Saved Bookmarks
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. |
If a stone dropped from the top of a tower travels half, of the height of the tower during last second of its fall, the time of fall is (in seconds) |
|
Answer» `3 + SQRT(2)` |
|
| 2. |
The P-V diagram for a cyclic process is a triangle ABC drawn in order. The co-ordinates of A, B, C are (4,1), (2,4) & (2,1). The co-ordinates are in order of P- V in which P is in "N/m"^(2) & volume in litres. The work done during the process from A to B is: |
|
Answer» `3xx10^(-3)J`  `W_(AB)`= Area under the curve AB. =Area of trapezium ABBA. `W_(AB)=(1)/(2)(Aa+Bb)xxBC` `=(1)/(2)(4+2)xx3xx10^(-3)` `[because" 1 LITRE "=1000 CM^(3)=10^(-3)m^(3)]` `=9xx10^(-3)J` |
|
| 3. |
A parallel plate capacitor of capacitane C is charged to a potential V. It is then connected to another uncharged capacitor with the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor. |
|
Answer» Solution : Initial energy ,`U=(1)/(2)Cv^(2)""...(i)` Common potential of COMBINED system V' `=(C-V+C*O)/(C+C)=(V)/(2)` Energy STORE in combined system U' `=(1)/(2)(C+C)V'2=(1)/(2).2C.(V^(2))/(4)""...(II)` From (i) and(ii)`(U)/(U)=1:2` |
|
| 4. |
The frequency of a tuning fork 'A' is 2% greater than that of a standard fork 'K'. The frequency of another tuning fork 'B'is 3% less than ‘K'. When 'A' and 'B' are vibrated together 6 beats per second are heard per second. Find the frequencies 'A' and 'B'. |
| Answer» SOLUTION :`122.4 HZ, 116.4 Hz` | |
| 5. |
(a) Draw a labellel diagram of a moving coil galvanometer. Describe briefly its principle and working. (b) Answer the following: (i) Why is it necessary to introduce a cyclindrical soft iron core inside the coil of a galvanometer? (ii) Increasing the current sensitivity of galvanometer may not necessarily increase its voltage sensitivity. Explain,giving reason. |
|
Answer» Solution :(a) Principle and working : A CURRENT carrying coil, placed in a UNIFORM magnetic field, (can) experience a torque Consider a rectangular coil for which no. of TURNS = N. (b) (i) The SOFT iron core not only makes the field radialbut also increases the strength of themagnetic field. (ii) |
|
| 6. |
Two long straight conductors AOB and COD are perpendicular to each other and carry currents I_1and I_2 The magnitude of the magnetic inductions at a point 'P' at a distance 'a' from the point in a direction perpendicular to the plane ACBD is |
|
Answer» `(mu_0 )/(2 PI a) ( I_1+I_2)` |
|
| 7. |
A particle is executing SHM on a straight line. A and B are two points at which its velocity is zero. It passes through a certain point P(AP lt BP) at successive intervals of 0.5 sec and 1.5 sec with a speed of 3 m/s. Then |
|
Answer» The maximum SPEED of PARTICLE is `3 SQRT2` m/s |
|
| 8. |
In the ideal double-slit experiment, when a glass-plate (refractive index 1.5) of thickness t is introduced in the path of onc of the interfering beams (wavelength lamda), the intensity at the position where the central maximum occurred previously remains unchanged. Find the minimum thickness of the glass-plate. |
| Answer» SOLUTION :`2lamda` | |
| 9. |
A physical quantity obtained from the ratio of the coefficient of thermal conductivity to the universalgravitational constant has a dimensional formula [M^(2a) L^(4b) T^(2c) K^(d) ], then the value of (a + b)/(c + b) - d is |
|
