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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 closed organ pipe and an open pipe of same length produce 4 beats when they are set intovibrations simultaneously. If the length of each of them were twice their initial lengths, the numberof beats produced will be__ |
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Answer» |
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| 2. |
A solid sphere of (radius = R) rolls without slipping in a cylindrical vessel (radius = 5R). Find the time period of small of oscillations of the sphere in s^(-1). Take R = (1)/(14)m and g = 10 m//s^(2). (axis is cylinder is fixed and horizontal). |
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Answer» `v = Romega` `omega'` = ANGULAR VELOCITY of `COM` of sphere `C` about `O` `= (v)/(4R) = (Romega)/(4R) = (omega)/(4)` `:. (domega')/(dt) = (1)/(4)(domega)/(dt) RARR alpha' = (alpha)/(4)` or `alpha' = (a)/(R)` for pure rolling where, `a = (gsintheta)/(1 + (1)/(mR^(2))) = (5gsintheta)/(7)` as, `I = (2)/(5)mR^(2)` For small `theta, sintheta approx theta`, being restoring in nature. `alpha' = -(5g)/(28g)theta` `:. T = 2pi SQRT((|(theta)/(alpha')|) = 2pisqrt((28R)/(5g))`. |
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| 3. |
The dielectric constant of air is 1.006. The speed of electromagnetic wave travelling in air is axx10^8 ms^(-1), where a is about: |
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Answer» 3 |
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| 4. |
Plum-pudding model of atom was suggested by: |
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Answer» Rutherford |
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| 5. |
Two equal negative charges -q are fixed at point (0, a) and (0, -a). A positive charge Q is released from rest at the point (2a, 0) on the x-axis. What type of oscillations does the charge Q execute? |
Answer» Solution : LET X be the DISPLACEMENT of charge Q at any time t. Now by finding the net force on Q and then its acceleration .a. `a= 2kQ q(-x)/(m(x^(2) + l^(2))^(3//2))` We can find that `a prop -x` not valid. Therefore charge Q will EXECUTE oscillations but not SHM. |
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| 6. |
The radiation corresponding to 3to2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3xx10^(-4)T. If the radius of largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to |
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Answer» 1.6 eV |
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| 7. |
When a parallel plate capacitor is connected across a d.c. battery. Explain briefly how the capacitor gets charged . |
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Answer» Solution :When a PARALLEL plate capacitor is connected to a BATTERY, positive charge from + ve terminal of battery flows and charge the plate A of the capacitor . As plate B of capacitor is close to plate A , an equal amount of negative charge is induced on inner surface of plate B and positive charge is induced on OUTER surface of plate B . However , negative charge coming from -ve terminal of battery nullifies thi positive charge of outer surface of plate B . Thus , effectively we GET equal and opposite charges on inner surfaces of plates A and B of the capacitor and we say that the capacitor is charged . 
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| 8. |
A coil of inductance 8.4 mH and resistance 6 Omega are connected to a 12 V battery. At what time the current in the coil will be 1.0 A ? |
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Answer» SOLUTION :`i_(0)=(E)/(R) =2`, Given current, `i= (i_(0))/(2)` `:. i=i_(0)(1-e^(-(t)/(tau))) implies (1)/(2)=1-e^(-(t)/(tau))` `implies e^((t)/(tau))=2"":. t=tau ln 2=(L)/(R)ln2` i.e., `t=(8.4 xx 10^(-3) xx 0.6931)/(6) ~~ 1 xx 10^(-3) s` |
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| 9. |
Show that the mass of radium (""_(88)^(226)Ra) with an activity of 1 curie is almost a gram. Given T_(1//2) = 1600 years. |
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Answer» Solution :The activity of the sample at any time t `R = lambda N` Here `lambda = (0.6931)/(T_(1/2))` ` R = 1 Ci = 3.7 xx 10^(10) dis s^(-1)` `T_(1/2) = 1600 year = 1600 xx 3.16 xx 10^(7)` dis `therefore` The amountof RADIUM, `N = (R)/(lambda) = (RT_(1/2))/(0.6931)` ` = (3.7 xx 10^(10) xx 1600 xx 3.16 xx 10^(7))/(0.6931)` ` (18707.2 xx 10^(17))/(0.6931)` `= 26990.62 xx 10^(17)` `N = 2.7 xx 10^(21)` atoms As 226 G of radium CONTAINS `6.023 xx 10^(23)` atoms so the amount of required strength. ` = (226 xx 2.7 xx 10^(21))/(6.023 xx 10^(23))` `= 101.311 xx 10^(-2)` ` = 1.013 g approx 1 g` |
