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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.
| 3251. |
For step-down transformer value of transformation ratio is ………. |
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Answer» Solution :Transformation ratio `r=N_2/N_1` and for step-downtransformer `N_2 lt N_1` `THEREFORE N_2/N_1 lt 1 "" therefore r lt 1` |
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| 3252. |
If a charge q is placed at the centre of a hemispherical body as shown below then the flux linked with the curved surface is |
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Answer» `(q)/(epsi_(0))` |
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| 3253. |
In the following compounds the order of acidity is- |
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Answer» `III gt IV gt I gt II` strength of `PROP -M` EFFECT `prop (1)/(+"M effect")` compound |
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| 3254. |
A ring of radius R carries a charge q uniformly distribited over it. A long thin wire carrying charge lambdaper unit length of it is held along its axis with one end coinciding with the centre of the ring. Find the interaction force between the ring and the thread. |
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Answer» |
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| 3255. |
Two nuclei have mass numbers in the ratio 8 : 125. Calculate the ratio of their nuclear radii. |
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Answer» Solution :Formula `R=R_(0) (A) ^(1//3)` `2:5` DETAILED Answer : We know that `R=R_(0) (A)^(1//3)` `rArr (R_1)/( R_2) = ((A_1)/( A_2))^(1//3)` `rArr (R_1)/( R_2) = ((8)/( 125))^(1//3) = (2)/(3)` |
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| 3256. |
n an oscillating system, a restoring force is a must. In an L-C circuit, the restoring force is provided by alan |
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Answer» inductor |
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| 3257. |
When a fish under water in a river observes a tree on the bank of the river, will the tree appear taller, same or shorter? Give reason. |
| Answer» SOLUTION :The tree APPEARS taller. This is due to refraction of LIGHT ray entering from AIR to water (rarer to denser MEDIUM). The light ray deviates towards the normal. | |
| 3258. |
Two large, parallel conducting plates X and Y , close to each other, are given charges Q_(1) and Q_(2)(Q_(1) gt Q-(2)) . The four surfaces of the plates are A, b, C and D, as shown |
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Answer» The charge on A is `(1)/(2) (Q_(1)+Q_(2))` |
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| 3259. |
Elucidate the formation of a N-type and P-type semiconductors. |
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Answer» Solution :N-type semiconductor: A n-type semiconductor is obtained by doping a pure Germanium (or Silicon) crystal with a dopant from group V PENTAVALENT elements like Phosphorus, Arsenic, and Antimony. The dopant has five VALENCE electrons while the Germanium atom has four valence electrons. During the process of doping, a few of the Germanium atoms are replaced by the group V dopants. Four of the five valence electrons of the impurity atom are bound with the 4 valence electrons of the neighbouring replaced Germanium atom. The fifth valence clectron of the impurity atom will be loosely attached with the nucleus as it has not formed the covalent bond.  The energy level of the loosely attached fifth electron from the dopant is found just below the conduction band edge and is called the donor energy level. At room temperature, these electrons can easily move to the conduction band with the ABSORPTION of thermal energy. It is shown in the figure (c). Besides, an external electric field also can set free the loosely bound electrons and lead to conduction. It is IMPORTANT to note that the energy required for an electron to jump from the valence band to the conduction band `(E_(g))` in an intrinsic semiconductor is 0.7 eV for Ge and 1.1 eV for Si, while the energy required to set free a donor electron is only 0.01 eV for Ge and 0.05 eV for Si. The group V pentavalent impurity atoms donate electrons to the conduction band and are called donor impurities. Therefore, cach impurity atom provides one extra electron to the conduction band in addition to the thermally generated electrons. These thermally generated electrons leave holes in valence band. Hence, the majority carriers of current in an n-type semiconductor are electrons and the minority carriers are holes. Such a semiconductor doped with a pentavalent impurity is called an n-type semiconductor.  