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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. |
What will be the effect on the velocity of the emitted photoelectrons if the wavelength of incident light is decreased ? |
| Answer» SOLUTION :The VELOCITY will .INCREASE. | |
| 2. |
An alpha particle and a proton travel with same velocity in a magnetic field perpendicular to the direction of their velocities. The ratio of the radii of their circular paths is |
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Answer» `4 : 1` |
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| 3. |
Wavelength of photon having 35 keV energy will be……..(h=6.625xx10^(-34))J-s,c=3xx10^(8)ms^(-1),1eV=1.6xx10^(-19)J). |
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Answer» `35xx10^(-12)m` `therefore =(hc)/(E )=(6.625xx10^(-34)xx3xx10^(8))/(35xx10^(3)xx1.6xx10^(-19))` |
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| 4. |
Two light beams of intensities I and 4I are used to make interference pattern on a screen. Two points P and Q are marked on the screen where resultant intensities are I_P and I_Q respectively. Phase difference between the waves at point P is pi//2 and the same at Q is pi. Calculate (I_P - I_Q)//I. {:(0,1,2,3,4,5,6,7,8,9):} |
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Answer» `I = I_1 + I_2 + 2sqrt(I_1I_2) cos DELTA` We can use the above relation to write the FOLLOWING : `I_P = I + 4I + 2sqrt(I xx 4I) cos pi/2 = 5I` `I_Q = I + 4I + 2sqrt(I xx 4I) cos pi = 5I - 4I = I` `implies I_P -I_Q = 5I - I` `implies (I_P - I_Q)/(I) = 4` |
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| 5. |
A : In pure rolling motion all the points of a rigid body have same linear velocity.R : Rolling motion is not possible on smooth surface. |
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Answer» If both Assertion & Reason are TRUE and the reason is the CORRECT explanation of the assertion, |
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| 6. |
In the accompanying circuit Fig. 5.10, if it is assumed that when that input voltage at the base resistance is 5 V, V_(BE) is zero and V_(CE) is also zero, what are I_(B),I_(C) and beta ? When the input is zero, I_(B) is zero. What will be the output waveform if the input waveform is as shown in the figure ? What is the practial use of this current ? |
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Answer» `(##NEP_NGM_PHY_MP_C05_E01_014_A01##)` |
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| 7. |
Show that if E is the electric intensity inside a conductor of electrical conductivity sigma, Ohm's law may be coveniently put in the form, vec(J) = sigmavec(E ) where J = current density in the conductor. |
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Answer» |
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| 8. |
A wave is described by the equation y = (1.0 mm)sin pi(x/(2.0cm) - 1/(0.01s)) Find the time period and the wavelength. |
| Answer» SOLUTION :20 m/s, 4.0 CM | |
| 9. |
In the two-dimensional collision in Fig., the projectile particle has mass m_1=m, initial speed v_u = 3v_0, and final speed v_(1f)=sqrt(5v_0). The initially stationary target particle has mass m_1 = 2m and final speed v_(2f)=v_2. The projectile is scattered at an angle given by tan theta_1 =2.0. (a) Find angle theta_2. (b) Find v_2in terms of v_0 .(c) Is the collision elastic? |
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| 10. |
Capacitance of spherical capacitor is 1 muF, so its diameter = ........ m. [(1)/(4pi epsilon_(0))=9xx10^(9) SI] |
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Answer» `1.8` `:. R = kC = 9XX10^(9)xx1xx10^(-6)=9xx10^(3) ` `:.` diameter =`2R = 2 xx9 xx10^(3)= 1.8xx10^(4)` m |
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| 11. |
(a) A conductor havingcavity C is chargedas shown in Fig. What is electricfied insidethe cavity ? Does the resultdependuponthe shape and size of the cavity ? (b)Can Gauss's law tell us exactly where teh charge is locatedinside the Gaussiansurface ? |
