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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.
| 38751. |
State Bragg's law. |
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Answer» Solution :Definition : Gauss.s law STATES that if a charge Q is enclosed by n arbitrary CLOSED surface, then the total electric flux `phi_(E)` through the closed surface is `phi_(E) = oint vec(E).d vec(A) = Q_("encl")/(epsilon_(0))` |
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| 38752. |
A uniform thin ring of radius R and mass m suspended in a vertical plane from a point in its circumference. Its time period of oscillation is |
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Answer» `2PI SQRT(R/g)` |
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| 38753. |
State de-Broglie relation for wavelength of matter waves. Show that the de-Broglie wavelength of electrons accelerated through a potential of V volt can be expressed as : lambda = h/sqrt(2meV) |
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Answer» Solution :The de-Broglie wavelength associated with a material particle of mass m moving with velocity V is given by `lambda = h/(mv)` Wavelength associated with a moving electron. Suppose an electron STARTING from rest falls through a potential difference V Let charge on electron = e Mass of electron = m, Velocity acquired by the electron in FALLING through a potential difference, V = v Work done on the electron = EV Kinetic energy of the electron `=1/2mv^(2)` OBVIOUSLY, `1/2mv^(2) = eV` or `v = sqrt((2EV)/m)`........(1) It `lambda` is the wavelength associated with the electron, then according to de-Broglie wave equation. `lambda =h/(mv)` or `lambda =h/(msqrt((2eV)/m)) =h/sqrt(2meV)` |
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| 38754. |
Assertion: For best contrast between maxima and minima in the interference pattern of young's double-slit experiment the intensity of light emerginig out of the two slits should be equal. Reason: The intensity of light is proportional to square of its amplitude. |
| Answer» SOLUTION :If `I_(1)=I_(2)=I(say)` than `I_(max)=4I and I_("min")=0`, so the contrast will be BEST. | |
| 38755. |
In a standing wave pattern obtained in an open tube filled with Iodine, due to vibrations between first node, eleventh node is found to be lm when the temperature of iodine vapour is 352°C. If the temperature is 127°C, the distance between consecutive nodes is (in centimeters) (approximately). |
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| 38756. |
Which of the following processes is reversible ? |
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Answer» Transfer of heat by RADIATION So correct CHOICE is (C ). |
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| 38757. |
Select incorrect unit of self inductance |
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Answer» Mho-second From `E=-L , (dI)/(dF) rArr L=-epsilon/(dI//dF)` unit of L `"Volt-sec"/"Amp"` `"Volt"/"Amp"`=ohm Unit of L = ohm.sec `THEREFORE` mho-sec is not unit of SELF inductance. |
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| 38758. |
The front wall of a drawer in a cabinet is a provider with two symmetrical handles. The distance between the handles is l and the length (i.e. depth) of the drawer is a. The maximum value of the coefficient of friction between drawer and cabinet, for which drawer can be pulled out by applyinga force on one handle perpendicular to the face of the drawer is : (Neglect the weight of the drawer) |
| Answer» Answer :A | |
| 38759. |
(a) State Huygen's principle . Using this principle draw a diagram to show how a plane wave front incident at the interface of the media gets refracted when it propagates from a rarer to a denser medium. Hence verify Snell's law of refraction. (b) When monochromatic light travels from a rarer to a denser medium, explain the following, giving reasons- (i) Is the frequency of reflected and refracted light same as the frequency of incident light? (ii) Does the decrease in speed imply a reduction in the energy carried by light wave? |
