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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. |
Assertion : An isolated radioactive atom may not decay at all whatever be its half- life. Reason : Radioactive decay is a statistical phenomenon. |
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Answer» If both, Assertion and Reason are TRUE and the Reason is the CORRECT EXPLANATION of the Assertion. |
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| 2. |
A particle of mass m and charge q is incident on xz plane with velocity v in a direction making angle theta with a uniform magnetic field applied along x-axis. The nature of motion performed by the particle is _______ . |
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Answer» helical |
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| 3. |
The lower frequencies used in TV broad casting are |
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Answer» `0.54` KHz to `1.6` KH, 80 MHZ to 108 MH |
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| 4. |
A : Total induced emf in a loop is not confined to any particular point but it is distributed around the loop in direct proportion to the resistances of its parts. R: In general when there is no change in magnetic flux, no induced emf is produced. |
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Answer» If both Assertion & Reason are TRUE and the reason is the correct EXPLANATION of the assertion. |
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| 5. |
A triangular loop as shown in the figure is started to being pulled out at t=0 forma uniform magnetic field with a constant velocity v. Total resistancee of the loop is constant and equal to R. Then the variation of power produced inthe loop with time will be : |
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Answer» POWER produced in the loop INCREASING linearly with time till whole loop COMES out |
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| 6. |
LEDS that can emit red, yellow, orange, etc. commercially available.Write any two uses of LED. |
| Answer» SOLUTION :LEDS find EXTENSIVE USE in remote controls, bur GLAR alarm systems, optical communication, etc. | |
| 7. |
A current is set up in a wire loop consiting of a semicircle of radius4.00 cm, a smaller concentric semicircle, and two radial straight lengths , all in the plane . Figure (a) shows the arrangment but is not drawn to scale. The magnitude of the magnetic field produced at the center of curvature si 74.25 muT . The smaller semicircle is then flipped over (rotated) until the loop is again entirely in the same plane figure (b). The magnetic field produced at the (same) center of curvature now has magnitude 15.75 muT and its direction is reversed from the initial magnetic field. What is the radius of the smaller semicircle? |
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Answer» |
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| 8. |
Compare width of slits with the intensity and hence amplitude of the waves. |
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Answer» SOLUTION :WIDTH `ALPHA a^2` and INTENSITY `alpha a^2`. HENCE, `(omega_1)/(omega_2)=(a_1^2)/(a_2^2)=(I_1)/(I_2)`. |
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| 9. |
Mosley's law for characteristic x rays is given by sqrt(f)=a(Z-b) Choose the correct statement(s). |
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Answer» Both a and B depend on the target material |
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| 10. |
Bottom surface of Kettles are : |
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Answer» blackened |
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| 11. |
A voltmeter of resistance R_(V) and an ammeter of resistance R_(A) are connected as shown in an attempt to measure the resistance R. The measured value of the resistance is R_(M) = (V)/(I_(0)) where V is reading of voltmeter andI_(0) is reading of the ammeter. find the true value of the resistance in terms of R_(M), R_(V) and R_(A). |
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Answer» |
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| 12. |
Water rises to a height of 5 cm in a glass capillary tube. If the area of cross section of the tube is reduced to (1/16)^(th) of its former value, the water rises to a height of |
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Answer» 80 cm |
