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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. |
In a measurement both positive and negative errors are found to occur with equal probability. The type of error responsible for this is |
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Answer» Proportional ERRORS |
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| 2. |
Two coherent sound sources A and B produce a sound of wavelength lambda. If at t=0 a detector starts moving with constant velocity v_(0) from pointO along y - direction, then the minimum time, after which it detects the sound of maximum intensity, is (D gt gt lambda) |
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Answer» `(sqrt7D)/(v_(0))` |
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| 3. |
A man cycles up a hill whose slope is 1 in 25 at the rate 7.2 km/h. The weight of the cycle and man is 150 kg. Find the power at which he is working . |
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Answer» 88.2 W |
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| 4. |
A can of paint with a mass of 6 kg hangs from a rope. If the can is to be pulled up to a rooftop with an acceleration of 1 m//s, what must be the tension in the rope ? |
| Answer» Solution :The phrase " constant velocity" automatically means a= 0 and, therefore, `F_(" NET")=0`. In the diagram above, `F_(T)` would need to have the same magnitude as `F_(w)` in order for the can to be moving at a constant velocity. Thus, in this case, `F_(T)=F_(w)=mg=(6)(10)=60 N`. | |
| 5. |
L is a smooth conducting loop of radius l=1.0 m & fixed in a horizontal plane. A conducting rod of mass m=1.0 kg and length slightly greater than l hinged at the centre of the loop can rotate in the horizontal plane such that the free end slide on the rim of the loop. There is a uniform magnetic field of strength B=1.0 T directed vertically downward. The rod is rotated with angular velocity omega_(0)=1.0 rad//s and left. The fixed end of the rod and the rimof the loop are connected through a battery of emf . E, a resistor of resistance R=1.0 Omega, and intially uncharged capacitor of capacitance C=1.0 F in series.Find : (i) the time dependence of emf . E such that the current I_(0)=1.0A in the circuit is constant. (ii) energy supplied by the battery by the time rod stops. |
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Answer» |
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| 6. |
A slab of a material of dielectric constaant 2 is placed between two identical parallel plates. One of the plates carries total charge density +2mC/m^(2) and the other plate carries total charge density -4mC/m^(2) the magnitude of electric field inside the dielectric in (in 10^(11)N//C) Take epsi_(0)=(80)/(9)xx10^(-12)Nm^(2)//C^(2)) |
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Answer» 2.25 `E=(sigma)/(epsi_(0))=(3)/((80)/(9)xx10^(-12))=3.38xx10^(11)N//C` |
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| 7. |
How can you increase the range of an ammeter? |
| Answer» Solution :By reducing the VALUE of SHUNT resistacne or by JOINING yet another shunt of appropriate vaue in parallel . | |
| 8. |
In a non-uniform magnetic field a diamagnetic material moves |
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Answer» from WEAKER to STRONGER PART of the FIELD. |
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| 9. |
Two resistors, X and R, are connected in the left gap and the right gap of a metre bridge, and the null point in obtained at 20 cm from the left end. With X + 24 Omega in the left gap and the same resistance R in the right gap, the null point is at the centre of the wire. Then, X is equal to |
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Answer» `4.8 Omega` |
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| 10. |
When an inductor L and a resistor R in series are connected across a 12 V, 50 Hz supply, a current of 0.5 A flows in the circuit. The current differs in phase from applied voltage by pi/3. radian. Calculate the value of R. |
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Answer» Solution :Here, `V_(rms) = 12 V, v = 50 Hz, I_(rms) = 0.5 A`, and phase difference `phi = pi/3` `therefore` IMPEDANCE `Z = (V_(rms)/I_(rms)) = (12 V)/(0.5 A) = 24 OMEGA` and `tan phi = tan pi/3 = X_(L)/R` or `sqrt(3) = X_(L)/R rArr X_(L) = sqrt(3) R` As `Z = sqrt(R^(2) + X_(L)^(2))`, hence `24 = sqrt(R^(2) + (sqrt(3)R)^(2)) = 2R` `rArr R = 24/2 = 12 Omega` |
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| 11. |
