45406 + Interview Questions in Physics in Class 12

1.	What type of mirror do dentists use ?
Answer» ANSWER :CONCAVE

Discussion

2.	A radio signal has a frequency of 10 M Hz. The least length of the antenna required for the trans mission of the signal is
Answer» `7.5` m 5 m `5.5` m `3.5` m Answer :A

Discussion

3.	A current is induced in coil C_(1) due to the motion of current carring coil C_(2) as shown in Fig. 6.32. (a) Write any two ways by which a large deflection can be obtained in the galvanometer G. (b) Suggest an alternative device to demonstrate the induced current in place of a galvanometer.
Answer» Solution :(a) Large deflection can be OBTAINED in the galvanometer G, if (i) the number of turns in coils `C_(1)` and `C_(2)` is increased. (ii) the coil `C_(2)` is moved at a faster rate, and (III) an iron rod is inserted into the coils along their AXIS. (b) We can use a low power indicator lamp to demonstrate the induced current in place of a galvanometer.

Discussion

4.	If 8Omega resistance and 6Omega reactance are present in an A.C. series circuit then what will be the impedence of the circuit ?
Answer» `20OMEGA` `5Omega` `10OMEGA` `14sqrt2Omega` Solution :`\|Z\|=SQRT(R^2+(X_L-X_C)^2)` `=sqrt((8)^2+(6)^2)` `=sqrt(64+36)` `=sqrt100` `=10Omega`

Discussion

5.	The wavelength of k_(alpha) line of X-ray spectrum of an ekonsent of atomic number z = 11 in lambda. The wavelength of same time of element of atomic number is z' is 4lambda. z' is
Answer» 4 6 44 11 Solution :By Mosley.s LAW `V=a^(2) (z-B)^(2)` `v_(1)/v_(2)=((z_(1)-b)^(2))/((z_(2)-b)^(2))` For `K_(alpha), b=1` `v_(1)/v_(2)=((z_(1)-1)/(z_(2)-1))^(2) rArr lambda_(2)/lambda_(1) =((z_(1)-1)/(z_(2)-1))^(2)` `(4LAMBDA)/(lambda)=((11-1)^(2))/((z-1)^(2))` Then z=6

Discussion

6.	A reference frame attached to the earth (a) is an inertial frame by definition (b) can not be an inertial frame because the earth is revolving around the sun(c ) is an inertial frame because Newton's laws are applicable in this frame(d) cannot be an inertial frame because the earth is rotating about its axis
Answer» a, B, C are CORRECT b only correct b and d are correct All are correct ANSWER :C

Discussion

7.	The focal length of a lens of power – 2D is :
Answer» + 50 CM - 50 cm + 2 cm - 2 cm. Answer :B

Discussion

8.	Two mutually perpendicular conductors carry currents i_(1) and i_(2) along x-axis and y-axis respectively. The locus of the points at which magnetic induction is zero.
Answer» `X=(i_(1))/(i_(2))y` `y=(i_(1))/(i_(2))x` `y=(i_(2)^(2))/(i_(1))x` `x^(2)+y^(2)=(i_(1))/(i_(2))` ANSWER :B

Discussion

9.	Two different loops are concentric and lie in the same plane. The current in the outer loop is clockwise and increasing with time. The induced current in the inner loop then, is
Answer» Clockwise Zero Counter CLOCK wise In a direction that DEPENDS on the ratio of the LOOP radii Answer :C

Discussion

10.	When alpha-particle and proton move perpendicular to uniform magnetic field with same speed then ratio of periodic times of their circular motion will be ______
Answer» `2:1` `1:2` `4:1` `1:4` Solution :For UNIFORM circular motion, `(MV^(2))/r=Bqv` `therefore(mv)/r=Bq` `therefore(m(romega))/r=Bq` `thereforeomega=(Bq)/m` `therefore(2pi)/T=(Bq)/m` `thereforeT=(2pim)/(Bq)(because2piandB" are constants")` `thereforeTpropm/q` `thereforeT_(ALPHA)/T_(p)=m_(alpha)/m_(p)xxq_(p)/q_(alpha)` `=(4m_(p))/m_(p)xxe/(2E)=2/1=2:1`

Discussion

11.	A body of mass 1 kg initially at rest explodes and breaks into three fragment of masses in ratio 1 : 1 : 3. The Pieces of equal masses fly of perpendicular to each other with a speed of 30 m/s each. What is the velocity of the heavier segment.
Answer» 10.142 ms 12.142 m/s 14.142 m/s None of these Answer :C

