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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 45001. |
Find the position of the image formed by the lens combination given in the figure. |
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Answer» SOLUTION :Image FORMED by the fist lens `(1)/(v_(1)) -(1)/(u_(1)) = (1)/(f_(1))` `(1)/(v_(1)) - (1)/(-30) = (1)/(10) or v_(1) = 15 cm ` The image formed by the first lens serves as the object for the second. This is at a DISTANCE of (15 -5) cm = 10 cm to the right of the second lens. though the image is real, it serves as a virtual object for the second lens, Which means that the rays appear to come from it for the second lens. `(1)/(v_(2)) - (1)/(10) = (1)/(-10) or v_(2) = infty` The virtual image is formed at an infinite distance to the left of the second lens. This ACTS as an object for the THIRD lens. `(1)/(v_(3)) - (1)/(u_(3)) = (1)/(f_(3))"" or "" (1)/(v_(3)) = (1)/(infty) + (1)/(30) "or" v_(3) = 30` cm The final image is formed 30 cm to the right of the third lens. |
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| 45002. |
Two identical charged spheres are suspended in air from a ceiling using strings of equal length. The point of suspension is common to both the spheres. Neglect the density of air. The density of material of spheres is 2 "g/cm"^3.Both spheres repel each other and the strings are inclined at a certain angle with the vertical when the system is in equilibrium. Now both the sphere are dipped into a liquid of specific gravity 1 and dielectric constant K, but angle that the strings are making with the vertical remains unchanged. Let T_1 be the tension in both the strings when the spheres are in air and T_2, be the tension in strings when spores are dipped in liquid. Let F_1 be the electrostatic force between charges when they were kept in air and F_2, be the electrostatic force when spheres are dipped in liquid. The magnitude of T_2//T_1 is |
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Answer» 1 |
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| 45003. |
A string 25 cm long and having a mass of 2.5g is under tension. A pipe closed at one end is 40 cm long. When the string is set vibrating in its firsy overtone and the air in the pipe in its fundamental frequency, 8 beats per sec are heard. It is observed that decreasing the tension in the string decreases the beat frequency. If the speed of sound in air is 320 m/s, find the tension in the string. |
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Answer» Solution :For first overtone (i.E., second harmonic) of thestring `f_(s)=40 sqrt(T)` and for FUNDAMENTAL frequency of closed organ pipe, `f_(c)=200 HZ` So `200-40 sqrt(T)=8 or 40 sqrt(T)-200=8` But decreasing the tension DECREASES the beat frequency. So `40 sqrt(T)=208 i.,e T=[(208)/(40)]^(2)=(5.2)^(2)=27N` |
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| 45004. |
A simple pendulum of length varphi oscillates with the time period T . Then which of the following graph correctly represents the relation? |
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Answer»
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| 45005. |
Two identical charged spheres are suspended in air from a ceiling using strings of equal length. The point of suspension is common to both the spheres. Neglect the density of air. The density of material of spheres is 2 "g/cm"^3.Both spheres repel each other and the strings are inclined at a certain angle with the vertical when the system is in equilibrium. Now both the sphere are dipped into a liquid of specific gravity 1 and dielectric constant K, but angle that the strings are making with the vertical remains unchanged. Let T_1 be the tension in both the strings when the spheres are in air and T_2, be the tension in strings when spores are dipped in liquid. Let F_1 be the electrostatic force between charges when they were kept in air and F_2, be the electrostatic force when spheres are dipped in liquid. The magnitude of F_2//F_1 is |
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Answer» 3 |
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| 45006. |
A person using a lens as a sample microscope sees an |
