45406 + Interview Questions in Physics in Class 12

1.	find outmotionof treebirdand oldman asseenbyboy ,
Answer» Solution :Withrespectto BOY: `V_("tree ") = 16 m//s( larr)` ` V_("bird ") = 12m//s( UARR)` `V_(oldman ") = 10 m//s ( larr)`

Discussion

2.	N fundamental charges each of charge 'q' are to be distributed as two point charges separated by a fixed distance, then the maximum to minimum force bears a ratio (N is even and greater than 2)
Answer» `((N - 1)^2)/(4N^2)` `(4N^2)/((N-1))` `(N^2)/(4(N-1))` `(2N^2)/((N-1))` ANSWER :C

Discussion

3.	Tritium and helium are put 2 xx 10^(-10)m apart in air. Magnitude of intesnity of electric field due to tritium at helium is in (N/C)
Answer» `3.6 XX 10^(10)` `36 xx 10^(-10)` `4 xx 10^(-2)` `7.2 xx 10^(-10)` Answer :A

Discussion

4.	A vertical circular coil of one turn and radius 9.42 cm is placed with its plane in the magnetic meridian and short magnetic needle is pivoted at the centre of the coil so that it can freely rotate in the horizontal plane. If a current of 6A is passed through the coil, then the needle deflects by (B_(H) = 4 xx 10^(-5)T)
Answer» `30^(@)` `45^(@)` `60^(@)` `90^(@)` ANSWER :B

Discussion

5.	Starting from rest a wheel rotates with uniform angular acceleration 2 pirad s^(-2). After 4s, if the angular acceleration ceases to act, its angular displacement in the next 4s is
Answer» `8 pi RAD` `16 pi rad` `24 pi rad` `32 pi rad` Answer :D

Discussion

6.	What is Toroid? Calculate the magnetic field at a point Open space exterior to the toroid
Answer» Solution :A slolenoid is bent in a way that both their ends are joined together to form a closed ring shape, is called as toroid . The magnetic FIELD has constant magnitude inside the toroid whereas in the interior region ( say , at POINT P ) and exteriorrigion ( say , at point Q ) , the magnetic field is zero. Inside the toroid : Let us calculate the magnetic field `B_(S) ` at point S by constructing an Amperian loop 2 of radius `r_(2)` around the point S as SHOWN in Figure. The length of the loop is`L_(2) = 2 pi r_(2)` Ampere's circuital law for the loop 2 is `underset("loop2")oint vecB_(s) . vec(dl) =mu_(0)I_("enclosed")` Let I be the current passing through the toroid and N be the NUMBER of turns of the toroid, then ` I_("enclosed") = NI ` and `underset("loop2")oint vecB_(s) . vec(dl) =underset("loop2")ointB dl cos theta = B 2 pi r_(2)` `underset("loop2")ointvecB_(s) . vec(dl) = mu_(0) NI ` ` B_(s) = mu_(0) (NI)/(2 pi r_(2))` The number of turns per unitlength is `n = N/(2 pi r_(2))` , then the magnetic field at pointS is ` B_(s) = mu_(0) n I`

Discussion

7.	Name the em waves which are suitable for radar systems used in aircraft navigation. Write the range of frequency of these waves.
Answer» SOLUTION :MICROWAVES FREQUENCY RANGE : `10^10` HZ to `10^12` Hz

Discussion

8.	Two co-axial identical circular current carrying loops are shown in figure, currents in them are in the same directions. Now, match the following two columns. {:(,"Column I",,"Colums II"),(A.,"Current i"_(1)" is increased",p.,"Loops will attract each other"),(B.,"Current i"_(2)" is decreased",q.,"Loops will repel each other"),(C.,"Loop-1 is moved towards loop-2",r.,"Current i"_(1)" will increase"),(D.,"Loop-2 is moved away from loop-1",s.,"Current i"_(2)" will increase"):}
Answer» Solution :When current `i_(2)` is decreased, then flux through A decreases. According to LENZ's law, ATTRACTION occurs between A and B. SIMILARLY, when current `i_(1)` is increased, then loops will repel each other. As distance between the loop decreases, M decreases. Hence, loops will repel each other and current `i_(1)` will INCREASE, when loop 1 is moved towards loop 2. When loop 2 is moved away, loops will attract and `i_(2)` will increase.

