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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. |
A sample of paramagnetic salt contains 2.0 xx 10^(24) atomic dipoles each of dipole moment 1.5 xx 10^(-23) J T^(-1) . The sample is placed under a homogeneous magnetic field of 0.64 , and cooled to a temperature of 4.2 K . The degree of magnetic saturation achieved is equal to 15% . What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K? |
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Answer» SOLUTION :Here number of atomic dipole `N = 2.0 XX 10^(24)`, magnetic moment of each dipole `m = 1.5 xx 10^(-23) J T^(-1)` . If degree of magnetic saturation be F, then total magnetic moment `m_0 = N m f` As at T = 4.2 K and B = 0.84 T, f = 15% = 0.15 , hence applying Curie.s law `m_0 = (CB)/(T)` , we have `C = (m_0 T)/(B) = (N m f T)/(B) = (2.0 xx 10^(24) xx 1.5 xx 10^(-23) xx 0.15 xx 4.2)/(0.64) = 22.5` Then for `T. =2.8 K ` and B.= 0.98 T, the total dipole moment of the sample will be `m._0 = (CB.)/(T.) =(22.5xx0.98)/(2.8) =7.88 A m^2` . |
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| 2. |
What had Anju said to Mini? |
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Answer» That the bird is beautiful |
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| 3. |
Consider two solid spherical asteriod of uniform density of mass M and radius R. In one asteroid a tunnel tof very small size of depth R is bored to the centre and in other asteroid a spherical into the cavities of both asteroids from the top most point P. If force experienced by particle is F_(I) and F_(II) |
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Answer» The RATIO of `F_(I)//F_(II)` is equal to `2x//R` |
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| 4. |
A circular coil of radius 10 cm, 500 turns and resistance 2 Omega 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. Estimate the magnitudes of the emf and current induced in the coil. Horizontal component of the earth’s magnetic field at the place is 3.0 ×x 10^(-5) T. |
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Answer» Solution :Initial flux through the coil, `Phi_(B("initial")) = BA cos theta ` `=3.0 xx 10^(-5) xx (pi xx 10^(-2)) xx cos 0^@` `=3pi xx 10^(-7)` Wb Final flux after the rotation. `Phi_(B("final")) = 3.0 xx 10^(-5) xx (pi xx 10^(-2)) xx cos 180^@ ` `= - 3pi xx 10^(-7)` Wb Therefore, estimated value of the induced emf is , `epsi =N (DELTA Phi)/(Deltat)` `= 500 xx (6 pi xx 10^(-7)) //0.25` `= 3.8 xx 10^(-3) V` `I = epsi//R = 1.9 xx 10^(-3) A ` Note that the magnitudes of `epsi` and I are the estimated values. Their INSTANTANEOUS values are different and depend upon the speed of rotation at the PARTICULAR instant. |
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| 5. |
A voltmeter reads 4 V when connected to a parallel plate capacitor with air as a dielectric. When a dielectric slab is introduced between plates for the same configuration , voltmeter reads 2V. What is the dielectric constant of the material ? |
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Answer» 0.5 |
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| 6. |
In Young.s experiment interference bands are produced on the screen placed at 1.5m from the two slits 0.15 mm apart and illuminated by light of wavelength 6000 A^(0). If the screen is now taken away from the slit by 50 cm the change in the fringe width will be |
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Answer» `2XX 10^(-4)m` |
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| 7. |
An electron in the hydrogen atom jumps from excited state n to the ground state. The wavelength so emitted illuminates a photo sensitive metal having work function 2.75 eV. If the stopping potential of the photoelectron is 10 V, the value of n is ... |
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Answer» 3 `phi=-2.75eV` Total incident energy `E=phi+KE_(max)=12.75eV` `:.DeltaE=E_(n)-E_(1)` `12.75=(-13.6)/(n^(2))-(-(13.6)/(12))` `12.75=(-13.6)/(n^(2))+13.6` `(13.6)/(n^(2))=13.6-12.75` `=0.85` `:.(13.6)/(0.85)=n^(2)` `:.n^(2)=16` `:.n^(2)=4` |
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| 8. |
Obtain the maximum kinetic energy of b-particles, and the radiation frequencies of g decays in the decay scheme shown in Fig. You are given that m(""_(8)^(19)Au) = 197.968233 u m(""_(8)^(19)Hg) =197.966760 u. |
