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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.
| 39251. |
What are nuclear force? |
| Answer» Solution :The NUCLEONS in a nucleus are held together by STRONG forces of attraction called nuclear forces. The nuclear force is an attractive force that acts between nucleons, PROTONS to proton, NEUTRON to neutron and proton to neutron. | |
| 39252. |
One second after being thrown straight down, an object is falling with a speed of 20 m//s. How fast will it be falling 2 seconds later ? |
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Answer» SOLUTION :Call down the positive direction, so `a = +g and v_(0) = +20 m//s` . We're given `v_(0),a, and t`, and asked for v. Since s is MISSING, we use Big Five #4. `v= v_(0)+at =(+20 m//s)+(+10 m//s^(2))(2 s)=40 m//s` |
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| 39253. |
What is the effect of induced dipole moment on the external electric field ? |
| Answer» SOLUTION :The induced dipole MOMENT PRODUCES ELECTRIC field which opposes the applied electric field. | |
| 39254. |
An electric lamp runs at 120 V d.c and consumes 12 A. It is connected to 200 V, 50 Hz a.c. Find the inductance required and the power factor. |
| Answer» SOLUTION :`4.24 XX 10^(-2) H, 0.6 ` | |
| 39255. |
When a capillary tube of radius 0.4 mm is dipped into water, the level of water inside the capillary rises to a height of 4cm above the surface of water. What is the surface tension of water if its angle of constant is zero and density is 1 gm//cm^2 |
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Answer» 0.0784 N/m |
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| 39256. |
Tow charges pm10 mu C are placed 5.0 mm apartdeterminethe electric field at (a)a pointp onthe axisof the dipole15 cm away from its centre o ont the side of the positivecharge as(a) and (b)a point q 15 cm away from o on a line passing through o and normal to the axis of the dipole as |
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Answer» <P> Solution :Here electric dipole moment of given electric dipole,`p = (2a)q` `=(5 xx 10^(-3)) (10 xx 10^(-6))` `therefore p = 5 xx 10^(-8)` cm Here, r = 15 cm = 0.15 m 2a = 0.005 m `RARR r gt gt gt gt 2a` Given dipole can be taken as point dipole. For a point dipole, electric field at far axial point is, `E_(a) =(2kp)/r^(3)` `=(2)(9 xx 10^(9))(5 xx 10^(-8))/(0.15)^(3)` `therefore E_(a) = 2.667 xx 10^(5) N//C` (in the direction of `vecp`) For a point dipole, electric field at far equatorial point is, `E_(e) = (kp)/r^(3)` `=(9 xx 10^(9))(5 xx 10^(-8))/(0.15)^(3)` `therefore E_(e) = 1.335 xx 10^(5) N//C` |
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| 39257. |
The wavelength of the first spectral line of the Lyman series is lambda then the wavelength of the first spectral line of the Paschen series is .........lambda |
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Answer» <P>`(108)/(7)` `(1)/(lambda_(L))=R[(1)/(1^(2))-(1)/(n^(2))]=R[(1)/(1^(2))-(1)/(2^(2))]=R[(1)/(1)-(1)/(4)]=R[(3)/(4)]` `:. lambda_(L)=(4)/(3R).....(1)` The wavelength of the first spectral line of the PASCHEN series `(1)/(lambda_(P))=R[(1)/(3^(2))-(1)/(4^(2))]= R[(1)/(9)-(1)/(16)]=R[(7)/(144)]` `:. lambda_(p)=(144)/(7R).....(2)` `:. (lambda_(P))/(lambda_(L))=(144)/(7R)xx(3R)/(4)=(108)/(7) :. lambda_(P)=(108)/(7)xx lambda` |
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| 39258. |
Obtain the resonant frequency and Q-factor of a series LCR circuit with L = 3.0 H, C = 27 muF , and R= 7.4Omega. It is desired to improve the sharpness of the resonance of the circuit by reducing its 'full width at half maximum' by a factor of 2. Suggest a suitable way. |
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Answer» Solution :`omega_r= (1)/(sqrtLC) = (1)/(sqrt(3 xx 27 xx 10^(-6) ) = 111 rad//s ` `Q = (X_L)/(R ) = (omega_r L)/ ( R) = (111 xx 3)/(7.4) = 45` For doubling Q for the same `omega_r` , R should be REDUCED to half and is EQUAL to`3.7 Omega ` |
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| 39259. |
The VI characteristics of four circuit elements are shown. Which of these is ohmic ? |
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| 39260. |
Four charges equal to - Q placed at the four corners of a sqare and a charge q is its centre. If the system is in equilibrium, the value of q is |
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| 39261. |
In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency. |
| Answer» SOLUTION :50 HZ for half-wave. 100 Hz for full-wave. | |
| 39263. |
The distance between the point P(3,4,-7) and Q (-2,5,10) is |
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Answer» 10.20 UNITS |
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| 39265. |
State laws of refraction. |
