Saved Bookmarks
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.
| 42701. |
Draw a plot showing the variation of (i) electric field (E ) and (ii) electric potential (V) with distance r due to a point charge Q. |
|
Answer» Solution :The electric FIELD is given by `E=(1)/(4pi epsilon_(0))(q)/(R^(2))` and electric POTENTIAL (V) with distance r due to a point CHARGE Q. `V=(1)/(4 pi epsilon_(0))(q)/(r )` 
|
|
| 42702. |
A block is pulled along a rough level surface at constant speed by the force vecP. The figure shows the free-body diagram for the block.vecF_(N) represents the normal force on the block, and vecf represents the force of kinetic friction If the coefficient of kinetic friction, u, between the block and the surface is 0.30 and the magnitude of the frictional force is 80.0 N, what is the weight of the block? |
| Answer» ANSWER :D | |
| 42703. |
A block is pulled along a rough level surface at constant speed by the force vecP. The figure shows the free-body diagram for the block.vecF_(N) represents the normal force on the block, and vecf represents the force of kinetic friction What is the magnitude of vec(F_N) ? |
|
Answer» What is the MAGNITUDE of ? |
|
| 42704. |
Two resistors when connected in series net resistance is 5 Omega and when they are connected in parallel net resistance is 1.2 Omega. What are these resistors ? |
|
Answer» 2` Omega , 3 Omega` SERIES connection `R_(1) + R_(2) = 5 Omega` .... (1) Parallel connection `(R_(1) R_(2))/(R_(1) + R_(2)) = 1.2 Omega ""` .... (2) By using VALUE of (1) in (2), `(R_(1)R_(2))/(5) = 1.2` `THEREFORE R_(1) R_(2) = 6 ""`.... (3) From given options , (A) 2 `Omega, 3 Omega` MATCHES with EQUATION (1) and (3) . |
|
| 42705. |
Which of the following is decomposed by sunlight |
|
Answer» NaCl |
|
| 42706. |
Two point sources oscillating in the same phase are on a straight line perpendicular to a screen. The nearest source is at a distance of D » lambda from the screen. What shape will the interference fringes have on the screen? What is the distance on the screen from the perpendicular to the nearest bright fringe if the distance between the sources is l=n lambda (n is an integer)? Note that lambda « l. |
|
Answer» Solution :KEY IDEA Here we first try to see why an interference pattern will be obtained. The interference pattern is observed because as seen on the screen, the paths of the waves coming from two sources are not equal. Calculation: The intensity on the screen will be maximum when `d_(2) -d_(1)= k lambda`. The locus of points on the screen REACHED by RAYS from both sources with this difference is a CIRCLE with its CENTER at point A. For this reason, the interference fringes will have the form of concentric circle. When `I = n lambda` the intensity will be maximum at point A (an interference maximum of the nth ORDER). The nearest bright interference fringe circle of the `(n - 1)^("th")` order is at a distance from point A determined from the equation, `d_(2)-d_(1)=sqrt((n lambda+D)^(2)+h_(n-1)^(2))-sqrt(D^(2)+h_(n-1)^(2))=(n-1)lambda` Bearing in mind the conditions of the problem that h «D and `lambda` « l, we get `h_(n-1) ~= sqrt((2D(D+n lambda))/(n))=sqrt(2D lambda((D)/(l)+1))` Learn: One interesting aspect of this problem is that the order of maxima at any point other than the center will be less than n. Recall your mathematics classes. In any triangle, the third side is always longer than the difference of the other two sides. Since in the triangle formed by the two sources and point on the screen, the distance between two sources is `n lambda`, the path difference at any arbitrary point should be less than that. 
