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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.
| 3851. |
A car is moving with a speed of 72 km "hour"^(-1)towards a roadside source that emits sound at a frequency of 850 Hz. The car driver listens to the sound while approaching the source and again while moving away from the source after crossing it. If the velocity of sound is 340 ms^(-1) the difference of the two frequencies, the driver hears is |
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Answer» 50 Hz `v_("cat") = 72 km "hour"^(-1) = 72 xx 5/18 ms^(-1) = 20 ms^(-1)` Velocity of sound, `v= 340 ms^(-1)` frequency of source, `v= 850 Hz` while APPROACHING the source `u.. = u((v+v_("car"))/v)` while moving AWAY from the source, `u.. = u((v-v_("car"))/v)` Difference in frequencies, `Deltau = u.-v..` `=(2uv_("car"))/v` `=(2 xx 850 xx 20)/340 = 100 Hz` |
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| 3852. |
Plane mirror is moving towards you at 5 cm//s. You can see your image in it. So at how much speed image moves towards you ? |
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Answer» 5 CM`//`s `vec(v_0)=5hati cm//s` Velocity of image towards the person, `vec(v_i)=-5 hati cm//s` `therefore` Velocity of image with RESPECT to person, `vec(v_(I0))=vec(v_i)-vec(v_0)=-5hati-5hati=-10hati` `therefore|vec(v_(i0))|=10" "cm//s` |
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| 3853. |
A parallel resonance a.c circuit : |
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Answer» acts like a resistance of very LOW value |
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| 3854. |
In a Young’s double slit experiment, the angular width of a fringe formed on distant screen is 0.1^@. The wave length of light used is 6000 overset@A. What is the spacing between the slits ? |
| Answer» SOLUTION :`3.4xx10^(-4)m`. | |
| 3855. |
Obtain Gauss's law from the flux associated with a sphere of radius rand charge 'q' at centre. |
Answer» Solution :Let US consider the total fiux through a SPHERE of radius r, which encloses a point charge q at its centre.  DIVIDE the sphere into small area elements, as shown in figure. The flux through an area element `DeltavecS` is, `Deltaphi = vecE.DeltavecS = vecE.hatr.DeltaS` where we have used Coulomb.s law for the electric field due to a single charge q. The unit vector `hatr`is along the radius vector from the centre to the area element. The area element `DeltavecS` and `hatr`have the same direction, `therefore Deltaphi =q/(4pi epsilon_(0)r^(2))DeltaS = (KQ)/r^(2) DeltaS (therefore k = 1/(4piepsilon_(0)))` The total flux through the sphere is obtained by adding up flux through all the different area elements. `therefore phi = sum_(DeltaS)(kq)/r^(2).DeltaS` `therefore phi = (kq)/r^(2) sum_(DeltaS).DeltaS = q/(4pi epsilon_(0)r^(2)) S (therefore sumDeltaS=S)` `therefore phi = q/(4pi epsilon_(0)r^(2)) xx 4pir^(2) = q/epsilon_(0)` (`therefore S = 4pir^(2)` for sphere) Gauss.s law: Electric flux through a close surface S. `phi = sum q/epsilon_(0)` `sumq` = total charge enclosed by S. The electric flux associated with any close surface is EQUAL to the ratio of net charg, enclosed by surface to `epsilon_(0)` `therefore phi = int_(S) vecE.dvecS = (sumq)/epsilon_(0)` |
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| 3856. |
Consider a metal ball of radius 'r' moving at a constant velocity 'v' in a unifrom magnetic field of induction vecB. Assuming that the direction of velocity forms an angle 'a' with the direction of vecB, the maximum potentail difference between points on the ball is |
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Answer» `r|vecB||vecv|sin ALPHA` |
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| 3857. |
Number of moles in 51 g ammonia is …………… |
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Answer» 1 |
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| 3858. |
What was Gafur’s condition when Tarkratna visited his house? |
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Answer» He was SHIVERING with HIGH fever |
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| 3859. |
A particleundergoes S.H.M. having time period T. The time taken in (3)/(8) th oscillation is : |
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Answer» `(3)/(8)`T Hence correct choice is ( C ). |
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| 3860. |
An infinitely long straight conduction is bent into the shape as shown in the figure . It carries a current of I ampere and the radius of the circular loop is r metre. Then the magnetic induction at its centre will be |
