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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. |
Find an expression for intensity of transmitted light when a polaroid sheet is rotated between I polaroids. In which position of the polaroid sheet will the transmitted intensity be maximum ? |
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Answer» Solution :Hydrogen atom LET r be the radius of the orbit of a hydrogen atom. Forces acting on electron are centrifugal FORCE and electrostatic attraction `(F_(e ))` At equilibrium. `F_(c )=F_(e )` `(mv^(2))/(r )=(1)/(4PI epsilon_(0))(e^(2))/(r^(2))` According to Bohr’s postulate `mvr = (nh)/(2pi)rArr u=(nh)/(2pi mr)` `m((nh)/(2pi mr))^(2)(1)/(r )=(1)/(4pi epsilon_(0))(e^(2))/(r^(2))rArr (mn^(2)h^(2))/(4pi^(2)m^(2)r^(2)r)=(1)/(4pi epsilon_(0))(e^(2))/(r^(2))` `r=(m^(2)h^(2)epsilon_(0))/(pi me^(2))rArr therefore r ALPHA n^(2)` |
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| 2. |
The total energy of electron E_(n)=(Z^(2)me^(4))/(8epsi_(0)^(2)n^(2)h^(2)) inatom is based on which hypothesis ? When is that true ? |
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Answer» Solution :To obtaine formula `E_(n)=(Z^(2)me^(4))/(8epsi_(0)^(2)n^(2)h^(2))`orbit of electron is ASSUMED to be circular but under the INFLUENCE of a inverse force, the orbit are elliptical shaped, LIKE the orbits of the planet AROUND the Sun are elliptical. However, it was shown by physicist Sommerfeld that, when the restriction of circular orbit is relaxed these equations continue to hold even for elliptic orbits. |
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| 3. |
Figure shows the three circular arcs centered on the origin of a coordinate system. On each arc, the uniformly distributed charge is given in terms of Q = 4.00 mu C. The radii are given in terms of R = 5.00 cm. What are the (a) magnitude and (b) direction (relative to the positive x direction) of the net electric field at origin due to the arc ? |
| Answer» Solution :(a) `1.30 XX 10^(7) N//C`, (B) `-45^(@)` | |
| 4. |
Consider aiming a beam of free electrons towards free protons. When they scatter, an electron and a proton can not combine to produce a H-atom, |
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Answer» because of energy conservation |
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| 5. |
Where was the race conducted? |
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Answer» NEHRU Stadium |
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| 6. |
A: The obstacle must be in the order of 10^(-7)m to see the differation of light. R: The wavelength of visible light is in the order of 10^(-7)m. |
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Answer» Both assertion and REASON are TRUE and the reason is correct explanation of the assertion. |
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| 7. |
A point charge +Q is placed at a point. Calculate the electric flux associated with one sphere of radius r , which the given point charge at its centre. |
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Answer» Solution :Electric lines of forces for point charge are directed radially OUTWARDS in all directions. The direction of area segment is also radially outwards. The angle between electric field intensity and area vector at any point of the sphere remains zero, though electric field intensity and area vector both are continuously changing direction from point to point . Electric flux can be written as follows : `phi=int_"sphere"vecE.d vecS` `phi=int_"sphere" EdS cos 0^@=int_"sphere" EdS` SINCE the magnitude of `vecE` is same everywhere on a spherical surface, so it can be taken out of the integration : `phi=E int_"sphere" DS = 1/(4piepsilon_0) Q/r^2 xx 4pir^2` hence, the electric flux through a spherical surface around point charge at its CENTRE is given by : `phi=Q/epsilon_0` 
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| 8. |
Which of the following are not electromagnetic waves |
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Answer» Cosmic RAYS |
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| 9. |
For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by, B= (mu_0 IR^2 N)/(2 (x^2 + R^2)^(3//2)) (a) Show that this reduces to the familiar result for field at the centre ofthe coil.(b) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by, B = 0.72 (mu_0 NI)/(R ) . approximately [Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.] |
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Answer» Solution : (b) In a small region of length 2d about the mid-point between the COILS `B = (mu_0 IR^2 N)/(2) xx [ { (R/2 + d)^2 + R^2 }^(-3//2)+ { (R/2- d)^2 + R^2 }^(-3//2)]` `= (mu_0 IR^2 N)/(2) xx ((5R^2)/(4))^(-3//2) xx [ ( 1 + (4d)/(5R)^(-3//2)) + ( 1 - (4d)/(5R) )^(-3//2) ]` ` = (mu_0IR^2N)/(2R^3)xx (4/5)^(3//2) xx [ 1 - (6d)/(5R) + 1 +(6d)/(5R) ]` where in the second and third steps above, TERMS containing `d^2 // R^2`and HIGHER powers of d/R are neglected since `d/R lt lt 1`.The terms linear in d/R cancel giving a UNIFORM field B in a small region: `B = (4/5)^(3//2) (mu_0 IN)/( R) = 0.72 (mu_0 IN)/(R )` |
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| 11. |