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Answer» Solution :(a) It is known that electric charges reside always on the surfacefo conductorsonly. Inside the conductors , charge is ZERO. The result is independent of the shapeand sizeof the cavity will always be zero irrespectivefo the size of the cavity. (b) Gauss's lawcannot determinethe position of charge. This is because, acccordingto Gauss's Law, electricflux overa Gaussian surface in vacumm is `1//in_(0)` TIMES the charge enclosedinsidethe surface . ThusGauss's law explains that the charge is insidethe CLOSED surface , and for EVERY positionof the charge inside TEH surface, the electricflux is the same. HenceGauss's Law. fails to specify the position of chargeinsidethe closed surface. |
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| 12. |
The bent wire shown in figure lies in a uniform magnetic field. Each straight section is 2.0 m long and makes an angle theta=60^(@) with the x-axis , and the wire carries a current of 2.0A . What is the net magnetic force on the wire in unit -vectorrotation if the magneticfield is given by (a) 4.0 hat k T and (b) 4.0hati T ? |
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| 13. |
Figure 2.32 shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on r for r//a gt gt 1 , and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge). |
| Answer» Solution :For large r, quadrupole POTENTIAL goes like `1//r^(3)`, dipole potential goes like `1//r^(2)` , MONOPOLE potential goes like 1/r. | |
| 14. |
Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then error in the value of resistance of the wire is : |
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Answer» `3%` `R=(V)/(I)` `(DeltaR)/(R)xx100%=(DeltaV)/(V)xx100%+(DELTAI)/(I)xx100%=6%` Hence correct choice is `(B)`. |
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| 15. |
Consider a tightly wound 100 turn coil of radius 10 cm, carrying a current of 1 A. What is the magnitude of the magnetic field at the centre of the coil ? |
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Answer» Solution :Magnetic FIELD PRODUCED by a current carrying circular coil of radius R with N no. of turns wound tightly at the centre of coil is, `B=(mu_(0)NI)/(2R)` `thereforeB=((4pixx10^(-7))(100)(1))/((2)(0.1))` `thereforeB=6.28xx10^(-4)T` (Perpendicular to plane of the coil) |
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| 16. |
In calcite crystal, double refraction does not take place |
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Answer» PERPENDICULAR to OPTIC axis |
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| 17. |
A circular of 500 turns of wire has anclosed area of 0.1m^(2) per turn. It is kept perpendicular to a magnetic field of induction 0.2T and rotated by 180^(@) about a diameter perpendicular to the field in 0.1s. How much charge will pass when the coil is connected to a galvanometer with a combined resistance of 50 Omega. |
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Answer» SOLUTION :Charge `= ("charge in FLUX")/("resistance")` `q= (phi_(i) - phi_(f))/(R )= (NBA- (-NBA))/(R )= (2NBA)/(R )` `q= (2 XX 500 xx 0.2 xx 0.1)/(50)= 0.4C` |
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| 18. |
Light of wavelength 5000 Å incidents on reflecting surface, then for ...... value of incidence angle, reflected and incident ray will be perpendicular. |
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Answer» `0^(@)` but `i=r so 2i=90^(@)` |
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| 19. |
Two large conducting platesX and Y. each having large surface area A (on one side), are placed parallel to each other as shown in figure (30-E7). The plate X ia given a charge Q whereas the other is neutral. Find (a) the surface charge dendity at the inner surface of the plates X, (b) the electric field at a point to the left of theplates, (c) the electric field at a point in between the plates and (d) the electric field at a point to the right of the plates. |
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Answer» Solution :(a) For the SURFACE charge density of a single plate. Let the surface charge density at both side be `sigma_1 and sigma_2`. Now ELECTRIC FIELD at both ends `= sigma_1/2 eplison_0 and sigma_2/2 eplison_0` Due to net balanced electric field on plate and `r_1 equal to r_2` `q_1= q_2 = Q/2 ` Net surface charge density = `Q/ 2A` (b) Electricfield to the left of the plates Since`r = Q/ 2A ` . Hence, electric field = `Q/2A(epsilon_0)` This must be directed towards left as 'X' is the charge plate. (c) and(d) here in both the cases the charged plate 'X' acts as the only SOURCE of electric field, with +ve in the inner side and'Y' attracts towards it with -ve in its innerside. So, for the middle portion E is towards right. (d) Similarly for extreme right the OUTER side of the 'Y' plate acts as +ve and hence it repels to the right with E = `Q/ 2A epsilon_0`. |
