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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.
| 2551. |
Draw a schematic sketch of a cyclotron. Explain briefly how it works and how it is used to accelerate the charged particles. (i) Show that time period of ions in a cyclotron is independent of both the speed and radius of circular path. (ii) What is resonance condition ? How is it used to accelerate the charged particles? |
| Answer» SOLUTION :Construction of a cyclotron : As SHOWN in labelled diagram `D_1 and D_2` are two semi-circular disc like metal containers called dees. The straight sections of the dees are open so that the ions can move freely from `D_1 ` to `D_2` and vice-versa. The whole assembly is evacuated to AVOID collisions between the ions and the air molecules. A high frequency oscillating ELECTRIC field (shown by dot marks in figure) by using an electromagnet. | |
| 2552. |
A beam of unpolarised light having flux 10^(-3) watt falls normally on a polarizer of cross sectional area 3 xx 10^(-4) m^(2). The polarizer rotates with an angular frequency of 3.14 rad/s. The energy of light passing through the polarizer per revolution will be |
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Answer» `10^(-4)` joule |
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| 2553. |
We put an arrow mark ( to) to show the direction of current flow. Why is the current not a vector quantity? |
Answer» Solution :The arrow of current only denotes the SENSE of current in which direction the charge flows. But current does not obey the laws of VECTOR addition and hence current is not a vector.  It is a scalar only. CONSIDER a current flowing along ABCA. If current is a vector, the resultant should be zero. However, the current remains the same i. Thus, current is not a vector. |
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| 2554. |
the zeros of polynomial x^2 +16x -2 are |
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Answer» -3,4 |
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| 2555. |
Speed c of e.m. waves through vacuum is equal to : |
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Answer» `SQRT(mu_(0) epsilon_(0))` |
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| 2556. |
An object is located in a fixed position in front of a screen. Sharp image is obtained on the screen for two positions of a thin lens separated by 10 cm. The size of the images in two situations are in the ratio 3:2. What is the distance between the screen and the object ? |
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Answer» 124.5 CM `(I_(1))/(I_(2))=(3)/(2)` `(I_(1))/(I_(2))=((d+x)/(d-x))^(2)` `(3)/(2)=((d+x)^(2))/(d-x)^(2)` `(3)/(2)=(d^(2)+x^(2)+2xd)/(d^(2)+x^(2)-2xd)` `3d^(2)+3X^(2)-6xd=2d^(2)+2x^(2)+2x^(2)+4xd` `d^(2)+x^(2)-10xd=0` `d^(2)-100d+100=0` `d=99cm` |
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| 2557. |
If gravitational force suddenly disappears, the weight of the body is maximum.Is it true? |
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| 2558. |
In figure , the plates of a parallel plate capacitors have unequal charges. Its capacitance is C. P is a point outside the capacitor and close to the plate of charge -Q. The distance between the plates is d. Then |
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Answer» a point CHARGE at point P will experience electric force due to the capacitors i.`E=(2Q)/(2Aepsilon_(0))+(Q)/(2Aepsilon_(0))=(3Q)/(2Aepsilon_(0))` or `E=(3)/(2)(Q)/(cd)` or `Ed=(3Q)/(2C)=V` ii. `F=E(-Q)=((2Q)/(2Aepsilon_(0)))xx((-Q))/(1)=(Q^(2))/(Aepsilon_(0))` `=(Q^(2))/(Aepsilon_(0))` ltbgt iii. Energy `=(1)/(2)epsilon_(0)E^(2)Ad=(1)/(2)epsilon_(0)((3Q)/(2Cd))^(2)Ad=(9)/(8)(Q^(2))/(C)` |
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| 2559. |
A line charge lambda per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Fig. 6.08). A uniform magnetic field extends over a circular region within the rim. It is given by, vecB = - B_(0)hatk (r le a, aItR) = 0 (otherwise) What is the angular velocity of the wheel after the field is suddenly switched off ? |