A heater coil rated at 1000W is connected to a 110V mains. How much time will take to melt 625 gm of ice at 0^@ C. (forice L= 80 cal/gm) |
| Answer» ANSWER :D | |
| 12. |
In which process maximum energy is released ? |
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Answer» `S_(G)^(-)+e^(-)rarrS_(g)^(-2)` |
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| 13. |
When a metal surface is illuminated by a light of wavelength 400nm and 250 nm . The maximum velocites of the photo electrons ejected are v and 2v respectively. The work function of the metal is (h= Planck's constant e= Velocity of light in air |
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Answer» `2hcxx10^(6)J` |
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| 14. |
On what factors the speed of electro-magnetic wave depands ? |
| Answer» Solution :It DEPENDS upon the electric and magnetic properties of the MEDIUM they TRAVEL in and not on the amplitude of FIELD variation. | |
| 15. |
Force and area are measured as 20N and 5 m^2with errors 0.05 N and 0.0125 m^2. The maximum error in pressure is (SI units) |
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Answer» `4 PM 0.0625` |
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| 16. |
An object with an initial velocity of 4 m//s moves along a straight axis under constant acceleration. Three seconds later, its velocity is 14 m//s. How far did it travel during this time ? |
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Answer» Solution :We're given `v_(0), t and v` and we're asked for `Delta s`. So a is MISSING : it isn't given and it isn't asked for, so we USE BIG Five #1. `Delta s= overline(v)t=(1)/(2)(v_(0)+v) t=(1)/(2) (4 m//s +14 m//s)(3 s)=27 m` |
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| 17. |
According to C.E. van der Waal, the interatomic potential varies with the average interatomic distance (R) as |
| Answer» Answer :D | |
| 19. |
The magnetic field at the centre O of the circular portion of radius 3 cm carrying current 4 ampere in the wire is |
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Answer» `(8pi//3)xx10^(-5)T` |
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| 20. |
How will the frequency of small oscillation of a simple pendulum changes. a point of suspension moves horizontally in the plane of oscillation with acceleration a. |
| Answer» SOLUTION :INCREASES | |
| 21. |
Do you find two magnetic field lines intersecting? Why? |
| Answer» SOLUTION :No, Because at the point of INTERSECTION the magnetic field will have two DIRECTIONS which is not possible. | |
| 22. |
Three long straight and parallel wires carrying currents are arranged as shown in fig. The wire 'B' which carries a current of 6A is placed so that it experiences no force. The distance of wire 'B' from A is then |
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Answer» 8cm |
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| 23. |
What force is required to stretch a steel wire of 1 cm^2 in cross - section to double its length? (Y = 2 xx 10^12 "dyne"/(cm^2)) |
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Answer» a)`2 xx 10^12` DYNE |
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| 24. |
A rectangular loop carrying a current i_(1), is situated near a long striaght wire carrying a steady current i_(2). The wire is parallel to one of the sides of the loop and is in the plane of the loop as shown in the figure. Then the current loop will |
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Answer» move away from the wire |
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| 25. |
Bacterial viruses usually have |
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Answer» SINGLE STRANDED RNA |
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| 26. |
Unit for a fundamental physical quantity is |
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Answer» defined as best of various REFERENCE standards |
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| 27. |
Magnetic intensity , at any point on the axis of a bar "magnet"/"dipole" is _____ to the direction of vecM . |
| Answer» SOLUTION :PARALLEL | |
| 28. |
The equivalent capacitance of the combination between A and B in the given figure is 4 mu F. What will be the potential drop across each capacitor ? |
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Answer» Solution :If potential drops across the `20 mu F` and `5 mu F ` capacitors be `V_1` and `V_2` respectively , then `(V_(1))/(V_(2)) = (C_(2))/(C_(1)) = (5 mu F)/(20 mu F) = (1)/(4) implies V_(2) = 4 V_(1)` MOREOVER , `V_(1) + V_(2) = 12 V implies V_(1) = 2 .4 V ` and `V_(2) = 9.6 V ` |
