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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 bar magnet of magnetic moment M is cut into two parts of equal lengths. The magnetic moment of each point will be : |
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Answer» zero |
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| 2. |
Note : Fill in the blanks with appropriate answer : Horizontal and vertical components of earth's magnetic field at a place are equal. The angle of dip at that place is _). |
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Answer» |
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| 3. |
In the given if point C is connected to the earth and a potential of +2000 V is given to the point A, the potential at B is |
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Answer» 1500 V |
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| 4. |
A potential barrier of 0.50 V exists across a p-n junction. a) If the depletion region is 5.0xx10^(-7)m wide, what is the intensity of the electric field in this region ? b) An electron with a speed of 5.0xx10^(5)m//s approaches the p-n junction from the n-side. With what speed will it enter the p-side ? |
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Answer» Solution :(a) The electric field is `E=V//d` `=(0.50V)/(5.0xx10^(-7)m)=1.0xx10^(6)V//m` (b)  Suppose the electron has a speed `v_(1)` when it enters the depletion layer and `v_(2)` when it comes out of it. As the potential energy increases by e x 0.50 V, from the PRINCIPLE of conservation of energy. `(1)/(2)mupsilon_(1)^(2)=exx0.50+(1)/(2)mupsilon_(2)^(2)` `(1)/(2)(9.1xx10^(-31))(5XX10^(5))^(2)` `=1.6xx10^(-19)xx0.5+(1)/(2)xx9.1xx10^(-31)xxupsilon_(0)^(2)` Solving this, `upsilon_(2)=2.7xx10^(5)m//s`. |
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| 5. |
The kinetic energy of a body is increased by 300 % .The momentum of the body would increase by |
| Answer» ANSWER :B | |
| 6. |
A boat which has a speed of 5 km/hr in still water crosses a river of width 1 km along the shortest path in 15 minutes. The velocity of the river water in kilometre per hour is |
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Answer» 1 |
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| 7. |
If the surface tension of a rectangular film of soap solution is T and length is l, then the total force acting on it is, |
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Answer» `2T//l` |
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| 8. |
Fill in the blank. _15^32P to .....+overline(e)+overline(v) (anti neutrino) |
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Answer» `_7^14N` |
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| 9. |
In which of the following cases do we obtain a spherical wave fron ? a) sunlight focussed by a convex lens b) light diverging from a straight slit c) light emitted by a point source in an isotropic medium d) a parallel beam of light reflected from a plane mirror |
| Answer» Answer :D | |
| 10. |
A 4.5cm needle is placed 12cm away from a convex mirror of focal length 15cm. Given the location of the image and the magniftication. |
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Answer» `6.7cm, 5//9` (focal length of convex mirror is taken as positive) DISTANCE of OBJECT `u=-12cm` Size of object `O=4.5cm` using the mirror formula `1/f=1/v+1/u` `1/15=1/v-1/12` `implies1/v=1/15+1/12=(4+5)/60=9/60`  Distance of image from the mirror `v=6.7cm` The positive sign SHOWS that the image is formed behind the mirror. Using the formula of magnification, `m=-v/u=L/O` `(-6.7)/(-12)=l/4.5` Size of image `l=2.5cm` As `l` is positive so image is erect and virtual. Magnification `m` is given by `m=l/O=2.5/4.5=25/45=5/9` |
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| 11. |
Statement - (A): The difference between a new torch light cell and an old one is due to increase in internal resistance. Statement - (B): At 0 kelvin specific resistance of perfect insulator is infinity. |
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Answer» Both A and B are TRUE |
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| 12. |
For a parallel beam of monochromatic light wavelength lambda diffraction is produced by single slit whose width 'a' is of the order of th wavelength of the light. If 'D' is the distance the screen from the slit, the width of th central maxima will be ...... |
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Answer» `(2Dlambda)/(a)` `:.` Condition for first minimum. In `d SIN theta= n lambda, d=a, n=1` `:.sin theta=(lambda)/(a)` `:.` Linear width of central maximum =`D|2 theta|` `=(2Dlambda)/(a)"" [ :.` for SMALL angle `sin theta ~~ theta]` |
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| 13. |
An alternating e.m.f. given by e = 200 sin 50 t is applied to a circuit containing only a resistance of 50 Omega. What is the value of r.m.s. current in the circuit ? |
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Answer» 2.828 A |
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| 14. |
Three resistors 1 Omega, 2 Omega and 4 Omega are combined in series. What Is the total resistance of the combination ? |
