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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. |
How many numbers of energy bands exist in solids ? What are they ?. |
| Answer» SOLUTION :THREE:CONDUCTION bandValence bandForbidden ENERGY BAND. | |
| 2. |
A body is projected at an angle 45° to the horizontal with K.E. ‘E’. The potential energy at the highest point of flight is : |
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Answer» Solution :Potential ENERGY at HIGHEST point =`mgH_(MAX)` `E_(p)=mgxx(u^(2)sin^(2)theta)/(2g)=1/2mu^(2)sin^(2)45^@=E/2` |
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| 3. |
In the steady state of the circuit shown in the figure the ratio of energy stored in the inductor to th energy stored in the capacitor is 10000. |
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Answer» |
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| 4. |
A battery of emf 12 V and internal resistance 2Omega is connected in series with a tangent galvanometer of resistance 4Omega . The deflection is 60^@ when the plane of the coil is along the magnetic meridian . To get a deflection of 30^@, the resistance to be connected in series with the tangent galvanometer is |
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Answer» `12OMEGA` |
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| 5. |
Fraction of the element decayed during its mean life will be: |
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Answer» e Now, `N=N_(0)e^(-lambdat)""rArr N=N_(0)e^(-lambdat)` `N=N_(0)e^(-1) =(N_(0))/e therefore tau =1/lambda` `N/(N_(0))=1/e` So fractional of ELEMENT decayed is `=1-1/e` |
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| 6. |
Out of red, blue and yellow lights, the scattering of __________ light is maximum. |
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Answer» |
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| 7. |
If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30^@ with the direction of applied field ? |
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Answer» Solution :We have MAGNETIC moment m = 0.6 A `m^2` . Further more B=0.25 T and `theta = 30^@` . `therefore ` Magnetide of torque `tau= m B sin theta - 0.6 xx 0.25 xx sin 30^@ = 0.075 N m= 7.5xx10^(-2) J`. |
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| 8. |
The S.T. of water is 7 xx 10^-2 N/m. The work required to break a drop of water of radius 0.5cm into identical drops, each of radius 1mm is, |
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| 9. |
In the circuit shown in figure all resistances are identical (each equal to R) and the cell has an emf of V_(0). The three voltmeters V_(1), V_(2) and V_(3) are identical and are nearly ideal. (a) find the reading of the voltmeter V_(1) when switch ‘S’ is open. (b) find the reading of the voltmeter V_(1) after the switch is closed |
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Answer» (B) `(2V_(0))/(9)` |
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| 10. |
A golfer hits a golf ball into the air over level ground. The velocity of the ball at a height of 10.3 m is vecv=(8.6hati+7.2hatj)m//s, with hati horizontal and hatj upward. Find (a) the maximum height of the ball and (b) the total horizontal distance traveled by the ball. What are the ( c) magnitude and (d) angle (below the horizonatl) of the balls velocity just before it touches the ground? |
| Answer» Solution :(a) `~~13M`, (b) `~~28M` ( C) 18 m/s, (d) `-62^(@)` (which is equivalent to `298^(@)`) | |
| 11. |
Which of the following ray diagram show physically possible refraction , , |
| Answer» Answer :a | |
| 12. |
A 1500 kg cable car moves vertically by means of a cable that connects the ground and the top of a hill. What is the tension in the supporting cable when the cab, originally moving downard at a speed of 9.0 m/s, is brought to rest with constant acceleration in a distance of 38 m. |
| Answer» SOLUTION :`1.6xx10^(4)N` | |
| 13. |
A 5 cm cube of a substance has its upper surface displaced by 0.5 cm, by a tangential force of 0.25 N. What is the modulus of rigidity of the substance ? |
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Answer» Solution :`theta = x/L = 0.5 XX 10^(-2)// 5 xx 10^2 = 0.1` `eta = F//A tetha = 0.25/25 xx 10^(-4) xx 10^(-1) = 1000 N//m^2` |
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| 14. |
A body of mass 2kg is moving along positive X-direction with a velocity of 5 ms^(-1). Now a force of 10sqrt2 N is applied at an angle 45^(@) with X-axis. Its velocity after 3s is, |
| Answer» Answer :C | |
| 15. |
Define RMS Value of Current. |
| Answer» Solution :RMS value of ac is defined as that value of direct current which produces the same heating EFFECT in a GIVEN RESISTOR as is produced by the given alternating current when passed for the same time. | |
| 16. |
State the success of the Bohr's atomic model. |
