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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. |
The SI units of magnetic susceptibility are |
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Answer» `Am` |
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| 2. |
An inductor 20 mH, a capacitor 50 muF" and a resistor "40omega are connected in series across a source of emf v=10sin340 t. The power loss in AC circuit is |
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Answer» 0.76 W |
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| 3. |
An ideal inductor of 5/(pi) H inductance is connected to a 200 V, 50 Hz a.c. supply. (a) Calculate the rms and peak value of current in the inductor. (b) What is the phase difference between current through the inductor and the applied voltage ? How will it change if a small resistance is connected in series with this inductor in the circuit ? |
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Answer» Solution :Here L=`5/piH,V_(rms)=200V` and frequency of a.c. V = 50 Hz (a) The nns value of current `I_(rms)=(V_(rms))/(X_(L))=(V_(rms))/(L.2piv)=200/(5/pixx2pixx50)=0.4A` and peak value of current `I_(m)=sqrt2I_(rms)=sqrt2xx0.4=0.57A` (b) The phase difference between current I through the conductor and the applied voltage is `pi/2` radian and I lags behind V. If a small RESISTANCE R is connected in SERIES with this inductor L, then current I lags behind the voltage V m phase by an phase angle `phi` where `tanphi=(X_(L))/R=(Lomega)/R`. Thus, `0^(@)ltphiltpi/2`. |
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| 4. |
An electron of mass m_(e ) and a proton of mass m_(p) are accelerated through the same potential. Then the ratio of their de Broglie wavelengths is |
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Answer» <P>1 |
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| 5. |
Usually it is the negative charge that is transferred when two bodies are rubbed together. Can you explain why? |
| Answer» SOLUTION :The electrons are very light and loosely bound to the ATOMS than the positive charges. That is why negative charge will transfer when two BODIES are rubbed TOGETHER. | |
| 6. |
A block of mass m=5kg is placed on the wedge of mass M=32kg as shown in the figure. Find the acceleration of wedge with respect to ground. (Neglect any type of friction. Spring and pulley are ideal) |
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Answer» `(1)/(2)m//s^(2)` FBD of wedge from GROUND frame `mg sintheta-ma costheta-T=ma` `RARR mg sintheta-T=ma(1+costheta)`……(`ii`) `N=m(gcostheta+asintheta)`…….(`III`) Using `(i) +(ii) (1+costheta)+ (iii) sintheta` `mg sintheta+ mg sintheta costheta=` `Ma+ma(1+costheta)^(2)+mg sintheta costheta+masin^(2)theta` `rArr a=(mg sintheta)/(M+2m(1+costheta))` given `theta=37^(@)`, `m=5kg` and `M=32kg` so, `a=(3)/(5)m//s^(2)` 
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| 7. |
When a source and observer move ini a parallel track as shown in the figure, the ratio of approx and actual frequency can be: |
| Answer» ANSWER :A::B | |
| 8. |
A short linear object of length l lies along the axis of a concave mirror of focal length f at a distance u from the pole of mirror. The size of image is nearly equal to |
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Answer» `l (U + F)^(2))` From eqn. (1), Longitudinal size of IMAGE, `dv = ((f)/(u-f))^(2) . L` |
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| 9. |
A liquid of refractive index 1.6 is placed between two planoconvex identical lenses, the medium of which has refrective index 1.5. Two possiblearrangements P and Q are shown. Then the system is |
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Answer» <P>DIVERGENT in P, CONVERGENT in Q |
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| 10. |
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be – |
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Answer» <P> Solution :Resistance of bulb `R= (V^2)/P =((220)^2)/(100) =22xx22 = 484 Omega`Power CONSUMED by the bulb when connected to mainsline `P =(V_1^2)/(R) = (110xx 110)/(484) =25 watts` |
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| 11. |
What for an inductor is used ? Given some examples. |
| Answer» Solution :INDUCTOR is a device used to store energy in a magnetic FIELD when an electric current flows through it. The typical EXAMPLES are coils, SOLENOIDS and toroids. | |
| 12. |
A car is negotiating a curved road of radius R. The road is banked at an angle theta . The coefficient of friction between the types of the car and the road is mu_s. The maximum safe velocity on this road is |
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Answer» `SQRT(G R ""(mu_(s) + TAN THETA)/(1 - mu_(s) tan theta))` |
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| 13. |
A bee is hovering in front of the concave spherical refracting surface of a glass sculpture. (a) Which part of Fig. 34-24 is like this situation ? (b) Is the image produced by the surface real or virtual and (c ) is it on the same side as the bee or the opposite side ? |
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Answer» |
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| 14. |
Show that Lenz's law is in accordance with the law of conservation of energy. |
