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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.
| 44751. |
State macroscopic form of Ohm's law. |
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Answer» Solution :Current densityJ at a point in a conductor is the amount of current flowing per unit area of the conductor around that point provided the area is held in a direction NORMAL to the current. `J=(I)/(A)` Drift velocity, `V_(d)=(eE)/(m)tau` `I=n eAV_(d)=n eA((eE)/(m))tau "" [J=(I)/(A), sigma =(n e^(2)tau)/(m)]` `(I)/(A)=((n e^(2)tau)/(m))E` `J=sigmaE` |
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| 44752. |
If one were to apply Bohr's model to a particle of mass 'm' and charge q moving in a plane under the influence of a magnetic field B, the energy of the charged particle in the n^th level will be: |
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Answer» `N((HQB)/(4pim))` |
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| 44753. |
Frequencies in the UHF range normally propagate by means of |
| Answer» Answer :D | |
| 44754. |
One side of the biconcave lens is silvered. It behaves like : |
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Answer» Convex mirror or `f = (R )/(2MU)` `=-[(4(mu-1))/(R)+(2)/(R)]=-(4mu-2)/(R)` `therefore `F = - ( R) = - (R )/(4mu - 2)` so it bahaves like a concave mirror |
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| 44755. |
Maximum intensity in YDSE is I_0. Find the intensity at a point on the screen where (a) the phase difference between the two interfering beams is pi//3 . (b) the path difference between them is lamda//4 |
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Answer» Solution :(a) We have `I = I_(max) cos^2 (PHI/2) `. Here , `I_(max)` is `I_0` (i.e., intensity due to independent sources is `I_0//4` ) (b) Phase difference corresponding to the given path difference `Delta X = lamda/4` is `phi = ((2PI)/(lamda)) (lamda/4) ( because phi = (2pi)/(lamda)Delta x) = pi/2 " or " phi/2 =pi/4` `I = I_0 cos^2 (pi/4) = (I_0)/(2)` |
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| 44756. |
The following question consists of two statements each, printed as assertion and reason. While an swering these questions you should choose any one of the following responses.Assertion : In a simple battery circuit, the point at the lowest potential is positive terminal of the battery. Reasons: The electrons flows from higher potential to lower potential. |
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Answer» Both ASSERTION and REASON are TRUE and the reason is a CORRECT EXPLANATION of the assertion. |
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| 44757. |
Two conducting concentric hollow sphers of radii a and 2a respectively are shown in the diagram. The inner shell has net charge Q on its surface. The outer shell is neutral. If the outer shell is connected to the earth the beat generated through the connecting wire is |
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Answer» ZERO |
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| 44758. |
When source is linear, the wavefront is |
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Answer» Spherical |
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| 44759. |
Two charge, each of q, are kept x = - a and x= a on the x axis. A particle of mass m and charge q_0 = q/2 is placed at the origin . If charge q_0 is given a small displacement(y let let a) along y axis . The net force acting on the particle is proportional to : |
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Answer» y |
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| 44760. |
(A): The coulomb force is the dominating force in the universe (R ): The coulomb force is stronger than the gravitational force |
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Answer» Both .A. and .R. are TRUE and .R. is the correct EXPLANATION of .A. |
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| 44761. |
There is a current of 1.0A in the circuit shown below. What is the resistance of P? |
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Answer» `1.5 OMEGA ` |
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| 44762. |
After being haunted by fear for many years Douglas decided to learn to swim. He took the help of |
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Answer» his mother |
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| 44763. |
An air bubble is formed inside water. Does it act as a converging lens or a diverging lens ? |