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Answer» Solution :`K_("MAX")(beta_(1)^(-))=0.284MeV, K_("max")(beta_(2)^(-))=0.960MeV` `v(gamma_(1))=2.627xx10^(20)Hz, v(gamma_(2))=0.995xx10^(20)Hz, v(gamma_(3))=1.632xx10^(20)Hz` |
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| 9. |
To increase magnifying power of a telescope, we should increase |
| Answer» SOLUTION :By INCREASING the DIAMETER `//` APERTURE of the OBJECTIVE. | |
| 10. |
The wave that are band used in the downlink of satellite communication |
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Answer» 9.5 to 2.5 GHz |
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| 11. |
A-photo detector used to detect the wavelength. 1700 nm, has energy gap about |
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Answer» `0.073ev` |
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| 12. |
The measured mass and volume of a body are 53.63 g and 5.8 cm^3respectively, with possible errors of 0.01 g and 0.1 cm^3. The maximum percentage error in density is about |
| Answer» ANSWER :B | |
| 13. |
A ray of light while travelling from a denser to a rarer medium undergoes total internal reflection. Derive the expression for the critical angle in terms of the speed of light in the respective media. |
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Answer» Solution : As shown in Fig. 9.36, if IC be the CRITICAL angle in a given denser medium number 2 with respect to a rarer medium number 1, then `sin i_( c) = n_(12) = p_(2)/p_(1) = ("speed of light in denser medium")/("speed of light in rarer medium")` If rarer medium be free space or air, then the RELATION is modified as: `sin i_( c) = ("speed of light in denser medium")/("speed of light in free space") = v/c` |
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| 14. |
A block is placed at distance of 2m from the rear on the floor of a truck (g=10ms^(-2)). When the truck moves with an acceleration of 8ms^(-2), the block takes 2 sec to fall off from the rear of the truck. The coefficient of sliding friction between truck and the block is |
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Answer» `0.5` |
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| 15. |
N_1atoms of a radioactive element emit N_2beta particles per second. Find the decay constant of the element (in s^(-1) ) |
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Answer» `N_(1)//N_2` |
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| 16. |
दाढ़ी बनाने के किस प्रकार के दर्पण का उपयोग किया जाता है ? |
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Answer» उत्तल |
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| 17. |
There is a steam of neutrons with a kinetic energy of 0.0327eV. If the half-life of neutrons is 700 seconds, the fraction of neutrons that will decay before they travel a distance of 10 netre is 6xx10^(-7), where x =______ (Given mass of neutron m=1.676xx10^(-27) kg ) |
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Answer» 8 |
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| 18. |
A solenoid 60 cm long and of radius 7.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside the solenoid (near its centre) normal to its axis, both the wire and the ais of the solenoid are in the hoorizontal plane. The wire is connected through two lead parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current (with appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire ? g=9.8 m s^(-2). |
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Answer» SOLUTION :`MG = Bi_("wire")` l but `B = μ_(0)n i_("SOLENOID")` `rArrmg=mu_(0)ni_("solenoid")xxi_("wire")xxl` `i_("solenoid")=(mg)/(mu_(0)ni_("wire")l)=108A` |
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| 19. |
What type of wave is suitable to represent the wave associated with a moving particle? |
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Answer» |
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| 20. |
Consider a 20 W bulb emitting light of wavelength 5000 Å and shining on a metal surface kept at a distance 2m.Assume that the metal surface has work function of 2eV and that each atom on the metal surface can be treated as a circular disk of radius 1.5 Å. (i)Estimate no. of photons emitted by the bulb per second.[Assume no other losses] (ii)Will much time would be required by the atomic disk to receive energy equal to work function (2eV)? (iv)How many photons would atomic disk receive within time duration calculated in (iii)above? (v) Can you explain how photoelectric effect was observed instantaneously? |
