Saved Bookmarks
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.
| 39951. |
A dip circle is placed in geographic meridian at a place where dip and declination are known be the respectively 30^(@) and 45^(@). What dip will be given by dip circle ? |
|
Answer» `TAN ^(-1) ((1)/(sqrt6))` |
|
| 39952. |
The graph between object distance P or u and image distance q or v for a plane mirror is a ____ slope equal to 1 |
| Answer» SOLUTION :STRAIGHT LINE | |
| 39953. |
Magnetic field is measured by |
|
Answer» PYROMETER |
|
| 39954. |
An a.c. voltage V=Vm sin ot is applied across an inductor L. Obtain an expression for current I. |
Answer» Solution :When an alternating voltage `V = V_(m)`sin OOT is applied across an inductor L, an alternating CURRENT flows through it and consequently an emf is induced across its ends which is `epsilon = -L (dl)/(dt)`. Thus, in equilibrium `V + epsilon=0`  or `V - L (dl)/(dt) =0` or `L (dl)/(dt) = V = V_(m) sin omega t` `therefore rArr I = V_(m)/L int sin omega t dt = V_(m)/L [(-cos omega t)/(omega)] = V_(m)/(L omega) cos omega t` `I_(m) = V_(m)/(L omega) = V_(m)/X_(L)` and `X_(L) = L_(omega)` is called the inductive REACTANCE of given inductor. |
|
| 39955. |
The resultant of two vectors vecP and vecQ is vecR If the magnitude of vecQ is doubled, the new resultant becomes perpendicular to vecP, the the magnitude of vecR is |
|
Answer» <P>`(P^(2)-Q)/(2PQ)` |
|
| 39956. |
When a U^(238) nucleus originally at rest, decays by emitting an alpha particle having speed 'u' the recoiled speed of residual nucleus is: |
|
Answer» `(-4U)/(238)` `=-(4)/(234)xxu=(-4u)/(234)` Hence choice is (c). |
|
| 39957. |
The atoms of lithium, sodium and potassium each contain a different number of electrons. Why then are all these elements monovalent? |
|
Answer» |
|
| 39958. |
Under what is the heat produced in an electric circuit: (i) directly proportional(ii) inversely proportional to the resistance of the circuit ? |
|
Answer» Solution :(i) If I in CIRCUITIS constantbecause` H = I ^2 Rt ` (ii)IfV in circuitisconstantbecause`H =(V^2 )/ (R )t ` |
|
| 39959. |
On decreasing the angular frequency of A.C. source used in L-C-R series circuit, the capacitive reactance ...... and inductive resistance ...... |
|
Answer» INCREASES, decreases |
|
| 39960. |
To shield an instrument from the external magnetic field it is placed insidea cabin made from |
|
Answer» WOOD |
|
| 39961. |
A moncohromatic source of light operating at 200 W emits 4xx10^(20) photons per second. Find the wavelength of the light. |
|
Answer» Solution :POWER `P=(N)/(t)(hc)/(lamda)` Energy of Photon `E=(P)/((N/(t)))=(200)/(4XX10^(20))=5xx10^(-19)J` `lamda=4xx10^(20)xx(6.6xx10^(-34)xx3xx10^(8))/(200)=3960Å` |
|
| 39962. |
An electron accelerated in an electric field of 20 kV enters a uniform magnetic field with induction 0.1 T. Its velocity vector makes an angle of 75^@ with the magnetic field vector. Find the shape of the path. |
|
Answer» In the lateral direction (i.e. in the my plane) the electron is acted upon by the Lorentz force, which makes the PROJECTION of its motion in this plane a circle with radius `R=(mv_(bot))/(e B)=sqrt(2me_(Psi) sin alpha)/(e B)` and with PERIOD `T 2pim//eB` In space the election moves along a helix which winds around the lines of induction. The radius of the circle, R, was given above. The pitch of the helix is `h=v_(1) T (2pi mv cos alpha)/(e B)-2pi R cot alpha`. 
