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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. |
Assertion : The positiveray particlesare more massive than electrons . Reason : Positive rays are reflected by a magnetic field to a greater extent than cathoderays . |
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Answer» If both the assertion and reason are true STATEMENT andreason is CORRECT explanation of the assertion . |
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| 2. |
A ball is dropped on a horizontal plate from a height.The total distance travelled by the ball before coming to rest is (e coeff, of restitution) |
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Answer» `H((1)/(1-e^2))` |
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| 3. |
A diode valve works in the region of space charge limited current . If the voltage is increased 4 times, how many times the space charge limited current will increase ? |
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Answer» will REMAIN same |
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| 4. |
An optical instrument used for angular magnification has a 25 D objective and a 20D eyepiece. The tube length is 25 cm when the eye is least strained. What is the angular magnification produced? |
Answer» SOLUTION :`m = (v_0)/(u_0). (D)/(f_e)` 
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| 5. |
A Zener diode is based on the principle of |
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Answer» THERMIONIC emission |
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| 6. |
In young's double slit experiment, the angular width of a fringe is found to be 0.2^@on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water as 4/3. |
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Answer» Solution :Formula :Angular fringe width =` beta/D = lamda/d` If APPARATUS is DIPPED in water, `lamda` changes to ` lamda_w = (lamda)/(n_w) = (lamda)/(4/3) = (3 lamda)/(4)` ` THEREFORE `NEW angular fringe width ` theta_w = (lamda_w)/(d)` `therefore (theta_w)/(theta) = (lamda_w)/(lamda)= ((3lamda)/(4))/(lamda) = 3/4` ` theta_w = 3/4 theta = 3/4 xx 0.2^@ = 0.15^@` |
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| 7. |
A potential difference of 2.5 V is applied across the faces of a germanium crystal plate. The face area of the crystal is 1 cm^(2) and its thickness is 1.0 mm. The free electron concentration in germanium is 2 xx 10^(19)m^(-3) and the electron and holes mobilities are 0.33m^(2)/V s and 0.17 m^(2)/V s respectively. The current across the plate will be |
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Answer» `0.2 A ` |
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| 8. |
Calculate the energy released in MeV in the following nuclear reaction : " "_(92)^(238)U to" "_(90)^(234)Th + " "_(2)^(4)He + Q mass of " "_(92)^(238)U = 238.05079 u, mass of " "_(90)^(234)Th = 234.043630 u, mass of " "_(2)^(4)He = 4.00260 u, 1 u=931.5 MeV/c^(2). |
| Answer» Solution :ENERGY released in the reaction `Q= [m (" "_(92)^(238)U) -m (" "_(90)^(234)Th) -m (" "_(2)^(4)He)]c^(2) =(238.05079 - 234.04363-4.00260) u xx c^(2)= (0.00456 u). c^(2) = 0.00456 xx 931.5 MeV= 4.25 MeV`. | |
| 9. |
An X-OR gate has following truth table: It is represented by following logic relation Y=barA.B+A.barB Build this gate using AND, OR and NOT gates. |
Answer» Solution : For above LOGIC circuit, Boolean EQUATION is `Y=Y_(1)+Y_(2)=barA*B+A*barB` which is CONSISTENT with the statement. |
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| 10. |
If the velocity of sound wave is 330m/s and wavelength of the wave is 0.5m, what is the frequency of the wave ? |
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Answer» SOLUTION :`V = nlamda` THEREFORE n = v/lamda` = 3300/0.5 = 660 HZ. |
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| 11. |
A time varying magnetic field B = Kr^(3)t is existing in a cylindrical region of radius 'R' as shown (where k is a constant). The induced electric field 'E' at r (R )/(2) is |
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Answer» `(KR^(4))/(20)` `E2pir = -(d)/(DT) [ - underset(0)overset(r )int B2pirdr]` `= (d)/(dt) [ underset(0)overset(r )int Kr^(3)t2pirdr]` `E2pir = (d)/(dt) (2pikt(r^(5))/(5))` `E = (Kr^(4))/(5)`, when `r le R` `:. At r = (R )/(2),E = (KR^(4))/(80)` 
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| 12. |
..........Was a spiritual leader than a politician. |
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Answer» NEHRU ji |
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| 13. |