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Answer» inverted virtual image |
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| 45007. |
What is half-life of nucleus? Give the expression. |
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Answer» SOLUTION :The half - life `T_(1/2)` is DEFINED as the time REQUIRED for the number of ATOMS initiallypresent toreduce to ONE half of the initial amount. `T_(1/2) = (In 2)/(lambda) = (0.6931)/(lambda)`. |
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| 45008. |
The velocity-time graphs of a car and a scooter are shown in the figure. (i) The difference between the distance travelled by the car and the scooter in 15 s and (ii) the time at which the car will catch up with the scooter are, respectively. |
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Answer» 112.5 m and 22.5 s |
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| 45009. |
In a Young's double slit experiment the intensity at a point where the path difference is (lambda)/(6)(lambda being the wavelength of the light used) is I. If I_(0) denotes the maximum intensity then (I)/(I_(0))=… |
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Answer» `(1)/(sqrt(2))` `=I.+2I. [ :. cos 0^(@)=1]` `=4I.` and intensity at any POINT, `I=I.+I.+2sqrt(I.I) cos PHI` where `phi=(2PI)/(lambda)xx` path DIFFERENCE `=(2pi)/(lambda)xx(lambda)/(6)=(pi)/(3)` rad `:.I=2I.+2I"cos"(pi)/(3)` `=2I.+I.=3I.` `:.(I)/(I_(0))=(3I.)/(4I.)=(3)/(4)` |
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| 45010. |
Molecular mass of amonia is.............. |
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Answer» 15u |
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| 45011. |
A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in free space. It is under continuous illumination of 200 nm wavelength of light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is Axx10^(z) (where 1ltAlt10). The value of 'Z' is |
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Answer» 2 |
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| 45012. |
Name any two characteristics of light which do not change on polarisation |
| Answer» SOLUTION :SPEED and FREQUENCY | |
| 45013. |
The slopes of anode and mutual characteristics of a triode are 0.02 mA V^(-1) and 1 mA V^(-1) respectively. What is the amplification factor of the valve |
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Answer» 5 |
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| 45014. |
A thin stick of length f/5 placed along the principal axis of a concave mirror of focal length f such that its image is real and elongated just touch the stick. What is the magnification? |
Answer» Solution : One end of image concides with one end of stick i.e., either A or B is centre of curvature. Since image is elongated i.e., point B is centre ofcurvature. Image of B is formed on B itself. Image of A `u=-(2f-F//5)=(-8f)/3,f=-f,v=?` `1/v+1/u=1/f RARR 1/v+1/(-8f//5)=1/(-f)` `1/v=-1/f+5/(8f)=(-8+5)/(8f)=(-3)/(8f)rArrv=-(8f)/3`  Image length, A'B'=`(8f)/5-2f=-2f//5` Longitudinal MAGNIFICATION, `m_(L)=(-A'B')/(AB)=(2f//5)/(f//5)=2` |
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| 45015. |
Long distance radio broadcasts use short - wave bands. Why ? |
| Answer» SOLUTION :Because in SW band, radio WAVES of frequency 4.75 MHz to 9.9 MHz whose very LONG distance transmission is POSSIBLE by ionosphere. | |
| 45016. |
Condition for maximum diffraction is |
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Answer» `(LAMBDA)/(a)=1` |
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| 45017. |
Mention the three types of energy loss in a transformer. |
| Answer» Solution :(a) COPPER loss (B) MAGNETIC flux LINKAGE loss (c) EDDY current loss. | |
| 45018. |
Let us suppose that the earth had a net surface charge density of 1 electron per m^(2). What would its potential be? Also what is the field just outside the earth's surface? |
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Answer» SOLUTION :DATA supplied, `sigma=1e//m^(2) =1.6 xx 10^(-19)C//m^(2), " RADIUS of the earth , "R=6.4 xx 10^(6)m` Potential `V=(1)/(4pi epsi_(0) ) q/R=(1)/(4pi epsi_(0)). (sigma 4pi R^(2))/(R) =(sigma R)/(epsi_(0)) =(1.6 xx 10^(-19) xx 6.4 xx 10^(6))/(8.854 xx 10^(-12)) =0.1156V` Field, `E=(1)/(4pi epsi_(0)).q/R^(2)=(1)/(4pi epsi_(0)) xx (sigma 4piR^(2))/(R^(2)) =(sigma)/(epsi_(0)) =(1.6 xx 10^(-19))/(8.854 xx 10^(-12))=1.81 xx 10^(-8) N//C` |