Discussion

9.	One cylindrical bar magnet, kept on the axis of one circular coil. Now if this magnet is rotated about this axis, then ____
Answer» CURRENT will FLOW in the COIL. current will not flow in the coil. only EMF will be induced. emf and current both will be induced. Solution :Because here, magnetic flux associated with the coil, does not change.

Discussion

10.	In a p-n junction diode , the current I can be expressed as , I=I_(0)[e^(eV//kT)-1] where I_(0) is called reverse saturation current, V is the voltage across the diode and is positive for forward bias and negative for reverse bias and I is the current through the diode, k is the Boltzmann constant (8.6xx10^(-5)eVK^(-1)) and T is the absolute temperature. If for a diode I_(0)=5xx10^(-12)A and T=300K, thenWhat will be the forward current at a forward voltage of 0.6V? What will be the increase in current, if the voltage across the diode is increase to 0.7V? What is the dynamic resistance? What will be the current, if the reverse bias voltage changes from 1V to 2V?
Answer» Solution :Here, `k=8.6xx10^(-5) eVK^(-1)` `=8.6xx10^(-5)xx1.6xx10^(-19)JK^(-1)`, `I_(0)=5xx10^(-12)A, T=300K`. If `V=0.6V`, then forward current is `I=I_(0)[e^(eV//kT)-1]` `=5xx10^(-12)[e^(((1.6xx10^(-19))xx0.6)/((8.6xx10^(-5)xx1.6xx10^(-19))xx300))-1]` `=5xx10^(-12)[e^(23.255)-1]` `=5xx10^(-12)[(2.718)^(23.255)-1]=0.063A` If `V=0.7V`, then `I_(1)=5xx10^(-12)[e^(((1.6xx10^(-19xx0.7)))/((8.6xx10^(-5)xx1.6xx10^(-19))xx300))-1]` `=5xx10^(-12)[e^(27.1318)-1]=3.035A` `:.` Increase in current `DeltaI=I_(1)-I=3.035-0.063=2.972A` Now `DeltaV=0.7-0.6=0.1V, DeltaI=2.972A` Dynamic resistance, `=(DeltaV)/(DeltaI)=0.1/2.972=0.0336Omega` For a change in VOLTAGE from 1 to 2 V, the current will remain to saturation stae, `I_(0)=5xx10^(-12)A`. It shown that the diode possesses practically infinite resistance in REVERSE biased.

Discussion

11.	In the case of single slit Franuhoffer diffraction the angle of diffreaction is theta for the second secondary maximum. If 'a' is the width of the slit, the wavelength of the light is
Answer» `(2a SIN theta)/(3)` `(3a sin theta)/(2)` `(2a sin theta)/(5)` `(2a cos theta)/(3)` Answer :C

Discussion

12.	In a college of 300 students, every student reads5 newspapers and everynewspaper is read by 60 students. The number of newspaper is?
Answer» at LEAST 30 At most 20 exactly 25 None of these Answer :C

Discussion

13.	A sand scorpion can detect the motion of a nearby beetle (its prey) by the waves the motion sends along the sand surface (Fig, 16-50). The waves are of two types transverse waves traveling at v_(1)= 50 m/s and longitudinal waves traveling at V_(1)=150 m//s. If a sudden motion sends out such waves a scorpion can tell the distance of the bestle from the difference ar in the arrival times of the waves at its leg nearest the beetle. What is that time difference if the distance beetle is 37.5 cm?
Answer» ANSWER :5.0 MS

Discussion

14.	As shown in the figure, P and Q are two coaxial conducting loops separated by some distance. When the switch S is closed, a clockwise current Ip flows in P as seen by E) and an induced current Iq_1, flows in Q. The switch remains closed for a long time. When S is opened, a current Iq_2 flows in Q. Then the directions of Iq_1 and Iq_2(as seen by E) are
Answer» RESPECTIVELY CLOCKWISE and Anti-clockwise both clockwise both Anti-clockwise respectively anti-clockwise and clockwise Answer :D