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Answer» Solution :(i)`P=(E_(n))/(t)=(nhf)/(t)=(NHC)/(tlambda)` `therefore ((n)/(t))=(lambda)/(nc)` =`((20)(5000xx10^(-10)))/((6.625xx10^(-34)(3xx10^(8))` `=5.03xx10^(19)` photon/second (ii)Energy of one photon, `E_(1)=hf=(hc)/(lambda)` `therefore E_(1)=((6.625xx10^(-34))(3xx10^(8)))/((5000xx10^(-10))(1.6xx10^(-19)))eV` `therefore E_(1)=2.49 eV` Here `phi_(0)=2eV` `therefore E_(1)gtphi_(0)` `implies` Photo emission will take place. (III)  Intensity of radiation ,`I=(E_(0))/(A_(0)t_(0))=(P)/(A)` `therefore t_(0)=(E_(0)A)/(PA_(0))` `therefore t_(0)=((2xx1.6xx10^(-19))(4xx3.14xx(2)^(2)))/((20)(3.14xx(1.5xx10^(-10))^(2)))` `therefore t_(0)=11.38 s` (iv) Area `to` (No.of photons made incident in one second) `therefore piR^(2)to(?)` `implies` No. of photons made incident on atomic disc in one second, `=(piR^(2)xx((n)/(t)))/(4pir^(2))=(1)/(4)xx(R^(2))/(r^(2))xx(n)/(t)` `implies` No. of photons made incident in time `t_(0)`, `=(1)/(4)xx(R^(2))/(r^(2))xx((n)/(t))xxt_(0)` `=(1)/(4)xx((2)^(2))/((1.5xx10^(-10))^(2))xx5.03xx10^(19)xx11.38` (v)Above calculation is based upon the assumption "Photon behaves like a wave".If this assumption is true then we should get emission of photoelectrons,approximately after 11.38 second.But experimentally we get almost instantaneous emission of why we have to believe that in this phenomenon,photons must be behaving like particles. |
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| 21. |
The strain produced in a wire of length 60 cm is 1% . The change in the length of the wire is |
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Answer» a)0.6 cm |
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| 22. |
When unpolarized light is incident on a Tourmaline crystal of proper thickness a) it exhibits dichroism b) it absorbs ordinary ray and transmits extraordinary ray c) it absorbs extraordinary ray and transmits ordinary ray. |
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Answer» only a and B are true |
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| 23. |
The a.c. main supply is given to be 220 V. What would be the average e.m.f. during a positive half cycle ? |
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Answer» 386 V |
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| 24. |
For a photoelectric cell the graph showing the variation of cut of voltage (V_(0)) with frequency (v) of incident light is best represented by |
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Answer»
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| 25. |
At a distance of 10 cm from a long straight wire carrying current, the magnetic field is 0.04 T. At the distance of 40 cm, the magnetic field will be ______ . |
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Answer» 0.01 T For long CURRENT carrying wire, `B=(mu_(0)I)/(2PI)rArrBprop1/y` (For constant I) `thereforeB_(1)/B_(2)=y_(2)/y_(1)rArrB_(2)=B_(1)xxy_(2)/y_(1)` `thereforeB_(2)=(0.04xx10)/40=0.01T` |
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| 26. |
Maximum value of magnetic field of progressive electromagnetic wave is 20 nT, then maximum value of electric field is ……… |
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Answer» `12(V)/(m)` `THEREFORE E_(0)=CB_(0)=3xx10^(8)xx20xx10^(-9)` `therefore E_(0)=6(V)/(m)` |
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| 27. |
पानी से भरी बाल्टी की गहराई कम मालूम पड़ने का कारण है |
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Answer» प्रकाश का परावर्तित होना |
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| 28. |
The torque on the magnet in the final orientation |
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Answer» SOLUTION :Torpue ` =| oversetto m xx oversetto B | = mB SIN THETA ` For `= 180^(@) , ` we have Torpue `= 6XX 0.44 sin 180^(@) =0` [If the student straight away writes that the torque is zero since MAGNETIC moment and magnetic field are anti parallel in this orientation , award full mark |
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| 29. |
The linear momentum of a particle varies with times asp=a_(0)+at+bt^(2). Which of the following represents force and time relation ? |
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Answer» Solution : `p=a_(0)+at+bt^(2)` `therefore F=(DP)/(dt)=a+2bt` At t = 0, F = a and F varies LINEARLY with TIME. 