In the given circuit. E_(1) = 6 volts, E_(2) = 2 volts, E_(3) = 6 volts R_(1) = 6OmegaR_( 2)=2Omega R_(3) = 4Oemga, R_(4) = 3Omega and C=5muF. Find the current in R_(3) and energy stored in the capacitor at steady state |
| Answer» SOLUTION :`1.5A,14.4xx10^(6)J` | |
| 12. |
A particle of mass m is moving along the line y-b with constant acceleration a. The areal velocity of the position vector of the particle at time t is (u=0) |
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Answer» constant |
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| 13. |
The energy gap in germanium is 0.75eV. Compare the intrinsic conductivities of germanium at 300K and 330K. Take k_(B)=8.6xx10^(-5)eVK^(-1) |
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Answer» |
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| 14. |
What is meant by diffraction of light? Write the conditions for maxima and minima of diffraction pattern in terms of the wavelength of light used for the diffraction at single slịt. |
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Answer» Solution :DIFFRACTION of light : The phenomenon of bending of light waves around the sharp EDGES or corners of an obstacle. CONDITION for maxima is given by ` theta = (2n +1) lamda/(2A)` where`n=1,2,3,…….` Condition for MINIMA is given by ` theta= (2n) ( lamda )/( 2a) ` where`n=1,2,3,........` |
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| 15. |
A proton and helium nucleus are shot into a magnetic field at right angles to the field with same kinetic energy. Then the ratio of their radii is : |
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Answer» <P>`1:1` `or R=(mv)/(q B)=sqrt((2mE)/(q B)) therefore r ALPHA sqrtm/q` `r_(p)/r_(He)=1/1` |
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| 16. |
Two metallic spheres of same radii, one solid, are charged to the same potential. Which will hold more charge? |
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Answer» SOLID SPHERE |
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| 17. |
भारत में कपास उत्पादन की दृष्टि से प्रथम स्थान पर कौन- सा राज्य है? |
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Answer» गुजरात |
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| 18. |
The resistance of the wire is 121 ohm. It is divided into .n. equal parts and they are connected in parallel, then effective resistance is 1 ohm. The value of .n. is |
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Answer» 12 |
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| 19. |
The ratio of longest wavelength and the shortest wavelength as observed in the five spectral series of emission spectrum of hydrogen is : |
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Answer» `(4)/(3)` |
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| 20. |
At t =0S_w is closed , if initially C_1 is uncharged andC_2is charged to a potential difference2inthen find out following(GivenC_1 =C_2 =C ) (a)Charge onC_1 and C_2 as a function of time (b) Find out current in the circuit as a function of time (c) Also plot the graphs for the relations derived in part (a) |
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Answer» Solution :Let qcharge flow in TIME 't' from the battery as shown.The charge on various plates of the capacitor is as shown in the figure Now applying KVL ` in -(q)/(c ) -iR ( q-2in C)/( C) =0 ` `in -( q)/(c) -( q)/(c) +2in -iR = 0` `3in =( 2q)/( C) + iR ` ` (##MOT_CON_JEE_PHY_C25_SLV_042_S01.png" width="80%"> ` 3in -iR=(2q)/(C) ` ` (dq)/( dt) RC=3in C - 2q` ` int _0^(q) (dq)/( 2in C -2q) =int _0^(t)(dt)/( RC) ` ` - (1)/(2)ln ((3Cin -2q)/( 3Cin )) =(t)/(RC) ` ` 1n ((3in C- 2q)/( 3in C) ) =-( 2T)/(RC) ` ` 3in C - 2q=3in C e^(=2 tau//RC)` ` 3in C( 1-e^(-2tau//RC)) = 2q` ` q= (3)/(2) in C(1-e^(-2tau//RC))` `i =(dq)/( dt) =(3in )/(R) e^(-2t//RC)` ` i =(dq)/( dt) =(3in)/(R) e ^(-2tau//RC)` on the plate B `q'=2in C -q` ` 2in C -(3)/(2) in C +(3)/(2) in C e^(-2tau//RC) ` ` (##MOT_CON_JEE_PHY_C25_SLV_042_S02.png" width="80%"> ` =(in C)/(2) + (3)/(2) in Ce^(-2t//RC) =(in C)/( 2) [1+ 3e^(-2tau//RC) ]` |
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| 21. |
The average power dissipated in a pure inductor |
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Answer» zero `P_("avg") = V_("rms") I_("rms") COS PHI` `P_("avg") = 0` |
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| 22. |
A voltmeter of resistance 20000 Omega reads 5 volt. To make it read 20 volt, the extra resistance required is |
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Answer» `40000 OMEGA` inparallel |
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| 23. |
In figure the current supplied by the battery is |
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Answer» `0.1 A` |
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| 24. |
In the study of transistor as amplifier, if alpha=I_C/I_E and beta=I_C/I_B, where I_C,I_B and I_E are the collector, base and emitter currents, then |
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Answer» `BETA =((1-ALPHA))/alpha` |
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| 25. |
In common - emitter amplifier the ratio (I_C)/(I_E) 0.98. The current gain will be |
