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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.
| 6351. |
एक दर्पण से किसी भी दूरी पर रखी एक वस्तु का सीधा प्रतिबिंब ही प्राप्त होता है दर्पण हो सकता है। |
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Answer» समतल |
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| 6352. |
Two capacitors 3muF and 6muF are first connected in series with a battery of emf 240V. The connection is broken and then connected in parallel. Find the final charges on the two capacitors. |
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| 6353. |
Three point charges are placed at the corners of an equilateral triangle . Assume that only electrostaticforcesare acting . |
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Answer» The SYSTEM will be in equilibriumif the chargeshave the same magnitude but not all have the same sign. |
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| 6354. |
A metal disc rotates freely, between the poles of a magnetic in the direction indicated. Brushes P and Q make contact with the edge of the disc and the metal axle. What current, if any, flows through R? |
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Answer» <P>a CURRENT from `P` to `Q` 
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| 6355. |
STATEEMNT-1 : Work done by frictional force is always either negative or zero. because STATEMENT-2 : Work done by the static friction on a system may be zero. |
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| 6356. |
The minimum velocity (in ms^-1) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction0.6 to avoid skidding is |
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Answer» 60 |
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| 6357. |
When light travels from one medium to the other, whose refractive index is different, then which of the following will change |
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Answer» frequency, WAVELENGTH and speed. |
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| 6358. |
By photoelectric effect, Einstein proved |
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Answer» `E = h upsilon` |
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| 6359. |
Blowing air with open mouth is an example of |
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Answer» isobaric process |
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| 6360. |
A beam of light of wavelenght 600 nm from a distant source falls on a single slit 1.00 nm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between the first dark fringes on either side of the central bright fringe is |
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Answer» 1.2 cm `=(2Dlambda)/(d)=(2xx2xx600xx10^(-9))/(1.00xx10^(-3))=2.4xx10^(-3)m=2.4mm` |
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| 6361. |
Explain why high frequency carrier waves are needed for effective tranmission of signals. |
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Answer» Solution :(i) To reduce the length of transmitting antenna. To INCREASE the power of antenna. To CONVERT a WIDE band signal into narrow band. To transmit SIMULTANEOUSLY a large number of audio, SIGNALS from the same antenna without getting them mixed up. |
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| 6362. |
The bending of beam of light around corners of obstacles is called. |
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Answer» reflection By DEFINITION |
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| 6363. |
70 Calories of heat are required to raise the temperature of 2 moles at constant pressure from 30^(@)C" to "35^(@)C, what is the amount of heat required in calories to raise the same amount through same range at contant volume ? (gamma=1.4) : |
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Answer» 70 Now `nC_(p) xx DELTA T=70,` `nxxC_(v)xxDelta T =Q` `therefore (Q)/(70)=(C_(p))/(C_(v))=(10)/(14)" or "Q=(70xx10)/(14) =50` cals. `therefore` Correct choice is (c ). |
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| 6364. |
A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam coverage if the lens isa concave lens of focal length 16 cm? |
