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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 400 mH coil of negligible resistance is connected to an AC circuit in which an effective current of 6 mA is flowing. Find out the voltage across the coil if the frequency is 1000 Hz. |
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Answer» SOLUTION :`L=400xx10^(-3)H,I_(eff)=6xx10^(-3)A,f=1000Hz` Inductive REACTANCE, `X_(L)=Lomega=Lxx2pif=2xx3.14xx100xx0.4=2512Omega` Voltage across L, `V=IX_(L)=6xx10^(-3)xx2512=15.072V`(RMS) |
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| 2. |
In the previous question if the current is I and the magnetic field at D has magnitude B, |
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Answer» `B = (mu_(0) i)/(2 sqrt(2) PI)` |
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| 3. |
What are the essential components in a tuning circuit? |
| Answer» SOLUTION :RESISTOR CAPACITOR and INDUCTOR | |
| 4. |
A horizontal wire AB which is free to move in a vertical plane and carries a steady current of i_(2) A, is in equilibrium at a height of h over another parallel long wire CD, which is fixed in a horizontal plane and carries a steady current of i_(1) A as shown. Show that when AB is slightly depressed and released it executes simple harmonic motion. Find the period of oscillations. |
Answer» Solution : Since upper wire is in EQUILIBRIUM, for this currents in wires should be UN opposite directions. Magnetic force per UNIT length between wires `(mu_(0)i_(1)i_(2))/(2pi h)`, upwards Let mass of unit length of wire `= lambda` Weight of upper wire per unit length `= lambda g` downward In equilibrium, `(mu_(0)i_(1)i_(2))/(2pi h) = lambda g`...(i) Let `AB`be SLIGHTLY depressed by `x`  Restoring force `F = -[(mu_(0)i_(1)i_(2))/(2pi(h - x)) - lambda g]` `= -[(mu_(0)i_(1)i_(2))/(2pi h(1 - x//h)) - lambda g]` `= -[(mu_(0)i_(1)i_(2))/(2pih)(1 - (x)/(h)) - lambda g]` `= - [lambda g(1 - (x)/(h))^(-1) - lambda g]` `= - lambda g[(1 - (x)/(h))^(-1) - 1]` ` = - lambda g[1 - (x)/(h) - 1] = (lambda g)/(h)x` (since `x ltlt h`) `a = (F)/(lambda) = -(g)/(h)x` `a alpha - x` (CASE of `S.H.M`), `(a = -omega^(2)x)` `T = 2pi sqrt((h)/(g))` |
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| 5. |
Glycerine (mu = 1.4) is poured into a large jar of radius 0.2m to a depth of 0.1 m. There is a small light source at the centre of the bottom of the jar. Find the area of the surface of glycerine through which light passes. |
| Answer» SOLUTION :`0.0327 m^2` | |
| 6. |
(A) : An electric field changing with time gives rise to a magnetic field and vice - versa. (R) : The displacement current is a source of magnetic field. |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the CORRECT EXPLANATION of 'A'. |
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| 7. |
What is Farady's first law of electromagnetci induction? |
| Answer» SOLUTION :Whenever MAGNETIC flux LINKED with a circuit changes, an e.m.f. is induced in it. | |
| 8. |
Three unequal resistors in parallel are equivalent to a resistance 1 ohm. If two of them are in the ratio 1:2 and if no resistance value is fractional, the largest of the three resistance in ohm is |
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Answer» 4 |
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| 9. |
The temperature at which r.m.s. velocity of oxygen molecules is equal to r.m.s. velocity of nitrogen molecules at N.T.P. is |
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Answer» `312^@K` |
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| 10. |
If the blue light is replaced by red light illuminating the object in a microscope, the resolving power of the microscope : |
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Answer» is INCREASED |
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| 11. |
A carnot engine first works between 100^@C and 0^@C and then between 0^@C and -100^@CWhat is the ratio of its efficiency in these two cases ? |
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Answer» 0.51 |
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| 12. |
A UFO flies directly over a football stadium at a speed of 0.50c. If the proper length of the field is 100 yards, what field length is measured by the crew of the UFO? |
