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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 angle theta to the horizon will be formed by the surface of petrol in the tank of a motor car moving horizontally with a constant acceleration of 2.44 m//s^2 ? |
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Answer» `theta=14^@` |
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| 2. |
A lamp of 100 W works at 220 volts. What is its resistance and current capacity? |
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Answer» <P> Solution :Power of the lamp, P = 100 W |
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| 3. |
For mixed connection shown in figure derive equation of equivalent resistance. |
Answer» Solution :`rArr` As SHOWN in figure `R_(2) and R_(3)` are connected in PARALLEL with B and C and with this `R_(1)` is connected between A and B.  where R. = EQUIVALENT resistance of `R_(2) and R_(3)` connected in parallel. `THEREFORE (1)/(R.) = (1)/(R_(2)) + (1)/(R_(3)) ` `R. = (R_(2) R_(3))/(R_(2) + R_(3)) ""` ... (1) `rArr` Equivalent resistance of all three RESISTORS, `R_(eq) = R_(1) + R.` `= R_(1) + (R_(2) R_(3))/(R_(1) + R_(3)) "" ` ... (2) `rArr` Let voltage between A and C be v, then current through`R_(1)`, `I = (V)/(R_(eq)) = (V)/(R_(I) + (R_(2) R_(3))/(R_(2) + R_(3))) ` `therefore I = (V(R_(2) + R_(3)))/(R_(1) R_(2) + R_(1) R_(3) + R_(2)R_(3))` 
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| 4. |
An aeroplane flying horizontally at an altitude of 490m with a speed of 180 kmph drops a bomb. The horizontal distance at which it hits the ground is |
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Answer» 500 m |
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| 5. |
For steady current to be extablished in a closed circuit, there must be some part circuit in which…… increases in the direction of current |
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Answer» |
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| 6. |
Secondary rainbow in the atmosphere is |
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Answer» the result of polarization and dispersion of light |
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| 7. |
Two identical sonometer wires have a fundamental frequency of 500 Hz when kept under the same tension. What fractional increase in the tension of one wire would cause an occurrence of 5 beats per sec, when both wires vibrate together |
| Answer» ANSWER :B | |
| 8. |
A whistle of frequency f_0 = 1300 Hz is dropped from a height H = 505 m above the ground. At thesame time, a detector is projected upwards with velocity v = 50 ms^( -1)along the same line. If the velocity of sound is c = 300ms^(-1)find the frequency detected by the detector after t = 5s. |
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Answer» 500 Hz |
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| 9. |
When a dielectric is introduced between the plates of a capacitor, which is connected to a battery of voltage V, the charge on the plates remains unchanged. |
| Answer» Solution :False – The charge RISES to K times of its PREVIOUS VALUE, where K is the dielectric CONSTANT of given dielectric | |
| 10. |
Read the following passage and then answer questions on the basis of your under standing of the passage and the related studied concepts. Passage: For an LCR series circuit driven with an alternating voltage of amplitude Vmand angular frequency oo, the current amplitude is given as I_(m) =V_(m)/z = V_(m)/sqrt(R^(2) + (X_(L)-X_(C))^(2)) =V_(m)/sqrt(R^(2) + (Iomega -1/(c omega))^(2)) If omega is varied then for a particular frequency omega_(0), X_( C) = X_(L) and then Z = R and hence, I_(m) = V_(m)/R is maximum. This frequency is called the resonant frequency. The resonant frequency omega_(0) = 1/sqrt(LC) = Resonance of a LCR series a.c. circuit is said to be sharping current amplitude Im falls rapidly on increasing/decreasing the angular frequency from its resonant value 0. Mathematically, sharpness of resonance is measured by the quality factor of the circuit, which is given as: Q = (omega L)/R = 1/R sqrt(L/C) (e) An alternating series LCR circuit consists of aninductance of 10 mH, a capacitance of 100 uF and a resistance of 5 Omega.Compute its resonance frequency w, as well as the Q-factor. |
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Answer» Solution :Here, L = 10 mH = 0.01 H, C = 100 `muF = 100 XX 10^(-6) F` and R = `5 Omega` `therefore` Resonant frequency `omega_(0) = 1/sqrt(LC) =1/sqrt(0.01 xx 100 xx 10^(-6)) = 100 s^(-1)` and Q-factor `= (omega_(0)L)/R = (1000 xx 0.01)/5 =2` |
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| 11. |