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Answer» `(mu_(0))/(4pi)(2i)/(r)(pi+1)` |
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| 3861. |
Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (1/2)QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor 1/2. |
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Answer» Solution :Force `=1/2"QEWork DONE"=1/2V XX Q` `F xx d =1/2 xx E.d xx Q""F=1/2 QE` The physical ORIGIN of the factor `1/2` in the force formula lies in the fact that just outside the conductor, field is E and insde it is zero. So, the average VALUE `E/2` CONTRIBUTES to the force. |
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| 3862. |
The resultant of two vectors vecA and vecB is perpendicular to vecA and equal to half of the magnitude of vecB. Find the angle between vecA and vecB ? |
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Answer» SOLUTION :Since `vecR` is perpendicular to `vecA`. figure SHOWS the three vectors `vecA, vecB and vec`R. ANGLE between `vecA and vecB` is p-q `"SIN" q=(R)/(B)=(B)/(2B)=(1)/(2)` `rArr q=30^(@) rArr` angle between A and B is `150^(@)` 
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| 3863. |
A series R-C circuit is connected to an alternating voltage source. Consider two situations (a) When capacitor is air filled. (b) When capacitor is mica filled. Current through resistor is / and voltage across capacitor is V then |
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Answer» `V_(a) = V_(b)` |
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| 3864. |
What is formula of Barium Phosphate? |
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Answer» `BaPO` |
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| 3865. |
When the electrical conductivity of a semiconductor is due to breaking of its covalent bonds, the semiconductor is said to be a ______ semiconductor. |
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Answer» donor |
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| 3866. |
Calculate the effective capacity in the following combination. Given C_(1)=3muF and C_(2)=2muF |
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Answer» |
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| 3867. |
Measurements carried out by the Soviet "Venus" space-probes with the aid of their landing modules have shown that from an altitude of 50 km above the surface of Venus the temperature of the planet's atmosphere changes linearly as the altitude decreases. Using the data given below prove that this layer of the atmosphere consists mainly of carbon dioxide gas. {:("Altitude above surface h, km",50,42,37,15,0),("Pressure p,atm",1,3.3,6,37,90),("Temperature t,"^@C,80,160,200,360,485),("T,K",353,433,473,633,758):} |
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Answer» `M = (alpha RT_0 "log" (p_0//p))/(G "log" (T_0//T)) , "where" alpha = (T_0 - T)/(hT_0) = 1.05 xx 10^(-5) m^(-1)` Noting that for Venus `g = 8.52 m//s^2`. , we can obtain the values of the molar mass at four altitudes. The average value is `M_(av) = (40.0 + 43.2 + 44.4 + 46.0)/(4) = 43.4` kg/mole. The molar mass of `CO_2`is 44 kg/mole . This leads to the CONCLUSION that the atmosphere of Venus consists mainly of carbon dioxide gas. Other experimental methods corroborate this conclusion. |
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| 3868. |
A small electric dipole is placed at origin with its dipole moment directed along positive x-axis. The direction of electric field at point (2,2sqrt(2),0)is |
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Answer» ALONG POSITIVE x-axis |
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| 3869. |
In the reaction ._1^2H+._1^3H to ._2^4He + ._0^1n. If the binding energies of ._1^2H,._1^3H and ._2^4Heare respectively a, b and c (in MeV), then the energy (in MeV) released in this reaction is |
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Answer» c+a-B |
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| 3870. |
(a) Define self inductance. Write its S.I. units (b) Derive and expression for self inductance of a long solenoid of length l cross-sectional area A having N number of turns. |
