This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
In a series RLC circuit, the fequency. Passing through a coil as shown in figure-5.253. The time variation of the magnitude of EMF generated across the coil during one cycle is: |
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Answer» CAPACITIVE |
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| 3. |
A space station is at a height equal to the radius of the Earth . If v_Eis the escape velocity on the surface of the Earth , the same the spastation v_E |
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Answer» `1/2` K.E = P.E `1/2mv_(s)^(2)=(GMm)/(2R)` `v_(s)^(2)=(GM)/R` `v_(s)=sqrt(gR)` But `v_(e)=sqrt(2gR)=sqrt2sqrtgR` `v_(e)=sqrt2v_(s)` `v_(s)=v_(e)/sqrt2` |
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| 4. |
State Brewster's law. |
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Answer» Solution :Brewster's law : The tangent of the POLARISING angle is equal to the REFRACTIVE index of the reflecting medium with RESPECT to the surrounding. If `i_(p)` is the polarising angle, tan `i_(p)=(n_(2))/(n_(1))` where `n_(1)` is the ABSOLUTE refractive index of the surrcoundingand `n_(2)` is that of the reflecting medium. |
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| 5. |
A mass of 0.2 kg is attached to the lower end of a massless spring of force constant 200 N m^(-1) , the upper end of which is fixed to a rigid support. Which of the following statement is not true? (Take g=10 m s^(-2)) |
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Answer» The frequency of oscillation will be nearly 5 Hz. `=1/(2pi) sqrt(200/0.2)`= 5 Hz. In equilibrium , kx = mg or `x=(mg)/k=(0.2 xx 10)/200`=0.01 m When mass is raised till the spring is unstretched, the work `=1/2kx^2=MGX` When the mass is released from the unstretched position of spring, then TOTAL work done mgx.=(mgx)+`1/2kx^2=2mgx` or x.=2x=2 x 0.1 = 0.02 m As `upsilon` of spring is INDEPENDENT of g so that the frequency of oscillation will be the same as that on the earth. |
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| 6. |
If the plates of a parallel plate capacitor are not equal in area then quantity of charge |
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Answer» On the plates will be same but nature of charge will differ As per the LAW of CONSERVATION of charge |
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| 7. |
A gas (gamma=1.5) is taken through adiabatic process in which volume is changed from 16000" cm"^(3) to 400" cm"^(3). If the initial pressure is 150 kPa, how much work is done by the gas in the process ? |
| Answer» Answer :D | |
| 8. |
Statement I : Half life for certain radioactive element is 5 min . Four nuclei of that elementare observed at a certain instant of time. After five minutes it can be definitely said that two nuclei will be left undecayed . Statement II : Half life is defined as time in which population is halfed . |
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Answer» Both STATEMENT -1 and statement -2 are true and statement -2 is the correct EXPLANATION of statement -1. |
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| 10. |
A thin film of thickness 4xx10^(-5) cm and mu=1.5 is iluminated by white light incident normal to its surface. What wavelength in the visible range be intensified in the reflected beam ? |
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Answer» |
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| 11. |
Mention the power factor of a pure inductor or a capacitor. |
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Answer» <P> Solution :In PURELY CAPACITIVE circuit, power FACTOR P = 0. |
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| 12. |
Proceeding from Maxwell's equation shown that in the case of a plane electromagnetic wave ( figure ) propagating in vacuum the following relations hold :(deltaE)/(deltat)=-c^(2)(deltaB)/( deltax), (deltaB)/(deltat)-( deltaE)/( deltax). |