|
|
| 39963. |
Bohr magnetonis given by (symbols have their usual meanings) |
|
Answer» `(4PI m _(e))/(EH^(2))` |
|
| 39964. |
Complete the nuclear reaction a. ""_(5)^(10)B + ………… to ""_3^7Li + ""_(2)^4He "b"_(……)^(63)Ni to ""_(29)^(…..)Cu + ""_(….)^(0)beta |
|
Answer» Solution :a. `""_(5)^(10)B + UL(""_0^1n) to ""_3^7Li + ""_(2)^4He` b. `""_30^(63)Ni to ""_(29)^63Cu + ""_1^(0)BETA` |
|
| 39965. |
A rectangular loop of wire ABCD is kept close to an infinitely long wire carrying a current I(t)=I_0 (1-t/T) for 0 le t le T and I(0)=0 for t gt T as shown in figure. Find the total charge passing through a given point in the loop, in time T. The resistance of the loop is R. |
Answer» Solution :Considering STRIP of thickness dr and LENGTH I at distance r from very long current CARRYING WIRE.  MAGNETIC flux linked with this strip, `dphi=BA=(mu_0I)/(2pir) (L_1dr)` `dphi=(mu_0IL_1)/(2pir)dr` Net flux linked with whole loop ABCD, `phi=int_x^(L_2+x) (mu_0IL_1)/(2pir)dr` `=(mu_0IL_1)/(2pi)[ln r]_x^(L_2+x)` `phi=(mu_0IL_1)/(2pi)ln ((L_2+x)/x)` `I(t)=I_0 (1-t/T)` `therefore phi=(mu_0L_1)/(2pi)I_0(1-t/T)ln ((L_2+x)/x)` Now induced charge, `Q=(Deltaphi)/R` `Q=(phi_1-phi_2)/R` For t=0 flux `phi_1=(mu_0L_1)/(2pi) I_0(1-0)ln ((L_2+x)/x)=(mu_0L_1I)/(2pi) ln ((L_2+x)/x)` For t=T time flux , `phi_2=(mu_0L_1)/(2pi) I_0(0)ln ((L_2+x)/x)=0` `rArr Q=(Deltaphi)/R =(mu_0L_1I_0)/(2piR)ln ((L_2+x)/x)` |
|
| 39966. |
a. When we look through a muslin cloth,we can see coloured spectrum. How will you explain this? b. Observe the shadow of your book when it is held a few centrimeters above as table with a lamp serveral centimeter above the book. Why is the shadow the book fuzzy at the edges?c.If you observe a distant street light between two fringers pinched together, you can visualise alternate bright and dark fringes. What is this due to? Explain. |
|
Answer» SOLUTION :a. It is a due to diffraction of light through the fine fibres, which act as slits. b. When light falls on the EDGES of the book, the phenomenon of diffraction is also produced. But the PATTERN is not sharply indentified.The CUMULATIVE effect is the formation of blurred image. c.It is to diffraction of light. The slits of formed by the two fingers are fine whose SIZE is comparable with the wavelength of light. |
|
| 39967. |
A magnetic field of vecB = 1 xx 10^(-2) hatk tesla applies a force of vec F = (36 hat I + 12 hat j) xx 10^(-23) newton on a proton. Calculate the velocity of proton. |
|
Answer» Solution :Let the velocity of PROTON be represented as `vec v = v _(x) hat I + v_(y) hat j + v_(z) hat k` Magnetic FORCE on a charge is given by `vecF = q vec v xx vecB` On substituting the VALUES we get : `vecF = (36 hat I + 12 hat j) xx 10^(-22)` `vecF = (1.6 xx 10^(-19)) (v_(x) hat i+ v_(y) hat j + v_(z) hat k) xx (1 xx 10^(-3) hat k)` ` rArr 36 hat i + 12 hat j = (1.6) (v_(x) hat i + v_(y) hat j + v_(z) hat k) xx ( HATK)` `rArr36/(1.6) hat i + 12/(1.6) hat j = - v_(x) hat j + v_(y) hat i ` ` rArr-v_(x) hat j + v_(y) hat i = 22.5hati + 7.5 hat j` `rArr v_(x) = - 22.5 m//s& v_(y) = 7.5 m//s` `rArrvec v = (-22.5 hati + 7.5 hat j ) m//s` |
|
| 39968. |
In geometrical optics a ray of light is defined as |
|
Answer» PATH of PROPAGATION of light |
|
| 39969. |
The total positive charge on a silver nucleus (atomic number :Z=47) is uniformly distributed on a sphere of radius 10^(-10) m . What will be the intensity of electric field at the surface ofthis sphere ? |
|
Answer» SOLUTION :Total CHARGE on sphere , q.=47e = `47xx1.6xx10^(-19)` C Radius of sphere , `r=10^(-10)` m The electric field intensity at the surface of the sphere is given by : `E=1/(4piepsilon_0)q/r^2` `=(9xx10^9xx47xx1.6xx10^(-19))/(10^(-10))^2` `=6.76xx10^12 NC^(-1)` |
|
| 39970. |