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| 13. |
If A and B are two non- zero vectors having equal magnitude, then angle between the vectors A and A-B is |
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Answer» `0^(@)` 
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| 14. |
At what angle to be horizon the sun be for its reflected rays from the surface of pond to be completely polarised (mu of water = 1.33) |
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Answer» `16^(@)56'` |
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| 15. |
The following table shows some measurements of the decay rate of a radionuclide sample. Find the disintegration constant. |
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Answer» Solution :`R-RO E^(-lambdat)` In R `=ln Ro-lambdat` ln `R=-lambdat+ln Ro` SLOPE of ln R v/s. t is `-LAMBDA` `-lambda=(0-1.52)/(210-254)` `-lambda =0.038"MIN ute"^(-1)` |
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| 16. |
A heavy cart is pulled by a constant force F along a horizontal track with the help of a rope that passes over a fixed pulley, as shown in the figure. Assume the tension in the rope and the frictional forces on the cart remain constant and consider motion of the cart until it reaches vertically below the pulley. As the cart moves to the right, its acceleration |
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Answer» DECREASES. |
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| 17. |
Lenz's law is a consequence of the law of conservation of |
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Answer» charge |
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| 18. |
In general viscous force is some what like friction as it opposes the motion and is a non-conservative force, but not exactly so, because. |
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Answer» a,b |
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| 19. |
A microscope consists of two convex lenses of focal lengths 2.0 cm and 6.25 cm placed 15.0 cm apart. Where must the object be placed so that the final virtual image is at a distance of 25 cm from the eye? What is the magnifying power of the microscope. |
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Answer» |
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| 20. |
The ionosphere is used for the propagation of : |
| Answer» Answer :A | |
| 21. |
What describes "tiger in a cell"? |
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Answer» LOCKED in CONCRETE cell |
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| 22. |
Calculate work done in raising a stone of mass 5 kg, specific gravity 3, lying at the bed of a lake through a height of 5 metre. |
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Answer» 163.32 J |
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| 23. |
The optical properties of a medium are governed by the relative permitivity (epsilon_(r)) and relative permeability (mu_(r)). The refractive index is defined as sqrt(mu_(1)epsilon_(r))=n . For ordinary sing is taken for the square root. In 1964, a Russian scientist V. Veselago postulated the existence of material with epsilon_(r)lt0 and mu_(r)lt0 . Since then such 'meta materials' have been produced in the laboratorised and their optical properties studied. For such materials n = -sqrt(mu_(r)epsilon_(r)). As light entes a medium of such refractive index the phases traval away front the direction ofpropagation.For light incident from air on the meta materials, then appropriate ray diagram is |
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Answer»
`-MU SIN theta R = sin theta i` |
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| 24. |
The optical properties of a medium are governed by the relative permitivity (epsilon_(r)) and relative permeability (mu_(r)). The refractive index is defined as sqrt(mu_(1)epsilon_(r))=n . For ordinary sing is taken for the square root. In 1964, a Russian scientist V. Veselago postulated the existence of material with epsilon_(r)lt0 and mu_(r)lt0 . Since then such 'meta materials' have been produced in the laboratorised and their optical properties studied. For such materials n = -sqrt(mu_(r)epsilon_(r)). As light entes a medium of such refractive index the phases traval away front the direction ofpropagation. Consider following statements i) Snall's slaw valid for these mater ials ii) Snell's slaw is not valid for these mater ials iii) Speed of light in these mater ial is v = (c)/(|n|) iv) Speed of light in these material is v = c |
