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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
What is doping in semiconductor ? Why is it done ? |
| Answer» Solution :Doping is a process of ADDITION of small AMOUNTS of certain SPECIFIC impurity atoms having valency difierent from that of host atoms to a pure semiconductor Doping is ■> done to increase the NUMBER of mobile electrons/holes and hence to increase the conductivity ofsemiconductor. | |
| 2. |
What are the forms, of energy into which the electrical energy of the atmosphere is dissipated during a lightning ? (Hint : The earth has an electric field of about 100 Vm^(-1) at its surface In the downward direction, corresponding to a surface charge density = - 10^(-9)cm^(-2). Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth). |
| Answer» Solution :The ELECTRIC energy of the ATMOSPHERE is DISSIPATED as light energy in lighting as WELL as heat and sound energy in the accompanying THUNDER. | |
| 3. |
Define the term self indutance of a solenoid. Obtained the expresssion for the magnetic energy stored in an inductor or self-inductanceL to a build up acertain I through it. |
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Answer» Solution :Self INDUCTANCE : It is EQUAL to themagnetic flux linked with the solenoid when a unit current PASSES through it. Alternatively : It is equal of the induced EMF in the solenoid when the current, though the solenoid itself changes at a unit RATE. `Here ""dU=LI(dI)/(dt)*dt " "," "U=int_(0)^(1)LIdI` |
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| 4. |
A solid cylinder of radius r_(1) = 2.5cm , length h_(1) = 5.5cm, emissivity 0.85, and temperature 30^(@)C is suspended in an environment of temperature 50^(@)C. (a) What is the cylinder's net thermal radiation transfer rate P_(1) ? (b) If the cylinder is stretched until its radius is r_(2) = 0.50 cm , its net thermal radiation transfer rate becomes P_(2). What is the ratio P_(2)//P_(1) ? |
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Answer» |
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| 5. |
Which of the following modulated signal has the best noise-tolerance ? |
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Answer» long-wave |
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| 6. |
In a single slit diffraction experiment first minimum for lambda_(1) = 660nm coincides with first maxima for wavelength lambda_(2). Then lambda_(2) is |
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Answer» 990 nm |
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| 7. |
Figure shows three arrangements of an electrons e and two protons p(where Dgtd) (a) Rank the arrangement according to the magnitude of the net electrostatic force on the electron due to the protons placing the largest first. (b) In situation (c), is the angle between the net force on the electrons and the line labeled horiontal less than or more than 45^(@)? |
Answer» SOLUTION :   (a) It is CLEAR from the diagram that `F_(a)gtF_(c)gtF_(b)` (b) SINCE `F_(1)gtF_(2), tan theta =F_(1)/F_(2)` or `thetagt45^(@)`. |
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| 8. |
A conducting ring of radius r is placed perpendicular inside a time varying magnetic field as shown in figure. The magnetic field changes with time according to B = B_0 + alphat where B_0 and alpha are positive constants. Find the electric field on the circumference of the ring. |
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Answer» `-PI alpha r^2` `(dphi)/(dt)=pir^2 d/(dt)(B_0 +alphat)` `THEREFORE (dphi)/(dt)=pir^2 (0+alpha)` `therefore -epsilon=pir^2alpha` `therefore epsilon=-pir^2alpha` |
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| 9. |
Water is heated from 0^(@)C" to "10^(@)C, then its volume |
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Answer» INCREASES  from `0^(@)C rarr 4^(@)C`, volume decreases and from `4^(@)C rarr 10^(@)C` volume increases. So, correct CHOICE is (B). |
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| 10. |
What is the meaning of 'smooth'? |
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Answer» Soft |
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| 11. |
As per phenomenon of electromagnetic induction an emf is induced in a coil when a magnetic flux linked with it changes. It is possible that emf is induced in single isolated coil due to change of magnetic flux through the coil by means of varying the current passing through the same coil. This phenomenon is called 'self-induction'. If we have coil of N turns closely wound together and a current I is passed through the coil then total magnetic flux associated with it Nphi_(B)propto I or Nphi_(B) = LI ...... (1) Here phi_(B) = flux linked with each turn of a coil. The proportionality constant L is known the self-inductance of the coil and its value depends on the dimensions of the coil and nature of core material. If the current flowing through the coil changes then an induced emfe is set up as : varepsilon =-d/dt(Nphi_(B))=-L(dI)/dt ......(ii) (a) Define self-inductance of a coil in terms of induced emf. (b) Obtain SI unit of self inductance. (c) Write its dimensions. (d) Do you agree with the statement "Self-inductance is electromagnetic analogue of mass in mechanics" ? (e) Why do we see some spark in the switch of an electric fan when a running fan is switched off? |
