45406 + Interview Questions in Physics in Class 12 Page 895 InterviewSolution

44701.	Estimate the minimum errors in determining the velocity of an electron, a proton, and a ball of mass of 1 mg if the coordinates of the particles and of the centre of the ball are known with uncertainly 1 mu m.
Answer» Solution :From the uncertainty principle (Eqn.(6.2b)) `DeltaxDelataP_(x)underset(~) gt ħ` THUS `Deltap_(x)=mDeltav_(x)underset(~) gt ( ħ)/(Deltax)` or `Deltav_(x)underset(~) gt ( ħ)/(mDelta_(x))` For an electron this means an uncertiainty in velocity of `116mm//s` if `Deltax=10^(-6)m=1 mu m` For a proton For a BALL `Deltav_(x)=1xx10^(-20)cm//s`

Discussion

44702.	(A): A FM signal is less susceptible to noise than an AM signal. (B) : In FM transmission, the message signal is in the form of frequency variations of carrier waves. During modulation process, the noise gets amplitude modulated.
Answer» Assertion and REASON are true and reason is the correct EXPLANATION of assertion Assertion and reason are true but reason is not correct explanation of assertion Assertion is true but reason is FALSE Assertion is false but Reason is true. Answer :A

Discussion

44703.	The two slits in Young's experiment have widths in a rations 100 :1 . Find the ratio of light intensity at the maxima and minima in the interference pattern .
Answer» ANSWER :`3:2`

Discussion

44704.	Two bodies of masses m and 4 m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is :
Answer» `-(4Gm)/(r )` `-(6Gm)/(r )` `-(9Gm)/(r )` `ZERO` Solution :`(GM)/(x^(2))=(4Gm)/((r-x)^(2)) rArr (1)/(x) =(2)/(r-x) "" THEREFORE r-x=2x` `x=(r )/(3)` Gravitational potential `V= -(Gm)/((r )/(3))-(4Gm)/((2r)/(3))= -(9Gm)/(r )` Correct choice : (C )

Discussion

44705.	Two concentric, coplanar circular loop of wire, with different diameter carry current in the same sense as shown in the figure. Which of the following statement(s) is //are correct?
Answer» the MAGNETIC force exerted by the outer LOOP on a short portion of the inner loop is radially outward the magnetic force exerted by the outer loop on a short portion of the inner loop is radially inward the net megnetic force exerted by the outer loop on a WHOLE inner loop is non zero and is radially outward the net magnetic force exerted by the outer loop on a whole inner loop is zero Solution :The magnetic field due to the outer loop is into the paper. The force on an element of the inner loop has the direction GIVEN by `vec(ds)xxvec(B)` where `vec(ds)` is the element length in the direction of current. Thus the force is radially outward. Since the force is symmetric i.e. radially OUTWARDS everywhere and the loops are concentric. There is no net force on the inner loop.

Discussion

44706.	If alpha, beta are the zeros of kx^2 - 2x + 3k such that alpha + beta = alpha.beta then k=?
Answer» 44199 -1/3 44230 -2/3 Answer :C

Discussion

44707.	Do these two meridians coincide anywhere ?
Answer» SOLUTION :Yes , What the two meridians COINCIDE, the DECLINATION becomes zero . Places with the PROPER tt are connected by the LINE called agonic lines.

Discussion

44708.	(a) A toroidal solenoid with an ain core has an average radlines of 0.15m, area of cross section 12 xx 10^(-4) m^(2) and 1200 turns. Obtain the self inductance of the foroid. Ignore field variation across the cross section of the foroid. (b) A second coil of 300 turns is wound closely on the toroid above. If the current in the primary coil is increased from zero to 2.0 A in 0.05 s, obruin the induced emſ in the secondary coil.
Answer» SOLUTION :(a) `B=mu_(0)n_(1)I=(mu_(0)N_(1)I)/(l)=(mu_(0)N_(1)I)/(2pi r)` TOTAL magnetic flux, `phi_(B)=N_(1)BA=(mu_(0)N_(1)^(2)IA)/(2pi r)` But `phi_(B)=LI"":. L=(mu_(0)N_(1)^(2)IA)/(2pi r)` `L=(4pi xx 10^(-7) xx 1200 xx 1200 xx 12 xx 10^(-4))/(2pi xx 0.15)H` `=2.3 xx 10^(-3)H=2.3` mH (b) `\|e\|=(d)/(dt)(phi_(2))`, where `phi_(2)` is the total magnetic flux linked with the second coil. `\|e\|=(d)/(dt)(N_(2)BA)=(d)/(dt)[N_(2)(mu_(0)N_(1)I)/(2pi r)A]` or `\|e\|=(mu_(0)N_(1)N_(2)A)/(2pi r)(dI)/(dt)` or `\|e\|=(2 xx 10^(-7) xx 1200 xx 300 xx 12 xx 10^(-4) xx 2)/(2 xx 0.15 xx 0.05)V` = 0.023V