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Answer» Solution :If the MAGNETIC field changes with TIME, then, Induced emf `varepsilon=-(dphi)/dt` This implies the presence of an electric field `vceE` tangential to all points of the periphery of the circular region denoted by a. If we move a test charge `qvarepsilon` round this periphery a work qe will be done once round it. The electric force on the test charge is q E and the work done by this force round this periphery is `(qE)(2pia)`. EQUATING the two values of work done, we have `qvarepsilon = (qE)(2pia) implies E = varepsilon/(2pia)` Substituting the value of `varepsilon`, we get `E = - 1/(2pia)(dphi_(B))/dt=-a/2 (dB)/dt [therefore phi_(B) = BA = pia^(2)B]` In the given problem, the total charge on the rim` q = lambda .2pia` Therefore, force on this charge `F = q E = -(lambda.2pia)(a/2(dB)/dt)`. According to Newton.s second law of motion, force F = M(acceleration) `= M (DV)/dt,` `-(lambda2pia)(a/2(dB)/dt)=M.d/dt(Romega)` `therefore -(lambda2pia^(2))/2.(dB)/dt=MR(domega)/dt impliesdomega=-(lambdapia^(2))/(MR)dB` On integrating, we get `omega=-(lambdapia^(2)B)/(MR) or `vec omega=-(Bpia^(2)lambda)/(MR)hatk` |
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| 2560. |
A straight wire carrying current I is made into a circular loop . If M is the magnetic moment associated with the loop , then length of the wire is |
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Answer» `SQRT((4piM)/I)` |
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| 2561. |
If the speed of a given charged particle moving on a circular path in a given magnetic field is doubled then the particle starts moving on a circular path of double radius. |
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| 2562. |
A band playing music at a frequency f is moving towards a wall at a speed v_b. A motorist is following the band with a speed v . If v is the speed of sound, the expression for the beat frequency heard by the motion is (n(V + V_(m))V_(b))/(V^(2)-V_(b)^(2)). Then n |
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| 2563. |
A faulty thermometer has it's fixed points marked 5 and 95 when this thermometer reads 68, the correct temperature in celsius is |
| Answer» Answer :B | |
| 2564. |
If the frequency in an A.C. circuit is 100 Hz, then the time taken by the voltageto cover from positive maximum to next positive maximum will be : |
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Answer» `5 xx 10^(-3) s` |
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| 2565. |
The rms value of a.c. signal in halfwave rectifier is …… |
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Answer» EQUAL to VALUE of D.C. |
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| 2566. |
Three resistorsOmega , 2Omega and3 Omega are combined in series. What is the total resistance of the combination? |
| Answer» SOLUTION :`6 OMEGA` | |
| 2567. |
When a beam of accelerated electrons collides with a target then continuous spectrum of X-rays is emitted. Which wavelength will be absent in the spectrum of X-ray emitted from X-ray tube operated at 40 kV ? |
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Answer» 1.5Å `lambda_(m) =(1240)/(40 xx 10^(3)) xx 10^(-9)` `=0.31 xx 10^(-10)=0.31 Å` Here 0.25 Å is LESS than `lambda_(m)`, THUS 0.25 Å is not emitted, |
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| 2568. |
A graph is drawn between stopping potential (y-axis) and incident frequency (x-axis). The intercept OA on x-axis gives 0.5xx10^(15) Hz. The K.E. of photo electron emitted from the metal | when the energy of incident photon is 5eV is. |
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Answer» 7 eV |
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| 2569. |
Lorentz-Fitzgerald contraction occus in "___________" direction of motion. |
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Answer» the same |
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| 2570. |
A conductor of length L is placed along the x - axis , with one of its ends at x = 0 and the other at x = L. If the rate of flow of heat energy through the conduct is constant and its thermal resistance per unit length is also constant , then Which of the following graphs is/are correct ? |
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Answer»
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| 2571. |
In 1959 Lyttleton and Bondi suggested that the expansion of the universe could be explained if matter carried a ent charge. Suppose that the universe is made up of hydrogen atoms with a number density N, which is maintained a constant. Let the charge on the proton be e_(p)=-(1+y)e where e is the electronic charge. (a). find the critical value of y such that expansion may start. ltBrgt (b) show that the velocity of expansion is proportional tot he distance from the centre. |