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Answer» Solution :EQUIVALENT resistance of series connection of given RESISTORS, `R_(s) = R_(1) + R_(2) + R_(3)` ` = 1 + 2 + 3 ` `THEREFORE R_(s) = 6 Omega` |
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| 15. |
A physics teacher tells his students in class that in paramagnetic materials, every atom has some permanent magnetic dipole moment. In the absence of an external magnetic fund, the atomic dipoles are randomly oriented so that average magnetic moment per unit volume of the material is zero. Therefore, in the absence of an external magnetic field, a paramagnetic material does not behave as a magnet. When an external magnetic field is applied, the torque developed tries to align the atomic magnetic dipoles in the direction of the field. That is why the specimen gets magnetised weakly in the direction of the field. Read the above passage and answer the following questions: (i) Name any three paramagnetic materials. (ii) Name any two ferromagnetic materials. How is their behaviour different from that of paramagnetic materials? (iii) The teacher asks the students how true is the famous saying: 'Spare the rod and spoil the child,' comment. |
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Answer» Solution :(i) Example of paramagnetic materials are Aluminium, chromium, oxygen. (II) Iron and cobalt are TWO ferromagnetic materials. The ferromagnetic substances show all the properties of paramagnetic substances, but to a much larger degree. For example, relative magnetic permeability `(mu_r)` of paramagnetic substances is SLIGHTLY greater than 1, whereas `mu_r` of ferromagnetic substances is much larger than 1. (iii) The saying SEEMS to be true. In the absence of external magnetic field (say, teacher) atomic DIPOLES (say, students) are randomly oriented (sitting helter skelter and gossiping). When external magnetic field is applied (teacher comes to class), atomic dipoles are aligned in the direction of the field (students orient themselves towards the teacher). Yet another proof of the same is that when a magnet is suspended freely from an unspun silk thread, it comes to rest along N-S direction only. This is because earth is behaving as a magnet. It disciplines the magnet forcing it to say along N-S direction always. |
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| 16. |
Self induction is the electrical analogue of |
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Answer» force |
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| 17. |
ABCD is a square of side 'a' metres and is made of wires of resistance x ohms//metre. Similar wires are connected across the diagonals AC and BD. The effective resistance between the ax corners A and C will be |
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Answer» `(2 - SQRT2) AX` |
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| 18. |
The duration of day is highest at : |
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Answer» EARTH THUS CORRECT choice is (c ). |
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| 19. |
Three component sinusoidal waves progressing in the same direction along the same path have the same period but their amplitudes are A, A/2 and A/3. The phases of the variation at any position x on their path at time t = 0 are 0, -pi/2 and -pi respectively. Find the amplitude and phase of the resultant wave. |
| Answer» SOLUTION :`5/6 A, -TAN^(-1) (3/4)` | |
| 20. |
……as a impurity, when added in Si or Ge P- type semiconductor is obtained. |
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Answer» Arsenic |
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| 21. |
Using elimination method, If 2x + 3y = 12 and 3x − 2y = 5 then |
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Answer» x = 2 ,y = 3 |
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| 22. |
What is the maximum wavelength for Balmer series in H atom. |
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Answer» SOLUTION :For MAXIMUM WAVE LENGTH of Balmer series `n_(2)=3,n_(1)=2` `(1)/(lamda)=R(1)^(2)[(1)/(4)-(1)/(9)]` On solving `lamda=6566.4A^(@)` |
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| 23. |
_____ has maximum resistance. |
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Answer» Ammeter |
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| 24. |
(a) If R = 15 cm, M = 350 g, and m = 50 g in Fig. 10-19, find the speed of the block after it has descended 50 cm starting from rest. Solve the problem using energy conservation principles. (b) Repeat (a) with R = 5.0 cm. |
| Answer» Solution :(a) `1.5xx10^(2)cm//s`, (B) `1.5xx10^(2)cm//s` = 1.5 m/s, SINCE it is independent of R. | |
| 25. |
A point mass m and charge Q is connectedwitha spring of negligible mass. Initially spring is in its naturallength L. Now a horizontaluniform field E is switched on as shown. Find (a) the maximumseperation between the mass and the wall (b) Find the seperation of the point mass andwall at theequilibriumposition of mass (c ) Find the energy storedinthe spring at theequi librium position of the point mass. |
