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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 refracting angle of a prism is A and refrac tive index of the material of the prism is cot(A/2) The angle of minimum deviation is : |
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Answer» `pi-A` `THEREFORE COT A/2=(sin((A+delta_m)/(2)))/(sin""A/2)` `therefore(cos A//2)/(sin A//2)=(sin((A+delta_m)/(2)))/(sin""A/2)` `therefore sin(9^@-A/2)=sin((A+delta_m)/(2))` `therefore (180^@-A)/(2)=(A+delta_m)/(2)` `therefore 180^@-A=A+delta_m` `therefore delta_m=180^@-2A` or `delta_m=pi-2A` |
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| 2. |
Calculate the amount of work done in moving a 1muC charge from one point to another if potential difference between both the points is 12V. |
| Answer» SOLUTION :`12XX10^(-6)J` | |
| 3. |
Two coils, held in fixed positions have a mutual inductance of 100muH. what is the peak emf in one coil when the current I is in amperes and t is in seconds? |
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Answer» 2.00V |
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| 4. |
Light takes of t_(1) to travel a distance x_(1) in vacuum and the same light takes time t_(2) to travel a distance x_(2) in a medium. The critical angle for that medium is |
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Answer» `SIN^(-1)((x_(2)t_(2))/(x_(1)t_(1)))` |
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| 5. |
In human female, ovulation occurs during menstrual cycle. |
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Answer» At the END of PROLIFERATIVE phase |
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| 6. |
Give the expression for the time-period of oscillation of an electric dipole in a uniform electric field. |
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Answer» SOLUTION :Time period, `T=2pi SQRT((1)/(p E))` where, I- MOMENT of INERTIA of the dipole p-dipole moment E=Electric FIELD strength |
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| 7. |
Protons are projected with an initial speed v_(i)=6kms^(-1) from a fileds-free region vec(E ) =-900hat jNC^(-1) present above the plane as shown in fig. The initial velocity vector of the protons makes an angle u with the plane. The proton are to hit a target that lies at a horizontal cross the plane and enter the electric field. Find the angle theta at which the protons must pass through the plane to strike the target. |
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Answer» Solution :`v_(i)=6.0xx10^(3)MS^(-1)` `a_(y)=(eE)/(m)=((1.6xx10^(-19))(900))/((1.6xx10^(-27)))=9.0xx10^(10)ms^(-2)` `R=(v_(i)^(2)sin 2 theta)/(a_(y))=2XX10^(-3)` `sin 2 theta =(1)/(2)` or `2 theta =30^(@)` or `theta =15^(@), 90^(@)-theta=75^(@)`. |
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| 9. |
If vec(A)=3hati + 4hatj and vec(B)=7hati + 24hatj. Find the vector having the same magnitude as b and is a parallel to vecA |
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Answer» `15HATI + 20hatj` `:. |B|hatA=25(3hati+4hatj)/5=15hati+20hatj` `|A|=sqrt(9+16)=5` |
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| 10. |
Give the circuit symbol and truth table for OR gate |
Answer» SOLUTION : 
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| 11. |
An electric field is uniform and in the positive x direction for positive x and uniform with the same magnitude but in the negative x direction for negative x. It is given that E=200 hat(i) N//C for x gt 0 and E=-200 hat(i) N//C for x lt 0. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one face is at x=+10 cm and the other is at x=-10 cm What is the net charge inside the cylinder? |
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Answer» Solution :The NET charge within the cylinder can be found by using Gauss's-law which GIVES `q=epsi_(0) phi` `=3.14xx8.854xx10^(-12) C` `=2.78xx10^(-11)C` |
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| 12. |
An electric field is uniform and in the positive x direction for positive x and uniform with the same magnitude but in the negative x direction for negative x. It is given that E=200 hat(i) N//C for x gt 0 and E=-200 hat(i) N//C for x lt 0. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one face is at x=+10 cm and the other is at x=-10 cm What is the net outward flux through the xylinder? |
| Answer» Solution :Net outward FLUX through the CYLINDER `phi=1.57+1.57+0=3.14 NM^(2) C^(-1)`. | |