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| 45019. |
A Neutron moving with velocity v collides with a stationary alpha particle. The velocity of the neutron after collision is |
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Answer» (A) `-3v//5` |
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| 45020. |
A uniform magnetic field B exists in a circular region of radius R. the field is perpendicular to the plane and is increasing at a constant rate of (dB)/(dt) = alpha. There is a straight conducing rod AB of length 2R. Find the emf induced in the rod when it is placed as shown in figure (a) & (b). Point C is midpoint of the rod in figure (b) |
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Answer» |
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| 45021. |
Ifv = c, the mass of the moving body, of rest mass 'm_(0)' is "_________________". |
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Answer» zero |
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| 45022. |
Monochromatic light falls on a right-angled prism at an angle of incidence 45^@. The emergent light is found to slide along the face AC. Find the refractive index of material of prism |
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Answer» Solution :Since the emergent light slides along the FACE AC, angle of emergence is `90^@`, as SHOWN. It implies that angle of INCIDENCE of the ray that falls on face AC is equal to the critical angle `theta_C`. `thereforer_2= theta_C ...(1)`  From the prism theory, we know ` r_1+r_2= A = 90^@ ` ` thereforer_2 = 90^@- r_1 ......(2)` fromequaiton(1) and (2)`,90^@ - r_1 =theta_c ` ` thereforesin( 90^@ -r_1) = sintheta_C(or)cos r_1= sintheta_C` Butsin ` theta_C = (1)/(mu )thereforecos r_1 = (1)/(mu)` Applyingsnell.slawat theboundaryAB , ` 1 xx sin45^@= mu sinr_1= mu sqrt(1- (1)/( mu^2))` ` therefore(1)/( sqrt(2)) = sqrt(mu^2 -1)or mu^2 = 3//2 = 1.5mu = sqrt(1.5)` |
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| 45023. |
Millikan's oil drop experiment established that |
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Answer» ACTS downwards |
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| 45024. |
Using Kirchhoff's rules determine the value of unknown resistance R in the circuit so that no current flows through 4 Omega resistance. Also find the potential difference between A and D. |
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Answer» Solution :Apply Kirchoof.s Voltage rule For loop ABEA `-9 + 6 + 4 xx 0 + 2 I =0` `2I -3 =0` `I = 3/2 = 1. 5 A` For loop ABCED `3 + IR + 4 xx 0-6 =0` `THEREFORE IR =3` Substituting the value of current I, `3/2 xx R =3 ` `therefore R = 2Omega` Potential DIFFERENCE between A & D through path ABCD: `+ 9V - 3 V - IR = V _(AD)` `+ 9- 3 - 3/2 xx 2 = V _(AD)` ` implies V _(AD) =3 V` Alternatively, through path AFD `3/2 xx 2 = V _(AD)` `implies V _(AD) = 3 V` |
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| 45025. |
Primitive unit Cell, atoms are present at.......... |
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Answer» CORNERS of the UNIT Cell |
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| 45026. |
STATEMENT-1 Angular width of central maximum in YDSE is independent of D i.e., distance betwen source and screen. STATEMENT-2 Fringe width of central maximum is double of the first maxima on the screen. |
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Answer» STATEMENT-1 is true Statemetnt-2 is True,Statement -2 is a CORRECT explanation for statement -1. |
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| 45027. |
Observe the following statements regarding isotones i)""^(39)K _(19) and ""^(40)Ca_(20)are isotones. ii)Nuclides having different atomic number (z) and mass number (A) but same number of neutrons (n) are called isotones iii) ""^(19)F_9 and ""^(23)Na_(11) are isotones |
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Answer» i,II and III are CORRECT |
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| 45028. |