| Answer» Solution : REAL, at 48 cm same SIDE as P | |
| 6365. |
A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam coverage if the lens isa convex lens of focal length 20 cm. and |
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Answer» <P> SOLUTION : REAL, at 7.5cm same SIDE as P, |
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| 6366. |
Two ice cubes of side 10cm, having cavity of volume 20cm^(3) at centre of cube but filled with different materials A and B respectively. The specific gravity of material B is 1.9 Now these cubes are placed in two different vessels of same base area as shown in figure. The water level before putting blocks in vessels are same Assume that ice melts uniformly from all sides and with same constant rate in both the vessels. (specific gravity of ice=0.9) Choose the correct statement: |
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Answer» Both CUBES sink simultaneously |
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| 6367. |
Infinite charges 'q' each are placed at x = 1, 2, 4, ... Find the electric potential at the origin. |
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Answer» |
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| 6368. |
Given fig. shows a coil bent with all edges of length 1m and carrying a current of 1A. There exists in space a uniform magnetic field of 2T in the positive y-direction. Find the torque on the loop. |
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Answer» SOLUTION :The forces on the edges FG and BC are zero. The forces on the other edges are : `F_(AB)=Bil(HATI)` `F_(CD)=Bil(-hati)`, `F_(HG)=Bil(-hati)` `F_(EF)=Bil(hati)`,  The torque of forces `F_(AB)` and `F_(HG)` is zero and that of `F_(CD)` and `F_(EF)` is zero. The forces on edges AH and DE are equal and OPPOSITE and so constitutes a net torque. Thus `F_(AH)=Bil(-hatk)` and `F_(DE)=Bil(hatk)` Torque `=Bil^(2)=2.1.1hati=2hatiN-m` |
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| 6369. |
Two ice cubes of side 10cm, having cavity of volume 20cm^(3)at centre of cube but filled with different materials A and B respectively. The specific gravity of material B is 1.9 Now these cubes are placed in two different vessels of same base area as shown in figure. The water level before putting blocks in vessels are same Assume that ice melts uniformly from all sides and with same constant rate in both the vessels. (specific gravity of ice=0.9) Find the ratio of initial submerged volumes of the blocks containing A and B respectively |
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Answer» `49/46` `v_(2)rho_(w)g=(v-Deltav)rho_(i)g+Deltav(rho_(m))g` `v_(1)/v_(2)=((v-Deltav)rho_(i)+Deltavrho_(m))/((v-Deltav)rho_(i)+Deltavrho_(m))=((v/(Deltav)-1)rho_(i)+rho_(m))/((v/(Deltav)-1)rho_(i)+rho_(m))` `=((1000/20-1)xx0.9+4.9)/((1000/20-1)xx0.9+1.9) =49/46` |
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| 6370. |
A block of mass 1 kg is connected with a smooth plank of the same mass is performing oscillations. The value of the spring constant is 200 N m^(-1)The block and the plank are free to move and there is no friction anywhere. The angular frequency of the oscillation is Omegarad s^(-1)Find the value of omega |
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| 6371. |
The given table is for {:("Input","Output"),(A""B,Y),("00",1),("01",1),("10",1),("11",0):} |
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Answer» OR GATE i.e., when A = 0 , B = 0 , then y = `bar(0.0)= bar0=1` `A=0,B=1y=bar(0.1)=bar0=1` `A=1,B=0y=bar(1.0)=bar0=1` `A=1,B=1,y=bar(1.1)= bar1=1` which is according to given TRUTH table. |
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| 6372. |
Rainbow is formed due to a combination of : |
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Answer» REFRACTION and absorption |
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| 6373. |
A flat insulating disc of radius 'a' carries an excess charge on its surface is of surface charge density sigma C//m^(2). Consider disc to rotate around the axis passing through its centre and perpendicular to its plane with angular speed omega rad/s. If magnetic field vecB is directed perpendicular to the rotation axis, then find the torque acting on the disc. |