Two points charges 20 mu C and 10muC are separated by 1m in air. Find the work done to separate them two metre apart. |
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Answer» Solution :Data SUPPLIED, `r_(1)=1m""r_(2)=2m ""q_(1)=20 xx 10^(-6)C""q_(2)=10 xx 10^(-6)C` For AIR, `epsi_(r)=1` Potential energy `U_(1)=(1)/(4pi epsi_(0)). (q_(1)q_(2))/(r_(1)) and U_(2)-(1)/(4pi epsi_(0)) .(q_(1)q_(2))/(r_(2))` WORK done `W=U_(1)-U_(2) =(1)/(4pi epsi_(0)).q_(1)q_(2) [1/r_(1)-1/r_(2)]=(1)/(4pi epsi_(0)).q_(1)q_(2) ((r_(2)-r_(1))/(r_(1)r_(2)))` `=9 xx 10^(9) xx 20 xx 10^(-6) xx 10 xx 10^(-6) xx ((2-1))/(1 xx 2) ="0.9 joules"` |
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| 12. |
A cyclist is going in a straight line at a unifrom velocity 18 km//h. The resistance force can be expressed as force kv^(2) where k=0.8 where velocity is in m//s and unit of force is newton. The mass of the cyclist with the bicycle is 100 kg. Neglect the rolling resistance force. If the power of the cyclist during the ride is 10^(x) watt, then what is x ? |
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Answer» SOLUTION :`P=kv^(2)(v)` `=kv^(3)=0.8xx(18xx(5)/(18))^(3)=100`WATT |
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| 13. |
A solid spere and a hollow sphere of the same diameter are charged to the same potential. Then |
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Answer» The charge on the hollow SPHERE will be GREATER |
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| 14. |
Read the following passage and then answer questions on the basis of your under standing of the passage and the related studied concepts. Passage : For many purposes it is necessary to change an alternating voltage from one value to another value of greater/smaller magnitude. This is done with a device called transformer. A transformer consists of two sets of coils of enamelled copper wire, insulated from each other, wound on a laminated soft iron core. The primary coil has N_(p)turns and an alternating input voltage V_(p)is applied to the primary coil. The secondary coil has N_(s)turns and produced an output voltage V_(s)when circuit of the secondary coil is an open circuit. For an ideal transformer, in which the primary has negligible resistance and all the flux in the core links both primary and secondary windings. For such a transformer V_(s)/V_(p) = N_(s)/N_(p) =k Why do we take the transformer core of soft iron and a laminated one ? |
| Answer» Solution :The core of a TRANSFORMER is made of SOFT IRON so as to minimise energy DISSIPATION due to magnetic hysteresis. The core is made LAMINATED one so as to minimise setting up of eddy cur rents in the core material. | |
| 15. |
An open pipe is in resonance in its 2nd harmonic with tuning fork of frequency f_(1).Now it is closed at one end. If the frequency of the tuning fork is increased slowly from f_(1)then again a resonance is obtained with a frequency f_(2) . If in this case the pipe vibrates nth harmonics then |
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Answer» `n=3, f_(2) = 3/4 f_(1)`  `therefore f_(1)`, the second harmonic of `f_(1) =(2V)/(2l) = v/l` , If one end is closed, it gives only odd harmonics. If fundamental `=v//4l`, The harmonic are `(3v)/(4l), (5v)/(4l)` etc. once, the frequency starts increasing, the FIRST higher harmonic that is resonated `=(3v)/(4l)` If n=3, `f_(2) =3/4.v/l = 3/4 f_(1)` However, here is a snag. The frequency increased from v/l. `therefore 3/4f_(1)` is not greater than `f_(1)`. `therefore 5/4 f_(1)` is the answer because this is greater than `f_(1)`. So, the ansser is (b) |
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| 16. |
Centrifuging may be used for the separation of isotopes. To do this a mixture of two gases is placed inside a cylindrical vessel rotating at a high speed. Because of the action of centrifugal forces, the isotope concentration near the cylinder wall will be different from that in the contre. Compare the concentrations of the light and the heavy uranium isotopes near the centrifuge walls, if the diameter of the cylinder is 10 cm, the rotation speed is 2.0 xx 10^3 r. p.s., the temperature of uranium hexafluoride is 27^@C. For concentrations in normal conditions see xi 25.6. Find the enrichment factor in the mixture of the heavy isotope near the walls of the vessel. The term enrichment factor applies to the quotient obtained by dividing the concentration ratio during rotation by the initial concentration ratio: x = ((n_2)/(n_1)) ((n_(02))/(n_(01))) |