Discussion

12.	Animals in Prison
Answer» 3 1 2 Solution :AUTUMN SEASON

Discussion

13.	Radioisotopes of phosphorus p^(32) and p^(35)are mixed in the ratio of 2:1 of atoms. The activity of the sample is 2 Ci. Find the activity of the sample after 30 days. T_(1//2)" of "P^(32) is 14 days and T_(1//2)" of "P^(35) is 25 days.
Answer» Solution :Let `A_(0)=` initial activity of sample. `A_(10)=` initial activity of isotope 1 and `A_(20)=` initial activity of isotope 2. `A_(0)=A_(10)=A_(20)` Similarly for final activity (Activity after time t) `A_(t)=A_(1t)=A_(2t)` `rArr""A_(t)=A_(10)e^(-lambda_(1)t)+A_(20)e^(-lambda_(2)t)` Now, in the givne equation, `A_(0)=2Ci rArr A_(0)=A_(10)+A_(20)=2"...(i)"` Initial RATIO of ATOMS of isotopes `=2:1` From definition of activity, `A=lambdaN` `rArr""(A_(10))/(A_(20))=(lambda_(1)N_(10))/(lambda_(2)N_(20))=(N_(10))/(N_(20))xx(T_(2))/(T_(1))` where T represents half life `rArr (A_(10))/(A_(20))=(2)/(1)xx(25)/(14)=(50)/(14)=(25)/(7)"...(ii)"` `A_(20)=(7)/(16) and A_(10)=(25)/(16)` `A_(t)=A_(10)e^(-lambda_(1)t)+A_(20)e^(-lambda_(2)t)` `rArr A_(t)=(25)/(16)e^(-(0.693)/(14)xx30)+(7)/(16)e^(-(0.693)/(25)xx30)` Consider the first exponential TERM : `e^(-(0.693xx30)/(14))=e^(-1.485)` Let `y=e^(-1.485)rArr ln y=-1.485` `rArr""logy=(-1.485)/(2.303)rArr y="antilog"((-1.485)/(2.303))` So, from above calculations you can derive a general result i.e., `e^(-X)=" antilog "((-x)/(2.303))` `A_(t)=(25)/(16)xx0.2265+(7)/(16)xx0.4365+(7)/(16)x0.4354=0.5444Ci`.

Discussion

14.	A travelling acoustic wave of frequency 500 Hz is moving along the positive x-direction with a velocity of 300 ms^(-1).The phase difference between two points x_(1)and x_2 is 60°. Then the minimum separation between the two points is
Answer» 1 mm 1 cm 10 cm 1 m Solution :The relation between phase DIFFERENCE and path differenceis GIVEN by. `Deltaphi = (2pi)/lambda. Deltax` , Here, `Deltaphi = 60^(@) = pi/3`, `V= 300 ms^(-1), v= 500 Hz` `therefore pi/3 = (2pi)/(300//500) xx Deltax (therefore v = v lambda)` `rArr Deltax = 3/(2 xx 5 xx 3) = 1/10 m = 10 cm`

Discussion

15.	Two cells epsi_1 and epsi_2 in the given circuit diagram have an emf of 5 V and 9 V and internal resistance of 0.3 Omega and 1.2Omega respectively. Calculate the value of current flowing through the resistance of 3 Omega .
Answer» Solution :Here net EMF of CIRCUIT `epsi = epsi_2 - epsi_1 = 9- 5 = 4 V` and total resistance of the circuit `R = (6 xx 3)/(6 + 3)+ 4.5 + 0.3 + 1.2 = 8Omega` ` therefore ` Main circuit current `I= epsi/R = (4V)/(8Omega) = 0.5A` If current flowing through `3OMEGA` resistance be `I_1` , then current flowing through `6Omega` resistance will be `(0.5 - I_1)` and hence `3I_1 = 6 xx (0.5 - I_1) rArr I_1 = 0.33 A`

Discussion

16.	T.V. transmission tower at a particular station has a height of 160 m. How much population is covered by transmission, if the average population density around the tower is 1200 per km?
Answer» Solution :Population covered = (population density) `XX` (area covered) `= (1200) xx (PID^(2))` `= (2400piRh) = 2400xx3.14 xx 6.4 xx 10^(3)xx 0.16= 77.17 lac`