Discussion

44709.	Answer carefully: (a) Two large conducting spheres carrying charges Q_(1)and Q_(2)are brought close to each other. Is the magnitude of electrostatic force between them exactly given by Q_(1),Q_(2)//4pi epsilon_(0)r^(2), where r is the distance between their centres? (b) If Coulomb’s law involved 1//r^(3) dependence (instead ofwould Gauss’s law be still true ? (c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point? (d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical? (e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there? (f) What meaning would you give to the capacitance of a single conductor? (g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).
Answer» Solution :Water has an unsymmetrical SPACE as compared to mica. Since, it has a permanent DIPOLE MOMENT it has a GREATER dielectric constant than mica.

Discussion

44710.	Name two elementary particles which have almost infinite life time.
Answer» SOLUTION :ELECTRON and proton have ALMOST INFINITE life time.

Discussion

44711.	Plane monochromatic light with intensity I_(0) falls normally on an opaque disc closing the first Fresnel zone for a point on the screen in front of the wave front. What would be the intensity I at the same point if (a) half of the disc along the diameter were removed, (b) half of the external half is removed along the the diameter ?
Answer» SOLUTION :`0, (l_(0))/2`

Discussion

44712.	An object approaches a convergent lens from the left of the lens with a uniform speed 5 m/s and stops at the focus. The image
Answer» MOVES AWAY from the lens with an uniform SPEED 5 m`//`s. moves away from the lens with an uniform acclcration. moves away from the lens with a non uniform acceleration, moves towards the lens with a non-uniform acceleration. Solution :When any OBJECT moves towards convergin lens (convex lens or positive lens) with constant speed, its image will move away froi the lens with non-uniform acceleration becaus for a lens, relation between object distance an image distance is not linear.

Discussion

44713.	A body is projected at an angle 30^(@) to the horizontal with a speed of 30 ms^(-1) The angle made by the velocity vector with the horizontal after 1.5 s is (g=10 ms^(-2))
Answer» ZERO `60^(@)` `45^(@)` `90^(@)` ANSWER :A

Discussion

44714.	Consider the arrangement shown in figure. By some mechanism , the separation between the slits S_(3) slitS_(4)can be changed . The intensity is measured at the point P which is at the common perpendicularbetween of S_(1),S_(2) Whenz = (Dlambda)/(2d) the intensity measured at P is l . Find the intensity when a is equalto 1 (Dlambda)/d 2 . (3dlambda)/(2d) (3) (2Dlambda)/d
Answer» O,I,2I I,O,2I 2I,O,I I,2I,3I Answer :A

Discussion

44715.	Considering the earth as ametallic sphere its capacitance would be nearly ...... mu F . (R = 6400 k.m epsilon_(0) = 8.85 xx 10^(-12) SI unit)
Answer» `70 MU F ` `7.0 xx10^(4) mu F ` `7.0 xx10^(3) mu F ` `700 mu F` Solution :` C = 4PI in_(0)R ` ` = 4xx3.14 xx8.85 xx10^(-12)xx6.4` `= 711xx10^(-6)=700xx10^(6) mu F `

Discussion

44716.	यदि (a,4) समीकरण 3x+y=10 के आलेख पर स्थित हो तो, a का
Answer» 3 1 2 4 Answer :C

Discussion

44717.	How we define magnetic lines of force ?
Answer» SOLUTION :It is DEFINED as the path followed by a unit NORTH pole in a magnetic FIELD.