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Answer» (b) In the relation `lvarepsilonl = L (dI)/, If (dI)/dt = 1 As^(-1) and lvarepsilonl = 1V`, then L = 1 and it is called a henry (H). So, SI unit of inductance is that in which 1 volt emf is induced on changing the current flowing through it at a rate of 1 ampere per second. (d) Self-inductance is electromagnetic analogue of mass in mechanics because induced emf due to self-inductance opposes any change of current in a circuit by itself i.e., self-inductance plays the role of electrical inertia. (e)When a running electrical fan is switched off, current flowing in the circuit falls to zero and consequently an induced emf is produced across the switch. As a result, a LIGHT spark takes place in the AIR gap of switch. |
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| 12. |
A block of mass 5kg is lying on a rough horizontal surface. The coefficient of static and kinetic friction are 0.3 and 0.1 and g=10ms^(-2). The frictional force on the block is |
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Answer» 25 N |
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| 13. |
Assertion: Two charged metallic surfaces parallel to each other always carry equal and opposite charges. Reason: Gauss's law: oint vecE*d vecS=(q_("In"))/(epsilon_(0)) |
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Answer» If both assertionand REASON are correct and reason is a correct explanation of the ASSERTION. |
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| 14. |
If the point source is moving towards right, spherical wave fronts of sound waves get modified to |
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Answer»
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| 15. |
The cause of potential barrier in a p-n diode is |
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Answer» DEPLETION of negative charges near the JUNCTION |
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| 16. |
Two parallel plate capacitos of capacitencesC and 2C are connected in parallel and charged to a potential difference V_(0) . The battery is then disconnected and the region between the plates of the capacitor C completely filled with a material of dielectric constant 2. The potential difference across the capacitors now becomes- |
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Answer» `V_(0)/4` |
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| 17. |
A ray of light is incident on one face of a transparent slab of thickness 15 cm. The angle of incidence is 60^(@). If the lateral displacement of the ray on emerging from the parallel plane is 5sqrt(3) cm, the refractive index of the material of the slab is |
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Answer» 1.414 |
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| 18. |
(a) Explain briefly the fact that electromagnetic waves carry energy. (b) Why do we not feel the pressure due to sunshine? |
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Answer» (b) Solar radiation exert radiation pressure of the ORDER of `7 xx 10^(-6) Pa`. As the pressure due to sunshine is EXTREMELY small, we do not feel it. |
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| 19. |
Assertion : Potentiometer si muchbetter than a voltmeter for measuring emf of cell Reason: A potentiometer drawn no current while measuring emf of a cell. |
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Answer» both A and R are TRUE and R is the correct EXPLANATION of A. |
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| 20. |
Describe an experiment to determine to determine the temperature co-efficient of a thermistor using a meter bridge. |