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| 10. |
Kichhoff's junction role represents……. |
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Answer» CONSERVATION of LINEAR momentum |
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| 11. |
In an L-R circuit connected to a battery of constant emf E, switch is closed at time t = 0. If e denotes the induced emf across the inductor and i the current in the circuit at any time t. Then the graph shows the correct variation of .e. with .i. is |
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Answer»
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| 12. |
During scattering of light, the amount of scattering is inversely proportional to .............. |
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Answer» SOLUTION :During scattering of light, the amount of scatteing is inversely proportional to fourth power of wavelength of light i.e., `I alpha (1)/(lambda^(4))` |
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| 13. |
Use the mirror equation to deduce that a.An object placed between f and 2f of a concave mirror produces a real image beyond 2f. |
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Answer» Solution :a.`(1)/(u) + (1)/(v) = (1)/(f) , (1)/(V) = (1)/(f)- (1)/(u) `............... (1) f and u are negative . i.e, f `lt` 0 and `u lt 0` . Object is placed betweenf and 2f, then `2f lt u lt f or (1)/(2f) gt (1)/(u) gt (1)/(f) ` i.e., - `(1)/(2f) lt - (1)/(u) lt - (1)/(f) ` `(1)/(f) - (1)/(2f) lt (1)/(f) - (1)/(u) lt (1)/(f) - (1)/(f)` Using EQUATION (1) `RARR (1)/(2f) lt (1)/(v) lt 0 ` i.e., .v. is negative, the IMAGE FORMED is REAL and beyond 2f. |
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| 14. |
The conduction current is same as displacement current when source as: |
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Answer» a.C. only |
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| 15. |
The flux of magnetic field through a closed conducting loop changes with time according to the equation phi= 0.2t^(2) + 0.4t + 0.6. Find the induced e.m.f at t= 2s. |
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Answer» SOLUTION :`phi= 0.2t^(2) + 0.4t + 0.6` `E= - (d phi)/(dt) rArr e= - 0.4t- 0.4` when `t= 2S, e= -0.8 - 0.4 = - 1.2V` |
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| 16. |
What is the force of repulsion between two charges of IC each kept at 1m apart in vacuum ? |
| Answer» SOLUTION :9 XX 10^9 N. | |
| 17. |
Which of the following transitiona in hydrogen atoms emit photons of highest frequency? |
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Answer» N=6 to n=2 `E_(1)=(-13.6)/(1^(2))=-13.6eV` `hv=E_(1)-(E_(2))=-3.4 +13.6=10.2eV("MAX")`. v is the highest i.E.` n_(2)=2+0 n=1`. |
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| 18. |
Determine the electric flux of a flat square having an area of 10m^2 is a uniform electric field of 8000 N/C passing perpendicular to it |
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Answer» `1.8 xx 10^5 Nm^2 /C` |
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| 19. |
In a nuclear fusion reaction, two nuclei, A & B , fuse to produce a nucleus C , releasing an amount of energy DeltaE in the process.If the mass defects of the three nuclei are DeltaM_A, DeltaM_B & DeltaM_C respectively , then which of the following relations holds ? Here, c is the speed of light. |
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Answer» `DeltaM_A + DeltaM_B = DeltaM_C - DeltaE//c^2` |
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| 20. |
Four combinations of two thin lenses are given in List I. The radius of curvature of all curved surface is r and the refractive index of all the lenses is 1.5. Match lens combinations in List I with their focal length in List II and select the correct answer using the code given below the lists. |
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Answer» P-1, Q-2, R-3, S-4 |
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| 21. |
A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetizing current of 1.2A ? |
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Answer» 2.5 T |
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| 22. |
