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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 the angular acceleration of a particle in circular motion, which slows down from 600 r.p.m to rest in 10s ? |
| Answer» SOLUTION :`ALPHA=(w_2-w_1)/t=(2PI(n_2-n_1))/t=(2pi(0-10))/10=-2pi(RAD)/sec^2` | |
| 2. |
(A): An AC can be transmitted over long distances without much power loss. (R): An AC can be stepped up or down with the help of a transformer |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the CORRECT EXPLANATION of 'A'. |
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| 3. |
The earth magnetic induction at certain point is 7xx10^(-5)this is to be annulled by the magneticradius 15 cm the required curretn in the loop is |
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Answer» 0.56 A |
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| 4. |
The electrostatic force between two charges Q_(1) and Q_(2) at separation r is given by F= (K.Q_(1)Q_(2))/(r^(2)) The constant K |
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Answer» depends on the system of UNITS only |
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| 5. |
Light of frequency v=1.5v_(0) (where v_(0)=threshold frequency) is incident on a photosensitive material and photoelectric current is I. if the frequency of incident light is halved and the intensity of light is doubled then the photoelectric current becomes_____. |
| Answer» SOLUTION :ZERO. [No photoelectric emission is possible if `v lt v_(0)`]. | |
| 6. |
Consider the following statements : If the same note is played on a flute and a sitar, one can still distinguish between them because they differ in 1. frequency 2 intensity 3. quality Which of the statements given above is/are correct? |
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Answer» 1 and only |
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| 7. |
Calculate the Debye temperature for iron in which the propagation velocities of longtitudinal and transverse vibrations are equal ot 5.85 and 3.23Km//s respectivel. |
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Answer» Solution :We proceed as in the prevoius example. The total number of modes MUST be `3n_(0)v` (total trnasverse and one longitudinal per ATOM). On the other hand of TRANSVERSE modes per unit frequency interval is by `dN^(_|_)=(Vomega^(2))/(pi^(2)V_(_|_)^(3))d omega` while the number of longitudinal moder per unit frequency interval is given by `dN^(||)=(V omega^(2))/(2pi^(2)v_(||)^(3))d omega` The total number per unit frequency interval is `dN=(V omega^(2))/(2pi^(2))((2)/(V_(_|_)^(3))+(1)/(V_(||)^(3)))d omega` If the high frequency cut off is at `omega_(0)=(kTheta)/( ħ)`, the total number of modes will be `3n_(0)V=(V)/(6 pi^(2))((2)/(V_(_|_)^(3))+(1)/(v_(||)^(3)))((K Theta)/( ħ))^(3)` Here `n_(0)` is thenumber of iron ATOMS per unit volume. Thus `: Theta=(ħ)/(k)[18 pi^(2)n_(0)//((2)/(v_(_|_)^(3))+(1)/(v_(||)^(3)))]^(1//3)` For iron `n_(0)=N_(A)//(M)/(rho)=(rhoN_(A))/(M)` `(rho=` density, `M=` atomic weight of iron `N_(A)=` Avogardo number). `n_(0)= 8.389xx10^(22)per c c` Substituting the data we get `Theta= 469.1K` |
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| 8. |
A canon of mass M is mounted on an east west frictionless railway track. The barrel of cannon faces the north east direction at an angle of 45^(@) with the vertical. The canon fires a shot of mass M with a speed V relative to the barrel. |
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Answer» The recoil speed of the cannon will be `V//4` `M(v cos^(2)45-v_(0))=Mv_(0)` `rArr 2MV_(0)=MV cos^(2)45^(@)` `rArr v_(0)=V//4` Vertical speed `=(v)/(sqrt(2))` North speed `=v cos^(2)45^(@)=(v)/(2)` East speed `=(V)/(2)-(V)/(4)=(V)/(4)` `rArr` Speed from ground `=vsqrt((1)/(2)+(1)/(4)+(1)/(16))=vsqrt((13)/(16))` |
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| 9. |
Using Faraday's law of electromagnetic induction, derive an equation for motional emf. |