Discussion

44718.	If the efficiency of an electric bulb of 1 watt is 10% what is the number of photons emitted by it in one second ?(The wavelength of light emitted by it is 500 nm.)(h=6.625xx10^(-34)J.s).
Answer» Solution :As the bulb is of 1 W,if its efficienct is 100% ,it may emit 1 J radiant energy in 1s.But here the efficiency is 10% ,HENCE it emits `(1)/(10)J=10^(-1)J` adiatn energy in 1s. Note:The efficiency of bulb is 10% .It MEANS it emits 10% of energy consumed in form of light and remaining 90% is WASTED in form of heat energy (due to the resistance of filamet.) `therefore` Radiant energy otained from the bulb in 1s=`10^(-1)`J If it consists of n PHOTONS, `nhf=10^(-1)J` `therefore n=(10^(-1))/(hf)` `=(0.1)/(6.625xx10^(-34)xx(c)/(lambda))` `=(lambdaxx1-^(-1))/(6.625xx10^(-34)xx3xx10^(8))(because f=(c)/(lambda))` `therefore n=(0.1xx500xx10^(-9))/(6.625xx10^(-34)xx3xx10^(8))` (`because` velocity of light ,`c=3xx10^(8)ms^(-1))` `therefore n=2.53xx10^(17)` photons.

Discussion

44719.	The time lag between two particles vibrating in a progressive wave seperated by a distance 20m is 0.02s. The wave velocity if the frequency of the wave is 500Hz, is
Answer» `1000ms^(-1)` `500ms^(-1)` `2000ms^(-1)` `250 MS^(-1)` ANSWER :A

Discussion

44720.	All surfaces are smooth. Find the horizontal displacements of the block and wedge when the block slides down from top to bottom.
Answer» Solution :When the block SLIDES down on the smooth wedge, the wedge moves backwards. In the horizontal direction there is no external FORCE. `vecF_(x)=0` , `therefore vecP_(x)` = constant `vecP_(f) = vecP_(i)` (along x-axis) `mvecu + MvecV =0` `x_(1)` = FORWARD DISTANCE moved by the block along x-axis. `x_2` = backward distance moved by the wedge along x-axis. `mvecu=-MvecV, m(x_(1)/t) =Mx_(2)/t` `mx_(1) =Mx_(2).x_(1) =(ML)/(M+m) =(Ml cos theta)/(M+m)` `x_(2) =(mL)/(M+m) =(mlcos theta)/(M+m)` L can also be written as `l cos theta`

Discussion

44721.	When the displacement in SHM is 0.40 times the amplitude x_(m), what fraction of the total energy is (a) kinetic energy and (b) potential energy? (c ) At what displacement, in terms of the amplitude, is the energy of the system half kinetic energy and half potential energy?
Answer» SOLUTION :(a) 0.84, (B) 0.16, (C ) `X= x_(m)//SQRT2`

Discussion

44722.	A sample of a radioactive element has a mass of 10 g at an instant t=0. The approximate mass of this element in the sample left after two mean lives is
Answer» 1.35 G 2.50 g 3.70 g 6.30 g Solution :Here, t=2 `tau=2xx1.44 T_(1//2)=2.88 T_(1//2)` As `N/N_0=m/m_0=(1/2)^(t//T_(1//2))=(1/2)^(2.88)` `THEREFORE m=m_0(1/2)^(2.88)=10(1/2)^2.88 ` g =10 X 0.1358 g = 1.358 g

Discussion

44723.	A point source kept at a distance of 1000 m has a illumination I. To change the illumination to 16I, the new distance should become
Answer» 250 m 500m 750 m 800 m Solution :`(I_(2))/(I_(1)) = ((r_(1))/(r_(2)))^(2)` `(161_(1))/(I_(1))=(1000)^(2)/(r_(2)^(2)),r_(2)=(1000)/(4)=250 m`

Discussion

44724.	Temperature coefficient of resistance and resistivity of a potentiometer wire must be
Answer» HIGH and LOW low and high low and low high and high ANSWER :B

Discussion

44725.	A parallel plate capacitor of plate area A has a charge q. The force on each plate of capacitor is :
Answer» `(q^(2))/(in_(0)A)` `(2Q^(2))/(in_(0)A)` `(q^(2))/(2in_(0)A)` ZERO. Answer :C