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Answer» Solution :(a) Huygen's Principle : It is based on the two assumptions: (i) Each point on the primary wave FRONT acts as a source of secondary wavelets, sending out disturbance in all directions in a similar manner as the original source of light does. (ii) The new position of the wave front at any instant is the envelope of the secondary wavelets at that instant. Refraction On the Basis of Wave Theory (i) Consider any point Q on the incident wave front. (ii) Suppose when disturbance from point P on incident wave front reaches point P' on the refracting surface XY. (iii) Since, P'A' representsthe refracted wave front, the time taken by lightto travel from a point on incident wave front to the corresponding point on refracted wave front should always be the same. Now, time taken by light to go from Q to Q' will be `t=(QK)/(c)+(KQ')/(v) "" `...(i) In right angled`Delta AQK, lt QAK =i` `therefore QK=AK sin i "" `...(ii) In right - angled `Delta P'Q'K, lt Q'P'K= r and KQ' =KP' sin r."" ` ...(iii) Substituting (ii) and (iii) in equation (i), we get `t=(AK sin i)/(c) +(KP'sin r)/(v)` `" Or" t=(AKsin i)/(c) =((AP'-AK)sinr)/(v) "" (thereforeKP'=AP'-AK)` ` "Or" t = (AP')/(c) sin r+AK((sinr)/(c)-(sinr)/(v)) "" ` ...(iv) The rays from different point on the incident wave front will take the same time to reach the corresponding points on the refracted wave from i.e., t given by equation (iv) is independent of AK. It will happen so, If`(sin i)/(c) - (sin r)/(v)=0 rArr (sin i)/(sin r) =(c)/(v)=mu =(sin i)/(sin r)` This is the Snell's Law for refraction of light. (b) (i) The FREQUENCY of reflected light remains same as the frequency of incident light. Frequency only depends on the source of light. (ii) Since, the frequency remains same, HENCE there is no reduction in energy. 
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| 38760. |
A steel wire of length (l) has a magnetic moment (M). It is then bent into a semicircular arc. The nem magnetic moment is : |
| Answer» ANSWER :B | |
| 38761. |
For an em wave, what is the relation between amplitude of electric and magnetic field in free space? |
| Answer» SOLUTION :They are TRANSVERSE in NATURE. | |
| 38762. |
The period of revolutions of Mars around th esun is 1.881 year and its mean distance from the sun is 1.524AU. Calculate the mass of sun (1AU=1.496xx10^(11)m). |
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| 38763. |
A simple harmonic oscillator has an amplitude A and time period T, the time required by it to travel from x = A to x=(A)/(2) is : |
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Answer» `(T)/(6)` `(A)/(2)=A cos omega timpliesomega t=pi//3`. `(2pi)/(T).t=pi//3impliest=T//6` Hence correct CHOICE is (a). |
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| 38764. |
A hole in a p-type semiconductor is ……. in a covalent bond. |
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Answer» an excess ELECTRON |
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| 38765. |
A ray of light passesthrough a glass slab of thickness t and refractive index mu. If velocity of light in vacuum is c then how much time will the light take to emerge from the slab? |
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Answer» Solution :VELOCITY of light in glass, `V = ( c )/(MU)` `THEREFORE "To overcome the glass slab of thickness t, time taken by light "` `= (t)/(v) = (t)/((c)/(mu))= (mut)/(c)` |
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| 38766. |
A body of mass m falls from a height h on a pan ( of negligiblemass ) of spring balance as shwon. The spring is ideal and has a spring contant k. Just after striking the pan, the body sticks with the pan and starts oscillatory motion in vertical direction. The velocity of the body at mean position will be |
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Answer» `SQRT( 2mg^(2) K +2gh ) ` |
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| 38767. |
Where on the earth.s surface is the magnetic dip zero ? |
| Answer» Solution :The MAGNETIC DIP is zero at magnetic equator on the earth SUFACE. | |
| 38768. |
In the given circuit, When switch S is opencd, the plates are uncharged and are separated by distance d=8.00mm. Battery is of 100V and springs are identical with spring constant K. Capacitance of capacitor with separation d=8mm is 2muF . When switch S is closed, the distance between plates is reduced to 4mm. Find the sum of spring constants of springs. |