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Answer» Solution :Consider a rectangular conducting loop of width 1 in a uniform magnetic field B which is perpendicular to the plane of the loop and is directed inwards. II. A part of the loop is in the magnetic field while the remainig par t is outside the loop as shown in Figure. When the loop is pulled with a constant velocity `vec(v)` to the right, the area of the portion of the loop within the magnetic field will decrease. iv. THUS, the flux linked with the loop will also decrease. According to Faraday's law, an electric current is induced in the loop which flows ina direction so as to oppose the pull of the loop. v. Let x be the length of the loop which is still within the magnetic field, then its area is lx. The magnetic flux linked with the loop is `phi_(B)=int_(A)B*dvec(A)=BAcostheta` Here`theta=0^(@)andcos0^(@)=1` = BA `phi_(B)=Blx` vi. As this magnetic flux DECREASES due to the movement of the loop, the magnitude ofthe induced emf is given by `EPSILON=(dPhi_(B))/(dt)=(d)/(dt)(Blx)` vii. Here, bothe B and l are constants. Therefore, `epsilon=Bl(dx)/(dt)` `epsilon=Blv` where `v=(dx)/(dt)` is the velocity of the loop. This emf is KNOWN as motional emf since it is produced due to the movement of the loop in the magnetic field. |
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| 10. |
Is reverse saturation current ofa diode independent of reverse bias voltage? |
| Answer» SOLUTION :Yes, the REVERSE saturation current of a diode isindependent of reverse BIAS voltate. | |
| 11. |
A permanent magnet in the shape of a thin cylinder of length 10cm has M=10^6 A//m. Calculate the magnetisation current lm…… |
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Answer» Solution :INTENSITY of magnetisation `M= 10^(6) A//m` Length `l=10 cm = 10^(-1)` m `I_(M) =` magnetisation current We know that, `M= (I_(M))/( l)` `therefore I_(M) = Ml` `= (10^6) (10^(-1) ) ` `=10^(5) A` |
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| 12. |
A galvanometer has a resistance of 30 Omega. It gives full scale deflection with a current of 2 mA. Calculate the value of the resistance needed to convert it into an ammeter of range 0 - 0.3 A. |
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Answer» SOLUTION :Here `R_G = 30 OMEGA, I_g = 2 mA = 2 XX 10^(-3) A and I = 0.3 A` `:.` VALUE of SHUNT resistance `r_(s) = (I_g cdot R_G)/((I - I_g)) = (2 xx 10^(-3) xx 30)/((0.3 - 2 xx 10^(-3))) = 0.06 Omega`. |
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| 13. |
A copper wire of cross - sectional area 3.4 m m^(2) and length of the wire 400m, specific resistivity of copper is 1.7xx10^(-8)Omega-m. Then the resistance of the wire is |
| Answer» Answer :C | |
| 14. |
A conductor of length l is placed perpendicular to a horizontal uniform magnetic field B.Suddenly, a certain amount of charge is passed throgh it, when it is found to jump to a height h. The amount of charge that passes through the conductor is |
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Answer» `(msqrt(gh))/(BL)` or `F/_\t=msqrt(2gh) or (ilB) /_\t=musqrt(2gh)` But `i/_\t=/_\Q` `:. (/_\q)(lB)=msqrt(2gh)` `HENCE /_\q=(msqrt(2gh))/(BI)` |
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| 15. |
A free neutron decays into a proton, an electron and |
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Answer» a neutrino |
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| 16. |
Classify the following statements into true or false. i. Radioactivity is unaffected by pressure and temperature. ii. Radioactivity is affected by large electric and magnetic fields. iii. The rate of disintegration of a radioactive substance at a particular time is directly proportional to the number of atoms present initially. iv. Nuclear force is charge independent but it depends on the relative orientation of the spins. |
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Answer» |
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| 17. |
Why are infrared waves often called as heat waves? Give their one application. |
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Answer» Solution :Infrared waves are called heat waves because molecules present in the MATERIALS READILY ABSORB the infra red rays get heated up. APPLICATION: They are USED in green bouses to warm the plants. |
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| 18. |
A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value? |
| Answer» SOLUTION :(a) 5 T YEARS (B) 6.65 T years | |
| 19. |
If photo-electric effect,when frequency of incident radiation is doubled ,maximum speed of electron emitted also becomes double .work function of metal will be…… |
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Answer» `(HV)/(4)` Now ,if frequency and speed both become DOUBLE, `(1)/(2)m(2v)^(2)=h(2v)-phi` `4((1)/(2)mv^(2))=2hv-phi` `4(hv-phi)=hv-phi` [From result (1)] `THEREFORE 4hv-4phi=2hv-phi` `therefore 2hv=3phi` `therefore phi=(2hv)/(3)` |