A roundcurrent-carryinglooplies in the plane boundary between magnetic and vacumm. The permeabilityof th emagnetic is equal to mu Find themagnetic induction Bat an arbitary point on the axis of the loop if in the absence of th e magnetic the magneticinduction at thesame pointbecomesequal to B_(0) . Genralize theobtainedresult to all poinsof the field. |
Answer» Solution :The medium `I` is vacumm and contains a circular current carryingcoil withcurrent `I`. The medium `II` is a megenticwith permeabitliy`mu`. The boundaryisthe plane`z = 0` and the coil is the plane`z = 1`. To findthe magnitudeinduction we notethat theeffectof the magenticmediumcan bewrittenas DUE to an image coil in `II` as faras the medim `I` isconcerned. Onthe other hand, theinduction is `II` as far asthe medium`I` isconcerned, Onthe other hand, the induction is `II` can bewrittenas due t the coil in `I`, carryinga differentcurrent. It issufficientto considerthe far awayfieldsand ensure that the BOUNDARY conditionsare SATISFIED there. Now for actualcoil in medium`I`. `B_(R) = - (2p_(m) cos theta')/(r^(3)) . ((mu_(0))/(4pi)). B_(0) = (p_(m) sin theta')/(r^(3)) ((mu_(0))/(4pi))` so,`B_(2) = (mu_(0) P m)/(4pi) (2 cos^(2) theta' - sin^(2) theta')` and`B_(x) = (mu_(0) p_(m))/(4pi) (-3 sin theta' cos theta')` where `p_(m) = I (PI a^(2)), a` = radiusof the coil. Similarlydue to the image coil, `B_(x) = (mu_(0) p')/(4pi) (2 cos^(2) theta' - sin^(2) theta), B_(x) = (mu_(0) p')/(4pi) (3 sin theta' cos theta'), p'_(m) = I' (pi a^(2))` As fas as the medium`II` is concered, we write similarly `B_(z) = (mu_(0) p''_(m))/(4pi) (2 cos^(2) - sin^(2) theta'), B_(x) = (mu_(0) p''_(m))/(4pi) (-3 sin theta' cos theta'). p''_(m)= I''(pi a^(2))` The boundaryconditionsare, `p_(m) + p'_(m) = p"_(m)` (from `B_(1n) - B_(2n)`) `p_(m) + p'_(m) = - (1)/(mu) p"_(m)` (from`H_(1t) = H_(2t)`) Thus,`I" = (2 mu)/(mu+ 1) I, I' = (mu - 1)/(mu + 1) I` Inthe limit, whenthe coil is on the boundary, the magnetic filedenvery where can beobtained by takingthe currenttobe `(2mu)/(mu + 1) I`. Thus, `vec(B) = (2mu)/(mu + 1) vec(B_(0))` |
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| 45029. |
A cricketplayer throws the ball to havethe maximum horizontal rangeof 120m. If thethrows the ball verticallywith same velocity whatis the maximum height it can reach ? |
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Answer» |
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| 45030. |
If the resistanceof coil is 3Omega at 20^(@)C and alpha=0.004//^(@)C then determine its resistance t 100^(@)C. |
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Answer» Solution :`R_(0)=3OMEGA, T=100^(@)C, T_(0)=20^(@)C` `alpha=0.004//^(@)C, R_(T)=?` `R_(T)=R_(0)(1+alpha(T-T_(0)))` `R_(100)=3(1+0.004xx80) rArr R_(100)=3(1+0.32)` `R_(100)=3(1.32) rArr R_(100)=3.96Omega` |
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| 45031. |
Match the column - I with column -II A certain amount of He+ ions are subjected to excitation and subsequently the emission spectrum for the same is observed. Column - I shows four different series to which the spectral lines different series to which the spectral lines belong and column - II shows five different energies of the photons emitted during the process . |
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Answer» a-r b-q c-s,t d-p |
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| 45032. |
Find the ratio of intensities of the voilet and red satellites closest to the fixed line, in the vibraiton sopecturm of Raman scattering by Cl_(2) molecule at a temperature T= 300K if the natural Vibration frequency of these molecules is omega= 1.06.10^(14)s^(-1),. By what factor will this ratio change if the temperature is doubled? |
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Answer» Solution :The violet satellite arise from the transition `1 rarr 0` in the vibrational STATE of the scattering molecule while the red satellite arise from the transition `0 rarr 1`. The intensities of these two transitions are in the RAITO of initial populations of the two states i.e., in the ratio `e^(-ħomega//kT)` THUS `(I_(v))/(I_(r ))=e^(-ħ omega//kT)=0.067` If the temperature is doubled,the ratio INCREASES to `0.259`, an INCREASE of 3.9 times. |
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| 45033. |