| Answer» SOLUTION :DIVERGING LENS | |
| 44764. |
Assertion: It is possible to transmit signals from one place on the earth to practically other place on earth. Reason: A geostationary satellite orbiting the earth is u sed for transmission. |
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Answer» |
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| 44765. |
Ina transistor the base is made very thinand a very hightly doped with an impurity : |
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Answer» to ENABLE the COLLECTOR to collect side . |
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| 44766. |
The position vectors of two point charges of 2 nC each are (2hati + 3hatj - hatk)m and (3hati + 5hatj + hatk)m respectively. Magnitude of the coulombian force acting between them is........ |
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Answer» `4 xx 10^(-3)` N `=(hati + 2hatj + 2hatk)` `therefore r = SQRT(1+4+4) = 3m` `therefore F = (kq^(2))/r^(2) = (9 xx 10^(9) xx 4 xx 10^(-12))/9` `=4 xx 10^(-3)` N |
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| 44767. |
Answer the following questions: (d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend aroundobstacles, how is it that the students are unable to see each other even though they can converse easily. |
| Answer» SOLUTION :For diffraction or bending of waves by obstacles/apertures by a LARGE angle, the size of the latter should be comparable to wavelength. If the size of the obstacle/aperture is much too large compared to wavelength, diffraction is by a small angle. Here the size is of the order of a few metres. The wavelength of light is about `5xx 10^(-7M)`, while sound waves of, say, 1 KHZ frequency have wavelength of about 0.3 m. Thus, sound waves can bend around the partition while light waves cannot. | |
| 44768. |
The dimension of an atom is of the order of an angstrom. Thus there must be large electric field between the protons and electrons. Why is then the electrostatic field inside a conductor is zero ? |
| Answer» Solution :Electric fields are CAUSED by excess electric CHARGE. There can be no excess charge inside an isolated conductor. Therefore, the electrostatic FIELD inside a conductor is ZERO. | |
| 44769. |
Which of the following combinations should be selected for better tuning of an L-C-R circuit used for communication ? |
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Answer» R=15 `Omega`, L=3.5 H , C=30 `muF` `Q=(omega_0L)/3=1/sqrt(LC)(1/R)=1/Rsqrt(L/C)[ because omega_0 =1/sqrt(LC)]` `therefore` For large Q value R and C should be SMALL and L should be large which is possible in OPTION (A). |
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| 44770. |
The load which is given to the pan in the determination of Young's modulus causing no deformation in the wire is called |
| Answer» Answer :D | |
| 44771. |
An a.c. voltage of 100 V, 50 Hz is connected across a 20 Omegaresistance and a 2 mH inductor in series. Calculate (1) impedance of the circuit, (ii) rms current in the circuit. |
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Answer» Solution :Here, `V_(RMS) = 100 V`, v= 50 Hz, `R = 20 Omega` and L = 2mH `=2 xx 10^(-3)` H (i) IMPEDANCE of the circuit `Z = sqrt(R^(2) + X_(L)^(2)) = sqrt(R^(2) + (2pi vL)^(2))` `= sqrt((20)^(2) + (2pi xx 50 xx 2 xx 10^(-3))^(2)) = 20 Omega` (ii) The rms current in the circuit `I_(rms) = V_(rms)/Z = (100 V)/(20 Omega) = 5A` |
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| 44772. |
The resistivity of a potentiometer wire is, if the area of cross section of the wire is 4cm^(2). The current flowing in the circuit is 1A, the potential gradient is 7.5 v/m _______ |
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Answer» `3XX10^(-3)Omega-m` |
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| 44773. |
When does a monochromatic beam of light incident on a reflective surface get completely transmitted ? |
| Answer» Solution :Let the light EMITTED the reflective surface with Brewster's ANGLE `(i_B)`. Now rotate the polariser at PARTICULAR alignment LASER source passes through polariser, and incident on the surface of the light incident on the surface is completely transmitted. | |
| 44774. |
Draw the graphs representing the variation of resistivity with temperature for (1) copper (2) nichrome (3) a typical semiconductor. |
Answer» SOLUTION :
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| 44775. |