A telescope uses lenses of focal lengths 75 cm and 5 cm. Calculate the minimum and maximum magnifying power which can be achieved with the instrument. |
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Answer» |
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| 14. |
The polarizing angle of glass is 57°. A ray of light which is incident at this angle will have an angle of refraction as |
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Answer» `38^(@)` `57^(@) + r = 90^(@) RARR r= 33^(@)` |
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| 15. |
Consider the decay scheme- A rarr B rarr C with lambda_(A) |
| Answer» Solution :Let `N_(1)` be the amount of parent atom of A at any instant `t-Delta t`.Therefore, ` N_(1)=N_(1)^(0) e^(-lambda_(A)(t-Deltat)`…..(1),Where `N_(1)^(0) ` is the number of parent atom of A at t=0.The number of daughter ATOMS at anytime t is given by `N_(2)=(lambda _(A)N_(1)^(0))/(lambda_(B)-lambda_(A))[e^(-lambda_(A)t)-e^(-lambda_(B)t)]`.......(2),For transient atom `lambda _(A) < lambda_(B)`, hence `e^(-lambda_(B))` , t term can be neglected in (2) `therefore N_(2)-(lambda _(A)N_(1)^(0))/(lambda_(B)-lambda_(A))e^(-lambda_(A)t)`........(3)Given thatactivity of A at `(t-Delta t) =activity of B at t.`implies``lambda _(A)N_(1)^(0) e^(-lambda _(A)(t-Delta t))=lambda_(B) (lambda_(A) N_(1)^(0) )/(lambda_(B)-lambda _(A))e^(-lambda_(A)t)`,`implies` `e^(-lambda _(A) Delta t)=(lambda_(B))/(lambda _(B)-lambda_(A))` `implies` `Delta t=(1)/(lambda_(A))In [(lambda_(B))/(lambda_(B)-lambda_(A))]`. The AVERGAE life of A `iota =(1)/(lambda_(A))` and that of B is `iota =(1)/(lambda_(B)) `,`therefore``Delta t =iota _(A)`In `[(1)/((1-lambda_(B))/(lambda_(A)))]`` implies` ` Delta t CONG iota _(A).(iota_(B))/(iota_(A))` as `iota _(B)/iota _(A) RARR 0` (given) `implies``Delta t= iota _(B)` | |
| 16. |
The change in kinetic energy of the particle is proportional to "____________". |
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Answer» CHANGE in its MASS and VELOCITY |
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| 17. |
The average number of neutrons produced in the fission of _92U^235 is |
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Answer» 1 |
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| 18. |
The photoelectric work function of potassium bis 2.3eV. If light having a wavelength of 2800Å falls on potassium, find (a) The kinetic energy in electron volts of the most energetic electrons emitted. (b) The stopping potential in volts are |
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Answer» SOLUTION :Given, `W=2.3eV""lamda=2800Å` `:.E("in "EV)=(12400)/(lamda("in"Å))=(12400)/(2800)=4.4eV` `K_(MAX)=E-W=(4.4-2.3)eV=2.1eV` (b) `K_(max)=eV_(0):.2.1eV=eV_(0)(or)V_(0)=2.1` VOLT |
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| 19. |
The migratory pattern of bird is one of the mysteries in the field of biology and indeed all of science. Explain this in the term of electromagnetic induction. |
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Answer» Solution :The migratory pattern of birds is one of the mysteries in the field of biology and indeed all of science. For example, every winter birds from Siberia fly unerringly to water spots in the Indian subcontinent. There has been a suggestion that electromagnetic induction MAY provide a clue to these migratory patterns. The Earth.s magnetic field has existed throughout evolutionary history. It would be of great benefit to migratory birds to use this field to DETERMINE the direction. As far as we know birds contain no ferromagnetic material. So electromagnetic induction seems to be the only reasonable MECHANISM to determine direction. CONSIDER the optimal case where the magnetic field `vecB` , the velocity of the bird `vecv` and two relevant points of its anatomy separated by a distance l, all the three are mutually perpendicular . From the formula for motional emf equation `epsilon`=Blv, `epsilon`=Blv Taking `B=4xx10^(-5)T`, l=2 cm wide and v=10 m/s, we obtain `epsilon=4xx10^(-5) xx2xx10^(-2) XX 10V= 8xx10^(-6)` V `=8 muV` This extremely small potential difference suggests that our hypothesis is of doubtful validity. Certain kinds of fish are able to detect small potential differences. However, in these fish, special cells have been identified which detect small voltage difference. In birds no such cells have been identified. Thus, the migration patterns of birds continues to remain a mystery. |
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| 20. |
What type of mirror can be used to obtain a real image of an object? Does this mirror produce a real imagefor all locationsof object? Explain. |
