractive index 1.55 at an angle of polarisation angle of refraction = ...... |
|
Answer» `75^(@)11'`  `n=tantheta_(p)` `1.55=tan theta_(p)` `:.theta_(p)=tan^(-1)(1.55)` `:. Theta_(p)=57^(@)10.` From FIGURE, `theta_(p)+90^(@)+r=180^(@)` `:.theta_(p)+r=90^(@)` `:.r=90-theta_(p)` `:.r=89^(@)60.-57^(@)10.` `:. r=32^(@)50.` |
|
| 24. |
A particle is moving in a straight line and passes through a point O with a velocity of 10ms^(-1). The particle moves with a constannt retardation of 5ms^(-2) for 3 s and there after moves with constant velocity. How long after leaving O does the particle return to O:- |
|
Answer» `3S` |
|
| 25. |
A capacitor of capacitance 5 muF is being charged from a d.c. source of 20V. The capacitance as a function of potential is given by (10V+4) volt. The energy stored on the capacitor is |
|
Answer» 10400 J |
|
| 26. |
Dimensions of (1)/(mu_(0) in_(0))where symbols have their usual meaning are |
|
Answer» `L^(-1) T` |
|
| 27. |
Statement-1: The pendulum shown in the figure consists of a bob suspended by a light rigid rod. The pendulum is suspended from the ceilling of a heavy wooden box which hcan freely fall on the guiding wires. Initially the box is at rest and the pendulum is released from the position shown. When the pendulum becomes vertical the box starts falling freely. The pendulum will move with a constant angular speed w.r.t box because Statement-2: The angular acceleration of the pendulum w.r.t. P once the box starts falling freely is zero. |
|
Answer» STATEMENT-1 is True, Statement-2 is true, Statement -2 is a CORRECT EXPLANATION for statement-1 |
|
| 28. |
What is meant by 'Robot'? Write its uses? |
| Answer» Solution :ROBOT is a mechanical device designed with electronic CIRCUITRY and programmed to perform a SPECIFIC task. These automated machines are highly significant in this robotic era where they can take up the role of humans in CERTAIN dangerous environments that are hazardous to people like defusing BOMBS, finding survivors in unstable ruins, and exploring mines and shipwrecks. | |
| 29. |
Two cylindrical rods, of different material, are joined as shown. The rods have same cross section (A) and their electrical resistivities are rho_(1) and rho_(2). When a current I is passed through the rods, a charge (Q) gets piled up at the junction boundary. Assuming the current density to be uniform throughout the cross section, calculate Q. |
|
Answer» Q is negative if `rho_(a) gt rho_(2)`] |
|
| 31. |
Two batteries of emf epsi_1 and epsi_2 (epsi_2 gt epsi_1)and internal resistancesr__1 and r_2 respectively are connected in parallel as shown in Fig. Then |
|
Answer» the EQUIVALENT EMF `epsi_(EQ)`of the two cells is between`epsi_1` and `epsi_2`i.e.,`epsi_1 lt epsi_(eq) lt epsi_2` |
|
| 32. |
In a balanced Wheatstone's network the galvanometer is replaced by another of lower resistance, then the network is |
|
Answer» Balanced |
|
| 33. |
Give an expression for band width in AM transmission. |
| Answer» Solution :In AM TRANSMISSION BAND width=2xmaximum FREQUENCY of the AUDIO SIGNAL. | |
| 34. |
A current of 2A is flowing in the sides of an equilateral triangle of side 9 cm. The magnetic field at the centroid of the triangle is |
|
Answer» `1.66xx10^(-6)T` |
|
| 35. |
A resistive wire is stretched till its length is increased by 100%. Due to the consequent decrease in diameter, the change in the resistance of a stretched wire will be |
|
Answer» 3 LET length of wire is l and area is A. R = `(rho l)/(A)` New length of wire l. = l + 100% l = 21 and area is A.. `THEREFORE (R.)/(R) = (rho l. )/(A.) (A)/(rho1)= (A)/(A.) (l.)/(l) = ((l.)/(l) )^(2)= ((2l.)/(l) )^(2) = 4` `R. = (rho l.)/(A.)` `therefore R. = 4R = R + 3R = ((4R - R)/(R) XX 100 ) % ` `= ((3R)/(R) xx 100 )%` = 300 % Percentage change in resistance = 300% |
|
| 36. |
The temperature of an ideal gas enclosed in a chamber is raised from 300 K to 600 K. The pressure becomes two fold because- |
|
Answer» Mean molecular velocity becomes `SQRT2` fold |
|
| 37. |
As the intensity of incident light increases |
|
Answer» KINETIC ENERGY of emitted photoelectrons increases |
|
| 38. |
A rectangular loop is placed near a current carrying straight wire as shown in the figure. If the loop is rotated about an axis passing throught one of its sides find the direction of induced current in the loop. |
|
Answer» |
|
| 39. |