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Answer» Solution :`VEC(E)=hat(j) E(t,x)` `vec(B)=hat k B (t_(1),x)` and Curl`vec(E)=hat(k) (deltaE)/( deltax)=-( delta vec(B))/( deltat)=-hat(k)(DELTAB)/( deltat)` so `-(deltaE)/(deltax)=(deltaB)/(deltat)` Also Curl`vec(B)= epsilon_(0)mu_(0)(delta vec(E))/(deltat)=(1)/(C^(2))(deltavec(E))/(delta t)` and Curl`vec(B)=-hat(j) (deltaB)/(deltax)` so `(deltaB)/(deltax)=-(1)/(c^(2))(deltaE)/( deltat).` |
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| 13. |
Prove that in a uniform field the work is independent of the path. |
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Answer» SOLUTION :Let the object move in the field from POINT M to point Nfirst along the rectilinear segment MN=l, and then along the broken line `MKN = l_(1)+l_(2)`, In the first case the work done is `A= Fl cos alpha =Fd` In the second case the work done is `A= A_(1)+A_(2)=Fl_(1) cosalpha_(1)+fl_(2)cosalpha_(2)=F(l_(1)cos alpha_(1)+l_(2)cosalpha_(2))=Fd` We see that the work done is independent of the PATH. |
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| 14. |
Show that the density of nuclear mass is constant for all isotopes. |
| Answer» SOLUTION :DENSITY = `("MASS")/("VOLUME")` | |
| 15. |
….. are used to kill cancerous cells. |
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Answer» X - RAYS |
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| 16. |
A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 5.0 xx 10^(-T). The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0 s^(-1). What is the moment of inertia of the coil about its axis of rotation ? |
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Answer» Solution :It given that for circular coil `N = 16 , R= 10 cm = 0.1 m and I = 0.5 A` `therefore ` Magnetic moment of current carrying coil m = N IA = `NI pi R^2` Frequency of small anuglar oscillations of the coil is given by `v = 1/(2pi) sqrt((MB)/(I))` where I is the moment of inerita of the coil about the axis of rotation . THUS, we have `I = (mB)/(4pi^2 v^2) = ((NI pi R^2)B)/(4pi^2 v^2) = (NIR^2 B)/(4pi v^2)` `B = 5.0 xx 10^(-2) T and v = 2.0 s^(-1)` `thereforeI = (16xx0.75xx(0.1)^2 xx (5.0 xx 10^(-2))/(4xx3.14xx(2.0)^2) = 1.2 xx10^(-4) KG m^2` . |
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| 17. |
A toaster operating at 240 V has a resistance of 120Omega. The power is ………. . |
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Answer» 400 W |
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| 18. |
A circuit diagram of a capacitor with two diodes ad an ammeter (Fig) is connected to a source of variable voltage producing rectangular pulses of constant amplitude and frequency. What current does ammeter show? Diodes and ammeter can be considered ideal. |
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Answer» |
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| 19. |
Explain the source of stellar energy. Explain the carbon - nitrogen cycle, proton cycle occurring in stars. |
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Answer» Solution :Scientists proposed two types of cyclic processes for the sources of energy in the sun and STARS. The first is know as carbon - nitrogen cycle and the scond is proton - proton cycle. 1. Carbon - Nitrogen Cycle : According to Bethe carbon - nitrogen cycle is mainly responsible for the production of solar energy. This cycle consists of a chain of nuclear reactions in which hydrogen is converted into Helium, with the help of Carbon and Nitrogen as catalysts. The nucleasr reactions are as given below. `._(6)^(12)C+ ._(1)^(1)H to ._(7)^(13)N + gamma` `._(7)^(13)N to ._(6)^(13)C + ._(+1)^(0)e + v` `._(6)^(13)C + ._(1)^(1)H to ._(7)^(14)N + gamma` `._(7)^(14)N + ._(1)^(1)H to ._(8)^(15)O + gamma` `._(8)^(15)O to ._(7)^(15)N + ._(+1)^(0)e + v` `._(7)^(15)N + ._(1)^(1)H to ._(6)^(12)C + ._(2)^(4)He` 2. Proton - Proton Cycle : A star is FORMED by the condensation of a large amount of matter at a point in space. Its temperature rises to `2,00,000 .^(@)C` as the matter contracts under the influence of gravitational attraction. At this temperature the thermal energy of the protons is sufficient to form a deuteron and a positron. The deuteron then COMBINES with another protobn to form LIGHTER nuclei of helium `._(2)^(3)He`. Two such helium nuclei combine to form a helium nucleus `._(2)^(4)He` and two protons releasing a total amount of energy 25.71 MeV. The nuclear fusion reactions are given below. `._(1)^(1)H+ ._(1)^(1)H to ._(1)^(2)H+ ._(1)^(0)e + v` `._(1)^(2)H + ._(1)^(1)H to ._(2)^(3)H + gamma` `._(2)^(3)He + ._(2)^(3)He to ._(2)^(4)He + 2._(1)^(1)H` The net result is `4 ._(1)^(1)H to ._(2)^(4)He + 2 ._(+1)^(0)e + 2v + 2gamma` |