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Answer» 59 YARDS |
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| 13. |
A magnet makes 30 oscillations per minute at a place where the field is 0.2xx10^(-4) Tesla , At another place it takes 1.5 sec to complete one oscillation . What is the field at this place. |
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Answer» |
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| 14. |
If the galvanometer reading is zero, in the given circuit. Then the value of 'X' will be |
| Answer» Answer :D | |
| 15. |
In the following circuit, 5Omega resistor develops 45 J/s due to current flowing through it. The power developed across 12Omega resistor is |
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Answer» 16 W |
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| 16. |
Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm. Approximately, what is the maximum distance at which these dots can be resolved by the eye ? [Take wavelength of light = 500 nm] |
Answer» Solution : `alpha_(MIN)=(1.22lambda)/(D)=(d_m)/(R)` `thereforeR=(Dxxd_m)/(1.22lambda)` `thereforeR=(3xx10^(-3)xx1xx10^(-3))/(1.22xx500xx10^(-9))` `thereforeR=4.918` m `THEREFORE` R=5 m |
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| 17. |
The two surfaces of a double concave lens are of curvature 10 cm and 40 cm. Find the focal length of the lens in water. The refractive index of water is 4//3 and that of the lens of the material is 3/2. |
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| 18. |
An electron ,an alpha-particle and a proton have the same kinetic energy .Which of these particles has the shortest de-Broglie wavelength? |
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Answer» Solution :Mass of electron `m_(E)`,mass of proton `m_(p)`, and mass of `alpha`-particle `m_(alpha)` `therefore m_(e)lt m_(p)ltm_(alpha)` `implies` Here KINETIC energy of all three is equal and relation between kinetic energy and momentum, `K=(p^(2))/(2m) therefore p=sqrt(2mK)` `implies` Now,the de-Broglie WAVELENGTH of particle, In `LAMBDA=(h)/(p)=(h)/(sqrt(2mK))`, h,2,k are equal `therefore lambda prop (1)/(sqrt(m))` Here mass of `alpha`-particle is maximum hence de-Broglie wavelength of `alpha` particle will be minimum. |
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| 19. |
The objective and eye piece of an astronomical telescope are double convexlenses with refractive jndex 1.5. When the telescope is adjusted to infinity the seperation between the lenses is 16 cm. If the space between the lenses is now filled with water and again telescope is adjusted for infinity. Then the present separation between the lenses is |
| Answer» Answer :D | |
| 20. |
If I_e is emitter current , I_b is base current and I_c is the collector current then : |
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Answer» `I_e =I_b -I_c` |
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| 21. |
A rectangular beam which is supported at its two ends and loaded in the middle with weight W sags by an amount delta such that delta=(Wl^(3))/(4Yd^(3)x), where l, d and Y represent length, depth and elasticity respectively. Guess the unknown factor using dimensional considerations : |
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Answer» BREADTH `:. x=(Wl^(3))/(4Yd^(3)delta)` `=([MLT^(-2)][L^(3)][L^(2)])/([MLT^(-2)][L^(3)][L])=[L].` Hence `(a)` is the CORRECT. |
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| 22. |
An ac generator G with an adjustable frequency of oscillation is used in the circuit, as shown Thermal energy produced by the resistance R in time duration 1 mus, using the source at resonant condition, is |
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Answer» 0J |
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| 23. |
A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by DeltaT in a time t. A number N of similar cells is now connected in series with a wire of the same material and cross - section but of length 2L. The temperature of the wire is raised by the same amount DeltaT in the same time. The volume of N is |