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Answer» I, iii are CORRECT `-MU SIN THETA R = sin theta i` |
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| 25. |
Anelectric dipole with dipole moment 4xx 10^(-9)C-mis aligned at30^(@)with the direction of auniform electric field of magnitude5xx 10^(4) NC^(-1) .Calculate the magnitude of the torque acting on the dipole. |
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Answer» Solution :Here `p=4 xx10 ^(-9) C-m ,theta =30^(@)and E=5xx 10^(4) NC^(-1) ` ` therefore ` Magnitude of torque ` tau =pE SIN theta =4xx 10^(-9)XX 5xx 10^(4)xx sin 30^(@) = 4xx 10^(-9) xx 5xx 10^(4)xx 0.5 =10^(-4)N-m ` |
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| 26. |
Plane microwaves from a transmitter are directed normallytowards a plane reflector . A detector moves along the normal to the reflector . Between positions of 14 successive maxima , the detector travelsa distance 0.13 m . If the velocity of light is3xx10^(8) m/s ,findthe frequency of the tranmitter . |
| Answer» Answer :A | |
| 27. |
A current carrying rod AB is palace perpendicular to an infinitely long current carrying wire as shown in figure. The point at which the conductor should be hinged so that it will not rotte (AC=CB) |
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Answer» A  Variation of magnetic force on wire `ACB` is as SHOWN in figure. POINT of application of net force lies some where between `A` and `C` |
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| 28. |
Two parallelvertivcalinetallicralisAB and CDare serparated by l= 1 m . Theyare connected at theendsby resistance R_(1) and R_(2) as show in the figure . A horizationlmetallic bar L of massm= 0.2 kg slides withoutfrictionvertically down the railsunderthe actionof gravity. Theseis a unifromhorizonal magneticfield of aB=0.6 Tperpendicular to the plane torails. Itis observedthat when the termainalveloctity is attained , the powerdisspated in R_(1) and R_(2)are P_(1)= 0.76W andP_(2) = 1.2 W respectively. Findthe terminal velocityof barL andthe values of R_(1) and R_(2) |
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Answer» V=(mg)/(B^(2)l^(2)).(R_(1)R_(2))/(R_(1)+R_(2))=0.997 m//s` |
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| 29. |
A simple harmonic wave has the equation y_1 = 0.3 sin (314 t - 1.57 x)and another wave has equation y_2= 0.1 sin (314t - 1.57x+1.57) where x,y_1 and y_2 are in metre and t is in second. |
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Answer» `v_1 = v_2 = 50Hz` |
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| 30. |
Particle 1 with mass m and velocity v and particle 2 with mass 2m and velocity -2v are moving toward each other along an x axis when they undergo a one-dimensional elastic collision. After the collision, what are the velocities of (a) particle 1 and (b) particle 2? What is the velocity of the center of mass of the two-particle system (c) before and (d) after the collision? |
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Answer» |
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| 31. |
A shell flying with velocity v=500m//s burts into three identical fragements so that the kinetic energy of the system increases eta=1.5 times. What maximum velocity can one of the fragments obtain? |
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Answer» Solution :From the conservation of linear momentum of the SHELL just before and after its fragementation `3vecv=vecv_1+vecv_2+vecv_3` (1) where `vecv_1`, `vecv_2` and `vecv_3` are the velocities of its fragments. From the energy conservation `3etav^2=v_1^2+v_2^2+v_3^2` (2) Now `overset~vecv_i` or `vecv_(iC)=vecv_i-vecv_C=vecv_i-vecv` (3) where `vecv_C=vecv`=velocity of the C.M. of the fragements the velocity of the shell. Obviously in the C.M. frame the linear momentum of a SYSTEM is equal to zero, so `overset~vecv_1+overset~vecv_2+overset~vecv_3=0` (4) Using (3) and (4) in (2), we get `3etav^2=(vecv+overset~vecv_1)^2+(vecv+overset~vecv_2)^2+(vecv-overset~vecv_1-overset~vecv_2)^2=3v^2+2overset~v_1^2+2overset~v_2^2+2overset~vecv_1*overset~vecv_2` or, `2overset~v_1^2+2overset~v_1overset~v_2costheta+2overset~v_2^2+3(1-eta)v^2=0` (5) If