| 13. |
Which of the following statemeni is correct |
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Answer» Intensity of X-rays DEPENDS upon the operating voltage of the tube |
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| 14. |
A T V transmitting antenna is 80 m tall.If the receiving antenna is on the ground .Find the service area. |
| Answer» SOLUTION :1024 `PI` SQ KM | |
| 15. |
Assertion :- A current is flowing through a coil with 5 volt battery . If self inductance of coil is increased then circuit takes more time to aquires its current peak. Reason :- Inductance strongly opposes the flow of current through it . |
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Answer» If the ASSERTION & Reason are True& the Reason is a correct EXPLANATION of the Assertion . |
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| 16. |
A sample of radioactive substance decays simultaneously by two processes A and B with half lives 1/2h and 1/4 h respectively. For the first half hour it decays with process A, next one hour with process B, and for further half an hour with both A and B If originally, there are N_0nuclei, then the Number nuclei after 2h of such decay |
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Answer» `N_0/(2^8)` |
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| 17. |
A point object 'O' approaches a biconvex lens of focal length 40 cm along its optic axis with a speed of 10 cm/s while the later receeds away from the former with a speed of 4 cm/s. Find the speed and direction of motion of the image when the object is at a distance of 60 cm from the lens. |
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Answer» 10 cm/s, LEFTWARDS |
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| 18. |
The resistance of a conductor is 5 ohm at 50^@C and 6 ohm at 100^@C .It resistance at 0^@C is : |
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Answer» `1OMEGA` |
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| 19. |
A dielectric slab is placed between the plates of a parallel plate capacitor. Its capacitance. |
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Answer» BECOMES zero |
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| 20. |
Define the term magnetic flux compute its dimensions. |
| Answer» SOLUTION :Magnetic flux through an area A PLACED in a uniform magnetic FIELD B is DEFINED as `phi= We have F=BILthat’s why `B=F/IL`,`phi=(F//IL)`, `A=(MLT^-2//AL)L^2`=`A^-1 ML^2 T^-2` | |
| 21. |
What is a NAND gate? |
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Answer» SOLUTION :It is an AND GATE followed by a NOT gate. Logic gate whose output zero (0) only when both the inputs are one. OR 
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| 22. |
A polariod (I) is placed infront of a monochromatic source. Anotherpolariod (II) is placed in front of this polaroid (I) and rotated till no light passes. A third polaroid (III) is now placed in between (I) and (II). In this case, wil light emerge from (II) . Explain. |
Answer» Solution :In the DIAGRAM shown, a monochromatic light is placed infront of polaroid (I) as shown below.  As per GIVEN question, monochromatic light emerging from polaroid (I) is plane polarised. When polaroid (II) is placed in FRONT of this polaroid (I), and rotated till no light passes through polaroid (II), then (I) and (II) are set in CROSSED positions, i.e., PASS axes of I and II are at `90^(@)`.  Consider the above diagram where a third polaroid (III) is placed between polaroid (I) and polaroid (II). When a third polaroid (III) is placed in between (I) and (II), no light will emerge from (II), if pass axis of (III) is prallel to pass axis of (I) of (II). In all other cases, light will emerge from (II), as pass axis of (II) will no longer be at `90^(@)` to the pass axis of (III). |
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| 23. |
In a thin spherical fish bowl of radius 10 cm filled with water of refractive index 4//3 there is a small fish at a distanceof 4 cm from the centre C as shown in the figure. Where will the image of fish appears, if seen from E. |
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Answer» `5.2 CM` |
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| 24. |
If two soap bubbles of different radii are connected by a tube then: |