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Answer» SOLUTION :(a) Let us suppose that universe is a perfect sphere of radius R and its constituent hydrogen atoms are distributed uniformly in the sphere. As hydrogen atom contains one proton and one electron, charge on each hydrogen atom. `e_(H)=e_(P)+e=-(1+gamma)e+e=-Ye=(Ye)` If E is electric field intenstiy at distance R, on the surface of the sphere, then ACCORDING to gauss' theorem, `ointE.ds=(q)/(epsi_(0))` i.e., `E(4piR^(2))=(4)/(3)(piR^(3)N|Ye|)/(epsi_0)` `E=(1)/(3)(N|Ye|R)/(epsi_(0))`. . .(i) Now, suppose, mass of each hydrogen atom `congm_(P)=` mass of a proton, `G_(R)=` gravitational field at distance R on the sphere, then `-4piR^(2)G_(R)=4piGm_(P)((4)/(3)piR^(3))N` `implieG_(R)=(-4)/(3)piGm_(P)NR`. . .(II) `therefore` Gravitational force on this atom is `F_(G)=m_(P)xxG_(R)=(-4pi)/(3)Gm_(P)^(2)NR`. . .(iii) Coulomb force on hydrogen atom at R is `F_(C)=(Ye)=E=(1)/(3)(NY^(2)e^(2)R)/(epsi_(0))` . . . [From Eq. (i)] Now, to start expansion `F_(C) lt F_(G)` ad CRITICAL value of Y to start expansion WOULD be when `F_(C)=F_(G)` `implies(1)/(3)(NY^(2)e^(2)R)/(epsi_(0))=(4pi)/(3)Gm_(P)^(2)NR` `impliesY^(2)=(4piepsi_(0))G((m_(P))/(e))^(2)` `=(1)/(9xx10^(9))xx(6.67xx10^(-11))(((1.66xx10^(-27))^(2))/((1.6xx10^(-19))^(2)))=79.8xx10^(-38)` `impliesY=sqrt(79.8xx10^(-38))=8.9xx106(-19)cong10^(-18)` Thus, `10^(-18)` is the required critical value of Y corresponding to which expansio of universe would start. (b) Net force experience by the hydrogen atom is given by `F=F_(C)-F_(G)=(1)/(3)(NY^(2)e^(2)R)/(epsi_(0))-(4pi)/(3)Gm_(P)^(2)NR` If acceleration of hydrogen atom is represent by `d^2R//dt^(2)`, then `m_(P)=(d^(2)R)/(dt^(2))=F=(1)/(3)(NY^(2)e^(2)R)/(epsi_(0))-(4pi)/(3)Gm_(P)^(2)NR` `=((1)/(3)(NY^(2)e^(2))/(epsi_(0))-(4pi)/(3)Gm_(P)^(2)N)R` `therefore(d^(2)R)/(dt^(2))=(1)/(m_(P))[(1)/(3)(NY^(2)e^(2))/(epsi_(0))-(4pi)/(3)Gm_(P)^(2)N]R=alpha^(2)R`. . .(iv) where, `alpha^(2)=(1)/(m_(P))[(1)/(3)(NY^(2)e^(2))/(epsi_(0))-(4pi)/(3)Gm_(P)^(2)N]` the general solution of Eq. (iv) is given by `R=Ae^(alphat)+Be^(-alphat)` We are looking for expansion here So `B=0 and R=Ae^(alphat)` `implies` Velocity of expansion `v=(dR)/(dt)=Ae^(alphat)(alpha)=alphaAe^(alphat)=alphaR)` Hence, `v prop R` i.e., velocity of expansion is proportional to the distance from the centre. |
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| 2572. |
If critical angle for total internal reflection in diamond is 24.5^@, then refractive index of diamond is ...... [Take sin 24.5^@ = 0.4147] |
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Answer» 1.41 Refractive INDEX N = `(1)/(SINC)` `thereforen=(1)/(0.4147)` `thereforen=2.41` |
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| 2573. |
How we name the continuous locus of vibrating particles which are in the same state of vibration or phase ? |
| Answer» SOLUTION :WAVEFRONT | |
| 2574. |
A force vector applied on a mass is represented asoverset(to)(F)=6overset(^)(i)-8overset(^)(j)+10overset(^)(k)and it accelerates it at 1 ms-2 What is the mass of the body? |
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Answer» `10sqrt(2) kg` `m=sqrt(6^(2)+(-8)^(2)+10^(2))/(1)=10sqrt2kg` Hence CORRECT choice is (a). |
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| 2575. |
An electron-positron pair is produced when a gamma -ray photon of energy 2.36MeV passes close to a heavy nuclens . Find the kinetic energy carried by each particle produced , as well as the total energy with each. |
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Answer» Solution :The reaction is represented by `gamma RARR (""_(-1)e^(0)) +(""_(+)e^0)` , so that `E=m_(0)C^2 +K.E_("ELECTRON")+m_(0)C^2+K.E_("positron")` `2.36MeV = 2m_(0) .C_(2) +K.E_(("electron"))+K.E_("(positror)")` `= 1.02 MeV +K.E_((e^(-)))+K.E_((e^(+)))` `:.` Kinetic ENEGY of `(e^(-))=K.E_((e^(+)))=1/2(2.36-1.02)MeV` (K.E carried each) = 0.67 MeV (motional energy) TOTAL energy shared by each particle is obviously `m_(0)C^2+K.E=0.51MeV +0.67 MeV = 1.18MeV.` |
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| 2576. |
A thin insulated wire form a spiral of N= 100 turns carryinga current of i=8 mA . The inner and outer radii are equal to a = 5cm and b=10 cm . Find the magnetic field at the centre of the coil. |