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Answer» Solution :In the first case `((3E)^(2))/(r).t=ms DeltaT"….(1)"` `[H=(V^(2))/(r).t]` When length of the WIRE is doubled, resistance and mass both are doubled THEREFORE, in the second case, `((NE)^(2))/(2R).t=(2m)sDeltaT"……..(2)"` Dividing Eq. (2) by Eq. (1) we get `(N^(2))/(18)=2 or N^(2)=36 or N=6` |
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| 24. |
The refractive index of glass is 1.5 the velocity of light in glass is: |
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Answer» `3xx10^8 m/sec` |
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| 25. |
Referring the circuit shown in Fig. match Column - I with Column - II : |
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Answer» |
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| 26. |
In a meter bridge, the null point is found at a distance of 40 cm from A. If now a resistance of 12 Omega is connected in parallel with S, the null point occurs at 60 cm from A. Determine the value of R. |
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Answer» `15OMEGA` |
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| 27. |
How many neutral points will be obtained when a bar magnet is kept with magnetic moment parallel to earth's magnetic field ? |
| Answer» ANSWER :D | |
| 28. |
A closed loop PQRS carrying a current is placed in a uniform magnetic field, If the magnetic forces on segments PS, SR, and RQ are F_(1), F_(2) and F_(1) respectively and are in the plane of the paper and along the directions shown, the force on the segment QP is |
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Answer» `F_(3)-F_(1)-F_(2)` |
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| 29. |
Find the limit of resolution of the human eye taking lambda = 6 xx 10^(-5) cm and diameter of the pupil as 2mm. |
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Answer» |
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| 30. |
Making use of the uncertainty principle,evaluate the minimum permitted energy of an electron in a hydrogen atom and its corresponding apparent distance from the nucleus. |
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Answer» Solution :HENCE we WRITE `R~Deltar,p~Deltap~ħ//Deltar` Then `E=(ħ^(2))/(2m r^(2))-(e^(2))/(r )` `=(1)/(2m)((ħ)/(r )-(me^(2))/(ħ))^(2)-(me^(4))/(2ħ^(2))` Hence `r_(eff)=(ħ^(2))/(me^(2))=53p m` for the equilibrium state. and then `E=(me^(4))/(2ħ^(2))= -13.6eV` |
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| 31. |
Why do you use a concave mirror for shaving purpose? |
| Answer» SOLUTION :A concave MIRROR produce erect and MAGNIFIED IMAGE when the object is PLACED between ole and focus. The principle helps for clean sharing | |
| 32. |
An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let lamda_(n),lamda_(g) be the de Broglie wavelength of the electron in the n^(th) state and the ground state respectively. Let A_(n) be the wavelength of the emitted photon in the transition from the n^(th) state to the ground state. For large n, (A, B are constant) |
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Answer» `A_(n)~~A+(B)/(lamda_(n)^(2))` but `K_(n)=(h^(2))/(2mlamdan^(2))` and `K_(g)=(h^(2))/(2mlamdag^(2))` `:.K_(g)-K_(n)=(h^(2))/(2m)[(1)/(lamda_(g)^(2))-(1)/(lamda_(n)^(2))]` but `K_(n)=-E_(n)` for emission of PHOTON and `K_(g)-E_(g)` `:.E_(n)-E_(g)=(HC)/(lamda_(n))` `K_(g)-K_(n)=(hc)/(A_(n))` `A_(n)=(hc)/(k_(g)-k_(n))=(hc)/((h^(2))/(2m)[(1)/(lamda_(g)^(2))-(1)/(lamda_(n)^(2))])` `A_(n)=(2mc)/(h[(lamda_(n)^(2)-lamda_(g)^(2))/(lamda_(n)^(2)lamda_(g)^(2))])=(2mclamda_(n)^(2)lamda_(g)^(2))/(h(lamda_(n)^(2)-lamda_(g)^(2)))` `A_(n)=(2mclamda_(g)^(2))/(h[1-(lamda_(g)^(2))/(lamda_(n)^(2))])=(2mclamda_(g)^(2))/(h)[1-(lamda_(g)^(2))/(lamda_(n)^(2))]^(-1)` `A_(n)=(2mclamda_(g)^(2))/(h)[1+(lamda_(g)^(2))/(lamda_(n)^(2))]` `A_(n)=(2mclamda_(g)^(2))/(h)+(2mclamda_(g)^(4))/(hlamda_(n)^(2))` `=A+(B)/(lamda_(n)^(2))` where `A=(2mclamda_(g)^(2))/(h)` and `B=(2mclamda_(g)^(4))/(h)` |
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| 33. |
Find the apparent depth of object of the object seenby observe A (in the figure shown) |
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Answer» |
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| 34. |
The velocities of light in two different media are 2 xx 10^8 m/s and 2.5 xx 10^8 m/s respectively. The critical angle for these media is |
