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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. |
In YDSE , the phase difference between two coherent sources s pi//3and constant . The intensity at a point which is at equal distance from the sources is ( I_(0) is maximum intensity ) |
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Answer» `(SQRT(3))/2 I_(0)` |
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| 2. |
A In question 3, If the dipole rotates from orientation 1 to orientation 2, is the work done on the dipole by field positive, negative, or zero ? B If, instead, the dipole rotates from the orientation 1 to orientation 4, is the work done by the field more than, less than, the same as in (a) ? |
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Answer» Solution :a. Electric FIELD will do positive work, if dipole rotates such that its ANGLE decreses with E. b.2 and 4 orientations are identical. Hence, work done will be same. |
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| 3. |
E is the electric field intensity at any point in a uniform electric field. : What is meant by uniform electric field? |
| Answer» Solution :REGION where magnitude and DIRECTION of ELECTRIC FIELD remain same. | |
| 4. |
The binding energy of " "_(1)^(2)H is 1.112 MeV per nucleon and an alpha-particle " "_(2)^(4)He has a binding energy of 7.047 MeV per nucleon. In the fusion reaction " "_(1)^(2)H + " "_(1)^(2)Hto " "_(2)^(4)He+Q, the energy released is |
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Answer» 1MeV |
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| 5. |
Two convex lenses of focal lengths f_(1) and f_(2) are placed coaxially, a distance d apart. If the axis of one of the lenses is lifted parallel to itself by Delta, find the distance by which the focal point is shifted and the distance of the focal point from the first lens. |
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Answer» |
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| 6. |
If a unit positive charge is kept in the air. Then the totat flux coming out ofunit charge is- |
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Answer» `4piepsilon_0^-1` |
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| 7. |
A police inspector is chasing a thief who is running away in a car with a speed 3 m/s. The speed of police jeep is 12 m/s. Then the speed of image of police jeep as seen by thief in the rear view mirror when the police jeep, is at a distance of 30 m is (value of focal length of the rear view mirror is 15 m) |
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Answer» 2m/s |
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| 8. |
in which of the following system will the radius of the second orbit be minimum ? |
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Answer» H-atom Orbital RADIUS `r_(N)prop(1)/(Z)` `:.` Radius of `Mg^(+2)` will be minimum. |
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| 9. |
A double convex lens of radius of curvature of each surface equal to 20 cm is made of glass of refractive index 1.5. Find the ratio of the powers of the lens when placed in air to its power when placed in water. Take refractive index of water = 1.33. |
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Answer» Solution :`"Given, first radius of CURVATURE, "R_(1)=20cm` `"Second radius of curvature, "R_(2)=-20cm` `"Refractive index of lens, "mu_(g)=1.5` `"Refractive index of water, "mu_(e)=1.33` For the lens placed in AIR : `P_(1)=(mu_(g)-1)[(1)/(R_(1))-(1)/(R_(2))]` `=(1.5-1)[(1)/(0.2)-((-1)/(0.20))]` `=0.5xx(2)/(0.2)=5D` For the lens placed in LIQUID : `P_(2)=((u_(g))/(mu_(e))-1)[(1)/(R_(1))-(1)/(R_(2))]` `=((1.5)/(1.33)-1)[(1)/(0.2)-((-1)/(0.20))]` `=0.128xx(2)/(0.2)=1.28D` `therefore""(P_(1))/(P_(2))=(5)/(1.28)=3.9:1` |
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| 10. |
A bar magnet of magnetic moment 10-J//T is free to rotate in a horizontal plane. The work done in rotating the magnet slowly from a direction parallel to a horizontal magnetic field of 4x×10 T to a direction 60^(o) from the field will be |