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Answer» SOLUTION :Stability and enrgy of hydrogen like atoms can be calcualted. which is the formula for the frequency of the Balmer series. `(1)/(lambda)=R[(1)/(2^(2))-(1)/(n^(2))], n=2,3,4..` can be applied to spectrum of very few elements such as the `H_(2) He^(+)` (singly ionised helium), `Li^(2+)` (Doubly ionized lithium, `Be^(+3)` (triply ionized berilium) can be explained. In hydrogen atom if electron transits from lower ENERGY STATE to higher energy state the wavelength of absorbed radiation and if electron transits from higher energy state to lower energy state, then wavelength of EMITTED radiation. given b `(1)/(lambda)=R[(1)/(m^(2))-(1)/(n^(2))]` where `n=2,3,4,.... and ,=m=-1` |
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| 17. |
Ionization in discharge tube us due to collision of ……..gas filled in tube. |
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Answer» electron and NEUTRAL atom/molecules |
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| 18. |
In above problem find the gravitational potential at a point whose co-ordinates are (5,4) in (J//kg) . |
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Answer» `-180` |
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| 19. |
The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which a neutron, would have the same the Broglie wavelength. |
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Answer» Solution :GIVEN `:` `lambda = 589 nm = 5.89 xx 10^(-7) m` The de Broglie wavelength `lambda - ( h )/( p )= ( h )/( sqrt( 2mE_(k))) ` `RARR lambda^(2) = ( h^(2))/(2mE_(k))` Kinetic energy `E_(k )=( h^(2))/( 2mlambda^(2))` For neutron `m = 1.67 xx 10^(-27) kg` `E_(k)= ((6.63 xx 10^(-34))^(2))/( 2 xx 1.67 xx 10^(-31)xx ( 5.89 xx 10^(-7))^(2))` `= 3.79 xx 10^(-28)J` |
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| 20. |
Three small , identical point charges lie on the circumference of a circle forming an equilateral triangle.Calculate the value of electric field at the centre of circle. |
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| 21. |
A photocell is illuminated by a small bright source placed 1 m away. When the same source of light is placed 1/2 m away, the number of electrons emitted by photo cathode would: |
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Answer» DECREASE by a FACTOR of 2 |
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| 22. |
A rate of radiation of aperfectly black body at 0^@ C is X j/sec. The rate of radiation of the same body at 273^@C is |
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Answer» a)16 X `j /SEC` |
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| 23. |
If a glass prism in dipped in water its dispersive power |
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Answer» increases |
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| 24. |
Sate the applications of Seeback effect. |
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Answer» Solution :APPLICATIONS of Seebeck effect: 1. Seebeck effect is used in thermoelectric generators(Seebeck generators). These thermoelecric generators are used in POWER plants to convert waste heat into electricity. 2. This effect is UTILISED in AUTOMOBILES as automotive thermoelectric generators for increasing fuel efficiency. 3. Seebeck effect is used in thermocouples and thermopiles to MEASURE the temperature difference between the two objects. |
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| 25. |
The stability of a nucleus can be measured by |
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Answer» AVERAGE BINDING energy |
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| 26. |
TV waves have a wavelength range of 1-10 meter. Their frequency range in MHz is |
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Answer» 30-300 and `v_(2)=(3 xx 10^(8))/(10)=3 xx 10^(7) Hz=30 MHz` |
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| 27. |
If real image obtained from concave mirror of focal length f is n times of object, then find object distance. |
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Answer» (n − 1)f `m=(f)/(f-u)` `-n=(f)/(f-u)` `THEREFORE u-f=f/n``thereforeu=f+f/n` `therefore u =((n+1)/(n))f` |
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| 28. |
Two positively charged particles each of mass is 9 xx 10^(-30)kg and carrying a charge of 1.6 xx 10^(-19)C are placed at a distance .r. apart. If each experiences a force equal to its weight, the value of r is (g = 10 ms^(-2)) |
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Answer» 1.6m |
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| 29. |
Maximum frequencyof the photon produced by the union of a proton and a antiproton is (given : m_(p)=1.67xx10^(-27)kg) : |
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Answer» `4.56xx10^(21)Hz` |
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| 30. |
In YDSE , the interfering waves have amplitude in the ratio 3 : 2. Find the ratio of Amplitude. |