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Answer» air flows from the bigger bubble to the smaller bubble till the sizes become equal `rArr` Excess pressure `prop1/r` Air goes from higher pressure to lower pressure `therefore` air flows from smaller bubble to bigger bubble. Thus correct choice is (c). |
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| 25. |
An aluminium wire of cross-sectional area 10^(-6)m^2 is joined to a copper wire of the same cross-section. This compound wire is stretched on a sonometer, pulled by a load of 10 kg. The total length of the compound wire between two bridges is 1.5 m of which the aluminium wire is 0.6 m and the rest is the copper wire. Transverse vibrations are set up in the wire in the lowest frequencyof excitation for which standing waves are formed such that the joint in the wire is a node. What is the total number of nodes qbserved at this frequncy excluding the twoat the endsofthe wire? Thedensity of aluminium is 2.6 xx 10^4 kg//m^3. |
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Answer» Solution :As the total LENGTH of the wire is 1.5 m and out of which `L_A =0.6m`, so the length of copper wire `L_c = 1.5-0.6 =0.9 m`.The tension in the whole wire is same `(= Mg =10g N)` and as FUNDAMENTAL frequency of vibrationof string is given by `f =1/(2L) sqrt(T/m) =1/(2L) sqrt(T/(rho A))` [ as ` m =rhoA`] So `f_A =1/(2L_A) sqrt(T/(rho_A A) and f_c = 1/(2L_c) sqrt(T/(rho_cA)`......(1) Now as in case of composite wire, the whole wire will vibrate with fundamental frequency `f = n_A f_A =n_c f_c`......(2) Substituting the values of `f_A and f_c` from Eqn. (1) in (2) `n_A/(2 xx 0.6) sqrt(T/(A xx 2.6xx10^3)) = n_c/(2 xx 0.9) sqrt(T/(A xx 1.0401 xx 10^4))` i.e., `n_A/n_c =2/3 sqrt(2.6/10.4) = 2/3 xx 1/2 = 1/3` So that for fundamental frequency ofcomposite string,`n_A = 1 and n_c =3`, i.e., aluminium string will vibrate in first HARMONIC and copper wire at second, overtone as shown in figure.  `:. f = f_A =3f_c` Thsi in TURN implies that total number of nodes in the string will be 5 and so number of nodes excluding the nodes at the ends = 5 - 2=3, and `f=f_A =2/(2 xx0.6) sqrt((10xx9.8)/(10^(-6) xx 2.6 xx 10^3)) ~~ 161 .8 Hz (= 3f_c)` |
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| 26. |
A coin is at rest at bottom of a tank filled with liquid. A ray of light coming towards surface will move on the surface after incident on it, then what is the speed of light in liquid ? |
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Answer» `2.4xx10^8 ms^(-1)`  Length of hypotenuse of RIGHT ANGLE `triangle`1ABC is 5. `therefore (1)/(mu)=(sini)/(SIN 90^@)` …... (`because` Refractive index of LIQUID =`mu`, refractive index of air =1) `therefore mu =(1)/(sini)=5/3` Now speed `v=(c)/(mu)=(3xx10^8)/(5/3)=1.8xx10^8` m`//`s |
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| 27. |
A resistor R_(1) dissipates the power P when connected to a certain generator. If the resistor R_(2) is put in series with R_(1) the power dissipated by R_(1) |
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Answer» DECREASES |
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| 28. |
Write the dimensional formula for resistivity. |
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Answer» `[ML^(2)T^(-2)A^(-2)]` `=(ML^(2)T^(-3)A^(-2)xxL^(2))/(L)=[ML^(3)T^(-3)A^(-2)]` which is CHOICE `(B)`. |
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| 29. |
A : The unpolarised light and polarised light can be distinguished from each other by using polaroid. R : A polaroid is capable of producing plane polarised beams of light. |
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Answer» Both A and R are TRUE and R is the CORRECT explanation of A |
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| 30. |
A thread having linear charge density lamda is in the shape of a circular arc of radius R subtending an angle theta at the centre. (a). Find the electric field at the centre. (b). Using the epression obtained in part (a) find the field at the centre if the thread were emicircular (c). Find the field at centre using the expression obtained in part (a) for the case thetato0. Is the result justified? (d). A thread having total charge Q (uniformly distributed is in the shape of a circular arc of radius R subtending an angle theta at centre. write the expression for the field at the center. Obtain the field when thetato0. Make sure you understand the difference in case. (c) and (d). |