|
|
| 42707. |
(a) Derive an expression for the elecrtric field E due to a dipole of length '2aat a point distant r from the centre of the dipole on the axial line. (b) Drawa graph of E versus r for r gt gt a. (c)If the diploe were kept in a uniform external electric fields E_(0)Diagrammatically represent the position of the dipole in stable and unstable equilibrium and write the expressions for the torque acting on the dipole in both the cases. |
|
Answer» Solution :(a) LET there be an ELECTRIC dipole consisting of charges -q and + q , separated by distance 2n and placed in vacuum . LetP be a point on the axial line at a distance r from centre O of the dipole on the dipole on the side of the charge `+ q . overset(to)(E)` due to charge- q at P is ` overset(to)(E)_(-q) = (-q)/(4piin_(0) (r + a)^(2)) hatP`(towards left) (along AB) Where, `hatP`unit vector along dipole AXIS from q to q ` overset(to)(E)_(-q) ` due to charge + q at P is ` overset(to)(E)_(-q) = (-q)/(4piin_(0) (r + a)^(2)) hatP`(along BP) ` overset(to)(E)_("axial")=overset(to)(E)_(+q) +overset(to)(E)_(-q)` ` overset(to)(E)_("axial")=(q)/(4piin_(0))= [(1)/((r- a)^(2)) - (1)/((r+ a)^(2)) ] hatP` ` overset(to)(E)_("axial")=(q)/(4piin_(0))= (4ar)/((r^(2) - a^(2)) )hatP = (1)/(4 piin _(0))XX (2Pr)/((r^(2) - a^(2))) hatp ` ( Directed towards `bar(BP)` product.) ` P = 2a xx q ` = Dipole moment For `r gt gt a , a^(2)` is neglected `rArroverset(to)(E)_("axial")=(1)/(4 pi in _(0)) (2P)/(r^(3))`(towardsright) (b)` E = (1) /(4 pi in _(0))xx (2p)/(r^(3))`  (c)Torque`tau = PE sin theta` In stable equillbrium ` theta = 0` (##SB_PHY_XII_17_OD_E01_026_S02.png" width="80%"> |
|
| 42708. |
Which quantity associated with the projectile motion remains constant during entire motion? |
|
Answer» kinetic ENERGY |
|
| 42709. |
Statement I: If the accelerating potential in an X-ray tube is increased, the wavelengths of the characteristic X-rays do not change. Statement II: When an electron beam strikes the target in an X-ray tube part of the kinetic energy is converted into X-ray energy. |
|
Answer» statement-1 and 2 are true and statement-2 is a CORRECT EXPLANATION for statement-2 |
|
| 42710. |
Relative permeability of iron is 5500. Its magnetic susceptibility is |
|
Answer» 5501 |
|
| 42712. |
A short pulse of white light is incident from air to a glass slab at normal incidence. After travelling through the slab, the first colour to emerge is |
|
Answer» blue `therefore v prop lambda` Since red light has maximum wavelength in the white light, its velocity WOULD be maximum and so it will COME out of glass slab first of all the remaining colours. |
|
| 42713. |
एकल समुच्चय क्या है ? |
|
Answer» एक से अधिक अवयव वाला समुच्चय |
|
| 42714. |
gamma-rays are also called electromagnetic waves. Why? |
| Answer» Solution :`GAMMA`-RAYS are photons of SHORT wavelength. They travel with the speed of em waves and exhibit those phenomena which are being exhibated by electromagnetic waves. HENCE `gamma`-rays are CALLED electromagnetic waves. | |
| 42715. |
In the previous question instead of spherical wall there is a vertical wall at a perpendicular distance d from the point & where the light is incident. |
Answer» SOLUTION : `TAN theta=x/d` `RARR x=d tan theta` `(DX)/(dt)=d sec^(2) theta (d theta)/(dt)` `=2 omega d sec^(@) theta""( :' (d theta)/(dt) =2 omega)` OR Considering an instantaneus cricle of radius `d sec theta`.  `v_(t)= 2 omega d sec theta` (`2 omega s cos theta` is a component of V.) `v cos theta=2 omega d sec theta` `rArr v= (2 omega sec theta)/(cos theta)=2 omega d sec^(2) theta` |
|
| 42716. |
The equation of a wave travelling on a string stretched along the x-axis is given by y = A ' where A, a and T are constants of appropriate dimensions |
|