Answer» Solution :Formula `alpha = ( 2.303 ( log R _(1) - log R _(2)))/( T _(1) - T _(2)) and R = ( Sl )/( 1-l ) Omega `  Ba-Battery K- Plug KEY S - Resistance BOX G - Galvanometer Procedure : The connections are made as shown in the circuit diagram. SUITABLE resistance S is unplugged in the STANDARD resistance box. The circuit is checked for opposite deflections by placing the slider at the two ends of the wire A & C. The thermistor is immersed in a beaker containing hot water. The slider is moved on the wire from the end A towards C till the galvanometer shown zero deflection at the point D, the balancing length l is measured. The corresponding temperature of the hot water is noted as `t_(1) ^(0) C (T_(1) = 273 + t _(1))` The resistance of thermistor `R_(1) at T_(1)` is calculated using the formula `R_(1) = Sl//1-l.` The experiment is repeated when the temperature of water decrease by a small value (about `5^(@)C)` and the corresponding resistance `R_(2)` is calculated. The temperature co-efficient of thermistor is CALCULATE using formula `alpha = ( 2.303 ( log R_(1) - log R _(2)))/( T _(1) - T _(2))` 
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| 21. |
While weight of a body depends directly on acceleration due to gravity g, the value of g depends upon many factors. It depends on the shape of the Earth, rotation of the Earth, etc. Weight of a body at a pole is more than that at a place on equator because g is maximum at poles and minimum on equator. Acceleration due to gravity g varies with latitude lambda according to the equation g_x = g - R omega^2 cos^2lambda, where R is radius of Earth and omega is angular velocity of Earth. A body of mass m weighs W_r in a train at rest. The train then begins to run with a velocity v around the equator from west to east. It conserved that weight W_m of the same body in the moving train is different from W_r"Let "v_e be the velocity of a point on equator with respect to axis of rotation of Earth and R be the radius of the Earth. Clearly, the relative velocity between Earth and train will affect the weight of the body. Difference in the weight in the two given states |
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Answer» <P>`m/Rv_e^2` |
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| 22. |
When a current through the medium, an electric field exists as well as a potential which varies in space. Suppose that there is a break in a high - voltage transmission line and the free end of a wire of length L is lying on the ground. An electric current flows through the regions of soil adjoining the conductor. If a man happens to be walking near by a potential difference, which is called the step voltage appears between the points where his feet touch the ground, Consequently, an electric current whose strength depends on this potential difference flows through the man. Let us calculate the step voltage. Since the conductor is quite long, we assume that the current flows from it to the ground in a direction perpendicular to the conductor. The equipotential surfaces are the surfaces of semi-cylinders whose axes coincide with the conductor. Suppose that the man is walking in a direction perpendicular to the conductor with a step of length 'b' the distance between the conductor and the foot closer to it being d. Assuming that the current flows uniformly from the conductor over the semi cylindrical region we obtain the following expression for the current density at a distance from the conductor: j=i/(pirL) In this case , the field strength along the radii perpendicular to the conductor is E_r=j/sigma=l/(pirLsigma) Consequently , the step voltage is V_(st)=int_d^(d+b) Edr =1/(pisigmaL)ln((d+b)/d) For example , If l=500A , d=1 m , b=65 cm and L=30 m we find that V_(st)=270 V. Much higher voltages may appear under other conditions and other shapes of conductors. When a part of a high- voltage transmission line falls on the ground, it creates a hazard |
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Answer» because there can be a DIRECT CONTACT between the CABLE and a human being |
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| 23. |
While weight of a body depends directly on acceleration due to gravity g, the value of g depends upon many factors. It depends on the shape of the Earth, rotation of the Earth, etc. Weight of a body at a pole is more than that at a place on equator because g is maximum at poles and minimum on equator. Acceleration due to gravity g varies with latitude lambda according to the equation g_x = g - R omega^2 cos^2lambda, where R is radius of Earth and omega is angular velocity of Earth. A body of mass m weighs W_r in a train at rest. The train then begins to run with a velocity v around the equator from west to east. It conserved that weight W_m of the same body in the moving train is different from W_r"Let "v_e be the velocity of a point on equator with respect to axis of rotation of Earth and R be the radius of the Earth. Clearly, the relative velocity between Earth and train will affect the weight of the body. Weight W_m of the body can be given as |
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Answer» `mg - m ((V_e + V)^2)/R` |
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| 24. |