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Answer» SOLUTION :(a) Self Inductance (or co-efficient of self induction) : Self inductance of a COIL is numerically equal to the amount of magnetic flux LINKED with the coil when unit current flows through the coil. Its S.I. units is Henry (H) (b) Self Inductance of a Long solenoid. Consider a long solenoid of length l, number of turns N and radius R. Suppose current I flows through it. Magnetic field set up in the coil is `B = (mu_(0)NI)/(l)` Flux through each turn `= BA = (mu_(0)NIA)/(l)` Total flux through N turns, `phi = N xx (mu_(0)NIA)/(l) = (mu_(0)N^(2)IA)/(l) " But " phi = LI` `:.` Self inductance, `L = (phi)/(I) = (mu_(0)N^(2)A)/(l)` |
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| 3871. |
When an electron makes transition from n = 4to n = 2, then emitted line spectrum will be |
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Answer» FIRST line of LYMAN series |
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| 3872. |
If the susceptibility for ion is 2000, the relativepermeability is |
| Answer» Answer :C | |
| 3873. |
A traffic policeman standing on a road sounds a whistle emitting the main frequency of 2.00 kHz. What could be the appparent frequency heard by a scooter-driver approaching the policeman at a speed of 36.0 km/h? Speed of sound in air = 340 m/s. |
| Answer» SOLUTION :2.06 KHZ | |
| 3874. |
Give an expression for range of an antenna interms of its height from ground. |
| Answer» SOLUTION :`d_(m)=SQRT(2Rh_(T))+sqrt(2Rh_( R ))` | |
| 3875. |
The two plates of a parallel plate capacitor are 4 mm apart . A slab of dielectric constant 3 and thickness 3 mm is introduced between the plates with its face parallel to them . The distance between the plates is so adjusted that the capacity of the capacitor becomes 2/3 rd of its original value . What is the new distance the plates ? |
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Answer» Solution :Here d = 4 mm , t = 3 mm , K = 3 and `C. = (2)/(3)` C . Let new distance between the capacitor PLATES be d. mm . Then `C = (in_0 A)/(d) = (in_(0) A)/(4 mm) … (i) and C.= (2)/(3) C (in_(0) A)/(d. - t (1 - (t)/(K))) = (in_(0) A)/(d.- 3 (1 - (1)/(3))) = (in_(0) A)/((d.-2)) .... (ii)` Dividing (i)by (ii) , we get `(3)/(2) = (d.-2)/(4 mm) implies d. = 8` mm |
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| 3876. |
A guitar string is 90 cm long and has a fundamental frequency of 124 Hz. Where should it be pressed to produce a fundamental frequency of 186 Hz ? |
| Answer» SOLUTION :The STRING should be PRESSED at 60 CM from one end | |
| 3877. |
Calculate battery current and equivalent resistance of the network shown in figure. |
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Answer» 20V, `5 Omega` |
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| 3878. |
Given below are certain quantities : Momentum, stress, pressure, force constant, work, moment of force, surface tension, Impulse form pair of quantities having same dimensions. |
| Answer» SOLUTION :(1) MOMENTUM and IMPULSE, (2) stress and Pressure (3) Force constant and surface TENSION (4) Work and moment of force. | |
| 3879. |
A full wave rectifier uses two diodes with a load resistance of 100Omega. Each diode is having negligible forward resistance. Find the efficiency of this full wave rectifier. |
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Answer» Solution :Forward resistance of the diode, `r_(f)=0Omega,`Load resistance, `R_(L)=100Omega, ETA=?` EFFICIENCY of full wave RECTIFIER `eta=(0.812R_(L))/(r_(f)+R_(L))=(0.812xx100)/(100)=0.812` The percentage efficiency of the full wave rectifier `= 81.2%` |
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| 3880. |
If a half-wave rectifier circuit is operating from 50Hz mains, the fundamental frequency in the ripple will be |
| Answer» ANSWER :B | |
| 3881. |
Consider the vernier calipers as shown, the instrument has no zero error. What is the length of the rod (in mm) shown, if "1 msd = 1 mm " ? Use "7 msd = 8 vsd." |
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Answer» |
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| 3882. |
A: The conductivity of a pure semiconductor increases on doping R: Doping causes the reduction in bond strength. |
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Answer» If both Assertion & Reason are true and thereasonis the not the CORRECTEXPLANATION of theassertion , then mark (1) |
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| 3883. |
A short bar magnet of magnetic moment 5.25 xx 10^(-2)JT^(-1) is placed with its axis perpendicular to the earth's field direction. At what distance from the centre of the magnet on the normal bisector is the resultant field inclined at 45^(@) with the earth's field. Magnitude of earth's field at that place is 0.42 G. [1G=10^(-4)T] |