The minimum number of cells in mixed grouping required to produce a maximum current of 1.5 A through an external resistance of 30 Omega , given the emf of each cell is 1.5 V and internal resistance is 1Omegais |
| Answer» Answer :B | |
| 45034. |
In the above question the centre of mass of plate is now shifted to which quadrant of plate for X - Y plane ? |
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Answer» First |
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| 45035. |
Figure illustrates the simplest ripple filter. A voltage V=V_(0)(1+ cos omega t) is fed to the left input. Find : (a) the output voltabe V^(')(t), (b)the magnitude of the product is eta =7.0 times less than the direct voltage component, if omega=314 s^(-1). |
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Answer» Solution :Here, `V=IR+(int_(0)^(t)Idt)/(C)` or ` Rdot(I)+(1)/(C)I=dot(V)=- omega V_(0) sin omegat ` Ignoring transients , a solution has the form `I=I_(0)sin ( omegat- ALPHA)` ` omega RI_(0)COS ( omegat- alpha)+(I_(0))/(C) sin ( omegat- alpha) =- omegaV_(0) sin omegat` `=-omegaV_(0){sin ( omegat- alpha) cos alpha + cos ( omegat- alpha ) sin alpha}` s o` RI_(0)=- V_(0) sin alpha` `(I_(0))/( omegaC)=-V_(0) cos alpha` ` alpha=pi+TAN^(-1) ( omega RC)` `I_(0)=(V_(0))/( sqrt(R^(2)+((1)/( omegaC))^(2)))` `I=I_(0) sin ( omegat- tan ^(-1)omegarc- pi)=- I_(0) sin ( omegat - tan^(-1)omegaRC)` Then`Q=int_(0)^(t) I dt=Q_(0)+(I_(o))/( omega)cos ( omegat- tan ^(-1) omega RC)` It satisfies `V_(0)(1+cos omegat)=R(dQ)/(dt)+(Q)/(C)` if `V_(0) (1+ cos omegat)=- RI_(0) sin ( omegat- tan ^(-1) omega RC)` `+ (Q_(0))/( C)+(I_(0))/( omegaC) cos ( omega t - tan ^(-1)omegaRC)` and `{:((I_(0))/(omegaC)=V_(0)//sqrt(1+( omegaRC)^(2))),(RI_(0)=(V_(0) omega RC)/(sqrt(1+ omegaRC))):}}checks` Hence`V^(')=(Q)/(C)=V_(0)+(V_(0))/( sqrt(1+( omegaRC)^(2)))cos ( omegat-alpha)` `(b)``(V_(0))/( eta)=(V_(0))/( sqrt(1+( omegaRC)^(2)))` or `eta^(2)-1= omega^(2) ( RC)^(2)` or `RC=sqrt( eta^(2)-1 )//omega = 22 ms.` 
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| 45036. |
A parallel plate capacitor C with plates of area 1 square unit and separation d is filled with a liquid of dielectric constant K=2. The level of liquid is d/3, initially. If the liquid level decreases at a constant speed v. Then, the capacity of capacitor after t second is |
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Answer» `(6epsilon_0)/(5d+3vt)` |
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| 45037. |
A conductor of length 3m moving in a uniform magnetic field of strength 100 T. It covers a distance of 70 m in 5 sec. Its plane of motion makes an angle of 30^(@) with direction of magnetic field. Calculate the emf induced in it. |
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Answer» Solution :`V= (x)/(t)= (70)/(5) =14ms ^(-1)` Component of velocity ALONG magnetic field `= V cos theta= V cos (90^(@) -30^(@))= V cos 60^(@)` `epsi =B L V cos theta ` `epsi =100 XX 3 xx 14 xx cos 60 ^(@)` `epsi =100 xx 3 xx 14 xx 1/2` ` epsi = 21000 V` |
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| 45038. |
The given arrangement carries a capacitor with capacitance 40mF and two inductors L_(1)=25H and L_(2)=100H. If the capacitor initially carries a charge of 10mC, then |
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Answer» the MAXIMUM current through the inductor `L_(1)` when key `K_(1)` is closed is 20mA |
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| 45039. |
Light falls on a prism grazing along first surface of a prism and the emergent ray is normal to the 2nd face of the prism. If D is angle of deviation then the refracting angle of the prism i |
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Answer» 90-2D |
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| 45040. |
A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength? |
| Answer» SOLUTION :E and B in x-y plane and are MUTUALLY perpendicular, 10 m. | |
| 45041. |
The threshold frequency for photo electric effect for a metal surface is found to be 4.8xx10^(16) Hz. The stopping potential required when the metal is irradiated by radiation of frequency 5.6xx10^(16) Hz is (taking h=6.6xx10^(-34)Js and c=1.6xx10^(-19)C ) |
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Answer» 22.4 V |