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Answer» SOLUTION :Let n = no. of turns PER unit length along the radial of spiral. Consider a ring radii X and x + dx. No. of turns in the ring = ndx. `n = (N)/((b-a))` Magnetic FIELD at the centre due to the ring . `dB = (mu_0(ndx)i)/(2x)` So net field `B = int dB = int_(a)^(b)(mu_0nidx)/(2x) = (mu_0ni)/(2) int_a^b (dx)/(x)` `or B = (mu_0 ni )/(2) ln"" (b)/(a) or B = (mu_0Ni)/(2(b-a)) "ln"(b)/(a)` ` = (4pixx 10^(-7) xx 100 xx 8 xx 10^(-3))/(2(10 -5)xx10^(-2)) "ln"(10)/(5)` `B = 6.96 xx 10^(-6) T` |
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| 2577. |
Given that force F is given by F=Pt^(-1)+Qt. Here t is time. The unit of P is same as that of : |
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Answer» ACCELERATION |
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| 2578. |
Statement - (A) : Thermistor can have only negative temperature coefficients of resistances. Statement - (B): Thermistors with negative temperature coefficients of resistance are used as resistance thermometers, to measure low temperatures of the order of 10 K. |
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Answer» both A and B are TRUE |
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| 2579. |
how many types of inequalities are there |
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Answer» 2 |
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| 2580. |
A ball is projected horizontally from an inclined plane with a velocity V, as shown in the figure. It will strike the plane after a time |
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Answer» `(v_(0))/(sqrt3g)` |
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| 2581. |
A dip needle indicates a dip of 60^(@) at a place. When a small magnet is placed on the horizontal line through the centre of the dip needle with its north pole pointing north at a distnace 0.2m from the needle, the dip changes to 45^(@) . Find the magnetic moment of the magnet if the horizontal component of the earth's field is 0.2xx10^(-4) tesla. |
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| 2582. |
In Fig., a nonconducting rod of length L = 8.15 cm has a charge -q = -4.23 fC uniformly distributed along its length. (a) What is the linear charge density of the rod ? What are the (b) magnitude and (c ) direction (relative to the positive direction of the x axis) of the electric field produced at point P, at distance a = 6.00 cm from the rod ? What is the electric field magnitude produced at distance a = 50 m by (d) the rod and (e) a particle of charge -q = -4.23 fC that we use to replace the rod ? (At that distance, the rod "looks" like a particle.) |
| Answer» Solution :(a) `-5.19 xx 10^(-14) C//m`, (B) `4.48 xx 10^(-3) N//C`, (c ) `-180^(@)` (d) `1.52 xx10^(-8) N//C`, (E) `1.52 xx 10^(-8) N//C` | |
| 2583. |
Formation of Eddy currents has desirable effects in I Electromagnetic damping II. Transformer III. Inductothcrmy |
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Answer» All are CORRECT |
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| 2584. |
Two protons A and B move parallel to the x-axis is opposite directions with equal speed v. At the instant shown, the ratio of magnetic force andelectric force acting on the proton A is (c=speed of light in vacuum) |
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Answer» `v/C` |
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| 2585. |
What do you mean by binocularvision ? |
| Answer» SOLUTION :Observing a 3-dimensional IMAGE of an object with help of TWO eyes is called binocular vision. So with binocular visionwe can predict not only the length and BREADTH but also the HEIGHT of an object. | |
| 2586. |
Six charges are placed at the vertices of a regular hexagon as shown in the figure. The electric field on the line passing through point O and perpendicular to the plane of the figure at a distance of x (gt gt a) from O is |
Answer» SOLUTION :This is basically a problem of finding the electric field due to three dipoles. The dipole moment of each dipole is P = Q(2a) Electric field due to each dipole will be `E= (KP)/(X^(3))`  The DIRECTION of electric field due to each dipole is as shown below: `E_("net") = E + 2E cos 60^(@)= 2E` `=2 ((1)/(4pi epsi_(0))) (2Qa)/(x^(3))= (QA)/(pi epsi_(0)x^(3))` |
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| 2587. |
Which of the two AM or FM is preferred for high fidelity reception? |