Answer» Solution :Figure shows the conditions of the problem. SUPPOSE P is the point on the NORMAL bisector of the magnet where the resultant of B (due to magnet) and H is inclined at `45^(@)` with H. This is possible if MAGNITUDES of B and H are the same.  `B = (mu_0)/(4pi) ( M)/(d^(3))`......... for a short magnet ` or d^(3) = (mu_0)/(4pi) (M)/(B) = 10^(-7) xx (5.25 xx 10^(-2))/( 0.42 xx 10^(-4)) = 125 xx 10^(-6)` `:. d = (125 xx 10^(-6))^(1//3) m = 5 xx 10^(-2) m = 5 cm ` |
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| 3884. |
A mass m moving horizontally with velocity v_0 strikes a pendulum of mass 'm'. If the two masses stick together after the collision, then the maximum height reached by the pendulum is |
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Answer» `v_0^2/(8G)` |
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| 3885. |
What is amplitude modulation? |
| Answer» SOLUTION :When the amplitude of CARRIER WAVE is varied in accordance with the modulation signal the PROCESS is CALLED amplitude modulation. | |
| 3886. |
Converging rays strike a spherical convex mirror such that they can form the Image (in the absence of mirror) between pole and focus. Now what can you say about final image formed by mirror? |
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Answer» real SINCE `x lt f`, so y is NEGATIVE and hence so image is real. `m=-(v)/(u)=-(y)/(x)rarr` positive so image is erect. Image will be enlarged 
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| 3887. |
Number of electric lines of force starting from 1C of positive charge in vacuum is |
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Answer» a)8.85xx10^(-12)` |
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| 3888. |
What are the factors on which the capacitance of a parallel plate air capacitor depends? |
| Answer» SOLUTION :AREA of PLATES, seperation between TWO plates and the MEDIUM which is placed in between the plates. | |
| 3889. |
The orbital velocity of a satellite in a circular orbit just above the earth's surface is v_(0). The orbital velocity for a satellite orbiting in a circular orbit at an altitude of half of earth's radius is: |
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Answer» `sqrt((3)/(2))v_(0)` and orbital speed at a height h above the surfaceof earth is `v=sqrt((GM)/(R+h))=sqrt((gR^(2))/(R+(R )/(2)))=sqrt((2)/(3)gR)` `v= sqrt((2)/(3))*v_(0)` So CORRECT choice is (b). |
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| 3890. |
Two transparent media A and B are separated by a plane boundary. The speed of light in medium A is 2.0xx10^(8)ms^(-1) and in medium B is 2.5xx10^(8)ms^(-1). The critical angle for which a ray of light going from A to B suffers total internal reflection is |
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Answer» `SIN^(-1)(1/(SQRT(2)))` |
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| 3891. |
A light source emitting three wavelength 5000Å, 6000Å, 7000Å has a total power of 10^(-3) m. W and the beam of diameter2xx10^(-3)m. The power density is distributed equally amongst the three wavelengths. The beam shines normally on a metallic surface of area 10^(4)m^(2) which has a work function of 1.9 eV. Assuming that each proton liberates an electron, calculate the charge emitted per unit area in one second : |
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Answer» 94 C `W=(hc)/(lambda_(0)) :. Lambda_(0)=6513Å` For PHOTOELECTRIC emission `lambda lt lambda_(0)`, so cut of the given wavelength `lambda_(1)=5000Å and lambda_(2)=6000Å` will causephoto emission. Power of source `=10^(-3)W` Diameter of beam `D=2xx10^(-3)m` Radius of beam `r=10^(-3)m` `:.` Area of cross section of beam `A=pi r^(2)=pi xx10^(6)m^(2)` Power DENSITY of beam `=(P)/(A)` `=(10^(-3))/(pixx10^(-6))=(1)/(pi)xx10^(3)W//m^(3)` Power density of each wavelength, `I.=(10^(3))/(3pi)W//m^(3)` Energy of photon of wavelength `lambda_(1)` `E_(1)=(hc)/(lambda_(1))=3.97xx10^(-19)J` Number of photons of wavelength `lambda_(1)` incident on metallic surface per unit area per sec. `N_(1)=(I)/(E_(1))=(10^(3))/(3pi xx3.97xx10^(-19))=2.67xx10^(20)` Energy of photon of wavelength `lambda_(2)` `E_(2)=(hc)/(lambda_(2))=3.31xx10^(-19)J` Then `N_(2)=(I.)/(E_(2))=3.21xx10^(20)` As each photon liberates an electron, the number of electrons EMITTED per sec unit area `N=N_(1)+N_(2)` `=2.67xx10^(20)+3.21xx10^(20)` `=5.88xx10^(20)` `:.` Charge emitted per sec. per unit area is `q=Ne` `=5.88xx10^(20)xx1.6xx10^(-19)` `or q=94C` |