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| 20. |
What is the average power dissipation in an ideal capacitor in A.C. circuit ? |
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Answer» `1/2CV^2` `delta =pi/2` rad `therefore` Power P = `V_(rms). I_(rms) COS delta` `therefore P=V_(rms) . I_(rms) xx cos pi//2` `therefore` P=0 |
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| 21. |
Matching {:("List - I","List - II"),("A)" K lt 1 ,"e) Critical state"),("B) K =1 ","f) Sub critical state"),("C)" Kgt1,"g) Super critical"):} |
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Answer» `A RARR f , B rarr E , C rarr G ` |
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| 22. |
Your starship passes Earth with a relative speed of 0.9990c. After traveling 10.0 y (your (time), you stop at lookout post LP13, turn, and then travel back to Earth with the same relative speed. The trip back takes another 10.0y (your time). How long does the round trip take according to measurements made on Earth? (Neglect any effects due to the accelerations involved with stopping, turning, and getting back up to speed.) |
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Answer» Solution :KEY IDEAS We begain by analyzing the outward trip: 1. This problem involves measurementsmade from two (inertial) referenceframes, one attached to Earth and the other (your reference frame) attached to your ship. 2. The outward trip involves two events: the start of the trip at Earth and the end of the trip at LP13. 3. Your measurement of 10.0 y for the outward trip is the proper time `Deltat_(0)` between those two events, because the events occur at the same location in your reference frame - namely, on your ship. 4. The Earth-frame measurement of the time interval `Deltat` for the outward trip must be GREATER than `Deltat_(0)`, according to `(Deltat=gammaDeltat_(0))` for time dilation. Calculations: Using Eq. 36-8 to substitute for `gamma` in Eq. 36-9, we find `Delta=(Deltat_(0))/(sqrt(1-(v//c)^(2)))=(10.0y)/(sqrt(1-(0.9990c//c)^(2)))` `=(22.37)(10.0y)=224y`. On the return trip, we have the same situation and the same data. Thus, the round trip requires 20y of your time but `Delta_("total")=(2)(224y)=448y ` of Earth time. In other words, you have aged 20 y while the Earth has aged 448 y. Although you cannot TRAVEL into the PAST (as far as we know), you can travel into the FUTURE of, say, Earth, by using high-speed relative motion to adjust the rate at which time passes. |
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| 23. |
Equal current 'i' flows in the two segments of a circular loop in the direction shown in fig. radius fo the loop is 'a'. Magnetic field at the centre of the loop is |
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Answer» ZERO |
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| 24. |
What is electric potential and what are it, s SI units ? |
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Answer» |
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| 25. |
If vec(A)=3hat(i) + 4hat(j) and vec(B)=7hat(i) + 24hat(j). Find the vector having the same magnitude as vec(B) and is parallel to vec(A): |
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Answer» `15hat(i) + 20hat(j)` Therefore `|B|hatA=25((3hati + 4hatj))/5=15hati+20hatj` |
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| 26. |
The figure shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm . The capacitor is beging charged by an external source (not shown in the figure. ) The charing is constant and equal to 0.15 A Obtain the displacement current across the plates. |