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| 20. |
Find the accelerating potential of the electron, when its de-Broglie wave length is 1Å. Data: lambda=Å, V = ? |
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Answer» SOLUTION :`LAMBDA=(12.27)/(sqrtV)` `sqrtV=(12.27xx10^(-10))/(lambda)` `(12.27xx10^(-10))/(1XX10^(-10))` `V=(12.27)^(2)=150.55V` |
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| 21. |
A nucleus X^232 initially at rest undergoes alpha decay, the alpha-particle produced in above process is found to move in a circular track of radius 0.22 m in a uniform magnetic field of 1.5 T. Find Q value of reaction. |
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Answer» |
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| 22. |
A lens is placed between a source of light and screen. It forms real image on screen for two different positions. If height of one image is 20 cm and the other is 80 cm, then the height of source of light (in cm) is_____. |
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Answer» `|m_1| = |(20)/(h_0)|` And for `2^(nd)` magnification`|m_2| = |(80)/(h_0)|` We know thant `m_1m_2 = 1 ` so , `h_0 = 40 CM` . We know thatSo, |
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| 23. |
Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg, [Neglect air resistance throughout].(a) Just after it is dropped from the window of a stationary train(b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h.(c ) just after it is dropped from the window of a train accelerating with 1 ms - 2(d) lying on the floor of a train which is accelerating with 1 ms - 2, the stone being at rest relative to the train. |
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Answer» Solution :(a) Here, m = 0.1 kg, a = +G = 9.8 m/s 2, Net force, `F = ma = 0.1xx9.8=0.98 N` This force acts vertically downwards. (b) When the train is RUNNING a constant VELOCITY its ACC. = 0. No force acts on the stone due to this motion. Therefore, force on the stone F = weight of stone `= mg = 0.1xx9.8=0.98 N` This force also acts vertically downwards. (c ) When the train is accelerating with 1 m/2, an additional force `F^(1) = ma=0.1xx1=0.1 N` acts on the stone in the horizontal direction. But once the stone is dropped from the train, `F^(1)` becomes zero and the net force on the stone is `F = mg = 0.1xx9.8 =0.98 N`, acting vertically downwards. (d) As the stone lying on the horizontal direction of motion of the train. Note the weight of the stone in this case is being balanced by the normal reaction. |
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| 24. |
Red tides are caused by |
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Answer» euglena |
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| 25. |
A: Stoke's formula for viscous drag is not really valid for oil -drops of extremely minute sizes. R: Stoke's formula is valid for motion through a homogeneous continuous medium and the size of the drop should be much larger than the intermolecular separation in the medium for this assumption to be valid. |
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Answer» If both ASSERTION & REASON are true and the reason is the correct EXPLANTION of the assertion , then mark (1) |
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| 26. |
The above configuration is |
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Answer» PLANE of symmetry |
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| 27. |
The electric field in a region of space is given by, vec(E ) = E_(0) hat(i) + 2 E_(0) hat(j) where E_(0)= 100N//C. The flux of this field through a circular surface of radius 0.02m parallel to the Y-Z plane is nearly. |
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Answer» `3.14 NM^(2)//C` |
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| 28. |
In Young's double-slit experiment using monochromatic light of wavelength lamda, the intensity of light at a point on the screen where path difference is lamda, is K units. What is the intensity of light at a point where path difference is lamda//3 ? |