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| 38769. |
The dominant mechanisms for motion of charge carriers in forward and reverse biased silicon P-N junction are |
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Answer» Drift in forward BIAS, DIFFUSION in REVERSE bias |
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| 38770. |
Unstable fission fragments decay by emitting neutrons and electrons, neutrons so emitted are called |
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Answer» PROMPT NEUTRONS |
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| 38771. |
Radius of curvature of a concave mirror is 20 cm, then its focal length is ...... cm. |
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Answer» - 20 cm R=2f `THEREFORE` 20=2f `therefore` F = -10 cm |
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| 38772. |
A 50 kg block rests on a rough horizontal surface. A force of 200 N is applied on the body. The block acquires a speed of4 m//s, starting from rest in 2 seconds. The value of coefficient of friction is |
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Answer» 0.38 |
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| 38774. |
The frequency of a FM transmitter without signal input is called |
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Answer» LOWER SIDE BAND frequency |
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| 38775. |
In photoelectric effect, the number of emitted photoelectrons is proportional to _____ of incident light. |
| Answer» SOLUTION :INTENSITY | |
| 38776. |
Derive mirror equation for a convex mirror. Using it, show that a convex mirror always produces a virtual image, independent of the location of object. |
Answer» Solution : Deduction of MIRROR formula `(1)/(v) + (1)/(u) = (1)/(f)` For a convex mirror f is always +ve. `therefore f gt c` Object is always placed in front of mirror HENCE `u lt 0` (for real object) `(1)/(v) + (1)/(u) = (1)/(f)` `rArr (1)/(v) = (1)/(f) -(1)/(u)` `"As" u lt 0 u -ve " hence"` `(1)/(v) gt 0` `rArr v gti.e.+ve " for all values of u. "` image will be formed behind the mirror and it will be VIRTUAL for all values of u. |
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| 38777. |
Consider the circuit given here. The potential difference V_(BC) between the points B and C is |
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Answer» `1V`  currentflowingin the circuit `i=(V )/( R_1+R_2 +R_3 ) = (3)/( (1 + 2+3 ) xx 10^3 ) = (3)/( 6 xx 10^3 )` `impliesi= 0.5xx 10^(-3)` A Potentialdropacrossthe armAD `V_(AD )= IR = 0.5xx 10^(-3)xx 3 xx 10^(-3) = 1.5 V ` thechargeis GIVENBY` Q=CV= ((1 xx 2 )/( 2 +1)) xx 1.5 ` `= 2/3xx 1.5= 1 mu C ` potentialatpointB ` V_B= V_(AD) -V_(AQ) ` ` = 1.5V -( 00.5m Axx 1 K Omega) =1 V ` Potentialatpoint` C , V_C= V_(AD) -V_(AP)` `=1.5V -(1 mu C )/( 1 mu F) = 1.5V- = 0.5V ` ` thereforeV_B-V_C- 1V- 0.5V = 0.5V ` |
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| 38778. |
Statement-I : In S.H.M., the velocity is maximum when acceleration is minimum. Statement-II : Displacement and velocity of S.H.M. differ in phase by pi//2. |
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Answer» STATEMENT-I is true, Statement-II is true and So correctchoice is (B). |
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| 38779. |
Using photon picture of light, show how Einstein's photoelectic equation can be established. Write two features of photoelectric effect which cannot be explained by b weve theory. |
Answer» Solution : Photolectons are EMITTED from plate P and are attracted towards C constituting photo current. He foundcurrent is a function of potenial difference `V_(CP)` for constant FREQUENCY and different intensity. If `V_(CP)`is large and + ve , the current becomes constant. The relation hv= `Phi_(0)+KE` (i) Weve theoty states the KE of photoelectrons sholudincrerase with intensity but `K_(max) = eV_(0)`it is indepandent of intensity. (ii) Weve theory STATE photoeletron should occure for any frequency of light, contrary, photoemission is possible for threshold and higher frequencies. |
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| 38780. |
A concave lens is in contact with convex lens. The ratio of magnitude of their power is 2/3. Their equivalent focal length is 30 cm. What are their individual focal lenghts ? |
