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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. |
The mutual inductance ofcoil does not depend on |
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Answer» NUMBER of TURNS of the coil |
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| 2. |
When was Evelyn's deafness confirmed? |
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Answer» by the AGE of 5 |
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| 3. |
A narrow neutron beam falls on a natural face of an aluminium single-crystal at a glancing angle of 5^(@) The distance between crystallographic planes parallel to the single-crystal face is 0.20 nm. What is the velocity and the energy of neutrons for which a first-order maximum is observed in this direction? What is the temperature responding to this neutron velocity? |
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Answer» `K=h^(2)/(2mlamda^(2)),v=h/(mlamda),T=h^(2)/(3mklamda^(2))` |
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| 4. |
What was the most important conclusion of Rutherford's experiment ? |
| Answer» Solution :The nuclear radius is of the ORDER of `10^-15m` and TAHT of the ATOM is about `10^-10m`. | |
| 5. |
A focal length of a lens is 10 cm. What is power of a lens in dioptre? |
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Answer» <P>0.1 D |
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| 6. |
10 cells, each of internal resistance 0.5 Omega and 1.2 V emf are connected (a) all in series and (b) all in paralle. Calculate the current sent in each case through a wire of resistance 0.8 Omega. |
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Answer» |
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| 7. |
When the voltage applied to an X-ray tube increased from V_(1)=10kV to V_(2)=20kV, the wavelength interval between the k_(alpha) line and the short-wave cut-off of the continuous X-rays spectrum increases by a factor n=3.0. Find the atomic number of the element of which the tube's anticathode is made. |
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Answer» SOLUTION :SUPPOSE `lambda_(0)=` wavelength of the characteristic `X`-rays line. Then using the FORMULA for short wavelength limit of limit of continous radiation `(lambda_(0)-(2piħc)/(eV_(1)))/(lambda_(0)-(2piħc)/(eV_(2)))=(1)/(N)` Hence `lambda_(0)=(2 piħc)/(eV_(1))((n-(V_(1))/(V_(2))))/(n-1)` Using also Moseley's law, we get `Ƶ=1+sqrt((8pic)/(3R lambda))=1+2sqrt((n-1)/(3 ħR_(n)-(V_(1))/(V_(2))))=29`. |
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| 8. |
Two parallel wires carry currents in the same direction will |
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Answer» do not EXPERIENCE a FORCE |
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| 9. |
Two identical point charges'of magnitude -q are fixed as shown in the figure below. A third charge +q is placed midway between the two charges at the point P. Suppose this charge is displaced a small distance from the point P in the directions indicated by the arrows, in which direction(s) will +q be stable with respect to the displacement? |
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Answer» `A_(1)` and `A_(2)` |
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| 10. |
The current gain of a transistor in common-emitter circuit is |
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Answer» 40 |
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| 11. |
Two particles are projected from the surface of the earth with velocities sqrt(5/7gR) and sqrt(2/5gR) where, R is the radius of the earth what should be the ratio of maximum heights attained? |
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Answer» `2/5` |
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| 12. |
Rakhi is a student of class 11th, she is strolling in a park near to her house. She saw a fat girl being teased by two thin playmates. They are playing see-saw. Rakhi enquired why they are teasing their playmate and why they are not allowing her to sit on see-saw. Two thin girls replied that it is not possible to play see-saw with the fat girl as she is very heavy for any of them. Rakhi told them if the fat girl sits slightly near towards the middle part then, they can see-saw even with her. And finally that fat girl is see-sawing with one of thin girl. They finally learn to play that game together (i) What technique is told by Rakhi to little girls? (ii) If mass of fat girl is 30 kg and thin girls are of 20 kg (each), then to operate see-saw properly what will be the ratio of their distances from the pivot of plank? (iii) What values are indicated by Rakhi? (iv) What values are indicated by little girls? |
