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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 figure shows a diagonal symetric arrangement of capacitors and a battery. If the potntial of C is zero, then |
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Answer» `V_A = +20V` |
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| 2. |
a.Every metal has a definite work function. Why all photoelectronsdo not come out with the same energy of incident radiation in monochromatie ? Why there is an energy distribution of photoelectrons? b. The energyand momentumof an electron are related to the frequency and wave-length of the associated matter wave by the relations, E=h upsilon, p=(h)/(lambda) But while the value of lambda is physically significant, the value of upsilon (and therefore the value of the phase speed upsilon lambdahas no physical significance. Why? |
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Answer» Solution :a. Work functionis MINIMUM amount of energy required for the electron in the topmost level of the CONDUCTION band to get out of the metal. All the electrons do not belong to this topmost level. Different electrons are knocked off from the metal from different levels, with different energyfor the same incident radiation. Thus the emitted electrons occupy a continuous band of ENERGYLEVELS. B. The absolute VALUE of energyE (but not momentum, p) of any particle is arbitraryto within an additive constant. Hence absolute value of `.upsilon.` of a metter wave has no direct physical meaning, while `.lambda.` is physically significant. The phase speed `(upsilon lambda)`is physically insignificant, while group speed is physicallysignificant. |
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| 3. |
Two wires AO and OC carry equal currents i as shown in the figure where |__AOC = theta ,Find the magnitude of the magnetic field at the point P on thebisector of anlge theta at a distance 'r' from O. Assume that the other ends of both wires extend to infinity |
Answer» Solution : The direction of magnetic field due to both the wires AO and OC are out of PAGE of paper Due to wire OC `B = (mu_0i)/(4pid){SIN phi_1 + sin phi_2}` `B = (mu_0i)/(4pisin""theta/2){sin (90 -theta/2) + sin 90}` `B = (mu_0i)/(4pi sin ""theta/2)(1 + cos"" (theta)/(2))` Due to both the wires, the net magnetic field is `B = (2 XX mu_0i)/(4pir sin""theta/2)(1 + cos""theta/2)` `B = (2 xx mu_0i)/(2pir sin""theta/2)(1 + cos""theta/2)` `B = (mu_0)/(2pi) (i)/(r) cot ""(theta)/(4)` |
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| 4. |
The substance having the least specific heat among the following is |
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Answer» Copper |
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| 5. |
In the circuit given A , B and C are inputs and Y is the output The output Y is |
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Answer» high for all the high INPUTS |
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| 6. |
The dipole moment of a hydrogen chloride molecule is 3.44 xx 10^(-30) C m, the separation of the dipole is 1.01 xx 10^(-10) m. Estimate the energy liberated in the course of formation of 1 kg of hydrogen chloride from the starting ma terials, if the number of molecules in 1 kg is 1.6 xx 10^(25). |
| Answer» Solution :The ENERGY liberated as the result of the formation of one HYDROGEN chloride molecule is about `U = 10.3 xx 10^(-20) J = 0.65 eV`. Per 1 kg we obtain `E = NU = 1.65 xx 10^(6)` J. (Actually, the energy liberated in the PROCESS of formation of 1 kg of hydrogen chloride is 2.5 MJ, this MEANS that our rough estimate GAVE the right order of magnitude.) | |
| 7. |
The figure shows the square face (of side 'a') of a transparent cuboidal block. The thickness or the third dimension of the block is negligible in comparison to 'a'. The block has uniform refrative index mu equal to 2. A point source S which can emit light in all directions can move inside the block. It is desired that no light of 'S' should pass through AB. Sketch the region in which S should be present to satisfy this condition. |
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Answer» If `S` is ANYWHERE in the shaded region, the LIGHT RAYS from `S` will strike `AB` MAKING an angle more than critical angle. 
