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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 angle of deviation (delta) vs angle of incidence (i) is plotted for a prism. Pick up the correctstatements. |
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Answer» The angle of PRISM is `60^@` |
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| 2. |
The element curium ""_(96)^(218)C_(m) has a mean life of 10^(13) second. Its primary decay modes are spontaneous fission and alpha- decay, the former with a probability of 8% and the latter with a probability of 92% Each fission releases 200MeV energy. The masses involved in the alpha- decay are as follows ""_(98)^(248)Cm=248.07220" amu ",""_(94)^(244)=244.064100" amu ",""_(2)^(4)He=4.002603" amu ",1" amu "=931MeV//e^(2) |
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Answer» The activity of FISSION `R=10^(7)` |
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| 3. |
By increasing temperature, the specific resistance of a conductor and a semi-conductor |
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Answer» INCREASE for both |
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| 4. |
A doubleslit interference pattern is producedon a screen , as shown in the figure , usingmonochromaticlight of wavelength500nm , Point P is the locationof the central bright fringe , that is produced when light waves arrives in phase withoutany path difference .A choice of the three stripsA,B and C of transparent materialswith different thickness and refractive indicates is available , as shown in the table . these are palcedoverone or both of the slits , singularityor in conjunction, causingthe interfernece pattern to be shiftedacross the screen from the origina pattern ,In the Column -I , how the stripshave been placed, is mentionedwhereas in the Clumn - II , order of the fringe at point P on the secreenthat will be produced due tothe placementof the strip (s) is shown. Correctly match both the column . {:("Film",A,B,C),("Thickness in" mu,5,1.5,0.25),("Refractive Index",1.5,2.5,2):} |
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| 5. |
An object is approaching a fixed plane mirror with velocity 3 m/s at an angle of 45^@ with normal, in a medium of refractive index 4/3. The speed of the image with respect to the mirror is ___ m/s. |
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| 6. |
A wooden cube (density of wood 'd') of side 'l' floats in a liquid of density 'rho' with its upper and lower surfaces horizontal. If the cube is pushed slightly down and released, it performs simple harmonic motion of period 'T'. Then, 'T' is equal to : |
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Answer» `2pi SQRT((ld)/(rho g))`  Restoring force, `F=mg-F_(B).` br> `F="mg"-rhoA(l_(0)+x)g` `DA l a=dAlg-rho A l_(0)g-rho g A x` `a=-(rho g)/(d l)x` `omega=sqrt((rho g)/(d l))` `T=2pi sqrt((l d)/(rho g))` So, correct CHOICE is (a). |
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| 7. |
Assume the dipole model for earth's magnetic field B which is given by B_V= vertical component of magnetic field = (mu_0)/( 4pi ) ( 2m cos theta )/( r^3) B_H= Horizontal component of magnetic field B_H= (mu_0)/( 4pi ) (m sin theta )/( r^3) theta =90^@ - latitude as measured from magnetic equator. (a) Find loci of points for which (i) |overset(to)(B) | is minimum, (ii) dip angle is zero, (iii) dip angle is pm 45^@. |
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Answer» Solution :(a) Given that, `B_V = (mu_0)/( 4pi ) (2m cos THETA)/( r^3) ""…(1)` `B_H = (mu_0)/( 4pi ) (m sin theta )/( r^3) ""…(2)` Squaring and ADDING equation (1) and (2), `B_(V)^(2) +B_(H)^(2) = ((mu_0)/( 4pi ))^(2) (m^2)/( r^6) [ 4 cos ^(2) theta + sin^(2) theta]` `therefore B^(2) = ((mu_0)/( 4pi ))^(2) (m^2)/( r^6) [ 4 cos^(2) theta +1 - cos^(2) theta ]` `therefore B^(2) = sqrt((B_(V)^(2) + B_(H)^(2) ) ` `= sqrt(((mu_0)/( 4pi ))^(2)(m^2)/(r^6) [ 3 cos^(2) theta +1] )` `= (mu_0)/( 4pi ) (m)/(r^3) [ 3 cos theta^(2) theta + 1]^(1//2) ""...(3)` From above equation, the value of B is minimum if `cos theta = 0`, hence `theta = (pi)/(2)` Thus, B is minimum at the magnetic equator. (b) For angle of dip, `TAN delta = (B_V)/( B_H) = (2 ((mu_0)/( 4pi )(m cos theta )/( r^3) ) )/( ( (mu_0)/( 4pi) (m sin theta )/( r^3) ) )` `-2 ( cos theta)/(sin theta)` `(B_V)/( B_H) = tan delta = 2 cot theta ""...(4)` For dip angle `delta = 0`, then `cottheta =0` `therefore theta= (pi)/(2)` Hence locus is on magnetic equator. (c) `tan delta = (B_V)/( B_H)` For `delta = pm 45^@` `(B_V)/( B_H) = tan (pm 45^@)` `(B_V)/( B_H) = 1` `therefore 2 cot theta =1 ""` [From equation (4)] `therefore cot theta = (1)/(2) ` ` therefore tan theta = 2` `therefore theta= tan^(-1) (2)` is the locus. |