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Answer» Solution :Suppose the DISC is PLACED in xy-plane and is rotated about the z-axis. Consider an annular ring of radius r and of thickness dr, the charge on this ring `DQ=sigma(2pirdr)` As the ring rotates with angular VELOCITY `omega`, so the CURRENT `i=(dq)/(dt)=(sigma(2pidr))/((2pi)/(omega))=sigmaomegardr` The torque on the current loop `vectau=ivecAxxvecB`.  Hence the torque on this annular ring `dvectau=i(dvecAxxvecB)` `=sigmaomegardr(pir^(2)Bsin90^(@))=pisigmaomegar^(3)Bdr` and `tau=pisigmaomegaBint_(0)^(a)r^(3dr)=(pisigmaomegaBa^(4))/4` |
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| 6374. |
The condition for obtaining secondary maxima in the diffraction pattern due to single slit is… |
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Answer» `a sin THETA= n lamda` NOTE: In the QUESTIONS they should give the values of n as n=1,2,3…… |
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| 6375. |
The angle between the magnetic meridian and geographical meridian is called |
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Answer» ANGLE of dip. |
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| 6376. |
Calculate the frequency of the 5th overtone, if the pipe is closed at one end. The fundamental frequency is 250Hz.(Neglect end correc tion). |
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Answer» 2750Hz |
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| 6377. |
(A):The v-t graph perpendicular to time axis is not possible in practice (R ):Infinite acceleration can be realized in practice |
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Answer» |
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| 6378. |
Explain briefly the following terms used in communication system: Transducer |
| Answer» Solution :TRANSDUCER: An energy transformation DEVICE. In communication systems, it is used to convert the electrical energy from the modulator into the energy CARRIED by ELECTROMAGNETIC waves. Also it produces the reverse transformation at the RECEIVING end. these transducers are essential parts of this transmitting and receiving antennas. | |
| 6379. |
निम्नलिखित में किसका दशमलव प्रसार सांत है? |
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Answer» 15/1600 |
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| 6380. |
Assertion : Electrons in the atom are hold due to coulomb forces. Reason : The atom is stable only because the centripetal force due to coulomb's law is balanced by the centrifugal force. |
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Answer» Assertion and Reason are CORRECT and Reason is the correct explanation of Assertion. |
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| 6381. |
Vehicles carrying inflammable materials usually have chains that hang down and drag on the ground. Why? |
| Answer» Solution :When a vehicle is in motion, its tyres RUB against the road and get charged due to friction. Further, due to friction of air, the BODY of the vehicle also get charged. If the accumulated charge BECOMES excessive, sparking may occur and the inflammable material may catch fire. Since the chain. ropes are touching the ground, the charge leaks to the earth. Hence, the DANGER of fire is avoided. | |
| 6382. |
Apply Biot-Savart law to find the magnetic field due to a circular current carrying loop at a point on the axis of the loop. |
Answer» Solution :1. Consider current carrying loop with radius R as SHOWN in figure.  2. Centre point of loop is O and loop lies on X-axis. 3. We wish to calculate the magnetic field at the point P on this axis. 4. Loop is kept perpendicular to the plane of loop. 5. Let x be the distance of P from the centre o of the loop. 6. The magnitude `dvecB` of the magnetic field due to `dvecl` is given by the Biot-Savart law, `dvecB=mu_(0)/(4pi)(Ivec(dl)xxvecr)/r^(3)` 7. Magnitude of this magnetic field is given by, `|dvecB|=mu_(0)/(4pi)(|Ivec(dl)xxvecr|)/r^(3)` = `mu_(0)/(4pi)(Idlrsintheta.)/r^(3)` where `theta.` = angle between `vec(dl)andvecr" but "vec(dl)_|_vecr` so, `sintheta.="sin"pi/2=1` `therefore|dvecB|=(mu_(0)I)/(4pi)(dl)/r^(2)""...(1)` 8. As shown in diagram `r^(2)=x^(2)+R^(2)` `|dvecB|=mu_(0)/(4pi)(dl)/((x^(2)+R^(2)))""...