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Answer» `SIN^(-1) (1/5)` |
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| 35. |
A : If the whole apparatus of Young.s experiment is immersed in liquid, the fringe width will decrease. R : The wavelength of light in water is more than that in air. |
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Answer» Both A and R are TRUE and R is the correct explanation of A |
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| 36. |
If a trolley filled with sand is sliding downwards on a smooth inclined inclined surface of inclination theta, then as the sand leaks out |
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Answer» the VELOCITY of the leaked out sand DECREASES. |
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| 37. |
Two condensers of capacities 4mF and 6mF are connected to two cells as shown. Then |
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Answer» P.D. of `6muF` condenser is 4V |
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| 38. |
A metal rod of resistance 20 Omega is fixed along a diameter of a conducting ring of radius 0.1m and lies on x-y plane. There is a magnetic field B =(50T)hatk. The ring rotates with an angular velocity omega=20 rad //s about its axis. An external resistance of 10 Omega is connected across the centre of the ring and rim. The current through external resistance is |
Answer» Solution : The equivalent circuit is  `E=(1)/(2)Bl^(2) omega =(1)/(2) XX 50 xx 0.1 xx 0.1 xx 20`, `:.` e = 5 V Hence the current through the EXTERNAL resistance is `i=(e)/(R)"":. i=(5)/(15)=(1)/(3)` AMP |
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| 39. |
WHO WROTE THE CHAPTER A PINCH OF SNUFF? |
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Answer» MANOHAR MALGAONKAR |
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| 40. |
The amplification factor of a triode is 50. If the grid potential is decreased by 0.20 V, what increase, in plate potential will keep the plate current unchanged ? |
| Answer» Answer :B | |
| 41. |
A small block m is placed at the top on the smooth inclined face AB of a wedge of mass M free to slide on a horizontal surface as shown in the figure and released from rest. (i) The total momentum of the block and the wedge is constant during motion. (ii) The total momentum of the block and the wedge along the horizontalis constant during motion. (iii) When the block reaches B the displacement of M to the left is(mh cot alpha)/(m+M). (iv) When the block reaches B the horizontal displacement of m to the right is (Mh cot alpha)/(m+M). |
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Answer» All are CORRECT EXCEPT (i) |
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| 42. |
Find the equivalent resistance of the circuit of the previous problem between the ends of an edge such as a and b in figure. |
Answer» Solution :Suppose a current `i` enters the circuit at the point a and the same current leaves the circuit at the point b. The current distribution is shown is figure.  The PATHS through .AD. and ah are equivalent and carry equal current `i_(1)`. The current through ab is then `i-2i_(1)`. The same distribution holds at the junction b. Currents in eb and cb are `i_(1)` each. The current `i_(1)` in ah is divided into a part `i_(2)` in he and `i_(1)-i_(2)` in hg. Similar is the division of current `i_(1)` in ad into dc and dg. The rest of the currents may be written easily using Kirchhoff.s junction law. The potential difference V between a and b may be written from the paths ab, aheb and ahgfcb as `V=(i-2i_(1))R, V=(i_(1)+i_(2)+i_(1))r` and `V=[i_(1)+(i_(1)-i_(2))+2(i_(1)-i_(2)+i(i_(1)-i_(2))+i_(1)]r` Which may be written as `V=(i-2i_(1))r, V=(2i_(1)+i_(2))r and V=(6i_(1)-4i_(2))r` Eliminating `i_(1) and i_(2)` from these equations, `(V)/(i)=(7)/(12)r` `therefore"The equivalent resistance "=7r//12` |
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| 43. |
Explain polarisation of light by reflection and write the Brewster's law and get the formula. |