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Answer» Solution :(i) RAKHI told them turning EFFECT of force is directly proportional to its lever arm. (II) For proper balance, `F_(i)x_(i) = F_(2)x_(2)` or `m_(1)gx_(1) = m_(2)gx_(2)` or `x_(1)/x_(2) = m_(2)/m_(1) = 20/30 = 2/3`  (iii) CONCERN for others, equality of all. (iv) Obedience to elders, adaptation to a better IDEA. |
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| 13. |
What are types of spectra ? |
| Answer» SOLUTION :EMISSION and ABSORPTION | |
| 14. |
A 50muF capacitor is connected to a 100 V, 50 Hz ac supply. Determine the rms value of the current in the circuit. |
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Answer» Solution :Here `C=50muF=50xx10^(-6)F=5xx10^(-5)F` `E_("RMS")=100V, v=50Hz`. Step 1. Capacitive REACTANCE, `X_(C)=(1)/(Comega)=(1)/(Cxx2piv)=(1)/(5xx10^(-5)xx2xx3.14xx50)=63.69Omega` Step 2. `I_("rms")=(E_("rms"))/(X_(C))=(100)/(63.69)=1.57A` |
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| 15. |
a. The peak voltage of an ac supply is 300V. What is the rms voltage ? b. The rms value of current in an ac circuit is 10A. What is the peak current ? |
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Answer» SOLUTION :a.`V_(RMS) = (V_(rms) )/(SQRT2)= (300)/(sqrt2) = 212.1 V` b. `I_(rms) = (I_(rms) )/(sqrt2)therefore I_(rms) = sqrt(2) I_(rms) = sqrt(2) xx 10 = 14.14 A` |
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| 16. |
a . What do you observe when the circuit is closed ? b. What will happen when the polarity of cell is reserved ? Why ? |
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Answer» SOLUTION :a. BULB A GLOWS B. Bulb B glows because the DIODE `D_2` is forward biased . |
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| 17. |
Explain, with the help of a schematic diagram, the principle and working of a Light Emiting Diode. What criterion is kept in mind while choosing the semiconductor material for such a device ? Write any two advantages of light emiting diode over conventional incandescent lamps. |
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Answer» Solution :A Light Emitting Diode (LED) is a heavily doped p-n junction encapsulated with a transparent cover. Under forward bias , a LED emits spontaneous radiation which are NEARLY monochromatic and can come out of the transparent cover. When the diode is forward biased, electrons are SENT from n-side to p-side across the junction and holes from p-side to n-side. carriers, the number of electrons increases. SIMILARLY on n-side number of minority holes increases. These excess minority carrierstend to recombine with majority carriers near the junction. Un recombination, the energy is released in the form of photons. These photons have energy either equal to or slightly less than the band gap energy E for that semiconductor material. The intensity of EMITTED photons (i.e., emitted light) depends on the forward current of the diode. We shall try to keep in mind the value of bandgap whilechoosing the semiconductor material for LED. For red LED, a band gap of 1.8 eV is required. Similarly for other colours band gap has a PREFERRED value. Some common advantages of LEDs over conventional incandescent lamp are as follows: (i) LEDs require less operational voltage and consume less power. (i) The light emitted is nearly monochromatic. (iii) Nowarm up time is required for LED and it has fast on-off switching capability. (iv) LEDs have long life and ruggedness. |
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| 18. |
What we call to the process of adding impurity to a pure semiconductor crystal (Si or Ge) ? |
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Answer» |
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| 19. |
A sphere rolling on a horizontal rough surface collides elastically with a smooth vertical wall, as shown in figure. During collision, angular momentum of the sphere is conserved about ____________. (Any point O on the surface/point of contact P) |
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Answer» <P> |
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| 20. |
Which of the following is an aviator? |
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Answer» A MARITIME officer |
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| 21. |
A convex lens is placed in contact with with a plane mirror . A point object at a distance of 20cm on the axis of this combination has its image coinciding with itself. What is the focal length of the lens? |