| Answer» ANSWER :A | |
| 11. |
Two identical springs are connected to mass m as shown (k = spring constant). If the period of the configuration (i) is 2 s, the period of the configuration (ii) is |
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Answer» `SQRT(2s)` Therefore, their effective spring constant is, `k_(s) =((K)(k))/(k+k) = k/2`  `therefore` Time period of oscillation, `T_(s) = 2pi sqrt(m/k_(s))`........(i) In figure, (ii) the two springs are connected in parallel. Therefore, their effective spring constant is `k_(p) = k + k = 2k` `therefore` time period of oscillation, `T_(P) = 2pi sqrt(m/k_(p))`...........(ii)  Divide (i) by (ii), we get `T_(s)/T_(p) = sqrt(k_(p)/k_(s)) = sqrt((2k)/(k//2)) =2` or `T_(p) = T_(s)/2 =(2s)/2 = 1s` |
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| 12. |
A capacitor, charged to 10 V , is being discharged through a resistance R. At the end of 1s the voltage across the capacitor is 5V. What will be the voltage after 2s ? |
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Answer» Solution :The equation for discharging of a capacitor through a resistance R is given by `q= q_0 E^(-t//CR)` or, `(q_0)/(q) = e^(t//CR)` or, `log_e (q_0)/(q) = t/(CR)` or , `t= CR log_e ((q_0)/(q))` As charge is proportional to voltage in case of a capacitor `t= CR log_e ((V_0)/(V))` Given that `V_0 = 10 ` volt At t = 1 sec, `V= 5 ` volt Hence `1= CR log_e ((10)/(5)) = CR log_e (2) ""....(1)` At t= 2 sec, the voltage will be given by : 2 = CR `log_e ((10)/(V)) "".....(2)` Dividing eq.(2) by eq. (1) , we get : `2= (log_e (10//V))/(log(2)) = (log_e 10- log_e V)/(log_e 2)` or , `log_e V= log_e 10-2 log_e 2` `=log_e 10 - log_e 4` or, `((10)/(4)) = (2.5) ` VOLTS |
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| 13. |
Where did ceremonies take place? |
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Answer» Rockstone amphitheatre |
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| 14. |
A parallel beam of light of wavelength 435 nm falls on a slit of width 0.4 mm. Find the distance between two dark bands on either side so the central maximum of the pattern produced on a screen placed at a distance of 2m from the slit. |
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Answer» |
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| 15. |
How does nuclear fusion occur, eventhough Coulombic repulsion exists? |
| Answer» SOLUTION :NUCLEAR FORCE is STRONGER than the COULOMBIC force. | |
| 16. |
A bird is flying to and fro between two cars moving with constant speeds on a straight smooth road. One car is moving at 18 km/h while other is moving towards the other at 27 km/h. The bird starts moving from the first car towards the other with a speed of 36 km/h when the distance of separation between the two is 36 km. What is the displacement of the bird when the two cars meet each other? |
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Answer» 18 km `v_1`=27km/h Speed of second `v_2`=18km/h As they are MOVING opposite to each other relavtive Velocity of ONE w.r.t. to other `=v_1+v_2=27+18`=45 km/h `:.` TIME required to meet `=(36)/(45)=0.8 hours` Speed of the bird =36 km/h `:.` Distance moved by the bird `=36xx0.8=28.8 km.` |
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| 17. |
Write the key components of robot. |
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Answer» Solution :The robotic system MAINLY consists of sensors, power supplies, control systems, manipulators and NECESSARY software. Most robots are composed of 3 main parts: 1. The Controller also known as the "brain" which is run by a computer program. It gives commands for the moving parts to perform the JOB. 2. Mechanical parts - MOTORS, pistons, grippers, wheels, and gears that MAKE the robot move, grab, turn, and lift. 3. Sensors - to tell the robot about its surroundings. It helps to determine the sizes and shapes of the objects around, distance between the objects, and directions as well. 