Discussion

15.	Light is refracted from air to a liquid. The angle of incidence is 60^@. The deviation produced is 15^@. The R.I. of liquid is:
Answer» 1.5 1.33 1.12 1.22 Answer :D

Discussion

16.	A mass of 1 kg is suspended by means of a thread. The system is (i) lifted up with an acceleration of 4.9 ms^(-2) (ii) lowered with an acceleration of 4.9 ms^(- 2). The ratio of tension in the first and second case is
Answer» `3:1` `1:2` `1:3` `2:1` SOLUTION :Case-I Using Newton.s SECOND LAW of motion, `T_(1)` - mg= ma `T_(1)=m(g+a)` …(i) Case-II Using Newton.s second law of motion, `mg-T_(2)=ma` `T_(2)=m(g-a)` …(ii) From EQN. (i) and (ii) `(T_(1))/(T_(2))=(g+a)/(g-a)=(9.8+4.9)/(9.8-4.9)=3/1`

Discussion

17.	During Rutherford, there was no particle acclerators to produce high energy particles, what he used ?
Answer» SOLUTION :`alpha-paricles`

Discussion

18.	Obtain an expression for the force between two straight parallel conductor carrying current. Hence define ampere.
Answer» Solution :X and Y are two long straight parallel conductors carrying currents `I_1` and `I_2` respectively, and PLACED close to each other. d is the separation between the two conductors and L is the length of the conductors. The magnetic field at any point on the conductor Y due to current `I_1` in the conductor X, is given by `B_1=mu_0/(4pi)(2I_1)/d. vecB_1`acts in a direction perpendicular to the plane containing the two conductors. The conductor Y which CARRIES current `I_2` experiences a mechanical FORCE due to `B_1` acting on it at this force is given by `F_1=B_1 I_2 L sin theta` . `thereforeF_1=(mu_0/(4pi) (2I_1)/d) L_2L sin theta` `F_1=mu_0/(4pi) (2I_1I_2)/d L`...(1) According to Fleming.s left hand rule, the direction of the force `F_1` on X is perpendicular to `B_2` and is towards the conductor Y. The magnetic field at any point on the conductor X due to the current `I_2` in y, is given by `B_2=mu_0/(4pi) (2I_2)/d` . `vecB_2`acts onX and oppositeto `B_1`. The mechanicalforce acts on Xdue to`B_2` is `I_1` L sin `theta`. `thereforeF_2=(mu_0/(4pi) (2I_2)/d) I_2L sin 90^@` `F_1=mu_0/(4pi) (2I_1I_2)/d L` ...(2) According to Fleming.s left hand side rule, the direction of the force `F_2` on X is perpendicular to X and it is towards the conductor y if the current `I_1` is inwards (or away from the conductor Y if the current `I_1` is outwards). The Force `F_1` acting on a certain length of the conductor Y due to the current in the conductor X is equal in magnitude to the force `F_2` acting on the same length of X due to the current in conductor Y. If the two conductors carry the currents in the same direction (parallel currents) then the forces attract each other. If the two conductors carry the currents in the opposite directions (anti parallel currents), then they are found to repel each other because the two forces act away from each other. The force per unit length on each conductor is `F_L=F_1/L=F_2/L=(mu_0/(4pi)(2I_1I_2)/d L)/L` i.e., `F_L=mu_0/(4pi) (2I_1I_2)/d` One AMPERE of current can be defined as that constant current which when maintained through each of the two infinitely long straight parallel conductors of negligible area of cross section in the same direction placed I metre apart in vacuum, causes an attractive force of `2 XX 10^(-7)Nm^(-1)`length on each conductor.

Discussion

19.	In the following question a statement of assertion (A) is followed by a statement of reason (R ) A : Electric field is discontinuous across the surface of a charged conductor. R : Electric potential is discontinuous across the surface of a charged conductor .
Answer» If both Assertion & Reason are true and the reason is the correct explanation of the assertion , then mark (1). If both Assertion & Reason are true but the reason is not correct explanation of the assertion then mark (2) If Assertion is true statement but Reason is false, then mark (3) If both Assertion and Reason are false statements , then mark (4). Answer :C

Discussion

20.	Why current is not formed in solid conductors in absence of electric field ?
Answer» Solution :`rArr` In solid conductors electric CURRENT can be explained as follows. `rArr` When METALLIC solid are formed, valence electron get separated from their parent atoms and become free electron positive ions are grounded in definite three dimensional geometric arran gement. `rArr`In ABSENCE of external electric field these positive ion, depending on their temperature oscillate about their mean position. Electron move randomly in space between positive ions. `rArr` During their moHon electron collide with positive ion and due to these collision their DIRECTION keeps on CHANGING. ` ` At a given time there is no unidirectional flow of electron in particular direction in given cross- section. Hence, current is not formed.

