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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.
| 37451. |
The electromagnetic waves do not transport. |
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Answer» energy |
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| 37452. |
A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. The resistance of the two wire line carrying power is 0.5Omega per km. The town gets power from the line through a 4000-220 V step-down transformer at a sub-station in the town. a. Estimate the line power loss in the form of heat. b. How much power must the plant supply assuming there is negligible power less due to leakage? c. Characterise the step up transformer at the plant. |
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Answer» Solution :Total line resistance ` = 0.5 xx 30 = 15` 200 A` a. Line power LOSS `=I_(rms)^2 R = (200)^2 xx 15 = 600 kW` b. Power supplied= 800 + 600 = 1400 kW c. VOLTAGE dropped ` = I_(rms) R = 200 xx 15 = 3000 V` Total voltage to be supplied= 4000 + 3000 = 7000 V ` therefore ` Step up transformer at the plant is 440 V/7000 V |
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| 37453. |
a man can see upto 100cm of the distant object. The power of the lens requiredto see far objects will be: |
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Answer» `+0.5D` |
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| 37454. |
If the object moves with a speed u towards a fixed mirror, how the image moves towards the mirror. |
| Answer» SOLUTION :same SPEED U | |
| 37455. |
Fixed frames of reference are used in "______________". |
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Answer» General THEORY of RELATIVITY |
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| 37456. |
A : Three capacitors having C_(1) lt C_(2) lt C_(3) are connected in parallel its effective capacitance will be C_(p)gt C_(3). R : (1)/(C_(P))=(1)/(C_(1))+(1)/(C_(2))+(1)/(C_(3)) |
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Answer» If both ASSERTION and REASON are true and reson is the CORRECT explanation of the assertion. |
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| 37457. |
Find the angular acceleration of a particle performing circular motion increases from 60 r.p.m to 180rpm in 20 seconds. |
| Answer» SOLUTION :`ALPHA=(2PI(n_2-n_1))/20=(2pi(3-1))/20=(2pi)/10`=0.6284 `rad/s^2` | |
| 37458. |
The radius of the innermost electron orbit of a hydrogen atom is 5.3 xx 10^(-11) m. What are the radii of the n = 2 and n = 3 orbits ? |
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Answer» SOLUTION :As `r_(n)=n^(2)R` `therefore r_(2)=4r_(1)=4xx5.3xx10^(-11)m=2.12 xx10^(-10)m` and `r_(3)=9r_(1)=9xx5.3xx10^(-11)m=4.77xx10^(-10)m` |
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| 37459. |
A charged wire is bent in the form of a semicircular arc of radius a. If charge per unit length is lambda coulomb/metre, the electric field at the centre O is |
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Answer» `(lambda)/(2PI a^(2)epsilon_(0))` |
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| 37460. |
The energy stored in a capacitor is given by |
| Answer» Answer :A::B | |
| 37461. |
Assertion: Mass of a body decreases slightly when it is charged Reason:Charging is only due to loss of electrons. |
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Answer» Both ASSERTION and Reason are true and Reason is the correct explanation of Assertion |
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| 37462. |
A magnetic needle oscillates in a uniform magnetic field of strength B_1 with a time period of 10 seconds.The same magnetic needle oscillates in another uniform magnetic field of strength B_2 with a time period of 20 seconds. Which magnetic field is strong, B_1 or B_2 ? Why ? |
| Answer» SOLUTION :` T = 2PI sqrtI/(MB)` i.e. `Tprop 1/sqrtB` HENCE, `B_1 GT B_2` | |
| 37463. |
Two copper spheres of the same radius, one of them being hollow and the other solid, are charged to the same potential. Which of them does contain a greater amount ofcharge? |
| Answer» SOLUTION :Both CONTAIN the same AMOUNT of CHARGE | |
| 37464. |
Write the block diagram of a detector for AM signal. |
Answer» Solution :Process of ( appropriate) superimposition of LOW frequency message signal , over a high frequency CARRIER WAVE , is called a MODULATION . 