Suppose that the radius of cross-section of the wire used in the previous problem is 2.the decreases in the radius of the loop is (ia^(2)BN)/(piY) if the magnetic field is swithed off.The Young's modulus of the material of the wire is Y.Then find value of N. |
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Answer» |
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| 44776. |
Consider uniform charged shell of surface charge density s (= 3e^(o) SI "units") and a dipole of dipole moment P (= 2pe_(n) SI "units"). Centre of the shell and the dipole lies at the origin, and dipole moment vector is along +x axis. If the field at a point on x-axis just over the shell is vec(E_(1)) and that at a point on y-axis just over the shell is vec(E_(2)). Find |vec(E_(1))|=|vec(E_(2))| in N/C. Radius of shell = 50 cm. |
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Answer» 6 |
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| 44777. |
Who eats the sandwich: |
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Answer» Omega |
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| 44778. |
For a given lens, the magnification was found to be twice as large as when object was 0.15 m distant from it as when the distance was 0.2 m. The focal length of the lens is |
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Answer» 1.5 m `(F)/(f-0.15)=2(f)/(f-0.20)` 2F - 0.30 = f - 0.02 , f = 0.10 m |
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| 44779. |
A wheel with 10 metallic spokes, each 0.50 m long, is rotated with a speed of 120 rev/minute in a plane normal to the earth.s magnetic field at the place. If the magnitude of the field is 0.40 gauss, what is the induced emf between the axle and the rim of the wheel? |
Answer» Solution :  Here each spoke of wheel ACT as a source of an induced emf (cell) and emf.s of all spokes are parallel. f = 120 rev/min = 2 rev/SECOND, B = 0.40 gauss `= 0.4 xx 10^(-4)T` , AREA swept, by each spoke per second, `A = pi r^2 f` Magnetic flux cut by each spoke per second, `(dphi_B)/(dt) = BA = B pi r^2 f ` Induced emf, `e = B pi r^2f`(numerically) ` e= 0.4 xx 10^(-4) xx 22/7 xx 0.5 xx 0.5 xx 2` ` e = 6.29 xx 10^(-5)` volt Induced emf in a wheel is independent of no. of spokes. |
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| 44780. |
A thin conducting ring of radius R is given a charge +Q. The electric field at the centre O of the ring due to the charge on the part ABCD of the ring is |
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Answer» 3E ALONG KO |
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| 44781. |
Is work to be done to move a charge on a surface of an isolated charged conductor ? |
| Answer» SOLUTION :No, as the SURFACE is EQUIPOTENTIAL | |
| 44782. |
Who nested in the tree? |
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Answer» Hawk-kite |
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| 44783. |
de- Broglie wavelength associated with an electron, accelerating through a potential |
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Answer» GAMMA rays `lamda=h/p=1.227/sqrtV nm` `lamda=1.227/sqrt100=1.227/10` `THEREFORE lamda=0.1227 nm` This wavelength FALLS in the region of X-rays. |
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| 44784. |
Two coils have a mutual inductance of 0.005H. The current changes in the first coil according to equation I=I_0 sin omegat ,where I_0 = 10A and omega = 100pirad/sec. The maximum value of emf in the second coil is ___ |
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Answer» SOLUTION :Induced emf `epsilon=-M(dI)/(dt) =-M d/(dt)(t^2 e^(-t))` `=-M I_0 omega cos omegat` When `cos omegat =-1` , then emf will be maximum `THEREFORE epsilon_"maximum"=0.005xx10xx100xx(-1)=5piV` |
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| 44785. |
In the combination of the following gates the output Y can be written in terms of inputs A and B as |
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Answer» `BAR(A.B)` |
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| 44786. |
Which one the following Carbide of s-block element produces ethyne when reacts with water |
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Answer» `CaC_2` `SrC_2 + 2H_2O to Sr(OH)_2 + CH =CH` |
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| 44787. |
Explain Curie temperature. |
