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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. |
A : The ammeters and volmetrs used for measuring alternating current and voltages have non-uniform divisons on their scales. R : The instruments used for measuring alternating current and voltage are based on heating effect of current. |
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Answer» If both ASSERTION & Reason are TRUE and the reason is the correct EXPLANATION of the assertion, then MARK (1) |
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| 2. |
A plano-convex lens when silvered in the plane side behaves like a concave mirror of focal length 30 cm. However, when silvered on the convex side it behaves like concave mirror of focal length 10 cm. Then the refractive index of its material will be: |
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Answer» `3.0` |
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| 3. |
In a communication system, noise is most likely to affect the signal |
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Answer» at the transmitter |
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| 4. |
Which of the following forces in non conservation one ? |
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Answer» `3hat(i) +4hat(j)` `(deltaF_(x))/(deltay)=(deltaF_(y))/(DELTAX)=(DELTA^(2)U)/(deltaxdeltay)` |
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| 5. |
An electron is in a state with l=3, (a) What multiple of h gives the magnitude of vecL? (b) What multiple of mu_(B) gives the magnitude of vecmu? (c ) What is the largest possible value of m_(l)? (d) what multiple of h gives the corresponding value of L_(z), and (e) what multiple of mu_(B) gives the corresponding value of mu_(orb,z)? (f) What is the value of the semiclassical angle theta between the direction of L_(z) and vecL? What is the value of angle theta for (g) the second largest possible value of m_(l) and (h) the smallest (that is, most negative) possible value of m_(l)? |
| Answer» SOLUTION :(a) 3.46, (b) 3.46, (C ) 3, (d) 3, (E) `-3`, (f) `30.0^(@)`, (g) `54.7^(@)`, (H) `150^(@)` | |
| 6. |
Assertion (A): The conventional direction of current is taken to be the direction of flow ofpositive charge. Reason (R) : Direction of current flow is opposite to the direction of flow of electrons in aconductor. |
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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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| 7. |
What are the tiny lights in traffic signals called and how do these operate? |
| Answer» Solution :Tiny lights in trafsic signals are LEDS. A LED emit LIGHT when it is connected in FORWARD BIAS. | |
| 8. |
The electric field in a certain region is acting radially outward and is given by E = Ar . A charge contained in a sphere of radius 'a' centred at the origin of the field, will be given by |
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Answer» `epsilon_0 Aa^3` |
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| 9. |
An under water swimmer is at a depth of 12 m below the surface of water. A bird is at a height of 18 m from the surface of water, directly above his eyes. For the swimmer the bird appears to be a distance from the surface of water equal to (mu_(w)=4//3) |
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Answer» 24 m |
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| 10. |
A circular disc of mass 'm' and radius '3r' is resting on a flat frictionless surface. Another circular disc of mass 'm'and radius '2r', moving with a velocity 'v_(0)' hits the first disc as shown in figure. The collision is elastic. The velocity of Bigger disc after collision is |
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Answer» `(4v_(0))/(5)` & plastic REGION of B is wider than plastic region of A implies B is more DUCTILE MATERIAL. |
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| 11. |
The surface which doesn.t reflect is given incidence light with 18 W//cm^(2) energy flux, than what will be the pressure? |
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Answer» `2 N//m^2` `=(18W //cm^2)/(3xx10^8 m//s)` `=(18xx10^4 W//m^2)/(3xx10^8 m//s)` `THEREFORE P=6xx10^(-4) N//m^2` |
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| 12. |
Explain the term .amplification. in communication system. Why is it necessary? |
| Answer» SOLUTION :Amplification : It is the process of increasing the strength of the SIGNAL and it is done using an electronic circuit is CALLED AMPLIFIER Amplification is required to COMPENSATE for attenuation of the signalin communication system, | |
| 13. |
A device to store electrical charge is called |
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Answer» transformer |
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| 14. |
Magnetic field does not cause deflection in |
| Answer» Answer :A | |
| 15. |
Which of the followin pair of compounds are nto isomers. |
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Answer» Propyne and cyclopropene |
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| 16. |
Why does the direction of motion of a projectile become horizontal at the highest point of trajectory ? |
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Answer» |
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| 17. |
Calculate the energy released, in MeV, in the reaction " "_(3)^(6)Li + " "_(0)^(1)n to " "_(2)^(4)He + " "_(1)^(3)H Given that mass (" "_(3)^(6)Li) = 6.015126 u, mass (n) = 1.008665 u, m(" "_(2)^(4)He) = 4.002604 and m(" "_(1)^(3)H) = 3.010000. Take 1 u = 931 MeV/c^(2). |
| Answer» SOLUTION :4.782MeV | |
| 18. |
The value of Brewster's angle for air-glass interface is (pi)/3 hence the refractive index of glass is ________. |
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Answer» Solution :`SQRT3` [Hint : N = TAN `THETA` = tan`pi/3=sqrt3`] |
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| 19. |
A screen is placed 50 cm from a single slit, which is illuminated with 6000 Å light. If distance between the first and third minima in the diffraction pattern is 3.00 mm, what is the width of the slit ? |