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| 18. |
The reference frame K^' moves in the positive direction of the x axis of the frame K with a relative velocity V^'. Suppose that at the moment when the origins of coordinates O and O^' coincide, the clock readings at these points are equal to zero in both frames. Find the displacement velocity x of the point (in the frame K) at which the readings of the clocks of both reference frames will be permanently identical. Demonstrate that overset(.)x lt V. |
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Answer» SOLUTION :SUPPOSE `x(t)` is the locus of points in the frame K at which the readings of the clocks of both reference system are permanently identical, then by Lorentz transformation `t^'=(1)/(sqrt(1-V^2//c^2))(t-(VX(t))/(c^2))=t` So differentiating `x(t)=c^2/V(1-sqrt(1-V^2/c^2))=c/beta(1-sqrt(1-beta^2)), beta=V/c` Let `beta=tan h theta, 0 le theta lt oo`, Then `x(t)=(c)/(tanhtheta)(1-sqrt(1-tan h^2theta))=c(coshtheta)/(sinhtheta)(1-(1)/(coshtheta))` `=c(coshtheta-1)/(sinhtheta)=csqrt((coshtheta-1)/(coshtheta+1))=ctanhtheta/2lev` (`tan htheta` is a monotonically increasing FUNCTION of `theta`) |
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| 19. |
When a magnetic field is applied on an electron moving with uniform velocity it is deflected in a direction |
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Answer» at RIGHT ANGLES to the field |
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| 20. |
The m agnetic moment of a bar magnet is 4 Am^2. If it is broken into two equal pieces, the magnetic moment of each piece is : |
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Answer» `2 Am^2` |
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| 21. |
The area of the coil in a moving coil galvanometer is 16 cm^(2) and has 20 turns. The magnetic induction is 0.2T and the couple per unit twist of the suspended wire is 10^(-6) Nm per degree. If the deflection is 45° calculate the current passing through it. |
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Answer» Solution :Given, A=`16CM^(2)=16xx10^(-4)m^(2)` `B=0.2T,N=20,C=10^(-6)Nm//`DEGREE, `theta=45^(@)` From, `Ctheta=BiAN` `i=(Ctheta)/("BAN")=(10^(-6)xx45)/(0.2xx16xx10^(-4)xx20)=7.03xx10^(-3)` |
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| 22. |
Assertion : A direct current flows through a metallic rod, produced magnetic field only outside the rod. Reason : There is no flow of charge carriers inside the rod. |
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Answer» If both ASSERTION and REASON are TRUE and the reason is the CORRECTEXPLANATION of the assertion. |
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| 23. |
The optical telescope in the VainuBappu observatory at Kavalur has an objective lens of diameter 2.3.m. What is its angular resolution if the wavelenght of light used is 589 nm? |
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Answer» Solution :`a=2.3m,lambda=589xx10^(-9)m,0=?` The equation for angular resolution is, `theta = (1.22 lambda)/(a)` SUBSTITUTING, `theta=(1.22xx589xx10^(-9))/(2.3)=321.4xx10^(-9)` `theta = 321.4 xx 10^(-7) RAD` NOTE: The angular resolution of human EYE is approximately. `3 xx 10^(-4)` rad |
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| 24. |
Consider a body, shown in figure, consisting of two identical balls, each of mass M connected by a light rigid rod. If an impulse J = MV is imparted to the body at one of its ends, what would be its angular velocity? |
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Answer» `V//L` `J=Iomega` `2M.(L/2)^(2).omega=MV.(L/2). Thereforeomega=V/L` |
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| 25. |
Two parallel wires carrying currents in the same direction attract each other, while two electron beams moving in parallel repel each other. Give reason. |
| Answer» SOLUTION :In the case of PARALLEL current carrying wires only magnetic force of attraction comes into PLAY. In the case of ELECTRON BEAMS, the electrostatic force of repulsion dominates over the magnetic force of attraction. | |
| 26. |
A particle is taken to a distance r (gt R) from centre of the earth. R is radius of the earth. It is given velocity V which is perpendicular to With the given values of V in column I you have to match the values of total energy of particle in column II and the resultant path of particle in column III. Here 'G' is the universal gravitational constant and 'M' is the mass of the earth. |
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Answer» |