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Answer» `n_(02) n_(01) = 99:28 : 0.715 = 139:1`. During the rotation of the centrifuge, the isotope of greater mass DUE to the action of centrifugal forces CONCENTRATES near the walls of the vessel. Using the result obtained in the previous problem, we have `n_1 = n_(01) exp ((m_1 omega^2 r^2)/(2kT)) = n_(01) exp ((M_1 omega^2 r^2)/(2RT))` `n_2 = n_(02) exp ((M_2 omega^2 r^2)/(2RT))` The concentration ratio close to the vessel.s walls is `(n_2)/(n_1) = (n_(02))/(n_(01)) exp [((M_2 - M_1)omega^2 r^2)/(2RT)]` Taking logarithms , we obtain `"log" ((n_2)/(n_1)) = "log" ((n_(02))/(n_(01))) +(0.434 (M_2 - M_1)omega^2 r^2)/(2RT)` `= "log " 139+ (0.434 xx 3 xx 4 pi^2 xx 4 xx 10^(6) xx 25 xx 10^(-4))/(2 xx 8.3 xx 10^3 xx 3 xx 10^2)` `2.143 + 0.103 = 2.246` Hence `n_2 n_1 = 176` 1. The enrichment factor is `x = ((n_2)/(n_1)) ((n_(02))/(n_(01))) = exp [((M_2 - M_1) omega^2 r^2)/(2RT)]` The obvious result is `log x = 0.103 `, given x = 1.27. |
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| 17. |
An object moving with a speed of 6.25 m/s , is decelerated at a rate given by: (dv)/(dt)= -2.5sqrt(v) where v is the instantaneous speedThe time taken by the object to come to rest would be: |
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Answer» 2s `:. (dv)/(sqrt(v)=-2.5 dt` or , `sqrt(v)` =- 2.5 =- 2.5 t +c: c=intergration constant. At t=0, v=6.25 as GIVEN in the question, puttings above `:. 2sqrt(6.25) =-2.5 xx0+c` or c=5.0 `:. 2sqrt(v)` = -2.5 t +5.0 When the OBJECT will COME to rest v will be zero putting above, `:.t=(5.0)/(2.5)=2s` |
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| 18. |
Read the following passage and then answer questions on the basis of your under standing of the passage and the related studied concepts. Passage : For many purposes it is necessary to change an alternating voltage from one value to another value of greater/smaller magnitude. This is done with a device called transformer. A transformer consists of two sets of coils of enamelled copper wire, insulated from each other, wound on a laminated soft iron core. The primary coil has N_(p)turns and an alternating input voltage V_(p)is applied to the primary coil. The secondary coil has N_(s)turns and produced an output voltage V_(s)when circuit of the secondary coil is an open circuit. For an ideal transformer, in which the primary has negligible resistance and all the flux in the core links both primary and secondary windings. For such a transformer V_(s)/V_(p) = N_(s)/N_(p) =k (a)Distinguish between a step up transformer and a step down transformer. |
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Answer» SOLUTION : A STEP up TRANSFORMER is that which transforms a strong current at low operating voltage into a weak current at high operating voltage, i.e.,`V_(s) gt V_(p)` or `N_(s) gt N_(p)` but `I_(s) lt I_(p)` A step down transformer transforms a weak current at high operating voltage into a strongcurrent at a low output voltage i.e., `V_(s) lt V_(p)` or `N_(s) lt N_(p)` but `I_(s) gt I_(p)`. |
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| 19. |
How can the resolving power of a telescope be increased ? |
| Answer» SOLUTION :By INCREASING the DIAMETER of the OBJECTIVE. R.P. `PROP` diameter of Objective. | |
| 20. |
Read the following passage and then answer questions on the basis of your under standing of the passage and the related studied concepts. Passage : For many purposes it is necessary to change an alternating voltage from one value to another value of greater/smaller magnitude. This is done with a device called transformer. A transformer consists of two sets of coils of enamelled copper wire, insulated from each other, wound on a laminated soft iron core. The primary coil has N_(p)turns and an alternating input voltage V_(p)is applied to the primary coil. The secondary coil has N_(s)turns and produced an output voltage V_(s)when circuit of the secondary coil is an open circuit. For an ideal transformer, in which the primary has negligible resistance and all the flux in the core links both primary and secondary windings. For such a transformer V_(s)/V_(p) = N_(s)/N_(p) =k What current is drawn by X-ray machine from the a.c. supply if there is no wastage of energy at all? |
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Answer» Solution :As operating current of X-ray machine l, = 100 mA = 0.1 A For an ideal transformer `V_(p)I_(p) = V_(s)I_(s)` `THEREFORE` Current drawn by X-ray machine from the 240 V a.c. supply `I_(p) = V_(s)/V_(p) xx I_(s) = (30 xx 10^(3))/240 xx 0.1 = 12.5` A |
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| 21. |