Discussion

17.	A smooth block is released at rest on a 45^(@) incline and then slides a distance 'd'. The time taken to slide is 'n' times as much to slide on rough incline than on a smooth incline. The coefficient of friction is
Answer» `mu_(k)=1-(1)/(n^(2))` `mu_(k)=SQRT(1-(1)/(n^(2)))` `mu_(s)=1-(1)/(n^(2))` `mu_(s)=sqrt(1-(1)/(n^(2)))` Solution :On smooth incline plane `d=(1)/(2)g SINN THETA xxt^(2)` On rough incline plane `d=(1)/(2)g(sin theta-mu_(k) COS theta) (nt)^(2)...(b)` SOLVING equations (a)and(b) `sin theta=n^(2)(sintheta-mu_(k) cos theta)`and Putting `theta=45^(@)` we get `mu_(k)=1-(1)/(n^(2))`

Discussion

18.	A magnetic field is established by a circular current i of radius a. Find the magnetic field gradient (i.e. the derivative of the magnetic field induction vector) in the direction of the circular current's axis at a point whose distance from the centre of the turn is x.
Answer» SOLUTION :The magnetic field induction at the point of interest is, `B=(2mu_(0) p_(m))/(4pir^2)=(mu_(0)p_(m))/(2pi) (a^(2)+x^2)^(-3/2)` The gradient is `(dB)/(DX) =(mu_(0)p_(m))/(2pi)=d/(dx) (a^2+x^2)^(-3/2)` `=-3/2 (mu_(0)p_m)/(2pi) (a^2+x^2)^(-5//2), 2x=-(3mu_(0) p_(m)x)/(2pir^3)`

Discussion

19.	Explain refraction of light from a denser toa rear medium, using the concept of wavefronts.
Answer» Solution : where, `BC=v_1 tau,AD=v_2 tau,v_2 gt v_1,tau=` time taken to travel optical path by light. AB - incident wave front DC = refracted wave front. i = angle of incidence in medium (1) For `I ge i_e` , total internal reflection takes PLACE at the INTERFACE PQ. For `i LT i_e`, there will be refraction of light fron denser to rarer medium From the right angled triangle ABC `sin i=(BC)/(AC)=(v_1 tau)/(AC)""...(1)` and from the right angled triangle ADC, `sin r =(AD)/(AC)=(v_2 tau)/(AC)""...(2)` `(1) div (2)` gives `(sin i)/(sinr) =(v_1)/(v_2)""...(3)` Since, `v_1 prop (1)/(n_1) and v_2 prop (1)/(n_2)` equation (3) may be written as `(sin i)/(sin r)=(n_2)/(n_1)` where , `n_2 lt n_1` THEREFORE n_1 sin i=n_2 sin r`.

Discussion

20.	In L-C oscillator, at ... time, energy in capacitor and energy in inductor are equal.
Answer» `T/8` `T/4` `T/2` T Solution :When energy in capacitor and energy in inductor are equal in L-C oscillations, `U=U_E+U_B` `therefore U=2U_E "" (because U_E=U_B)` `therefore q_0^2/(2C)=2(q^2/(2C))` `therefore q^2=q_0^2/2` `therefore q=q_0/sqrt2` `therefore q_0 COS omega_0t=q_0/sqrt2 ""(because q=q_0 cos omega_0t)` `therefore cosomega_0 t = 1/sqrt2` `therefore omega_0 t=pi/4` `therefore ((2PI)/T)t = pi/4 "" (because omega_0 =(2pi)/T)` `therefore t=T/8`

Discussion

21.	What is the ratio of magnetic moments of two bar magnets of identical dimensions and masses if there periods are in the ratio 1:2, when suspended freely ?
Answer» SOLUTION :`T=2pisqrt1/(MB_H)` ` THEREFORE T_1/T_2=sqrt(M_2/M_1)` `1/2=sqrt(M_2/M_1)` ` thereforeM_1/M_2=4/1`

Discussion

22.	Forward bias has to be provided for the ….. Semiconductor device to work.
Answer» PHOTODIODE zener diode varactor diode LIGHT emitting diode (LED) ANSWER :D

Discussion

23.	Determine the speed of sound waves in water, and findthe wavelength of a wave having a frequency of 242 Hz. Take B_("water") = 2xx 10^9 Pa.
Answer» Solution :SPEED of SOUND wave, `V = sqrt(B/rho) = sqrt(((2xx10^9))/10^3) =141m//s` Wevelength `lambda =v/f = 5.84 m`