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Answer» 4.9 `BETA=(alpha)/(1-alpha)=(0.98)/(1-0.98)=(0.98)/(0.02)=49` |
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| 26. |
The dimensions of magnetic field in M,LC ( Coulomb) are given by : |
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Answer» `[MLT^(-1)C^(-1)]` `:.B=(F)/(qvsintheta)=(ML^(1)T^(-2))/(Q^(1)L^(1)T^(-1))=MT^(-1)Q^(-1)` `:.(c )` is the correct choice. |
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| 27. |
A point object is placed at a distance of 24 cm from a convex surface of a medium (mu=1.5) and radius of curvature 24 cm in air. What is the distance of image position from the pole? |
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Answer» REAL image, 36 cm in AIR |
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| 28. |
किस भौतिक राशि का मात्रकों की सभी तीन पद्धति में समान मात्रक होता है? |
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Answer» द्रव्यमान |
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| 29. |
Figure shows a solid transport semi cylinder of radius 10 cm. A screen is placed at a distance 60 cm from O. A narrow beam is incident along X-axis at O. If cylinder starts rotating about O in clockwise direction with angular speed 6 rad//s spot formed on screen will move upward (Refractive index of material of cylinder=(5)/(3)) At waht distance from C bright spot on screen will disappear. |
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Answer» `100 CM` |
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| 30. |
Figure shows a solid transport semi cylinder of radius 10 cm. A screen is placed at a distance 60 cm from O. A narrow beam is incident along X-axis at O. If cylinder starts rotating about O in clockwise direction with angular speed 6 rad//s spot formed on screen will move upward (Refractive index of material of cylinder=(5)/(3)) What is initial angular velcity of ray refrected from plane surface. |
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Answer» `2 rad//s` For SMALL angles `r=mui` `theta=(mu-1)irArr(d theta)/(DT)=(mu-1)(di)/(dt)=((5)/(3)-1)6=4 rad//s` 
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| 31. |
A p-n junction diode has a potential difference 0.5 V across its junction which does not depend on the circuit current. A resistance of 200 Omega is connected in series with the junction and a current of 10 mA passes through it. The voltage of forward bias applied is _______. |
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Answer» |
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| 32. |
Consider an electron in the n^th orbit of a hydrogen atom in the Bohr model. The circumference of the orbit can be expressed in terms of de Broglie wavelength of that electron as |
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Answer» `[(0.529)N LAMBDA ]` |
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| 33. |
A conducting wire of lengthand mass m is placed on a horizontal surface with its length along y direction. There exists a uniform magnetic field B along positive x direction. With wire carrying a current I in positive y direc- tion, the least value of force required to move it in x and y directions are F_(1) and F_(2). Now the direction of current in the wire is reversed and the value of two forces becomes F' _(1) and F'_( 2). Find the ratio of forces F_(1) : F_(2) : F'_(1) : F'_(2) |
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Answer» |
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| 34. |
(a) Describe briefly how electromagnetic waves are produced by oscillating charges. (b) Give one use of each of the following : (i) microwaves. (ii) ultraviolet rays. (iii) infrared rays. (iv) gamma rays. |
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Answer» Solution :(a) SEE POINT Number 6 under the heading ..Chapter At A Glance... (b) (i) Microwaves are used in radar system, in microwave ovens and for satellite communications etc. (ii) Ultraviolet rays are used for LASIK eye surgery and to kill GERMS in water purifiers and hospitals etc. (iii) INFRARED rays are used in remote switches. (iv) Gamma rays are used for radio therapy in cancer patients and for CAUSING certain nuclear reactions. |
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| 35. |
The magnetic field B due to a straight conductor of uniform cross-section of radius 'a' and carrying a steady current varies with the distance 'r' from the centre of conductor as shown in graph . |
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Answer»
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| 36. |
The observation of the tracks of secondary electrons showed that a neutral pion decayed into two identical photons. The angle of separation of the photons is 90^(@) Find the kinetic energy of the pion and the energy of each photon. |
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Answer» `epsi_(pi)^(2)-epsi_(0)^(2)=4epsi_(gamma)^(2)cos^(2)45^(@),or,epsi_(pi)^(2)-epsi_(0)^(2)=epsi_(pi)^(2)cos^(2)45^(@)` It then follows that `epsi_(pi)=epsi_(0)sqrt2`. The KINETIC energy of the pion is `K_(pi)=epsi_(0)(sqrt2-1)` and the photon energy is `epsi_(gamma)=epsi_(0)//sqrt2`. 