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Answer» (B). `E=(2Klamda)/(R)` (C) `E=0` (d). `E=(KQ)/(R^(2))` |
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| 31. |
निम्नलिखित में से किसके स्राव में फ्रक्टोस, कैल्शियम तथा कुछ एन्जाइम प्रचुर मात्रा में पाए जाते हैं? |
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Answer» नर सहायक ग्रन्थियाँ |
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| 32. |
Discuss special cases of Biot-Savart law. |
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Answer» Solution :(1) If `theta=0^(@),sin0^(@)=0`, so that dB = 0 i.e. the magnetic field is zero at points on the AXIS of the current ELEMENT. (2) If `theta=90^(@),SIN90^(@)=1` so that dB = is maximum. i.e. the magnetic field due to a current element is maximum in a plane PASSING through the element and perpendicular to its axis. |
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| 33. |
The displacement x(in metres) of a particle in simple harmonic motion is related to time t (in seconds) as x=0.01 cos(pit+pi/4). Find the frequency. |
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Answer» SOLUTION :`omega=pi` `2pin=pi` `n=1/2=0.5Hz` |
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| 34. |
Two charge particles are moving such the distance between them remains constant. The ratio of their masses is 1 : 2 and they always have equal and opposite momentum. The particles interact only though electrostatic force and no other external force is acting on them. The electrostatic interaction energy for the pair of the particles is – U_0. Find the kinetic energy of the lighter particle. How does the kinetic energy change with time? |
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Answer» |
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| 35. |
(A ): In electromagnetic wave, electric and magnetic fields oscillate in the same plane and in the same phase. (R) Electric field is the primary energy carrier in the electromagnetic wave. |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the CORRECT EXPLANATION of 'A'. |
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| 36. |
A small coin is resting on thebottom of a beaker filled with liquid. A ray of light from the coin travels upto the surface of the liquid and moves alongits surface. How fast is the light travelling in the liquid? |
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Answer» `2.4xx10^8 m s^(-1)` (as `i_c` is the angle which the ray fromthe coin makes with 4 CM SIDE) As REFRACTIVE index , `mu=c/v_l` `therefore v_l=c/mu=(3xx10^8 m s^(-1))/((5//3))=1.8xx10^8 m s^(-1)` 
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| 37. |
Pluto has 1//500 the mass and 1//15 the radius of Earth. What is the value of g on the surface of Pluto ? |
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Answer» `0.3 m//s^(2)` `g_("Pluto")=G(M_("Pluto"))/(R_("Pluto")^(2))=G((1)/(500)M_("Earth"))/(((1)/(15)R_("Earth"))^(2))=(15^(2))/(500)* G(M_("Earth"))/(R_("Earth")^(2))=(225)/(500)(10 m//s^(2))=4.5 m//s^(2)` |
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| 38. |
In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of (nlamda)/a. Justify this by suitably dividing the slit to bring out the cancellation. |
| Answer» Solution :Divide the single slit in to n SMALLER slits of WIDTH `a.=a/n`. The angle `theta=(n lamda)/a=(lamda)/(a.)`. Each of the smaller slits sends ZERO intensity in the direction `theta`. The combination gives zero intensity as WELL. | |
| 39. |
(a) A toroidal solenoid with an air core has an average radius of 0.15m, area of cross section 12 xx 10^(-4) m^2and 1200 turns. Obtain the self inductance of the toroid. Ignore field variation across the cross section of the toroid. (b) A second coil of 300 turns is wound closely on the toroid above. If the current in the primary coil is increased from zero to 2.0 A in 0.05 s, obtain the induced emf in the secondary coil. |
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Answer» Solution :`(a) B = mu_0 n_1 I = (mu_0 N_1I)/(l) = (mu_0 N_1 I)/(2pi R)` Total magnetic flux ,` phi_B = N_1BA = (mu_0 N_1^2 IA)/(2pi r)` But `phi_B = LI therefore L = (mu_0 N_1^2 A)/(2pi r)` ` L = (4pi xx 10^(-7) xx 1200 xx 1200 xx 12 xx 10^(-4))/(2pi xx 0.15) H` ` = 2.3 xx 10^(-3) H = 2.3 mH` `(b) |E| = (d)/(DT) (phi_2) , ` where `phi_2` is the total magnetic flux linked with the second COIL , `|e| = (d)/(dt) (N_2BA) = (d)/(dt) [ N_2 (mu_0 N_1 I)/(2r)A]` or`|e| = (mu_0 N_1N_2 A)/(2r) (dI)/(dt)` or `|e| = (4pi xx 10^(-7)xx 1200 xx 300 xx 12 xx 10^(-4) xx 2)/(2 xx 0.15 xx 0.05) V` = 0.023 V |