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Answer» Solution :I law : The incident ray, the REFRACTED ray and the normal DRAWN at the point of incidence all lie in the same plane. II Law : The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for a given PAIR of media and given WAVE LENGTH (color) of light. |
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| 39266. |
A person standing between two parallel hills fires a gun. He hears the first echo after 3/2 sec, and a second echo after 5/2 sec. If speed of sound is 332 m/s, calculate the distance between the hills. When will he hear the third, fourth, fifth and sixth echo? |
| Answer» SOLUTION :10.664m,4s, 5.5s, 6.5s, 8 s | |
| 39267. |
What is the color code for a resistor of resistance 350 Omega with 5% tolerance? |
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Answer» Solution :Resistance of resistor `= 350 Omega` with 5% tolerance `= 350 xx 10^(-3) Omega` `= 35 xx 10^(-2) Omega` First significant figure (3) indicates `1^(st)` band Second significant figure (5) indicates `2^(nd)` band Third significant figure `(10^(-2))` indicates `3^(rd)` band we know that 0 1 2 3 4 5 6 7 8 BBROY of great BRITIAN has very good wife wearing Gold silver NECKLACE `{:(9,10^(-1), larr 10^(-2)):}` 3 indicates orange 5 indicates green `10^(-2)` indicates Silver 5% tolerance substance is gold. `:.` COLOR code of given resistor is orange, green silver, gold |
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| 39268. |
A composite rod (formed by joining two rods of equal length lbut different materials) is held between two fixed supports as shown. If temperature of the system is lowered byDeltatheta, then the displacement of the contact point R is (Symbols or having their usual meanings) |
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Answer» `2alpha l Deltatheta` |
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| 39269. |
If germanium is dopped with arsenic, that will result in |
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Answer» n-type SEMICONDUCTOR |
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| 39270. |
Four container are filled with monatomic ideal gases. For each container, the number of moles, the mass of an individual atom and the rms speed of the atoms are expressed in terms of n, m and v_(rms) respectively. If T_(A),T_(B),T_(C)andT_(D) are their temperature respectively then which one of the options correctly represents the order? |
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Answer» `T_(B)=T_(C)gtT_(A)gtT_(D)` |
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| 39271. |
For three situations, the initial and final positions, respectively, along the x axis for the block in Fig. 8-12 are (a) -3 cm, 2 cm, (b) 2cm, 3 cm , and (c) -2 cm, 2cm. In each situation, is the work done by the spring force on the block positive, negative, or zero? |
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| 39272. |
Assertion : A given transformer can be used to step-up ot step-down the voltage. Reason : The output voltage depends upon the ratio of the number of turns of the two coils of the transformer. |
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Answer» If both ASSERTION ans REASON are TRUE ans reaason is the CORRECT explanation of assertion. |
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| 39273. |
An ammeter and a voltmeter are connected in series to an ideal battery of emf E. If the ammeter reads I and voltmeter reading V, then |
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Answer» `V lt EPSILON` |
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| 39274. |
Meaning of the word dupe... |
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Answer» To lie to SOMEBODY in order OT make him/her believe something or do something |
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| 39275. |
Is it possible that there is no increase in the temperature of a body despite being heated ? |
| Answer» SOLUTION :Yes, during the change of state (such as MELTING of ICE, boiling of water etc.) the SYSTEM takes heat but its temperature does not change. Here the INTERNAL energy changes | |
| 39276. |
Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-exites from level n to level (n-1) . For large n, show that frequency equals the classicalfrequency of revolution of the electron in the orbit. |
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Answer» Solution :The energy of an ELECTRON in the N -th energy level of hydrogen atom `E_n =(-2pi^2me^4)/((4pi in_0)^2*n^2h^2)` `therefore E_n-E_(n-1)=(2pi^2me^4)/((4piin_0)^2*h^2)*(2n-1)/(n^2(n-1)^2)` `therefore` FREQUENCY, f `=(E_n-E_(n-1))/(h) =(2pi^2me^4)/((4piin_0)^2h^3) *(2n-1)/(n^2(n-1)^2)` |
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| 39277. |