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Answer» SOLUTION :i. Lenz's law can be established on the basis of the law of conservation of energy. ii. The EXPLANATION is as follows : According toLenz's lawl, when a magnet is moved EITHER towards or away from a coil, the induced current produced OPPOSES its motion. iii. As a result, there will always be a resisting force on the moving magnet. Work has to be DONE by some external agency to move the magnet against this resting force. iv. Here the mechanical energy of the moving magnet is converted into the electrical energy which in turn, gets converted into joule heat in the coil i.e., energy is converted from one form to another. |
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| 15. |
If the length of objective lens in increased, then magnifying power of |
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Answer» microscpoe SILI INCREASE but that of telescope decrease |
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| 16. |
In a potentiometer circuit, two wires of same matirial resistivity rho, one of coross-section 'a'and other of radius of cross-section '2a' are joined in series. They are of length l and 2 l respectively. This combination acts as the potentiomter wire of length 3l. The emf of the primary circuit is epsi and internal resistance is (rhol)/(2pia^(2)) This cell is conneted to the potentiometer wire by a conducting wire of negligible resistance wire positive terninal of the cell conneted to one end (call it A) of longer wire. The negative terminal of the cell is conneted to one end of the samller wire. The remaining ends of the two wires are joined together. Find : (i) The maximum voltage which can be balanced on the potentiometer wire. (ii) The length, measured from point A, where cell of emd epsi/2 will balance. (iii) if positive terminal of cell of emf epsi/2 and internal resistance (pl)/(2pia^(2)) is connetcted to pointA and other terminal is joined to the junction of the two wires, then find the current through this cell. terminal is joined to the junction of the two wires, then find the current through this cell. |
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Answer» `(5L)/(2)` (iii) `EPSI/(7R)," where R="(rhol)/(A)" and A="2pia^(2)` |
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| 17. |
Unpolarised light of intensity I_0 passes through five successive polaroid sheets,each of whose transmission axis makes an angle 45^0 With the previous one.The intensity of the finally transmitted light is |
| Answer» Answer :B | |
| 18. |
The dimensions of a copper plate are as shown in the diagram (Fig.) When the longitudinal voltage is Deltapsicurrent i flows in the conductor. If a magnetic field with induction B perpendicular to the plate is established with the current still flowing, a Hall voltage of Deltavarphi_(H) appears between the upper and the lower faces of the plate. Find the concentration and the mobility ofconduction electrons in copper, if l = = 60 mm, h = 20 mm, d = 1.0 mm, Deltapsi=0.51 mV, Deltavarphi_(H)=55 nV, i=10 A, B=0.1 T. |
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Answer» |
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| 19. |
A point charge is placed at the centre of a hollow conducting sphere of internal radius ' and outer radius '2r. The ratio of the surface charge density of the inner surface to that of the outer surface will be ___________ . |
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Answer» |
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| 20. |
In a thin spherical fish bowl of radius 10 cm filled with water of refractive index 4/3 there is a small fish at a distance of 4 cm from the centre C as shown in figure. Where will the image of fish appears, if seen from E |
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Answer» 5.2 cm |
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| 21. |
A coil of area A is held normal to the magnetic field B. The magnetic flux linked with it is: |
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Answer» -BA |
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| 22. |
What does SMS stands for? |
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Answer» SHORT MESSAGING service |
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| 23. |
Charge Q is uniformly distributed over a body. How should the body be divided into two parts, so that force acting between the two parts of body is maximum for a separation between them ? |
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Answer» SOLUTION :CHARGE Q is UNIFORMLY distributed over the body. Now, suppose the body is broken into two parts such that the charge on one part of body is q and on the other is Q - q. The force existing between the two parts separated by distance r will be, `F =k(q(Q-q))/r^(2)` Now, for the MAXIMUM force E. `q(Q-q) = Qq-q^(2)` should be maximum. `therefore q = Q/2` |
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| 24. |
Time period of a particle of mass m and charge q undergoing circular motion in a perpendicular magnetic field B is |
| Answer» Answer :C | |
| 25. |
Two charges - q and +q are located at points (0, 0, - a) and (0,0, a), respectively. Obtain the dependence of potential on the distance r of a point from the origin when r/a gt gt 1. |
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Answer» <P> SOLUTION :The given charges behave as an electric dipole and the dipole length is 2a, hence dipole MOMENT `p = q.2a`.If `R/a GT gt 1 or r gt gt a `, then electric potential due to thedipole is inversely proportional to `r^2` i.e., `1/r^2`. 