we have had used `overset~vecv_2=-overset~vecv_1-overset~vecv_3`, then Eq. 5 were CONTAIN `overset~v_3` instead of `overset~v_2` and so on. The PROBLEM being symmetrical we can look for the maximum of any one. Obviously it will be the same for each. For `overset~v_1` to be real in Eq. (5) `4overset~v_2^2cos^2thetage8(2oversetv_2^2+3(1-eta)v^2)` or `6(eta-1)v^2ge(4-cos^2theta)overset~v_2^2` So, `overset~v_2levsqrt((6(eta-1))/(4-cos^2theta))` or `overset~v_(2(max))=sqrt(2(eta-1))v` Hence `v_(2(max))=|vecv+overset~vecv_2|_(max)=v+sqrt(2(eta-1))v=v(1+sqrt(2(eta-1)))=1km//s` THUS owing to the symmetry `v_(1(max))=v_(2(max))=v_(3(max))=v(1+sqrt(2(eta-1)))=1km//s` |
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| 32. |
The electrical resistance of a conductor |
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Answer» varies directly proportional to its area of cross-section. [HINT: Resistance R =`(pl)/A=1/(SIGMA).1/a`, where `sigma`=conductivity of the MATERIAL of CONDUCTOR.] |
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| 33. |
The thin plastic rod of length L = 12.0 cm in Fig. 24-61 has a nonuniform linear charge density lambda= cx, where c=49.9p C//m^(2). With V = 0 at infinity, find the electric potential at point P_(2) on the y axis at y = D = 3.56 cm. (b) Find the electric field component E_(y) at P_(2). (c) Why cannot the field component E_(x) at P_(2) be found using the result of (a) ? |
| Answer» SOLUTION :(a) `4.02 XX 10^(-2) V,` (B) `0.321 N//C` | |
| 34. |
The critical angle for total internal reflection in diamond is 24.5^(@). The refractive index of diamond is |
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Answer» 2.41 |
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| 35. |
To obtain the X-rays of shortest wavelength of0.5Å, potential difference applied across the electrodes of the tube should be: |
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Answer» 13.89 kV `:.(hc)/(LAMBDA)=eV rArr V=(hc)/(e lambda)=24*8kV` |
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| 36. |
What is refraction of light ? Explain laws of refraction. |
Answer» Solution :The direction of propagation of an obliquely incident ray of light that enters the other MEDIUM, changes at the interface of the two media. This phenomenon is called refraction of light. `(0^@ lt i lt 90^@)` In figure, the surface separating two media PQ. AB is incident ray, BC is refracted ray NN. `bot` PQ at B. `angle`ABN = incidence angle i, `angle`N.BC refraction angle r. `n_1` and `n_2` are refractiv indices of medium-1 and medium-2 respectively Snell experimentally obtained laws of refraction (i) The incident ray, the refracted ray and th normal to the interface at the point of incidence, all lie in the same plane. (ii) The ratio of the sine of the angle o incidence to the sine of angle of refraction is constant. `THEREFORE(sini)/(sinr)=n_(21)` ... (1) Here, `n_(21)` is a constant called the refractive index of the second medium with respect to the first medium. Equation is the well KNOWN Snell.s law of refraction. `n_(21)` depends on the wavelength of light, but is independent of the angle of incidence. |
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| 37. |
Obtain an expression for average power of AC over a cycle. Discuss its special cases. |
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Answer» SOLUTION :i. POWER of a circuit is defined as the rate of consumption of electric energy in that circuit. It is given by the PRODUCT of the voltage and current. In an AC circuit, the voltage and current vary continuously with time. Let us first calculate the power at an instant and then it is averaged over a complete cycle. ii. The alternating voltage and altelrnating current in the series RLC circuit at an instant are given by `v=V_(m)sinomegat" and "i=I_(m)=(omega+phi)` iii. where `phi` is the phase angle between v and i. The instantaneous power is then written as P = vi `=V_(m)I_(m)sinomegatsin(omegat+phi)` `=V_(m)I_(m)sinomegatsinomegat` `[sinomegatcosphi-cosomegatdsinphi]` `P=V_(m)I_(m)` `[cosphisin^(2)omegat-sinomegatcosomegatsinphi]` iv. Here the average of `sin^(2)` over a cycle is `(1)/(2)` and that of `sinomegat cosomegat`zero. Substituting these values, we obtain average power over a cycle. `P_(av)=V_(m)I_(m)cosphixx(1)/(2)` `=(V_(m)I_(m))/(sqrt(2)sqrt(2))cosphi` `P_(av)=V_(RMS)I_(RMS)cosphi` v. where `V_(RMS)I_(RMS)` is called apparent power and `cosphi` is power factor. The average power of an AC circuit is also KNOWN as the true power of the circuit. Special Cases i. For a purely resistive circuit, the phase angle between voltage and current is zero and `cosphi=1.