There is a, and four charges +q, +q, -q and -q are fixed at the four corners of square. See the figure for the placement of charges. Calculate: (i) Electric potential at point O. (ii) Work done to move a charge +e from point O to point B. (iii) Work done to move a charge +e from point O to A. |
Answer» Solution :The given arrangement is shown below:  Let the distance of each point charge from O be r. Then `r=OP=OQ=OR=OS` Also, `PR^(2)=PS^(2)+SR^(2)` Here, `PS=SR=RQ=PQ=a` `therefore PR^(2)=a^(2)+a^(2) rArr PR=sqrt(2)a` `therefore r=PO=(sqrt(2)a)/(2)=(a)/(sqrt(2))` (i) Potential at the centre O is : `V_(O)=(1)/(4pi epsilon_(0))[(q)/(r )+(q)/(r )-(q)/(r )-(q)/(r )]=0` (ii) To find the work done to move a charge from O to B, we NEED to first find the electric potential at point B. Potential at point B will be given by: `V_(B)=(1)/(4pi epsilon_(0))[(q)/(PB)+(q)/(QB)-(q)/(RB)-(q)/(SB)]` Also, `PB^(2)+PD^(2)+DB^(2)` `=((a)/(2))^(2)+a^(2)=(5a^(2))/(4)` `rArr PB=(sqrt(5)a)/(2)` Also, `SB=PB=(sqrt(5)a)/(2) and QB=RB=(a)/(2)` `therefore V_(B)=9xx10^(9)[(q)/((sqrt(5a))/(2))+(q)/((a)/(2))-(q)/((a)/(2))-(q)/((sqrt(5)a)/(2))]=0` Since, `V_(B)=V_(O)=0` `therefore` No work will be done on moving a charge from O to B. (III) To find the work the work done to move a charge from O to A, we need to first find the electric potential at point A. Potential at point A will be given by: `V_(A)=(1)/(4pi epsilon_(0))[(q)/(PA)+(q)/(QA)-(q)/(RA)-(q)/(SA)]` Also, `RA^(2)=AC^(2)+CR^(2)=a^(2)+((a)/(2))^(2)=(5a^(2))/(2)` `rArr RA=(sqrt(5)a)/(2)` Also, `SA=RA=(sqrt(5)a)/(2) and PA=QA=(a)/(2)` `therefore V_(A)=9xx10^(9) [(a)/((a)/(2))+(q)/((a)/(2))-(q)/((sqrt(5)a)/(2))-(q)/((sqrt(5)a)/(2))]` `=9xx10^(9)[(2q)/((a)/(2))-(2q)/((sqrt(5)a)/(2))]` `=9xx10^(9)[(4q)/(a)-(4q)/(sqrt(5)a)]` `=(9xx10^(9)xx4q)/(a)[1-(1)/(sqrt(5))]` `=(36q)/(a)xx10^(9) (1-(1)/(sqrt(5)))` `therefore` Work done in moving a charge +E from O to A will be: (this work is done by an external agent so its potential differenceis taken as final minus initial) `V_(A)-V_(O)=(W_(OA))/(e )` `rArrV_(A)-0=(W_(OA))/(e )` `rArr W_(OA)=eV_(A)=e xx (36q)/(a) xx 10^(9)(1-(1)/(sqrt(5)))` `=(36qe)/(a)xx10^(9)(1-(1)/(sqrt(5)))` |
|
| 39971. |
Between the primary and secondary rainbows, there is a dark band known as Alexandar's dark band. This is because |
|
Answer» light SCATTERED into this region interfere destructively. |
|
| 39972. |
Two unequal soap bubbles are formed one on each side of a tube closed in the middle by a tap. What happens when the tap is opened to put the two bubbles in communication? |
|
Answer» air FLOWS from the BIGGER bubble to the smaller bubble TILL the sizes become equal |
|
| 39973. |
A skier is pulled by towrope up frictionless ski slope that makes an angle of 12^(@) with the horizontal . The rope moves parallel to the slope with a constant speed of 1.0 m/s. The force of the rope does 880 J of work on the skier as the skier moves a distance of 7.0 m up the incline. (a) If the rope moved with a constant speed of 2.0 m/s, how much work would the force of the rope do on the skier as the skier moved a distance of 8.0 m up the incline ? At what rate is the force of the rope doing work on the skier when the rope moves with a speed of (b) 1.0 m/s and (c) 2.0 m/s ? |
|
Answer» |
|
| 39974. |
Two progressive waves have equations y_1=3 sin (100 pit) and y_2 = 4 sin (100 pit + 2pi). The resultant is : |
|
Answer» 7 units |
|
| 39975. |
You are given four sources of light each on providing a light of a single colour - red, blue green and yellow. Suppose the angle o refraction for a beam of yellow ligh corresponding to a particular angle o incidence at the interface of two media is 90° Which of the following statements is correct i the source of yellow light is replaced with tha of other lights without changing the angle o incidence ? |