Answer» The SPEED of the wave is a/T. |
|
| 42717. |
In a direct vision spectroscope there are two flint glass prisms each of angle 5^(@) and dispersive power 0.36 and two crown glass prisms of dispersive power 0.24. Calculate the angle of each crown glass prism and the net dispersion produced by the system of prisms. (mu_("crown")=1.5 and mu_("flint")=1.68) |
|
Answer» |
|
| 42718. |
Find the force between 2C and -1C separated by a distance 1m in air(in newton). |
|
Answer» `18 X 10^9` |
|
| 42719. |
If the magnetic field is double through a coil of turn n then the induced e.m.f. will be: |
|
Answer» 4 times |
|
| 42720. |
The ratio of the force between two charges in air and that in a medium of dielectric constant K is |
|
Answer» K : 1 |
|
| 42721. |
______________are used to kill germs in water purifires. |
| Answer» SOLUTION :ULTRAVIOLET RAYS | |
| 42722. |
Two coils are being moved out of magnetic field, one coil is moved rapidly and the other slowly. In which case is more work done and why? |
| Answer» Solution :In the CASE of RAPIDLY MOVING COIL Because INDUCED e.m.f will be more in the rapidly moving coil as compared to slowly moving coil. | |
| 42723. |
(A) The electron in the hydrogen atom passes from energy level n = 4 to the n= I level. The maximum and minimum number of photon that can be emitted are six and one respectively. The photons are emitted when electron make a transition from the higher energy state to the lower energy state. |
|
Answer» Both 'A' and 'R' are true and 'R' is the CORRECT EXPLANATION of 'A'. |
|
| 42724. |
If vecA and vecB are two vectors then the correct statement is |
|
Answer» `VECA+VECB = vecB+vecA` |
|
| 42725. |
In young's double slit experiment the fringes width can be increased by decreasing : |
|
Answer» SEPARATION of the slits |
|
| 42726. |
The angular momentum is? |
|
Answer» SCALAR quantity |
|
| 42727. |
STATEMENT-1: A small metal ball is suspended in a uniform electric field with an insulated tea. Of high energy X-ray beam falls on the ball, the tall will be deflected in the electric field. because STATEMENT-2: X-ray emit photoelectrons and metal becomes negatively charged. |
|
Answer» Statement-1 is TRUE , Statement-2 is True , Statement-2 is a CORRECT explanation for Statement-1. |
|
| 42728. |
The mass of electrons , when it is accelerated to kinetic energy of 10000 eV is : |
|
Answer» `9.28 xx10^(-31) KG` |
|
| 42729. |
A man is walking on a road with a velocity of 3km/h . Suddenly rain starts falling . The velocity of the rain is 10 km/h in vertically downward direction. The relative velocity of the rain is |
|
Answer» Solution :Rel. VELOCITY of rain w.r.t MAN is. `sqrt(3^2+10^2) = sqrt(109) (CM)/H)` |
|
| 42730. |
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length ofthe wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what isthe emf of the second cell ? |
|
Answer» Solution :Here `epsi_1 = 1.25 V, l_1 = 35.0 CM " and" l_2 = 63.0 cm` If emf of the second cell be `epsi_2`, then `epsi_1/epsi_2= l_1/l_2` ` THEREFORE epsi_2 = epsi_1 .l_2/l_1 = (1.25 xx 63.0)/(35.0) = 2.25 V` |
|
| 42731. |
In the above question, if the disc executes rotatory motion, its angular acceleration will be :- |
|
Answer» `2.5 "rad/sec"^2` |
|
| 42732. |
A semiconductor has an electron concentration of 0.45 xx 10^(12)m^(-3)and a hole concentration of 5.0 xx 10^(20) m^(-3) Calculate its conductivity. Given electron mobility =0.135 m^2 V^(-1) s^(-1) , hole mobility =0.048m^2 V^(-1) s^(-1) |
|
Answer» Solution :The conductivity of a semiconductor is the SUM of the conductivities DUE to electrons and holes and is given by `sigma= sigma_e+ sigma_h= n_ee mu _e + n_he mu_h= e ( n_e mu_e+n_hmu_h)` As per given date, N is negligible as compared to n, so that we can write ` sigma= e n_hmu_h ` `= ( 1.6 xx 10^(-19)C )/ ( 5.0 xx 10^(20) m^(-3) ) ( 0.048 m^2 V^(-1)S^(-1))` `= 3.84 Omega^(-1)m^(-1)= 3.84S m^(-1)` where S (siemen) stands for `Omega^(-1)` |