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| 3892. |
A circuit involving five ideal cells, three resistors (R_(1), R_(2) and 20 Omega) and a capacitor of capacitance C=1 mu F is shown. At stead state match the proper entries from column -2 to column -1 using the codes given below the columns. {:(,"Column" I,,"Column" II,),(,(P)K_(2) "is open and" K_(2) "is in position" C,,(1)"Potential at point A is greater than potential at B",),(,(Q)K_(2)"is open and" K_(1) "is in position D",,(2)"Current through" R_(1) "is downward",),(,(R)K_(2)"is closed and" K_(1) "is in positionC",,(3)"Current through" R_(2) "is upward",),(,(S)K_(2) "is closed and" K_(1) "is in position"C,,(4)"Charge on capacitor is" 10muC.,):} |
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Answer» `{:((P),(Q),(R),(S),),(4,3,1,2,):}` Assume the potential at point P to be zero. When Key `K_1` is in position C: `V_A`=16 Volt and `V_B`=6 volts. Hence current in both `R_1 and R_2` will flow downwards. When Key `K_1` is in position D : `V_A`=2 Volt and `V_B`=-8 volts. Hence current through `R_1` wil flow downwards and through `R_2` will flow upwards. 
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| 3893. |
When a force acceleration a body immersed in a fluid, some of the fluid must also be accelerated , since it must be pushed out of the way of the body and flow aroundit . Thus , the force must overcome not only the inertia of the body, but also the inetia of the fluid pushed out of the way. it can be shown that for a spherical body completely immersed in a nonviscous fluid, the extra inertia is that of a mass of fluid half as large as the fluid displaced by the body. The acceleration ("in" m//s^2)of small spherical air bubble in water is ("take" g= 9.8 m//s^2). |
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Answer» |
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| 3894. |
Three identical cells each of emf 2 V and unknown internal resistance are connected in parallel. This combination is connected to a 5 Omega resistor. If the terminal voltage across the cells is 1.5 V, what is the internal resistance of each cell? |
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Answer» |
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| 3895. |
The spectacular array of South African jets was a display of - |
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Answer» MILITARY's precision |
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| 3896. |
M(g)rarrM_(g)^(+)+e^(-),DeltaH1=100 KJ//mol M(g)rarrM_(g)^(+2)+2e^(-),DeltaH2=300 KJ//mol M(g)rarrM_(g)^(+3)+3e^(-),DeltaH3=650 KJ//mol Select incorrect sattlement: |
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Answer» `IE_(3) of M is 350 KJ//mol` |
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| 3897. |
Two plane mirrors A and B are aligned parallel to each other, as shown in the figure. A light ray is incident at an angle 30^@ at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. Find the maximum number of times the ray undergoes reflections (including the first one) before it emerges out |
Answer» Solution :`d= 0.2tan 30^@ = (0.2 )/(sqrt3)`  if there are N REFLECTIONS, then `Nd=2 SQRT(3)THEREFORE(0.2 )/( sqrt(3)) = 2 sqrt(3)implies N=30` The maximum number of reflections are 30. |
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| 3898. |
When a plane electromagnetic wave travels in vacuum, the average electric energy density is given by (here E_(0) is the amplitude of the electric field of the wave) : |
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Answer» `(1)/(4) epsilon_(0)E_(0)^(2)` |
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| 3899. |
The figure shows a potentiometer arrangement. D is the driving cell. C is the cell whose emf is to be determined. AB is the potentiometer wire and G is a galvanometer. J is a sliding contact whichcan touch any point on AB. Which of the following are essential conditions for obtaining balance? |
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Answer» The EMF of `D` must be greater than the emf of `C` |
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| 3900. |
A lorry and a car moving with same K.E. are brought to rest by the application of brakes which provide equal retarding forces. Which of them will come to rest in a shorter distance. |
| Answer» Solution :Work done by VEHICLE = loss of K.E. i.e.F.S= loss in K.E. Here K.E. and retarding FORCE f are same. So the distance S travelled by them before COMING to rest would also be same. | |