InterviewSolution
Saved Bookmarks
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.
| 4702. |
If a hole is made at the centre of a bar magnet, then its magnetic moment will |
|
Answer» Increase |
|
| 4703. |
A dielectric of relative permittivity (dielectric constant ) k completely fills the space betweenthe plates of a parallel-plate capacitor with a surface charge density sigma. Show that (i) the induced density of surface charge on the dielectric is sigma_(p)=sigma(1-(1)/(k)) (ii) the capacitance of the capacitor is increased by a factor equal to the ratio of the electric field without the dielectric to that with the dielectric. |
|
Answer» Solution :CONSIDER a parallel-plate capacitor without a dielectric, of plate area A, plate separation d and capacitance `C_(0),` charged to a potential difference V and then isolated. Suppose the free charges on its conducting plates are +Q and -Q, FIG. (a). The surface density of free charge is `sigma =(Q)/(A) ""` ...(1) If A is very large and d is very small, the electric field in the region between the plates is almost uniform, except at the edges. The magnitude of the electric field intensity is `E_(0)=(V)/(d)=(sigma)/(epsilon_(0))=(Q)/(epsilon_(0)A) "" ` (2) Now, suppose a dielectric completely fills the space the charged plates. A polarisation charge - `Q_(p)` appears on the exterior surface of the dielectric nearer to the POSITIVE platewhile a polarisation charge `+Q_(p)` appears on its opposite face. Since the capacitor was isolated after charging, the free charge Q on the plates is the same as earlier.  Consider the Gaussian surface S as shown in Fig. (b). It encloses free charge +Q on the left plate and the bound polarisation charge - `Q_(p)` on the surface of the dielectric. The net charge enclosed by `S=Q-Q_(p).` `therefore` By Gauss's theorem, the TNEI over the Gaussian surface is `epsilon_(0) underset(s) int vecE* d vecS=epsilon_(0)EA=Q-Q_(p)` `therefore` The magnitude of the electric field intensity in the dielectric is `E=(Q-Q_(p))/(epsilon_(0)A)=(Q)/(epsilon_(0)A)-(Q_(p))/(epsilon_(0)A)=(sigma)/(epsilon_(0))-(sigma_(p))/(epsilon_(0)) "" ` ...(3) Writing `E=(E_(0))/(k),` where k is the relative permittivity (dielectric CONSTANT) of the MEDIUM, `(E_(0))/(k)=(sigma)/(epsilon_(0))-(sigma_(p))/(epsilon_(0))` ` therefore (sigma)/(k epsilon_(0))=(sigma)/(epsilon_(0))-(sigma_(p))/(epsilon_(0))` `therefore (sigma)/(k)=sigma-sigma_(p)` `therefore` The surface density of induced charge is `sigma_(p)=sigma-(sigma)/(k)=sigma(1-(1)/(k)) "" ` ...(4) Without the dielectric, the capacitance of the capacitor is, `C_(0)=Q//V.` `therefore Q=C_(0)V=C_(0)E_(0)d "" ` ...(5) Let the capacitancewith the dielectric be C. Since the free charge Q remains the same,`Q=CEd "" `...(6) Equating the right hand sides of Eqs. (5) and (6), `CEd = C_(0)E_(0)d` `therefore C=C_(0)((E_(0))/(E))=kC_(0) "" ` ...(7) Thus, the capacitance increases by the factor of `k=E_(0)//E.` |
|
| 4704. |
Surface churgc densitieson the spheres of radius r_(1) and r_(2) are same, then ratio of their electric potential will be ...... . |
|
Answer» `(r_(1)^(2))/(r_(2)^(2))` `:. In V = (4pir^(2)sigma)/(4pi in_(0)r)=(sigmar)/(in_(0)), (sigma)/(in_(0))` is same `:. V PROP r ` `:. (V_(1))/(V_(2))=(r_(1))/(r_(2))` |
|
| 4705. |
A moving snowball is completely melted by its impact against a wall. The speed (in ms^(-1)) of snow ball is: |