P-type semiconductor: Here, a trivalent atom from group Ill elements such as Boron. Aluminium, Gallium and Indium is ADDED to the Germanium or Silicon substrate. The dopant with three valence electrons are bound with the neighbouring Germanium atom as shown in Figure (a). As Germanium atom has four valence electrons, one electron position of the dopant in the Germanium crystal lattice will remain vacant. The missing electron position in the covalent bond is denoted as a hole.  To make complete covalent bonding with all four neighbouring atoms, the dopant is in need of one more electron. These dopants can accept electrons from the neighbouring atoms. Therefore, this impurity is called an acceptor impurity. The energy level of the hole created by each impurity atom is just above the valence band and is called the acceptor energy level, as shown in Figure (b).  For each acceptor atom, there will be a hole in the valence band in addition to the thermally generated holes. In such an extrinsic semiconductor, holes are the majority carriers and thermally generated electrons are minority carriers. The semiconductor thus formed is called a p-type semiconductor. |
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| 3260. |
The effective resistance in series combination of two equal resistance is 's'. When they are joined in parallel the total resistance is p. If s = np then the minimum possible value of 'n' is |
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Answer» 4 |
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| 3261. |
The radius of gyration of a uniform square plate of mass M and side a about its diagonal is |
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Answer» `a/SQRT(6)` |
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| 3262. |
For a given a.c. circuit, distinguish between resistance, reactance and impedance. |
Answer» SOLUTION : DISTINCTION between RESISTANCE, REACTANCE and IMPEDANCE: 
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| 3263. |
(A): In the phenomenon of mutual induction, self induction of each of the coils persists.(R) : Self induction arises when strength of current in one coil changes. In mutual induction, current is changing in both the individual coils. |
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Answer» Both .A. and .R. are TRUE and .R. is the CORRECT EXPLANATION of .A. |
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| 3264. |
The total magnification produced by a compound microscope is 20. The magnification produced by the eye piece is 5. The microscope is focussed on a certain object. The distance between the objective and eye piece is observed to be 14 cm. If least distance of distinct vision is 20 cm, calculate the focal length of objective and eye piece. |
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Answer» SOLUTION :Here, `m=-20 , m_e=5 , v_e=-20` cm For EYEPIECE , `m_e=v_e/mu_e` `rArr 5=(-20)/mu_e rArr =(-20)/5`=-4 cm Using lens formula, `1/v_e-1/u_e=1/f_e` `-1/20+1/4=1/f_e` `(-1+5)/20=1/(f_e) rArr f_e=5cm` Now , TOTAL MAGNIFICATION `m=m_exxm_0` `-20=5xxm_0` `m_0=1-v_0/f_0` `-4=1-10/f_0` `-5=10/(f_0) rArr f_0` = 2cm |
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| 3265. |
The current I in an inductance coil varies with time according to the graph given in the figure. Which one of the following graphs gives the variation of voltage with time across the inductor? |
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Answer»
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| 3266. |
A magnetic field of 2 xx 10^(-2)T acts at right angles to a coil of area 100cm^2with 50 turns. If the average emf induced in the coil is 0.1V when it is removed from the field in time t. The value of.t. is |
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Answer» 0.1s |
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| 3267. |
Which of following causes production of heat, when current is set up in a wire |
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Answer» Fall of ELECTRON from HIGHER orbits to lower orbits |
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| 3268. |
An alpha-particle of energy 4 Me V is scattered by gold foil (Z = 79) .Calculate the maximum volume in which positive charge of the atom is likely to be concentrated. |
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Answer» `3.3 xx 10^(-40) m^(3)` |
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| 3269. |
The correct order in which the dimensions of 'Length' increases in the following physical quantities is (a) Permittivity(b) Resistance (c) Magnetic permeability (d) Stress |
| Answer» Answer :C | |
| 3270. |
5 gm of air is heated from 273 K to 275 K. the change in internal energy of air will be [C_(V)=172cal//kg" K and J"=4.2J//cal] |
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Answer» 7.22 J `=5XX10^(-3)xx172xx2=4.2` `=7.22J`. |
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| 3271. |
If hatn is a unit vector in the direction of vector vecA, then |
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Answer» `HATN=VECA/(absvecA)` |
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| 3272. |