|
| 42733. |
A ray of light is incident at an angle of 60^(@) on one face of a prism of angle 30^(@). They ray emerges normally from the other face of the prism. The refractive index of the prism material is |
| Answer» Answer :B | |
| 42734. |
Total internal reflection of light ray from cylindrical optical fiber An optical fiber consists of a cylindrical core of unknown refractive index kept in air. A ray of light is incident on the flat face of the core so that it gets totally reflected at the curved surface (Fig.34-19). What is the possible refractive index if the ray is totally reflected at any angle of incidence phi from air ? |
|
Answer» Solution :When a ray refracts from air to glass, total internal reflection cannot occur. However, when this ray strikes curved surface of the fiber, the incidence is from denser to rarer medium. If n is such that total internal reflection occurs at this point, it will continue to occur as the ray strikes the curved surface repetedly at the same angle. Calculation : LET us assume that the ray is INCIDENT at an angle `phi` from air and is refracted into the cylinder at an angle `theta`. It is then again incident on the air cylinder boundary at the curved surface as shown in Fig. 34-19 . The condition is that the ray should undergo total internal reflection at the face. For this the angle of incidence here should be greater than critical angle. `90^(@)-theta gt theta_(c )` Since sine is an increasing function, taking sine of both sides will not CHANGE the inequality. `sin(90^(@)-theta) gt sin theta_(c )=(1)/(n)(nsintheta_(c )=1xxsin90^(@))` `costheta ge (1)/(n)` Again, applying Snell.s law at flat face, we get `1xxsinphi=nxxsintheta` Squaring and adding , we get `n^(2)sin^(2)theta+n^(2)cos^(2)theta ge 1+sin^(2)phi` `n^(2) ge 1 +sin^(2) phi` But the maximum value of `sin phi` is `1` when the angle of incidence is `90^(@)`. For this condition, `n^(2) ge 2` or `n ge sqrt(2)` Learn : Interestingly this yields `mu ge 1.41`. Thus, even ordinary glass can be used in optical fiber. But for practical reasons, this core is coated with a cylindrical SHELL called cladding. If cladding has refractive index of `mu_(1)`, the condition on `mu_(2)` is `mu_(2) ge sqrt(mu_(1)^(2)+1)` 
|
|
| 42735. |
The number (0) zero is required for |
|
Answer» TRANSISTOR |
|
| 42736. |
Imagine a man swimming along a current carrying conductor in a direction opposite to that of current and facing the conductor. A magnetic needle free to rotate in a horizontal plane is mounted on a stand under the wire. Then |
|
Answer» The NORTH pole of the NEEDLE will deflect towards his LEFT hand |
|
| 42737. |
What change is observed in interference pattern of Young's double slit experiment , if one of the two slits is painted. So that it transmits half the light intensity of the other? |
|
Answer» Solution :When the slit is not painted `a_1^2=a_2^2=1` `therefore I_max=(sqrtI+sqrtI)^2 =4I`and ` I_min=0`. When the slit is painted `a_1^2=2,a^2_2=I/2` `therefore I_max=(sqrtI+sqrtI/2)^2=2.915 ` I I_min =(sqrtI-sqrtI/2)^2=0.086 I. THUS , upon painting the maximum intensity DECREASES and minimum intensity increases. HENCE , there is decrease in CONTRAST. |
|
| 42738. |
A wireof resistancex ohm is drawout, so thatits lengthin increased totwiceits original length,and its new resistancebecomes 20 Omegathen x will be . |
| Answer» Answer :A | |
| 42739. |
A body is falling freely from a point A at certain height from the ground and passes through point B, C and D (vertically as shown) so that BC=CD. The time taken by the particle to move from B to C is 2 seconds and from C to D 1 second. Time taken to move from A to B in second is |