Discussion

24.	A transvers sinusoidal wave move along a string in the positive x-direction at a speed of 10 cm/s. The wavelength of the wave is 0.5 m and its amplitude is 10 cm . At a particular time t, the snap-shot of the wave is shown in figure . The velocity of point P when its displacement is 5 cm is :
Answer» <P>`(sqrt(3 pi))/(50 ) hatj` m/s `- (sqrt(3 pi))/(50 ) hatj` m/s `(sqrt(3 pi))/(50 ) hati` m/s ` - (sqrt(3 pi))/(50 ) hati` m/s Solution :Velocity of point P = `(dy)/(dt ) = - ` (velocity of wave ) `xx (dy)/(dx) ` Here `(dy)/(dx) ` is negative . Therefore,velocity at point P ispositive and is along y-axis only. ALITER Equations of a wave moving in positive x-a-axisis given as y = A sin `(OMEGA t - phi ) or V_(p) = A omega cos (omega t - phi) ` Here y= 5 cm , A = 10 cm, `therefore 5= 10 sin (omega t - phi) rArr omega t - phi = 30^(@)` Subsitituting this value in the quation of velocity we get `V_(p) = 0.10 xx omega cos 30^(@)` Now V = v `LAMBDA "" therefore v= (v)/(lambda) = (0.10)/(0.5)= 0.2` `therefore omega = 2pi v= 2pi xx 0.2 = 0.4 pi` `rArr v_(P) = 0.1 xx 0.4 pi xx (sqrt(3))/(2) = (sqrt(3))/(50 ) pi ` It has to be in positive y direction. So correct choice is (a).

Discussion

25.	How many image will be formed if two mirrors on the cellings?
Answer» 2 3 11 7 Answer :D

Discussion

26.	Describe schematically the equipotential surfaces corresponding to (a) a constant electric field in the z-direaction . (b) a field that uniformaly increases in magnitude but reamins in a constant (say,z) direction. (c) a single positive charge at the origin, and (d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.
Answer» Solution :By defination an equipotential SURFACE is that EVERY point of the which potential is the same. In the four cases given above, (a) Equipotential surfaces are planes PARALLEL to x - y plane. These are EQUIDISTANT. (b) Equipotential surfaces are places parallel to x - y plane. As the FIELD increases uniformly distance between the planes decreases. (c ) Equipotential surfaces concentric spheres with origin at the centre. (d) Equipotential surfaces have the shape which changes periodically at far off distance from the grid.

Discussion

27.	The objective of a microscope has a focal length of 1 cm and diameter 0.5 cm. Using light of 6000 Å, is it possible to resolve the two point objects if they are separated by a distance of 10^(-6) m ?
Answer» Solution : Suppose the two objects are a distance `Delta x `apart. Since these objects are almost at the focal LENGTH of the objective, the angle SUBTENDED by `Delta x, i.e., Delta phi = (Delta x)/(f_0)` For the two objects to be clearly resolved ,` Delta phi GT Delta theta` `(Delta x)/(f_0) gt (1.22 lamda)/(a) (Delta theta = (1.22 lamda)/(a)) " or " Delta x gt (1.22 f_0 lamda)/(a)` `Delta x gt ((1.22)(1 xx 10^(-2) m)(6000 xx 10^(-10) m) )/((0.5 xx 10^(-2)m) ) = 1.46 xx 10^(-6) m` Because `10^(-6)m` is smaller than `1.46 xx 10^(-6)`m, the two objects will not be resolved .

Discussion

28.	A very long straight conductor has a circular cross- section of radius R and carries a current density J. Inside the conductor there is a cylindrical hole of radius a whose axis is parallel to the axis of the conductor and a distance b from it. Let the z-axis be the axis of the conductor, and let the axis of the hole be at x = b. Find the magnetic field on the x=axis at x = 2R
Answer» Solution :`B=(mu_(0)J)/(2)((a^(2))/(2R-b)-(R)/(2))`

Discussion

29.	A glass rod when rubbed with silk cloth acquires a charge of 1.6 xx 10^-13C. What is the charge on the silk cloth ?
Answer» SOLUTION :1.6 XX 10^(-13)C.