Discussion

44726.	Two conductors arc made of the same material and have the same length. Conductor A is a solid wire of diameter 1 mm. Conductor B is a holJow tube of outer diameter 2 mm and inner diameter lrrun. Find the ratio of resistanceR_(A)" to " R_(B).
Answer» SOLUTION :Resistance of A, `R_(A) =(rho l )/(A) = (rho l)/(PI r^(2))` `therefore R_(A)= (rho l )/(pi xx ((1 xx 10^(-3))/(2) )^(2) ) =(4 rho l )/(pi xx 10^(-6)) "" `... (1) Resistance of B, `R_(B) = (rho l )/(pi [ ((2 xx 10^(-3))/(2) ) - ((1 xx 10^(-3))/(2) )^(2) ] )` `= (4 rho l)/( pi [ 4 xx 10^(-6) - 1 xx 10^(-6) ] ) ` `therefore R_(B) = (4 rho l)/(pi xx 3 xx 10^(-6))` `rArr(R_(A))/(R_(B)) = (4 rho l)/(pi xx 10^(-6)) xx (pi xx 3 xx 10^(-6))/(4 rho l)` `therefore (R_(A))/(R_(B)) = (3)/(1)` `therefore R_(A) : R_(B) = 3 : 1 `

Discussion

44727.	A small coin is placed on a stationery horizontal disc at a distance r from its centre. The disc starts rotating about a fixed vertical axis through its centre with a constant angular acceleration alpha. Determine the number of revolutions N, accomplished by the disc before the coin starts slipping on the disc. Find the coefficient of static friction between the coin and the disc is mu_(9).
Answer» Solution :we draw the `FBD` of the coin `f_(T)=` tangential component of frictional force on coin `f_(R )=` RADIAL component of frictional force on coin `s` `f_(T)=malphar` `f_(r )=momega_(f)^(2)r` `SQRT(f_(R )^(2)+f_(T)^(2)) LE mu_(g)mg` `2piN=(omega_(f)^(2))/(2alpha)` Solving the equations, we get `N=(((mu_(g)g)/(r )-alpha^(2))^(1//2))/(4pialpha)`

Discussion

44728.	In the following figure, the p.d. between the points M and N is balanced at 50 cms length. The length in cms, balancing for the p.d. between points N and C will be – (##MOT_CON_NEET_PHY_C22_E01_034_Q01.png" width="80%">
Answer» 40 100 75 25 Answer :B

Discussion

44729.	How p-n junction diode is formed ? And explain depletion layer and barrier potential.
Answer» Solution :When in one REGION of Si wafer acceptor impurity (Al) is added the p-type semiconductor and another region donor impurity (As) is added the n-type semiconductor is form. Thus, p-n junction is formed and there is metallic junction between p and n region. Important processes occur during the formation of a p-n junction. Thus, is diffusion and drift. The number of electrons or holes per unit volume is called number density of electron or holes. During formation of p-n junction difference in number densities of p and n semiconductor diffusion of hole takes place from p-semiconductor to n-semiconductor `(p to n)` and diffusion of electron takes place from n-semiconductor to p-semiconductor `(n to p)`. This motion of charge carriers gives rise to diffusion current across the junction. When an electron DIFFUSES from `n to p` it leaves behind an ionised donor on n-side. This ionised donor (positive charge) is immobile as it is bonded to thesurrounding atoms. As the electrons continue to diffuse from `n to p`, a layer of positive charge (or positive SPACE charge region) on n-side of the junction is DEVELOPED. Similarly, when a hole diffuses from `p to n` due to the concentration gradient, it leaves behind an ionised acceptor (negative charge) which is immobile. These ionized charges are immobile means they cannot move. The space-charge region on EITHER side of the junction together is known as deplection region or depletion layer. The n-region of depletion layer does not have majority charge carrier electron and in p-region does not have charge carrier holes. These regions are empty in terms of their majority charge carriers. The thickness of region is of the order of one-tenth of a micrometer in p-n junction. (Means approximately 0.5 `mum`). Due to the positive space-charge region on n-sideof the junction and negative space charge region on p-side of the junction an electric field directed from positive charge towards negative charge develops. Due to this field, an electron on p-side of the junction moves to n-side and a hole on n-side of the junction moves to p-side. The motion of charge carriers due to the electric field is called drift. Initially, diffusion current is large and drift current is small. As the diffusion process continues, the space charge regions on either side of the junction extend thus increasing the electric field strength and hence drift current. This process continues until the diffusion current equals the drift current. Thus a p-n junction is formed. Since p-n junction has two connection terminals, it is called p-n junction diode. In p-n junction under equilibrium there is no net current.

Discussion

44730.	Who many type of solids are there ? What are they ?
Answer» Solution :Amorphous Ex. RUBBER, Plastic, Paraffin Wax, CEMENT etc.Crystalline Ex. QUARTZ, Calcite, mica, DIAMOND etc.