Discussion

21.	A galvanometer coil has a resistance of 15 Omegaand the metre shows full scale deflection for a current of 4 mA. How will you convert the metre into an ammeter of range 0 to 6 A?
Answer» SOLUTION :SHUNT RESISTANCE ` = 10 m OMEGA `

Discussion

22.	What is dimension of [L]?
Answer» SOLUTION :`M^1L^2T^-2A^-2`

Discussion

23.	AU-tube is partly filled with a liquid A. Another liquid B, which does not mix with A, is poured into one side until it stands a height h above the level of A on the other side, which has mean while risen a heigh l. The density of B relative to that of A is :
Answer» `l/(h+l)` `l/(h+2L)` `(2l)/(h+2l)` `l/(h+2l)` SOLUTION :POINTS `C_(1)` and `C_(2)` are at the same pressure. Therefore, the pressure drop from `C_(1)` to `D_(1)` is the same as that from `C_(2)` to `D_(2)` both `D_(1)` and `D_(2)` being at atmospheric pressure. Let `p_(a)` and `p_(b)` be the densities of A and B, respectively. We then have (h+2l)`p_(b)g=2lp_(a)g` or `(p_(b))/(p_(a))=(2l)/(h+2l)` CORRECT choice is (c).

Discussion

24.	What is the minimum thickness of a thin film required for constructive interference in the reflected light from it? Given, the refractive index of the film = 1.5, wavelength of the light incident on the film = 600nm
Answer» 100 NM 300 nm 50 nm 200 nm Solution :Condition for constructive interference is `2mu t = [2n + 1](lambda)/(2)` where ,n = 0, 1, 2, 3,... For minimum THICKNESS, n = 0 `2mu t = (lambda)/(2)` `rightarrow t = (lambda)/(4mu) = (600 xx 10^(-9))/(4 xx 1.5)` = 100 nm

Discussion

25.	Which one of the waves cannot be polarized
Answer» RADIO waves X-rays ultra VIOLET rays sound waves Answer :D

Discussion

26.	Wavelength of K_(a) line of X-ray spectra varies with atomic number as
Answer» `lambda prop Z` `lambda prop sqrtZ` `lambda prop (1)/(Z^(2))` `lambda prop (1)/(sqrtZ)` Solution :According to moseley.s law, `sqrtV = a(Z-b) or V = C/(lambda) = a^(2)(Z-b)^(2) THEREFORE lambda INFTY (1)/(Z^(2))`

Discussion

27.	If in the infinite series circuit, C = 9 mu F and C1 = 6 mu F then the capacity across AB is
Answer» `1.25 mu F ` `6 mu F ` `3 mu F ` `12 mu F ` Answer :C

Discussion

28.	Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).
Answer» Solution :Water MOLECULES are having unsymmetric SHAPE and, thus, possess permanent electric dipole moment. Consequently, POLARISATION of charge takes place EASILY in water and HENCE dielectric constant of water is extremely high. Mica is non-polar in nature and consequently, its dielectric constant is not so high.

Discussion

29.	Define radioactive decay constant and half-life and derive a relation between them.
Answer» Solution :Law of radioactive DECAY. The number of atoms disintegratedper second is directly proportional to the number of radioactive atom actually present at that instant. Decay constant i.e. `-(dN)/(dt) prop N_(0)` Decay constant is defined as the TIME after which the number of radioactive atoms reduce to 1/e times the original number ofatoms. Relation between half LIFE T and `lambda`. Half-life period The half-life of a radioactive substance is defined as the time during which half of the atoms of radioactive substancewill disintegrate. Let us denote the half-time of a substance by T. Then, by definition, after time T, number of atoms left behind will be `(N_(0))/(2)`. ...(1) Setting this CONDITION i.e. when `t=T, N=(N_(0))/(2)` in the given equation (1), we GET `(N_(0))/(2)=N_(0)e^(-lambda T)` or `e^(-lambdaT)=1/2 ` or `e^(lambdaT)=2` or `lambda T=log_(e)2=2.303 log_(10)2` `=2.303 xx 0.3010=0.693` Therefore, `T=(0.693)/(lambda)` ...(2) Thus, half life of radioactive substance is inversely proportional to its decay constant and is a characteristic property of its nucleus and cannot be altered by any known method.