| Answer» SOLUTION :FREQUENCY MODULATION is prefered for high FIDELITY RECEPTION. | |
| 2588. |
A point source is sinking down in a liquid of refractive index u with a constant velocity . Find the rate of change of the area through which light will escape from the liquid when the source is at a depth h_0 |
| Answer» SOLUTION :`(2 pih_0v_0)/( (mu^2 -1))` | |
| 2589. |
A sample of gas expands from volume to V_1 to V_2 The amount of work done by the gas is maximum when the expansion is |
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Answer» Adiabatic |
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| 2590. |
A galvanometer having 50 divisions provided with a variable shunt s is used to measure the current when connected in series with a resistance of 90Omega and a battery of internal resistance 10Omega . It is observed that when the shunt resistance are 10Omega,50Omega, respectively the deflection are respectively 9 & 30 divisions . What is the resistance of the galvanometer ? Further if the full scale deflection of the galvanometer movement is 300 mA , find the emf of the cell. |
| Answer» SOLUTION :`2.33.3Omega144V` | |
| 2591. |
A body is released from the top of a tower of height metre. It takes time T seconds to reach the ground. Where is the body at time T/2seconds? |
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Answer» At `H/2` m from the ground In timeT/2`h=(1)/(2)g. (T^(2))/(4)=(1)/(2)g.(1)/(4),(2h)/(g)=(h)/(4)` from top `:.` Body is at `(h-(h)/(4))=(3h)/(4)` from BOTTOM |
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| 2592. |
Suppose now that the fish sees the boy. Find (a) apparent distance of eye as seen by fish. (b) apparent distance of image of eye as seen by fish. |
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Answer» |
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| 2593. |
If electric dipole is placed in non-uniform electric field, then...... |
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Answer» RESULTANT force on dipole is 0. Knowledge BASED QUESTION. |
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| 2594. |
Magnetic field lines |
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Answer» cannot INTERSECT at all |
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| 2595. |
Statement I : To float, a body must displace liquid whose weight is grater than weight of body itself. Statement II : In case of floating, the body will experience no net downward force. |
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Answer» STATEMENT-I is true, Statement-II is true and Statement-II is CORRECT EXPLANATION for Statement-I. |
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| 2596. |
A ball is dropped from tower of height 45 m. At the same moment another ball is thrown up with a speed of 40 ms from base. What is the relative speed of the balls as a function of time? |
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Answer» The relative speed with RESPECT to TIME is zero |
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| 2597. |
A radar operates at wavelength 50.0 cm. If the bbeat frequency between the transmitted signal and the signla reflected from aircraft (Deltav) is 1 kHz, then velocity of the aircraft will be : |
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Answer» `800 (km)/(H)` `v. = (C + a)/(c) v` when surce moves towards observer at rest `v.. = (c)/(c-a) v. = (c + a)/(c - a) v = [1 + (a)/(c)]/(1 - a/c) v` `= (1 + (a)/(c))(1 - (a)/(c))^(-1) v` `therefore Deltav = v.. - v = (2av)/(c) = (2A)/(lambda)` `therefore a = lambda .(Deltav)/(2) = 250 m s^(-1) = 900(km)/(h)`. |
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| 2598. |
A critical angle for a medium is 60^@. Then the refractive index of the medium will be |
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Answer» `sqrt3` |
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| 2599. |
In a YDSE, the central beight fringe can be indentified: |
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Answer» as it has GREATER intensity than the other BRIGHT fringes |
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| 2600. |
In experiment for measuring surface tension by capillary rise method, reading for positions A, B, C and D for internal diameter of capillary tube are given as under Mean internal radius of capillary is |
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Answer» `0.002 CM` |
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