| Answer» SOLUTION :`(A_("MAX"))/(A_("min"))= ((A_(1)+A_(2))^2)/((A_(1)-A_(2))^2)= (3+2)/(3-2)= 5`. | |
| 31. |
Deive an expression for the electric field at any point on the equatorial line of an electric dipole. |
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| 32. |
The field normal to the plane of a coil of 'n' turns and radius 'r' which carries a current 'i' is measured on the axis of the coil at a small distance 'h' from the centre of the coil. This is smaller than the field at the centre by the fraction. |
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Answer» `3/2( h^2)/(r^2)` |
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| 33. |
Which of the following is S.H.M. ? |
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Answer» Earth spinning about its axis |
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| 34. |
What is inductance dominated circuit ? |
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Answer» Solution :The circuit for which `X_(L) gt X_(C)` . In these CIRCUITS, VOLTAGE leads the CURRENT. |
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| 35. |
Define tesla. using the expression for the force on a charged particle moving in a magnetic field. |
| Answer» Solution :When a charge of1 C, moving with VELOCITY 1 m/s, normal to themagnetic FIELD, experience a force 1N, the magnetic field is said to be one tesla. | |
| 36. |
The platesof a parallelplate capacitorhave an area of 100 cm^2each and are separatedby 3mm.The capacitor is charged by connectingit to a 400 V supply. if a dielectricis dielectricconstant2.5 isintroducedbetween the platesof the capacitor,then findthe electrostaticenergy storedand also change in theenergy stored. |
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Answer» SOLUTION :`E^1=epsilon_r E=2.5xx2.361xx10^(-6)` `=5.903xx10^(-6)` J `DeltaE=E^1-E=(5.903-2.361)xx10^(-6)` J DIFFERENCE in theenergystored `DELTAF` , `DeltaE=3.542xx10^(-6)` J |
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| 37. |
The focal lengths of the objective and eye-lens of a microscope are 1 cm and 5 cm respectively. If the magnifying power for the relaxed eye is 45, then the length of the tube is |
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Answer» `30 CM` |
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| 38. |
The platesof a parallelplate capacitorhave an area of 100 cm^2each and are separatedby 3mm.The capacitor is charged by connectingit to a 400 V supply. Calculate the electrostatic energy stored in the capacitor. |
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Answer» Solution :GIVEN `A=100 cm^2 =10^2 xx10^(-4)=10^(-2) m^2` , d=3 mm = `3XX10^(-3) ` m , V=400 V Electricalenergy `.E.=1/2CV^2=(epsilon_0AV^2)/(2d)` i.e., `E=(8.854xx10^(-12)xx10^(-2)xx(400)^2)/(3xx10^(-3))` i.e., `E=2.361xx10^(-6)` J |
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| 39. |
Twpdipoles thatareback to back form a linear quadrapole i) Calculate E_(x) for points on the X-axis such that x gt gt a ii) Calculate E_(y) for points on the Y-axis such that y gt gt a |
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Answer» |
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| 40. |
A radioactive sample at any instant has its disintegration rate 5000 disintegrations/minute. After 5 minutes, the rate is 1250 disintegrations/min. Then decay constant (per minute) is : |
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Answer» 0.2 LN 2 |
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| 41. |
If the earth of radius R, while rotating with angular velocity to become standstill, what will be the effect on the weight of a body of mass m at a latitude of 45^(@)? |
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Answer» REMAINS unchanged thus `g.=g-R omega^(2)((1)/(sqrt(2))^(2))=g(Romega^(2))/(2)` When earth stops rotating, `omega=0`, so g.=g Increase in weight of body `=g-(g-(R omega^(2))/(2))=(R omega^(2))/(2)` |
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| 42. |
The magnetic field in a plane electromagnetic wave is given byB_(y)=2xx10^(-7)sin (0.5xx10^(3)x + 1.5 xx 10^(11) t)T. Write an expression for the electric field. |
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Answer» Solution :`E_(o) =B_(o)c =2xx10^(-7) TXX 3XX10^(8) m//s =6XX10^(1) V//m` The electric field component is perpendicular to the direction of propagation and the direction of MAGNETIC field. Therefore, the electric field component along the z-axis is obtained as `E_(z) =60 sin (0.5 xx10^(3) x +1.5 xx10^(11) t) V//m` |
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| 43. |
Identify the part of the electromagnetic spectrum to which the following wavelengths belong : 10^-1m |
| Answer» SOLUTION :MICROWAVE | |
| 44. |
A wooden block is placed on an inclined plane. The block just begins to slide down when the angle of the inclination is increased to 45^0. The coefficient of the friction is |