|
Answer» `8 cdot 2xx10^(2) m//s` `therefore` Work done `W=(1)/(2) MV^(2)` Heat PRODUCED `Q=(W)/(J) =(mv^(2))/(2J)` Heat REQUIRED to melt ice ball Q.=mL `therefore (mv^(2))/(2J) =mL rArr v=sqrt(2JL)` `=sqrt(2xx4 cdot 2xx80xx10^(3))=8 cdot 19 xx10^(2) m//s`. `=8 cdot 2xx10^(2) m//s`. Thus, correct choice is (a). |
|
| 4706. |
Suppose the sun expands, so that its radius becomes 100 times its present radius and its surface temperaturebecomes half its present value. The total energy emitted by it will increases by a factor of : |
|
Answer» 16 `E=sigmaT^(4)A=sigmaT^(4).4piR^(2)` `:.(E_(2))/(E_(1))=((T_(2))/(T_(1)))^(4).((R_(2))/(R_(1)))^(2)=(((T)/(2))/(T))^(4).((100)/(R))^(2)` `=(1)/(16)xx1000` `E_(2)=625E_(1)` correct CHOICE is (b). |
|
| 4707. |
In Young's experiement the width of the fringes obtained with light of wavelength 6000Å is 2mm. Calculate the fringe width if the entire apparatus is immersed in a liquid of refractive index 1.33. Data : lambda=6000Å=6xx10^(-7)m, beta=2mm=2xx10^(-3)m mu=1.33, beta=? |
|
Answer» SOLUTION :`beta=(Dlambda.)/(d)=(lambdaD)/(MUD)=(beta)/(MU)` `[:.mu=(lambda)/(lambda.)]` `:.beta.=(2XX10^(-3))/(1.33)=1.5xx10^(-3)m` (or) `1.5mm`. |
|
| 4708. |
The distance x covered by a body moving in a straight line in time is given by the relation 2 + 3x = t. If is the velocity of the body at a certain instant of time, its acceleration will be |
|
Answer» `-v^(3)` Differetaiating both side w.r.t time, `:. 4X(dx)/(dt)+3(dx)/(dt)=1` Since `v=(dx)/(dt) :. 4x v+3v=1` `(4x+3)v=1 implies 4x+3=(1)/(v)` `implies a=(-4v^(2))/(1//v)=-4v^(3)` |
|
| 4709. |
Light of wavelength 500 nm is incident on a metal with work functin 2.28 eV.The de-Broglie wavelength of the emitted electron is :[h=6.6xx10^(-34Js) and c=3xx10^(8)m//s] |
|
Answer» `le2.8xx10^(-12)m` `therefore E=(hc)/(lambda)to` in eV `=(6.6xx10^(-34)xx3xx10^(8))/(5xx10^(-7)xx1.6xx10^(-19))=2.475` `~~2.48 eV` Work FUNCTION `phi_(0)=2.28eV` `therefore`ACCORDING to Einstein.s equation, `K_(max)=E-phi_(o)=(2.48-2.28)eV` `lambda=(h)/(sqrt(2mE))` `(6.6xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx0.20xx1.6xx10^(-19)))` `therefore lambda=(6.6xx10^(-34))/(sqrt(5.825xx10^(-50)))` `therefore lambda=(6.6xx10^(-34))/(2.4132xx10^(-25))` `therefore =2.788xx10^(-9)` m `therefore lambda~~28xx10^(-10)m=28Å` `therefore lambda ge2.8 xx10^(-9)`m |
|
| 4710. |
Consider a two slit interferennce arrangements such that the distance of the screen from the slits is half the distance between the slits.That the first minima on the screen falls at a distance D from the centre O, then D is |
|
Answer» `(lambda)/(3)` `S_(1)P = sqrt((s_(1)T_(1))^(2) + (PT_(1))^(2))` =`[D^(2) + (D - x)^(2)]1/2` `S_(2)P = [D^(2) + (D + x)^(2)]1/2` MINIMA will occur when `[D^(2) + (D + x)^(2)]1/2 - [D^(2) + (D - x)^(2)]1/2 = (lambda)/(2)` If x = D `[D^(2) + 4D^(2)]1/2 = (lambda)/(2)` `(5D^(2))1/2 = (lambda)/(2), therefore D = (lambda)/(2 sqrt 5)` |
|
| 4711. |
Figure shows lines of force for a system of two point charges. The possible choice for the charges is |
|
Answer» `q_(1)= 4muC, q_(2) = -1.0 muC` |
|
| 4712. |
Color of precipitate of Barium sulphate is... |
|
Answer» Blue |
|
| 4713. |
In a moving coil galvanometer, a. why is horse-shoe magnet used? b. why is phosphor bronze fibre used? c. why is a soft iron cylinder used? |
| Answer» SOLUTION :a. To produce a cylindrical FIELD with RADIAL LINES | |