| Answer» Solution :A concave mirror can produce a real image. It can be used to produce a real image when OBJECT is BEYOND the FOCUS. For all the POSITIONS between the focus and POLE the image is virtual. | |
| 21. |
Two charges of magnitude -2Q and +Qare located at points (a,0) and (4a, 0) respectively ,What is the electric flux due to these charges through a sphere of radius 3a with its centre at the origin? |
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Answer» Solution :As shown in accordance with Gauss.law the ELECTRIC flux through the SPHERE surface. ` phi_E =-( 2Q )/( in_0) ` 
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| 22. |
A magnet 20 cm long with its polesconcentrated at itsends is palced vertically with itsnorth pole on the table at a pointdue 20 cmif horizontal compoent of the earth 's field is H=0.3 gauss then the polestrength of the magnet is nearly |
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Answer» 1.85 Am Magnetic induction at P due to N pole `=(m)/(20)^(2)` along NO `therefore` Resultantinduction at o at the horizontaldirection `=(m)/(20^(2))-(m)/(20sqrt(2)^(2) cos theta` `=(m)/(400)xx0.6465`  At NEUTRAL point `(m)/(400) xx0.6465=H=0.3` then m=185 ab-A-cm |
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| 23. |
A nuclear af rest undergoes a decay emitting and alpha- particle of de-Broglie wavelength lamda=5.76xx10^(-15) meter. he mass of the daughter nucleus is 223.610 amu and that of the alpha- particle is 4.002 a.m.u. (1"amu"=931.470(MeV)/(C^(2))) Mass of the nucleus is |
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Answer» 227.605 amu |
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| 24. |
A nuclear af rest undergoes a decay emitting and alpha- particle of de-Broglie wavelength lamda=5.76xx10^(-15) meter. he mass of the daughter nucleus is 223.610 amu and that of the alpha- particle is 4.002 a.m.u. (1"amu"=931.470(MeV)/(C^(2))) The total kinetic energy of the system in the final state is ( m_(d)= mass of daughter nucleus) m_(alpha)= mass of alpha- particle |
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Answer» <P>`(P_(ALPHA)^(2)(m_(alpha)+m_(d)))/(2m_(alpha)m_(d)),(P_(d)="linear momentum of "alpha-"PARTICLE")` |
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| 25. |
One metallic ring of radius 2 m has a conducting rod connected between centre and rim of wheel. It rotates in a plane perpendicular to uniform magnetic field 1 T with frequency 25"rev"/s. Find emf induced across the rod. |
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Answer» |
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| 26. |
A box with a mass of 2kg accelerates in a straight line from 4 m/s to 8m/s due to the application of a force whose duration is 0.5 s. find the average strength of this force. |
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Answer» 2N `overline(F)Deltat=Deltap=p_(f)-p_(i)=m(v_(f)-v_(i))implies overline(F)=(m(v_(f)-v_(i)))/(Deltat)=((2kg)(8m//s-4m//s))/(0.5s)=16N` |
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| 27. |
A cathode ray tube has potential difference V between the cathode and anode. The speed v of electrons ejected is : |
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Answer» `vpropV` |
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| 28. |
(a) Using the phenomenon of polarisation, show how transverse nature of light can be demonstrated. (b) Two polaroids P_1 and P_2 are placed with their pass axes perpendicular to each other. Unpolarised light of intesity I_0 is incident on P_1. A third polaroid P_3 is kept in between P_1 and P_2 such that its pass axis makes an angle of 30^@ with that of P_1. Determine the intensity of light transmitted through P_1, P_2 and P_3. |
Answer» SOLUTION : Intensity of LIGHT coming from `P_1 = (I_0)/2` Intensity of light coming from `P_2 = (I_0)/2 cos^2 30^@` `= (I_0)/2 ((SQRT3)/(2))^(2) = (3I_0)/(8)` Intensity of light coming from `P_3 = (3I_0)/8 cos^2 60^@` `= (3I_0)/(8) (1/2)^2= (3I_0)/(32)`. |
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| 29. |
When ironor steel retain their magnetism so long maagnetising force is applied is called ? |
| Answer» SOLUTION :TEMPORARY MAGNET | |
| 30. |
"Electric current has both direction and magnitude" - Even though current has both magnitude and direction it is not a vector quantity. Why? |
| Answer» Solution :AVERAGE VELOCITY of themmal MOTION is ZERO. Hence thermal motion does not PRODUCE current | |
| 32. |
When ordinary light is made incident on a quarter wave plate, the emergent light is |
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Answer» LINEARLY polarised |