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Answer» Solution :(a) `C=epsi_(0)A//d =80.1 pF` `(DQ)/(dt) =C ""(dV)/(dt)` `(dV)/(dt) =(0.15)/(80.1xx10^(-12))=1.87xx10^(9) V s^(-1)` (b) `i_(d) =epsi_(0) ""(d)/(dt) Phi _(E)`. Now across the capacitor `Phi_(E)=EA`, ignoring and CORRECTIONS. Therefore, `i_(d)=epsi_(0)A""(d Phi_(E))/(dt)` Now, `E=(Q)/(epsi_(0)A)`. Therefore, `(dE)/(dt)=(i)/(epsi_(0) A)`, which implies `i_(d)=i=0.15 A.` (c) Yes, provided by ‘CURRENT’ we mean the sum of conduction and displacement currents. |
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| 27. |
Electron of mass m and intial velocity vecV=v_(0)hati (V_(0)gt0) enter electric field vecE=-E_(0)hati(E_(0)) constantat t=0 initial de-Broglie wavelength of electron is lambda_(0) then at time t=t its de-Broglie wavelength will be….. |
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Answer» `lambda_(0)` `therefore veca=(vecF)/(m)=(qvecE)/(m)` `((-e)(-E_(0))hati)/(m)=(eE_(0))/(m)hati`….(1) From equation of motion with constant acceleration, v=u+at [From equation (1)] `therefore v=v_(0)+(eE_(0))/(m)t` Now,de-Broglie wavelength, `LAMBDA=(h)/(mv)=(h)/(m[v_(0)+(eE_(0))/(m)t])` `(h)/(mv_(0)[1+(eE_(0))/(mv_(0))t])` `=(lambda_(0))/(1+(eE_(0))/(mv_(0))t) [because (h)/(mv_(0))=lambda_(0)]` |
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| 28. |
A uniform rod AB of mass m = 2 kg and length l = 1.0 m is placed on a sharp support P such that a = 0.4 m and b = 0.6 m. A. spring of force constant k= 600 N//m is attached to end B as shown in Fig. To keep the rod horizontal, its end A is tied with a thread such that the spring is B elongated by 1 CM. Calculate reaction of support P when the thread is burnt. |
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Answer» |
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| 29. |
A square conducting coil of area 100 "cm"^2 is placed normally inside a uniform magnetic field of 10^3 "Wbm"^(-2). The magnetic flux linked with the coil is ___Wb . |
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Answer» Solution :`phi=AB cos theta =100xx10^(-4)xx10^(+3)xx cos 0^@` =10 WB |
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| 30. |
A current I flows around a closedpath in the horizontal plane of the circle as shown in the figure. The path consists of eight arcs with alternating radii r and 2r. Each segment of arc subtends equal angle at the common centre P. Find the magnetic field produced by current path at point P |
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Answer» Solution :B due to STRAIGHT parts of the WIRE is zero. Each arc SUBTENDS an angle `theta = (2PI)/(8) = pi //4` `therefore B_("net") = 4 ((mu_0 I)/(4pi r) xx pi//4) + 4 ((mu_0 I)/(4pi (2r)) xx pi // 4 )` ` = (3 mu_0 I)/(8r)`, perpendicular to the plane fo the paper and DIRECTED inward. |
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| 31. |
The radius of the first orbit of hydrogen atom is 0.5 Å. The electron moves is an orbit with a uniform speed of 2.2xx10^(6) ms^(-1). What is the magnetic field produced at the centre of the uncleus due to the mation of this electron ? Use mu_(0)= 4pi xx 10^(-7) Hm^(-1) and electric charge=1.6 xx 10^(-19) C |
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Answer» Solution :Here, `"" r = 0.5 Å xx 10^(-10) `m V = 2.2 `xx 10^(6) ms^(-1)` period of revolution of electron T = `(2pi r)/(v) = (2 xx 22 xx 0.5 xx 10^(-10))/(7 xx 2.2 xx 10^(6)) = (1)/(7) xx 10^(-15)`S Equivalent CURRENT, I = `("charge")/("time") = (E)/(T) = (1.6xx 10^(-19) xx 7)/(10^(-15)) ` Magnetic FIELD produced at the centre of the nucleus, `B = (mu_(0)I)/(2r) = (4pi xx 10^(-7) xx 1.12 xx 10^(-3))/(2 xx 0.5xx 10^(-10))` B = 14.07 T |
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| 32. |
वन्य जीव सुरक्षा कानून कब बनाया गया ? |
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Answer» 1970 |
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| 33. |
Huygen's wave theory of light cannot explain |
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Answer» RECTILINEAR PROPAGATION of light |
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| 34. |
A cell of emf 1.6 V is connected across a potentiometer wire of length 500 cm. The cell has negligible resistance. A thermo couple whose cold junction is at 0^@ C and hot junction at 80^@ C is connected in the secondary circuit. The balancing length is found to be 125 cm. Find the thermo emf per degree centigrade difference of temperature of the junctions. |