A ringof radius R is with uniformly distributed charge Q on it. A charge q is now placed at the centreof the ring. Find the increment in tension in the ring. |
Answer» SOLUTION : CONSIDER an element. Its enlarged view is as shown. For equilibrium of this segment, we can write. `F= 2DeltaT sin ((d THETA)/(2))` Here F is the repulsive FORCE between q and elementalcharge dQ `dQ=(Q)/( 2pi R) (Rd theta)` The electric outward force on element is `F=(1)/(4pi in_(0)) (qdQ)/( R^(2))` From the above three equations, we can write `(1)/(4pi in_(0)) (q)/(R^(2))(QRd theta)/(2piR) ~~ 2DeltaT((d theta)/(2))` (`:. sin alpha = alpha` for small ANGLE) `Delta = (Qq)/( 8pi^(2) in_(0) R^(2))` |
|
| 40. |
Calculate the height of the potential barrier for a head on collision of two deuterons. |
|
Answer» Solution :Hint : The height of the POTENTIAL BARRIER is given by the COULOMB repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm. For head on collision DISTANCE between centres of two deuterons `= r = 2xx` radius `r = 4 fm = 4xx10^(-15)m` Charge of each deuteron `e = 1.6xx10^(-10)C` Potential energy `(e^(2))/(4pi epsilon_(0)r)=(9xx10^(9)(1.6xx10^(-19))^(2))/(4xx10^(-15))` Joule `= (9xx1.6xx1.6xx10^(-14))/(4xx1.6xx10^(-16))` KeV PE = 360 KeV P.E `= 2xx` K.E of each deuteron = 360 KeV K.E of each deuteron `= (360)/(2)=180` KeV This is a measure of height of Coulomp barrier. |
|
| 41. |
The two vectors are vec(A) = hat(i) + hat(j) and vec(B) = hat(i) - hat(j) . What is the angle between them ? |
|
Answer» 45° `:. theta=90^@` |
|
| 42. |
In optical fibre, refractive index of inner part is 1.68 and refractive index of outer part is 1.44. The numerical aperture of the fibre is |
|
Answer» 0.5653 |
|
| 43. |
The sensitivity of a galvanometer is 50 division//amp.When a shunt is used, its sensitivity becomes 10 division//amp. If galvanometer is of resistance 30Omegathe value of stunt used is, |
|
Answer» `4Omega` |
|
| 44. |
A uniform. rod of length l can rotate without friction about an axis passing through its upper end(Fig). The rod is deflected by an angle a_0and let go. Find the speed of the lower end of the rod as a function of the angle alpha |
|
Answer» `K=U_0-U` where `K = 1/2 Iomega^2, U_0 =mgh_0 and U = mgh` The moment of inertia of a rod about an axis passing through one end is `I = 1/3ml^2` .In the course of oscillations the centre of gravity of the rod rises to a height `h_0=1/2l(1-cosalpha_0),h=1/2l(1-cosalpha)` Substituting the values obtained into the equation for the energy balance, we obtain `omega= sqrt((3G)/l(cos alpha-cosalpha_0))` the speed of the end of the rod is `v= omega l` . |
|
| 45. |
Which of the following statement is incorrect for the depletion layer of a diode? |
|
Answer» Here the MOBILE CHARGES exist |
|
| 46. |
What is step down transformer ? |
| Answer» Solution :Step down TRANSFORMER converts HIGH ALTERNATING voltage to LOW alternating voltage. | |
| 47. |
The contactforce exerted by a body A onanother bodyB is equal to the normal forcebetween the bodies . We conclude that the surfacesmustbefrictionaless the force of frictionbetween the bodies is zero the magnitude of normal force equals that of friction the bodies may be rough but they don't slip on each other . |
|
Answer» a and C are TRUE |
|
| 48. |
What would be maximum wavelength for Brackett series of Hydrogen spectrum? |
|
Answer» 74583 Å |
|
| 49. |
A particle performing S.H.M. has an acceleration of 100 cm/s^2, when it is at a distance of 25 cm from the mean position. Calculate the period of SHM. |
|
Answer» SOLUTION :"Acceleration"`=omega^2X=((2PI)/T)^2x` T^2=(4pi^2x)/(Ac c^n)= (4xx9.87xx25)/100=9.87` `therefore T=pi sec.`. |
|
| 50. |
A high powered laser beam (lambda= 630 nm) with a beam diameter of 0.1 m is aimed at the Moon, 3.8 xx 10^8 m distant. The beam spreads only because of diffraction, The angular location of the edge of the central diffraction disc is given by sin theta=(1.22 lambda)/(d) where d is the diameter of the beam aperture. What is the diameter of the central diffraction disc on the Moon's surface ? |
|
Answer» 1010.2m Now angular speed `theta=("Diameter of central disc on moon")/("Distance of moon")` Diameter of central dic `=theta xx " Distance of moon"` `=(1.22 LAMBDA)/(d). D [therefore THET approx sin theta=(1.22lambda)/(d)]` Now `D=3.8 xx 10^(8) m` and d=0.1m Diameter of the central disc `=(1.22 xx 650 xx 10^(-9) xx 3.8 X 10^(8))/(0.1)` =3013.4 m |
|