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Answer» <P> Solution :A scientist named Malus discovered that the reflected RAY is partically plane polarised when light rays incident on the surface of the transparent medium and the refracted ray is also partially polarised.State of polarization of reflected ray depends on angle of incidence.  When a ray of light is incident on a surface of transparent medium at some definite angle of incidence, reflected ray is found to be totally plane polarized. In this state all the light vectors in the reflected ray of light are parallel to each other and perpendicular to the plane of incidence. This is called `sigma` components and since it contains 15% component of incidence ray, hence the reflected ray is fade. In refracted rays there are 85% `sigma` components of incident ray and are all tt components and hence it slightly fade compare to incident ray. When a ray of light is incident on a surface of transparent medium at some define angle of incidence, reflected ray is found to be totally plane polarised. Such an angle of incidence is called angle of polarisation of the given transparent medium. This angle is denoted by `i_(B)` or `theta_(P)`. From Smells. law at a given point, `mu=(sini_(B))/(sinr)" but "r=(pi)/(2)-i_(B)` `:.mu=(sini_(B))/(sin((pi)/(2)-i_(B)))=(sini_(B))/(cosi_(B))` `:.mu=tani_(B)` This is called Brewster.s law. In WORDS, when a ray reflected from a surface of transparent object is totally plane polarized the TANGENT of the angle of incidence is equal to the refractive index of the material of the object. |
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| 44. |
Graphs are very useful to represent a physical situation. Various quantities can be easily represented on graphs and other quantites can be determined from the graph. For example the slope of velocity-time graph represents instantaneous acceleration. For a motion with constant acceleration slope of velocity-time graph is constant. If acceleration is changing with time, slope with change and thus velocity-time graph will be a non-linear curve. Further the area of velocity-time graph gives displacement. The average velocity of the car for the entire trip is |
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Answer» `v_(0)` |
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| 45. |
(A) : Reduction factor (K) of a tangent galvanoinete helps in reducing deflection for a given current. (R) : Reduction factor has no unit. |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the correct explanation of 'A'. |
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| 46. |
Graphs are very useful to represent a physical situation. Various quantities can be easily represented on graphs and other quantites can be determined from the graph. For example the slope of velocity-time graph represents instantaneous acceleration. For a motion with constant acceleration slope of velocity-time graph is constant. If acceleration is changing with time, slope with change and thus velocity-time graph will be a non-linear curve. Further the area of velocity-time graph gives displacement. The correct relation between v_(0),a_(1),a_(2) and t_(0) is |
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Answer» `v_(0)=((a_(1)a_(2))/(a_(1)+a_(2)))t_(0)^(2)` |
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| 47. |
What is meant by .detection. of a modulated carrier wave? Descibe briefly the essential steps for detection. |
Answer» Solution :The transmitted signal loses energy travelling through a medium (attenuation). The receiver consists of an amplifier which amplifies the modulated carrier wave. The LOCAL oscillator produces a variable frequency range so that on selection of broadcast signal frequency, the receiver produces a beat frequency called intermediate frequency which remains a constant irrespective of broadcast FREQUENCIES. The detector separates the carrier frequency from the low frequency which is then sufficiently AMPLIFIED before CONNECTING to the TRANSDUCER. 
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| 48. |
Compare the radii of two nuclei with mass number 1 and 27 respectively. |
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Answer» |
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| 49. |
In nuclear fission process , energy is released because |
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Answer» mass of PRODUCTS is more than mass of nucleus |
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