Answer» `+ (3)/(2)` [K] = [`M^(1) L^(1) T^(-3) K^(-1) `]. Dimensional formula of universal gravitational constant, [G] = `[m^(-1) L^(3) T^(2) ]` Now, `([K])/([G]) = [M^(2) L^(-2) T^(-1) K^(-1) ] ` compare above equation with `[M^(2A) L^(4b) T^(2b) K^(d) ]` This will give us,a = 1 , b = - `(1)/(2), C = - (1)/(2) `and = -1 This will give us, a =1 , b = - `(1)/(2), c = - (1)/(2)` and d = -1 Now, ` (a + b)/(c + b) - d= (1 - (1)/(2))/(- (1)/(2) - (1)/(2))- (-1) or (a + b)/(c + b) - d = (1)/(2) ` |
|
| 10. |
What -ve sign implies. |
| Answer» Solution :The `-ve` SIGN MEANS the energy of ELECTRON increases as n increases. | |
| 11. |
A wheel with 10 metallic spokes, each 0.5 m long, is rotated with an angular speed of 120 rpm in a plane normal to the earth's magnetic field at the place. If the magnitude of the field is 0.4G, the magnitude of induced emf between the axle and the rim of the wheel is equal to |
|
Answer» ` 1.256 XX 10^(-3)V` `epsi=1/2 Bl^(2) omega=1.2 0.4 xx 10^(-4)0 xx (0.5)^(2) xx (4pi) =6.28 xx 10^(-5)V` |
|
| 12. |
The voltage applied to a supply inductive coil of self inductance 15.9 mH is given by the equation V=100 sin314t+75 sin 942 t+50sin 1570t. Find the equation for current wave. |
|
Answer» Solution :The standard equation of the voltage can be written as `V=V_(01)sin omega_(1)t+V_(02)sinomega_(2)t+V_(03) sin omega_(3)t` On comparing this with the GIVEN equation, we have `omega_(1)=313("rad")/(s),X_(L1)=omega_(1)L=314xx15.9xx10^(-3)=5Omega` `Omega_(2)=942("rad")/(S),X_(L2)=omega_(2)L=942xx15.9xx10^(-3)=15Omega` `and omega_(3)=1570("rad")/(S),X_(L3)=omega_(3)L=1570xx15.9xx10^(-3)=25Omega` Hence : `i_(01)=(V_(01))/(X_(L1))=(100)/(5)=20A`, `i_(02)=(V_(02))/(X_(L2))=(75)/(15)=5A, and i_(03)=(V_(03))/(X_(L3))=(50)/(25)=2A` THUS `i=i_(01)sin(omega_(1)t-phi)+i_(02)sin (omega_(2)t-phi)+i_(03)sin (omega_(3)t-phi)` `=20sin(314t-(pi)/(2))+5SIN(942t-(pi)/(2))+2sin (1570t-(pi)/(2))` |
|
| 13. |
Explain why a horse cannot pull a cart and run in empty space |
|
Answer» SOLUTION :While TRYING to pull a cart, a horses pushes the ground BACKWARDS with a certain force at an angle. The ground offers an EQUAL reaction in the opposite direction on the FEET of the horse. The forward component of this reaction is responsible for the motion of the cart. In empty space, there is no reaction and hence a horse cannot pull the cart and run. |
|
| 14. |
What is remote control? How has it been used in T.V. working? Name the main types of remote control receivers. |
|
Answer» SOLUTION :Remote control is an electronic device which HELPS the observer to operate the television from a distance without leaving his or her seat. In a remote control, there are three main UNITS. (i) A transmitting unit: which trasmits the command signal, (II) a frequency controller, which selects the suitable frequency for a particular function to be performed and (iii) a mechanical drive or electronic colur unit, which changes the VOLUME or colour according to the signal. There are two mains types of remote control receivers : (i) Electromechanical system (ii) Electronic system. |
|
| 15. |
If the magnetic field is parallel to the positive y-axis and the charged particle is moving along the positive x-axis (Fig. 4.4), which way would the Lorentz force be for (a) an electron (negative charge), (b) a proton (positive charge). |
| Answer» Solution :The velocity v of particle is along the x-axis, while B, the MAGNETIC field is along the y-axis, so v × B is along the Z-axis (screw rule or right-hand thumb rule). So, (a) for electron it will be along –z axis. (b) for a positive charge (proton) the FORCE is along +z axis | |