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| 26. |
The resistance of a carbon resistor having bonds of colours brown, black and brown is____. |
| Answer» SOLUTION :`100 OMEGA` | |
| 27. |
The energy of electron in the nth orbit of hydrogen atom is expressed as E_m=(-13.6)/(n^(2)) eV. The shortest and longest wavelength of Lyman series will be |
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Answer» 910Å, 1213Å `and (1)/(lambda_("min"))=R[(1)/((1)^(2))-(1)/OO] rArr lambda_("min")=(1)/(R) approx 910Å` |
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| 28. |
A rectangular coil of sides 'l' and 'b' carrying a current I is subjected to a uniform magnetic field vecB, acting perpendicular to its plane. Obtain the expression for the torque acting on it. |
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Answer» SOLUTION :Equivalent magnetic MOMENT of the coil, `vecm = Iahatn` `thereforevecm = I_(l)bvecn` `(I_(l)bvecn` = unit vector `BOT` to the plane of the coil) `therefore` TORQUE ` = vecm xx vecB` ` = I_(l)bvecn xx vecB` =0 (as `I_(l)bvecn and vecB` are PARALLEL or antiparallel other) |
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| 29. |
A prism of RI, sqrt2, has refracting angle of 60^@. At what angle the ray must be incident on it so as to suffer a minimum deviation through the prism? |
| Answer» ANSWER :A | |
| 30. |
Two satellites revolve round the earth with orbital radii 4R and 16 R, if the time period of first satellite is T then that of the other is |
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Answer» 4 T |
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| 31. |
In negative logic, logic state 1 corresponds to a |
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Answer» NEGATIVE VOLTAGE |
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| 32. |
A particle is placed at rest inside a hollow hemisphere of radius 'R'. The co-efficient of friction between the particle and the hemisphere is mu = (1)/(sqrt3) The maximum height upto which the particle can remain stationary is |
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Answer» `R/2` |
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| 33. |
Which of the following logic gates the given truth table represents |
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Answer» NOT GATE |
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| 34. |
Which of the following logic gates the given truth table represents |
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Answer» XOR GATE |
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| 35. |
Find the kinetic energy that must be imparted to a positron for a proton-antiproton pair to be obtained as a result of its collision with a stationary electron. |
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Answer» `evarphige(epsilon.0^2)/(2epsilon_0)-2epsilon_0~~(epsilon._0^2)/(2epsilon_0)=3.46TeV` At present there are no such accelerators. |
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| 36. |
Figure shows the results of an experiment involving photoelectric effect. The graphs A, B, C, D arerelated to the light beams having different wavelength. Choose incorrect option :- |
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Answer» BEAM B has highest frequency |
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| 37. |
At two different places the angles of dip are respectively 30^(@) and 45^(@). At these two places the ratio of horizontal component of earth's magnetic field is |
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Answer» `sqrt3 : sqrt2` `therefore B_(H) = B_(e) cos^(30) and B_(H_(2)) = B_(e) cos45^(@)` Thus, `B_(H_(1))/B_(H_(2))=(cos30^@)/(cos45^@)=(sqrt3/2)/(sqrt1/2)=sqrt3/sqrt2` |
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| 38. |
The switch in figure Iis connected to position a for a long time interval. At t = 0, the switch is thrown to position b. After this time , what are |
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Answer» the frequency of OSCILLATION of the LC CIRCUIT is `200 Hz` |
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| 39. |
What will be angle of first order maxima obtained by Fraunhoffer diffraction by single slit of width0.50 mm ,using light of wavelength 500 nm |