Read the following passage and then answer questions on the basis of your under standing of the passage and the related studied concepts. Passage : For many purposes it is necessary to change an alternating voltage from one value to another value of greater/smaller magnitude. This is done with a device called transformer. A transformer consists of two sets of coils of enamelled copper wire, insulated from each other, wound on a laminated soft iron core. The primary coil has N_(p)turns and an alternating input voltage V_(p)is applied to the primary coil. The secondary coil has N_(s)turns and produced an output voltage V_(s)when circuit of the secondary coil is an open circuit. For an ideal transformer, in which the primary has negligible resistance and all the flux in the core links both primary and secondary windings. For such a transformer V_(s)/V_(p) = N_(s)/N_(p) =k A X-ray machine rated as 30 kV, 100 mA is to be operated on an a.c. supply of 240 volts. Specify the configuration of transformer used for successfully running the X-ray machine. |
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Answer» SOLUTION : Here output voltage `V_(s) = 30 kV = 30 xx 10^(3)`V and input supply voltage `V_(p) = 240 `V. So, we must use a step up transformer WHOSE transformation ratio is : `k = N_(s)/N_(p) = V_(s)/V_(p) = (30 xx 10^(3))/240 =125` |
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| 22. |
Four open organ pipes of different lengths and different gases at same temperature are as shown in figure. Let f_(A), f_(B), f_( C) and f_( D)be their fundamental frequencies then :[Take gamma_(CO_(2)) = 7//5] |
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Answer» `f_(A)//f_(B)=2` |
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| 23. |
A canon shell explodes in mid air, its total |
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Answer» MOMENTUM increases |
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| 24. |
The north pole of a very strong electromagnet is brought near the meniscus of a liquid contained in a narrow. U - tube . The liquid is seen to rise towards the north pole. This indicates that the liquid is |
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Answer» ferromagnetic |
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| 25. |
Read the following passage and then answer questions on the basis of your under standing of the passage and the related studied concepts. Passage : For many purposes it is necessary to change an alternating voltage from one value to another value of greater/smaller magnitude. This is done with a device called transformer. A transformer consists of two sets of coils of enamelled copper wire, insulated from each other, wound on a laminated soft iron core. The primary coil has N_(p)turns and an alternating input voltage V_(p)is applied to the primary coil. The secondary coil has N_(s)turns and produced an output voltage V_(s)when circuit of the secondary coil is an open circuit. For an ideal transformer, in which the primary has negligible resistance and all the flux in the core links both primary and secondary windings. For such a transformer V_(s)/V_(p) = N_(s)/N_(p) =k Does working of a transformer violate the law of conservation of energy? |
| Answer» SOLUTION :No, working of a TRANSFORMER does not violate the LAW of CONSERVATION of energy. | |
| 26. |
A beam of electrons having energy 28 keV each strikes a target generating X rays. The minimum wave length lambda_("min") (called cut off wavelength ) of the X rays generated is |
| Answer» ANSWER :C | |
| 27. |
If the solenoid in the above problem is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30' with the direction of applied field? |
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Answer» `6.5 XX 10^(-2) J` |
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| 28. |
Two identical coils have· a common centre and their planes are at right angles to each other and carry equal currents. If the magnitude of the induction field at the centre due to one of the coil is 'B ', then the resultant magnetic induction field due to combination at their common centre is |
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Answer» B |
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| 29. |
If the gravitational force is assumed to vary as the nth power of the distance, then the time period of a planet round the sun will be proportional to : |
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Answer» `R^(N)` `therefore (mv^(2))/(R )=(GMm)/(R^(n)) rArr V= sqrt((GM)/(R^(n-1)))` Now `T=(2PI R)/(v)= (2pi R)/(sqrt((GM)/(R^(n-1))))= 2pi sqrt((R^(n+1))/(GM))` `rArr T prop R^(((n+1)/(2)))` Thus correct choice is (c ). |
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| 30. |