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| 40. |
What is a phasor ? What is its use in a.c. circuit ? Draw a phasor diagram for pure resistive a.c. circuit. |
Answer» Solution : A phasor is a vector which ROTATES about the origin with angular speed `omega` . Notion of phasors is used for voltage and current in an a.c. circuit to SHOW phase relationship between them. The amplitudes of phasors `vecV` and `vecl`. represent the peak values `V_(m)`and `I_(m)`of these oscillating quantities. The vertical components of phasors V and I represent the INSTANTANEOUS values of FIG. 7.14 voltage and current respectively. Phasor diagram for a pure resistive a.c. circuit has been shown in Fig. 7.14. Here, V and I are in same phase. |
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| 41. |
Which party did Mandela join? |
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Answer» INDIAN NATIONAL Congress |
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| 42. |
If N is number of closed packed structures, Then number of Octahedral Voids is |
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Answer» N |
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| 43. |
In the figure shown a conducting rod of length l, resistance R and mass m is moved with a constant velocity v. The magnitude field B varies with time t as B=5t, what t is time in second. At t=0 the area of the loop containing capacitor and the rod is zero and the capacitor is uncharged. the rod started moving at t=0 on the fixed smooth conducting rails which have negligible resistance. Find the current in the circuit as a function of time t. |
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Answer» |
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| 44. |
Can a lens be used in a medium of which it is made? |
| Answer» Solution :No, In that case it will BEHAVE LIKE a plane GLASS plate. | |
| 45. |
Consider the following statements A) If time constant is small the condenser discharges slowly.B) For small values of inductance the rate of decay of current will be large |
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Answer» Both are CORRECT |
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| 46. |
A current carrying conductor is in the form of a sine curve as shown, which carries current I. If the equation of this curve is Y=2 sin((pix)/(L)) and a uniform magnetic field B exists in space. |
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Answer» Force on wire is BIL if field is along Z axis |
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| 47. |
A set of 'n' equal resistors, of value 'R' each, are connected in series to a battery of emf 'E' ,and internal resistance 'R'. The current drawn is I. Now, the 'n' resistors arc connected in parallel to the same battery. Then the current drqwn from battery becomes 10 I. The value of 'n' is |
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Answer» 9 First SITUATION  `THEREFORE I = (E)/(R + nR) "" `....(1) Second situation  `therefore I. = (E)/((R)/(N) + R) = (NE)/(R + nR) "" `... (2) But I. = 10 I `therefore (nE)/(R + nR) = (10 E)/(nR + R)` `thereforen =10` |
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| 48. |
Select the correct statement about rainbow: |
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Answer» We can see a RAINBOW in the western SKY in the LATE afternoon |
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| 49. |
What are advantages of AM over FM transmission? |
| Answer» SOLUTION :The FM transmission requres MUCH wider channels as COMPARED to AM transmisson. (II) The reception of FM signal is limited to line-of-sight. As a result, the area in which FM transmission can be received is much smaller in case of AM transmission. | |
| 50. |
If two opposite electric charges having same magnitude are 10 cm away from each other, they experience 0.9N attractive force then magnitude of electric charges will be...... |
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Answer» `1pC` `F = 0.9 N, k = 9 XX 10^(9) SI, gamma = 10 cm = 10^(-2) cm` `therefore Q^(2) = (Fr^(2))/k = (0.9 xx 10^(-2))/(9 xx 10^(9)) = 10^(-12)` `therefore q = 1 xx 10^(-6) C, therefore q = 1muC` |
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