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| 27. |
Suppose that the cantaloupe and table of Fig. 5-6 are in an elevator cab that begins to accelerate upward. (a) Do the magnitudes of vec(F)_(TC) and vec(F)_(CT) increase, decrease, or stay the same ? (b)Are those two forces still equal in magnitude and opposite in direction? ( c) Do the magnitudes of vec(F)_(CE) and vec(F)_(EC) increase, decrease, or stay the same? (d) Are those two forces still equal in magnitude and opposite in direction? |
| Answer» Solution :(a) equal, (b) greater (ACCELERATION is UPWARD, thus net FORCE on body MUST be upward) | |
| 28. |
Light of wavelength 4500 Åin air is incident on a plane boundary between air and another medium at an angle 30° with the plane of boundary. As it enters from air into the other medium, it deviates by 15^@ towards the normal. Find refractive index of the medium and also the wavelength of given light in the medium. |
Answer» Solution : Angle of incidence ` i=90^@ - 30^@ =60^@ .`Asthe raybendstowardsthe normal,itdeviatesby anangle `i-r =15^@ `( given ) ` thereforer=45^@ ` Applyingsnell.slaw `mu_("sir") sin i=mu_("MED") sin r ` ` therefore1 xx SIN60^@= muxx sin45^@` ` thereforemu = ( sin 60^@)/(sin 45^@) = ( sqrt( 3) //2)/( 1//2)(or)mu = sqrt(1.5 )` in termsof wavelengths , `mu = sqrt(1.5 ) = ( lamda_("air"))/( lamda_("med"))(or)lamda_("med") = ( lamda_( "air")/( sqrt(1.5 )) = ( 4500 )/( sqrt( 1.5 ))` ` lamda_("med")= 3674 Å` |
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| 29. |
State Lenz's law |
| Answer» Solution :According to Lenz.s law, INDUCED emf or current is ALWAYS PRODUCED in a direction such that it opposes the CAUSE of its generation. | |
| 30. |
A cone of base radius R and height h is located in a uniform electric field vecE parallel ot its base. The electric flux entering the cone is : |
| Answer» ANSWER :D | |
| 31. |
In a problem number 34 energy dissipated in resistance per unit time, once the terminal speed is attained is |
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Answer» 20J |
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| 32. |
Which of the following is not true for a region with a uniform electric field? |
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Answer» It can have FREE charges. |
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| 33. |
A diode is connected to 200 V (rms) ac in series with a capacitor as shown in figure. The voltage across the capacitor is |
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Answer» 220 V `V_(rms)=V_0/2=(200sqrt2)/2=200/sqrt2` |
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| 34. |
Give three differences between intrinsic and extrinsic semiconductors |
Answer» SOLUTION :
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| 35. |
A block of mass m=15g is suspended in an elevator with the help of three identical light elastic cords (spring constant k=100N//m each) attached vertically. One of them cord 1 is tied to the ceilling of the elevator and the other two cords 2 and 3 are tied to the elevator floor as shown in the figure. When the elevator is stationary the tension force is each of the lower cords is T=7.5N. Take g=10m//s^(2). Now the elevator starts moving with given four accelerations shown in column I. Column II given the displacement ofthe block with respect to elevator when its is accelerating. Column III gives tension in the cords when elevator is accelerating. Tension in cord -1 |
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Answer» (I)(i)(S) So, `T_(1)-150=(15)(1.5)` `implies` Tension in cord `1, T_(1)=172.5N` |
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| 36. |
A beam of light consisting of two wave lengths 4200 Å and 5600 Å is used to obtain interference fringes in Young's double slit experiment. The distance between the slits is 0.3 mm and the distance between the slits and the screen is 1.5 m. Compute the least distance of the point from the central maximum, where the bright fringes due to both the wavelengths coincide. |
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Answer» Solution :Here , `lambda_1`=4200Å=`4200xx10^(-10)` m `lambda_2`=5600Å = `5600xx10^(-10)` m d=0.3 mm = `0.3xx10^(-3)` m D=1.5 m Let `n^(th)` bright fringe due to `lambda_1`= 4200 Å COINCIDE with `(n-1)^(th)`bright fringe due to `lambda_2`= 5600 Å `nlambda_1=(n-1)lambda_2` i.e., `nxx4200xx10^(-10)=(n-1)5600xx10^(-10)` `n=(n-1)8/6` 6n=8n-8 n=4 `THEREFORE` The least distance REQUIRED `X=nlambdaD/d` `=(4xx4200xx10^(-10) xx1.5)/(0.3xx10^(-3))=(25200xx10^(-10))/(0.3xx10^(-3))=84000xx10^(-7)` X=840 nm |