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Answer» Solution :In Young.s double-slit EXPERIMENT two SLITS are identical and intensity of light coming out of them is exactly same i.e., `I_(1)=I_(2)=I` (say). At a point on the screen where path difference is `lamda`, the phase difference `phi=(2pi)/(lamda),lamda=2pi" rad"` and HENCE resultant intensity of light there will be `I_(res)=4I"COS"^(2)(phi)/(2)=4Icos^(2)pi=4I=K` At the other point, where path difference is `(lamda)/(3)`, the corresponding phase difference `phi.=(2pi)/(lamda)xx(lamda)/(3)=(2pi)/(3)`, rad and hence resultant intensity at this point will be `I_(red).=4Icos^(2)((2pi)/(3))=4I(-(1)/(2))^(2)=I` Comparing (i) and (ii), we GET `(I_(res).)/(I_(red))=(I_(res).)/(K)=(I)/(4I)=(1)/(4)impliesI_(red).=(K)/(4)`. |
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| 29. |
The figure here shows three arrangements of an electron e and two protons p. (a) Rank the arrangement according to the magnitude of the net electrostatic force on the electron due to the protons, largest first. (b) In situation c, is the angle between the net force on the electron and the line labelled d less than or more than 45^(@) ? |
| Answer» SOLUTION :(a) a, C, B, (b) LESS than | |
| 30. |
Natural radioactivity is a spontaneous and self disruptive activity exhibated by a number of heavy elements in nature. Thus a heavy element disintegrates by itself without being forced by any external agent to do so. According to radioactive decay law, the number of atoms disintegrated per second (i.e., rate of disintegration of radioactive atoms) at any instant is directly propotional to the number of radioactive atoms actually in the sample at that instant, i.e., -(dN)/(dt) propN or R=-(dN)/(dt)=lambdaN, where lambda is decay constant. Read the above passage and answer the following questions: (i) The count rate form a radioactive sample containing 10^(16) atoms is 4xx10^(6) per second. What is the value of decay constant? (ii) Name the three types of radiactive radiations. Which one of them is most penetrating? (iii) What does radioactive decay law imply in day to day life? |
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Answer» Solution :(i) Here, `N=10^(16) and (dN)/(dt)=4xx10^(6)per sec. lambda=?` FORM `(dN)/(dt)=lambdaN` `lambda=(dN//dt)/N=(4xx10^(6))/(10^(16))=4xx10^(-10)sec^(-1)` (ii) The three types of radiactive radiations are : `alpha`-rays , `beta`-raysand `gamma`-rays. Out of then, `gamma`-rays have MAXIMUM penetrating power. (iii) Radioactive decay LAW is the law of mass action, i.e., larger the available mass, greater is its rate of decay. The same applies in day to day LIFE. Larger the population, greater is their decay rate, i.e., under GIVEN conditions, when number of people is large, the rate at which they the die is also large. The reverse is also true. |
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| 31. |
Which of the following statement is not correct when a junction diode in forward bias? |
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Answer» The width of depletion region decreases |
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| 32. |
A real gas behaves like an ideal gas if its : |
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Answer» pressure and TEMPERATURE are both high |
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| 33. |
When exposed to sunlight, thin films of oil on water often exhibit brilliant colours due to the phenomenon of ..... |
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Answer» interference |
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| 35. |
In Figure, a light ray is incident at angle theta_(1)= 55^(@) on a series of five transparent layers with parallel boundaries. For layers 1 and 3. L_(1) = 20 mu m, L_(3) = 25 mu m, n_(1) = 1.6, and n_(3) = 1.15. (a) At what angle does the light emerge back into air at the riglit? (b) How much time does the light take to travel through layer 3? |
| Answer» SOLUTION :(a) `55^(@)` (same as incident ANGLE), (b) `1.4 XX 10^(-13)` s | |
| 36. |
An achromatic doublet of focal length 50 cm is used as an objective of a telescope. The refractive indices of the glasses of the lenes for yellow are 1.6 and 1.5. The radius of curvature of the sides on contact is 15 cm. Find the radii of curvature of the other surfaces. The dispersive powers of the glasses are 0.33 adn 0.24. |
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Answer» |
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| 37. |
Suppose that A, B, C, D represent the energies of1erg,1"joule",1"Calorie",1"Kcalorie. What is the correct order of magnitudes of their energies ? |