What is a capacitors ? |
| Answer» Solution :Capacitor is a device used to store electric CHARGE and ELECTRICAL energy. Capacitors are widely used in many electronic CIRCUITS and have applications in many AREAS of scienece and technology. | |
| 23. |
Abus accelerates from rest at a constant rate alpha for some time,after which it decerates at a constant rate beta to come to rest.If the total time elapes is t seconds then ,evaluate. (a)the maximum velocity achieved and (b)the total distance travelled graphically. |
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Answer» Solution :Let `t_(1)` be the time of acceleration and `t_(2)` that of decelration of the bus. The TOTAL time is `t=t_(1)+t_(2)`. Let `v_(max)` be the maximum velocity. As the acceleration and deceleration are CONSTANTS the velocity time graph is a straight line as shown in the figure with +ve slope for acceleration and -ve slope for deceleration.From the graph,the slope of the line OA gives the acceleration `alpha`. `THEREFORE alpha` =slope of the line `OA=(v_(max))/(t_(1))impliest_(1)=(v_(max))/(alpha)` The slope of AB gives the deceleration `beta` `thereforebeta=slope of AB=(v_(max))/(t_(2))impliest_(2)=(v_(max))/(beta)` `t=t_(1)+t_(2)=(v_(max))/(alpha)+(v_(max))/(beta)` `t=V_(max)((alpha+beta)/(ALPHABETA))` `therefore v_(max)=((alphabeta)/(alphabeta))` `thereforev_(max)=((alphabeta)/(alpha+beta))t` Displacement =area under the v-t graph  |
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| 24. |
What is this story- 'Guitar Player' about? |
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Answer» A GIRL who is a musician |
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| 25. |
(A) : Quasar emits radiowaves more than radio galaxy. (R) : Quasar has very small size. |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the correct EXPLANATION of 'A'. |
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| 26. |
Find the number of ratational levels per unit energy interval, dN//dE, for a diatomic molecule as a function of rotational energy E. Calculate of an magnitude for an iodine molecule in the state with rotational quantum number J=10. The distance between the nuclei of that molecule is equal to 267 p m. |
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Answer» Solution :From the FORMULA `J(J+1)( ħ^(2))/(2I)= E` we get `J(J+1)- 2IE// ħ^(2)` or `(J+(1)/(2))^(2)-(1)/(4)= (2IE)/( ħ^(2))` Hence `J=-(1)/(2)+sqrt((1)/(4)+(2IE)/( ħ^(2)))` writing `J+1= -(1)/(2)+sqrt((1)/(4)+(2I)/( ħ^(2))(E+DeltaE))` we find `1=sqrt((1)/(4)+(2I)/(ħ^(2))E+(2I)/(ħ^(2))DeltaE)-sqrt((1)/(4)+(2IE)/(ħ^(2)))` `sqrt((1)/(4)+(2I)/(ħ^(2))E)[(1+(DeltaE)/(E+(ħ^(2))/(8I)))^(1//2)]` `sqrt((1)/(4)+(2I)/(ħ^(2))E).(DeltaE)/(2(E+(ħ^(2))/(8I)))` `sqrt((2I)/(ħ^(2)))(DeltaE)/(2sqrt((E+(ħ^(2))/(8I))))` The QUANTITY `(dN)/(dE)is (1)/(DeltaE)`. For large `E` it is `(dN)/(dE)= sqrt((I)/(2 ħ^(2)E))` For an iodine molecule `I= M_(I)d^(2)//2= 7.57xx10^(-38)gm cm^(2)` Thus for `J= 10` `(dN)/(dE)=sqrt((I)/(2ħ^(2).(ħ^(2))/(2I)J(J+1)))=(I)/(sqrt(J(J+1)ħ^(2)))` Substitution gives `(dN)/(dE)= 1.04xx10^(4)` levels per `eV`. |
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| 27. |
What is Lenz's law? |
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Answer» Solution :The DIRECTION of INDUCED E.m.f. is in such a direction so as to OPPOSE the effect of change. `therefore E = -(dphi)/dt` where E = Induced e.m.f. and `dphi/dt` RATE of change of magnetic flux. |
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| 28. |
As the age of star increases |
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Answer» Helium quantity increases |
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| 29. |
A ball tied to a rope making an angle 30^(@) with the horizontal is held so that the rope is just If the ball is released, calculate the heat generated in joule when it crosses 30^(@) down the horizontal (dotted)Givenm=Kg,L=2m,g=10m//s^(2). |
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Answer» SOLUTION :The speed just after JERKING `v=sqrt(2gL) cos 30^(@)=sqrt((3gL)/(2))` `therefore` heat=10 SEC in kinetic ENERGY `=(1)/(2)m(2gL-(3gL)/(2))` `=(MGL)/(4)=5J` |
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| 30. |
The maximum deviation of sound carrier by audio signal is _____. |
| Answer» SOLUTION :`pm50kHz.` | |
| 31. |
Whare Bohr was confined ? |
| Answer» Solution :ALTHOUGH Bohr gave his theory of ATOMS. Yet he CONFINED his disccusion exclusively to hydrogen atom. | |