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| 8. |
(a) For a given a.c. I = I_(m) sin omega tshow that the average power dissipated in a resistor over a complete cycle is R1/2 I_(m)^(2)R (b) A light bulb is rated 100 W for a 220 V a.c. supply. Calculate the resistance of the bulb. |
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Answer» Solution :(a) When an alternating current flows in the circuit is given by `I = I_(m) sin omega t` `therefore`AVERAGE power dissipated per COMPLETE cycle of a.c. is given by `P_(av) = 1/T int_(0)^(T) V I dt = 1/T int_(0)^(T) V_(m) sin omegat. I_(m) sin omega t dt = (V_(m) I_(m))/T int_(0)^(T) sin^(2) omega Tdt` `(V_(m)I_(m))/(2) = 1/2 I_(m)^(2)R [ therefore V_(m) = RI_(m))`] (b) Here P = 100 W, `V_(RMS) = 220V`. As `P= I_(rms)^(2) R = (V_(rms)^(2))/R` `rArr` Reactance `R = (V_(rms)^(2))/P = (220)^(2)/100 = 484 Omega` |
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| 9. |
The half life of a cobalt radio istope is 5.3 years. What strength will a milli curic source of isotope have after a period of one year? |
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Answer» 0.10 mci `N/N_(0) =R. =e^(-lambda//t) or log_(10)(N)/(N_(0)) =(lambda t)/(2.3026)` `or log_(10) (1/R.)=(0.693 xx t)/(2.3026 xx T)=(0.693 xx 1)/(203026 xx 5.3) [lambda=(0693)/(T)]` `or 1/(R.)=1.14 THEREFORE R.=(1)/(1.14) =0.87 m Ci` |
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| 10. |
Internal resistance of cell having emf 24 V is 0.12 Omega. If cell is joined with external resistance 3 Omega, so terminal voltage of cell is ....... |
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Answer» 23.08 V In `EPSILON = V + IR` I = `(epsilon)/(R + r) = (24)/(3.0 + 0.12) = (24)/(3.12) = 7.6 `A V = `epsilon - Ir` =` 24 - 7.6 XX .12` = 24 - 0.912 23.08 V |
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| 11. |
An elevator and it's load weight a total of 1600 kg. If the elevator originially moving downward at 20 m/sec is brought to rest with constant acceleration in a distance of 50 m, then tension T is how much |
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Answer» 1800 poundals |
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| 12. |
In an adiabatic change the, pressure and temperature of a monoatomic gas are related as PalphaT^c What is the value of c ? |
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Answer» `5/2` |
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| 13. |
When a charged capacitor is made to discharge through an inductor, the charge Q on the capacitor and the current i in the inductor vary sinusoidally. The phase difference between Q and i is |
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Answer» `PI` |
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| 14. |
If the critical angle for total internal reflection from a medium to vaccum is 30^(@), the speed of light in the mediu is |
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Answer» `3xx10^(8)ms^(-1)` `therefore"Speed of light in the medium V"=(c)/(n)=(3xx10^(8))/(2)=1.5xx10^(8)ms^(-1)` |
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| 15. |
An infinite number of electric charges each equal to 5 nano - coulombs (magnitude) are placed along Xaxis at x = 1 cm, x = 2 cm, x = 4 cm, x = 8 cm ............... and so on. In this setup if the consecutive charges have opposite sign, then the electric field at x = 0 ((1)/( 4 piepis_0) = 9 xx 10^9N-m^2 //C^2) |
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Answer» `12xx 10^4N//C` |
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| 16. |
Pick out the statement which is incorrect |
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Answer» Field lines never intersect |
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| 17. |
A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads 49 N, when the lift is stationary. If the lift moves downward with an acceleration of 5 m//s^(2), the reading of the spring balance will be : |
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Answer» 49 N When the lift is STATIONARY, then `49 = m xx 9.8` or `m= 5 KG` Hence `R = 5(9.8 - 5) = 24 N` Hence choice is (b). |