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| 20. |
A block rides on a piston (a squatcylindrical piece) that is moving vertically with simple harmonic motion. (a) If the SHM has period 1.0s, at what aplitude of motion will the block and piston separate? (b) If the piston has an amplitude of 3.0cm. What is the maximum frequency for which the block and piston will be in contact continuously ? |
| Answer» SOLUTION :(a) 0.25, (B) 2.9Hz | |
| 21. |
Spectrum of X-rays is |
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Answer» continuous |
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| 22. |
A beam of light converges to a point P , A lens is placed in the path of the convergent beams 12 cm from P. At what point does the beam converge if the lens is a.A convex lens of focal length 20 cm. b.A concave lens of focal length 16 cm ? |
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| 23. |
The electric current in a circuit is given by i=3t. Find the rms current for the period t=0 to t=1 sec. |
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Answer» SOLUTION :`i=3t, i^(2)=9T^(2)` `i_("rms")^(2)=(int_(0)^(t)i^(2).dt)/(int_(0)^(1)dt)=(int_(0)^(t)9t^(2)dt)/(int_(0)^(1)dt)=3,""I_("rms")=sqrt3A` |
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| 24. |
The dimensional formula for flux in terms of mass, length, time and charge (Q) will be |
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Answer» `ML^2T^-1Q^-1` |
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| 25. |
Corresponding to the path difference of 171.5 times the wavelength of interfering waves, the two waves differ by 0.01029overset@A cm. Find the wavelength ? |
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Answer» `6000overset@A` |
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| 26. |
Light traveling from a transparent medium to air undergoes that internal reflection at an angle of incident of 45^(@). Then refractive index of the medium may be |
| Answer» SOLUTION :`MU = (1)/(sin i_(c)) = (1)/(sin 45^(@))= sqrt(2) = 1.414 approx 1.5` | |
| 27. |
Keeping forward bias voltage from 0.6 V to 0.7 V in a p-n junction diode its current become 1 mA to 3mA. The dynamic resistance of diode is ……. |
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Answer» `50Omega` RESISTANCE in forward BIAS (dynamic resistance) `r_(fb)=(DeltaV)/(DeltaI)` `r_(fb)=(0.1)/(2XX10^(-3))[because DeltaV=0.7-0.6=0.1V and DeltaI-3-1=2mA]` `THEREFORE r_(fb)=50Omega` |
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| 28. |
The plates of a parallel plate capacitor have an area of100 cm^(2) each and are separated by 3 mm . The capacitor is charged by connecting it to a 400 V supply . If a dieletric of dielectric constant 2.5 is introduced between the plates of the capacitor , then find the electrostatic energy stored and also change in the energy stored . |
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Answer» Solution :Given ` A = 100 cm^(2) = 10^(2) xx 10^(-4) = 10^(-2) m^(2)` d = 3 mm = ` 3xx 10^(-3) m`, V = 400 V ` E^(1) = in _(r) E = 2.5 xx 2.61 xx 10^(-6) = 5.903 xx 10^(-6) J` `Delta E = E^(1) - E = (5.903 - 2.361 ) xx 10^(-6) J ` Difference in the energy STORED ` Delta E ` , ` Delta E = 3.542 xx 10^(-6) J ` |
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| 29. |
A red LED emits light at 0.1 watt iniformly around if the amplitude of the electric field of the light at a distance of 1m from the diode is: |
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Answer» 1.73 v/m |
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| 30. |
The plates of a parallel plate capacitor have an area of100 cm^(2) each and are separated by 3 mm . The capacitor is charged by connecting it to a 400 V supply . Calculate the electrostatic energy stored in the capacitor . |
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Answer» Solution :Given ` A = 100 cm^(2) = 10^(2) xx 10^(-4) = 10^(-2) m^(2)` d = 3 MM = ` 3xx 10^(-3) m`, V = 400 V Electrical energy.E. `= 1/2 CV^(2) = (in _(0) AV^(2))/(2D) ` i.e., ` E = (8.854 xx 10^(-12) xx 10^(-2) xx (400)^(2))/(3 xx 10^(-3)) ` i.e., ` E = 2.361 xx 10^(-6) J` |