| 16. |
Which of the following is not the unit of intensity of magnetisation ? |
|
Answer» T |
|
| 17. |
The internal resistance of a primary cell is 4Omega. It generates a current of 0.2A, in an external resistance of 21Omega. The rate at which chemical energy is consumed which is providing the current is: |
|
Answer» 0.42 J/s |
|
| 18. |
Define the terms (a) threshold frequency, and (b) stopping potential. How were these tenns incorporated in Einstein's photoelectric equation? |
|
Answer» (b) Stopping potential is that minimum negative (retarding) potential `(V_(0))` given to the plate of a photoelectric tube for which the PHOTOCURRENT STOPS or becomes zero. Stopping potential is a measure of MAXIMUM kinetic energy of photoelectrons being PRODUCED for a given incident radiation. As per Einstein.s photoelectric equation, we have `K_("max")=hv-phi_(0)` but work function `phi_(0)=hv_(0)` and `K_("max")=eV_(0)` so the equation becomes `eV_(0)=hv-hv_(0)` `rArrH(v-v_(0))=eV_(0)` In this way, `v_(0)` and `V_(0)` are incorporated in Einstein.s photoelectric equation. |
|
| 19. |
A cubical vessel of side 1 metre contains one gram molecule of nitrogen at pressure of 2 atmospheres and 300 K. If the molecules are assumed to move with their rms velocity find the number of collisions per second which he molecules can make with the wall of vessel. Further if the vessel now thermally isolated moved with a constant speed V and then suddenly results in a rise of temperature 2^@C. Find V. |
|
Answer» |
|
| 20. |
Employing the uncertainty principle, evaluate the indetreminancy of the velocity of an electron in a hydrogen atom if the size of the atom is assumed to be l = 0.10nm. Compare the obtained magnitude with the velocity of an electron in the first Bohar orbit of the given atom. |
|
Answer» Solution :As in the PREVIOUS problem `Deltav underset(GT)~(ħ)/(ml)=1.16xx10^(6)m//s` The acutal velocity `V_(1)` has been CALCULATED in problem 6.21. It is ltBrgt `v_(1)=2.21 xx10^(6)m//s` Thus `Delta~v_(1)` (They are of the some order of MAGNITUDE) |
|
| 21. |
A boy completes one round of a circular track of radius "R' in 40 What will be his displacement at the end of 2 minutes 20 second |
|
Answer» zero Displacement of Boy is 2R. |
|
| 22. |
In Young's double slit experiment, the fringes are formed at a distance of 80 cm from double slit of separation 0.1 mm. The distance of 4th dark band from centre will be (lambda = 6000overset@A) : |
|
Answer» `1.68xx10^(-4) m` |
|
| 23. |
If force acting on a point charge 6.4 xx 10^(-3)C placed in uniform electric field is 0.128N, then electric field at point is ...... N/C. |
|
Answer» 2 |
|
| 24. |
Barretter B is a conductor whose idealized characteristic. The barretter is connected in series with a resistor of resistance R to a power supply of e.m.f. epsi. Find the current in the circuit and the voltage drop across the barretter and the resistor. Neglect the internal resistance of the power supply. |
|
Answer» |
|
| 25. |
What is an electromagnetic wave ? |
| Answer» Solution :ELECTROMAGNETIC waves are PERIODICALLY changing ELECTRIC and magnetic fields, which propagate through the SPACE. The electric and magnetic field vectors are at right angles to each other and at right angles to the direction of propagation. | |
| 26. |
Briefly explain its working. Draw its V-I characteristics for two different intensities of illumination. |