Drawgraphs showing the distribution of charge in a capacitor and current through an inductor during LC oscillations with respect to time. Assume that the charge in the capacitor is maximum initially. |
Answer» SOLUTION :For capacitor :  The CHARGE decays EXPONENTIALLY with TIME. When be capacitor discharge. |
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| 3273. |
A charged 8mF capacitor having charge 5mC is connected to a 5mH inductor. What is : (iii) the maximum current in the inductor? |
| Answer» Solution :(iii) MAXIMUM CURRENT in the circuit `I_(0) = (q_(0))/(SQRT(LC) ) ` | |
| 3274. |
Explain how Corpuscular theory predicts the speed of light be greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water? If not, which alternative picture of light is consistent with experiment ? |
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Answer» Solution :In Newton's corpuscular picture of refraction, particles of light INCIDENT from a RARER to a denser MEDIUM experience a FORCE of attraction normal to the surface. This results in an increase in the normal component of velocity but the component along the surface remains unchanged. This means `c sin i =v sin r or v/c =(sin i)/(sin r)=mu_(0) GT 1, therefore vgt c` The prediction is opposite to the experimental result : `(v lt c).` The wave picture of light is consistent with experiment. |
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| 3275. |
Audio signal cannot be transmitted, because |
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Answer» the signal has more NOISE |
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| 3276. |
A thin superconducting(zero resistance) ring a held above a vertical long solenoid, as shown in the figure. The axis of symmetry of the ring is same to that of the solenoid. The cylindrically symmetric magnetic field around the ring can be described approximately in terms of the vertical and radial component of the magnetic field vector as B_(z) = B_(0)(1-alpha z) and B_(r) = B_(0) beta r, where B_(0),alpha and beta are positive constants, and z & r are vertical and radial position coordinates, respectively, Initially plane of the ring is horizontal has no current flowing in it. When relreased, it starts to move downwards with its axis axisstill vertical. Initial coordinates of the centre of the ring 'O' is z = 0 and r =0. In the given diagram point O is on the axis and slightly above the solenoid having vertical and radial position coordinates as (0,0) Ring has mass m, radius m, radius r_(0) and self inductance L. Assume the acceleration due to gravity as g. Find the vertical coordinates z for equilibrium position of the ring. |
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Answer» `(mgL)/(2 B_(0)^(2)alpha beta pi^(2)r_(0)^(4))` `PHI =B_(z)pi r_(0)^(2)-LI` SINCE, `R=0`, so `phi=B_(0)(1-alphaz)pi r_(0)^(2)-LI=` constant From initial condition `(z= 0,I=0)`, the value of constant is `phi = B_(0) pi r_(0)^(2)` Using the above equation the current in the ring `I = (1)/(L) B_(0) alpha pi r_(0)^(2)z` The lorentz foce acting on the ring (which can only be vertical because of the symmetry of the ASSEMBLY) can be expressed as) `F_(z)=-B_(0)I(z)2pi r_(0)=-(2 B_(0)^(2)alpha beta pi^(2)r_(0)^(4)z)/(L)=-kz` Equation of motion of the ring is `ma_(z) = F_(z) - mg = -kz - mg` Equilibrium position `z_(0) = -mg//k omega_(0) = sqrt((k)/(m))`. |
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| 3277. |
Mention the shape of wavefront for the portion of wave front og light from a distant star intercepted by the earth. |
| Answer» SOLUTION :PLANE WAVEFRONT | |
| 3278. |
A thin superconducting(zero resistance) ring a held above a vertical long solenoid, as shown in the figure. The axis of symmetry of the ring is same to that of the solenoid. The cylindrically symmetric magnetic field around the ring can be described approximately in terms of the vertical and radial component of the magnetic field vector as B_(z) = B_(0)(1-alpha z) and B_(r) = B_(0) beta r, where B_(0),alpha and beta are positive constants, and z & r are vertical and radial position coordinates, respectively, Initially plane of the ring is horizontal has no current flowing in it. When relreased, it starts to move downwards with its axis axisstill vertical. Initial coordinates of the centre of the ring 'O' is z = 0 and r =0. In the given diagram point O is on the axis and slightly above the solenoid having vertical and radial position coordinates as (0,0) Ring has mass m, radius m, radius r_(0) and self inductance L. Assume the acceleration due to gravity as g. Find the time period of SHM (for small displacement alon z-axis) of the ring. |