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| 8. |
There are n electrons of charge on a drop of oil of density rho. It is in equilibrium in an electric field E. Then radius of drop is |
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Answer» `[(2n e E)/(4pi RHO G)]` |
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| 9. |
How does the saturation current in a n-p-n junction diode change. with increase in temperature ? |
| Answer» Solution :The saturation CURRENT DOUBLES for every `10^@C` rise. | |
| 10. |
if 'a' is the number of distributions of six different balls into three different boxes . Also, 'b' is the number of distributions of five different balls into three identical boxes . Further , 'c' is the number of distribution of four identical balls into three different boxes. In each case no box remains empty. Then a/(b-2c-1) is equal to |
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Answer» 3 b=25 c=3 |
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| 11. |
For the experimental determination of Young.s modulus of elasticity Y of the material of a given wire by using Searle.s apparatus, the spherometer reading while load increasing and decreasing are tabulated below: The mean extension for 2kg load is |
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Answer» `2.240` |
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| 12. |
For the situation shown in the figure, what should be the value of force required to move the rod with constant velocity v = 2 m/s ? |
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Answer» `3.75xx10^(-3)` N `THEREFORE I= (Bvl)/R` Now, magnetic force `F_m=BI L` `=(B(Bvl))/R. l=(B^2vl^2)/R` `therefore F_m=((0.15)^2xx2xx(0.50)^2)/3` `=0.00375=3.75xx10^(-3)` N |
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| 13. |
On inverstigation of light from three different stars A,B and C it was found that in the spectrum of A, the intensity of blue colour is maximum and in C the intensity of yellow colour is maximum in A. the intensity of yellow colour is maximum in A. from the observation it can be concluded that |
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Answer» the TEMPERATURE of A is MAXIMUM, B is MINIMUM and C is intermediate |
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| 14. |
In a metere bridge. The balance length from left end (standard resistance of 1Omega is in the right gap) is found to be 20cm the length of resistance in left gap is 1/2 m and radius is 2mm its specific resistance is |
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Answer» `pixx10^(-6)ohm-m` |
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| 15. |
Hydrogen atoms are excited from ground state to the state of priciple quantum number 4.Then the number of spectral lines observed will be |
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Answer» 3 `therefore` the NUMBER of spectral lines EMITTED `= (n(n-1))/(2) = (4 xx 3)/(2) = 6` |
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| 16. |
A permanent magnet |
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Answer» attaract all SUBSTANCES |
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| 17. |
(a) Show that a current carrying solenoid behaves like a small bar magnet . Obtain the expression for the magnetic field at an external point lying on its axis. (b) A steady current of 2 A flows through a circular coil having 5 turns of radius 7 cm. The coil lies in x - Y plane with its centre at the origin. Find hte magnitude and direction of the magnetic dipole moment of the coil. |