Answer» Solution :(i) Working : When reverse biased photo DIODE is illuminated with light of energy greater than the FORBIDDEN energy gap (`E_g`), electron hole pair are generated in, or near, the depletion region. DUE to junction field, electrons are COLLECTED on the n-side and holes on p-side, giving rise to a potential difference. (ii) 
|
|
| 27. |
Suppose a car is molded as a cylinder moving on earth at a speed v. If A is the area of cross section of the car and phi is the density of air then |
|
Answer» Power LOSS due to air resistance is `(A rho v^(3))/(2)` |
|
| 28. |
An archaeologist extracts a sample of carbon from an ancient ax handle and finds that it emits an average of 10 beta emissions per minute. She finds that the same mass of carbon from a living tree emits 40 beta per minute. Knowing that the half life of carbon -14 is 5730 years, she concludes that the age of the ax handle is about |
|
Answer» 2865 years |
|
| 29. |
The FM radio broadcasting band is, |
|
Answer» 5 MHz to 30 MHz |
|
| 30. |
Three tuning forks with unknown frequencies f_1,f_2 and f_3 are vibrated. 5 beats per second are heard when f_1 and f_2 are vibrated, 6 beats per second forf_1 and f_3 while 7 beats per second forf_2 and f_3. If f_2 isloaded with wax, number of beats for f_2and f_3decreases while for f_1 and f_2 increases. Find tuning forks shaving maximum frequency and minimum frequency in terms of f_2. |
|
Answer» |
|
| 31. |
Derive an experession for the RMS value of AC. |
|
Answer» Solution :i. The term RMS REFERS to time-varying sinusoidal CURRENTS and voltage and not used in DC systems. ii. The root mean square value of an alternating current is defined as the square root of the mean of the squares of all currents over one cycle. It is denoted `I_(RMS).` For alternating voltages, the RMS value is given by `V_(RMS).` iii. The alternating current `i=I_(m)sinomegatori=I_(m)sintheta,` is represented graphically in Figure. The corresponding squared lines. iv. The sum of the squares of all currents over one cycle is given by the area of one cycle of squared wave. Therefore, `I_(RMS)=sqrt(("Area of one cycle of squared wave")/("Base length of one cycle"))` v. An ELEMENTARY area of thickness `d""theta` is considered current wave as shown in Figure. Let `i^(2)` be the mid-ordinate of the element. Area of the element `=i^(2)d""theta` Area of one cycle of squared `"wave"=int_(0)^(2pi)i^(2)d""theta`  `int_(0)^(2pi)I^(2)sin^(2)thetad""theta=I_(m)^(2)int_(0)^(2pi)sin^(2)thetad""theta` `=I_(m)^(2)int_(0)^(2pi)[(1-cos2theta)/(2)]d""theta` `" since "sin^(2)theta=(1-cos2theta)/(2)` `=(I_(m)^(2))/(2)[int_(0)^(2pi)d""theta-int_(0)^(2pi)cos2thetad""theta] `(I_(m)^(2))/(2)[theta-(sin2theta)/(2)]_(0)^(2pi)` `=(I_(m)^(2))/(2)[(2pi-(sin2xx2pi)/(2))-(0-(sin0)/(2))]` `=(I_(m)^(2))/(2)xx2pi=I_(m)^(2)pi[becausesin0=sin4pi=0]` Substituting this in equation (1), we get `I_(RMS)=sqrt((I_(m)^(2)pi)/(2pi))=(I_(m)^(2))/(sqrt(2))[" Base length on cone cycle is "2pi]` `I_(RMS)=0.707I_(m)` |
|
| 32. |
A carrier wave of peak voltage 12 V is used to transmit a message signal. The peak voltage of modulating signal in order to have a modulation index as 0.75 is |
|
Answer» 12V |
|
| 33. |
A Light wave travels from a rarer medium to a denser medium. What is its frequency in the denser medium of R.I. 1.5, if its frequency in the rarer medium is 4.5 xx 10^14 Hz ? |
|
Answer» `3 XX 10^14 HZ` |
|
| 34. |