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Answer» `(1)/(B_(0)r_(0)^(2))sqrt((2mL)/(alpha beta))` `phi =B_(z)pi r_(0)^(2)-LI` Since, `R=0`, so `phi=B_(0)(1-alphaz)pi r_(0)^(2)-LI=` constant From initial condition `(z= 0,I=0)`, the value of constant is `phi = B_(0) pi r_(0)^(2)` Using the above equation the current in the ring `I = (1)/(L) B_(0) alpha pi r_(0)^(2)z` The lorentz foce acting on the ring (which can only be vertical because of the symmetry of the assembly) can be EXPRESSED as) `F_(z)=-B_(0)I(z)2pi r_(0)=-(2 B_(0)^(2)alpha beta pi^(2)r_(0)^(4)z)/(L)=-kz` Equation of MOTION of the ring is `ma_(z) = F_(z) - mg = -kz - mg` Equilibrium position `z_(0) = -mg//k omega_(0) = sqrt((k)/(m))`. |
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| 3279. |
Can we a transformer be used to step up or step down a D.C. voltage? Explain. |
| Answer» SOLUTION :In STEADY current no induction phenomenon will TAKE place. | |
| 3280. |
The first known collision between space debris and a functioning satellite occurred in 1996: At an altitude of 700 km, a year-old French spy satellite was hit by a piece of an Ariane rocket. A stabilizing boom on the satellite was demolished, and the satellite was sent spinning out of control. Just before the collision and in kilometers per hour, what was the speed of the rocket piece relative to the satellite if both were in circular orbits and the collision was (a) head-on and (b) along perpendicular paths? |
| Answer» SOLUTION :(a) `5.4 XX 10^4 km//h, (B) 3.8 xx 10^4 km//h` | |
| 3281. |
Radiation of wavelength 180 nm eject photoelectrons from a plate whose work function is 2.0 eV. If a uniform magnetic field of flux density 5.0xx10^(-5)T is applied parallel to plate, what should be the radius of the path followed by electrons ejected normally from the plate with maximum energy: |
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Answer» 0.074 m Here `, lambda=180 nm=180xx10&^(-9)m` `w=2.0eV=2.0xx1.6xx10^(-9)J` `:.E=7.8xx10^(-19)J` ALSO `(1)/(2)mv^(2)=E:. V= sqrt((2E)/(m))` so `v=sqrt((2xx7.8xx10^(-19))/(9.1xx10^(-31)))=1.31xx10^(6)m//s` Radius R in a magnetic field in INDUCTION B is given `BEV=(mv^(2))/(r) or r=(mv)/(eB)` Using `B=5xx10^(-5)T` Then `r=0.149 m` |
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| 3282. |
A capacitor is half filied with a dielectric (K = 2 ) as shown in figure A. if the same capacitor is to be filied with same , what would be the thickness of dielectric so that capacitor still has same capacity ? |
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Answer» 2d/3 |
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| 3283. |
Anelectromagnetic eddy current braka consistsof adiscof conductivitysigma andthickness drotating about axisthrough its centrebetweenrectangularpoles offace area A ata distancefrom the centre from thecentre . Calculate the torque tending to show downthe disc . |
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Answer» |
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| 3284. |
The magnetic susceptibility of a ferromagnetic substance is : |
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Answer» LARGE and positive |
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| 3285. |
Total K.E. of a solid sphere of mass M rolling with velocity v is |
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Answer» `7/10Mv^2` |
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| 3286. |
Substances which are strongly attracted and move from weaker to stronger part of the field are: |
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Answer» Ferromagnetic |
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| 3287. |
A point charge causes an electric flux of -1.0 xx 10^(3) Nm^(2)//C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. a. If the radius of the Gaussion surface were doubled, how much flux would passes through the surface? b. What is the value of the point charge? |
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Answer» SOLUTION :a. Flux does not CHANGE since the total charge q enclosed by the surface does not change b. `phi=q/epsi_(0), q=epsi_(0) phi=8.85 xx 10^(-12) xx -1 xx 10^(3) =-8.85 xx 10^(-9)C` |
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| 3288. |
The number of chromosomes in embryo is 30 then what will be number of chromosomes in aleurone layer |
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Answer» 30 |
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| 3289. |
Tw3o pendulum difer in lengths by 22 cm. They oscillate att the same placeos that one of them makes30 oscillationsand the othermakes 36 oscillationsduring the same time. The lengths (in cm) of thependulums are |
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Answer» 72 and 50 `T = 2pi sqrt(L/g) therefore T_(1)/T_(2) = sqrt(l_(1)/l_(2))` Also , ` T_(1)/T_(2) = N_(2)/N_(1) " Where N_(1) = 30 and N_(2) = 36 ` ` Sqrt(l_(1)/l_(2)) = N_(2)/N_(1) = 36/30` GIVEN ` l_(1) - l_(2) = 22 cm ` Solvingthe eqns , (i) and (II) we got ` l_(1) = 72 cm and l_(2) = 50 cm ` |
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| 3290. |
Find the current in each cell considering circuit given in fig. |