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| 30. |
A rectangular coil of 20 turns and area of cross section 25 cm^2 has a resistance of 100 ohm. If the magnetic field which is perpendicular to the plane of coil changes at the rate of 1000 tesla/sec, the current in the coil is |
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Answer» 1amp |
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| 31. |
(A) : A current carrying conductor produces only an electric field. (R) : Electrons in motion in a current carrying wire give rise to a electric field |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the CORRECT EXPLANATION of 'A'. |
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| 32. |
निम्न मे से कौन सी संख्या एक अपरिमेय संख्या है :- |
| Answer» Answer :B | |
| 33. |
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 inot the interior of the inductor. The glow of the ligth bulb (a) increases (b) decreases (c) is unchanged as the iron is inserted. Give your answer with reasosns. What will be your answer if ac source is replaced by dc source ? |
| Answer» SOLUTION :Decreases, DUE to increase in INDUCTIVE reactance. | |
| 34. |
निम्न मे से कौन सी संख्या एक परिमेय संख्या है :- |
| Answer» Answer :D | |
| 35. |
निम्न में से किसमें सबसे कम समान लक्षण होते है ? |
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Answer» कुल |
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| 36. |
The resistance of all the wires between any two adjacent dots is R. Then equivalent resistance between A and B as shown in figure is : |
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Answer» `7//3 R` |
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| 37. |
A bulb connected in series with a solenoid is lit by a.c. source. If a soft iron core is introduced in the solenoid, will bulb glow brighter ? |
| Answer» Solution :No, the BULB will glow dimmer. This is because, on INTRODUCING soft iron core in the solenoid, its inductance L increases, the inductive REACTANCE `X_(L) = omega L` increases and hence the current through the bulb decreases. | |
| 38. |
The magnifying power of an astronomical telescope for normal adjustment is 10 and the length of the telescope is 110cm. Find the magnifying power of the telescope when the image is formed at the least distance of distinct vision for normal eye. |
| Answer» ANSWER :1 | |
| 39. |
एक शून्येतर परिमेय संख्या और एक अपरिमेय संख्या का योगफल होता है : |
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Answer» परिमेय |
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| 40. |
In a region, electric field depends on X-axis as E = E_0x^2. There is a cube of edge a as shown. Then find the charge enclosed in that cube. |
| Answer» Answer :A | |
| 41. |
A pipe open at both ends produces a note of frequency f_(1). When the pipe is kept with (3)/(4)th of its length in water, it produces a note of frequency f_(2) The ratio (f_(1))/(f_(2)) is : |
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Answer» `(3)/(4)` `f_(2) = (v)/(4xx(1)/(4) l ) = (v)/(l) ` ` therefore(f_(1))/(f_(2))= ( (V)/(2l))/((V)/(l)) = (1)/(2) ` . correct choice is (c). |
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| 42. |
Coulomb's law for electrostatic force between two point charges and Newton's law for gravitational force between two stationary point masses, both have inverse-square dependence on the distance between the charges and masses respectively, (a) Compare the strength of these forces by determining the ratio of their magnitudes (i) for an electron and a proton and (ii) for two protons. (b) Estimate the accelerations of electron an proton due to the electrical force of the mutual attraction when they are =10^(-10)m apart? (m_(p) = 1.67 xx 10^(-27) kg, m_(e) = 9.11 xx 10^(-31) kg) |