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| 45042. |
If 50% of energy released during fission would be converted into electrical energy, then the number of fissions in a second in a nuclear reactor of 6.4MW output is (Energy per fission is 200MeV) |
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Answer» `4XX10^(15)` |
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| 45043. |
Calculate the de Broglie wavelength of a proton whose kinetic energy is equal to 81.9 xx 10^(-15) J. (Given : mass of proton is 1836 times that of electron ) . |
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Answer» Solution :`{:("de-Broglie wavelength of proton,",,m_(p) = 1.67 xx 10^(-27)kg),(lambda = (H)/(sqrt(2 mK)),,K.E = 81.9 xx 10^(-15) J):}` `= (6.6 xx 10^(-34))/(sqrt(2 xx 1.67 xx 10^(-27) xx 81.9 xx 10^(-15))) =(6.6 xx 10^(-34))/(1.6539 xx 10^(-20)) = 3.99 xx 10^(-14)` `lambda = 4 xx 10^(-14) m` |
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| 45044. |
Two bodies A and B of masses m and 2mrespectively are placed on a smooth floor. They are connected by a light spring of stiffness k.A third body C of mass m moves with velocity v_(0) along the line joining A and B and collides elastically with A. If l_(o) be the natural length of the spring then find the minimum separation between the blocks. |
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Answer» Solution :Initially there will be collision between C and A which is elastic, therefore by using conservation of momentum we obtain, `mv_(0)=mv_(A)+mv_(c ), "" v_(0)=v_(A)+v_(C )` Since `e=1, v_(0)=v_(A)-v_(C )` Solving the above two EQUATIONS, `v_(A)=v_(0)` and `v_(c )=0`. Now A will move and compress the spring which in turn accelerate B and retard A and finally.both A and B will move with same VELOCITY v. (a) Since net EXTERNAL force is zero, therefore momentum of the system (A and B) is conserved. `rArr v=v_(0)//3` (b)If `x_(0)` is the maximum compression, then using energy conservation `(1)/(2)mv_(0)^(2)=(1)/(2)(m+2m)v^(2)+(1)/(2)kx_(0)^(2)` `rArr (1)/(2)mv_(0)^(2)=(1)/(2)(3m)(v_(0)^(2))/(9)+(1)/(2)kx_(0)^(2)` `rArr x_(0)=v_(0)sqrt((2m)/(3k))` Hence minimum distance `D=l_(0)=x_(0)==l_(0)-v_(0)sqrt((2m)/(3k))` |
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| 45045. |
A conductor AB of length 10 cm at a distance of 10 cm from an infinitely long parallel conductor carrying a current 10A. What work must be done to move AB to a distance of 20cm if it carries 5A? |
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Answer» Solution :FORCE on a conductor at a distance X is `F=(mu_(0)i_(1)i_(2)l)/(2pix)` WONE done to displace it through a small distance `dx=dW=vecF.vec(dx)` `dW=(mu_(0)i_(1)i_(2)l)/(2pix)dx` `W=int_(0.1)^(0.2)(mu_(0)i_(1)i_(2)l)/(2pix)dx` `W=(mu_(0)i_(1)i_(2)l)/(2PI)[log_(E)x]_(0.1)^(0.2)` `W=(3pixx10^(-7)xx10xx5xx10xx10^(-2))/(2pi)log_(e)2` `W=0.693xx10^(-6)J` |
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| 45046. |
The horizontal component of earth's magnetic field is 3 xx 10^(-4) T. The magnetic dip angle is 45°. Find the vertical component. |
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Answer» `sqrt(3) xx 10^(-4)` T `tan phi = (B_v)/( B_h) ""[phi = 45^(@) , B_(h) = 3 xx 10^(-4) ]` `therefore B_(V) = tan phixx B_(h) ` `= tan 45^(@) xx 3 xx10^(-4)` `= 1 xx 3 xx 10^(-4) ""[tan 45^(@) =1]` `= 3 xx 10^(-4)` T |
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| 45047. |
An engine of one metric ton is going up an inclined plane, 1 in 2 at the rate of 36 kmph. If the co-efficient of friction is 1//sqrt3, the power of engine is |
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Answer» 9.8W |
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| 45048. |
Whether the switch K is open or closed, the reading of galvanometer is the same. If I denotes the current then : |
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Answer» `I_(R_4)` = `I_G` |
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| 45049. |
Frequency in the UHF range normally propagate by means of: |
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Answer» GROUND waves |
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| 45050. |
The value of Rydberg constant is _____ m^-1. |
| Answer» SOLUTION :`1.097xx10^7` | |