Discussion

30.	A primary cell has an e.m.f, of 1.5 V, when short circuited it gives a current of 3A, then the internal resistance of the cell is
Answer» `4.5Omega` `2OMEGA` `0.5Omega` `2/5Omega` ANSWER :C

Discussion

31.	यदि बहुपदf(x)=5x^2+13x+k का एक मूल दूसरे का व्युत्क्रम हो तो k=
Answer» 0 5 `1/6` 6 Answer :B

Discussion

32.	Photoelectric effect and black body radiation tell us light has ___ from.
Answer» SOLUTION :PARTICLE

Discussion

33.	A : If the phase difference between the light waves emerging from the slits of the Young.s experiment is pi-raidan, the central fringe will be dark R : Phase difference is equal to (2pi"/"lambda) times the path difference.
Answer» Both A and R are true and R is the correct EXPLANATION of A Both A and R are true and R is not the correct explanation of A A is true and R is false Both A and R are false Answer :B

Discussion

34.	What is the refractive index of material of a plano-convex lens, if the radius of curvature of the convex surface is 10 cm and focal length of the lens is 30 cm?
Answer» Solution :ACCORDING to lens-maker.s formula, ` (1)/(F)= (mu-1) [(1)/(R_1) -(1)/(R_2)]` Heref=30cm,`R_1 = 10cmand R_2= oo` so ` (1)/(30) = (mu-1) [(1)/(10) - (1)/(oo)]` ` i.e.,3 mu -3=1 ormu =(4 //3)`

Discussion

35.	Optical fibres are based on
Answer» TOTAL INTERNAL reflection refraction scattering dispersion Answer :C

Discussion

36.	If the behaviour of light rays is as shown in the figure. The relation between refractive indices mu, mu_(1) "and" mu_(2) is
Answer» `mugtmu_(2)gtmu_(1)` `multmu_(2)ltmu_(1)` `multmu_(2), MU=mu_(1)` `mu_(2)ltmu_(1), mu=mu_(2)` SOLUTION :`mu=mu_(1), multmu_(2)`

Discussion

37.	A perfectly conducting ring of radius L is kept fixed on a horizontal surface. A vertical uniform magnetic field B_(0) exists in the region. A conducting rod (AB) of length L is hinged at the centre of the ring at A and its other end (B) touches the ring. The ring and the end A of the rod are connected to an external circuit having resistance R and capacitance C. The rod is made to rotate at a constant angular speed omega_(0). Neglect friction and self inductance of the circuit. (i) Find work done by the external agent in rotating the rod by the time the capacitor acquires a charge q_(0). (ii) Find heat generated in resistance R by the time the capacitor acquires a charge q_(0).
Answer» ANSWER :(i) `(1)/(2) q_(0)B_(0) omega_(0)L^(2)` (ii) `(1)/(2) q_(0) [B_(0)omega_(0)L^(2) -(q_(0))/(C)]`

Discussion

38.	What type of mirror is to be used for getting parallel rays from a small source of light ?
Answer» ANSWER :CONCAVE

Discussion

39.	A plane electromagnetic wave of frequency v=10 MHz propagates in a poorly conducting medium with conductivity sigma=10 mS//m and permittivity epsilon=9. Find the ratio of amplitudes of conduction and displacement current densities.
Answer» Solution :CONDUCTION current density `= sigmavec(E)` Displacement current density `=(deltavec(D))/( deltat)=epsilonepsilon_(0)(deltavec(E))/( delta t ) = i omega EPSILON epsilon_(0)vec(E)` Ration of magnitudes `=( sigma)/( omega epsilon epsilon_(0))=(j_(c))/( j_(dis))=2`,on putting the values.