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| 29. |
According to Bohr's theory, relation between n and radius of orbit is: |
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Answer» `r prop 1/N` |
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| 30. |
A long straight wire AB carries a current I.A proton P travels with a speed v, parallel to the wire , at a distance d from it in a direction opposite to the current as shown in figure what is the forceexperienced by the proton and what is its direction? |
Answer» Solution :Magnetic field DUE to current carrying WIRE is perpendicular to plane of paper- downward.  i.e `vecB = - (mu_(0)I)/(2 pi f) hatv` FORCE `vecF = q ( vecv xx vecB) = e [(-v hatj) xx ((mu_(0)I)/(2 pi d) hat v)]` ` = (mu_(0)ev I)/(2 pi d) hati ` That is the magnetic force has magnitude `(mu_(0)e v I)/(2 pi d) ` and is DIRECTED along `+ `ve X - AXIS i.e in the planeof paper perpendicular to direction of `vecv` and to the right. |
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| 31. |
If '0' is at equilibrium then the values of the tension T_(1)and T_(2)are x, y, if 20 N is vertically down. Then x, y are |
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Answer» 20 N, 30 N |
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| 32. |
The north pole of horizontal bar magnet is being brought closer to vertical, conducting plane, along perpendicular direction the induced current in the conducting plane is _____ |
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Answer» clockwise  Here TAKING N pole of MAGNET PERPENDICULAR to coil, N pole produces in the coil, so in the coil induced current produced in anticlockwise. |
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| 33. |
Two polaroids P_(1) and P_(2) are set up so that their pass axis are crossed with respect to each other. A third polaroid P_(3) is now introduced between these two so that its pass axis makes an angle theta with the pass axis of P_1. A beam of unpolarised light of intensity I_(0) is incident on P_(1). if the intensity of light, that gets transmitted through the combination of three polaroids, be I find the ratio (I)/(I_(0)) when theta equals. (i) 30^(@), (ii) 45^(@). |
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Answer» Solution :The intensity of light, getting TRANSMITTED through the combination of three polaroids, is GIVEN by the relation `I=(I_(0))/(8)sin^(2)(2theta)` (i) `theta=30^(@)`, then `I_(30^(@))=(I_(0))/(8)sin^(2)(60^(@))=(I_(0))/(8)((sqrt(3))/(2))^(2)=(3I_(0))/(32)` (ii) If `theta=45^(@)`, then `I_(45^(@))=(I_(0))/(8)sin^(2)(90^(@))=(I_(0))/(8)*1=(I_(0))/(8)` `implies(I_(30^(@)))/(I_(45^(@)))=(3I_(0)//32)/(I_(0)//8)=(3)/(4)` |
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| 34. |
Why do we have better reception of short wave transmission at night than in day? |
| Answer» Solution :Short waves, or, high FREQUENCY waves, getting reflected from the upper and more ionised layer of ionosphere, come BACK to earth surface. During day, these waves are partialy absorbed by the lower and less ionised layers. As a result, TRANSMITTED message through short wave become UNCLEAR. At night, in the absence of sunlight, the lower and less ionised layers are almost absent, so there is almost no absorption of the short waves. For this reason, the clarity of RECEPTION of transmitted message increases greatly. | |
| 35. |
A 100 mH inductor, a 20 mu F capacitor and a 10 Omega resistor are connected in series to a 100 V a.c. source. Calculate (i) impedance of circuit at resonance (ii) current at resonance (iii) resonant frequency. |
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Answer» Solution :(i) IMPEDANCE of CIRCUIT at resonance `Z = R = 10 Omega` (ii) Current at resonance `= (E)/(R) = (100)/(10) = 10 A` (III) Resonant frequency, `v_(0) = (1)/(2pi sqrt(LC))` `v_(0) = (1)/(pisqrt((100 xx 10^(-3))(20 xx 10^(-6)))) = 112.5 Hz` |
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| 36. |
Two waves with the same amplitude and wavelength interface in three different situation to produce resultant waves with the following equations: (1) y' (x,t) =4 sin (5x-4t) (2) y' (x,t) =4 sin (5x) cos (4t) (3) y' (x,t) =4 sin (5x+4t) In which stituation are the two combing waves traveling (a) toward positive x, (b) toward negative x, and (c) in opposite directions? |
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Answer» |
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| 37. |