Upon expansion, the pressure of a gas rose linearly (Fig. ). What work did the gas perform? By how much did its internal energy increase? What quantity of heat has been supplied to it? The gas was monoatomic. What was the molar leat of the gas in this process? Compare with the specific heals al constant pressure and at constant volume. |
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Answer» `DeltaU = m/M C_(MV) (T_2 - T_1) = (C_(mv))/(R) (p_2V_2 - p_1V_1)` The quantity of heat is `Q = DeltaU + A`. The molar heat is `C_m = Q/(T_2 - T_1) = (QR)/(p_2V_2 - p_1V_1)` and its ratio to the GAS constant is `(C_(m))/(R) = (Q)/(p_2V_2 - p_1V_1) = 2.06 ` The corresponding VALUE for a monoatomic gas is `C_(mp)//R = 2.5, C_(mv)//R = 1.5`. Clearly `C_(mv) LE C_m le C_(mp)`. |
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| 31. |
An infinite collection of current carrying conductor each carrying a current I outward perpendicular to paper are placed at x = a, x= 3a, x = 5a, .... on the x-axis. Another infinite collection of current carrying conductor each carrying current inward perpendicular to paper are placed at x = 2a, x = 4a x = a... where a is a positive constant. Then the magnetic field at the origin due to the above collection of current carrying conductor is |
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Answer» ZERO |
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| 32. |
Assertion: A spherical surface surrounds a point charge. If the volume of the sphere is doubled then flux through the sphere also doubled. Reason:When the shape of a closed surface enclosing a fixed charge is changed flux through the surface also changes. |
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Answer» Both Assertion and REASON are TRUE and Reason is the CORRECT EXPLANATION of Assertion |
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| 33. |
A plano convex lens fits exactly into a plano concave lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices y, and y, and R is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is ...... |
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Answer» `(R)/(2(mu_1+mu_2))` but `f_1=(R)/(mu_1-1)` and `f_2=-(R)/(mu_2-1)` `THEREFORE 1/f=((mu_1-1))/(R)-((mu_2-1))/(R)` `therefore 1/f=([mu_1-1-mu_2+1])/(R)=((mu_1-mu_2))/(R)` `therefore f=(R)/((mu_1-mu_2))` |
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| 34. |
यदि A={1, 2, 3} , B={3, 4} , C={4, 5, 6} हो तो Auu(AuuB)होगा - |
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Answer» {3} |
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| 35. |
A polythene piece rubbed with wool is found to have a negative charge of 3 xx 10^(-7)C. a. Estimate the number of electrons transferred (from which to which?) b. Is there a transfer of mass from wool to polythene? |
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Answer» Solution :`q= 3 xx 10^(-7)C` a. `q=nc, n=q/c =(3 xx 10^(-7))/(1.6 xx 10^(-19)) =2 xx 10^(12)` (SINCE polythene has negative charge, electrons are TRANSFERRED from wool to polythene). b. YES, Mass transferred `=n xx m_(c)=9.1 xx 10^(-31) xx 2 xx 10^(12)=1.82 xx 10^(-18)kg` |
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| 36. |
List-IList-II a) Einsteine) velocity of light b) Huygenf) diffraction of light c) Focaultg) wave nature of light d) Fresnelh) particle nature of light |
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Answer» `A-H, B-G, C-F, D-E` |
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| 37. |
Which one among the following shows particle nature of light? |
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Answer» PHOTOELECTRIC effect |
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| 38. |
The energy required to shift a satellite from orbital radius r to orbital radius 2r is E. What energy will be required to shift the satellite from orbital radius 2r to orbital radius 3r? |
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Answer» E `E=(GMm)/(2r)` Also `E.=(-GMm)/(3r)-((-GMm)/(2r))=(GMm)/(r )[(1)/(2)-(1)/(3)]` `E.=(GMm)/(6r)` Thus `E.=(E )/(3)` So the CORRECT CHOICE is (d). |
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| 39. |
A stationary detector measures the frequency of a sound source that first moves at constant velocity directly toward the detector and then (after passing the detector) directly away from it. The emitted frequency is f. During the approach the detected frequency is f'_(app) and during the recession it is f'_(rec) If (f_(app) - f'_(rec)) //f= 0.200, what is the ratio v/v of the speed of the source to the speed of sound? |
| Answer» SOLUTION :`9.90 XX 10^(-2)` | |
| 40. |