` `thereforeP_(av)=V_(RMS)I_(RMS)` ii. For a purely inductive or CAPACITIVE circuit, the phase angle is `pm(pi)/(2)andcos(pm(pi)/(2))=0thereforeP_(av)=0` iii. For series RLC circuit, the phase angle `phi=tan^(-1)((X_(L)-X_(C))/(R))` `thereforeP_(av)=V_(RMS)I_(RMS)cosphi=1.` iv. For series RLC circuit at resonance, the phase angle is zero and `cosphi=1.` `thereforeP_(av)=V_(RMS)I_(RMS)` |
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| 38. |
What's volume of the nucleus is related to atomic number ? |
| Answer» SOLUTION :`VpropA` | |
| 39. |
An electrostatic field line is a continuous curve. That is , a field line cannot have sudden breaks. Why not? (b)Explain why two field lines nevercross each other at anypoint. |
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Answer» Solution :Electric FIELD is CONTINUOUS and exists at all POINTS around a charges distribution. Hence an electrostatic field line is a continuous curve and cannot have sudden breaks. (b) TWO field lines nevercross each other , because if they do so then at the point. Of intersection there will be two possible directions of electric field, which is IMPOSSIBLE. |
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| 40. |
Doping a semiconductor results in |
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Answer» The DECREASE in mobile charge carriers |
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| 41. |
What isthe ratio of nuclear densities of two nucleus having mass numbers in the ratio 1 : 2 ? |
| Answer» Solution :The nuclear DENSITY is INDEPENDENT of MASS NUMBER. So the ratio is ONE. | |
| 42. |
How did the explosion affect the ship? |
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Answer» A TORRENT of green and WHITE WATER broke over the ship |
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| 43. |
What do you mean by half life period of a radio active substance ? |
| Answer» Solution :Half life is the TIME TAKEN to REDUCE half of its INITIAL VALUE. | |
| 44. |
What is the maximum flux when current l_(l) = l_(0)sinomega t? |
| Answer» SOLUTION :`phi_(MAX) = -(mu_(0)l_(0)L)/(2pi)log_(e)""(a+l)/(a)` | |
| 45. |
What is the impedance of a capacitor of capacitance C in an ac circuit using source of frequency n Hz? |
| Answer» Solution :Impedance of a CAPACITOR `X_(C )=(1)/(C.2pi n)` | |
| 46. |
A beaker contains water (mu=4//3) filled to a height of 32 cm. A concave mirror is fixed 6 cm above the surface of water as shown in figure. An object is placed at the bottom of the beaker and its image is formed 14 cm below the surface of water. The focal length of the mirror is |
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Answer» |
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| 47. |
The time period (T) of small oscillations of the surface of a liquid drop depends on its surface tension (s), the density (rho) of the liquid and it's mean radius (r) as T = cs^(x) rho^(y)r^(z). If in the measurement of the mean radius of the drop, the error is 2 %, the error in the measurement of surface tension and density both are of 1%. Find the percentage error in measurement of the time period. |
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Answer» |
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| 48. |
A dipcicrleis at rightanglesto the magneticmeridian. Whatwill be theapparent dip ? |
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Answer» `0^(@)` |
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| 49. |
In the figure shown if the object 'O' moves towards the plane mirror, then the image (which isformed after successive reflections from M_1 & M_2 respectively) will move |
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Answer» TOWARDS RIGHT |
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| 50. |
An oil drop of 12 excess electrons is field stationary under constant clectric field of 2.55 xx 10^(4) NC^(-1) in Millikan's oil drop experiment. Then density of the oil is 1.26g cm^(-3). Estimate the radius of the drop. (g=9.81 ms^(-2), e=1.60 xx 10^(-19)C). |
| Answer» SOLUTION :(a) ZERO ,(B) zero(C ) 1.9 N/C | |