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| 37. |
A block of mass m=15g is suspended in an elevator with the help of three identical light elastic cords (spring constant k=100N//m each) attached vertically. One of them cord 1 is tied to the ceilling of the elevator and the other two cords 2 and 3 are tied to the elevator floor as shown in the figure. When the elevator is stationary the tension force is each of the lower cords is T=7.5N. Take g=10m//s^(2). Now the elevator starts moving with given four accelerations shown in column I. Column II given the displacement ofthe block with respect to elevator when its is accelerating. Column III gives tension in the cords when elevator is accelerating. Tension in cord -2 |
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Answer» (I)(iv)(P) So `165+DeltaT-2xx(7.5-DeltaT)-150=(15)(1)` `implies DeltaT=5N` `implies` Tension in cord 2, `T_(2)=2.5N` |
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| 38. |
At thetop of mountain ,a thernomete reads 7^(@)Cand a barometerreadas 70 cmof Hg . At thebottom of themountain , theyread27^(@)Cand76 cmof Hg . Calculatethe rationthe densityof theairat thetopwith thatat the bottom. |
| Answer» SOLUTION :[0.9868] | |
| 39. |
What is the shape of fringes in Youngs double slit experiment ? |
| Answer» SOLUTION :HYPERBOLIC | |
| 40. |
Derive relation between induced charge and change in magnetic flux. |
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Answer» Solution :From Faraday.s LAW, we have learnt that the magnitude of the induced emf is, `|epsilon|=(DeltaPhi_B)/(Deltat)`…(1) However, `|epsilon|=Ir=(DELTAQ)/(Deltat)R`…(2) Here r is resistance of coil. Thus, from equations (1) and (2) `(DeltaPhi_B)/(Deltat)=(DeltaQ)/(Deltat)r` `THEREFORE DeltaPhi_B=DeltaQr` `therefore DeltaQ=(DeltaPhi_B)/r` So, induced charge is only depends on change in flux but it does not DEPEND on rate of change of flux. |
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| 41. |
A block of mass m=15g is suspended in an elevator with the help of three identical light elastic cords (spring constant k=100N//m each) attached vertically. One of them cord 1 is tied to the ceilling of the elevator and the other two cords 2 and 3 are tied to the elevator floor as shown in the figure. When the elevator is stationary the tension force is each of the lower cords is T=7.5N. Take g=10m//s^(2). Now the elevator starts moving with given four accelerations shown in column I. Column II given the displacement ofthe block with respect to elevator when its is accelerating. Column III gives tension in the cords when elevator is accelerating. Tension in cord -3 |
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Answer» (I)(i)(P) So, `T_(1)-150=(15)_(2)` `implies` Tension in cord `1,T_(1)=180N` So DISPLACEMENT of block `=15cm` downward |
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| 42. |
The drift velocity of the electrons in a copper wire of length 2m under the application of a potential difference of 100 V is 0.2 ms^(-1). Their mobility is (in m^2 v^(-1) s^(-1)) |
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Answer» `2.5xx10^(-3)` |
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| 43. |
The Bohr's model of atoms: |
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Answer» ASSUMES that the angular momentum of electrons is quantized |
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| 44. |
In the broadcast of communication modulation is necessary:-Explain any two reason why modulation is necessary? |
| Answer» Solution :MODULATION is the process of superimposing low frequency baseband SIGNAL with HIGH frequency carrier wave | |
| 45. |
Linespassing through places of equal declination are called |
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Answer» Isogonal |
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| 46. |
A uniform circular disc A of radius r is made from A copper plate of thickness t and another uniform circular disc B of radius 2r is made from a copper plate of thickness t/2. The relation between the moments of inertia I_A and I_B is? |
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Answer» `I_AgtI_B` |
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| 47. |
A circuit of logic gte made from two AND gate is shown in following figure. Draw output waves for point X. For A and for Y. |
Answer» Solution : From the given COMBINATION according to work of output of FIRST AND gate if input is .1. then output is .1.. Input is .0. then output is also .0.. Therefore at point X and Y output is equal. 