Discussion

40.	A plane wavefront is incident on a plane reflectingsurface at an angle of 30^(@). What angle will the reflected rays make with the reflecting surface ?
Answer» Solution :Here, `I = 30^(@)` According to the laws of reflection, `r = I = 30^(@)` :. the reflected WAVEFRONT will make an ANGLE of `30^(@)` with the plane surface. The refracted RAYS will make an angle of `(90^(@) + 30^(@)) = 120^(@)` with the plane surface.

Discussion

41.	What is Balmer Series ?
Answer» Solution :Balmer SERIES in the SPECTRA of hydrogen atom is due to jump of electron from all POSSIBLE from HIGHER orbits to second `(n = 2)` orbit.

Discussion

42.	In the above question, the equivalent resistance between the point A and B if V_AltV_B is
Answer» `10OMEGA` `20OMEGA` `30Omega` `40Omega` ANSWER :A

Discussion

43.	A coil of reactance 8 Omega and resistance 6 Omega is connected in DC circuit then the effective resistance of the circuit …..Omega.
Answer» 14 8 6 `24/7` Solution :Here magnitude of INDUCTIVE REACTANCE `=X_L=omega_L=0` `implies` For DC, effective RESISTANCE of given circuit, `sqrt(R^2+X_L^2)=sqrtR^2(becauseX_L=0)` =R

Discussion

44.	The critical angle for total internal reflection from a medium to vacuum is 30^@ . The velocity of light. In the medium is
Answer» `6xx10^B MS^(-1)` `2xx10^B ms^(-1)` `3xx10^B ms^(-1)` `1.5xx10^8 ms^(-1)` ANSWER :A

Discussion

45.	What is the story of 'Svayamvara' about?
Answer» A WOMAN's freedom A FOOLISH man Both a and b None of the above Answer :A

Discussion

46.	The equation of a transverse wave travelling on a rope is given by y = 10 sin pi (0.01x - 2.00t) where y and x are in cm and t in seconds. The maximum transverse speed of a particle in the rope is about
Answer» 62.8 CM/s 75 cm/s 100 cm/ s 121 cm/s Answer :A

Discussion

47.	The material of a permanent magnet has
Answer» high retentivity, LOW COERCIVITY low retentivity, low coercivity low retentivity , low coercivity high retentivity, high coercivity ANSWER :C

Discussion

48.	ParticlesP and Q of mass 20 gand 40grespectivley are simultaenously projectedfrompoints A andB on the ground . Theinitial velocites of P andQ make 45^(@) and 135^(@) angles respectively withthe horizontal ABas sownin the Fig. Each particlehasan initial speed of 49m//s . TheseparationAB is 245 m. Bothparticletravlein the samevertical planeand undergo a collision. Aftercollision P retraces its path Determine the positon of Q when it hitsthe ground. How much time after the collision does the particle Q taketo reach the ground? (Takeg = 9.8 m//s^(2))
Answer» SOLUTION :`61.25m , 3.565 s`

Discussion

49.	Doppler shift for the light of wavelength 6000 Å emitted from the sun is 0.04 Å.If radius of the sun is 7 xx 10^(8) m then time period of rotation of the sun will be :
Answer» 30 dyas 365 dyas 24 hour 25 days Solution :Doppler.s shift is given by `d lambda = (v)/(c) lambda = R.OMEGA ((lambda)/(c))` `omega = (2pi)/(T)` `THEREFORE R((2pi)/(T)).(lambda)/(c) = d lambda` `T = (R . 2 PI)/(d lambda).(lambda/c) = 25 days`.