Discussion

30.	A very long straight conductor has a circular cross- section of radius R and carries a current density J. Inside the conductor there is a cylindrical hole of radius a whose axis is parallel to the axis of the conductor and a distance b from it. Let the z-axis be the axis of the conductor, and let the axis of the hole be at x = b. Find the magnetic field on the y = axis at y = 2R.
Answer» Solution :`B_(x)=mu_(0)JR((1)/(4)-(a^(2))/(4R^(2)+B^(2))),B_(y)=-(mu_(0)J)/(2)((a^(2)b)/(4R^(2)+b^(2)))`

Discussion

31.	A body is placed in an insulating container at temperature 227^(@)C. In order to keep its temperature 727^(@)C, power of 60 W is required, the power required to keep its temperature 1227^(@)C will be :
Answer» 304 321 90 120 Solution :Here E = power `(E_(1))/(E_(2))=(T_(1)^(4)-T_(0)^(4))/(T_(2)^(4)-T_(0)^(4))` `(60)/(E_(2))=((1000)^(2)-(5000)^(4))/((1500)^(4)-(500)^(4))=(16-1)/(81-1)=(15)/(80)` or `E_(2)=(60xx80)/(15)=320W` Correct choice is (b).

Discussion

32.	The displacement current is found
Answer» between the PLATES of a condenser when it is being charged between the plates of a condenser when it is being discharged between the plates of a condenser when AC is applied to the condenser in the above 1, 2, 3 cases Answer :D

Discussion

33.	What is interference of light?
Answer» Solution :The PHENOMENON of addition on superpostion of TWO light waves which produces INCREASE in INTENSITY at some points and decrease in intensityat some other points is CALLED interfreene of light.

Discussion

34.	Two point charges +5muC and -2 muC are kept at a distance of 1m in free space. The distance between the two zero potential points on the line joining the charge is
Answer» a. `2/7m` b. `2/3m` C. `(22)/(21)m` d. `(20)/(21)m` Answer :D

Discussion

35.	80 microC charge is given on sphere of 4 cm radius and 40 muCcharge is given on sphere of 6 cm radius. If they are connected by a conducting wire, then charge transferred from sphere of 4 cm radius to sphere of 6 cm radius will be......
Answer» `48 muC` `72 muC` `32 muC` `36 muC` SOLUTION :If sphere of radii 4 cm and 6 cm are called 1 and 2 respectively, then TOTAL charge, `Q = Q_(1) + Q_(2)` =40 + 80 = `120 muC` Charge on both sphere after connecting by wire, `Q_(1) = 48 muC` and `Q_(2) = 72 muC` by USING formula `Q(r_(2)/(r_(1) +r_(2)))` `therefore (80 muC - 48muC) = 32 muC` will be transferred from sphere of 4 cm to sphere of 6 cm radius.

Discussion

36.	A focal length of a lens is 10 cm. What is power of a lens in dioptre ?
Answer» 0.1 D 10 D 15 D 1 D. Solution :(b) Given, F = 10 CM power P = ? The power of LENS `P = (1)/("focal length"(m))` `P = (1)/(f(m))` `or "" P = (100)/(f(cm)) = (100)/(10) = 10 D`.

Discussion

37.	Uncertainty in position of electron is of the order of de-Broglie wavelength .By using Heisenberg's principle uncertainty in velocity will be….
Answer» 1 v 2 v `(v)/(2PI)` `2piv` Solution :`DELTAX.Deltap=(h)/(2pi)` `therfore Delta x[mDeltav]=(h)/(2pi)` `therefore Deltax[mDeltav]=(h)/(2pi)` `therefore Deltav=(h)/(2pimDeltax)=(h)/(2pimlambda)[because Deltax=lambda]` `therefore Deltav=(p)/(2pim)=(mv)/(2pim)=(v)/(2pi)[because (h)/(PI)=p]`

Discussion

38.	Express dimension of L/R ? (where L = Self inductance and R = resistance of the coil)
Answer» SOLUTION :DIMENSION of `L/R` = `[M^0L^0T^1]`

Discussion

39.	How does the magnetisation vector of a paramagnetic sample vary with temperature ?
Answer» SOLUTION :Magnetisation of a PARAMAGNETIC sample is INVERSELY proportional to its TEMPERATURE.