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| 48. |
Compare the electrical capacitance of a spherical capacitor with and without the presence of dielectric medium. |
| Answer» Solution :`C_(m) : C_(a) : : epsilon_( R ) :1` where `epsilon_(r ) ` is a DIELECTRIC constant. | |
| 49. |
In a chamber, a uniform magnetic field of 6.5 G (1 G = 10^(-4) T) is maintained. An electron is shot into the field with a speed of 4.8 x× 10^6 m s^(-1) normal to the field. Explain why thepath of the electron is a circle. Determine the radius of the circular orbit. (e = 1.5 x× 10^(-19) C, m_e = 9.1x×10^(-31) kg) |
Answer» Solution :1.  Consider a uniform magnetic field `vecB`, WHOSE field lines are perpendicularly outward from the plane of figure shown by symbol "." in above figure. 2. Consider an electron entering above magnetic field perpendicularly with velocity `vecv_(1)` at point `A_(1)`. At this point magnetic force acting on this electron is `vecF_(1)=e(vecBxxvecv_(1))` which is in the direction `vec(A_(1)C)`. This force deflects this electron along curved path from `A_(1)` to `A_(2)`. 3. At point `A_(2)`, magnetic force exerted on electron is `vecF_(3)=e(vecBxxvecv_(3))` which is in the direction `vec(A_(3)C)`. Now, if same magnetic field is extended on the left side of above magnetic field then as explained earlier, electron will complete a circle of radius r and then it will repeat this circular motion with constant speed. 4. Here, from each point on this circular path, direction of magnetic force exerted on electron is FOUND to be centripetal because it is radially inward pointing towards point C which is centre of circle. Thus, under the influence of this centripetal force, electron follows circular trajectory. 5. Expression of radius of circular path : When a particle with charge q performs circular motion in a plane PERPENDICULAR to uniform magnetic field `vecB`, necessary centripetal force is provided in the form of magnetic force acting on it. Hence, `F_(m)=F_(c)` `thereforeBqv_(_|_)=(mv_(_|_)^(2))/r`  (Where `v_(_|_)=vsintheta` is the component of velocity of particle perpendicular to magnetic field, v = constant speed of particle, `theta` = angle between initial velocity `vecv` and magnetic field `vecB`). `thereforer=(mv_(_|_))/(Bq)` `thereforer=(mvsintheta)/(Bq)` 6. At present `vecv_|_vecBand" so "theta=90^(@)`. Substituting other VALUES, `r=((9.1xx10^(-31))(4.8xx10^(6))SIN90^(@))/((6.5xx10^(-4))(1.6xx10^(-19)))` `thereforer=4.2xx10^(-2)m` |
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| 50. |
Two identical blocks are pulled along a rough surface as suggested in the figure. Which one of the following statements is false? |
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Answer» The coefficient of kinetic FRICTION is the same in each case. |
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