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Answer» |
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| 22. |
Which experiment showed that the positive charge of an atom is concentrated at the centre of the atom |
| Answer» SOLUTION :`ALPHA`-PARTICLES | |
| 23. |
In Young's double slit experiment the 7th maximum with wavelength lambda_1 is at a distance d_1 and that with wavelength lambda_2 is at distance d_2. Then ((d_1)/d_2) = : |
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Answer» `((lambda1)/lambda_2)` |
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| 24. |
The property of halogen acids , that indicated incorrect is - |
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Answer» `HF GT HCl gt HBr gt HI…..` acidic STRENGTH `HI gt HBr gt HCl gt HF` |
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| 25. |
Eels are able to generate current with biological cells called electro plaques. The electroplaques in an eel are arranged in 100 rows, each row stretching horizontally along the body of the fish containing 5000 electroplaques. The arrangement is suggestively shown here. Each electroplaques has an emf of 0.15V and internal resistance of The water surrounding the eel completes a circuit between the head and its tail. If the water surrounding it has a resistance of the current an eеl can produce in water is about |
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Answer» 1.5A |
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| 26. |
A pipe of 30 cm long and open at both the ends produces harmonics. Which harmonic mode of pipe resonates a 1.1 kHz source ? Given speed of sound in air = 330 ms^(-1). |
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Answer» Fifth harmonic `V = (v)/(2l) = (330)/(2 xx (30)/(100)) = (3300)/(6) ` = `550` Hz. Let frequency of nth harmoonic resonantes with 1.1 kHz. n.v = 1100 n= `(1100)/(550) = 2 `. so correct choice is d . |
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| 27. |
The number of electrons to be put on a spherical conductor of radius 0.1 m to produce an electric field of 0.036 N/C just above its surface is |
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Answer» `2.7 XX 10^5` |
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| 28. |
A small telescope has an objective lens of focal length 150 cm and eyepiece of focal length 5 cm. What is the magnifying power of the telescope for viewing distant obj ects in normal adjustment ? If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens ? |
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Answer» SOLUTION :Here, `f_(0) = 150 cm ` and `f_(e) = 5 cm` Magnifying power of telescope in normal adjustment `m=-f_(0)/f_(e) = -150/5 = -30` The -ve sign signifies that the image formed is an inverted image. Again height of tower H = 100 m and DISTANCE of tower d = 3 km = 3000 m. `therefore`Angle subtended by the tower at the objective lens a is given by `alpha = h/d = h^(.)/f_(0)`, where h. is the height of the image of the tower formed by the objective lens. `rArr h. = (h.f_(0))/d = ((100 m) xx (150 cm))/(3000 m) =((100 m) xx (1.5 m))/(3000 m) = 0.05` or `5.0 cm` |
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| 29. |
The trajectory of a projectile projected from origin is given by the equation y = x - (2x^(2))/(5) .The initial velocity of the projectile is |
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Answer» `25 ms^(-1)` ` b=2/5 =(G)/(2v^2 cos^2 theta)` ` U^2= (5 g)/( 4 cos ^2 theta)` ` U^2= ( 5G )/( 4 cos45^@) = (5g)/(2)= 2.5 g = 25 m//s` |
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| 30. |
State and obtain Malus' law. |
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Answer» Solution :The proof of Malus's law : (i) Consider the plane of polariser and ANALYSER are inclined to each other at an angle `theta`. Let `I_0`be the intensity and 'a' be the amplitude of the electric vector transmitted by the polariser. (ii) The amplitude 'a' of the incident light has two rectangular components, `(a cos theta)` and `(a sin theta)` which are the PARALLEL and perpendicular components to the axis of transmission of the analyser. (iii) Only the COMPONENT `(a cos theta)` will be transmitted by the analyser. The intensity of light transmitted from the analyser is proportional to the square of the component of the amplitude transmitted by the analyser. `Iprop ( a cos theta)^2` `I = k (a cos theta)^2` Where k is constant of proportionality. ` I = Ka^2 cos^2 theta` ` I = I_0 cos^2 theta` (iv) Where, `I_0 = ka^2` is the maximum intensity of light transmitted from the analyser. The following are few special cases. Case (i) When ` theta = 0^@ , cos 0 = 1 , I = I_0` When the transmission axis of polariser is ALONG that of the analyser, the intensity of lighttransmitted from the analyser is equal to the incident light that falls on it from the polariser. Case (ii) When ` theta= 90^@ , cos 90^@ = 0, I = 0 ` When the transmission axes of polariser and analyser are perpendicular to each other, the intensity of light transmitted from the analyser is ZERO. |