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| 37465. |
Calculate the height which paraffin oil of density 800 kg/m^3 will rise in a capillary tube of radius 0.5mm, if surface tension of paraffin oil is 24.5 xx 10^-3 N/m and its angle of contact with the tube is 28^@27'(g = 9.8 m//s^2) |
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Answer» 1.1 cm |
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| 37466. |
The density of an electron - hole pair in a pure germanium is 3xx10^(-16) m^(-3)at room temperature. On doping with aluminium , the hole density increase to 4.5xx10^(22) m^(-3)Now the electron density ( in m )in doped germanium will be |
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Answer» `1XX10^(10)` |
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| 37467. |
Total energy of one LC circuit is xJ. Then at t=0, energy of fully charged capacitor and inductor are respectively …..and ……… |
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Answer» x,0 `rArr` At t=0 if `U_E` =MAX. then `U_B=0`. Hence , `U_E+0=x`. THUS at t=0, `U_E=x` and `U_B=0` |
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| 37468. |
What is the name given to the part of the electromagnetic spectrum which is used for taking photograph of earth under foggy conditions from great heights: |
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Answer» ULTRAVIOLET rays |
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| 37470. |
A positively ahargedparticel (q, m) enters in a uniform magnetic field 'B' at an angle of 60^(@) as shown in figure with speed v_(o). Collision between the charge particle and the wall is perfectly elastic. Then find the time in which the charge particel comes out from the magnetic field. ("take"d = ((sqrt3 - 1)mv)/(2qB)) |
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Answer» `(pim)/(2qB)` `cos theta - (1)/(2) = (sqrt3)/(2) - (1)/(2)` `theta = 30^(@)` So, total angle rotates inside the magnetic FIELD `= 30^(@) + 30^(@) = 60^(@)` Time period `= (m)/(qb) (pi)/(3) = (pi m)/(3qB)` 
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| 37471. |
Figure below shows a bulb connected in an electrical circuit. ( ## EXP_SPS_PHY_XII_C07_E03_001_Q01 .png" width="80%"> :When the key is switched ON the bulb obtains maximum glow only after a shorter interval of time which property of the solenoid is responsible for the delay? |
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Answer» SELF induction |
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| 37472. |
Audio sine wave of 3kHz frequency are used to amplitude modulate a carrier signal of 1.5MHz. Which of the following statements are true ? |
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Answer» The side BAND frequencies are 1506 kHz and 1494 kHz |
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| 37473. |
The total power contain of an AM wae is 2.64 kW at a modulation factor of 80% .Calculate (i)Power of carrier wave (ii)Power in USB (iii)Power of LSB |
| Answer» SOLUTION :(i)2000 WATT, (II)320 watt, (iii)320 watt | |
| 37474. |
The total flux ( in S.I units ) through a closed surface constructed around a positive charge of 0.5 C placed in a dielectric medium of dielectric constant 10 is |
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Answer» `5.65xx10^9` |
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| 37475. |
The plate current i_(p) in a triode valve is given i_(p)=K(V_(p)+mu V_(g))^(3//2) where i_(p) is in milliampere and V_(p) and V_(g) are in volt. If r_(p)=10^(4) ohm and g_(m)=5xx10^(-3)mho, then for i_(p)=8mA and V_(p)=300 volt, what of K and grid cur off voltage |
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Answer» `-6V,(30)^(3//2)` |
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| 37476. |
An electron enters an electric field with its velocity in the direction of the electric lines of force. Then |
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Answer» the path of the electron will be a CIRCLE |
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| 37477. |
The rms value of electric field of the light coming from the sun is 720 NC^(-1). The average total energy density of the electromagnetic wave is : |
