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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 angular velocity due to the earth's spin is, |
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Answer» `5.27xx10^(-5)`rad/sec |
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| 2. |
Rankpaths 1,2 and3infig 21 .14accordingto theenergytransferto thegasas heat, greatestfirst. |
| Answer» SOLUTION : 1,2,3(`Q_3 =0 Q_2 ` goes into work `W_2`, but` Q_1` goes into GREATER work `W_1` and increases GAS temperature) | |
| 3. |
What is the value of inductance L for which the current is maximum in a series LCR circuit with C=10muF and omega=1000s^-1 ? |
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Answer» 100mH |
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| 4. |
The minimum distance between the object and its real image for a concave mirror of focal length 2f is |
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Answer» A) f |
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| 5. |
STATEMENT-1: The temperature of a metallic rod is raised by a temperature At so that its length becomes double. The value of a (coefficient of linear expansion) is given (ln2)/(Deltat) because STATEMENT-2: Coefficient of linear expansion is defined as 1/l (dl)/(dt) |
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Answer» Statement-1 is TRUE , Statement-2 is True , Statement-2 is a CORRECT EXPLANATION for Statement-1. |
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| 6. |
Due to economic reasons, only the upper sidebards of an AM wave is transmitted, but at the receiving station there is a facility for generating the carrier. |
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Answer» Solution :Let the two signals be represented by `A_(1) cos(omega_(c)+ omega_(m))t and A_(c ) cos omega_( c)t` MULTIPLYING the WAVES we GET, `A_(1)A_(c) cos(omega_(c) + omega_(m))t cdot cos omega_(c)t` `= (A_(1)A_(c))/(2) cos omega_(m)t + (A_(1)A_(c))/(2) cos (2omega_(c) + omega_(m))t` The separation of the relationship clearly indicates that the modulatingsignal `(A_(1)A_(c))/(2) cos omega_(m)t` can be easily recovered at the receiving station. |
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| 7. |
Zener diode is used in reverse bias. When its reverse bias is increased, how does the thickness of the depletion layer change? |
| Answer» SOLUTION :THICKNESS of DEPLETION LAYER INCREASES. | |
| 8. |
In the unbalanced reaction, CrO_(5)+SnCI_(2)rarrCrO_(4)^(2-)+SnCI_(4), the element undergoiing oxidation and reduction respectively are: |
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Answer» `CrSn` |
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| 9. |
An electron moving with uniform velocity enters a uniform electric field perpendicular to its direction of motion . The path of the lelectron will be ________ . |
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Answer» Circular |
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| 10. |
The work function of a metal is 6 eV. If two photons each having energy 4 eV strike with the metal surface will the emission is possible? Why? |
| Answer» SOLUTION :NO, ENERGY of one PHOTON is less than WORKFUNCTION | |
| 11. |
A hydrogen atom in the normal state is locatred at a distance r=2. 5cm from a long straigh condctor carrying a current I= 10A. Find the force acting on the atom. |
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Answer» Solution :The magnetic field at a distacne `R` from a long current carrying wire is mostly tangential and GIVEN by `B_(varphi)=(mu_(0)I)/(2 pi r)=(mu_(0))/(4pi)(2I)/(r )` The force on a magnetic dipole of mement `VEC(mu)` due to this magnetic iels is also tangential and has magnitude `(vec(mu).grad)B_(varphi)` This force is nonvanishing only when the COMPONENT of `vec(mu)` along `vec(r )` non zero. Then `F=mu_(r )(del)/(del r)B_(varphi)= -mu_(r )(mu_(0))/(4pi)(2I)/(r^(2))` Now the maximum value of `mu_(r )=+-mu_(B)` Thus the force is `F_(max)=mu_(B)(mu_(0))/(4pi)(2I)/(r^(2))= 2.97xx10^(-26)N` |
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| 12. |
Draw a neat labelled diagram of a typical X-ray spectrum. State and explain its important features. |
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Answer» Solution :Important features of a typical X-ray SPECTRUM : (1) There is abroad continuous distribution of wavelenghts of the X-rays, called as continuous X-rays.  (2) Superimposed on the continuous distribution are PEAKS of sharply defined wavelengths, called the CHARACTERISTIE X-rays, because they are unique characteristics of the element used as the target metal. (3) There is a sharply defined minimum or cutoff wavelength,`lamda _(min)` below which the continuous spectrum does not exist. On increasing the kinetie energy E of the electrons striking the target,`lamda _(min)`decreases but the wavelengths of the characteristic X-rays remain unchanged. The intensity at all wavelengths increases. |