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Answer» `- 75, 50` `therefore "" f_(2) = - 2//3 f_(1)` `(1)/(f_(1)) + (1)/(f_(2)) = (1)/(30)` Solving we get `f_(1) = - 15 CM , f_(2) = 10 cm` |
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| 38781. |
If an electron in hydrogen atom jumps from an orbit of level n= 3 to an orbit of level n =2, the emitted radiation has a frequency (R= Rydberg constant, c= velocity of light) : |
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Answer» `(RC)/(25)` `c/lambda=Rc (1/n_(f)^(2)-1/n_(i)^(2))` `v=Rc (1/2^(2)-1/3^(2))` `=Rc (1/4-1/9)` `v=5/(36)Rc`. |
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| 38782. |
At which angular frequency of source, current in series LCR A.C. circuit would be maximum ? |
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Answer» `sqrt(LC)` `|Z|=sqrt(R^2+(X_L-X_C)^2)`= minimum `rArr X_L-X_C=0` `rArr X_L=X_C` SUPPOSE above condition is satisfied at `omega=omega_0` then `omega_0L=1/(omega_0C)` `THEREFORE omega_0^2=1/(LC)` `therefore omega_0=1/sqrt(LC)` Above is the expression of series resonant angular frequency . |
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| 38783. |
The volume of a cube in m^(3) is numerically equal to its surface area in m^(2). The volume of cube is : |
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Answer» `1000m^(3)` `:.` Volume`=216m^(3)` or `(d)` is the answer. |
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| 38784. |
An alternating current emf device has a smaller resistance than that of the resistive load, to increase the transfer of energy from the device to the load, a transformer will be connected between two.Then |
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Answer» `N_(S)` should be greater than `N_(P)` |
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| 38785. |
What is peak value of a 240 volt AC supply ? |
| Answer» SOLUTION :`I_0=240 SQRT 2 VOLT`. Or `I_0 = 339 volt`. | |
| 38786. |
Four metallic plates each with a surface area of one side A are placed at a distance d from each other as shown. Then, the capacitance of the system between X and Y is |
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Answer» `(2 epsi_(0)A)/(d)` |
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| 38787. |
There are two radio nuclei A and B. A is an alpha emitter and B a beta emitter. Their disintegration constants are in the ratio of 1:2. What should be the ratio of number of atoms of A and B at any time t so that probabilities of getting alpha and beta particles are same at that instant ? |
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| 38788. |
(A) : Magnetic lines forms closed loops in nature. (R) : Mono-magnetic pole does not exist in nature. |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the correct EXPLANATION of 'A'. |
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| 38789. |
An ideal gas undergoes a circular cycle as shown in the figure. Find the ratio of maximum temperature of cycle to minimum temperature of cycle. |
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Answer» `((1+sqrt(2))/(sqrt(2)-1))^(2)` At point `A` `(P_(A))/(P_(B))=(V_(A))/(V_(B))=2+cos45^(@)=(2sqrt(2)+1)/(sqrt(2))=(4+sqrt(2))/(2)` `nRT_(A)=P_(A)V_(A)=((4+sqrt(2))/(2))^(2)P_(0)V_(0)` Similarly at point `B` `(P_(B))/(P_(0))=(V_(B))/(V_(0))=2-cos45^(@)=(4-sqrt(2))/(2)` `nRT_(B)=P_(B)V_(B)=((4-sqrt(2))/(2))^(2)P_(0)V_(0)` `rArr (T_(A))/(T_(B))=((4+sqrt(2))/(4-sqrt(2)))^(2)` 
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| 38790. |
If the air resistance causes a vertical retardation of 10% of value of acceleration due to gravity, then the time of flight of an oblique projectile will be decreased by nearly: |
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Answer» 0.06 and `t.=(2vsintheta)/(g + (gxx10)/100)=(2vsintheta)/(11/10.g)` or `t/(t.)=11/10` Then %AGE INCREASE =`(t-t.)/txx100` `=(11-10)/11xx100approx9%` |
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| 38791. |
Both light and sound waves produce diffraction. It is more difficult to observe the diffraction with light waves because |
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Answer» Light WAVE do not REQUIRE medium |
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| 38792. |