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Answer» `3.3xx10^(-3) Jm^(-3)` |
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| 37478. |
When a plane wave travels ina medium in the positive X direction , the displacement of the particle is given by y(x,t)=0.01sin2pi(t-0.1x) where x and y are measured in meters an t in seconds. What is the wavelength of the wave ? |
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Answer» 0.1m |
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| 37479. |
The work function of a metal surface is 4.2 eV. The maximum wavelength which can eject photo-electron from this surface is =…….Å |
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Answer» 2956 `THEREFORE lambda_(0)=(hc)/(phi_(0))=(6.6xx10^(-34)xx3xx10^(8))/(4.2xx1.6xx10^(-19))` `therefore lambda_(0)=2956xx10^(-10)m` `therefore lambda_(0)2956Å` |
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| 37480. |
Elinstein got Nobal prize in physics in 1921 for hisw explanation of photo electric effect.Draw a graph to show the variation of stopping potential with frequency of incident radiation. How can the values of Planks constant be determined from the graph? |
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Answer» SOLUTION :`hV=hV_0+1/2mv^2` `1/2mv^2=h(V-V_0)eV_0=h(V-V_0)` `h=eV_0/(V-V_0)`from the GRAPH, `h=exxslope=e((V_0)/(V-V_0))` |
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| 37481. |
An object is kept at a distance of 60 cm from a concave mirror. For getting a magnification of (1)/(2), Focal length of the concave mirror required is |
| Answer» Answer :C | |
| 37482. |
A uniform electric field of 1 MV/m and a unifordi magnetic field of 10^(-2) T were set up in some region of space. The electric field strength vector is perpendicular to the magnetic induction vector. A muon beam moves in a direction perpendicular to both vectors and passes without being deflected by the combined action of both fields. What is the velocity of the particles? Can the charge of the particle and its sign be determined in this experiment? |
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Answer» |
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| 37483. |
Therefore are four light - weight -rod samples A , B , C , D separately suspended by threads . A bar magnet is slowly brought near each sample and the following observations are noted(i) A is feebly repelled(ii) B is feebly attracted(iii)Cis strongly attracted(iv)D remains unaffected |
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Answer» B is of a paramagnetic material |
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| 37484. |
A p-n junction diode can be used as |
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Answer» Rectifier |
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| 37485. |
Ifmu_s=coefficient of static or limiting friction and mu_k=coefficient of kinetic friction, then |
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Answer» `mu_sgtmu_k` |
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| 37486. |
In a young.s Double slit experiment ,first maxima is observed at a fixedpoint Pon the screenNow the screenis continuously moved away from the plane of slits . The ratio of intensityat point P to the intensity at pointO (centre of the screen ) |
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Answer» remains constant |
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| 37487. |
Three concentric spherical conductors A, B, and C of radii r, 2R, and 4R, respectively. A and C is shorted and B is uniformly charged (charge +Q). Potential at A is |
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Answer» `Q/4piepsilon_0R`  `q_(1)+q_(2)=0` `v_(A)=(kq_(1))/(R)+(kQ)/(2R)+(kq_(2))/(4R)` `v_(C)=(kq_(1))/(4R)+(kQ)/(4R)+(kq_(2))/(4R)` `v_(A)=v_(C)` `q_(1)=-(Q)/(3)` and `q_(2)=(Q)/(3)` `v_(A)=K[(-Q)/(3R)+(Q)/(2R)+(Q)/(12R)]=(Q)/(12piepsilon_(0)R)]=(Q)/(16piepsilon_(0)R)` `v_(B)=k[(-Q)/(6R)+(Q)/(2R)+(Q)/(12R)]=(5Q)/(48piepsilon_(0)R)` |
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| 37488. |
Three concentric spherical conductors A, B, and C of radii r, 2R, and 4R, respectively. A and C is shorted and B is uniformly charged (charge +Q). Potential at B is |