| 32. |
A bucket of water of mass 21 kg is suspended by a rope wrapped around a solid cylinder 0.2 m in diameter. The mass of the solid cylinder is 21 kg. The bucket is released from rest. Which of the following statement(s) is/are correct? |
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Answer» The tension in the rope is 70 N. 210- T= 21A ….(i) Again `tau- =Ialpha` `rArrTR=(la)/R` `rArrT=(IA)/(R^(2))=(1/2xx21xxR^(2)xxa)/(R^(2))` `=1/2xx21a` `21a=2T` From eq.(i) 210-T=2T or 3T= 210 or T= 70 N Again from eq.(i), 210 - 70 = 21a `a=(140)/21ms^(-2)=20/3ms^(-2)` 
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| 33. |
Consider the motion of a charged particle of mass m and charge q moving with velocity vecv in a magnitude field vecB. If vecv is perpendicular to vecB, show that it describe a circular path having angular frequency omega = (qB)/(m) |
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Answer» Solution :When a charged particle having charge q and moving with a velocity `vecv` enters a uniform magnetic field `vecB`, it experiences a Lorentz force `VECF = q(vecv xx vecB)` whose direction is given by Fleming.s left hand rule. The force is maximum when charged particle is moving perpendicular to the magnetic field. In that CASE `F = q v B sin 90^@ = q v B` As the force is acting perpendicular to both `vecv and vecB`, it does not change the magnitude of velocity but changes its direction from a STRAIGHT from a straight line path to a circular path of radius r such that the magnetic force - centripetal force NEEDED. `:. B q v = (mv^2)/(r) implies r = (mv)/(Bq)` The time period and frequency of the charged particle in its circular path will be `T = (2 pi r)/(v) = (2pi ((mv)/(Bq)))/(v) = (2 pi m)/(Bq) , ` Frequency `v = 1/T = (Bq)/(2 pi m)` and angular frequency `OMEGA = 2 pi v = (qB)/(m)`. From this relation it is clear that the time period (as well as frequency) is independent of the velocity (energy) of the charged particle as well as the radius of the circular path. 
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| 34. |
An electric tea kettle has two heating coils. When one of the coils is switched on, boiling begins in 6 min. When the other coil is switched on, the boiling begins in 8 min. In what time, will the boiling begin if both coils are switched on simultaneously (i) in series and (ii) in parallel. |
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Answer» Solution :Let power of first COIL is `P_(1)` and that of SECOND coil is `P_(2)`. Let H is the amount of heat required to oil water. Then `H=P_(1)t_(1)=p_(2)t_(2)` where `t_(1)="6 MIN,"t_(2)="8 min"` i) When the coils are connected in series : `P=(P_(1)P_(2))/(P_(1)+P_(2))""t=(H)/(P)=H[(1)/(P_(1))+(1)/(P_(2))]=H[(t)/(H)+(t_(2))/(H)]` `=t_(1)+t_(2)=6+8="14 min"` ii) When the coils are connected in parallel `P=P_(1)+P_(2)` `t=(H)/(P)=(H)/(P_(1)+P_(2))=(H)/((H)/(t_(1))+(H)/(t_(2)))=(t_(1)t_(2))/(t_(1)+t_(2))=(6xx8)/(6+8)="3.43 min."` |
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| 35. |
The density of the given gas at constant pressure and temperature is rho and its rateof diffusion isr. If density of the gas becone rho//3 then rate of diffusion becomes |
| Answer» ANSWER :D | |
| 36. |
Frequency of electromagnetic wave is 25MHz. The value of electric field intensity at any point at any time in wave in wave is 6.3 Vm^(-1). The value of magnetic field at that point is …… Wb//m^(2). |
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Answer» `3.9xx10^(-2)` `therefore C=(E )/(B)` `therefore B=(B)/(c )=(6.3)/(3xx10^(8))` `therefore B=2.1xx10^(-8)T` |
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| 37. |
Rutherford model shows that ...... |
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Answer» in central part of atom positive charge resides. |
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| 38. |
For photo senstive surface A and B graph of stopping potential versus frequency (v) is shown in figure .From graph work function of A surface ….. |
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Answer» More than that of B.  If given graph are EXTRAPOLATED then graph A will INTERACT V-axis in `V_(0)` Here `(V_(0))_(A)lt(v_(0))_(B)` hence, `(hv_(0))_(A)lt(hv_(0))_(B)` `THEREFORE phi_(A)ltphi_(B)` |
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| 39. |
When a proton moves opposite to electric field, work done on it by electric field is ..... . and electrostatic potential energy of proton ......, |