|
Answer» 0.6 |
|
| 42740. |
Draw the acceleration-time graphs for the given velocity-time graphs of a particle moving along x-axis. |
Answer» SOLUTION :
|
|
| 42741. |
A particle is vibrating in simple harmonic motion with an amplitude of 4 cm. At what displacement from the equilibrium position is its energy potentil and half kinetic ? |
| Answer» Answer :C | |
| 42742. |
Derive radioactive decay equation using laws of radioactivity. |
|
Answer» Laws of Radioactive disintegration 1. Radioactivity is spontaneous process which does not depend upon external factors. During disintegration either `alpha` or `beta`-particle is emitted. Both are never emitted simultaneously. 3. Emission of `alpha`-particle decreases atomic number by two and mass number by 4. 4. Emission of `beta`-particle increases atomic number by one but mass number remains the same. 5. Emission of `gamma`-ray does not change atomic or mass number. 6. The number of atoms disintegrated per second is directly proportional to the number of radioactive atoms actually present at that instant. This law is called radioactive decay law. i.e. `-(dN)/(dt) prop N` or `-(dN)/(dt)= lambda N`,...(i) where `lambda` is a constant called disintegration constant and depends upon the NATURE of the radioactive substance. Now from (i) , we have `(1)/(N) dN = - lambda dt` or `INT 1/N dN = lambda int dt` or `log_(e)N=lambda t +C`,...(ii) where C is constant of integration. To determine its value. Let `N=N_(0)` initially, i.e. when `t=0, N=N_(0)` `log_(e)=N=0+C` Substituting the value of C in (ii) , we have `log_(e)N=-lambda t + log_(e)N_(0)` or `log_(e)N-log_(e)N_(0)=-lambda t` or `log_(e). (N_(0))/(N)= -lambdat` or `(N)/(N_(0))=e^(-lambda t)` or `N=N_(0)e^(-lambda t)` which is the required equation. Disintegrationconstant `(lambda)` is defined as the time after which the number of radioactive atoms reduce to 1/e times the ORIGINAL number of atoms. HALF life period (T) is the time during which the number of atoms of a radioactive material reduces to half of the original number . |
|
| 42743. |
The horizontal component of the earth's magnetic field at any place is |
|
Answer» `B_(V) = B_(H)tandelta` `=(0.36 xx 10^(-4) xx tan60^(@)` `=0.36 xx 10^(-4) xx SQRT(3) = 0.622 xx 10^(-4)` T `=0.622 xx 10^(-4)Wb//m^(2)` |
|
| 42744. |
A spherical copper weight of 3.0 cm radius submerged in olive oil hangs from a spring whose elasticity coefficient (force constant) is 1.0xx10^(2)N//m Fig. Find the natural frequency of the oscillatory system, its Q-factor and the time the oscillations will take to practically damp out. |
|
Answer» |
|
| 42746. |
A plane electromagnetic wave propagating along x - direction can have the following pairs of vec(E ) and vec(B). |
|
Answer» `E_(X), B_(y)` In given question (from data) electromagnetic waves propagate in x - direction, hence, electric field and magnetic field vector can be in y - direction or z - direction. Hence, option (B) and (D) are correct. |
|
| 42747. |
Two capillary tubes made of glass are dipped in a beaker containing water. The diameter of A is twice that of B. Which one of the following will be observed? |
|
Answer» Rise of WATER in 'A' is twice that in 'B' |
|
| 42748. |
A stone is dropped from a certain height which can reach the ground in 5 sec. It is dropped after 3 second of it's fall and then is again released . The total time taken by the stone to reach the ground |
|
Answer» 4 s |
|
| 42749. |
Polaroid sheets are often used for making sunglasses. This is because polarised glasse |
|
Answer» Cut off glare |
|
| 42750. |
A centripetal force is considered as- |
| Answer» Answer :A | |