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| 33. |
What is responsible for depletion barrier in depletion layer? |
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Answer» Ions DEPLETION barrier is formed due to accumulation of +ve and -ve ions near both the SIDE of p-junction. |
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| 34. |
A slit of width 'd' is illuminuted by red light of wavelength 6500 Å. For what value of 'd' will (i) the fiirst minimum fall at an angle of defraction of 30^(@) and (ii) the first maximum fall at an angle of diffraction of 30^(@). |
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Answer» |
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| 35. |
For the given transitions of electron, obtain the relation betweenlambda_(1), lambda_(2) & lambda_(3). |
Answer» Solution :For given CONDITION `E_(3)-E_(1)=(E_(3)-E_(2))+(E_(2)-E_(1))`  `rArr (hc)/(lambda_(3))=(hc)/(lambda_(2))+(hc)/(lambda_(1)) rArr (1)/(lambda_(3))=(1)/(lambda_(2))+(1)/(lambda_(1))` THEREFORE `lambda_(3)=(lambda_(2)lambda_(1))/(lambda_(2)+lambda_(1))` |
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| 36. |
The potential difference applied to an X-ray tube is 5kV and current is 3.2 mA. Then no. of electrons striking the target per sec is : |
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Answer» `2xx10^(16)` `i=3*2xx10^(-3)A t=1s` `:.n=(3*2xx10^(-3))/(XX1*6xx10^(-19))=2xx10^(16)` |
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| 37. |
A thin rod of length 1 m is suspended from its end and is made to oscillate in a vertical plane. The distance between the point of suspension and centre of oscillation will be : |
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Answer» `(1)/(2)m` |
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| 38. |
A positive point charge Q is placed (on the axis of disc) at a distance of 4 R . Above the centre of a disc of radius R as shown in situation I. the magnitude of electric flux through the disc is (phi). Now a hemis pherical shell of radius R is placed over the disc. such that it forms a closed surface as shown in situationII. The flux through the curved suface in situation II taking direction of area vector along outward normal as positive is |
| Answer» ANSWER :C | |
| 39. |
A coil of resistance 20Omegaand inductance 0.5 henry is switched to dc 200 volt supply. Calculate the rate of increase of current: a) at the instant of closing the switch and b) after one time constant. c) Find the steady state current in the circuit. |
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Answer» Solution :a) This is the case of GROWTH of current in an L-R circuit. Hence , currentat TIME t is given by `i=i_0(l-e^(-t//tau_L))` RATE of increase of current, `(di)/(dt) = (i_0)/(tau_L) e^(-t//tau_L)` At t` = 0 (di)/(dt) = (i_0)/(tau_L) = (E//R)/(L//R) = E/L` `(di)/(dt) = (200)/(0.5) = 400 A//s` b) At `t = tau_L` `(di)/(dt) = (400)e^(-1) = (0.37)(400) = 148A//s` c) The steady state current in the circuit , its `i_0 = E/R = 200/20 = 10A` |
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| 40. |
In fig. a rescue plane flies at 198 km/h ( = 55.0 m/s)and constant height h = 500 m toward a point directly over a victim , where a rescue capsule is to land . (a) what should be the angle phi of the pilots line of sight to the victim when the capsulerelease is made ?(b) as the capsule reaches the water , what is the velocity vecv? |
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Answer» Solution :One released , the capsule is projectile , so the horizontal and verical MOTIONS can be considered separately (we need not consider the actual curved path of the capsule) In fig. we see that `PHI` is given by `phi = tan^(-1) x/h`  where x is the horizontal coordinate of the victim (and of the capsule when it hits the water ) and h= 500 m . we should be able to find x with eq. ` x - x_0 (v_ cos theta_0) y` To find t, we next consider the vertical motion and specifically eq. ` x- y_0 = (v_0 sin theta_0) t - 1/2 "gt"^2` Here verical displacement y - `y_0` of the capsule is -500 m( the negative value indicates that the capsule MOVES downward). so ` - 500m = (55.0 m//s) sin 0^@) y - 1/2 (9.8 m//s^2) t^2` solving for t . we find t = 10.1 s. using that valuein eq. yields `x - 0 = (55.0 m//s) (cos 0^@) (10.1 s)` x = 555.5 m The eq. gives us `phi = tan^(-1) (555.5m)/(500 m) = 48.0^@` (b) when the capsule reaches the water ` v_x = v_0 cos theta_0 = (55.0 m//s) (cos 0^@) = 55.0 m//s` using eq. and the capsule.s time of fall t = 10.1 s , we also find that when the capsule reaches the water , `v_y = v_0 sin theta_0 - "gt"` `= (55.0 m//s) (sin 0^@)- (9.8 m//s^2)(10.1 s)` = -99.0 m/s Thus , at the water `VECV = (55.0 m//s)hati - (99.0 m//s)HATJ ` From eq. the magnitude and angle of `vec v` are `v = 113 m//s andtheta = - 60.9^@` |