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Answer» NEGATIVE, INCREASES `:. W lt 0 (because "Here" vecF ` and `vecd` are in opposite directions ) Here potential energy of proton will increase. `( because ` its kinetic energy decreases ) |
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| 40. |
A current carrying loop in the form of a right angle isosceles triangle ABC is placed in a uniform magnetic field acting along AB. If the magnetic force on the arm BC is F, what is the force on the arm AC? |
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Answer» `-sqrt2vecF` |
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| 41. |
An alternating current of frequency 200 rad/sec and peak value 1 A as shown in the figure, is applied to the primary of a transformer. If the coefficient of mutual induction between theprimary and the secondary is 1.5 H , the voltage induced in the secondary will be |
| Answer» ANSWER :B | |
| 42. |
Is Ohm's law universally applicable for all condcuting elements? If not, give examples of elements which do not obey Ohm's law. |
| Answer» SOLUTION :No. EXAMPLES of non-ohmic ELEMENTS, vacuum DIODE, SEMICONDUCTOR diode. | |
| 43. |
A square plate of edge d and a circular disc of diameter d are placed touching each other at the midpoint of an edge of plate as shown. Then center of mass of the combination will be (assume same mass per unit area for the two plates): |
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Answer» `(2d)/(2 + pi)` left to the center of the DISC `= d/(1 + (pi)/4) = (4d)/(4 + pi) "right to centre of disc".` 
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| 44. |
A long jump athlete jumps at an angle theta for maximum distance . Value of theta is |
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Answer» `30^@` |
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| 46. |
What is principle of working of a potentiometer. |
| Answer» Solution :It works on the principle that when a steady current flows through a WIRE of UNIFORM cross section and composition, the potential drop ACROSS any length of the wire is directly PROPOTIONAL to it.s length. | |
| 47. |
IF X joule of work must be done to move electric charge to 4C from a place, where potential is -10V to another place. Where potential is 5V, then the value of X….J. |
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Answer» 30 `=4[5-(-10)]=5 times 15` `=60J=XJ` `therefore X=60` |
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| 48. |
How will a dia , para ferro magnetic material behave when kept in a non uniform external magnetic field ? Give one example of each of these materials. |
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Answer» SOLUTION :In a non - uniform magnetic field, a DIAMAGNETIC substance tends to move slowly from stronger to weaker parts of the field , a paramagnetic substance tends to move slowly from weaker to stronger part of the field, while a ferromagnetic substance MOVES easily from weaker to stronger parts of the field. Diamagnetic MATERIAL : BISMUTH Paramagnetic material : Aluminium Ferromagnetic material : Iron |
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| 49. |
Protons with kinetic energy T strike a stationary hydrogen target. Find the threshold values of T for the following reactions: (a) p+prarr p+p+p+ overset~p , (b) p+prarr p+p+pi^(0). |
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Answer» SOLUTION :(a)For `p+prarrp+p+p+barp` `TgeT_(TH)=(16m_(p)^(2)-4m_(p )^(2))/(2m_(e ))C^(2)=6m_(p)c^(2)= 5.63 GeV` (b) For `p+prarrp+p+pi^(o)` `Tge T_(th)=((2m_(p)+m_(pi^(0))^(2))-4m_(p)^(2))/(2m_(p))c^(2)` `=((2m_(pi)+m_(pi^(o))^(2))/(2m_(p)))c^(2)= 0.280GeV` |
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| 50. |
Magnetic needle (compass) point 3.5 ^(@) west of geographic north. Another needle free to rotate in vertical plane parallel to magnetic meridian has its north tip pointing down at 18^(@) with the horizontal. The magnitude of the horizontal component of the earth's magnetic field at the place is 0.40 G. What is the magnitude of earth's magnetic field at that place? (Given cos 18^(@) = 0.95, sin 18^(@) = 0.31) |
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Answer» Solution :`B _(H) = B_(e) COS del` ` 0.40 = B_(e) cos 18^(@)` `0.40 = B_(e) XX 0.95` `B _(e) = (0.40)/( 0.95) G` `B _(e) = 0.42 G` 
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