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Answer» `0.25` |
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| 45. |
State Fleming's right hand rule. |
| Answer» Solution :The thumb, index finger and middle finger of right hand are stretched out in mutually perpendicular DIRECTIONS. If the index finger points the direction of the MAGNETIC field and the thumb INDICATES the direction of motion of the conductor, then the middle finger will indicate the direction of the INDUCED CURRENT . | |
| 46. |
Let in_(0) denote the dimensional formula of the permittivity of vacuum. If M = mass L = length, T = time and A = electric current, then |
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Answer» `in_(0) = M^(-1) L^(2) T^(-1)`A |
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| 47. |
What is meant by the statement that the current through an inductor lags behind the emf across it by pi/2. |
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Answer» SOLUTION :This means that in an inductive a.c. CIRCUIT , whatever value emf ATTAINS , CURRENT attains a similar value a quarter of cycle later. For EXAMPLE,if the emf attains it.s maximum value at t=0. then , current attains it.s maximum value at t=T/4 and so on. |
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| 48. |
A boat at anchor is rocked by waves whse crests are 100 cm aprt and whose velocity is 25 cm/s. these waves reach the boat once every : |
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Answer» Solution :`lambda `= 100 cm `"" `v = 25 cm/s. `v = (v)/(lambda) = (25)/(100) = (1)/(4)` `THEREFORE T = (1)/(v)` = 4 sec Hence the correct choice is (b) . |
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| 49. |
Explain the formation of stationary waves in stretched strings and hence deduce the laws of transverse wave in stretched strings. |
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Answer» Solution :A string is a metal WIRE whose length is large when compared to its thickness. A stretched string is fixed at both ends, when it is plucked at mid point, two reflected waves of same amplitude and frequency at the ends are travelling in opposite direction and overlap along the length. Then the resultant waves are known as the standing waves ( or) stationary waves. Let two transverse progressive waves of same amplitude a, wave length `lambda ` and frequency 'v', travelling in opposite direction be given by `y_(1) = a sin ( kx - omega t ) ` and `y_(2) =a sin ( kx +oemga t ) ` where ` omega = 2piv ` and `k = ( 2pi)?( lambda)` The resultant wave is given by`y= y _(1) + y _(2) ` `y = a sin ( kx + omega t )+ a sin (kx + omega t ) ` `y = ( 2a sin kx ) COS omegat ` ` 2 a sin kx =` Amplitudeof resultant wave. It dependson 'kx' . If ` x = 0 , ( lambda )/(2) , ( 2lambda)/(2) , ( 3lambda)/(2) ,.....`etc,the amplitude =ZERO. These positions are known as "Nodes". If `x= ( lambda)/(4), ( 3lambda)/(4) , ( 5 lambda)/(4),......` etc, the amplitude= maximum (2a) These positions are called "Antinodes".  If the string vibrates in 'P' segments and'l' is its length then length of each segment `= ( l)/( p)` Which is equal to `( lambda)/(2)` ` :. ( l)/(p) = ( lambda)/(2) implies lambda = ( 2l)/( P)` HARMONIC frequency `v =(upsilon)/(lambda) =( upsilonP)/( 2l)` `v = ( upsilonP)/(2l)`...(1) If'T' is tension( streching force ) in the string and '`mu`' is liner density thenvelocity of transverse wave (v) in the string is ` v =sqrt((T)/(mu))`...(2) From the EQS. (1) and (2) Harmonic frequency `v= (P )/( 2l) sqrt((T)/(mu))` P=1 then it is called fundamental frequency ( or) first harmonic frequency `:.` Fundamental Frequency `v = (1)/(2l) sqrt((T)/(mu))`...(3) Law of Transverse Waves Along StretchedString `:` Fundamental frequency of the vibrating string `v = (1)/(2l) sqrt((T)/(mu))` First Law `:`When the tension (T) and linear density `( mu )`are constant, the fundamental frequency (v) ofa vibrating string is inversely proportional to its length. ` :.v prop (1)/(l) implies vl` =constant, when T and `' mu'` constant. Second Law `:` When the length (l) and its, linear density (m) are constant the fundamental frequency of a vibratingstring is directly proportional to the square root of the stretchingforce (T). `:. v prop sqrt(T) implies (v)/( sqrt(T))` = constant , when 'l' and 'm' are constant. Third Law `:` When the length ( l)and the tension (T) are constant , the fundamental frequency of a vibrating string is inversely proportional to the square root of the linear density ( m ). `:.v prop (1)/(sqrt(mu)) implies v sqrt( mu ) `= constant , when'l' and 'T'are constant. |
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| 50. |
Equipotential at a great distance from a collection of charges whose total sum is not zero are approximately [NCERT Exemplar] |
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Answer» spheres |
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