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Answer» `Q/4piepsilon_0R`  `q_(1)+q_(2)=0` `v_(A)=(kq_(1))/(R)+(kQ)/(2R)+(kq_(2))/(4R)` `v_(C)=(kq_(1))/(4R)+(kQ)/(4R)+(kq_(2))/(4R)` `v_(A)=v_(C)` `q_(1)=-(Q)/(3)` and `q_(2)=(Q)/(3)` `v_(A)=K[(-Q)/(3R)+(Q)/(2R)+(Q)/(12R)]=(Q)/(12piepsilon_(0)R)]=(Q)/(16piepsilon_(0)R)` `v_(B)=k[(-Q)/(6R)+(Q)/(2R)+(Q)/(12R)]=(5Q)/(48piepsilon_(0)R)` |
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| 37489. |
Three concentric spherical conductors A, B, and C of radii r, 2R, and 4R, respectively. A and C is shorted and B is uniformly charged (charge +Q). Charge on conductor A is |
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Answer» `Q//3`  `q_(1)+q_(2)=0` `v_(A)=(kq_(1))/(R)+(kQ)/(2R)+(kq_(2))/(4R)` `v_(C)=(kq_(1))/(4R)+(kQ)/(4R)+(kq_(2))/(4R)` `v_(A)=v_(C)` `q_(1)=-(Q)/(3)` and `q_(2)=(Q)/(3)` `v_(A)=K[(-Q)/(3R)+(Q)/(2R)+(Q)/(12R)]=(Q)/(12piepsilon_(0)R)]=(Q)/(16piepsilon_(0)R)` `v_(B)=k[(-Q)/(6R)+(Q)/(2R)+(Q)/(12R)]=(5Q)/(48piepsilon_(0)R)` |
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| 37490. |
A particle of mass3 xx 10^(-6)g has the same wavelength as an electron moving with a velocity 6 xx 10^(6) ms^(-1). The velocity of the particle is |
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Answer» `1.82xx10^(-18)m//s` `lambda=h/(MV)=(6.63xx10^(-34))/(9.1xx10^(-31)xx6xx10^(6))=1.214xx10^(-10)m` Velocity of the PARTICLE `v=h/(mlambda)=(6.63xx10^(-34))/(3xx10^(9)xx1.214xx10^(-10))=1.820xx10^(-15)ms^(-1)` |
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| 37491. |
A block A of mass m connected with a spring of force constant k is executing SHM. The displacement time equation of the block is x =x_(0) + a sin omega t.An identical block B moving towards negative x-axis with velocity vo collides elastically with block A at time t = 0. Then, |
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Answer» displacement time equation of A after collision will be `x=x_(0) -v_(0) sqrt(m/k) sin omegat` `therefore v=(DX)/(DT) = aomega cos omega g` (for block A) at t=0, `x=x_(0)` and `v= aomega` i.e., block A is at `x =x_(0)` (mean position) and its velocity is `a omega`in positive x-direction. It collides elastically with an identical block B moving towards negative x-direction with velocity `v_(0)`. So, the blocks will INTERCHANGE their velocities i.e., `V_(a) =v_(0)`(in negative x-direction) and `v(B)= aomega`(in positive x-direction) Let A be the new AMPLITUDE of block A, then from conservation of mechanical energy `1/2mv_(A)^(2) =1/2kA^(2)` `therefore A =v_(A) sqrt(m/k) = v_(0) sqrt(m/k)` `therefore` New displacement-time equation for block A will be: `x=x_(0) -v_(0) sqrt(m/k) sin omega t` |
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| 37492. |
Consider Experiment: (a) What would you do to obtain a large deflection of the galvanometer ? (b) How would you demonstrate the presence of an induced current in the absence of a galvanometer ? |
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Answer» Solution :(a) FOLLOWING are DIFFERENT ways to increase the deflection in the galvanometer. (i) We can increase no. of turns of a coil. (ii) We can have turns of a coil with greater area. (iii) We can keep FERROMAGNETIC core at the centre of coil. (iv) We can increase the speed of bar mag net shown in experiment 6.2. (b) We can REPLACE galvanometer by a small bulb. Whenever induced current flows through this bulb, it glows which indirectly proves the existence of induced current. |
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| 37493. |
Do free electrons travel to region of higherpotential or lower potentail ? |
| Answer» SOLUTION :As FREE electronsare negativelycharged, they wouldmove to REGIONSOF higherpotential. | |
| 37494. |
Two metallic rods, each of length L, area of cross A_(1) and A_(z), having resistivitiesrho_(1) and rho_(2)are connected in parallel across a d.c. battery. Obtain the expression for the effective resistivity of this combination. |