Discussion

44731.	What are matter waves? Derive an expression for the de Broglie wave length.
Answer» SOLUTION :The wave associated with material PARTICLES in inotion are CALLED inatter waves or de-Broglie waves. de – Broglie wave length of a particie is given by `lamda = (h)/(P) =(h)/(MV)` Where h`to` plank.s constant. m `to` mass of the PARTICLE v` to` velocity of the particle.

Discussion

44732.	(A) : Increasing the current sensitivity of an MCG may not necessarily increase the voltage sensitivity. (R) : Increasing the number of turns of an MCG, increases the resistance of the coil of MCG.
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 'A' is false and 'R' is false Answer :A

Discussion

44733.	A block takes n times as much time to slide down a rough incline of 45^(@) as it takes to slide down a perfectly smooth 45^(@) incline. Coefficient of kinetic friction is :
Answer» `(1)/(1-n^(2))` `1-(1)/(n^(2))` `(1)/sqrt(1-n^(2))` `sqrt(1-(1)/(n^(2)))` Solution :TIME taken to slide smooth inclined plane of length `l=(1)/(2)gsintheta.t^(2)impliesl=(1)/(2)g(1)/(sqrt2)t6(2)..(i)` Time taken to slide ROUGH inclined plane is `l=(1)/(2)g(sintheta-mucostheta)t^(2)` `l=(1)/(2)g(1)/(sqrt2)-muxx(1)/(sqrt2)t^(2)...(2)` From (1) and (2), `(1)/(2)g((1)/(2))((1)/((sqrt(2)-muxx(1))/(sqrt(2))))t^(2)=(1)/(2)gxx(1)/(sqrt(2))t^(2)` As t. = nt ` implies(1 - mu) n^(2).t^(2) = t^(2) n^(2) - n^(2) mu = 1` `n^(2)mu=n^(2)-1` `mu=1-(1)/(n^(2))` HENCE CORRECT choice is (b).

Discussion

44734.	An alternating voltage, of angular frequency omega is induced in electric circuit consistin of inductance L and capacitance C, connected in parallel. Then across the inductance coil
Answer» current is maximum when `OMEGA^(2)=(1)/(LC)` current is minimumwhen `omega^(2)=(1)/(LC)` voltageis MINIMUM when `omega^(2)=(1)/(LC)` voltageis maximum when `omega^(2)=(1)/(LC)` Answer :BD

Discussion

44735.	Raed the following passage and and answer the following questions. Stefan's law states that the ernergy emitted per second per unit area of the perfect black body is directly proportional to fourth power of its absolute temperature i.e. EpropT^(4) or E=sigmaT^(4) where sigma is called Stefan's constant. If the body is placed in enclosure maintained at temperature T_(0) then E=sigma(T^(4)-T_(0)^(4)). If temperature difference between the body and the surroundings is very small then energy emitted per second per unit area of the body and surrounding i.e Eprop(T-T_(0)). This is known as Newton's Law of cooling and is a special case of Stefan's Law.Two bodies P and Q are having equal surface areas and are maintained at temperature 17^(@)C and 27^(@)C respectively. Then radiant energy emitted per second are in the ratio of :
Answer» `16:81` `8:27` `2:3` `7:8` SOLUTION :`(E_(1))/(E_(2))+(sigmaT_(1)^(4))/(sigmaT_(2)^(4))=((290)/(300))^(4)= 87=7/8` So correct CHOICE is (d).

Discussion

44736.	The power of a AM transmitter is 100 W. If the modulation index is 0.5 and the transmission is having single side band, the percentage of useful
Answer» 1.1 W 11 W 2.2 W 22 W Answer :B

Discussion

44737.	Photons of wavelength lambda are incident on a metal. The most energetic electrons ejected from the metal are bent into a circular arc of radius R by a perpendicular magnetic field having magnitude B. The work function of the metal is……………………
Answer» `(hc)/(lambda) - me_(e) + (e^(2) B^(2) R^(2))/(2m_(e))` `(hc)/(lambda) + 2m_(e) [(eBR)/(2m_(e))]^(2)` `(hc)/(lambda) - m_(e)c^(2) - (e^(2) B^(2) R^(2))/(2m_(e))` `(hc)/(lambda) - 2m_(e) [(eBR)/(2m_(e))]^(2)` Solution :Magnetic lorentz FORCE = Centripetal force `BqV = (mv^(2))/(R)` `therefore""V = (qBR)/(m) = (eBR)/(m)` From Einstein.s photo electric equation `KE_(max) = (hc)/(lambda) - phi` `phi = (hc)/(lambda) - (1)/(2) m_(e) V^(2) = (hc)/(lambda) - (1)/(2) m_(e) ((eBR)/(m_(e)))^(2)` `phi = (hc)/(lambda) - 2m_(e) ((eBR)/(2m_(e)))^(2)`