(2)` 9. Direction of `dvecB` is given by perpendicular to plane of `vec(dl)andvecr`. 10. `dvecB` can be divide in two components, parallel component `dB_(x)=dBsintheta` perpendicular component `dB_(_|_)=dBcostheta` 11. When the components perpendicular to the ring axis are summed over, they cancel out and we obtain null result. Here, `dB_(_|_)=dBcostheta` component due to `vec(dl)` is cancelled by the contribution due to the diametrically opposite `vec(dl)` element shown in figure. Thus, only the parallel component survives. 12. The net contribution along X-direction can be obtained by integrating `dB_(x)=dBsintheta` over the loop, `dB_(x)=dBsintheta""...(3)` From diagram geometry, `sintheta=R/((x^(2)+R^(2))^(1//2))""...(4)` 13. Substitute dB and `costheta` in equation (3) `dB_(x)=mu_(0)/(4pi)(IDL)/((x^(2)+R^(2)))*R/((x^(2)+R^(2))^(1//2))` `thereforedB_(x)=mu_(0)/(4pi)(IdlR)/((x^(2)+R^(2))^(3//2))""...(5)` 14. To integrate this term, we can get total magnetic field in x - direction. `B=intdB_(x)` `B=(mu_(0)IR)/(4pi(x^(2)+R^(2))^(3//2))intdl` `B=(mu_(0)IR)/(4pi(x^(2)+R^(2))^(3//2))(2piR)" "[becauseintdl=2piR]` `thereforeB=(mu_(0)IR^(2))/(2(x^(2)+R^(2))^(3//2))""...(6)` 15. Vector form of magnetic field, `vecB=(mu_(0)IR^(2))/(2(x^(2)+R^(2))^(3//2))hati" "[becausehati" is UNIT vector of x-axis"]` |
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| 6384. |
Calculate the photoelectric workfunction of a metal whose threshold wavelength is 5200 Å. |
| Answer» SOLUTION :`3.822xx10^(-19)` | |
| 6385. |
What is the nature of electromagnetic waves ? |
| Answer» SOLUTION :The WAVES are transverse in nature.i.e. the electric and magnetic FIELDS are perendicular to the DIRECTION of PROPAGATION. | |
| 6386. |
A conductor of length l is connected to a.d.c, source of potential V. If the length of the conductoris tripled by stretchingit, keeping V constant, explain how do the following factors vary in the conductor ? (i) Drift speed of electrons (ii) Resistance (iii) Resistivity. |
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Answer» Solution :When the WIRE is stretched to thrice its LENGTH the area is reduced to thrice its length, the area is reduced to ONE - thrid. `:.` Let `l_1 = l, l_2 = 3l, A_1 = A, A_2 = A//3` Now `R_1 = rho(l_1)/A_1 and R_2 = rho (l_2)/A_2` `R_2/R_1 = (l_2/l_1) xx(A_1/A_2) = 3 xx 3 = 9` or `R_2 = 9R_1` i.e. resistance will become 9 times. Resistivity depends upon the nature of the material of the conductor and TEMPERATURE. It does not depend upon the dimensions of the wire. Since meterial of the wire and the temperature remain constant,HENCE resistivity will remain constant. Drift vel., `v_d = (eEtau)/m = (eV tau)/(ml)` Since, e, V, `tau` and m are constant, hence when lenght is tripled, the drift velocity becomes `1/3rd` that of the original length. |
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| 6387. |
Statement-1 : Two tuning forkes having frequencies 410 Hz and 524 Hz are kept closed and made to vibrate. Beats will not be heard. because Statement-2 , Sound waves superimpose only when the frequencies of superposing waves are equal or nearly equal. |
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Answer» Statement -1 is True, statement -2 is True, Statement -2 is a CORRECT EXPLANATION for Statement -1. |
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| 6388. |
The LC product of a tuned amplifier circuit require to generate a carrier wave of 1 MHz for amplitude modulation is |
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Answer» `1.5xx10^(-14)s` `V=(1)/( 2pisqrt(LC))` As per question, `v=1 MHz=10^(6)Hz` or `10^(6)=(1)/(2pisqrt(LC))` or `LC=(1)/(4pi^(2)(10^(6))^(2))=2.5xx10^(-14)s` |
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| 6389. |
Calculate the electric field due to a dipole on its equatorial plane.(OR) Electric field due to an electric dipole at a point on the equatorial plane |