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Answer» Solution :Resistances `R_(1)` and `R_(2)` of two given rods will be: `R_(1)= (rho_(1)L)/(A_(1))` and `R_(2)= (rho_(2)L)/(A_(2))` `:. ` EFFECTIVE resistance of PARALLEL combination `R = (R_(1)R_(2))/(R_(1)+R_(2))=((rho_(1)L)/(A_(1))xx (rho_(2)L)/(A_(2)))/(((rho_(1)L)/(A_(1))+rho_(2)L))impliesR=(rho_(1)rho_(2)L)/((rho_(1)A_(1)+rho_(2)A_(2)))` Moreover, since length of both rods is same, effective EQUIVALENT AREA A `= (A_(1)+A_(2))` `:.` Effective resistivity `rho- (RA)/(L) - (rho_(1)rho_(2)(A_(1)+A_(2)))/((rho_(1)A_(2)+rho_(2)A_(1)))` |
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| 37495. |
A p-n diode is used in a half wave rectifier with a load resistance of 1000Omega . If the forward resistance (r_f) of diode is 10 Omega , calculate the efficiency of this half wave rectifier. |
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Answer» SOLUTION :LOAD resistance `R_L= 1000 Omega `, Forward resistance of the diode `= r_f= 10 Omega `. Efficiency of half wave rectifier `[( 0.406 R_L)/( r_f+R_L)]= ( 0.406 XX 1000)/( 1010) = 0.4019.` The percentage efficiency of the half wave rectifier `eta ` = 40.19% |
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| 37496. |
The electron in a hydrogen atom moves in a circular orbit of radius 5 xx 10_(-11) m with a speed of 0.67 xx 10_(8) m//s Then |
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Answer» the frequency of revolutions of the electron is `6 xx 10^(-15) rev//s` `i = FE, where f = frequency =(omega)/(2pi)=(v)/(2pir)=(0.6xx10^(6)xxpi)/(2pixx5xx10^(-11))=6xx10^(15)rev//s(0.6xx10^(6)xxpi)/(2pixx5xx10^(-11))=6xx10^(15)rev//s` `i=6xx10^(15)xx1.6xx10^(-19)`= `0.96 xx 10^(-3)A = 0.96 mA` therefore Hence, all the options are correct |
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| 37497. |
A clock moves along an x axis at a speed of 0.700c and reads zero as it passes the origin of the axis. (a) Calculate the clock's Lorentz factor. (b) What time does the clock read as it passes x = 180m ? |
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Answer» |
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| 37498. |
Give two different properties of current carrying coil and state ampere's suggestion. |
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Answer» Solution :1. We have shown that a current loop : (1) Produces a magnetic field and behaves like a magnetic dipole at large distances. (2) It is subject to TORQUE like a magnetic needle. 2. This led Ampere to suggest that all MAGNETISM is due to circulating currents. 3. This SEEMS to be partly true and no magnetic MONOPOLES have been seen so far. 4. However, elementary particles such as an electron or a proton also carry an intrinsic magnetic moment not ACCOUNTED by circulating currents. |
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| 37499. |
A wire of 15 Omega resistance is gradually stretched to double its original length. It is then cut into two equal parts. These parts are then connected in parallel across a 3.0 volt battery. Find the current drawn from the battery. |
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Answer» Solution :Here R = 15 `Omega`. When the wire is gradually stretched to double its original length, its cross-sectionarea is reduced to half of its original value so that the volume remains same. Hence l = 21 and A. = A/2. ` therefore ` New resistance of wire`R. = (rho l.)/(A.) = (rho . (2l))/((A/2)) = 4 (rho l)/(A) = 4R = 4 XX 15 = 60 Omega` On cutting the wire into 2 equal parts, resistance of each part `R_1 = R_2= (R.)/(2) = 30 Omega` On joining the two parts in PARALLEL, equivalent resistance `R_(eq) = (R_1 R_2)/(R_1 + R_2) = (30 xx 30)/(30 + 30) = 15 Omega` If emf of battery be `epsi`= 3.0 V, then current DRAWN from the battery `I = (epsi)/(R_(eq)) = (3.0 V)/(15Omega) = 0.2 A` |
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| 37500. |
An elevator in which a man is standing is moving upwards with a speed of 10m/sec . If the man drops a coin from a height of 2.45 metre, after what time it reaches the floor of the elevator. |
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Answer» SOLUTION :Hereu=0 beacause lift is moving. RELATIVE velociy of COIN w.r.t lift is zero, a `a = + G = 9.8 m/s^2 t = ? s = 2.45 m` `As S = ut+1/2xx9.8t^2=4.9t^2` `t^2 = 2.45/4.9 = 1/2 or t = 1/(sqrt2sec)` |
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