Discussion

50.	A steel plate of thickness h has the shape of a square whose side equals l, with h lt lt l. The plate is rigidly fixed to a vertical axle OO which is rotated with a constant angular acceleration beta (figure). Find the deflection lambda, assuming the sagging to be small.
Answer» Solution :The deflection of the plate can be noticed by going to a co-rotating frame. In this frame each element of the plate experiences a pseudo force proportional to its mass. These forces have a moment which CONSTITUTES the bending moment of the problem. To calculate this moment we note that the acceleration of an element at a distance `xi` from the axis is `a=xibeta` and the moment of the force exerted by the section between x and l is `N=rholhbetaunderset(x)overset(l)intxi^2dxi=1/3rholhbeta(l^3-x^3)`. From the fundamental equation `EI(d^2y)/(dx^2)=1/3rholhbeta(l^3-x^3)`. The moment of inertia `I=underset(-h//2)overset(+h//2)intz^2ldz=(lh^3)/(12)`. Note that the natural surface (i.e. the surface which contains lines which are neither stretched nor compressed) is a vertical PLANE here and z is perpendicular to it. `(d^2y)/(dx^2)=(4rhobeta)/(Eh^2)(l^3-x^3)`. Integrating `(dy)/(dx)=(4rhobeta)/(Eh^2)(l^3x-x^4/4)+c_1` Since `(dy)/(dx)=0`, for `x=0`, `c_1=0`. Integrating again, `y=(4rhobeta)/(Eh^2)((l^3x^2)/(2)-(x^5)/(20))+c_2` `c_2=0` because `y=0` for `x=0` THUS `lambda=y(x=l)=(9rhobetal^5)/(5Eh^2)`

Discussion

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In the following question a statement of assertion (A) is followed by a statement of reason (R ) A : Electric field is discontinuous across the surface of a charged conductor. R : Electric potential is discontinuous across the surface of a charged conductor .

Why current is not formed in solid conductors in absence of electric field ?

A galvanometer coil has a resistance of 15 Omegaand the metre shows full scale deflection for a current of 4 mA. How will you convert the metre into an ammeter of range 0 to 6 A?

What is dimension of [L]?

AU-tube is partly filled with a liquid A. Another liquid B, which does not mix with A, is poured into one side until it stands a height h above the level of A on the other side, which has mean while risen a heigh l. The density of B relative to that of A is :

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If in the infinite series circuit, C = 9 mu F and C1 = 6 mu F then the capacity across AB is

Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).

Define radioactive decay constant and half-life and derive a relation between them.

A primary cell has an e.m.f, of 1.5 V, when short circuited it gives a current of 3A, then the internal resistance of the cell is

यदि बहुपदf(x)=5x^2+13x+k का एक मूल दूसरे का व्युत्क्रम हो तो k=

Photoelectric effect and black body radiation tell us light has ___ from.

A : If the phase difference between the light waves emerging from the slits of the Young.s experiment is pi-raidan, the central fringe will be dark R : Phase difference is equal to (2pi"/"lambda) times the path difference.

What is the refractive index of material of a plano-convex lens, if the radius of curvature of the convex surface is 10 cm and focal length of the lens is 30 cm?

Optical fibres are based on

If the behaviour of light rays is as shown in the figure. The relation between refractive indices mu, mu_(1) "and" mu_(2) is

What type of mirror is to be used for getting parallel rays from a small source of light ?

A plane electromagnetic wave of frequency v=10 MHz propagates in a poorly conducting medium with conductivity sigma=10 mS//m and permittivity epsilon=9. Find the ratio of amplitudes of conduction and displacement current densities.

A plane wavefront is incident on a plane reflectingsurface at an angle of 30^(@). What angle will the reflected rays make with the reflecting surface ?

What is Balmer Series ?

In the above question, the equivalent resistance between the point A and B if V_AltV_B is

A coil of reactance 8 Omega and resistance 6 Omega is connected in DC circuit then the effective resistance of the circuit …..Omega.

The critical angle for total internal reflection from a medium to vacuum is 30^@ . The velocity of light. In the medium is

What is the story of 'Svayamvara' about?

The equation of a transverse wave travelling on a rope is given by y = 10 sin pi (0.01x - 2.00t) where y and x are in cm and t in seconds. The maximum transverse speed of a particle in the rope is about

The material of a permanent magnet has

Doppler shift for the light of wavelength 6000 Å emitted from the sun is 0.04 Å.If radius of the sun is 7 xx 10^(8) m then time period of rotation of the sun will be :

A steel plate of thickness h has the shape of a square whose side equals l, with h lt lt l. The plate is rigidly fixed to a vertical axle OO which is rotated with a constant angular acceleration beta (figure). Find the deflection lambda, assuming the sagging to be small.