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| 18. |
In double refraction |
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Answer» only the 0-ray is polarised |
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| 19. |
In the experiment of interference of light the intensity at a point where the path difference is lambda is I. The intensity at a point where the path difference is lambda/3, is: |
| Answer» Answer :A | |
| 20. |
Copper and Germanium are cooled from room temperature to 100 K , Then the resistance of |
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Answer» GERMANIUM INCREASES , COPPER increases |
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| 21. |
A radio transmitter, microwave transmitter and a microwave receivers are placed in a hilly area. In this situation, what are your suggestions for proper communication? Are Microwaves suitable there? |
| Answer» Solution :To have LARGE coverage, INCREASE the HEIGHT of antenna or stations. Microwaves are not SUITABLE for this kind of transmission. | |
| 22. |
The number of spectral lines in hydrogen atom are |
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Answer» infinite |
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| 23. |
Consider a transparent homogenous hemisphere of refractive index n=2 in front of which a small object is placed in air as shown in figure. |
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Answer» if value of `x=3R` then the final image of the small object OA will be real & inverted. `(2)/(V)-(1)/(-3R)=(2-1)/(R)implies(2)/(V)=(1)/(R)-(1)/(3R)` `(2)/(V)=(2)/(3R)impliesV=3R` for second surface `u=2R` `(1)/(V)=(2)/(2R)impliesV=R` Real & inverted. (B) for first surface `(2)/(V)-(1)/(-2R)=(2-1)/(R)implies(2)/(V)=(1)/(R)-(1)/(2R)` `impliesV=4Rimpliesm_(1)=-1` for second surface `u=3R` `(1)/(V)=(2)/(3R)impliesV=(3R)/(2)` real & inverted and same SIZE. 
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| 24. |
A thin rod of copper is being moved in a uniform magnetic field such that it is cutting the magnetic field lines. An emf is induced between the ends of the rod. |
| Answer» SOLUTION :TRUE - Motional emf is INDUCED between the ends of rod whose magnitude is `lvarepsilonl = BLV`. | |
| 25. |
A road runs midway between two parallel rows of buildings. A motorist moving with a speed of 36 Km/h sounds the horn. He hears the echo one second after he has sounded the horn. Then the distance between the two rows of buildings is.(Velocity of sound in air is 3330 m/S) |
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Answer» `80sqrt(17)m` |
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| 26. |
What is the length of the telescope when it forms the final image at infinity? |
| Answer» Solution :The objective LENS collects parallel rays from a DISTANT object. So, it forms the image at the FOCUS. `f_0` of the objective. As the final image is at infinity, the image formed by the objective must be at the focus `f_0` of the EYE PIECE. In other words when final image is formed at infinity, `f_0` and `f_e` coincide. Therefore, length of the tube of the telescope = `f_0+f_e` where `f-0` and `f_e` are focal lengths of objective and eye piece respectively. | |
| 27. |
When a small 2.0 g coin is placed at a radius of 5.0 cm on a horizontal turntable that makes three full revolution is in 3.14s, the coin does not slip. What are (a) the coin's speed, the (b) magnitude and (c) direction (radially inward or outward) of the coin's acceleration, and the (d) magnitude and (e) direction(inward or outward) of the frictional force on the coin? The coin is on the verge of slipping if it is placed at a radius of 10 cm. (f) What is the coefficient of static friction between coin and turntable? |
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| 28. |
Show mathematically that the rotation of a coll in a magnetic field over one rotation Induces an alternating emf of one cycle. |