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| 37. |
The quantity [PV/kT] represent |
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Answer» mass of the GAS `nKT.N_(a),(K=(R )/(N_(A)))` `(PV)/(KT)=nN_(A)= "number of molecule"` |
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| 38. |
If both the number of protons and the number of neutrons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice-versa) in a nuclear reaction ? Explain. |
| Answer» SOLUTION :When neutrons and protons are conserved, then after every `ALPHA and beta` deacy electrons get back to LOWER ENERGY states from higher energy states and difference of the two energy LEVELS appear as radiation of high frequency like radiation. | |
| 39. |
How can the defect of astigmatism be corrected ? |
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Answer» |
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| 40. |
A body is projected horizontally from the top of a tower with a velocity of 10m/s. If it hits the ground at an angle of 45^@, the vertical component of velocity when it hits ground in m/s is |
| Answer» ANSWER :D | |
| 41. |
An electric potential difference will be induced between the ends of the conductor shown in the diagram, when the conductor moves in the direction |
| Answer» Solution :cionductor CUTS the flux only when, if it moves in the direction of `M`. | |
| 42. |
Explain - ''Changing electric field produces magnetic field''. Explain importance of this statement. |
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Answer» Solution :Maxwell logically showed that by changing electric field magnetic field can be obtained. This phenomena is very IMPORTANT. This phenomena first time was shown by Oersted who was PHYSICS teacher. This phenomena explain radiowaves, gamma waves, electromagnetic waves and other FORMS of electromagnetic waves. Production of magnetic field by changing electric field can be EXPLAINED charging of CAPACITOR by using Ampere circuital law. |
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| 43. |
The density of electron and holes in an intrinsic semiconductor is n_( e) and n_(h) respectively. Which of the following options are true? |
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Answer» `n_(h) gt n_(E )` |
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| 44. |
Find the Q-value and the kinetic energy of the emitted alpha - particle in the alpha - decay of ._(88)^(226)Ra Given m (._(88)^(226)Ra)=226.02540 u, m (._(86)^(222)Rn)=222.01750 u, m (._(86)^(222)Rn)=220.01137 u, m (._(84)^(216)Po) = 216.00189 u. |
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Answer» Solution :`._(88)Ra^(226)rarr ._(85)RN^(222)+._(2)He^(4)` Q VALUE `[m(._(88)Ra^(226))-m(._(86)Rn^(222))-m_(ALPHA)]xx931.5 MeV` `= [226.02540-222.01750-4.00260]xx931.5 MeV` `Q = 0.0053xx931.5 MeV = 4.94 MeV` K.E of a particle `= ((A-4)Q)/(A)=(226-4)/(226)xx4.94 = 4.85 MeV` |
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| 45. |
The x-component of the resultant of several vectors (a) is equal to the sum of the x-components of the vectors (b) may be smaller than the sum of the magnitudes of the vectors (c) may be greater than the sum of the magnitudes of the vectors (d) may be equal to the sum of the magnitudes of the vectors |
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Answer» only a |
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| 46. |
Two audio speakers are kept some distance apart and are driven by the same amplifier system. A person is sitting at a place 6.0m from one of the speakers and 6.4m from the other. If the sound signal is continuously varied from 500 Hz to 5000Hz, what are the frequencies for which there is a destructive interference at the place of the listener ? Speed of sound in air = 320 m/s. |
| Answer» SOLUTION :1200HZ, 2000Hz, 2800Hz, 3600Hz and 4400Hz. | |
| 47. |
In Young's double slit experiment the distance between the slits is d. An interference fringe is obtained on a screen placed at a distance D from the slits . A dark band is noticed just opposite to one of the slits. The wavelength of the light used is |
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Answer» `D^(2)/(2D)` |
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| 48. |
An ideal inductor takes a current of 10A when connected to a 125V, 50 Hz ac supply. A pure resistor across the same source takes 12.5 A. If the two are connected in series across a 100V, 40 Hz supply, the current through the circuit will be |
| Answer» Answer :A | |
| 49. |
Shown below is a graph of current versus applied voltage for a diode . Approximately , what is the resistance of the diode for an applied voltage of – 1 V ? |
| Answer» ANSWER :B | |
| 50. |
Sensitivity of potentiometer can be increased by a) increasing series resistance in the primary circuit b) decreasing the length of potentiometer wire c) using thin and high resistivity wire as potentiometer wire d) increasing the length of the wire |
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Answer» a and C are CORRECT |
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