| 4714. |
An inductor of 3H is connected to a battery of emf 6V through a resistance of 100 Omega. Calculate the time constant. What will be the maximum value of current in the circuit? |
|
Answer» |
|
| 4715. |
A block of mass 1 kg is released from the top of a rough incline havingmu= (1)/(sqrt (3)). The initial speed of the block is 2 ms^(-1). The inclined plane has unknown length and has a spring of spring constant k = 1Nm^(-1) connected at the base as in the figure. Find the maximum compression of spring. (answer in meter). |
|
Answer» |
|
| 4716. |
Assertion: Gaussian surface is considered carefully Reason:The point where electric field to be calculated should be with in the surface are law : |
|
Answer» Both Assertion and Reason are TRUE and Reason is the correct explanation of Assertion |
|
| 4717. |
A bar magnet of moment of inertia 9xx10^(-5) "kg m"^2placed in a vibration magnetometer and oscillating in a uniform magnetic field 16pi^2 xx 10^(-5) T make 20 oscillations is 15 s . The magnetic moment of the magnet is |
|
Answer» a) `3 Am^2` |
|
| 4718. |
Two identical small spheres carry charge of Q_(1) and Q_(2) with Q_(1) gt gt Q_(2). The charges are d distance apart. The force they exert on one another and then spheres are made to touch one another and then separated to distance d apart. The force they exert on one another now is F_(2). Then F_(1)//F_(2) is :- |
|
Answer» `(4Q_(1))/(Q_(2))` |
|
| 4719. |
Calculate the gain of (i) a positive feedback amplifier and (ii) a negative feedback amplifier with an internal gain of A = 100 and feedback factor beta=(1)/(1000). |
|
Answer» SOLUTION :i) For positive feedback `A_(f)=(A)/(1-Abeta)=(100)/(1-100xx((1)/(1000)))=(100)/(1-0.1)=(100)/(0.9)=111.1` II) For NEGATIVE feedback, `A_(f)=(A)/(1+Abeta)` `therefore A_(f)=(100)/(1+100xx((1)/(1000)))=(100)/(1+0.1)=(100)/(1.1)=90.909~~90.91` |
|
| 4720. |
Two sound waves having the same frequency ,having the intensity 1 and 91 . When they interfere, the resultant intensity at certain points is 131 , then the phase difference between two sound waves will be |
| Answer» Answer :C | |
| 4721. |
Resistance of a wire is 8 Omega . It is drawn in such away that it experiences a longitudinal strain of 400 %. The new resistance is |
|
Answer» `100 Omega ` |
|
| 4722. |
Obtain expression for the radius of the circular path and its frequency, of a charged particle entering into an uniform magnetic field. |
|
Answer» Solution :Let B represent the uniform magnetic applied perpendicular to the plane of the paper and pointing inwards. Let q represents a +ve charge entering the field with a velocity v at right angles to the field. Let F represent the magnetic force on the charged particle, which ACTS perpendicular to the direction of velocity. No WORK is done by the force. Magnetic force `vec(F)=qvec(v) times vec(B)`. This force acts as a CENTRIPETAL force and produces a circular motion (as per the Fleming's Left Hand Rule).  i.e., `""(mv^(2))/r=qvBsin90^(@)` i.e., `""r=(mv)/(qB)` Therefore, radius of the circular PATH `r propto m/q` for a given uniform speed and magnetic field. Period of revolution `T=(2pir)/v=(2pim)/qB " and frequency " v=l/T=(qB)/(2pim)` In the presence of the component of velocity along B, the path described by the charged particle will be helix and whose pitch `p=v_(h)T, v_(h)` is the horizontal component of velocity along B. 