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Answer» SOLUTION :Applying kirchhoff.s law for the loop ABCDA `6= 40i + (i_1 + i_2)10` `6= 40i_1 + 10i_1 + 10i_2` `6= 50i_1 + 10i_2""….(1)` For the loop DCFED  `2=10(i_2 + i_2) + 50i_2` `2 = 10i_1 + 10i_2 + 50i_2` `2 = 10 i_1 + 60 i_2""......(2)` From (1) and (2) `290 i_1 = 34` `i_1 = 0.1172 A ""......(3)` Substituting (3) in (2) `2 = 10 xx 0.1172 + 60i_2 , 2= 1.172 + 60i_2` `60i_2 = 2 - 1.172 , i_2 = (2-1.172)/(60) =0.0138 A` CURRENT in the 6V cell = 0.1172 A Current in the cell = 0.0138 A |
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| 3291. |
A small object is enclosed in a sphere of solid glass 8cm in radius. It is situated 2cm from the centre and is viewed from the side of which it is nearer. Where will it appear to be if mu of glass =1.5/ |
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Answer» `6cm` from the CENTRE `R=+8cm` `(MU(2))/v-(mu_(1))/u=(mu_(2)-mu_(1))/r` `1/v-1.5/6=(1.5-1)/8` `:. V=3 1/5 cm`  |
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| 3292. |
Determine the age of ancient wooden item if it is known that the specific activity of C^(14) nuclide in them amounts to 3//5 of that in lately felled trees. The hlaf-life of c^(14) nuclei is 5570 years |
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Answer» Solution :In OLD wooden atoms the number of `C^(14)` nuclei steadily decreases because of radioactive DECAY. (In live trees biological process keep replenshing `C^(14)` nuclei maintaining a balance. This balance starts GETTING disrupeted as soon as the tree is felled.) If `T_(1//2)` is the half life of `C^(14)` then `e^(-TXX(In 2)/(T_(1//2)))=(3)/(5)` HENCE `t=T_(1//2)(In 5//3)/(In 2)= 4105 years ~~ 4.1 xx10^(3)years` |
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| 3293. |
A hollow metal sphere is filled with water and a small hole is made at its bottom. It is hanging by a long thread and is made to oscillate. How will the time perfod change if water is allowed to flow through the hole till the sphere is empty? |
| Answer» Solution :The time PERIOD will increase first and reaches to its maximum VALUE, and then decreases to its ORIGINAL value. | |
| 3294. |
The difference between phase and frequency modulation |
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Answer» PRACTICALLY they are same but THEORETICALLY they differ |
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| 3295. |
How much work must done by a force on 50 kg body in order to accelerate it from rest to 20 m/s in 10 seconds |
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Answer» `10^6 J` |
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| 3296. |
A copper wire of length 3 m and radius r is nickel plated till its radius becomes 2r. What would be the effective resistance of the wire, if specific resistance of copper and nickel arerho _c andrho_nrespectively. |
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Answer» `R = (R_C R_n)/(R_C + R_n)`. |
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| 3297. |
Draw the graph showing thervariation of binding energy per nucleon with mass number. Give therason for the decrease of binding energy per nucleon for nuclei with high mass numbers. The binding energy per nucleon for nuclei with high mass number is low due to the large coulomb repulsion between the protons inside these nuclei. |
Answer» Solution :The graph showing the variation of BINDING energy per nucleon with mass NUMBER is SHOWN below.  The binding energy per nucleon for nuclei with high mass number is LOW DUE to the large coulomb repulsion between the proton inside these nuclei. |
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| 3298. |
Consider two concentric conducting spheres. The outer sphere is hollow and the inner sphere is solid. Initially they are given a charge - 7Q and a charge + 2Q respectively then the charges on the outer surface and inner surface of the outer spher is |
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Answer» `-2Q, -7Q` |
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| 3299. |
A string of length '2a' has been tied at A and B such that AB='a'. The string is passing through a bead C and initially the bead is very close toA. Now the bead is allowed to fall. Let h be the height by which the bead falls when the string becomes taut and v be the speed of the bead just after the string becomes taut. Pick correct option (s):- |
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Answer» `V=SQRT(3ag)sin(26.5^(@))` `v_(0)=sqrt(2gh)` `v_(0)=sqrt(1.5ag)` and `v=v_(0)COS(90^(@)-26.5^(@))` `implies v_(0)sin(26.5^(@))` `implies v=sqrt(1.5ag) sin (26.5^(@))` 
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| 3300. |
According to Gandhi, happiness is largely a |
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Answer» MENTAL Condition |
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