Classify the following properties of the waves into de Broglie wave, e.m wave and sound wave. i. Associated with the moving moving particle ii. Longitudinal wave iii. Electric field and magnetic field are perpendicular to each otheriv. Can produce photoelectric effect v. Wavelength is inverselyproportional to mass of the moving particle vi. Velocity in vacuum is 3xx10^(8)m//s |
|
Answer» Solution :• de BROGLIE wave - i, v • e.m wave - III, IV, vi • SOUND wave - ii |
|
| 35. |
The maximum distance upto which TV transmission from a TV tower of heighth can received is proportional to |
| Answer» SOLUTION :`d=sqrt(2Rh),dproph^(1//2)` | |
| 36. |
In hot summer as we move up refractive index of air in atmosphere ........ |
|
Answer» DECREASES |
|
| 37. |
An electronic circuit which converts an alternating voltage into a unidirectional pulsating voltage is called |
|
Answer» a TRANSISTOR |
|
| 38. |
An exothermic reaction is indicated by writing? |
| Answer» Answer :D | |
| 39. |
A wave y = A cos(omega t – kx)passes through a medium. If V is the particle velocity and a is the particle acceleration then, |
|
Answer» y,V and .a. all are in the same PHASE |
|
| 40. |
What is full wave rectification? Explain the working of a full wave rectifier. Indicate the wave forms of input and output voltage. |
|
Answer» <P> Solution :FULL wave rectification: The process in which both the halves of a.c. are rectified to obtain a pulsating DIRECT current is known as full wave rectification.A transformer with a centre TAP is used for a full wave rectification. Input contains ac. A minimum of two diode are used at the secondary of the transformer. A load resistor is connected between the junction of the N type semiconductor and the centre tap. Duringthe first half CYCLE, diode `D_(1)` will be forward biased and `D_(2)` will be reverse biased so that the current flows from B to centre tap. During the second half cycle, diode, `D_(2)` will be forward biased while `D_(1)` will be reverse biased. The current once again takes the same direction from B to centre tap. The I/P and the O/P Waveforms are shown in fig. the average DC is `2((V_(p))/(pi))`  
|
|
| 41. |
A horizontal wire is free to slide on the verticle rails of a conducting frame as shown in figure. The wire has a mass m and length l and the resistance of the circuit is R. If a uniform magnetic field B is directed perpendicular to the frame, the terminal speed of the wire as it falls under the force of gravity is In the above problem if m=1kg and teminal velocity attained by its is 4m/s after falling a height of 1m, the energy dissipated as heat till then is (g=10m//s^(2)) |
|
Answer» 10J |
|
| 42. |
An electromagnetic wave going through vacuum is described by E=E_(0) sin (kx - omegat) which of the following "is" // "are" independent of the wavelength ? |
| Answer» ANSWER :B | |
| 43. |
The decay constant, for a given radioactive sample is (0.3465)/(day). What percentage of this sample will get decayed in a period of 4 years? |
|
Answer» Solution :Here `lambda = 0.3465/day, t = 4 years` `T_(1/2) = (0.6931)/(lambda) = (0.6931)/(0.3465) = 2 days` `n = (t)/(T_(1/2)) = (4)/(2) = 2` Hence sample left undecayed after a PERIOD of 4 years, `(N)/(N_(0)) = ((1)/(2))^(n) = ((1)/(2))^(2) = 1/4 = 25 %` |
|
| 44. |
A train takes f s to perform a journey. It travels for (t)/(n) s with uniform acceleration, then for (n -3) (t)/(n) with uniform spced v and finally it comes to rest with unifom retardation. The average of the speed of the train is |
|