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Answer» `0.005 V//^@ C` |
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| 35. |
The 1-V characteristic of an LED is shown as |
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Answer»
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| 36. |
Find the energy Q liberated in beta^(-_(-)) and beta^(+)- decays and in K- caputure if the masses of the parent atom m_(p), the daughter atom M_(d) and an electron m are known. |
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Answer» Solution :In `beta^(-)` decay `Z^(X^(A))rarr_(Z+1)Y^(A)+e^(_)+Q` `Q=(M_(x)-M_(y)-m_(e ))C^(2)` `=[(M_(x)+ZM_(e )c^(2))-(M_(y)+Zm_(e )+m_(e ))]c^(2)` `=(M_(p)-M_(d))c^(2)` SINCE `M_(p),M_(d)` are the masses of the atoms. The binding energy of the ELECTRONS in ignored. In `K` capture `e_(k)^(-)+._(Z)X^(A) rarr_(z-1)Y^(A)+Q` `Q=(M_(X)-M_(Y))c^(2)+m_(e )c^(2)` `=(M_(x)^(c^(2))+Zm_(e )c^(2))-(M_(Y)c^(2)+(Z-1)m_(e )c^(2))` `=c^(2)(M_(p)-M_(d))` In `beta^(+) decay _(Z)X^(A) rarr_(Z-1)Y^(A)+e^(+)+Q` Then `Q= (M_(x)-M_(y)-m_(e ))c^(2)` `[M_(x)+Zm_(e )]c^(2)-[M_(y)+(Z-1)m_(e )]c^(2)-2m_(e )c^(2)` `=(M_(p)-M_(d)-2m_(e ))c^(2)` |
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| 37. |
Name the electromagnetic radiations used for (a) water purification, and (b) eye surgery. |
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Answer» SOLUTION :(a) ULTRA VIOLET RAYS Ultra violet rays /LASER |
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| 38. |
If a sphere is rolling, the ratio of the translational energy to total kinetic energy is given by |
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Answer» `7:10` `E=_("trans")+E_("not")=1/2mv^(2)+1/2lomega^(2)` `=1/2mv^(2)+1/2xx(2/5mr^(2))omega^(2)=1/2mv^(2)+1/5mv^(2)=7/10mv^(2)` `THEREFORE(E_("trans"))/(E_("total"))=(1/2mv^(2))/(7/10mv^(2))=5/7` |
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| 39. |
How many electron states (including spin states) are possible in a hydrogen atom if its energy is -3.4 eV? |
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Answer» 2 |
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| 40. |
A conducting rod is bent as a parabola y=Kx^(2), where Kis a constant and it is placed in a unifrom magnetic field of induction B. At t=0 a conductor of resistance per unit length lambda starts sliding up on the parabola with a constant acceleration a and the parabolic frame starts rotating with constant angular frequency omega about the axis of symmetry, as shown in the figure. Find the instantaneous current induced in the rod, when the frame turns through pi//4rad. |
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Answer» Solution :When the frame has turned through an angle `theta`, `Phi=BA costheta` where `A=int_(0)^(y)2xdy=(2)/(sqrt(k))int_(0)^(y)sqrt(y)dy=(4)/(3sqrt(k))y^(3//2)` Since `y=(1)/(2)at^(2)` `:. Phi=(B)/(3)sqrt((2)/(K))a^(3//2)t^(3)costheta` By FARADAY's law, `E_("ind")=-(dPhi)/(dt)` or `E_("ind")=(Ba^(3//2))/(3)sqrt((2)/(K))[t^(3)sintheta((d theta)/(dt))-3t^(2)costheta]` When the frame thus through `pi//4`, `t=(theta)/(omega)=(pi)/(4omega)` `E_("ind")=(Ba^(3//2))/(3)sqrt((2)/(K))t^(2)[OMEGAT sintheta-3costheta]` or `E_("ind")=(pi^(2)Ba^(3//2))/(48)sqrt((2)/(K))[(pi)/(4)-3](1)/(sqrt(2))` `E_("ind")=(pi^(2)(pi-12)Ba^(3//2))/(192omega^(2)sqrt(k))` `I=(E_("ind"))/(R )=(pi^(2)(pi-12)Ba^(3//2))/(192omega^(2)Rsqrt(k))` `R=lambda2x=lambda.2sqrt((y)/(k))=(2lambda)/(sqrt(k))sqrt((a)/(2))t=sqrt((2a)/(k))(lambda pi)/(4omega)` `I=(E_("ind"))/(R )=(pi^(2)(pi-12)Ba^(3//2)sqrt(k)4omega)/(192omega^(2)sqrt(k)sqrt(2)pilambda)=(BAPI(pi-12))/(48sqrt(2)omegalambda)` |
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| 41. |
Two wavelengths lamda_(1) and lamda_(2) are used in Young's double slit expt. lamda_(1)=430nm. What is the value of lamda_(2) if 4th bright band of one coincides with 6th bright band of the other? |