|
Answer» The beam of red light would undergo tota. internal reflection. `implies sinC prop lambda)` `implies` If `lambda` decreases then critical angle C will also decrease. Here among given beams of light, blue light has MINIMUM wavelength and so critical angle for blue light would be minimum and so blue light will undergo TIR. (Total Internal Reflection). |
|
| 39976. |
Find integrals of given functions int cos theta(tan theta + "sec" theta) d theta |
|
Answer» |
|
| 39977. |
A parallel - plate capacitor, filled with a dielectric of dielectric constant k, is charged to a potential V_0 .It is now disconnected from the cell and the slab is removed. If it now discharges, with time constant tau , through a resistance, then find time after which the potential difference across it will be V_0 ? |
|
Answer» SOLUTION :When the slab is removed , the potential difference ACROSS capacitor increases to `k V_0` `CV_0 = kCV_0 e^(-t/tau) as q_0 = KCV_0` `1/k = e^(-t/tau) THEREFORE ln k = t/tau RARR t= tau ln k ` |
|
| 39978. |
On coolidge tube when 25 kV potential is applied current is obtained 2.2 mA. If the distance between the electrode is 1 m calculate (i) Heat produce at anode in cal/sec. (ii) no. of electron striking the anode per second (iii) Velocity of electron striking the anode (iv) Energy of electron |
|
Answer» Solution :`(i) (5.5)/(4.2)"CAL"//s"" (ii) (11)/(8)xx10^(16)e` `(iii)9.4 xx 10^(7) m//s"" (iv) 25 KEV` |
|
| 39979. |
Three point charges q , -4q and 2q are placed at the vertices of an equilateral triangle ABC of side 'l' as shown in the Fig. Obtain the experssion for the magnitude of the resultant electric force acting on the charge q. |
Answer» Solution :As shown in Fig , force `vecF_(AC) = (1)/(4pi in_(0)) (q (2q))/(L^(2))` along `CA = (1)/(4pi in_(0)) * (2q^(2))/(l^(2))` along CA and `vecF_(AB) = (1)/(4pi in_(0)) * (q(4q))/(l^(2))` along AB `= (1)/(4pi in_(0)) * (4q^(2))/(l^(2))` along AB.  If the magnitude of the resultant electric force acting on the charge q placed at A be `F_(A)` , then `F_(A)= SQRT((F_(AC))^(2) + (F_(AB))^(2) + 2 F_(AC) . F_(AB) cos 120^(@))` `= sqrt(((2q^(2))/(4pi in_(0) l^(2)))^(2) + ((4q^(2))/(4 pi in_(0) l^2))^(2) + 2 ((2q^(2))/(4pi in_(0) l^(2)))^(2) ((4q^(2))/(4pi in_(0) l^(2))) (- (1)/(2)) ) vecF_(A)` `= (2q^(2))/(4 pi in_(0) l^(2)) sqrt(1 + 4 - 2) = (SQRT3 q^(2))/(2pi in_(0) l^(2))` |
|
| 39980. |
Calculate de Broglie wavelength of neutrons whose energy iseV.(Given mass of electrons (1.67 xx 10^(-27)kg) |
|
Answer» SOLUTION :We know that `lambda = h/SQRT(2ME) = (6.6 XX 10^-34)/sqrt(2 xx 1.67 xx 10^-27 xx 1.6 xx 10 ^-10` = `0.2857 xx 10^-10 = 0.2857 A^@` |
|
| 39981. |
Two point charges q_(A)=3muC and q_(B)=-3muC are located 0.2 m apart in vacuum. a. What is the electric field at the mid point O of the line AB joining the two charges? b. If a negative test charge of magnitude 1.5xx10^(-9)C is placed at this point, what is the force experienced by the test charge? |
Answer» Solution :(a) The situation is represented in the given figure . O is the MID - pointof line AB. Distance between the two charges , AB =20 cm`thereforeAO=OB=10cm` Net electric field at point O=E Electric field at point O caused by `+3muC` charge , `E_(1)=(3xx10^(-6))/(4piepsilon_(0)(AO)^(2))` `=(3xx10^(-6))/(4piepsilon_(0)(10xx10^(-2))^(2))N//C`alongOB where , `epsilon_(0)=`Permittivity of free space `(1)/(4piepsilon_(0))=9xx10^(9)Nm^(2)C^(-2)` Magnitude of electric field at point O caused by `-3muC` charge. `E_(2)=|(-3xx10^(-6))/(4piepsilon_(0)(OB)^(2))|` `=(3xx10^(-6))/(4piepsilon_(0)(10xx10^(-2))^(2))N//C` along OB `thereforeE=E_(1)+E_(2)` `=2xx(9xx10^(9))xx(-3xx10^(-6))/((10xx10^(-2))^(2))` [ Since the value of `E_(1)andE_(2)`are same , the value is multiplied with 2] `=5.4xx10^(6)N//C`along OB Therefore , the electric field at mid - point O is `5.4xx10^(6)NC^(-1)` along OB. (b) ATEST chrge of amount `1.5xx10^(-9)C` is PLACED at mid - point O. `q=1.5xx10^(-9)C` Force experienced by the test charge=F `thereforeF=qE` `=1.5xx10^(-9)xx5.4xx10^(6)` `=8.1xx10^(-3)N` The force is directed along line OA . This is because the NEGATIVE test charge is repelled by the charge placed at point B but attracted towards point A. Therefore , the force experienced by the test charge is `8.1xx10^(-3)N` along OA. |