In which of the following photocell is not used ? |
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Answer» Burglar ALARM |
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| 38793. |
According to Bohr's theory of hydrogen atom, total energy of electron in a stationary orbit is E=-13.6/(n^(2)) eV, Where n is the number of orbit. Clearly, total energy of electron in a stationary orbit is negative, which means the electron in bound to the nucleus and is not free to leave it. An n increases, value of negative energy decreases, i.e., energy is progessively larger in the outer orbits. Read the above passage and answer the following questions: (i) What is total energy of electron in ground state of hydrogen atom? What does it imply? (ii) Energy required to remove an electron is smaller when atom is in any one excited state. Comment. (ii) How is this concept translated in day to day life? |
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Answer» Solution :(i) for ground state, N=1 `:. E=(-13.6)/(n^(2))eV=(-13.6)/(1^(2)) eV=-13.6eV` It implies that 13.6eV energy is required to remove an electron form hydrogen atom in its groud state. (ii) In first excited state, n=2, `:. E=-13.6/(2^(2))eV=-3.4eV` It means that energy required to remove an electron an electron form hydrogen atom in first excited state is 3.4eV, which is less than 13.6eV. Therefore, the statement is true. (iii) Negative energy of electron indicates that it is bound to the nucleus and cannot leave it until energy equal to its negative energy is SUPPLIED form outside. The same is true in day to life. A person intending to leave the country has to show that nothing is pending against himin a COURT of law, and he has cleared income tax/sales tax payments due form him. A person formwhom why type of payment is due, is not FREE to leave the country. HEIS bound till he clears all his dues. |
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| 38794. |
An-type and P-type silicon can be obtained by doping pure silicon with |
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Answer» ARSENIC and PHOSPHOROUS |
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| 38795. |
In Van der wal's equation (p+a/V^2)(V-b)=RT dimension of 'a' will be |
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Answer» `ML^5T^-2` |
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| 38796. |
How does a charge q oscillating at certain frequency produce electromagnetic waves ? Sketch a schematic diagram depicting electric and magnetic fields for an electromagnetic wave propagating along the x-direction. |
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Answer» Solution :A CHARGE Q OSCILLATING at a certain frequency produces an oscillating electric field in space, which produces an oscillating magnetic field, which in turn is a source of oscillating electric field and so on. The oscillating electric and magnetic fields thus REGENERATE each other i.e., an electromagnetic wave propagates through the space. The frequency of the electromagnetic wave EQUALS the frequency of oscillation of the charge. Schematic diagram depicting electric and magnetic fields for an e.m. wave propagating along the x-direction is given in Fig. |
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| 38797. |
When an X-ray tube is operated with an accelerating potential difference V, the cutoff wavelength is proportional to |
| Answer» ANSWER :A | |
| 38798. |
A hole is drilled along the diameter of the earth and pen is dropped into it. The time taken by it is reach other end of the earth is |
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Answer» 162.2 MIN |
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| 38799. |
For the explanation of reflection of light by Newtons, it is assumed that the direction of force exerted by reflecting surface is along : |
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Answer» REFLECTED ray |
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| 38800. |
What is the difference between light waves and matter waves ? |
| Answer» Solution :LIGHT waves are ELECTROMAGNETIC waves, matter waves are probabilities waves. In VACUUM light waves of DIFFERENT WAVELENGTH all have the same velocity . But , the velocity of matter waves of different wavelengths are different. | |