While weight of a body depends directly on acceleration due to gravity g, the value of g depends upon many factors. It depends on the shape of the Earth, rotation of the Earth, etc. Weight of a body at a pole is more than that at a place on equator because g is maximum at poles and minimum on equator. Acceleration due to gravity g varies with latitude lambda according to the equation g_x = g - R omega^2 cos^2lambda, where R is radius of Earth and omega is angular velocity of Earth. A body of mass m weighs W_r in a train at rest. The train then begins to run with a velocity v around the equator from west to east. It conserved that weight W_m of the same body in the moving train is different from W_r"Let "v_e be the velocity of a point on equator with respect to axis of rotation of Earth and R be the radius of the Earth. Clearly, the relative velocity between Earth and train will affect the weight of the body. Difference between weight W_r and the gravitational attraction on the body can be given as |
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Answer» `(mv^2)/R` |
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| 25. |
On comparing the oscillation of damped oscillations of mechanical system with those in parallel RLC circuit connected to external source. The damping coefficient is found to be analogous to :- |
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Answer» `(1)/( R)` |
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| 26. |
Twelve wires of equal length and same cross-section are connected in the form of a cube. If the resistance of each of the wires is R , then the effective resistance between the two diagonal ends would be |
| Answer» Answer :C | |
| 27. |
Distance of closest approach when a 5.0MeV proton approaches a gold nucleus is (Z=79) |
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Answer» 16 fermi |
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| 28. |
Two identical looking iron bars A and B are given, one of which is definitely known to be magnetised (We do not know which one). How would one ascertain whether or not both are magnetised ? If only one is magnetised, how does one ascertain which one ? [Use nothing else but the two bars A and B] |
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Answer» Solution :Bring different ends of two bars A and B CLOSER. If in some situation repulsive force is EXPERIENCED then both bars are magnets. However, if force is attractive every time then only one BAR is MAGNETISED. If only one bar is magnetised then take a bar (SAY A) and bring its one end first on one end of other rod B and then on the middle of rod B. If bar A does not experience any force in the middle of B, it means that B is magnetised but A is non-magnetised. However, if there is no change in force at the end or on the middle then A is magnetised and bar B is non-magnetised. |
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| 29. |
Ray optics is valid, when characteristic dimensions are |
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Answer» of the same ORDER as the wave LENGTH of light |
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| 30. |
A prism of angle 60^(@) is made of glass of refractive index 1.50 for red and 1.56 for violet. Find the angular seperation of these rays when a narrow pencil of composite light is incident at minimum deviation. [ Hint : u=(sin(A+delta_(m))//2)/(sinA//2)] |
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Answer» |
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| 31. |
Arrange the dipoles in desending order of their potential energy . |
| Answer» SOLUTION :`U_(1) =U_(3) gt U_(2)=U_(4)` | |
| 32. |
A Nicol prism is made of |
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Answer» Quartz |
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| 33. |
Discuss the composition of two S.H.M.s along the same path having same period. Find the resultant amplitude and initial phase. A sonometer wire is in unison with a tuning fork of frequency 125Hz when it is stretched by a weight. When the weight is completely immersed in water, 8 beats are heard per second. Find the sepecific gravity of the material of the weight. |