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| 32. |
For which arrangement of two coils, shown in the below figure, coefficient of mutual inductance is maximum and for which it is minimum? |
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Answer» For (a) and (d), respectively |
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| 33. |
What energy should be spent to fill a balloon with a lifting force of 3000 N with hydrogen under normal conditions? How much will it cost at the price of 4 copecks per kW.h? Ignore the heating of the solution in the course of electrolysis. |
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Answer» `m=rho_(H)V=rho_(H)F//g(rho_(0)-rho_(H))`. Applying Faraday.s law, we find the charge Q passing through the electrolyte solution. The energy required to produce the hydrogen is  where Q represents the Joule heat loss. The polarization e.m.f. has already been FOUND in Problems 32.5. |
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| 34. |
If x times momentum is work, then thedimensions of x are |
| Answer» Answer :A | |
| 35. |
Assertion: If a gas container in motion is suddenly stopped, the temperature of the gas rises. Reason:The kinetic energy of ordered mechanical motion is converted in to the kinetic energy of random motion of gas molecules. |
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Answer» If both ASSERTION and REASON are true and the reason is the correct EXPLANATION of the assertion. |
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| 36. |
The ratio (in S.I. units) of magnetic dipole moment to that of the angular of electron of mass m kg and charge e coulomb in Bohr's orbit of hydrogen atom is : |
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Answer» `(c)/(2m)` `=(e(pi r^(2)))/(2pi) omega=er^(2) omega=(er^(2) omega)/(2) ........(1)` Angular momentum `=mvr =mr^(2) omega` `("magnetic dipole moment")/("Angular momentum")=(er^(2)omega)/(2) XX (1)/(mr^(2) omega)=(e)/(2m)` |
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| 37. |
A particle of mass m having negative charge q moves along an ellipse around a fixed positive charge Q so that its maximum and minimum distances from fixed charge are equal to r_(1) "and " r_(2) respectively. Calculate angular momentum L of this particle: |
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Answer» |
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| 38. |
Rate of reaction (r ) is plotted against temperature (T) for an enzyme catalysed reaction. What of the following is correct representation ? |
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Answer»
 For enzyme catalyst reaction UPTO `24^(@)`C rate `uparrow` and then `downarrow` |
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| 40. |
In Thomson atomic model mass of atom is ..... mass of atom in Rutherford model. |
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Answer» very LARGE COMPARED to |
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| 41. |
A circular ring of mass 10 kg rolls along a horizontal floor. The center of mass of the ring has a speed 1.5 m/s. The work required to stop the ring is |
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Answer» Solution :Given mass of a circular ring, m = 10kg speed of CENTRE of mass of the ring i.e., linear speed of ring,u = 15 m/s `K_i = K_("rotational") + K_("linear") = 1/2 I omega^2 + 1/2 mv^2` Here , `I = mR^2 " and " omega= v/R` ` = 1/2 mR^2 (v/R)^2 + 1/2 mv^2 = 1/2 mv^2 + 1/2 mv^2 = mv^2` `= 10 xx (1.5)^2 = 22.5 J` According to work-energy theorem, work required to stop the ring ` omega`change in kinetic energy ` = K_f - K_i = 0 - 22.5 = -22.5J` |
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| 42. |
A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much " "_(92)^(235)U did it contain initially ? Assume that the reactor operaters 80% of the time, that all the energy generated arises from the fission of " "_(92)^(235)U and that this nuclide is consumed only by the fission process. |