|
| 39982. |
The total energy of an electron in an atom in an orbit is -3.4 eV. Its kinetic and potential energies are, respectively: |
|
Answer» `3.4 EV, 3.4 eV` But `K.E =-E` `:. K.E=-(-3.4)=3.4eV` `P.E.=2E` `2(-3.4)` `-6.8eV` |
|
| 39983. |
Name two advantages of alternating current. |
| Answer» Solution :a.c voltage can be stepped up/stepped down. TRANSMISSION power loss can be minimised.a.c. can be controlled USING chocks, CAPACITOR with MINIMUM power loss. | |
| 39984. |
Based on Huygen's construction, draw the shape of a plane wavefront as it gets refracted on passing through (a) a thin prism, and (b) a thin convex lens. |
Answer» SOLUTION :(a) and (B) RESPECTIVELY. 
|
|
| 39985. |
एक गोलीय दर्पण द्वारा बने प्रतिबिंब का आवर्धन m ऋणात्मक (negative) है। इसका अर्थ यह है कि प्रतिबिंब |
|
Answer» वस्तु (बिम्ब ) से छोटा है |
|
| 39986. |
In an a.c. circuit containing capacitance only |
|
Answer» voltage is AHEAD of current by `pi/2` |
|
| 39987. |
What is cyclotron ? Discuss the principle of cyclotron. |
|
Answer» Solution :1. Cyclotron is a device used to accelerated charged particles like protons, DEUTERONS, `alpha`-particles etc., to very high energies. It was invented by E.O.Lawrence and M.S.Livingston in 1934. 2. The cyclotron uses both electric and magnetic fields in COMBINATION to increase the energy of charged particles. As the fields are perpendicular to each other they are called crossed fields. Cyclotron uses the fact that the frequency of revolution of the charged particle in a magnetic field is independent of its energy. 3. PRINCIPLE : A charged particle can be accelerated to very high energies by MAKING it pass through a moderate electric field a number of times. This can be done with the help of a perpendicular magnetic field which throws the charged particle into a circular motion, the frequency of which does not depend on the speed of the particle and the radius of the circular orbit. |
|
| 39988. |
Separation between the slits of YDSE set-up is 0.1 cm. Films of same thickness 0.5 mm is pasted on both the slits. Refractive index of one of the films 1.52 and for the other it is 1.48 Wavelength of the light used is 600 nm. Sereen is placed at a distance 1 m from the slits. Calculate fringe width. What will be the distance of first available maxima from the centre? |
|
Answer» Solution :FRINGE width is not affected by the films and is WRITTEN as follows : `omega = (lambda)/d` `omega = (600 xx 10^(-9) xx 1)/(0.1 xx 10^(-2)) = 6.0 xx 10^(-4) m` Optical PATH difference introduced at the centre due to thin films can be written as follows: `Delta x = (mu_1 - mu_2)t` `= (1.52 - 1.48) xx 0.5 xx 10^(-3)` `= 0.04 xx 0.5 xx 10^(-3) = 2 xx 10^(-5) m` Now equating `Deltax = n lambda`, we get the following: `n = (Delta x)/(lambda) = (2 xx 10^(-5))/(600 xx 10^(-9)) = 33.33` From value of n obtained, we can understand that on one side maxima is at distance `y_1 = 0.33 W` and on the other side it is at a distance `y_2 = 0.67 w`, where `w` is fringe width of the pattern. `y_1 = 0.33 xx 6 xx 10^(-4) = 1.98 xx 10^(-4) m` `y_2 = 0.67 xx 6 xx 10^(-4) = 4.02 xx 10^(-4) m`. |
|
| 39989. |
The core material of an optical fibre has a refractive index of 1.654 while the cladding material has refractive index of 1.645. Find the value of the critical angle in the core material. |