Answer» SOLUTION :(a) A current carrying solenoid behaves like a small bar magnet field of solenoid is similar to that of a bar magnet. Magnet field lines inside a current carrying solenoid are parallel straight lines along its axis . Outside the solenoid, is cut into two pieces then we get two smaller solenoids with weaker magnetic properties just like a bar magnet. Consider a solenoid of length .2l. and radius .a. `(a lt lt 2l)` and having .n. turns per unit length . Let us calculate magnetic field B at a point P on its axis situated at a distance .r. form the centre `O(r gt l)`For this PURPOSE , we consider a circular element of solenoid coil situated at a distance x from centre and of thinness .dx. so that it contains .ndx. number of turns.  This element may be considered a small circular coil and magnetic field due to it at point P will be `dB=(mu_0(ndxI)a^2)/(2[(r-x)^2+a^2]^(3//2))` However , as `a lt lt l` (and hence `a lt lt r ` ALSO) , the above relation may be simplified as : `dB=(mu_0(ndxI)a^2)/(2(r-x)^3)` `:.` Magnitude of total magnetic field at P due to entire solenoid. `B = intdB=int_(-l)^(+l)(mu_0nIa^2dx)/(2(r-x)^3)=(mu_0nIa^2)/2.[(1)/(2(r-x)^2)]_(-l)^(+l)` `=(mu_0)/(4pi).(4pia^(2)nIrl)/((r^(2)-l^2))=(mu_0)/(4pi).(2(nIpia^(2)2l)r)/((r^2-l^2)^2)` `=(mu_0)/(4pi).(2mr)/((r^2-l^2)^2)`, where `nIpia^(2).2l=m=`magnetic dipole moment of solenoid coil. If `r gt gt l`, then we have `B=(mu_0)/(4pi) .(2m)/r^3` (b) Here current I = 2 A, no of turns in the coil N = 5 and RADIUSA= 7 cm = 0.07 cm `:.` Magnetic dipole moment m NIA `NIpia^2=5xx2xx(22)/7xx(0.07)^2=0.308Am^(2)=0.31A m^2` Since coil lies in X - Y PLANE, the magnetic dipole moment is alongZ - axis i.e., `vecm=0.31 hatkAm^2` |
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| 18. |
A Copper atom has 29 electrons revolving around the nucleus. A copper ball contains 4 xx 10^(23) atoms. What fraction of the electrons be removed to give the ball a charge of +9.6 mu C? |
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Answer» `~1.8 XX 10^(-13)` |
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| 19. |
A body is rotating ? Is it necessarily being acted upon by an external torque. |
| Answer» SOLUTION :No, TORQUE is REQUIRED only for ANGULAR ACCELERATION | |
| 20. |
A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge or 3.5 muC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends). |
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Answer» SOLUTION :`l=15 cm =15 xx 10^(-2)m ""a=1.4cm =1.4 xx 10^(-2)m"b=1.5cm =1.5 "xx 10^(-2)m Q=3.5muC` `C=(2pi epsi_(0) l)/(2.303 LOG(b/a))=(1 xx 15 xx 10^(-2))/(2 xx 3.303 log ((1.5)/(1.4)) xx 9 xx 10^(9))=1.2 xx 10^(-10)F` `V=Q/(2pi epsi_(0)l) log (b/a) =(2 xx 9 xx 10^(9) xx 3.5 xx 10^(-6))/(15 xx 10^(-2)) log""((1.5)/(1.4))=2.9 xx 10^(4)V` |
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| 21. |
Monochromatic light of frequency 6.0xx10^(14)Hz is produced by a laser. The power emitted is 2.0xx10^(-3)W. (a). What is the energy of a photon in the light beam? (b) How many photons per second, on an average are emitted by the source? |
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Answer» <P> Solution :a.Each PHOTON has an energy`E=h upsilon=(6.63.10^(-34)Js)(6.0xx10^(14)Hz)=3.98xx10^(-19)J` b. If N is the number of photons emitted by the source per second, the power P transmitted in the BEAM equalsN times the energyper photon E, so that P = N E. Then `N=(P)/(E )=(2.0xx10^(-3))/(3.98xx10^(-19))=5.0xx10^(15)` photons per second. |
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| 22. |
Give the action of diode as valve? |
| Answer» Solution :A VALVE open only on way i.e.,valve allow FLOWS in onlyone direction.Consider an ac applied to a junction DIODE with P is +ve and N is -ve.During the next half cycle,when p is -ve and N is +ve the diode is reverse biased and it will not CONDUCT. | |
| 23. |
The displacement x and time t for a particle are related to each other as t-3=sqrt x . What is work done in first six seconds of its motion. |
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Answer» A. 0J |
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| 24. |
A point charge is positioned at the centre of the base of a square pyramid as shown. The flux through one of the four identical upper faces of the pyramid is, (q)/(Pxepsilon_(0)) then P is |
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Answer» |
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| 25. |
Two long and parallel straight wires A and B carrying currents of 10.0 A and 4.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A. |