Nuclei of radioactive element X. are being produced at a constant rute .K. and this element decays to a stable nucleus .Y. with decay constant .lamda. with half lite ..t_(0)... At time t=0, there are ..N.._(0).. number of nucle of the element .X. then The number N_(x) of nuclei of .X. at time t=t_(0) is |
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Answer» `(K+lamdaN_(0))/(2lamda)` |
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| 39278. |
The position of a particle along x-axis at time t is given by x=1 + t-t^2. The distance travelled by the particle in first 2 seconds is |
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Answer» 1 m |
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| 39279. |
The electromagnetic waves travel with |
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Answer» the same speed in all MEDIA |
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| 39280. |
Suppose a coil of 1000 turns of wire is wound around a book, and this book is lying on a table. The vertical component of earth's magnetic field is about 0.6 xx 10^(-4) Wb//m^(2). The area of the coil is 0.15 xx 0.20 m^(2). If this book is turned through 90^(@) about a horizontal axis in 0.1 sec, what average emf will be induced in the coil? |
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| 39281. |
How will the frequency of small oscillation of a simple pendulum changes. in which case, if at all, is analogous to a pendulum mounted on a cart rolling down an inclined plane. |
| Answer» SOLUTION :CASE (I) | |
| 39282. |
Nuclei of radioactive element X. are being produced at a constant rute .K. and this element decays to a stable nucleus .Y. with decay constant .lamda. with half lite ..t_(0)... At time t=0, there are ..N.._(0).. number of nucle of the element .X. then The number N_(Y) of nuclei .Y. at time t=t_(0) is |
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Answer» `K(LN2)/(lamda)+(3)/(2)((K-lamdaN_(0))/(lamda))` |
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| 39283. |
A wire is moving with velocity 18xx10^-2m/sec in magnetic field of 0.50T. If the induced e.m.f. is 9xx10^-3V, the length of wire is : |
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Answer» 5 cm |
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| 39284. |
50% of X–rays obtained from a Coolidge tube pass through 0.3 mm. thick aluminium foil. If the potential difference between the target and the cathode is increased, then the fraction of X–rays passing through the same foil will be :– |
| Answer» ANSWER :B | |
| 39285. |
The frequency band of medium wave Radio broad casting transmission is |
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Answer» 540 KHZ to 1600 KHz |
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| 39286. |
Two identical small balls, made of insulting material are attached to the ends of a light insulating rod of length d. The rod is suspended from the ceiling by a thin torsion free fibre as shown in figure. Each ball is given a charge q. There is a uniform magnetic field B_(0), pointing vertically down, in a cylindrical region of radius R. The fibre is along the axis of the cylindrical region. The system is initially at rest. Now the magnetic field is suddenly switched off. Calculate the angular velocity acquired by the system. Each ball has mass m. |
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| 39287. |
For a common emitter amplifier , current gain is 70 . If the emitter is 8.8 mA , calculate the collector and base current . Also calculate current gain , when transistor is worked on common base amplifier . |
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Answer» SOLUTION :Data SUPPLIED , `beta = 70 , I_(E) = 8.8 mA , beta = (I_(C))/(I_(B))` `I_(C) = beta I_(B) = 70 xx I_B` `I_(E) = I_(C) + I_(B) = 70 I_(B) + I_(B)` `I_(B) = (8.8)/(71) = 0.124 mA` `I_(C) = beta I_(E) = 70 xx 0.124 = 8.68` mA `beta = (ALPHA)/(1 - alpha)` `70 = (alpha)/(1 - alpha)` `alpha = (70)/(71) = 0.986` |
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| 39288. |
A source of sound of frequency 600 Hz is placed inside water. The speed of sound in water is 1500 m/s, and in air it is 300 m/s. the frequency of sound recorded byan observer who is standing in air is : |
| Answer» Solution :The frequency of wave does not CHANGE with MEDIUM . So correct choice is (D). | |
| 39289. |
What is the radius of the path of an electron (mass 9 × 10-31 kg and charge 1.6 ×x 10^(–19) C) moving at a speed of 3 ×x10^7m/s in a magnetic field of 6 x× 10^(-4) T perpendicular to it?What is its frequency?Calculate its energy inkeV. (1eV = 1.6 xx 10^(-19) J) |
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Answer» Solution :Using eq. we FIND `r = mv// (qB) = 9 xx 10^(-31) kg xx 3 xx 10^7 ms^(-1) // (1.6 xx 10^(-19) C xx 6 xx 10^(-4) T)` ` = 28 xx 10^(-2) m = 28 cm` V `= v//(2pi r) = 17 xx 10^6 s^(-1) = 17 xx 10^6 Hz = 17 MHz` `E = (1//2) mv^2 = (1//2) 9 xx 10^(-31) kg xx 9 xx 10^14 m^2 // s62 = 40.5 xx 10^(-17) J` `= 4 xx 10^(-16) J = 2.5 KEV. ` |