|
|
| 4723. |
The graph shown here figure, shows the variation of the total energy(E) stored in a capacitor against the value of the capacitance (C) itself. Which of the two - the charge on the capacitor or the potential used to 4 charge it is kept constant for this graph ? |
|
Answer» Solution :From the graph, it is CLEAR that charge remains CONSTANT because then `E = Q^2/(2C) rArr E prop 1/C andE - C` graph is a hyperbola. |
|
| 4724. |
Modern trains are based on Maglev technology in which trains are magnetically elevated which runs its EDS Maglev system. There are coils on both sides of wheels.Due to motion of train current induces in the coil of track which elevate it.This is in accordance with Lenz's law.If trains lower down then due to Lenz's law a repulsive force increases due to which train gets uplifted and if it goes much high then there is a net downward force due to gravity. The advantage of maglev train is that there is no friction between the train and the track, thereby reducing power consumption and enabling the train to attain very high speeds. Disadvantage of maglev train is that as it slows down the electromagnetic forces decreases and it becomes difficult to keep it elevated and as it moves forward according to Lenz law there is an electromagnetic drag force. What is the advantageof this system ? |
|
Answer» No friction hence no POWER consumption |
|
| 4725. |
Modern trains are based on Maglev technology in which trains are magnetically elevated which runs its EDS Maglev system. There are coils on both sides of wheels.Due to motion of train current induces in the coil of track which elevate it.This is in accordance with Lenz's law.If trains lower down then due to Lenz's law a repulsive force increases due to which train gets uplifted and if it goes much high then there is a net downward force due to gravity. The advantage of maglev train is that there is no friction between the train and the track, thereby reducing power consumption and enabling the train to attain very high speeds. Disadvantage of maglev train is that as it slows down the electromagnetic forces decreases and it becomes difficult to keep it elevated and as it moves forward according to Lenz law there is an electromagnetic drag force. What is the disadvantage of this system? |
|
Answer» Train experiences UPWARD force according to Lenz's law |
|
| 4726. |
The refractive index of water is 4/3 and that of glass is 5\3. What will be the critical angle of ray of light entering water from glass ? |
|
Answer» `sin^-1(4/5)` |
|
| 4727. |
Modern trains are based on Maglev technology in which trains are magnetically elevated which runs its EDS Maglev system. There are coils on both sides of wheels.Due to motion of train current induces in the coil of track which elevate it.This is in accordance with Lenz's law.If trains lower down then due to Lenz's law a repulsive force increases due to which train gets uplifted and if it goes much high then there is a net downward force due to gravity. The advantage of maglev train is that there is no friction between the train and the track, thereby reducing power consumption and enabling the train to attain very high speeds. Disadvantage of maglev train is that as it slows down the electromagnetic forces decreases and it becomes difficult to keep it elevated and as it moves forward according to Lenz law there is an electromagnetic drag force. Which force causes the train to elevate up? |
|
Answer» ELECTROSTATIC force |
|
| 4728. |
Mark the correct statements for a particle go ing on a straight line: (a) If the velocity and acceleration have opposite sign, the object is slowing down. (b) If the position and velocity have opposite sign, the particle is moving towards theorigin. (c) If the velocity is zero at an instant, the acceleration should also be zero at that instant. (d) If the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval. |
|
Answer» a, b & C are CORRECT |
|
| 4729. |
Velocity of electron in hydrogen atom in its first second and third orbit are in the ratio: |
|
Answer» `1:2:3` |
|
| 4730. |
Four light sources produce the following four waves(i) y_(1) = a sin (omegat + phi_(1))(ii)y_(2)= asin2 omegat(iii)y_(3) = a^(‘) sin (omegat + phi_(2))(iv) y_(1) = a sin (omegat + phi_(1))Superposition of which two waves give rise to interference? |
|
Answer» (i) and (II) |
|
| 4731. |
The diameter of an optical fibre is : |
| Answer» Answer :B | |
| 4732. |
Consider a non conducting plate of radius a and mass m which has a charge q distributed uniformly over it, The plate is rotated about its own axis with an angular speed omega. Show that the magnetic moment M and the angular momentum L of the plate are related as M/L = q/(2m). |
| Answer» SOLUTION :`M/L = ((qomegaa^2)/(4))/((omegama^2)/(2)) = Q/(2M)` | |
| 4733. |
A charge q is placed at the point of intersection of body diagonals of a cube. The electric flux passing through any one of its face is |
|
Answer» `(q)/(6in_(0))` |
|
| 4734. |