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| 13. |
Equal volumes of water and alcohol are taken in two similar colorimeter at temperature 50°C. Water filled calorimeter takes 1 00s to cool from 50°C to 40°C and alcohol filled tapes 70s and col from 50°C to 40°C. If the thermal capacity of each calorimeter is numerically equal to the volume to either liquid, then the specific heat capacity of alcohol is: (take relative density of alcohol 0.8 and specific heat capacity of water is 1 both in CGS units) |
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Answer» `0.4 cal//gm^(@) C` |
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| 14. |
A network of resistance is constructed with R_(1)" and R_(2) as shown in thefigure. The potential at the points 1, 2, 3,………., N are V_(1), V_(2), V_(3)",……..," V_(n) respectively each havig a potential K time smaller than privious one. Find : (R_(1)/R_(2))xx(R_(2)/R_(3))" in terms of K". |
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Answer» K-1 |
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| 15. |
In a current carrying toroid , the field produced does not depend upon |
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Answer» NUMBER of turns per unit LENGTH |
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| 16. |
Why are Si and GaAs are preferred materials for solar cells ? |
Answer» Solution :The solar radiation spectrum received by US is shown in figure.  The maxima is near 1.5 eV . For photo -excitation. `hv gt E_(g ) `. Hence, semicondutorwith band gap -1.5eVor lower is likely to give BETTER (in spite of its higher band gap ) than Si because of itsrelatively higher absorption coefficient.If we choose materials like CdS or `CdSe ( E_(g)~ 2.4eV)`,we can use only the high ENERGY component of the solar energy for photo-conversion and a significant part of energy will be no use. The QUESTION arises `:` why we do not use material like pbS ( `Eg ~0.4eV)` which satisfy the condition `hv gt E_(g )` for v maxima corresponding to the solar radiation SPECTRA ? if wedo so, mostof the solar radiation will be absorbed on the top-layer of solar cell and will not reach in or near the deplection region.For effective electron -hole separation,due to the junction field, we want the photo-generation to occur in the junctionregion only. |
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| 17. |
The amount of heat generated in 500 Omega resisttance, when the kye is thrown over from contanct 1 to 2, as shown in figure, is |
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Answer» `6.25xx10^(-2)J` |
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| 18. |
A car travels 100 km cast an then 100 km south. Finally, it comes back to the starting point by the shortest route. Throughout the journey, the speed is constant at 60 km/h. The average velocity for the whole of the journey is |
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Answer» 60 km/h |
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| 19. |
The wrong statement about electric lines of force is |
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Answer» These originate from positive charge and END on NEGATIVE charge |
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| 20. |
How can you explain the formation of interference fringes in accordance with law of conservation of energy ? |
| Answer» Solution :It is found that intensity or bright points is `4^2` and at dark points it is ZERO. According to LAW of conservation of ENERGY, energy can be NEVER be created or destroyed. Here also the energy is not destroyed but onbly referred from point bof minimum intensity the AVERAGE being `2a^2`. Therefore,the formation of interference fringes is in accordance with the law of conservation of energy. | |
| 21. |
What is alpha-particles scattering experiment arrangement ? |
| Answer» Solution :Rutherford USED a narrow BEAM of HIGH energy `alpha-particle` from a radioactive SOURCE and made it incident on a THIN sheet of gold. | |
| 22. |
As the quantum number increases, the difference of energy between consecutive energy levels ....... |
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Answer» DECREASES |
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| 23. |
The wavelength of electromagnetic waves employed for space communication lie in the range of |
| Answer» Answer :D | |
| 24. |
If two charges +q and +4q are separated by a distance 'd' and a point charge Q is placed on the line joining the above two charges and in between them such that all charges are in equilibrium. Then the charge Q and it's position are |
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Answer» `(4Q)/(9)`at a distance `d/3`from 4q |