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| 39290. |
The strength of a dielectric is defined as the maximum electric intensity which it can withstand. A paralel-plate capacitor of plate area A on one side is filled with a dielectric of strength E and permittivity epsilon. The maximum charge which can be given to the capacitor is |
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Answer» `epsilonEA` |
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| 39291. |
An antenna is : |
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Answer» Inductive |
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| 39292. |
Velocity of light in rarer medium is v_(1) it passes through denser medium with velocity v_(2) incidents at 'i' angle and refracts by 'r angle, then ..... |
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Answer» `v_(1)=v_(2)` `(sin i)/(sin r)=(v_(1))/(v_(2))` REFRACTED rau ends towards normal. `r lt i` `:.(sin i)/(sin r) gt 1` `:.(v_(1))/(v_(2)) gt 1` `:.v_(1)gt v_(2)` |
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| 39293. |
A horizontal power line carries a current of 90A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5m below the line? |
Answer» Solution : According to FORMULA, `B=(mu_(0)I)/(2pir)` `thereforeB=((4pixx10^(-7))(90))/((2pi)(1.5))` `thereforeB=1.2xx10^(-5)T` (TOWARDS south from point P, as shown in the figure according to right HAND THUMB rule). |
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| 39294. |
A body of displaced from position (hati+hatj+hatk) m to position (4hati+5hatj+6hatk) m under the action of force (5hati+4hatj-3hatk) N. The angle between the force and displcement vector is |
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Answer» `cos^-1(FRAC{8}{25})` |
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| 39295. |
A ray of light travelling in a rarer medium strikes a plane boundary between the rarer medium and a denser medium at an angle of incidence 'i' such that the reflected and the refracted rays are mutually perpendicular. Another ray of light of same frequency is incident on the same boundary from the side of denser medium. Find the minimum angle of incidence at the denser-rarer boundary so that the second ray is totally reflected |
Answer» Solution : Figure SHOWS incidence of a ray at the rarer-denser BOUNDARY such that reflected and REFRACTED RAYS are mutually perpendicular. ` i.e.r+90^@ +r’=180^@`. or `r = 90^@ - r = 90^@- i`[r=i, law of reflection] Apply Snell.s law at the boundary, ` therefore mu_R SINI= mu _D sin r .` ` mu_Rsin i = mu _D sin (90^@-i)= mu_Dcos i` ` or ( mu_D)/(mu_R) = tan i` `sin theta_C =(1)/( ""_(R )mu_D )= (1)/(mu_D //mu_R) = (mu_R)/( mu_D) ` Using equation (1), ` sin theta_c =(1)/( tani )=cot i` ` = theta_c =sin^(-1) ( coti)` |
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| 39296. |
Describe the motion of a charged particle in a cyclotron if the frequency of the radio frequency (rf) field were doubled. |
| Answer» Solution :When the frequency w of ELECTRIC field (oscillator) is doubled the time PERIOD `(T=(2pi)/omega)` becomes half. So, the charged PARTICLE will take half time to REACH between dees. Hence, a charged particle accelerates as it moves in circular path between the dees during motion in dee.s the RADIUS of moving charged particle remain same. | |
| 39297. |
Consider the circuit shown is after switch S is closed. What amount of charge will flow through the battery ? |
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Answer» `20 muC` Figure (b) corresponds to when switch is closed.  .  . Varous charges FLOWING after closing the switch are as follows: `Deltaq_(1)=60-40=20 muC` `Deltaq_(2)=20-0=20 muC` `Deltaq_(3)=20-0=20 muC` `Deltaq_(4)=Deltaq_(1)+Deltaq_(2)+Deltaq_(3) =60muC`. |
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| 39298. |
How can the concept of electromagnetic induction be used to apply breaks to fast moving automobile. |
| Answer» Solution :To RETARD the motion of FAST moving automobile, a strong magnetic FIELD should be applied ACROSS the wheels . | |
| 39299. |
A changing magnetic field pierces the interior of a circuit containing three identical resistors. Two ideal voltmeters are connected to the same points, as shown in the following figure. If V_(1) reads 1 mV, then V_(2) reads |
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Answer» 0 |
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| 39300. |
A perfectly black bloody radiates 81j/sec at 300^@K. The amount of energy radiated per sec by an ordinary surface of same area having emissivity od 0.8 at a temperature of 500^@K will be |
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Answer» a)100 `j/sec` |
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