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Answer» Solution :We have `beta=(Dlamda)/d` and `beta=("DISTANCE of nth bright band")/("No. of bright BANDS")=(x("say"))/n` `:.x/n=(D lamda)/d` i.e. `x=(ND lamda)/d` Her `(n_(1)Dlamda_(1))/d=(n_(2)Dlamda_(2))/d"":.n_(1)lamda_(1)=n_(2)lamda_(2)` `lamda_(2)=(n_(1)lamda_(1))/(n_(2))=(4xx430)/6=286.7nm` |
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| 42. |
A convex lens is made of three different materials which are symmetrically distributed as shown in figure. If some point object is placed on its principal axis then what will be the number of images formed ? |
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Answer» |
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| 43. |
Consider a conductor with a spherical cavity in it. A point charge q_(0) is placed at the centre of cavity and a point charge Q is placed outside conductor. Statement-1: Total charge induced on cavity wall is equal and opposite to the charge inside Statement-2: If cavity is surrounded by a Gaussian surface, where all parts of Gaussian surface are located inside the conductor, ointvec(E)dvec(A)=0 hence q_(induceed)=-q_(0) |
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Answer» STATEMENT-1 is true, statement-2 is true and statement-2 is correct EXPLANATION for statement-1. |
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| 44. |
The electric field of a plane electromagnetic wave travelling in the positive z direction is described by |
| Answer» SOLUTION :`E = E_0 SIN (KZ- OMEGAT)` | |
| 45. |
If the external resistance is equal to the internal resistance of a cell of emf E. The p.d. across its terminal is |
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Answer» `E//2` |
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| 46. |
Lenz's law follows the principleof Conservation of |
| Answer» ANSWER :D | |
| 47. |
The radio waves of frequency 300 MHz to 3000MHz does not belong to |
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Answer» high FREQUENCY band |
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| 48. |
Light with wavelength lambda falls normally on a long rectangular slit of width b. Find the angular distribution of the intensity of light in the case of Fraunhofer diffraction, as well as the angular position of minima. |
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Answer» Solution :If a plane wave is incident normally from the left on a slit of width `b` and the diffracted wave is observed at a large distance, the resulting pattern is called Fraunhofer diffraction. The condition for this is `b^(2) lt lt l lambda` where `l` is the distance between the slit and the screen. In paractice light MAY be focussed on the screen with the HELP of a lens(or a telescope). Consider an element of the slit which is an infinite strip of width `dx`. We use the formula of problem`5.103` with the following modifications. The factor `(1)/(r )` characteristic of spherical waves will be comiited. The factor `K (varphi)` wil also be dropped if we confine overselves to not too large `barphi`. In the DIRECTION defined by the angle `varphi` the extra path difference of the wave emitted from the element at `x` relative to the wave emitted from the centre is `Delta =- sin varphi` Thus the amplitude of the wave is given by `a UNDERSET(-b//2)overset(+b//2)int e^(ik sin varphi) dx = (e^(i(1)/(2)kb sin varphi)-e^(-i(1)/(2)kb sin varphi))//ik sin varphi` `= sin((pib)/(lambda)sin varphi)/((pib)/(lambda)sin varphi)` Thus `I = I_(0)(sin^(2)alpha)/(alpha^(2))` where `alpha = (pib)/(lambda) sin varphi` and `I_(0)` is a constant Minima are observed for `sin alpha = 0` but `alpha ~~ 0` Thus we find minima at angles given by `b sin varphi = k lambda, k = +-1,+-1, +-2,+-3,........`
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| 50. |
The first & second stage of two stage rocket separately weigh 100 kg and 10 kg and contain 800kg and 90kg fuel respectively. If the exhaust velocity of gases is 2 km/sec then velocity of rocket after second stage is (nearly) (log_(10)5=0.6990) (neglect gravity) |
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Answer» `7.8 XX 10^(3) m//s` |
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