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Answer» Solution :At maximum seperation,velocity of POINT mass is zero. From work energy theorem, `W_("spring") +W_("field")=0` `qEx_(0) -(1)/(2) Kx_(0)^(2) =0 ` (`x_(0)` is maximum elongation) `rArr x_(0)=(2qE)/(K)` `therefore` seperation `=L+(2qE)/(K)` (b) At EQUILIBRIUM position, `Eq=KX rArr x =(qE)/(K)` `rArr ` seperation `=L+(qE)/(K)` (c ) `U=(1)/(2) Kx^(2) =(1)/(2) K((qE)/(K))^(2) =(q^(2)E^(2))/(2K)` |
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| 26. |
Assertion:The speed of light in a rarer medium is greater than that in a denser medium Reason: One light year equals to 9.5 × 10^(12) km |
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Answer» If both assertion and reason are true and the reason is the CORRECT EXPLANATION of the assertion |
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| 27. |
What is atomic nucleus? |
| Answer» Solution :The CENTER core of the ATOM which CONTAINS all of the ATOMS positive charge and most of it.s mass is called atomic nucleus. | |
| 28. |
Read the following passage and then answer questionson the basis of your understanding of the passage and the related studied concepts. A small township with a demand of 880 kW of electric power at 220 V is situated 20 km away from a mini hydropower generating station generating power at 440 V. The resistance of the two wire line carrying power is 0.5 ohm per km. The town gets power from the line through a 11 kV-220 V step down transformer at a substation at the township. If the electric substation of township starts supplying power to the township, using a 33 kV - 220 V step down transformer then estimated value of line power loss in the form of heat will change to |
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Answer» 384 kW [Hint: If line voltage is V = 33 kV, then line current `I. = P/(V.) = (880 kW)/(33 kV) = 80/3 A` `therefore` Line power LOSS `=I^(2)R =(80/3)^(2) xx 20 = 14200 W = 14.2 kW` |
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| 29. |
Two large metal sheets having surface charge density +sigma and –sigma are kept parallel to each other at a small separation distance d. The electric field at any point in the region between the plates is |
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Answer» `SIGMA/epsilon_0` |
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| 30. |
Suppose that the lower half of a concave mirror's reflecting surface is covered with an opaque non-reflecting material. What effect will this have on the image of an object placed in front of the mirror ? |
| Answer» Solution :TAKING the laws of reflection to be true for all POINTS of the remaining (UNCOVERED) PART of the mirror, the image will be that of the whole object. As the area of the reflecting surface has been reduced, the intensity of the image will be low (in this case HALT). | |
| 31. |
The angular resolution of the telescope is determined by the |
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Answer» image PRODUCED by the telescope |
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| 32. |
Read the following passage and then answer questionson the basis of your understanding of the passage and the related studied concepts. A small township with a demand of 880 kW of electric power at 220 V is situated 20 km away from a mini hydropower generating station generating power at 440 V. The resistance of the two wire line carrying power is 0.5 ohm per km. The town gets power from the line through a 11 kV-220 V step down transformer at a substation at the township. (a) The current flowing through the two wire supply line is |
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Answer» 160 A |
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| 33. |
Two media of refractive indices mu_(1) and mu_(2)(nu_(1) gt mu _(2)) are separated by a plane surface. If some part of a plane wavefront is in first medium and other part of wavefront in the second medium show the shape of the wavefront in this position diagramatically. |
Answer» SOLUTION :AB and BC are two PLANE WAVEFRONTS inclined to each other 
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| 34. |
Read the following passage and then answer questionson the basis of your understanding of the passage and the related studied concepts. A small township with a demand of 880 kW of electric power at 220 V is situated 20 km away from a mini hydropower generating station generating power at 440 V. The resistance of the two wire line carrying power is 0.5 ohm per km. The town gets power from the line through a 11 kV-220 V step down transformer at a substation at the township. (b) Estimated value of line power loss in the form of heat is |
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Answer» 128 kW [Hint: Reistance of TWO wier LINE `R = 0.5 xx 2 xx 20 = 20 Omega` |
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| 35. |
The two I-H graphs A and B in the adjacent figure are for . |
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Answer» a paramagnetic and a FERROMAGNETIC SUBSTANCE, RESPECTIVELY |
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| 36. |