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| 31. |
Two points separated by a distance of 0.1 mm can just be separated by a microscope, when light of wavelength 6000 A is used. What will be the limit of resolution if light of wavelength 4800 A is used? |
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Answer» 0.05 mm |
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| 32. |
What is the region of wavlengths of visible region ? |
| Answer» SOLUTION :`3.9xx10^-3` to `7.8xx10^-5`CM | |
| 33. |
A current of 3A flows in a circuit the potential difference between points A and D is |
| Answer» ANSWER :C | |
| 34. |
The work done by a force is defined by W=vecF cdot vecS. In certain situation vecF and vecS are not equal to zero BUT workdone is zero.From this we conclude that |
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Answer» `vecF and VECS` are in same direction |
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| 35. |
A circuit containing a 80 mH inductor and a 60 uF capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible. (a) Obtain the current amplitude and rms values. (b) Obtain the rms values of potential drops across each element. (c) What is the average power transferred to the inductor ? (d) What is the average power transferred to the capacitor ? (e) What is the total average power absorbed by the circuit? ['Average implies' average over one cycle] |
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Answer» Solution :Here, L = 80 mH `=80 xx 10^(-3) H = 0.08 H, C = 60 muF = 60 xx 10^(-6) F = 6 xx 10^(-5) F, V_(rms) = 230 V` and v = 50 Hz or `omega = 2pi v = 2 xx 3.14 xx 50 = 314 s^(-1), R=0` (a) `therefore` Impedance `Z = X_(L) - X_(C) = L. omega -1/(C omega) = 0.08 xx 3.14 - 1/(6 xx 10^(-5) xx 314) = 25.1 - 53.1 = 28 Omega` (capacitance) `therefore I_(rms) = 8.24 A` and `I_(m) = sqrt(2) xx I_(rms) = sqrt(2) xx 8.23 = 11.6 A` (B) `therefore V_(L) = I_(rms) = sqrt(2)xx 8.23 = 11.6` A and `V_( C) = I_(rms) xx X_(C ) = 8.24 xx 53.1 = 437 V` `rArr V_( C) - V_(L) = 437 - 207 = 230 V` ( c) As `V_(L)` leads the current by `pi/2` hence, averae power transfered to the inductor `=V_(L). I_(rms) cos phi = V_(L).I_(rms) cos pi/2 =0` (d) As Vc LAGS behind the current by `pi/2` , hence average power transferred to the capacitor `=V_(C) I_(rms) cos phi = V_( C) I_(rms). cos pi/2=0` (e) Total average power absorbed by the circuit = 0 |
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| 36. |
When two coherent waves interfere, the minimum and maximum intensities are in the ratio 16 : 25. Then a) the maximum and minimum amplitudes will be in the ratio 5 : 4 b) the amplitudes of the individual waves will be in the ratio 9 : 1 c) the intensities of the individual waves will be in the ratio 41 : 9 d) the intensities of the individual waves will be in the ratio 81 : 1. |
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Answer» a, B and C are true |
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| 37. |
Which of the following has a negative temperature gradient ? |
| Answer» SOLUTION :Carbon | |
| 38. |
What should be the length of a half wave dipole antenna required to transmit audio frequency signals? |
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Answer» 5 km |
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| 39. |
Given sets of elements are phosphorus , arsenic , indium and bismuth . The addition of which in pure semiconductor will result in p - type semiconductor : |
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Answer» PHOSPHORUS , ARSENIC and INDIUM |
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| 40. |
A uniform rod of length L and mass M is hinged at its upper point and is at rest at that moment in the vertical plane.A current i flows in it.A uniform magnetic field of strength B exists perpendicular to the rod and in horizontal direction is (iLB)/N.Then find the value of N. |
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Answer» |
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| 41. |