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| 25. |
When light is reflected from a surface it can be either regular reflection or diffuse reflection. The essential difference between regularly and diffusely reflecting surfaces is that |
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Answer» regularly reflecting surfaces are smoother than DIFFUSELY reflecting surfaces |
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| 26. |
A : Evaporation occurs at any temperature whereas the boiling point depends on the external pressure. R : Evaporation of a liquid occurs from the surface of a liquid at all temperature whereas boiling takes place at a temperature determined by the external pressure. |
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Answer» If bothAssertion & Reason are true and the reason is the correct explanation of the assertion, then MARK (1) |
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| 27. |
The displacement y in x direction is given by y = 10^(-4) sin (600t - 2x + pi/3)m where x is in m ,t in s, then speed of wave motion is xx102 m/s , what is the value x ? |
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Answer» |
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| 28. |
What does Deep Water signify? |
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Answer» Beauty |
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| 29. |
Two identical pieces of ice fly towards each other with equal and opposite velocities and are converted into water upon impact. If their initial temperature is -12^(@)C, then the minimum possible velocity of either piece before impact is (S_("ice") = 0.50 "cal" gm^(-1) ""^(@)C^(-1),1"cal" = 4.2J) |
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Answer» 850 m/sec i.e., Loss in K.E. = Latent heat + specific heat `rArr 2 xx (1)/(2) mV^(2) = 2ML + 2m S_("ice") Delta theta rArr V= sqrt(2(L + S_("ice") Delta theta))` `therefore V= sqrt(2(80 xx 1000 + 0.5 xx 1000 xx 12) 4.2)m//s= 850 m//s` |
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| 30. |
Explain with the help of a diagram the formation of depletion layer and potential barrier in a p-n junction. |
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Answer» SOLUTION :Formation of depletion layer : When a p-n junction is PREPARED, electrons from n-region diffuse into p-region and holes diffuse from p-region to n-region. When an electron diffuses from n`RARR` p, it leaves behind an ionised donor on n-side . This ionised donor (positive charge) is immobile as it is bonded to the surrounding atoms. Thus, due to diffhusion of electrons from n`rarr` p, a layer of positive space charge is developed on n-side of the junction. Similarly due to diffusion of holes from p`rarr` n a layer of negative space charge on the p-side of the junction is developed. This space charge region on either side of the junction together is KNOWN as the "depletion region" or "depletion layer". Barrier potential: Due to diffusion of holes fromp-region to n-region and difusion of electrons in the reverse direction, part of depletion layer on n-side of junction becomes positively charged and the part of depletion layer on p-type of junction becomes negatively charged. Thus, a junction potential is developed, which opposes further diffusion of holes/electrons. Hence, this potential acts as a barrier and is known as "barrier potential" `V_(B)` 
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| 31. |
Two protons of equal kinetic energies enter a region of uniform magnetic field. The first proton enters normal to the field direction while the second enters at 30^@to the field direction. Name the trajectories following by them. |
| Answer» Solution :Trajectory of 1st proton is a CIRCULE trajectory of 2ND proton is a HELIX. | |
| 32. |
In a metre bridge when the resistance in the left gap is 2Omega and an unknown resistance in the right gap, the balance point is obtained at 40 cm from the zero end. On shunting the unknown resistance with 2Omega, find the shift of the balance poiont on the bridge wire. |
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Answer» |
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| 33. |
In the given distribution, what is the value of I ? |
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Answer» 3A |
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| 34. |
Find the magnitude of magnetic induction at a point 0.06m from the centre and along the axis of a circular coil carrying a current of 2 A. Also calculate the magnitude of magnetic induction at the centre of the coil. Given: Number of turns in the coil = 20 Mean radius of the coil = 0.05 m. |