Answer» Solution : Situation given in the STATEMENT is shown in above diagram. Here, MAGNETIC force exerted on L length of first wire is, `F=(mu_(0)I_(1)I_(2)l)/(2piy)` `thereforeF=((4pixx10^(-7))(8)(5)(0.1))/((2pi)(0.04))` `thereforeF=2xx10^(-5)N` (Direction : towards wire B, attractive in nature) |
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| 26. |
Give the circuit symbol of AND-gate. |
Answer» SOLUTION :
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| 27. |
Consider a harmonic wave travelling on a string of mass per unit length mu The wave has a velocity v, amplitude A and frequency f. The power transmitted by a harmonic wave on the string is proportional to (take constant of proportionality as 2pi^2) |
| Answer» ANSWER :A::B::C::D | |
| 28. |
A closely wound solenoid 80 cm long has 4 layers of windings of 500 turns each. The diameter of the solenoid is 1.7 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre. |
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Answer» Solution :MAGNETIC field INSIDE an extremely long current carrying solenoid is, `B=mu_(0)nI` `thereforeB=mu_(0)((Np)/l)I` (Where p = no. of layers in a solenoid) `thereforeB=((4pixx10^(-7))(400xx5)(8))/((0.8))` `thereforeB=2.5xx10^(-2)T` (Parallel to AXIS of solenoid) |
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| 29. |
Who is the poet of this poem? |
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Answer» JOHN Keats |
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| 30. |
If the speed of light were 2/3 of its present value, what would be fractional decrease in the energy released in a given atomic explosion |
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Answer» Solution :ENERGY `prop c^2` FRACTIONAL decrease in energy =`(E_1-E_2)/E_1=1-(C_2/C_1)^2` `=1-(2/3)^2=5/9` |
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| 31. |
If the distance between two equal charges is doubled and their individual charges are also doubled. What will happen to the force between them ? |
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Answer» SOLUTION :`F= 1/4pi epsilon_0 xx Q xx q/r^2` ` F = 1/ 4pi epsilon_0 2q.2q/(2R)^2 = F` |
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| 32. |
Draw a neat Labelled diagram of Cyclotron. |
Answer» SOLUTION :
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| 33. |
The frequency of oscillation of current in the indcutor is |
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Answer» `(1)/(3sqrt(LC))` |
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| 34. |
We call each point on a wavefront is a source of fresh disturbance called . |
| Answer» SOLUTION :SECONDARY WAVEFRONT | |
| 35. |
Show that the wavelength of electromagnetic radiation is equal to the de-Broglie wavelength of its quantum (photon). |
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Answer» Solution :For PHOTON de-Broglie wavelength `lambda=(h)/(p)` , Momentum of electromagnetic RADIATION having v frequency and `lambda` wavelength, `p=(e )/(C )=(hv)/(c )` `THEREFORE p=(h)/(c ).(c )/(lambda.)[because v=(c )/(lambda.)]` `therefore p=(h)/(lambda.)` `therefore lambda.=(h)/(p)` but `(h)/(p)=lambda` de-Broglie wavelength of photon `therefore lambda.=lambda` Thus `lambda.` is wavelength of electromagnetic radiation which is equl to de-Broglie wavelength of phton with wavelength `lambda`. |
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| 36. |
A radioactive sample has 2.6 mu g of pure ""_(7)^(13)"N" which has a half - life of 10 minutes. (a) How many nuclei are present initially? (b) What is the activity initially?(c) What is the activity after 2 hours? (d) Calculate mean life of this sample. |