Discussion

40.	Planck's constant, has dimensions of Energy x time. In a system of units, used in particle physics, Planck's constant and the velocity of light c are both put equal to 1 and energy is a fundamental unit. What will be the unit for measurement of angular momentum? What will be the dimension of length? If the unit of energy is 1 MeV, what will be the unit of length?
Answer» ANSWER :H,`E^(-1)`,`[1MEV]^(-1)`

Discussion

41.	One thousand small water droplets of equal size combine to form a big drop. Find the ratio of the final surface energy to the initial surface energy of water droplets.
Answer» Solution :`4/3 PI R^2 = 1000 XX 4/3 pi r^3` `therefore R^3 = 1000r^3` ` therefore` R = 10r` `E_1 = 1000T (4pir^2)` `E_2 = T xx 4pi (10r)^2` `therefore E_2/E = (100r^2)/(1000r^2)` = 1/10

Discussion

42.	Outer flexible layer in euglenoids is called
Answer» CELL wall pellicle glycocalyx None of the above Answer :B

Discussion

43.	Writer Bohr's postulates for the hydrogen atom model.
Answer» Solution :(i)Bohr.s first POSTULATE was that an electron in an atom could resolve in certain stable orbits without the emissionof radiant energy. ACCORDING to this postulate each atom has certain definite stable orbits in which it can exists and each possible state has definite total energy . These are called the stationary states of the atom. (ii)Bohr.s second postulate DEFINES these stable orbits. This postulate states that the electrons resolves AROUND the nucleus only in those orbits for which the angular momentum is integral multiple of `h/(2pi)` where h is the plank.s constant. (iii)Bohr.s third postulates states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy . When it does so a photon is emitted having energy equal to the energy difference between the initial and final states. The frequency of the emitted photon is then given by `hv=E_r-E_i RARR v=(E_f-E_i)/h`

Discussion

44.	Subtract 2.5xx10^4 and 3.9xx10^5 with due regard to significant figures.
Answer» Solution :LET `xx=2.5xx10^4=25000` `y=3.9xx10^5=390000` `y=x=39000-25000=3.65xx10^5=3.6xx10^5` ROUNDED up to ONE place of DECIMAL.

Discussion

45.	What are linear isotropic dielectrics ?
Answer» SOLUTION :SUBSTANCES whose induced dipole MOMENT act in the direction of the electric FIELD are called linear ISOTROPIC dielectrics.

Discussion

46.	Consider the following statement A and B and identify the correct answer A) Radio waves diffract around buildings but light waves does not B) To cut down glare of incident light we prefer sun glasses made from polaroids
Answer» A is FALSE but B is TRUE A is true but B is false Both A and B are true Both A and B are false Answer :C

Discussion

47.	Draw a ray diagram to show the image formation by a concave mirror when the object is kept between its focus and the pole. Using this diagram, derive the magnification formula for the image formed.
Answer» Solution :RAY diagram, showing the image formation by a concave mirror when the object AB is KEPT between its focus F and the pole P, is given in Fig. 9.33. The image A.B. is a virtual, erect and magnified image. CONSIDER `DeltaABP`and `triangleA.B.P`. Obviously, the triangles are SIMILAR triangles. Hence, we have PB `=-u` `PB. =+v, AB =h` and `A.B. =h.` `THEREFORE h^(.)/h=(+v)/(-u)` `therefore` Linear magnification of image formed `mh^(.)/h =-v/u`

Discussion

48.	A voltage V = V_(0) sin omega t is applied to a series LCR circuit. Derive the expression for the average power dissipated over a cycle. Under what condition is (i) no power dissipated even though the current flows through the circuit, maximum power dissipated in the circuit ?
Answer» Solution :APPLIED voltage `= V_(0) sin omega t` Current in the circuit `= I_(0) sin (omega t - phi)` where `phi` is the phase lag of the current with respect to the voltage applied, Hence,instantaneous power dissipation `= V_(0) sin omega t xx I_(0) sin(omega t - phi)` `= (V_(0)I_(0))/(2)[2 sin omega t. sin (omega t - phi)]` `= (V_(0)I_(0))/(2)[cos phi - cos (2 omega t - phi)]` Therefore average power for ONE complete cycle = average of `[(V_(0)I_(0))/(2) {cos phi - cos (2 omega - phi)}]` The average of the SECOND TERM over a complete cycle is zero. Hence, average power dissipated over one complete cycle `= (V_(0)I_(0))/(2) cos phi` Conditions : (i) No power is dissipated when R = 0 (or `phi = 90^(@))` [NOte : We can also WRITE, this condition cannot be satisfied LCR circuit..] (ii)Maximum power is dissipated when `X_(L) = X_(C)` or `omega L = (1)/(omega C)` (or `phi = 0`)