Read the following passage and then answer questionson the basis of your understanding of the passage and the related studied concepts. A small township with a demand of 880 kW of electric power at 220 V is situated 20 km away from a mini hydropower generating station generating power at 440 V. The resistance of the two wire line carrying power is 0.5 ohm per km. The town gets power from the line through a 11 kV-220 V step down transformer at a substation at the township. How much power must the hydropower plant supply to the township to fulfill its entire power demand ? Assume that there is no power loss due to leakage during transit. |
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Answer» Solution :(ii) 1008 kW [Hint: Total power to be SUPPLIED by hydropower plant = 880 kW + 128 kW = 1008 kW] |
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| 37. |
The structure they created formed the basis of one of the harshest, most inhumane, societies the world has ever known. What structure is Mandela talking about? |
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Answer» RACIAL DOMINATION against the BLACK skinned |
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| 38. |
A charge particle is moving on a circular path of radius R in a uniform magnetic field under the Lorentz' force F. How much work is done by the force in one round ? Is the momentum of the particle changing ? |
| Answer» SOLUTION :PARTH of the MOTION | |
| 39. |
A hydrogen atom emits a photon corresponding to an electron transtion from n= 5 to n = 1 . The recoil speed of hydrogen atom in nearly equal to |
| Answer» Answer :C | |
| 40. |
How is rm.s voltage of a.c related to peak value of a.c voltage? |
| Answer» SOLUTION :`V_("RMS") = (V_(0))/SQRT2 "or" V_(rms) = V_(m)/sqrt2` | |
| 41. |
Solve the differential equation of L-C circuit and obtain the expression of current. |
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Answer» Solution :The differential equation of L-C circuit is, `(d^(2)Q)/( dt^(2))+(1)/(LC) q = 0` The solution of the equation is, `q= q_(m) COS ( omega_(0) t+ phi)`…(1) where `q_(m)` is the maximum VALUE OFQ and `phi` is phase constant. At t= 0, `q= q_(m)` `:. q_(m) = q_(m) cos phi` `:. 1 = cos phi` `:. phi = 0^(@)` `:.` From equation (1), `q = q_(m) cos ( omega_(0) t)` Differentiating w.r.t. time t, `(DQ)/(dt) = q_(m) omega_(0)( - sin omega_(0)t)` but `(dq)/(dt)= -1` `:. -I = - q_(m) omega_(0) sin omega_(0) t ` `:. I = I_(m) sin omega_(0) t ` where,` q_(m) omega _(0) =I_(m) ` `:. I = I_(m) sin omega_(0) t` |
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| 43. |
Draw the frequency spectrum of amplitude modulated signal |
Answer» SOLUTION :
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| 44. |
In a young's double slit experiment a very large screen is used. When a transparent film is kept infront of one of the slits then |
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Answer» all maxima will shift by same distance |
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| 45. |
Let f be a differentiable function with range (0,oo) and g(x)=(f(x))^2 -(f(x))^3 + (f(x))^4, then |
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Answer» g(x) is always INCREASING `=[2-3(f(x))+4(f(x))^2]xxf(x).f(x)` Let `H(x)=4. (f(x))^2 -3 f(x)+2` D=9-4.2.4=-23 `rArr` h(x) is always(+)ve Also f(x) is always (+)ve `rArr` g'(x) will have same sign as of f(x) `rArr` g(x) is increasing , if f(x) is increasing . |
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| 46. |
शुद्ध जल की मोलरता है - |
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Answer» 55.5 |
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| 47. |
Young.s doubleslit experiment is carriedout usingmicrowaves of wavelength lambda=3 cm. Distance between the slitis d = 5 cm and the distance between the plane of slits and the screenis D = 100 cm . Find te disatnceof positions from centralmaximum wheremaximas are formed . |
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Answer» `0,+ 25 cm , - 25` cm |
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| 48. |
If the temperature of a metallic conductor is increased, the drift velocity of electrons in theconductor decreases. |
| Answer» SOLUTION :True - On increasing the temperature of a conductor, RESISTIVITY of conductor decreases andrelaxation time of ELECTRONS decreases. As a RESULT, the DRIFT velocity of electrons decreases. | |
| 49. |
If the coefficient of mutual induction of the primary and secondary coils of an induction coil is 6 H and a current of 5A is cut off in 1//5000 second, calculate the conf. induced in the secondary coil. |
| Answer» Solution :`|e|=M(dI)/(dt), e=6 xx (5)/(1 // 5000)V=15 xx 10^(4)V` | |
| 50. |
Two magnets of equal mass are joined at 90^(@) each other as shown in figure. Magnet N_(1)S_(1) has a magnetic moment sqrt3 times that of N_(2)S_(2). The arrangement is pivoted so that it is free to rotate in horizontal plane. When in equilibrium, what angle should N_(1)S_(1) make with magnetic meridian? |
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Answer» `75^(@)` |
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