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Answer» Solution :Induction of emf by changing relative orientation of the coil with the magnetie FIELD: Consider a rectangular coil of N turns kept in a uniform magnetic field `vecB` figure (a). The coil rotates in anti-clockwise direction with an angular VCLOCITY o about an axis, perpendicular to the field. At time =0, the plane of the coil is perpendicular to the field and the flux linked with the coil has its maximum value `phi_(m)` = BA (where A is the area of the coil).  In a time t seconds, the coil is ROTATED through an angle `theta (=omegat)` in anti-clockwise direction. In this position, the flux linked is `phi_(m) cos omegat,` a COMPOUND of `phi_(m)` normal to the plane of the coil figure (b)). The component parallel to the plane `(phi_(m) sin omegat)` has no role in electromagnetic induction. Therefore, the flux at the flux linkage linkage at this deflected position is `N phi_(B)=N phi_(m) cos omegat`. According to Faraday.s law, the emf inducded at that instant is   When the coil is rotated through `90^(@)` from initial position, `sin omegat=1`. Then the maximum value of induced emf is `epsi_(0) m=N phi_(m)omega= NBA omega "since "phi_(m)=BA` Therefore, the value of induced emf at that instant is then GIVEN by `epsi=epsi_(m) sin omegat` It is seen that the induced emf varies as sine function of the time angle wt. The graph between induced emf and time angle for one rotation of coil will be a sine curve and the emf varying in this manner is called sinusoidal emf or alternating emf. |
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| 29. |
If a half-wave rectified voltage is fed to a load resistor , which part of a cycle the load current will fl ow ? |
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Answer» `0^(0)-90^(0)` |
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| 30. |
The rate at which the wave transports energy is : |
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Answer» Solution :Power `P = (1)/(2)muomega^(2)A^(2)V` = `(1)/(2) xx 0.525 xx (754)^(2) xx (8.5 xx 10^(-3))^(2) xx 9.25` = 100 W |
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| 31. |
INSAT-IA satellite is used for : |
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Answer» RADIO communication |
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| 32. |
Differentiate between a ray and a wavefront. |
| Answer» SOLUTION :A wavefront is a SURFACE of constant phase. A light ray is a STRAIGHT line path drawn perpendicular to the wavefront. | |
| 33. |
The magnetic dipole moment for circular coil having n turns and carring current I "nIA". The factor --------- in this expression is called ampere turns. |
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Answer» nI |
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| 34. |
Two particles are thrown horizontally in opposite directions from the same point from a height h with velocities 4 ms^(-1) and 3 ms^(-1) . What is the separation between them when their velocities are perpendicular to each other? |
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Answer» Solution :Let us choose a coordinate system with x-axis along horizontal and the y-axis along vertical direction. Then, USING EQUATION `vec v = vec u + vec at`, we GET velocities as `vec v_(1) =4 hat i-gt hat j and vec v_(2)=3 hati -gt hat j`. For velocities to be perpendicular `vec v_(1) cdot vec v_(2) =0` i.e., `(4hat i -g hati j) =0" or "12- g^(2)t^(2)=0` `rarr" "t= sqrt((12)/(g^(2)))=0.35 s` In vertical direction both have same initial velocities and THUS has no separation. Hence, there is only horizontal separation. Therefore, separation `=(v_(1)+v_(2))t =7(0.35) =2.45 m`. |
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| 35. |
In the above problem, which wavelength (s) will have a strong intensity ? |
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Answer» 400 nm |
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| 36. |
The current sansitivity of a moving coil galvanometer can be increased by reducing the ____________. |
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Answer» |
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| 37. |
Flute is what? |
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Answer» an open organ PIPE |
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| 38. |
Differentiate between evaporation and boiling. |
| Answer» Solution :EVAPORATION is a slow change from the LIQUID to the gaseous STATE which takes at the surface of a liquid and at all temperatures. Boiling is the rapid change of a substance from the liquid to the gaseous state which takes PLACE throughout the mass of the liquid at a definite temperature. | |
| 39. |