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Answer» Solution :Consider a point C at a distance r from the midpoint O of the dipole EQUATORIAL plane on as shown in Figure. Since the point C is equidistant from `+q` and `-q` the MAGNITUDE of the electric fields of `+q` and `-q`are the same. The direction of `vecE(+)`is along BC and the direction of `vecE_(-)` is alongCA. `vecE_(+)` and `vec_(-)` are resolved into twocomponents, one component parallel to the dipole axis and the other perpendicular to it. The perpendicular components of `|vecE_(+)| sin theta`and `vecE_(-)sin theta` are oppositely directed and cancel each other. The magnitude of the total electric field at point C is the sum of the parallel components of `vecE_(+)` and `vecE_(-)` and its direction is along `hatp` `vecE_(tot)=-|E_(+)|costhetahatp-|vecE_(-)|costhetahatp`ldots(1) The MAGNITUDES `vecE_(+)`and `vecE_(-)` are the same and are given by `|vecE_(+)|=|vecE_(-)|=(1)/(4piepsilon_(0))(q)/((r^(2)+a^(2)))ldots(2)`By substituting equation (1) into equation (2), we get `vecE_(tot)=-(1)/(4piepsilon_(0))(2qcostheta)/((r^(2)+a^(2)))hatp=(1)/(4piepsilon_(0))(2qa)/(r^(2)+a^(2))^((3)/(2))hatp` `sinncos=(a)/(sqrt(r^(2)+a^(2))` `vecE_(tot)=-(1)/(4piepsilon_(0))(vecp)/((r^(2)+a^(2))^((3)/(2)))` `sincevecp=2qahatpldots(3)`, At very large distances `(r gt gt a)`, the equation becomes`vec_(tot)=-(1)/(4piepsilon_(0)r^(3)(rgt gta)` 
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| 6390. |
""_(+1)e^(0)+""_(-)e^(@) rarr2gammaThe above equation satisfies the law of conservation of |
| Answer» Answer :D | |
| 6391. |
Two ice cubes of side 10cm, having cavity of volume 20cm^(3) at centre of cube but filled with different materials A and B respectively. The specific gravity of material B is 1.9 Now these cubes are placed in two different vessels of same base area as shown in figure. The water level before putting blocks in vessels are same Assume that ice melts uniformly from all sides and with same constant rate in both the vessels. (specific gravity of ice=0.9) Choose the correct graph showing the variation of heights of water- level in two vessels with time |
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Answer»
LET at `t=t_(0),` cube A sinks. `vrho_(w)G=(v-Deltav)rho_(i)g+Deltavrho_(m)g` v is volume of cube which is changing linerarly with time at `tgtt_(0)` `vrho_(w)g+N_(A)=(v-Deltav)rho_(i)g+Deltavrho_(m)g` after sinking water level decreases due to melting of ice, `1/10(dv)/(dt)=A(dh_(1))/(dt),` A-cross-sectionof vessel Let at t=`t_(0)`cube B sinks `v'rho_(w)g=(v'-Deltav)rho_(i)g+Deltavrho'_(m)g` `1/10 (dv)/(dt)=A (dh_(2))/(dt)` `:. (dv)/(dt)=(dv)/(dt) Rightarrow (dh_(1))/(dt)=(dh_(2))/(dt)` final heights are same in both reach. |
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| 6392. |
Explain experiment which produced magnetic field due to straight long current carrying wire. |
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Answer» Solution :1. One straight long current carrying wire which is kept perpendicular to the PLANE of paper. 2. The ring of compass needles surrounds the wire.  The magnetic field due to a straight long current-carrying wire. The wire is perpendicular to the plane of the paper. A ring of compass needles surrounds the wire. The orientation of the needles is shown when (a) the current emerges out of the plane of the paper, (b) the current moves into the plane of the paper. (c) The arrangement of iron filings around the wire. The DARKENED ends of the needle REPRESENT north poles. The EFFECT of the earth.s magnetic field is neglected. 3. When current emerges out of the plane of the paper the orientation of the needles is shown in figure (a). 4. When current moves into the plane of paper the orientation of the needle is shown in figure (b). 5. The darkened ends of the needle represent north poles. 6. If iron ore is spreaded around it, then the arrangement of iron fillings around the wire as shown in figure (c). 7. So, from this experiment we can conclude that when current is passing through conducting wire, magnetic field induced around it. |
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| 6393. |
A body is projected up with velocity u.It reaches a point in its path at times t_(1) and t_(2) seconds from the time of projection .Then (t_(1)+t_(2)) is |
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Answer» `(2U)/(G)` |
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| 6394. |
Two point charges of +1muC and +4muC are kept 30 cm apart. How far from the +1muC charge on the line joining the two charges, will the net electric field be zero? |