Discussion

44738.	If E_0 and mu_0 respectively denote the permittivity constant and permeability constant ,then which of the following expression is dimensionally correct ?Here'C' is velocity of light in vacuum, then C=
Answer» `SQRT(mu_0E_0)` `1/sqrt(mu_0E_0` `sqrt(mu_0/E_0` `sqrt(E_0/mu_0` ANSWER :B

Discussion

44739.	The electric potential at a certain distance from a point charge is 600 V and the electric field is 200 NC^(-1). Which of the following statements will be true?
Answer» The work done in moving a POINT charge of `1muC` from the given point to a point at a distance of 9 m will be `4 xx 10^(-4)J`. The distance of the given point from the charge is 3 m. The potential at a distance of 9 m will be 200 V. The magnitude of charge is `0.2 xx 10^(-3)`C. SOLUTION :a.,b.,c. `V=(1)/(4piepsilon_(0))(q)/(r)=600V` And `E=(1)/(4epsilon_(0))(q)/(r^(2))=200 NC^(-1)` Hence, `(V)/(E)=r=(600)/(200)=3m` Now, `(q)/(4piepsilon_(0)r)=600,(q)/(4piepsilon_(0)r))=V^(-1)'` `THEREFORE (v')/(600)=(r)/(r')=(3)/(9)` or `V'=200V` `W=q(V-V')=10^(-6)(600-200)` `=4xx10^(-4)J`

Discussion

44740.	A proton accelerated through a potential difference of 100V, has de-Broglie wavelength lamda_0 . The de-Broglie wavelength of an alpha-particle, accelerated through 800V is
Answer» `(lamda_0)/(sqrt2)` `(lamda_0)/(2)` `(lamda_0)/(4)` `(lamda_0)/(8)` SOLUTION :`lamda_p/lamda_alpha = sqrt( (m_alpha q_alpha V_alpha)/(m_p q_p V_p) ) = sqrt(( (4m_p) (2e) XX (800) )/( (m_p) xx (e ) xx (100) ) ) = 8` `rArr lamda_alpha = (lamda_p)/(8) = (lamda_0)/(8)`

Discussion

44741.	A rod is hinged vertically at one end and is forced to oscillate in a vertical plane with hinged end at the top, the motion of the rod:
Answer» is SIMPLE HARMONIC is OSCILLATORY but not simple harmonic is pericolic but not oscillatory MAY be simple harmonic ANSWER :D

Discussion

44742.	A hollow cylinder has a charges q coulomb within it. If phiis the electric fluxin units of V-m associated with the curved surface B, the flux linked with the plane surface A in units of V-m will be
Answer» `(q)/(2in _0) ` `(phi)/(3) ` ` (q)/( in_0) -phi` ` (1)/(2) [( q)/(in _0) -phi ]` Solution :TOTAL FLUX over entire cylinder `=(q)/(in_0) ` From symmetry `phi_A =phi_C and phi_B= phi ` ` therefore ""phi_A+phi_B+phi_C=phi_A+phi +phi__A=phi_A=(q)/(in_0) RARR phi_A=(1)/(2)[(q)/(in_0)-phi]`

Discussion

44743.	The power factor of a series L-C-R circuit is unity if it is operated at a frequency of :
Answer» `(2PI)/(LC)` `(LC)/(omega)` `1/(2pisqrtLC)` `1/(omega^2LC)` ANSWER :C

Discussion

44744.	Tuning fork A of frequency 258 Hz gives 8 beats with a tuning fork B. When the tuning fork A is filed and again A and B are sounded the number of beats heard decreases. The frequency of B is
Answer» 250 HZ 266 Hz 258 Hz 242 Hz Answer :B