At a certain distance from a point charge the electric field is 500 V m^(-1)and the potential is 3000 V. The distance is |
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Answer» 6 m |
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| 42. |
Identify the part of the electromagnetic spectrum used in (i) radar, and (ii) eye surgery. Write their frequency range. |
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Answer» SOLUTION :(i) Microwaves, `10^(@) Hz` to `10^(@) Hz`, (II) Ultraviolet rays, `7xx10^(14)Hz" to "5XX10^(17)Hz`. |
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| 43. |
Two boys are standing at the ends A and B of a ground where AB = 'a'. The boy at B starts running in a direction perpendicular to AB with a velocity The boy at A also starts running simultaneously with velocity v in straight line and catches the B in time then t is : |
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Answer» `a/SQRT(V^(2)+v_(1)^(2))` `(v_(1)t)^(2) + a^(2)=(VT)^(2)`or `a^(2)=t^(2)(v^(2)-v_(1)^(2))` `:.t=sqrt(a^(2)/(v^(2)-v_(1)^(2)))` |
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| 44. |
Evaluate the maximum value of energy and momentum of a phonon (acousite quantum) in copper whose Debyed temperature is equal to 300K. |
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Answer» SOLUTION :The maximum energy of the phonon is `ħ omega_(m)= kTheta= 28.4meV` On substituting `Theta= 330K`. To get corresponding value of the maximum moment we must have the dispersion relation `omega= omega (vec(K))`. For small `(vec(k))` we know `omega=|vec(k)|` where `v` is the velocity of sound in the crystal. For an order of magnitude estimate we continue to use this RESULT for HIGH`|vec(k)|`. Then we estimate `v` from the values of the models of elasticity and density `v~sqrt((E )/(rho))` We write `E~ 100GPa, rho= 8.9xx109^(3)kg//m^(3)` Then `v~3xx10^(3)m//s` Hence `ħ|vec(k)|_(max)~( ħ omega_(m))/(v)~ 1.5xx10^(-19) gm cms^(-1)` |
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| 45. |
A solenoid is connected to a battery so that a steady current flows through it. If an iron core is inserted into the solenoid, will the current increase or decrease? Explain. |
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Answer» Solution :The current will DECREASE because as the iron rod is inserted in the solenoid the magnetic field and hence the magnetic FLUX increase. According to Lenz.s law, the INDUCED emf RESIST this increase, which can be ACHIEVED by a decrease in current. |
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| 46. |
Four light sources produce the following waves (I) y_(1)= a sin (omega t+ phi_(1)) (II) y_(2)= a sin (2omega t) (III) y_(3) = a^(1) sin (omega t+phi_(2)) (IV) y_(4)= a^(2) sin(3omega t+phi) Superposition of which two waves gives rise to interferece |
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Answer» I and II |
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| 47. |
The graph of y = 5 is a line parallel to the |
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Answer» X- axis |
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| 48. |
A scooter, which is starting from rest moves with a con-stant acceleration for a time Deltat_(1) then with a constant velocity for the next Deltat_(2) and finally with a constant decelcration for the next Deltat_(3) to come to rest. A 500 N man sitting on the scooter behind the driver manages to stay at rest with respect to the scooter without touching any other part. The force exerted by the seat on the man is |
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Answer» 500 N THROUGHOUT the JOURNEY |
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| 49. |
Assume that two waves of light in air, of wavelength 585 nm, are initially in phase. One travels through a glass layer of index of refraction n_(1) = 160 and thickness L. The other travels through an equally thick plastic layer of index of refraction n_(2)= 1.50. (a) What is the smallest value L should have if the waves are to end up with a phase difference of 5.65 rad? (b) If the waves arrive at some common point with the same amplitude, is their interference fully constructive, fully destructive, intermediate but closer to fully constructive, or intermediate but closer to fully destructive? |
| Answer» SOLUTION :(a) 5.26 mm, (b) intermediate CLOSER to FULLY constructive | |
| 50. |
The electromagnetic waves used in the telecommunication are : |
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Answer» ultraviolet |
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