|
Answer» Solution :Refractive INDEX of CORE MATERIAL, `muunderset1` = 1.654 Refractive index of cladding material, `muunderset2`=1.654 `thereforesin iundersetc` = muunderset2/muunderset1 = .645/1.654=0.9946 `thereforeiundersetc =sin^-1(0.9946)=84^@` |
|
| 39990. |
हिमालय से निकलने वाली नदियां, जहां विलुप्त हो जाती हैं उसे कहते हैं? |
|
Answer» भांगर |
|
| 39991. |
A primary of transformer has 100 turns an secondary has 500 turns. If input voltage 20 and frequency 50Hz, so the output voltage an frequency are... V and ....... Hz |
|
Answer» 200,500 `THEREFORE epsilon_2=epsilon_1 XX N_1/N_2` `=20xx5000/500` `therefore epsilon_2`= 200 V `therefore` Frequency of TRANSFORMER remains unchanged. |
|
| 39992. |
Explain the three modes of propagation of electromagnetic waves through space. Propagation of electromagnetic waves: |
|
Answer» Solution :The electromagnetic wave transmitted by the transmitter travels in three different modes to reach the receiver according to its frequency range: (i) Ground wave propagation (or) surface wave propagation (nearly 2 kHz to 2 MHz) (ii) Sky wave propagation (or) ionospheric propagation (nearly 3 MHz to 30 MHz) (iii) Space wave propagation (nearly 30 MHz to 400 GHz) (i) Ground wave propagation If the electromagnetic waves transmitted by the transmitter glide over the surface of the earth to reach the receiver, then the propagation is called ground wave propagation. The corresponding waves are called ground waves or surface waves. Both transmitting and receiving antennas must be close to the earth. The size of the antenna plays a major role in deciding the efficiency of the radiation of signals. During TRANSMISSION, the electrical signals are attenuated over a distance. Some reasons for attenuation are as follows:  • Increasing distance:: The attenuation of the signal depends on (i) power of the transmitter (ii) frequency of the transmitter, and (iii) condition of the earth surface. Absorptionof energy by the Earth: When the transmitted signal in the form of En dve is in contact with the Earth it induces charges in the Earth and constitutes a current Due to this, the earth behaves like a leaky capacitor which leads to the attenuation of the wave. Tilting of the wave: As the wave progresses, the wavefront starts gradually tilting according to the curvature of the Earth. This increase in the tilt decreases the electric field Strength of the wave. Finally, at some distance, the surface wave dies out due to energy loss. The frequency range of EM waves in this mode of propagation is 3 to 30 MHz. EM waves of frequency more than 30 MHz can easily penetrate through the ionosphere and does not undergo reflection. It is used for short wave broadcast services. Medium and high frequencies are for long-distance radio communication. Extremely long distance communication is also possible as the radio waves can undergo multiple reflections between the earth and the ionosphere. A single reflection helps the radio waves to travel a distance of approximately 4000 km. Ionosphere acts as a reflecting surface. It is at a distance of approximately 50 km and spreads up to 400 km above the Earth surface. Due to the absorption of ultraviolet rays. cosmic ray, and other high energy radiations like a, BRAYS from sun, the air molecules in the ionosphere getionized. This produces charged ions and these ions provide a reflecting medium for the reflection of radio waves or communication waves back to earth within the permitted frequency range. The phenomenon of bending the radio waves back to earth is nothing but the TOTAL internal reflection.  (iii) Space wave propagation: The process of sending and receiving information signal through space is called space wave communication. The electromagnetic waves of very high frequencies above 30 MHz are called as space waves. These waves travel in a straight line from the transmitter to the receiver. Hence, it is used for a line of sight communication (LOS).  For high frequencies, the transmission towers must be high enough so that the transmitted and received signals (direct waves) will not encounter the curvature of the earth and hence travel with less attenuation and loss of signal strength. Certain waves reach the receiver after GETTING reflected from the ground. |
|
| 39993. |
Dimensions of 1 // (mu_(0) epsilon_(0)) is: |
|
Answer» `L^(2) // T^(2)` `:. c^(2)=(1)/(mu_(0) epsilon_(0))` or `L^(2)T^(-2)=(1)/(mu_(0) epsilon_(0))` |
|
| 39994. |
A particle moves so that position vector is given by r = coscotx + sincotywhere cois a constant. Which of the following is true? |
|
Answer» Velocity and ACCELERATION both are PERPENDICULAR to `VECR`. |
|
| 39995. |
Explain the concept of intensity of electromagnetic waves. |
| Answer» Solution :The energy crossing PER unit AREA per unit time and perpendicular to the direction of propagation of ELECTROMAGNETIC WAVE is called the intensity। Intensity, I = (u)C | |
| 39996. |
Draw the I - V characteristics of Ge and Si pn- junction diodes and explain the same . |
Answer» Solution : Junction diode CHARACTERISTICS, Ge and Si Explanation : (1) Forward bias characteristics : There is forward current only after the barrier potential of the pn-junction is overcome. Then there is an exponential RISE in the current beyond the knee region. The forward - bias voltage required to reach theregion of upward swing iscalled the threshold voltage `V_(T)` or cut in voltage. When rounded offto the nearest , thethreshold voltage is 0.3 V for a germanium diode and 0.7 V for a silicon diode. (2) Reverse bias characteristics : In the reverse bias, the potential barrier at the junction is large and the current due to majority CARRIERS in each region is zero. However, minority charge carriers are able to cross the junction and CONSTITUTE a very small current in thereverse direction. This reverse current quickly reaches its maximum or saturation value and remains fairlyconstant with increase in thereverse - bias voltage. It is calledthereverse saturation current `I_(s)`, which is typically a few nanoamperes for a silicon diode and a few microamperes for a germanium diode. Too high a negative voltage results in a sharp change in the reverse - bias characteristics. At a CERTAIN characteristic negative voltage, called the peak inverse voltage (PIV) or breakdown voltage, thecurrent in the reverse direction increases very rapidly . In general, a silicon diode has a higher PIV rating `(~1000 V)` than a germanium diode `(~ 400 V)`. |
|
| 39997. |
The polarization of electromagnetic waves is in |
|
Answer» the DIRECTIONS of ELECTRIC and magnetic field |
|
| 39998. |
In an amplitude modulation with modulation index 0.5 the ratio of the carrier wave to that of side band in the modulated wave is |
|
Answer» `4:1` Amplitude of the carrier wave `=A_(c)` Amplitude of the SIDE band `=mu(A_(c))/(2)` Their ratio `=(2)/(mu)=(2)/(0.5)=(4)/(1)` |
|
| 39999. |
A uniform rope of mass M length L hangs vertically from the ceiling, with lower end free. Adistbance on the rope trvelling upwards starting from the lower end has a velocity v. At a point P at distance x from the lower end. |
|
Answer» TENSION at point P is mg |
|
| 40000. |
Let rho(r) = Q/(pi R^4) r be the charge density distribution for a solid sphere of radius R and total charge Q for a point 'p' inside the sphere at distance r_1 from the centre of the sphere, the magnitude of electric field is |
|
Answer» `(Qr_1^2)/(3 PI epsilon_0 R^4)` |
|