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| 41. |
Soap helps in cleaning clothes because |
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Answer» It increases the angle of CONTACT between the SOAP and the clothes |
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| 42. |
To calculate the rate of flow of a liquid, which of the following is used? |
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Answer» STOKE's law |
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| 43. |
Find the work done by an external agent in lowering a block of mass m slowly ona light of stiffness k, till it comes to rest. |
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Answer» Solution :`W= int F_("ext").DX=int_(0)^(x,0)(mg-kx)dx cos pi` `=-(mgx_(0)-(kx_(0)^(2))/(2))`...(a) Where `kx_(0)=(mg)/(k)`...(b) using (a) & (b) . `W_("ext")=-(m^(2)g^(2))/(2k)` 
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| 44. |
Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. Why ? |
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Answer» SOLUTION :(a) Ionosphere reflects waves in these bands. (b) Television signals are not properly REFLECTED by the ionosphere (see text). Therefore, reflection is effected by satellites. (c) Atmosphere absorbs X-rays, while visible and radiowaves can penetrate it. (d) It absorbs ultraviolet radiations from the SUN and prevents it from reaching the earth’s surface and causing damage to life. (e) The temperature of the earth would be lower because the Greenhouse EFFECT of the atmosphere would be absent. (f) The clouds produced by global NUCLEAR war would perhaps cover substantial parts of the sky preventing solar light from reaching many parts of the globe. This would cause a ‘winter’. |
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| 45. |
At the surface of a charged conductor electrostatic field must be normal to the surface at every point'. Explain. |
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Answer» Solution :If `vecE` were not normal to the surface, it WOULD have some non-ZERO component along the sudace. Free charges on the surface of the conductor would then experience FORCE and move. Hence, conductor does not remains in stable siluation. Therefore `vecE` should have no TANGENTIAL component pardllel lo the surface in stable situation. Thus, electrostatic field at the surface of a charged conductor MUST be normal to the surface at every point. (For a conductor without any surface charge density, field is zero even at the surface). 
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| 46. |
A particle starts its S.H.M. from the mean position at t=0. If the time period is T, then it will beat the extreme position after a time |
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Answer» `T/4` |
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| 47. |
In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment. |
| Answer» SOLUTION :`LAMBDA=(1.2xx10^(-2)xx0.28xx10^(-3))/(4xx1.4)m=600nm` | |
| 48. |
The ratio of momenta of an electron and an alpha particle which are accelerated from rest through a potential difference of 100 volts is : |
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Answer» <P>1 `U=(P^(2))/(2m)` (p is momentum) `:.(p_(e)^(2))/(2m_(e))=q_(e)V""....(1)` ` and (p_(alpha)^(2))/(2m_(alpha))=q_(alpha)V ""....(2)` DIVIDING we get `((P_(e))/(P_(alpha)))^(2)((m_(alpha))/(m_(e)))=(q_(e))/(q_(alpha))` `:.(p_(e))/(p_(alpha))= sqrt((q_(e))/(q_(alpha))*(m_(e))/(m_(alpha)))` `:. (q_(e))/(p_(alpha))=(1)/(2)` `:. (P_(e))/(P_(alpha))= sqrt((m_(e))/(2m_(alpha)))` |
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| 49. |
A disc of metal is melted and recast in the form of solid sphere. What will happen to the moment of inertia about a vertical axis passing through the centre ? |
| Answer» SOLUTION :The MOMENT of INERTIA will DECREASE/ | |
| 50. |
Order of magnitude of the following values can be determined as follows : |
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Answer» Solution :(a) `49 = 4.9 xx10^(1) ~~ 10^(1)` `:.`Order of magnitude `= 1` (b) `51 = 5.1 xx10^(1) ~~ 10^(2)` `:.` Order of magnitude `= 2` (C) `0.049 = 4.9 xx 10^(–2) ~~10^(–2)` `:.` Order of magnitude `= –2` (d) `0.050 = 5.0 xx 10^(–2) ~~ 10^(–1)` `:.`Order of magnitude `= –1` (e) `0.051 = 5.1 xx10^(–2) ~~ 10^(–1)` `:.`Order of magnitude `= –1` |
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