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Answer» `AgtCgtBgtD` |
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| 38. |
A galvanometer can be converted into a voltmeter by connecting |
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Answer» high RESISTANCE is SERIES |
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| 39. |
Name the experiment which establishes the wave nature of a particle. |
| Answer» SOLUTION :Davisson-Germer EXPERIMENT. | |
| 40. |
Two light waves having the same wavelength lambda in vocume are in phase initially. Then the first ray travels a path of length L_(1) through a medium of refractive index mu_(1). The second ray travels a path of length L_(2) through a medium of refractive index mu_(2). The two waves are then combined to observe interference effects. The phase difference between the two when they interfere, is |
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Answer» `(2pi)/(lambda)(L_(1) - L_(2))` |
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| 41. |
A thin prism P-i with angle 4^(@)and made from glass of refractive index 1.54 is combined with another thfn prism P_(2) . made from glass of refractive index 1.72 to produce dispersion without deviation. The angle of the prism P_(2) is |
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Answer» `5.33^(@)` |
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| 42. |
The work function of zinc is 6.8 xx 10^(-19) J. What is the threshold frequency for emission of photoelectrons for zinc ? Data: W = 6.8 xx 10^(-19)J, v_(0) = ? |
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Answer» SOLUTION :Work Function `W=hv_(0)` `hv_(0) = 6.8 xx 10^(-19)` `v_(0)=(6.8 xx 10^(-19))/(6.626 xx 10^(-34))` `v_(0) = 1.026 xx 10^(15) HZ`. |
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| 43. |
For the same objective , the ratio of the least separation between two points to be distinguisghed by a microscope for light of 5000 Å and electrons accelerated through 100 V used as the illuminating substance is |
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Answer» `10^(-3)` where `beta` is the angle subtended by the OBJECTIVE atthe object. For light of `5500 Å` `d_(min) = (1.22 xx 5.5 xx 10^(-7))/(2 sin beta) m` For electrons accelerated through 100 V the de Brogile WAVELENGTH is `lambda = (h)/(p) = (1.227)/(SQRT(100)) = 0.13 nm = 0.13 xx 10^(-9)`m `THEREFORE d._(min) = (1.22 xx 1.3 xx 10^(-10))/(2 sin beta)` `(d._(min))/(d_(min)) = (1.3 xx 10^(-10))/(5.5 xx 10^(-7)) approx 0.2 xx 10^(-3)` |
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| 44. |
The resistance of galvanometer is 50Omega and can measure a maximum current of 1.5 A.In order to convert it into voltmeter upto 1500 V, the value of series resistance should be : |
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Answer» `9.5Omega` |
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| 45. |
The energy flux of sunlight reaching the surface of the earth is 1.388 xx 10^(3) W//m^(2) . How many photons (nearly) per square metre are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm. |
| Answer» SOLUTION :`4XX10^(21)" photons/m"^(2)s` | |
| 46. |
The incorrect structure of glucine at given pHare- |
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Answer» `H_(2)overset(o+)(N)CH_(2)-underset(O)underset(||)(C)-OH` at `PH = 2.0` |
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| 47. |
Communication is an act of _____ of informaion. |
| Answer» SOLUTION :TRANSMISSION | |
| 48. |
Assertion: Coulomb force between charges is central force Reason:Coulomb force depends on medium between charges |
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Answer» Both ASSERTION and Reason are true and Reason is the CORRECT EXPLANATION of Assertion |
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| 49. |
A full-wave p-n diode rectifier uses a load resistor of 1500Omega. No filter is used. The forward bias resistance of the diode is 10Omega. The efficiency of the rectifier is |
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Answer» `81.2%` |
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| 50. |
Holes are charge carries in A) Intrinsic semiconductors B) Ionic solids C) P-type semiconductors D) Metals |
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Answer» A and B |
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