Answer» `(3n-2) (nu)/(2N)`  Displacement = AREA under `nu-t` graph `=(1)/(2)(t+(N-3)(t)/(n))nu` `"Average speed " = ("TOTAL displacement ")/("Total time")` `=((1)/(2)(t+(n-3)(t)/(n))nu)/(t)` `=nu/(2)(1+(n-3)(1)/(n))=(nu)/(2n)(2n-3)` |
|
| 45. |
A parallel plate capacitor is charged by a DC supply of 500V. Plate separation is 1mm. If capacitoris lowered into water with water filling in the gap between the plates, then change in pressure at any point between the plates is kepsilon_(0)xx10^(13)(N//m)^(2).Find the value of K, given epsilon_r(for water)=81. |
|
Answer» |
|
| 46. |
A log of wood floats in water with 1/5th of its volume above the surface of water. The density of wood is : |
|
Answer» `0.8xx10^(3)KGM^(-3)` `Vxxpg=4/5V.d.g` `p=4/5.d=0.8xx10^(3)kgm^(-3)` Correct CHOICE is (a). |
|
| 47. |
Two moles of an ideal monoatomic gas are confined within a cylinder by a massless spring loaded with a frictionless piston of negligible mass and of cross-sectional area 4 xx 10^(-3) m^2. The springis initially in its relaxed state. Now the gas is heated by a heater for some time. During this time the gas expands and does 50 J of work in moving the piston through a distance of 0.1 m. The temperature of the gas increases by 50 K. Calculate the spring constant and the heat supplied by the heater. |
Answer» Solution : A is the INITIAL (equilibrium) position of the piston when the spring is relaxed. When the gas is heated, it expands and pushes the piston up by a distance, say,x. The spring is COMPRESSED, if K is the force constant of the spring and A the area of cross-section of the piston (which is equal to the cross-sectional area of the cylinder),the force in the spring is F = kxand the pressure exerted on the gas by the spring is `P_S=F/A(kx)/A` If `P_0` is the atmospheric pressure, at equilibrium of piston the pressure of the gas in the cylinder is `P=P_0+P_S=P_0+(kx)/A` The INCREASE in the volume of the gas by infinitesimal movement dx of the piston is dV=Adx Thus work done is given as `W=intPdV=int_0^x(P_0+(kx)/A)Adx` `=P_0Aint_0^x dx+k int_0^x xdx` or `W=P_0Ax+1/2kx^2` Given `A=4xx10^(-3)m^2`,x=0.1 m , W=50 J.The atmospheric pressure`P_0=0.76` mof Hg = 0.76 x 9.8 x 13600 = `1.013xx10^5 Nm^(-2)`. Using these values in aboveexpression ofwork and solvingfor k, we get `k=1896 Nm^(-1)` To find heat energyQ suppliedby the heater , we use the firstlaw of thermodynamics. `Q=DeltaU+W` Now `DeltaU=f/2nRDeltaT=nC_VDeltaT` `DeltaU=3/2nRDeltaT` `=3/2xx2xx8.31xx50`=1246.5 J[ For a MONOATOMIC gas we have `C_V=3/2R` ] Thuswe have ,heat suppliedgiven as Q=1246.5+50=1296.5 J |
|
| 48. |
Two coils each of 100 turns are held at right angles such that one lies in the vertical plane with their centres coinciding. The radius of the vertical coil is 20 cm and that of the horizontal coil is 30 cm. How would you neutralize the magnetic field of the earth at their common centre ? What is the current to be passed through each coil? Horizontal component of earth's magnetic induction=3.49 xx 10^(-5) T and angle of dip = 30^@ |
| Answer» SOLUTION :`i_(1) = 0.1110 A, i_2 = 0.096A` | |
| 49. |
For a sinusoidally varying alternating current, what is the ratio of the average value and rms value ? |
|
Answer» |
|
| 50. |
If z is a complex number such that amp("z-2"/"2z+3i")=0 and z_0=3+4i, then the value of z_0z +barz_0 barz(whenever defined) |
|
Answer» is 6 `(z-2)/(2z+3i)=k` (k gt 0) z-2=2zk+3ik `z=(3ik+2)/(1-2k)` `z_0z+barz_0barz=2Re(z_0z)` `=2Re((3+4i).((3ik+2)/(1-2k)))` `2("6-12k"/"1-2k")` `=2.6"(1-2k)"/"(1-2k)"` =12 |
|