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Answer» `1.5XX10^(-4)` radian `d sin theta_(m)=(2m+1)(lambda)/(2)` Taking `m=1` `sin theta_(1)=(3lambda)/(2d)` `=(3xx5xx10^(-7))/(5xx10^(-4))=3xx10^(-3)` RAD if `theeta_(1)` is too small `sin theta_(1)~~ theta_(1)` `theta_(1)=3xx10^(-3)` radian |
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| 40. |
The gravitational field due to mass distribution is E=K/x^2 in the X direction (K is stationary). Taking the gravitation potential be zero at infinity. Find the value of the gravitationl potential at a distance x. |
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Answer» Solution :`E=(-dv)/DX thereforedv=-dx` and `E=K/x^3 V=undersetxoversetoointK/x^3dx=K[x^(-3+1)/(-3+1)]_x^oo=-K/2[1/X^2]_x^ooV=-^K//_2[-1/x^2]=K/(2x^2)` |
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| 41. |
Give an example that impurities decreases the surface tension of liquid? |
| Answer» Solution :A SMALL paper BOAT is provided with campher piece as it.s bottom and is allowed to float in WATER. SURFACE tension of water decreases as camphor dissolved in it and the boat moves from a PLACE of lower surface tension to region of higher surface tension. | |
| 42. |
A vertical straight conductor carries a current vertically upwards. A point P lies to the east of it as a small distance and another point Q lies to the west at the same distance. The magnetic field at P is |
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Answer» GREATER than at Q |
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| 43. |
A copper rod moves parallel to the horizontal direction. The emf. Induced will be maximum at the |
| Answer» Answer :D | |
| 44. |
The reactasnce of a capacitor C is . If both frequence and the capacitance be doubled, the new reactance will be |
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Answer» X |
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| 45. |
The last member of Lymann series of Hydrogen atom is912ÅCalculate The wavelength of series limit of blamer series. |
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Answer» Solution :Rydberg formula for the wavelengths of SPECTRAL LINES in HYDROGEN spectrum is `(1)/(lamda)=R((1)/(n_(f)^(2))-(1)/(n_(i)^2))` `:.R=(1)/(913.4Å)` `:.` The short wavelength limit `lamda_(B)` for the LYMAN series would be `(1)/(lamda_(B))=R((1)/(oo^(2))-(1)/(2^(2)))=(R)/(4)` `:.lamda_(B)=(4)/(R)=4xx913.4Å` |
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| 46. |
lnternal resistance of a battery of 2V terminal voltage is 0.2Omega and currrent flowing through is 0.5 A . So emf of battery will be ..... . |
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Answer» 1.9 V `epsilon = V + IR= 2 + 0.5 XX 0.2 = 2.1 ` V |
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| 47. |
The minimum intensity of audibility of sound is 10^(-2) watt/m^(2) intensity of sound is 10^(-9) watt/m^(2) the intensity level of this sound in decibels is : |
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Answer» 1000 L = 10 `log_(10)"" (I)/(I_(0)) = 10 log_(10)" "(10^(-9))/(10^(-12)) = 10 XX 3 ` L = 30 dB. |
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| 48. |
One mole of monoatomic gas is mixed with three moles of diatomic gas. What is molecular specific heat of the mixture at constant volume ? R = 8.31 J/mole/K: |
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Answer» 18.7 J/mole/K For diatomic gas `C_(v).=5//2 R` `(C_(v))_("MIX")=(nC_(v)+n.C.v)/((n+n.))` `=(1xx(3)/(2)R +3xx(5)/(2) R)/(1+3)` `=(9)/(4) R=(9)/(4) xx8.31` `=18.7` J/mole/K. `therefore` Correct choice is (a). |
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| 49. |
An electron and alpha particle have the same de-Broglie wavelength associated with them. How are their kinetic energies related to each other ? |
| Answer» SOLUTION :As `LAMDA=(h)/(sqrt(2mK)) and lamda_(e)=lamda_(alpha)`, hence, we CONCLUDE that `(K_(e))/(K_(alpha))=(m_(alpha))/(m_(e))`. | |
| 50. |
Power of a convex lens of refractive index 3/2 is 2.5 D in air. If it is inerted in liquid of refractive index 2, then new power will be ...... |
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Answer» -1.25 `THEREFORE 2.5=(3/2-1)((1)/(R_1)-(1)/(R_2))` … (1) for air Also,`(1)/(f.)=(""_lmu_g-1)((1)/(R_1)-(1)/(R_2))` [`""_lmu_g=(3/2)/(2)=3/4` `therefore (1)/(f.)=(3/4-1)((1)/(R_1)-(1)/(R_2))` …(2) By TAKING RATIO of equation (1) and (2), `2.5f.=(0.5)/(-0.25)` `therefore f.=(-5)/(25xx0.25)` `therefore(1)/(f.)=-1.25 D` |
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