A transformer having efficiency 90% is working on 100 V and at 2.0 kW power. If the current in the secondary coil is 5A, calculate (i) the current in the primary coil and (ii) voltage across the secondary coil. |
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Answer» Solution :Here `eta=90%=(9)/(10),I_(s)=5A,E_(p)=100V`, (i) `E_(p)I_(p)=2kW=2000W` `I_(p)=(2000)/(E_(p)) or I_(p)=(2000)/(100)=20A` (ii) `eta=("Output POWER")/("Input power")=(E_(s)I_(s))/(E_(p)I_(p)) or E_(s)I_(s)=eta xxE_(p)I_(p)` `=(9)/(10)xx2000=1800W` `therefore E_(s)=(1800)/(I_(s))=(1800)/(5)="360 volt"` |
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| 38. |
A cube is placed in front of a large concave mirror. Will the image of the cube be a cube? |
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Answer» |
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| 39. |
Oil film of thickness 1 mu m is deposited on a glass plate. Refractive index of oil is 1.25 and that of the glass is 1.5. Which wavelength in the visible region (400nm to 700 nm) will be strongly reflected by this film? |
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Answer» Solution :There is denser medium (glass) on one side of the FILM, hence condition for strong reflection in this case will be written as follows: `2 mu t = n lambda` `implies n lambda = 2 xx 1.25 xx 1 xx 10^(-6)` `implies n lambda = 2.25 xx 10^(-6)` Now we can substitute MAXIMUM and minimum wavelengths of given range to get corresponding minimum and maximum values of n. `n_("min") xx 700 xx 10^(-9) = 2.5 xx 10^(-6)` `implies n_("min") = 3.57` `n_("max") xx 400 xx 10^(-9) = 2.5 xx 10^(-6)` `n_("max") = 6.25` Integers in the above-calculated range are 4,5 and 6. Corresponding wavelengths can be calculated as follows: `n lambda = 2.5 xx 10^(-6) implies 4 lambda_1 = 2.5 xx 10^(-6) implies lambda_1 = 625 nm` `n lambda = 2.5 xx 10^(-6) implies 5 lambda_2 = 2.5 xx 10^(-6) implies lambda_2 = 500 nm` `n lambda = 2.5 xx 10^(-6) implies 6 lambda_3 = 2.5 xx 10^(-6) implies lambda_3 = 417 nm`. |
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| 40. |
An electric dipole moment vec(p)=(2.0hat(i)+3.0hat(j))muC m is placed in a uniform electric fiel vec(E)=(3.0hat(i)+2.0hat(k))xx10^(5)NC^(-1). |
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Answer» The torque that `vec (E )` exerts on `vec(P)` is `(0.6 hat(i) - 0.4hat(j) - 0.9 hat(K))` Nm. |
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| 41. |
The hybridization states of the nitrogen atom in pyridine, piperdine and pyrrole are respectively |
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Answer» `SP^(2),sp^(3)and sp^(2)` |
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| 42. |
If a body under the action of force (6i+8j-10k)N , acquires an acceleration of magnitude 5 ms^-2 , then the mass of body is |
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Answer» `SQRT2`KG |
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| 43. |
Two capacitors of capacitances C_(1) and C_(2) are connected in parallel. If a charge Q is given to the combination, the ratio of the charge on the capacitor C_(1) to the charge on C_(2) will be |
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Answer» `(C_(1))/(C_(2))` [Hint: When two capacitors are connected in parallel, potential DIFFERENCE across them is same i.e., `V_(1)=V_(2)=V`(SAY), So `(Q_(1))/(Q_(2))=(C_(1)V_(1))/(C_(2)V_(2))=(C_(1)V)/(C_(2)V)=(C_(1))/(C_(2)).`] |
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| 45. |
आवेशित खोखले गोलीय चालक के अन्दर विधुत-तीव्रता होती है |
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Answer» `epsilon_o.SIGMA` |
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| 46. |
If E_(a) be the electric field strength of a short dipole at a point on its axial line and E_(e ) that on the equatorial line at the same distance, then |
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Answer» `E_(E ) = 2E_(a)` |
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| 47. |
Surface tension of the liquid is the property by virtue of which the free surface of the liquid tends to- |
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Answer» BECOME maximum |
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| 48. |
Magnetic length is slightly _____ than _____ length of the magnet . |
| Answer» SOLUTION :LESS, GEOMETRICAL | |
| 49. |
What are the different transistor configurations ? |
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Answer» SOLUTION :i. Common base CONFIGURATION . ii. Common emitter configuration. III. Common collector configuration |
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