Discussion

49.	Determine potential energy of the charge configuration shown in figure.
Answer» ANSWER :`(Q^(2))/(4piepsi_(0)a)(-SQRT(2))`

Discussion

50.	Which one is an incorrectly matched pair?
Answer» Phycomycetes-Mucor, Albugo Ascomycetes- PENICILLIUM, Aspergillus Basidiomycetes-Puccinia, Agaricus Deuteromycetes- USTILAGO, Colletotrichum Answer :D

Discussion

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Animals in Prison

Radioisotopes of phosphorus p^(32) and p^(35)are mixed in the ratio of 2:1 of atoms. The activity of the sample is 2 Ci. Find the activity of the sample after 30 days. T_(1//2)" of "P^(32) is 14 days and T_(1//2)" of "P^(35) is 25 days.

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T.V. transmission tower at a particular station has a height of 160 m. How much population is covered by transmission, if the average population density around the tower is 1200 per km?

A smooth block is released at rest on a 45^(@) incline and then slides a distance 'd'. The time taken to slide is 'n' times as much to slide on rough incline than on a smooth incline. The coefficient of friction is

A magnetic field is established by a circular current i of radius a. Find the magnetic field gradient (i.e. the derivative of the magnetic field induction vector) in the direction of the circular current's axis at a point whose distance from the centre of the turn is x.

Explain refraction of light from a denser toa rear medium, using the concept of wavefronts.

In L-C oscillator, at ... time, energy in capacitor and energy in inductor are equal.

What is the ratio of magnetic moments of two bar magnets of identical dimensions and masses if there periods are in the ratio 1:2, when suspended freely ?

Forward bias has to be provided for the ….. Semiconductor device to work.

Determine the speed of sound waves in water, and findthe wavelength of a wave having a frequency of 242 Hz. Take B_("water") = 2xx 10^9 Pa.

A transvers sinusoidal wave move along a string in the positive x-direction at a speed of 10 cm/s. The wavelength of the wave is 0.5 m and its amplitude is 10 cm . At a particular time t, the snap-shot of the wave is shown in figure . The velocity of point P when its displacement is 5 cm is :

How many image will be formed if two mirrors on the cellings?

The objective of a microscope has a focal length of 1 cm and diameter 0.5 cm. Using light of 6000 Å, is it possible to resolve the two point objects if they are separated by a distance of 10^(-6) m ?

A glass rod when rubbed with silk cloth acquires a charge of 1.6 xx 10^-13C. What is the charge on the silk cloth ?

A body is placed in an insulating container at temperature 227^(@)C. In order to keep its temperature 727^(@)C, power of 60 W is required, the power required to keep its temperature 1227^(@)C will be :

The displacement current is found

What is interference of light?

Two point charges +5muC and -2 muC are kept at a distance of 1m in free space. The distance between the two zero potential points on the line joining the charge is

80 microC charge is given on sphere of 4 cm radius and 40 muCcharge is given on sphere of 6 cm radius. If they are connected by a conducting wire, then charge transferred from sphere of 4 cm radius to sphere of 6 cm radius will be......

A focal length of a lens is 10 cm. What is power of a lens in dioptre ?

Uncertainty in position of electron is of the order of de-Broglie wavelength .By using Heisenberg's principle uncertainty in velocity will be….

Express dimension of L/R ? (where L = Self inductance and R = resistance of the coil)

How does the magnetisation vector of a paramagnetic sample vary with temperature ?

One thousand small water droplets of equal size combine to form a big drop. Find the ratio of the final surface energy to the initial surface energy of water droplets.

Outer flexible layer in euglenoids is called

Writer Bohr's postulates for the hydrogen atom model.

Subtract 2.5xx10^4 and 3.9xx10^5 with due regard to significant figures.

What are linear isotropic dielectrics ?

Consider the following statement A and B and identify the correct answer A) Radio waves diffract around buildings but light waves does not B) To cut down glare of incident light we prefer sun glasses made from polaroids

Draw a ray diagram to show the image formation by a concave mirror when the object is kept between its focus and the pole. Using this diagram, derive the magnification formula for the image formed.

A voltage V = V_(0) sin omega t is applied to a series LCR circuit. Derive the expression for the average power dissipated over a cycle. Under what condition is (i) no power dissipated even though the current flows through the circuit, maximum power dissipated in the circuit ?

Determine potential energy of the charge configuration shown in figure.

Which one is an incorrectly matched pair?