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Answer» Solution :Given `I=2A, r=5 times 10^(-2)m, n=20, d=6 times 10^(-2)` We know that the magnetic field at the centre of the COIL, `""B=(mu_(0)/(4pi))((2pini)/r)T`. i.e., `""B=(10^(-7) times 2 times 3.142 times 20 times 2)/(5 times 10^(-2))=50.272 times 10^(-5)T` Magnetic fields at point `B^(')=(10^(-7)2pini)/(r(1+(x/r)^(2))^(3//2))` i.e., `""B^(')=(B)/((1+(x/r)^(2))^(3//2))` `""B^(')=(50.272 times 10^(-5))/(1+(0.06/0.05)^(2))^(3//2)=(50.272 times 10^(-5))/3.811` `""B^(')=13.19 times 10^(-5)T.` |
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| 35. |
A and B are two conductors carrying a rarr A current I in the same direction. X and rarr B y are two electron beams moving in …..rarr X the same direction …..rarr Y |
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Answer» There will be repulsion between A and B ATTRACTION between X and y |
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| 36. |
The oscillating magnetic field in a plane electromagnetic wave is given by B_(y)=(8xx10^(-6))sin[2xx10^(11)t+300pix]T (i) Calculate the wavelength of the electromagnetic wave. (ii) Write down the expression for the oscillating electric field. |
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Answer» SOLUTION :(i) Comparing the EQUATION of oscillating electric field with the STANDARD wave equation, we `therefore ` Wavelength of the electromagnetic wave `lambda=(2pi)/(k)=(2pi)/(300pi)=(1)/(150)m` (ii) The electromagnetic wave is propagating long `-ve`direction of x - axis and magnetic field is oscillating along x - axis, hence electric field must be oscillating along z- axis with an amplitude. `E_(0)=c.B_(0)=3xx10^(8)xx8xx10^(-6)=2.4xx10^(3)Vm^(-1)` Hence, expression for oscillating electric field is `E_(z)=2.4xx10^(3)SIN[2XX10^(11)t+300pix]Vm^(-1)`. |
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| 37. |
There are two identical vessels containing same quantity of an ideal gas at same pressure vessel A is placed in a train moving with constant speed and the vessel B is placed on the platform in the frame of an observer standing on the platform select the correct statement |
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Answer» TEMPERATURE of gas in both VESSEL is same where as KINETIC energy of gas in the vessel is different |
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| 39. |
The graph between the stopping potential (V_(0)) and wave number (1//lamda) is as shown in the figure. phi is the work function, then: |
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Answer» `phi_(1):phi_(2):phi_(3)=1:2:4` `:.(hc)/(LAMBDA)=phi+eV_(s)` `:. V_(S)=(hc)/(elambda)-(phi)/(c)` `:.` slope `"tan" theta=(hc)/(e)` `(1)/(lambda0_(1))=0.001 mm^(-1) rArr lambda0_(1)=1000Å` `(1)/(lambda0_(2))=0.002 nm^(-1) rArr lambda0_(2)=5000Å` `(1)/(lambda0_(3))=0.004 nm^(-1) rArr lambda0_(3)=2500 Å` Here ultraviolet light can be used to eject photoelectrons from all 1,2 and 3 metals. |
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| 40. |
Two coherent sources of intensity ratio beta interfere, the (I_(max) - I_(min))/(I_(max) + I_(min)) is: |
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Answer» `(BETA)/(1 + beta)` `therefore I_(max)/(I_(min) = (I_(max) - I_(min))/(I_(max) + I_(min))` ` = ((sqrtI_(1) + sqrtI_(2))^(2) - (sqrtI_(1) - sqrt I_(2))^(2))/((sqrtI_(1) + sqrtI_(2))^(2) + (sqrtI_(1) - sqrt I_(2))^(2))` `(4.sqrtI_(1). sqrtI_(2))/(2(I_(1) + I_(2))` `2sqrt(I_(1) + I_(2))/(I_(1) + I_(2))= 2 sqrt((I_(1))/(I_(2)))/((I_(1))/(I_(2)) + 1) = (2 sqrtbeta)/(beta + 1)` `= (2 sqrt beta)/(1 + beta)` |
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| 41. |
A particle performing linear S.H.M. has a period 2pi seconds and an amplitude of 10 cm. What is its velocity when its displacement is 6 cm from the mean position ? |
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Answer» SOLUTION :`v=omegasqrt(a^2-x^2)=(2PI)/Tsqrt(a^2-x^2)` `=(2pi)/(2pi)SQRT(10^2-6^2)=8=8 cm//s` |
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| 42. |
1% of light of a source with luminous intensity 50 candela is incident on a circular surface of radius 10 cm . The average illuminance of surface is |
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Answer» 100 lux |
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| 43. |
Two identical charges are placed at the two corners of an equilateral triangle. The potential energy of the system is U. The work done in bringing an identical charge from infinite to the third vertex is |
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Answer» U |
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| 44. |
Consider a wire carrying a steady current, I placed in a uniform magnetic field B perpendicular to its length. Consider the charges inside the wire. It is known that magnetic forces do no work. This implies that |