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Answer» Solution :The ferromagnetic property DEPENDS on temperature. Increasing the temperature of ferromagnetic substance, dipole moment of its molecules become random and DESTROYED. At high enough temperature a ferromagnetic substance BECOMES PARAMAGNETIC. The temperature of transition from ferromagnetic to paramagnetism is called the Curie temperature `(T_C)`. Curie temperature that is in the paramagnetic phase is described by `CHI = (C )/( T- T_C) (T gt T_C)` The Curie temperature `(T_C)` of some of the ferromagnetic material : 
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| 44788. |
Calculate the wavelength of de Broglie waves associated with a beam of protons of kinetic energy 5xx10^(2) eV. (Mass of each proton =1.67xx10^(-27)kg,h=6.62xx10^(-34)Js ) |
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Answer» `2.28xx10^(-12)m` |
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| 44789. |
A device ‘X’ is connected to an AC source. The variation of voltage, current and power in one complete cycle is shown in fig. (b) What is the average power consumption over a cycle? |
| Answer» SOLUTION :(B) ZERO. | |
| 44790. |
A particle with mass m and positive charge q released from rest at the origin as shown in figure. There is a uniform electric field E_(0) in +y direction and uniform magneticfield B_(0) directed out of the page. The path of the particle as shown is called cycloid. The particle always moves in x-y plane. The velocity at any time t after the start is given as The y- coordinate at the highest point of trajectory is |
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Answer» `(4M)/(qB_(0)^(2))` `y=(mE_(0))/(qB_(0)^(2))[1-cos((qB_(0))/(m)t)],y_(max)=(mE_(0))/(qB_(0)^(2))` |
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| 44791. |
What happens to the resistivity of thematerial if the conductor is streched to double its length? |
| Answer» SOLUTION :REMAIN the same | |
| 44792. |
A device ‘X’ is connected to an AC source. The variation of voltage, current and power in one complete cycle is shown in fig. (a) Which curves shows power consumption over a full cycle? |
| Answer» Solution :(a) A (a curve of power have a MAX. AMPLITUDE of V and I) | |
| 44793. |
For preparing a permanent magnet, the material used should be _____. |
| Answer» SOLUTION :FERROMAGNETIC | |
| 44794. |
A device ‘X’ is connected to an AC source. The variation of voltage, current and power in one complete cycle is shown in fig. (c) Identify the device X if curve B shows voltage. |
| Answer» Solution :(c) as AVERAGE power is zero the DEVICE is a capacitor. | |
| 44796. |
When a charged particle enters perpendicular to uniform magnetic field, path followed by it is ______ |
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Answer» linear |
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| 44797. |
Define the term 'work function' of a metal. The threshold frequency of a metal is f_(0). When the light of frequency 2f_(0) is incident on the metal plane, the maximum velocity of electrons is v_(1). When the frequency of the incident is increased to 5f_(0), the maximum velocity of electrons emitted is v_(2). find the ratio of v_(1) to v_(2). |
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Answer» |
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| 44798. |
Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n–1). For large n,show that this frequency equals the lassical frequency of revolution of the electron in the orbit. |
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Answer» Solution :The first orbit Bohr’s model has a radius `a_0` given by `a_0 = (4 pi epsilon_0 (h//2pi)^2)/(m_e e^2)`. If we consider the atom bound by the gravitational force `(Gm_pm_e//r^2).` We should replace `(e^2/4 pi epsilon_0)` by `Gm_p m_e`. That is, the radius of the first Bohr orbit is given by `a_0^G = ((h//2pi)^2)/(Gm_p m_e^2) ~= 1.2 XX 10^(29) m`. This is MUCH greater than the estimated size of the whole universe!. |
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| 44799. |
A man of 50 kg is standing at one end on a boat of length 25 m and mass 200 kg. If he starts running and when he reaches the other end, he has a velocity 2 ms^(-1) with respectto the boat. The final velocity of the boat is (in ms^(-1)) |
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Answer» `2/5` Let L be the velocity of the boat. Then, `(200 + 50) K+50 XX 2 = 0 ` `250 V+ 100 = 0` `V=-100/250= -2/5 ms^(-1)`. |
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| 44800. |
Plutonium decays with a half-life of 24,000 years. If plutonium is stored for 72,000 years, what fraction of it remains ? |
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Answer» `1//8` |
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