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Answer» Solution :(a) (Electrons) `F_(e), F_(e)` (Proton) `|F_(e)| = k((e)(e))/r^(2)`………..(1)  `|F_(g)| =G(m_(e).m_(p))/r^(2)`…….(2) Required ratio, `|F_(e)/F_(g)| = (ke^(2))/(Gm_(e).m_(p))` `=(9 xx 10^(9) xx (1.6 xx 10^(-19))^(2)/(6.67 xx 10^(-11) xx (9.1 xx 10^(-31))(1.67 xx 10^(-27))))` `=2.273 xx 10^(39)` `|F_(e)| GT gt gt gt |F_(g)|`  `|F_(e)| = k((e)(e))/r^(2)`..........(3)  `|F_(g)| = G.(m_(p)^(2))/r^(2)`......(4) Required ratio, `|F_(e)|/|F_(g)| = (ke^(2))/(Gm_(p)^(2))` `=(9 xx 10^(9) xx (1.6 xx 10^(-19))^(2))/((6.67 xx 10^(-11))(1.67 xx 10^(-27))^(2))` `=1.349 xx 10^(36)` `|F_(e)| gt gt gt gt gt |F_(g)|`  Electric force exerted between electron and proton, `F_(e) =k((e)(e))/r^(2)` `=(9 xx 10^(9) xx (1.6 xx 10^(-19))^(2))/(10^(-10))^(2)` `thereforeF_(e) = 2.304 xx 10^(-8)` N.........(5) Now, here `F_(e) = m_(e)a_(e) = m_(p)a_(p)` (i) `a_(e) =F_(e)/m_(e) = (2.304 xx 10^(-8))/(9.1 xx 10^(-31))`............(6) `=2.532 xx 10^(22) m//s^(2)` (II) `a_(p) = F_(e)/m_(p) = (2.304 xx 10^(-8))/(1.67 xx 10^(-27))` `=1.380 xx 10^(19) m//s^(2)` |
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| 43. |
The area of each plate of a parallel capacitor is 100 cm^(2) and the distance between them is 0.05 cm. It is filled with a dielectric and its capacitance becomes 3.54 xx 10^(-4) muF. Find the dielectric constant of the substance. |
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Answer» Solution :Data supplied, `A=100 cm^(2)=100 xx 10^(-4) m^(2)` `C=3.54 xx 10^(-4) muF= 3.54 xx 10^(-4) xx 10^(-6) F =3.54 xx 10^(-10) F` `d=0.05 cm =0.05 xx 10^(-2)m ""epsi_(0) =8.854 xx C^(-12) C^(2) N^(-1) m^(-2)` `C=(epsi_(0) KA)/(d) ""K=(Cd)/(epsi_(0) A)=(3.54 xx 10^(-10) xx 0.05 xx 10^(-2))/(8.854 xx 10^(-12) xx 100 xx 10^(-4)) K=2` |
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| 44. |
Explain hard ferromagnetic and soft ferromagnetic materials. |
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Answer» Solution :Hard ferromagnetic material : In some ferromagnetic MATERIALS, the magnetisation persists on removal of external magnetic field. Such material are called hard magnetic materials. Alnico, an alloy of iron, aluminium, nickel, cobalt and copper is one such material. The naturally occurring lodestone is also ferromagnetic material. Its relative permeability is `lt 1000.` Such material forms permanent magnet to be used AMONG other thing as a COMPASS needle. Soft magnetic materials : There are some ferromagnetic material in which the magnetisation disappears on removal of external magnetic field. These material are called soft magnetic materials. These materials are used in electric BELL, CRANE, transformers etc. |
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| 45. |
In the above question size element will raise the power factor to unity? |
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Answer» an INDUCTOR should be PLACED in series |
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| 46. |
द्विघात बहुपद x^2 -3x-4 के शून्यक alpha ,betaहो तो alpha+beta-alphabetaका मान होगा - |
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Answer» 3 |
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| 47. |
Find the apparent depth and apparent shift of an object placed at the bottom of the vessel containing a glass slap of thickness 10 cm and water of height 50 cm. as shown in the figure. |
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Answer» Solution :Apparent DEPTH = `50/n_(W) + 10/n_(g) = 50/(4//3) + 10/(3//2) = (75/2 + 20/3) cm = 44.17 cm` `therefore` Apparent SHIFT = (50+ 10 - 44.17) cm = 15.83 cm. |
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| 48. |
The potential barrier of a semiconductor is 0.6 V at room temperature. What is the. approximate value of its potential barrier if the temperature is increased by 20^@C ? |
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Answer» `0.7V` |
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| 49. |
A broad source of light lambda = 6800 Å illumintates normally two glass plates 12 cm long. They touch at one end and are separated by a wire 0.048 mm. In diameter at the other end.How many bright fringes appear over the 12 cm. Distance. |
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Answer» 141  For bright fringes FORMED by a gap of wedge due to LIGHT incident normally `2d = (n+1/2)lambda` Here d = 0.0048 CMS diameter of WIRE (an gap) at the point where NTH fringe is formed. `therefore n = (2d)/(lambda) -1/2` `= (2 xx 0.0048)/(6800 xx 10^(-8) -1/2` = 141 |
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