Discussion

44745.	From the output characteristics shown in Figure , calculate the values of beta_(ac)and beta_(dc) of the transistor when V_(CE) is 10 V and I_(C) = 4.0 mA .
Answer» Solution :`beta_(ac) = ((Delta I_(C))/(DeltaI_(B)))_(V_(CE)) , beta = ((I_(C))/(I_(B)))` For DETERMINING `beta_(ac)` and `beta_(dc)`at the stated values of `V_(CE)` and `I_(C)` consider any TWO charateristics for two values of `I_(B)` which lie above and below the given value of `I_(C)` . Here `I_(C) = 4.0` mA . (Choose characteristics for `I_(B) = 30` and `20 mu A`) . At `V_(CE) = 10 V` , we read the two values of `I_(C)` from the graph . Then , `Delta I_(B) = (30 - 20) mu A = 10 mu A , Delta I_(C) = (4.5 -3.0) mA = 1.5 mA` THEREFORE , `beta_(ac) = (1.5 mA)/(10 muA) = 150` For determining `beta_(ac)`and `beta_(dc)` at the stated values of `V_(CE)` and `I_(C)` consider any two characteristics for two values of `I_(B)` which lie above and below the given value of `I_(C)`. Here `I_(C) = 4.0` mA . (Choose characteristic for `I_(B) = 30` aove `20 mu A`) At `V_(CE) = 10 V` , we read the two values of `I_(C)` from the graph . Then , `Delta I_(B) = (30 - 20) mu A = 10 mu A , Delta I_(C) = (4.5 - 3.0) mA = 1.5 mA` Therefore , `beta_(ac) = (1.5 mA)/(10 mu A) = 150` For determining `beta_(dc)` . either estimate the value of `I_(B)` corresponding to `I_(C) = 4.0 mA` at `V_(CE)= 10 V` or calculate the two values of `beta_(dc)` for the two characteristicschosen and find their mean. Therefore , for `I_(C) = 4.5` mA and `I_(B) = 30 mu A` . `beta_(dc) = (4.5 mA)/(30 mu A) = 150` and for `I_(C) = 3.0` mA and `I_(B) = 20 mu A` `beta_(dc) = (3.0 mA)/(20 mu A) = 150 ` Hence , `beta_(dc) = ((150 + 150))/(2)= 150`

Discussion

44746.	Two conducting spheres of radius r_1 and r_2 are equally charged. The ratio of their intensities will be :
Answer» `r_2/r_1` `r_1/r_2` `(r_1/r_2)^2` `(r_2/r_1)^2` ANSWER :D

Discussion

44747.	In which telescope the final image is erect ?
Answer» SOLUTION :In TERRESTRIAL telescope and Galilean telescope the FINAL imageis ERECT.

Discussion

44748.	Which inpurity is doped in Si to form N-type semicondutor ?
Answer» A B As None of these Solution :Because AS in PENTAVALENT IMPURITY.

Discussion

44749.	Identify the highest energy photons of the following
Answer» Industrial strength microwave cookers HIGH power AM radio TRANSMITTERS that cover THOUSANDS of square MILES Waves from atomic electron TRANSITIONS among outer orbits Waves from atomic transitions to inner orbits Answer :D

Discussion

44750.	The circuit diagram shown in fig consists of a very large (infinite) number of elements . The resistances of the resistors in each subsequent element differ by a factor k from the resistance of the resistors in the previous element. Determine the resistance R_(AB) between points A and B if the resistances of the first element are R_1 and R_2
Answer» Solution :From symmetry considerations , we can remove the first element from the circuit, the resistance of the remaining circuit between point C and D will be `R_(CD)= kR_(AB)` . Therefore , the EQUIVALENT circuit of the INFINITE chain will have the form shown in figure. Thus `R_(AB) = R_1 = (R_2(kR_(AB)))/(R_2 + (kR_(AB)))` `R_(AB_[R_2 + KR_(AB))] = R_1 [R_2 + kR_(AB)] + kR_2 R_(AB)` `R_2 R_(AB) + kR_(AB)^2 = R_1 R_2 + kR_1 R_(AB) + kR_2 R_(AB)` or `kR_(AB)^2 + (R_2 + kR_1 - kR_2)R_(AB) - R_1 R_2 = 0` `therefore R_(AB) = (-(R_2 - kR_1 - kR_2)PM sqrt((R_2 - kR_1 - kR_2)^2 + 4kR_R_2))/(2k)` As resistance cannot be negative `R_(AB) = (kR_1 + kR_2 - R_2+ sqrt((R_2 - kR_1 - kR_2)^2 + 4kR_1 R_2))/(2k)`

Discussion

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