In the single-slit diffraction experiment, let the wavelength of the light be 500 nm, the slit width be 6.00 mum, and the viewing screen be at distance D=4.00 m. Let a y axis extend upward along the viewing screen, with its origin at the center of the diffraction pattern. Also let I_(P) represent the intensity of the diffracted light at point P at y = 15.0 cm (a) What is the ratio of I_(P) to the intensity I_(m) at the center of the pattern? (b) Determine where point P is in the diffraction pattern by giving the maximum and minimum between which it lies, or the two minima between which it lies. |
| Answer» SOLUTION :(a) 0.487, (B) 0.60 | |
| 41. |
A test charge is moved from lower potential point to a higher potential point. The potential energy of test charge will |
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Answer» REMAIN the same |
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| 42. |
In the given circuit , the two cells have no internal resistance. Calculate the potential difference across the 20 Omega resistor . |
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Answer» Solution :Applying Kirchoff.s law to loop PRQP : 60I + `20 I_1 = 4 ""….(1)` For Loop PSRP : `200(I-I_1) - 20 I_1= 5 ""….(2)` Solving (1) and (2) , `I_1 = (5)/(172) A` `impliesV_(20) = (5)/(172) xx 20 =(25)/(43) ` or 0.594 VOLT |
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| 43. |
What is total energy of an electron ? |
| Answer» Solution :By Bohr.s theory, the total ENERGY of an ELECTRON in the nth ORBIT of hydrogen is GIVEN by, `E_n =-13.6/n^2eV`. | |
| 44. |
A particle having a charge of 10.0 muC and mass 1 mug moves in a circle of radius 10 cm under the influence of a magnetic field of induction0.1 T. When the particle is at a point P, a uniform electric field is switched on so that the particle starts moving along the tangent with a uniform velocity. The electric field is |
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Answer» 0.1 V`//`m |
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| 45. |
X-rays are produced by jumping of: |
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Answer» electrons from LOWER to higher ENERGY orbit of atom |
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| 46. |
A straight wire of length L and radius a has a cur- rent I. A particle of mass m and charge q approaches the wire moving at a velocity v in a direction anti parallel to the current. The line of motion of the particle is at a distance r from the axis of the wire. Assume that r is slightly larger than a so that the magnetic field seen by the particle is simi- lar to that caused by a long wire. Neglect end effects and assume that speed of the particle is high so that it crosses the wire quickly and suffers a small deflection theta in its path. Calculate theta. |
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Answer» |
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| 47. |
The electrons in a particle beam each have a kinetic energy of 1.6 xx 10^(-17)J. What are the magnitude and direction of the electric field that stops these electrons in a distance of 10.0cm |
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Answer» `10^(3)` v/m in the DIRECTION of VELOCITY of electrons |
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| 48. |
A double slit S_(1)S_(2) is illuminated by a coherent light of wavelength lambda. The slits are separated by a distance d. A plane mirror is placed in front of the double slit at a distance D_(1) from it and a screen Sigma is placed behind the double slit at a distance D_(2) from it. The screen Sigma receives only the light reflected by the mirror. Find the fringe width of the interference pattern on the screen. |
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Answer» `(LAMBDA)/(d)(D_(1) + D_(2))` |
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| 49. |
A coaxial cable consists of a thin inner current carrying conductor fixed along the axis of a hollow current conductor, Let B_(1) and B_(2) be the magnetic fields in the region between the conductors, and outside the conductor respectively. |
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Answer» `B_(1) != 0, B_(2) != 0` for CONDUCTORS carrying equal CURRENTS in opposite DIRECTIONS |
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| 50. |
A metallic wire of 1 m length has a mass of 10xx10^(-3)kg. If a tension of 100 N is applied to a wire, what is the speed of transverse wave ? |
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Answer» `100 ms^(-1)` The speed of the transverse WAVE is `V = sqrt((T)/(m)) = sqrt((100)/(10 xx 10^(3))) = 100 m//s` |
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