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Answer» Solution :In one act of fission, the ENERGY liberated is 200 MEV `therefore` NUMBER of atoms in 1g of `" "_(92)^(235)U = N_(A)/A= (6.023xx10^(23))/235` `therefore` Energy generated by 1g of `" "^(235)U =(200MeV xx6.023xx10^(23))/235=(200 xx 1.6xx 10^(-13) xx 6.023xx10^(23))/235 Jg^(-1)= 8.20 xx 10^(10) J g^(-1) = 8.20 xx 10^(13) Jkg^(-1)` As the generator operates only 80% of the time, in 5 years it will effectively operate for a time `f=5 xx80/100 year= 4 year = 4 xx 3.154 xx 10^(7)s` `therefore` Total energy generated by the reactor in 5 years E=P.t=`1000MWxx4xx3.154xx10^(7)s=1000xx10^(6)Wxx4xx3.154xx10^(7)J=12.616xx10^(16)J` `therefore` AMOUNT of `" "_(92)^(235)U` consumed`=(12.616xx10^(16))/(8.20xx10^(13))kg=1538kg` As the reactor consumes half of the fuel in 5 years, hence amount of fuel present in the reactor initially`=2xx1538=3076kg`. |
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| 43. |
A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band? |
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Answer» Solution :WAVELENGTH corresponding to frequency of 7.5 MHz `(or 7.5xx10^(6)Hz)` is `lambda_(1)=(C )/(v)=(3xx10^(8))/(7.5xx10^(6))=40m` and wavelength corresponding to frequency of 12 MHz `(or 12xx10^(6)Hz)` `lambda_(2)=(3xx10^(8))/(12xx10^(6))=25m` Thus, wavelength band is `25m-40m`. |
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| 44. |
Suppose we want to verify the analogy between electrostatic and magnetostatic by an explicit experiment. Consider the motion of (i) electric dipole overset(to)(p) in an electrostatic field overset(to) (E ) and (ii) magnetic dipole overset(to)(M) in a magnetic field overset(to)(B). Write down a set of conditions on overset(to)(E) , overset(to)(B), overset(to)(M) so that the two motions are verified to be identical. (Assume identical initial conditions.) |
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Answer» Solution :Suppose the ANGLE between `OVERSET(to)(p) and overset(to) (E )` is `theta`. Torque on electric dipole of MOMENT `overset(to)(p)` in an electric field `overset(to)(E )`, `tau = p E sin theta ".…(1)` Let us assume that the angle between `overset(to)(M ) and overset(to)(B)` is `theta`. The torque on magnetic dipole moment `overset(to)(M)` in magnetic field `overset(to)(B)`, `therefore tau. = M B sin theta ""...(2)` If there two motion are identical, then `tau= tau.` `therefore p E sin theta = MB sin theta` `therefore PE= MB""...(3)` but `E= cB` Putting this value in equation (3), `pcB = MB` `therefore p= (M)/(c )` |
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| 45. |
In the circuit as shown in the figure, the heat produced by 6 ohm resistance due to current flowing in it is 60 calorie per second. The heat generated across 3 ohm resistance per second will be |
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Answer» 30 calorie |
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| 46. |
A circular coil of radius 3cm has 50 turns. It is placed on the horizontal plane and a current of 3A flows through it in clockwise direction as seen from above. Calculate the magnetic field at a point on the axis of the coil, at a distance of 4cm from its centre. Also indicate the direction of the magnetic field. |
| Answer» Solution :`6.78 XX 10^(-4) T`, vertically DOWNWARD | |
| 47. |
The correct expression for bandwidth is: |
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Answer» `X=(lambdaD)/d` |
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| 48. |
When the electric flux associated with closed surface becomes positive, zero or negative ? |
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Answer» Solution :If electric flux ASSOCIATED with area s due to electric field E is `vecE`, then `phi = vecE. vecS` `therefore vecphi = ES cos theta`……(1) Where `theta` is angle between `vecE` and `vecS`. (i) If `vecS bot vecE` MEANS surface is parallel to `vecE` then, `theta = 90^(@)` `therefore` from equation (1) `phi = ES cos 0^(@)=0` `therefore` Flux associated with surface is zero. (ii) If `theta lt 90^(@)`, then `cos theta GT 0` (negative) hence flux is POSITIVE. (iii) If `theta gt 90^(@)`, then `cos theta lt 0`. (negative) hence flux is negative. All three are shown in FIGURE (a), (b) and (c) respectively. 
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| 49. |
A full wave p-n junction diode rectifier uses a load resistance of 1300 Omega . The internal resistance of each diode is 9Omega . Find the efficiency of this full wave rectifier. |
| Answer» ANSWER :B | |