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Answer» Solution :(a) To find `N_(0)` we have to find the number of `""_(7)^(13)"N"` atoms is `2.6 mu g`. The atomic mass of nitrogen is 13. Therefore, 13 g of `""_(7)^(13)"N"` contains Avogadro number `(6.02 xx 10^(23))` of atoms. In 1 g, the number of `""_(7)^(13)"N"` is equal to be `(6.02 xx 10^(23))/(13)` atoms. So the numberof `""_(7)^(13)"N"` atoms in `2.6 mug` is `N_(0) = (6.02 xx 10^(23))/(13) xx 2.6 xx 10^(-6) = 12.04 xx 10^(6)` atoms. (b) To find the initial activity `R_(0)`, we have to evaluate decay CONSTANT `lambda` `lambda = (0.6931)/(T_(1/2)) = (0.6931)/(10 xx 60) = 1.155 xx 10^(-3) s^(-1)` Therefore `R_(0) = lambda N_(0) = 1.155 xx 10^(-3) xx 12.04 xx 10^(16)` `= 13.90 xx 10^(13) (decays)/(s)` `= 13.90 xx 10^(13)Bq` In terms or a curie, `R_(0) = (13.90 xx 10^(13))/(3.7 xx 10^(10)) = 3.75 xx 10^(3)Ci` since 1Ci = `3.7 xx 10^(10)Bq` (C) Activity after 2 hours can be calculated in TWO different ways: Methid 1 : `R = R_(0)e^(-lambda t)` At t = 2 hr = 7200 s `R = 3.75 xx 10^(3) xx e^(-7200) xx 1.155 xx 10^(-3)` `R = 3.75 xx 10^(3) xx 2.4 xx 10^(-4) = 0.9 Ci` Method 2 : `R = ((1)/(2))^(n) R_(0)` Here `n = (120 min)/(10 min) = 12` ` R = ((1)/(2))^(12) xx 3.75 xx 10^(3) approx 0.9 Ci` (d) mean life `tau = (T_(1/2))/(0.6931) = (10 xx60)/(0.6931) = 865.67 s` |
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| 37. |
Two point source S_(1) and S_(2) separated by a distanc 10 mu m in phase. A circular wire of radius 40mu m is placed around the sources as shown in figure, then (O is the centre of the circle and OS_(2) = OS_(2)) |
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Answer» point A and B are dark and points C and D are BRIGHT `Delta x_(B) = S_(1)B - S_(2)B = 0` and at point D, `Deltax_(D) = S_(1)D - S_(2)D = 0` Phase difference, `phi = (2pi)/(lamda) xx (0) = 0` So, intensity at points B and D `I = I_("max") cos^(2) (phi//2) = I_("max")` So, points B and D are bright. Phase difference at point A, `Deltax_(A) = 35mu m [40-5]` But, FIGURE shows `Deltax_(A) = Delta X_(C )` is maximum for given circle. When path difference is maximum, intensity will be MINIMUM. So, point A and C are dark. 
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| 38. |
In which atomic model positive charge contributes to almost all mass of atom? |
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Answer» Thomson |
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| 39. |
Draw a graph showing variation of potential with r distance for a uniformly charged spherical shell. |
Answer» Solution :GRAPH of V `RARR`r for spherical SHELL
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| 40. |
Two point charges q_(A)= 3mu C and q_(B) = - 3muC are located 20cm apart in vacuum. What is the electric field at the mid point of the line joining the two charges. |
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Answer» `54 xx 10^(6)N//C` ALONG OA |
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| 41. |
In Young's double slit experiment, the 10th maximum of wavelength lambda_(1) is at a distance y_(1) from its central maximum and the 5th maximum of wavelength lambda_(2) is at a distance y_(2) from its central maximum. The ratio y_(1) // y_(2) will be: |
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Answer» `(2 lambda_(1))/(lambda_(2))` |
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| 42. |
A 10 kg mass is attached to a string that has a breaking strength of 1,500 N. if the mass is whirled in a horizontal circle of radius 90 cm, what is the maximum speed it can have ? (Neglect the effects of gravity). |
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Answer» Solution :The first thing to do in a problem LIKE this is to identify what force (s) provide the centripetal force. In this example, the tension in the string provides the centripetal force: `F_(T)` provides `F_(c)=IMPLIES F_(T)=(MV^(2))/(v)implies v=sqrt((rF_(T))/(m)) implies v_(MAX)=sqrt((rF_(T,max)))/(m)=sqrt(((0.90m)(1500N))/(10kg))` `=12m//s` |
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| 43. |
Which of the following particles will describe the smallest circle when projected with the same velocity in a perpendicular magnetic field ? |
| Answer» Answer :D | |
| 44. |
The primary coil of an ideal step up transformer has 100 turns and the transformation ratio is also 100. The input voltage and the power are 220 V and 1100 W respectively. Calculate (i) number of turns in the secondary (ii) the current in the primary (iii) voltage across the secondary (iv) the current in the secondary (v) power in the secondary. |