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Answer» Solution :Let charges of `q_(1)=1muC and q_(2)=+4muC` be kept fixed at point A and B RESPECTIVELY situated at a distance 30 cm apart. Let net electric field be zero at a point O situated at a distance x cm from CHARGE A as shown in figure. Then, at point O `vecE_(A)+vecE_(B)=vec0 or |vecE_(A)|= |vecE_(B)| and vecE_(A) and vecE_(B)` MUST be in MUTUALLY opposite DIRECTIONS  `rArr (1)/(4pi in_(0)). (q_(1))/(x^(2))=(1)/(4pi in_(0)). (q_(2))/((30-x)^(2))` `rArr (1muC)/(x^(2))=(4muC)/((30-x)^(2))rArr x=10cm or 0.1m` |
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| 6395. |
If the current in the primary circuit is decreased, then balancing length for same measuring potential difference is obtained at |
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Answer» LOWER LENGTH |
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| 6396. |
A radar has a power of 1kW and is operating at a frequency of 10 GHz. It is located on a mountain top of height 500m. The maximum distance upto which it can detect object located on the surface of the earth (Radius of earth = 6.4 xx 10^(6) m) is : |
| Answer» Answer :A | |
| 6397. |
Inductive reactance (w L ) and capacitive reactance [1/wC] depend upon _____ of a.c. |
| Answer» SOLUTION :FREQUENCY | |
| 6398. |
Frequency of a wave is 6 xx 10^(15) Hz. The wave is |
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Answer» Radiowave |
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| 6399. |
Draw the ray diagram for simple microscope. Write expression for its magnifying power. |
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Answer» Solution :Microscope. Microscope is a device which is used to see very small objects which cannot be seen with naked eye. Simple Microscope Principle. When an OBJECT is placed between principal focus and optical centre of a convex lens, a vertical and erect image will be formed on the same side of the object. Construction and Action An ordinary convex lens of small focal length held close to the eye can be used as magnifying glass or a simple microscope. The object is made to lie between the lens and its focus so that image formed is virtual, erect and magnified. As the object is to be seen clearly, the image is made to lie at the distance of distinct VISION from the eye. The course of rays indicating the formation of image is shown in the figure.  Magnifying Power Magnifying Power of a magnifying glass is defined as the ratio of the angle subtended at the eye by the image to the angle subtended by the object when both are placed at the least distance of distinct vision and seen directly. In Fig., the image AB is fomed at the distance of distinct vision and subtends an `/_A.CB. = beta` on the eye as it is held close to the lens. If the object AB is also moved to the distance of distinct vision i.e., to the position EB., it would SUBTEND an `/_ECB. = alpha` at the eye. Magnifying power, `M=(beta)/(alpha)` `=(tan beta)/(tan alpha)` (`:. beta` and `alpha` are small) `=((AB)/(CB))/((EB.)/(CB.))` `=(AB)/(CB) xx (CB.)/(EB.)` `=(CB.)/(CB)` (`:.` AB = EB.) `M=(v)/(u)` Using the lens formula, `(1)/(v)-(1)/(u)=(1)/(f)` `:. (1)/(u)=(1)/(v)-(1)/(f)` or `(v)/(u)=(v)/(v)-(v)/(f)` or `(v)/(u)=1-(v)/(f)` Since image is formed at least distance of distinct vision D and is virtual, `:.`v = - D So `M=1-(-D)/(f)` or `M=1+(D)/(f)` This shows that SMALLER the focal length of the lens, greater will be the magnifying power of microscope. |
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| 6400. |
The displacement of a particle moving in one dimension under the action of a constant force is related to timer by the equation t= sqrt(x + 3), where x is in metres and in second. What is the displacement of particle when its velocity is zero? |
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Answer» Solution :`tsqrt(x)+3 implies SQRT(x)=t-3 implies x=t^(2)+9-6t` `(dx)/(dt)=2t-6 implies 0=2t-6 implies t=3 s` `:. X=(3)^(2) +9-6xx3=0` or `(dt)/(dx)=(1)/(2)x^(-1//2)+0 implies (dx)/(dt)=2x^(1//2)` `implies 0=2x^(1//2)impliesx=0` |
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