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Answer» Solution :Consider two S.H.M.s having same period and travelling along the same straight line but the they are having different initial PHASES and different AMPLITUDES Equations of DISPLACEMENT of two S.H.M.s travelling along same straight line (`X`-axis) are `x_(1)=a_(1)sin(omegat+alpha_(1))`............`(i)` and `x_(2)=a_(2)sin(omegat+alpha_(2))` ...........`(ii)` Here `a_(1)` and `a_(2)` are amplitudes and `alpha_(1)` and `alpha_(2)` are initial phases of two S.H.M.s. Resultant displacement at any instant is equal to vector sum of its displacement due to both S.H.M.s at that instant. Resultant displacement, `x=x_(1)+x_(2)` `:. x=a_(1)sin(omegat+alpha_(1))+a_(2)sin(omegat+alpha_(2))` By using `sin(A+B)=sinAcosB+cosAsinB` and expanding above equations, we get `x=a_(1)[sinomegat*cosalpha_(1)+cosomegat*sinalpha_(1)]+a_(2)[sinomegat*cosalpha_(2)+cosomegat*sinalpha_(2)]` `:.x=sinomegat[a_(1)cosalpha_(1)+a_(2)cosalpha_(2)]+cosomegat[a_(1)sinalpha_(1)+a_(2)sinalpha_(2)]`........`(iii)` Let, `a_(1)sinalpha_(1)+a_(2)sinalpha_(2)=Rsindelta` .............`(iv)` and `a_(1)cosalpha_(1)+a_(2)cosalpha_(2)=Rcosdelta`.........`(v)` where `R` is resultant amplitude of S.H.M. and `delta` is resultant phase of S.H.M. Substitute this value in equation `(iii)`, we get `x=sinomegat*Rcosdelta+cosomegat*Rsindelta` `x=R(sinomegat*cosdelta+cosomegat*sindelta)` `:. x=Rsin(omegat+delta)`..........`(vi)` This equation shows that resultant motion is also S.H.M. whose amplitude is `R` and initial phase angle is `delta` and having same period as that of individual S.H.M.'s. Resultant amplitude (R ) of the resultant motion : SQUARING and adding equation `(iv)` and `(v)`, `R^(2)sin^(2)delta+R^(2)cos^(2)delta=(a_(1)sinalpha_(1)+a_(2)sinalpha_(2))^(2)+(a_(1)cosalpha_(1)+a_(2)cosalpha_(2))^(2)` `:. R^(2)[sin^(2)delta+cos^(2)delta]` `=a_(1)^(2)sin^(2)alpha_(1)+a_(2)^(2)sin^(2)alpha_(2)+2a_(1)a_(2)sinalpha_(1)sinalpha_(2)+a_(1)^(2)cos^(2)alpha_(1)+a_(2)^(2)cos^(2)alpha_(2)+2a_(1)a_(2)cosalpha_(1)cosalpha_(2)` `:. R^(2)=a_(1)^(2)[sin^(2)alpha_(1)+cos^(2)alpha_(1)]+a_(2)^(2)[sin^(2)alpha_(2)+cos^(2)alpha_(2)]+2a_(1)a_(2)`[sinalpha_(1)sinalpha_(2)+cosalpha_(1)csalpha_(2)]` But `sin^(2)alpha_(1)+cos^(2)alpha_(1)=1`, `sin^(2)alpha_(2)+cos^(2)alpha_(2)=1` By using `cos(A-B)=sinA*sinB+cosA*cosB` in the above equation, we get `R^(2)=a_(1)^(2)+a_(2)^(2)+2a_(1)a_(2)cos(alpha_(1)-alpha_(2))` Taking square root on both sides, we get `R=sqrt(a_(1)^(2)+a_(2)^(2)+2a_(1)a_(2)cos(alpha_(1)-alpha_(2)))`........`(viii)` This expression of resultant amplitude of the S.H.M. Divide equation `(iv)` by equation `(v)`, we get, `(Rsindelta)/(Rcosdelta)=(a_(1)sinalpha_(1)+a_(2)sinalpha_(2))/(a_(1)cosalpha_(1)+a_(2)cosalpha_(2))` `tandelta=(a_(1)sinalpha_(1)+a_(2)sinalpha_(2))` `a_(1)cosalpha_(1)+a_(2)cosalpha_(2))` `delta=tan^(-1)((a_(1)sinalpha_(1)+a_(2)sinalpha_(2))/(a_(1)cosalpha_(1)+a_(2)cosalpha_(2)))` This is the expression of initial phase angle `delta`. Numerical : Given : `n_(1)=125Hz`, Beats per second `=8` `:. n_(2)=125-8=117Hz` `(n_(1))/(n_(2))=(sqrt((sigma)/(sigma-1)))` `(125)/(117)=sqrt((sigma)/(sigma-1))` `1.141=(sigma)/(sigma-1)` `sigma=8.09` |
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| 34. |
It is a common belief that the exposure of radiations for long time is dangerous to health. Are the meals cooked in microwaves oven not dangerous to health? |
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Answer» SOLUTION :As is KNOWN, the expoure of HIGH energy radiation like `gamma-`rays, `X-`rays and ultraviolet rays for long time is dangerous to health as they spoil the blood cells of the body. Microwaves are low energy waves with energy much less than that of `gamma`-rays and X-rays. The microwavesdo not have the energy necessary to ionise molecules in the WAY, that `gamma`-rays ,X-rays can do. Thereofore, the meals cooked in microwave oven are not dangerous to health. |
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| 35. |
Derive an expression for magnetic potential energy .for a magnetic dipole kept in a uniform magnetic field. |
Answer» Solution :As shown in FIGURE a bar magnet is held at an angle `theta` with the direction of a uniform magnetic field `overset(to) (B)`. The potential energy of this bar magnet is indicated by `U_(m)`, `U_(m) = int tau (theta) d theta` `= int m B sin theta d theta` `= - m B cos theta` `U= - overset(to) (m). overset(to)(B)...(1)` SPECIAL cases : (1) If bar magnet MAKES an angle `0^@` with magnetic field `overset(to) (B)`. U = - MB which is the minimum value of potential energy. The magnet is in most stable position. (2) If bar magnet makes an angle 180° with magnetic field `overset(to) (B)`. U = + mB which is the maximum value of potential energy. The magnet is in most unstable position. (3) If bar magnet is perpendicular to `overset(to)(B)`, `U= m B cos 90^@` `U=0` |
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| 36. |
During favorable conditions bacteria mainly reproduce by |
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Answer» Budding |
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| 37. |
A charged particle oscillates about its mean equilibrium position with a frequency of 10^(9) Hz. What is the frequency of the electromagnetic waves produced by the oscillator ? |
| Answer» Solution :According to Maxwell, a charged particles oscillating with a FREQUENCY PRODUCES electromagnetic waves of same frequency waves of same frequency . Hence frequency of EM wavs produced is, `10^(9)` HZ. | |