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Answer» SOLUTION :It is given that `N_(p) = 100`, transformation ratio `K = 100, V_(p) = 220 V` and `P_("input") = 1100 W` (i) `N_(s) = k, N_(p) = 100 xx 100 = 10000` (ii) `I_(p) = P_("input")/V_(p) -(1100 W)/(220 V) = 5A` (iii) `V_(s) = k, V_(p) = 100 xx 220 = 22000 V` (iv) `I_(s) = I_(p)/k = (5A)/(100) = 0.05 A` (v) Output POWER `= V_(s).I_(s) = 22000 xx 0.05 = 1100 W` |
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| 45. |
An alternating current is given by the equation , i=i_1 cos omegat + i_2 sin omegat. What is r.m.s. current ? |
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Answer» `1/2 (i_1^2 + i_2^2)^(1//2)` |
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| 46. |
The earth's magnetic field at geomagnetic poles has a magnitude 6.2 xx 10^(-5) T. The radius makes an angleof 135^(@) with the axis of the carth's assumed magnetic dipole. Then, which of the following statement(s) is/are correct? |
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Answer» At a point where `theta = 135^(@)`, magnetic field0 `4.9 xx 10^(-5) T`. The magnetic field B at geomagnetic poles is `B_(p) = (mu_(0))/(4pi)(2M)/R^(3)` The magnetic field due to a dipele at a distance R away from its centre has a magnitude, `B = (mu_(0))/(4pi)M/R^(3)(1+3cos^(2) theta)^(1//2)=1/2B_(p)(1+3cos^(2)theta)^(1//2)` This field is in a direction making an angle `alpha` with the radial direction such that `tanalpha = (tantheta)//2,` as shown in the FIGURE. At a point where `theta = 135^(@)`, the field B is `B = (B_(p))/2(1+3cos^(2)135^(@))^(1//2)` `=1/2xx6.2xx10^(-5) T xx 1.58 = 4.9 xx 10^(-5)T ` The angle `alpha` of this field with the vertical the given by `tan alpha = (tantheta)/2 = (tan 135^(@))/2 = - 0.5` `therefore alpha = 153^(@)` The inclination (dip) is the angle made by the earth.s magnetic field with the horizontal plane. Here it is `153^(@) - 90^(@) = 63^(@)` below the horizontal. 
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| 47. |
What are the possible harmful effects of usage of Nanoparticles? Why? Possible harmful effects of usage of Nanoparticles: |
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Answer» Solution :Possible harmful effects of usage of Nanoparticles: The research on the harmful impact of application of nanotechnology is also equally important and fast developing .The major concern here is that the nanoparticles have the dimensions same as that of the biological MOLECULES such as proteins.They may easily get absorbed onto the surface of living organisms and they might enter the TISSUES and fluids of the body . The adsorbing nature depends on the surface of the nanoparticles ,Indeed it is possible to deliver a drug directly to a specific cell in the body y designing the surface of a nanoparticle so that it adsorbs specifically onto the surface of the target cell. The interation with living systems is also affected by the dimensions of the nanoparticles,for INSTANCE ,nanoparticle of a few nanometer size may reach well INSIDE biomolecules,which is not possible for LARGER nanoparticles. Nanoparticles can also cross cell membranes.It is also possible for the inhaled nanoparticles to reach the blood ,to reach other sites such as the liver ,heart or blood cells. Researchers re trying to understand the response of living organisms to presence of nanoparticles of varrying size,shape,chemical composition and surface characteristics. |
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| 48. |
Mark the correct statement(s) from the following: |
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Answer» Image formed by a convex mirror can be real  For a virtual object, lying between focus and pole, image formed by a convex mirror would be real, magnified, erect, and lying in front of mirror.  For a virtual object beyond focus, image formed by a convex mirror would be virtual, inverted, magnified or diminished (depending on the LOCATION of object), and will lie on the same of mirror. 
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| 49. |
The resistance of a galvanometer is 999 Omega. A shunt of 1 Omega is connected to it. If the main current is 10^(-2) A, what is the current flowing through the galvanometer? |
| Answer» SOLUTION :`10^(-5)A` | |