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Answer» motion of CHARGES inside the CONDUCTOR is unaffected by B since they donot absorb energy |
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| 45. |
(a) Deduce an expression for the frequency of revolution of a charged particle in a magnetic field and show that it is independent of velocity or energy of the particle. (b) Draw a schematic sketch of a cyclotron. Explain, giving the essential details of its construction, how it si used to accelerate the charged particles. |
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Answer» Solution :(a) When a charged particle with charge q MOVES inside a magnetic field `vecbeta` with velocity v, it experiences a force, which is given by : `VECF=q(vecv^(xx)vecbeta)` Here `vecv` is perpendicular to `vecbeta`,`vecF` is the force on the charged particle which ACTS as the centipetal force and makes it move along a circular path.  Let m be the mass of the charged particle and r be the radius of the circular path. `therefore q(vecv^(xx)vecbeta)=mv^(2)` V and B are at right angles. `therefore "" qvB=(mv^(2))/(r)""therefore"" (mv)/(Bq)` TIME periodof circular MOTION of the charged particle can be calculated as shown below : `T=(2pir)/(v)=(2pi)/(v)(mv)/(Bq)rArrT=(2pim)/(Bq)` `therefore " Angular frequency is" omega=(2pi)/(T)"" therefore " " omega=(Bq)/(m)`. Therefore, the frequency of the revolution of the charged particle is independent ofvelocity or energy of the particle. (b)Cyclotron : |
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| 46. |
A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45° with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east. (a) Determine the horizontal component of the earth's magnetic field at the location. (b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90° in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero. |
Answer» Solution : Here, plane of the coil is passing through de, VERTICAL and perpendicular to plane of figure. Magnetic meridian is passing through ab and vertical also perpendicular to the plane of figure. `angle acd=435^@` is given. Let `overset(to) (B_c ) = ` magnetic field at the centre of given coil due to current passing through it. `overset(to) (B_h)=` horizontal component of Earth.s magnetic field. `overset(to) (B_R) = overset(to) (B_c) + overset(to)(B_h)=` resultant magnetic field at the centre `c` of given circular coil. As per the statement, magnetic needle PLACED at the centre of coil, free to rotate in horizontal plane remains horizontal from west to east and so its magnetic dipole moment `overset(to) (m)` and resultant magnetic field `overset(to)(B_R)` both MUST be pointing from west to east. In this situation for the right angled `Delta CPQ`. `sin45^@= (B_h)/( B_c)` `therefore B_h= 0.7071 B_c` `=0.7071 xx (mu_0)/( 2R)` `=0.7071 xx ((4pi xx 10^(-7) )(30) (0.35) )/( (2) (0.12))` `therefore B_h = 3.88 xx 10^(-5)` T (b) Now, the coil is rotated by 90° about vertical axis passing through its plane and its centre, anticlockwise (as seen from top) then magnetic moment of the needle and resultant magnetic field `overset(to) (B_R)` both reverse their directions and once again needle comes in stable EQUILIBRIUM condition. See the figure given below.  Here, `overset(to)(B._(R)) = overset(to)(B._(C ) ) + overset(to) (B_(h) ) ""(because overset(to) ( B_(h))= "constant" )` Also, `overset(to) (m.) || overset(to) (B._(R) ) ` |
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| 47. |
Calculate the specific activities of Na^(24) and U^(235) nuclides whose half lives are 15 hrs and 7.1 xx 10^8 yrs : |
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Answer» `3.2 xx 10^(20), 7.9 xx 10^(7) Bq//kg` No. of atom in 1 kg of `NA^(24)=(6.03 xx 10^(26))/(24)` `A_(sod) ("activity of sodium")=lambda N` `=(0.6931)/(15 xx 3600) xx (6.03 x 10^(26))/(24)` `=3.2 xx 10^(20) Bq kg^(-1)` `A_("uran")=(0.6931)/(7.1 xx 10^(8) xx 365 xx 86400)=(6.03 xx 10^(26))/(235)` `=7.9 xx 10^(7) Bq kg^(-1)` |
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| 48. |
Which of the following statements are true in the context of a Compound Microscope ? (a) Each lens produces a virtual and inverted image. (b) The objective has a very short focal length. (c) The eyepiece is used as a simple magnifying glass. (d) The objective and eyepiece are convex and concave lenses respectively. |
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Answer» (a), (B) and (d) |
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| 49. |
Free electron number density in one wire is n, its area of cross-section is A, and drift velocit of electrons is v_(d)Then electric current forme in this wire is ...... |
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Answer» `"ne"v_(d)` We have J = `(I)/(A) = "nev"_(d)` `therefore I = Av_(d) ` ne |
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