| 38. |
The angle between electric field and equipotential surface is |
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Answer» `90^@` |
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| 39. |
Themass defect for the nucleus of helium is 0.0303 amu. The binding energy per nucleon in MeV is nearly |
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Answer» 28 |
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| 40. |
The earth takes 24 h to rotate once about its axis. How much time does the sun take to shift by IS when viewed from the earth ? |
| Answer» SOLUTION :t= 6 MIN | |
| 41. |
A metallic rod of length .L. is rotated with angular frequency of .omega. with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius R, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf between the centre and the metallic ring. |
Answer» Solution : The magnitude of the emf, GENERATED across a length dr of the rod, as it moves at right angles to the MAGNETIC field, is given by `d EPSILON = B v dr`. Therefore, `epsilon= ointd epsilon= int_(0)^(R) Bvdr = int_(0)^(R)int B omegardr= (B omegaR^(2))/(2)` |
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| 42. |
For a projectile 'R' is range and 'H' is maximum height |
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Answer» a - G , b-h, C-E, d-f |
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| 43. |
An electron accelerated under a potential difference V volt has a certain wavelength lamda. Mass of the proton is 2000 times the mass of the electron. If the proton has to have the same wavelength lamda, then it will have to be accelerated under a potential difference of: |
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Answer» V volt `:.200xxV.=V xx 1 rArr V.=(V)/(2000)` Volt |
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| 44. |
For a real object, which of the following can produce a real image? |
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Answer» a) plane mirror |
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| 45. |
A lorry carries a tank of water. If the lorry moves with a uniform acceleration 'a' in the horizontal direction, determine the angle made by the surface of water in the tank with the horizontal. |
Answer» Solution : Let the LORRY move towards the left. The lorry moving with uniform acceleration a is a non - inertial FRAME of reference. An observer in this frame of reference. An observer in this frame considers a pseudo force on the water in tank. If the lorry moves towards the left as shown in fig. the pseudo force on the water of mass m is ma to the right is REPRESENTED by `vec(OA)`. The gravitational force mg is represented by `vec(B)`. The resultant force is represented by `vec(OC)`. For equilibrium, the water surface in the tank must be perpendicular to the resultant `vec(OC)`. So, the liquid surface tilts, being raised at the rear. If `theta` is the angle made by the liquid surface with the horizontal, `theta = angle BOC`. `tan theta = (BC)/(OB)=(a)/(g) i.e., theta = tan^(-1)(a//g)` 
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| 46. |
(a) Define the term magnetic susceptibility and write its relation in terms of relative magnetic permeability. (b) Two magnetic materials A and B have relative magnetic permeabilities of 0.96 and 500. Identify the magnetic materials A and B. |
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Answer» Solution :Magnetic susceptibility `X= mu,-1`, where `mu`, is the relative permeability of GIVEN material. (b) The magnetic material A, WHOSE relative permeability is 0.96, is a DIAMAGNETIC substance. The magnetic material B, whose relative permeability is 500, is a ferromagnetic substance. |
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| 47. |
In Young's double slit experiment, the distance of the screen from the two slits in 1m. When a light of wavelength 600nm is allowed to fallon the slits width of the fringes obtained on the screen is 2mm. Calculate the width of the fringe if the wavelength of the incident light is 400nm. Calculate band width in each case if the arrangement is immersed in water of refractive index 1.33. |
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Answer» |
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| 48. |
If half life of radioactive atom is 2.3 days, then its decay constant would be : |
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Answer» `0.1 "DAYS"^(-1)` |
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| 49. |
A magnetic field in a certain region is given by barB = (40 hati - 18 hatk)G . How much flux passes through a 5.0cm^2area loop if the loop lies flat on xy - plane? |
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Answer» `-900 nwb` |
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