An electron starting from rest takes14 xx 10 ^(-9)secto reach from one plate to other of a capacitor placed 2 cm apart. If charge to mass ratio of electronis1.8 xx 10 ^(11) CI kg. Then find the potential difference between the plates. |
| Answer» SOLUTION : ` V = 2400 `VOLT | |
| 40. |
Energy generation in stars is mainly due to |
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Answer» CHEMICAL reactions |
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| 41. |
Obtain an expression for torque acting on a rectangular current loop. |
Answer» Solution :Let a and b represent breadth and length of a rectangular loop.  Let I be the current in the coil. Let m be the MAGNETIC moment, `tau` be the torque on the coil and B by the magnetic flux linked with the coil. Let B and s represent brush and slip ring. By applying Fleming's LEFT Hand Rule, we note that force on AB and CD remains the same but act opposite to each other. These two and opposite forces constitute couple. DEFLECTING couple or torque is given by the product of magnitude of any one force and perpendicular distance between the forces. Torque, `""tau=Fa` where, `""BIbsintheta=F " on " AB " or " CD.` `""=(BIb)asintheta=IB(ab)sintheta` i.e., `""tau=IBAsintheta ""` where area of the loop = A = ab If `theta=0, sintheta=0` then torque `tau=0` But magnetic moment of the current loop ABCD = IA = m Hence torque on the current loop `=tau=mBsintheta` i.e., `""vec(tau)=vec(m) TIMES vec(B)` Note : As an analogue to torque on electric dipole `""vec(tau)=vec(p)_(e) times vec(E)` where, `p_(e)` - dielectric moment, `vec(E)=` Electric field 
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| 42. |
The shape of interference fringe obtained on screen by using monochromatic light in Young's double slit experiment is ...... |
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Answer» PARABOLIC |
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| 43. |
In a semiconductor, free electron desnity is about |
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Answer» `10^(6)//m^(3)` |
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| 44. |
A steel wire of length 1 m and diameter 0.2mm is elongated by 1mm due to a weight of 3.14kg. The Young's modulus of steel wire is |
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Answer» a)`9.8 XX 10^9 N/m^2` |
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| 45. |
Explain the significance of negative energy of an electron in an orbit. |
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Answer» |
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| 46. |
Does the current in an a.c circuit lag lead or remain in phase with the applied voltage. When (i)gamma=gamma_(r)(ii)gammaltgamma_(r)(iii)gammagtgamma_(r)." Where "gamma_(r) is the resonant frequency ? |
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Answer» Solution :i. `gamma=gamma_(R)" occurs when "X_(L)=X_(C).` Then the circuit becomes purely RESISTIVE. So current and voltage will be in same PHASE. ii. `X_(L)=2pigammaLandX_(c)=(1)/(2pigammac)` When `gammaltgamma_(r),X_(L)" is small and "X_(C)` is large the circuit is capacitive, so current leads the voltage in phase. The circuit is inductive. So current LAGS behind the VOTAGE in phase. |
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| 47. |
A magnetic field B = 2t + 4t^(2) (where t = time) is applied perpendicular to the plane of a circular wire of radius and resistance R. If all the units are in SI the electric charge that flows through the circular wire during t = 0s to t = 2 s is |
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Answer» `(6pir^(2))/R` At t = 0, initial magnetic field `B_(i) = 0T` At t= 2 s, final magnetic field `B_(f)= 2(2) +4(2)^(2) = 20T` As `q= (Delta) q = (Delta phi)/R=(pi r^(2)(B_(f)-B_(i)))/R = (pir^(2)(20-0))/R = (20pir^(2))/R` |
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| 48. |
In the previous question, if the bird is diving vertically down with speed =6m/s , his appearent speed as seen by a stationary fish underwater is |
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Answer» 8m/s |
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| 49. |
If the current flowing through a coil is reduced by 50%, then what is the energy in the coil? |
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Answer» SOLUTION :`E_2/E_1 = 1/2L(I/2)^2//1/2LI^2 = 1/4` `THEREFORE E_2 = 1/4E_1` `therefore E_1-E_2= 3/4E_1` |
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