The speed of a body is constant. Can it have a path other than a circular or straight line path ? |
| Answer» SOLUTION :All CURVED PATHS are POSSIBLE. | |
| 4735. |
Find the dispersive power of a prism, given that n_(v) = 1.657 and n_(R) = 1.631. |
|
Answer» Solution :`n_(v) = 1.657, "" n_(R) = 1.631"" OMEGA = (n_(v) - n_(R))/(n - 1)` ` n = (n_(v) + n_(R))/(2) = (1.657 + 1.631)/(2) = 1.644` `THEREFORE omega(1.657 - 1.631)/(1.644 - 1) = (0.026)/(0.644) = (0.0404)` |
|
| 4736. |
In the circuit given below , V(t) is the sinusoidal voltage source , voltage drop V_(AB)(t) across the resistance R is |
|
Answer» Is half wave RECTIFIED Since `R_1 and R_2` are different , hence the peaks during positive half and negative halfof the input signal will be different. |
|
| 4737. |
What is the magnetic moment associated with a current loop of area 2 xx 10^(-3) m^2 and carrying current of 0.5A? |
|
Answer» Solution :`A= 2 xx 10^(-3)m^2 , I=0.5A ` MAGNETICMOMENT`m=IA ` ` m=2 xx 10^(-3) xx 0.5` ` m=1 xx 10^(-3)Am^2` |
|
| 4738. |
यदि A और B दो अरिक्त समुच्चय हो ताकि AUB=A तोAnnB= |
|
Answer» `varphi` |
|
| 4739. |
In the previous problem, suppose that another force in addition to the electrical force acts on the particle so that when it is released from rest, it moves to the right. After it has moved 5 cm, the addition force has done 9xx10^(-5 J of work and the particle has 4.5xx10^(-5) J of kinetic energy. (a) What work was done by the electrical force ? (b) What is the magnitude of the electric field ? (c) What is the potential of the starting point with respect to the end point ? |
|
Answer» (B) `3XX10^(5) N//C` (c) `-1.5xx10^(-4) V` |
|
| 4740. |
A thin metallic ring of radius R and area of cross section A carries an electric charge q that is uniformly distributed. A point charge Q is placed at the centre. The signs of charges on the ring and that of Q are same. Young's modulus for the material of the ring is Y. What is the tension developed in the ring? |
|
Answer» `(qQ)/(4piepsilon_0R^2)` Charge on the segment : `q/(2pi)Deltatheta` Force between point charge Q at the centre and this segment can be written as follows :`F=1/(4piepsilon_0)((Q/(2pi)Deltatheta)Q)/R^2 =(qQ)/(8pi^2epsilon_0 R^2) Deltatheta` …(i) The components of tension T cos `(Deltatheta)/2` are cancelled from each other and T sin `(Deltatheta)/2` are added to act against electric force F. For EQUILIBRIUM : 2 T sin `(Deltatheta)/2=F` 2T sin `(Deltatheta)/2 = (qQ)/(8pi^2 epsilon_0 R^2)Deltatheta` For small values of `Deltatheta` we can approximate sin `(Deltatheta//2)` as `Deltatheta//2` , hence we GET the following :`2T (Deltatheta)/2 = (qQ)/(8pi^2 epsilon_0 R^2) Deltatheta rArr T= (qQ)/(8pi^2 epsilon_0 R^2)` 
|
|
| 4741. |
80 g of impuresugar when dissolved in a litre of water gives an optical rotation of 9.9^(@) when placed in a tube of length 20 cm.If the specific rotation of sugar is 66^(@), find percentage purity of sugar solution: |
|
Answer» Solution :`C= (theta)/(l xx ALPHA), theta = 9.9^(@), l - 2.0` decimetre and `alpha = 66^(@)` `therefore C = (9.9)/(2.0 xx 66^(@)) = 0.075 g/cc = 75 g/litre` `therefore` purity `= (75)/(80) xx 100 = 93.75 %`. |
|
| 4742. |
Relation betweengives pair is - |
|
Answer» ENANTIOMER 
|
|
| 4743. |
The integral multiples of fundamentals frequency in case of vibrating string are called- |
|
Answer» harmonics |
|
| 4744. |
What is magnetic substance ? Write its types. |
|
Answer» Solution :The substance is called magnetic substance if the substance is placed in external magnetic field and induced magnetic field PRODUCED in it. There are three types of magnetic SUBSTANCES : (i) Diamagnetic (II) Paramagnetic (iii) Ferromagnetic |
|
| 4745. |
Two whistles A & B produces notes of frequencies 660 Hz and 596 Hz respectively. There is a listenerat the middle point of the line joining them. Both the whistles B and the listener start moving with speed 30 m/s away from the whistle A. If speed of sound be 330 m/s, how many beats will be heard by the listener ? |
|
Answer» 2 `f_(A) = (330 - 30)/(330 - 0) xx 660 ` = 600 HZ. BEAT freq. = 600 - 596 = `4s^(-1)`. correct choice is (b). |
|
| 4746. |
The ratio of SI units to the CGS units of coefficient of viscosity will be |
|
Answer» `10^-1` |
|
| 4747. |
A parallel plate capacitor is filled with a uniform dielectric , maximum charge that can be given to it does not depends upon |
|
Answer» DIELECTRIC constant of the dielectric |
|
| 4748. |
There are two nuclides ""_(10)^(22)Ne and ""_(11)^(22)Na. a. What are these nuclides ? ""b. How do they differ? |
|
Answer» Solution :a. They are isobars. B. Same number of NUCLEOUS but DIFFERENT number of protons. |
|
| 4749. |
ABCD is a square of side 4 cm. +16 esu, -16 esu and +32 esu charges are placed at the points A, C and D respectively. Determine the resultant intensity at B. |
|
Answer» |
|
| 4750. |
In (Q.18)problem, the minimum number of photons emitted by the H-atom is |
| Answer» SOLUTION :When transition takes place from n=4to n=1. ONE photon will be emitted. | |