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Answer» Solution :In case of diffraction at single slit, the position of minima is given by `d SIN theta= n lamda`. where d is the aperture size and for small `theta : sin theta = theta =(y//D)` `THEREFORE d(y/D) = n lamda, i.e., y = D/d (n lamda)` so that `y_3 - y_1 = D/d (3lamda -lamda) = D/d (2 lamda)` and hence ` d = (0.50xx(2 xx 6 xx 10^(-7)) )/(3 xx 10^(-3)) = 2 xx 10^(-4) m = 0.2 mm` |
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| 20. |
State the principle of working of a galvanometer. A galvanometer of resistance G is converted into a voltmeter to measure upto V volts by connecting resistance R_(1) in series with the coil. If a resistance R_(2) isconnected in series with it, then it can measure V/2 volts. Find the resistance, in terms of R_(1) and R_(2), required to be connected to convert it into a volt meter that can read upto 2 V. Also find the resistance G of the galvanometer in terms of R_(1) and R_(2). |
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Answer» Solution :Energy stored in a CAPACITOR, `E=(1)/(2)CV^(2)` In parallel,`0.25=(1)/(2)(C_(1)+C_(2))(100)^(2)` In series,`0.045 =(1)/(2)((C_(1)C_(2))/(C_(1)+C_(2)))(100)^(2)` From (i)`C_(1)+C_(2)=0.25xx2xx10^(-4)` `C_(1)+C_(2)=5xx10^(-5)` From (II)`(C_(1)C_(2))/(C_(1)+C_(2))0.045xx2xx10^(-4)` `(C_(1)C_(2))/(C_(1)+C_(2))=0.09xx10^(-4)=9xx10^(-6)` Form (iii) `C_(1)C_(2)=(2xx0.045xx5xx10^(-5))/(10^(4))=4.5xx10^(-10)` `C_(1)-C_(2)=sqrt((C_(1)+C_(2))^(2)-4C_(1)C_(2))` `C_(1)-C_(2)=2.64xx10^(-5)` SOLVING (ii) and (iv)`C_(1)=38.2 mu F` `C_(2)=11.8 mu F` In parallel`Q_(1)=C_(1)V=38.2xx10^(-6)xx100=38.2xx10^(-4)C` `Q_(2)=C_(2)V=11.8xx10^(-6)xx100=11.2xx10^(-4)C` |
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| 21. |
Show that the tangential component of electrostatic field is continuous from one side of acharged surface to another. |
| Answer» SOLUTION :There can be no discontinuity in tangential COMPONENT. The SURFACE is equipotential and the work done along an equipotential surface is ZERO. The reason being there is no potential DIFFERENCE and work done W = q.dV = 0. Since E dx = dV, there is no discontinuity in tangential component of electric field. | |
| 22. |
64 charged water droplets each with a diameter 1 mm and charge 2 xx 10^(-12)C coalesce to form a single group. Calculate the potential of the bigger drop. |
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Answer» |
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| 23. |
Computethe firstthreeenergylevelsof doublyionizedlithium. Whatis theionisationpotential |
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Answer» Solution :energy `E_n = (Z^2 mc^2 alpha ^2)/(2n^2 )` forhydrogen ,` z=1andE_n= (- 13.6eV )/( n^2)` forlithium, z= 3`thereforee_n= ( - 112.4 )/( n^2) eV ` ` thereforeE_1=- 122.4 ` eV Ionizationpotential`= 122.4 ` volts ` E_2=(-122.4 )/( 4 ) =- 30.8eV,E_3= (-122.4 )/(9) =- 13.6 eV , E_4= (-122.5 )/( 16)=- 7.65eV ` |
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| 24. |
A stream of electrons moving with a velocity of 3xx10^(7) ms^(-1) is deflected by 2 mm in traversing a distance of 0.1 m in a uniform electric field of strength 18 Vm^(-1) in Fig. 1.42. An electron enters the field symmetrically between the plates with a speed upsilon_(0). The length of each plate is l. Find the angle of deviation of the bath of the electron as it comes out of the field. |
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Answer» |
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| 25. |
The ratio of magnetic field at the centre of a circular current carrying coil to its magnetic moment is 'x'. If the current in the wire is doubled, the ratio will become (radius is constant) |
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Answer» 2x |
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| 26. |
A radioactive sample at any instant has its disintegration rate 5000 disintegrations per minute. After 5 minutes, the rate becomes 1250 disintegration per minute. Then its decay constant (per minute) is |
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Answer» (a)`0.8 log_e 2` |
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| 27. |
In an experiment on alpha- particle scattering, alpha-particles are directed towards a gold foil and detectors are placed in postition P,Q and R . What is the distribution of alpha-paritcles as recored at P,Q and R? |
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Answer» <P>`{:(P,,Q,,R),("all",,"none",,"none"):}` |
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| 28. |
If the two slits in Young's double slit experiment are of unequal width , then |
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Answer» the bright FRINGES will have unequal spacing . `I_("min") = (A - B)^(2) ne 0` i.e, the dark fringes are not perfectly dark . |
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| 29. |
An electron is moving at 10^(6) ms^(-1) in a direction parallel to a current of 5A flowing through an infinitely long straight wire , separated by a perpendicular distance of 10 cm in air. Calculate the magnitude of the force experienced by the electron. |
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Answer» Solution :Magnetic field of the straight wire carrying a current of 2A, at a DISTANCE of 10cm or 0.1m from it is `B = (mu_(0) I)/(2pi r) = (4pi xx 10^(-7)xx 5)/(2i xx 0.1) = 10^(-5)` T this field ACTS perpendicular to the direction of the electron . So magnetic FORCE on th electron is F = qv B sin `90^(@) = 1.6 xx 10^(-19) xx 10^(6) xx 10^(-5) xx 1` F = `1.6 xx 10^(-18)` N |
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| 30. |
A wooden block of mass 1 kg rests on a soft horizontal floor. When an iron cylinder of mass 20kg is placed on top of the block, the floor yields steadily and booth go down with an acceleration of 0.1ms^(-2). What is the action of the block on the floor (a) before and (b) after the floor yields? (Take g=10ms^(-2)). Identify the action and reaction pairs in the problem. |
Answer» Solution : (a) The block is at REST on the floor. Its FBD shows two forces on it, Weight = mg `=1xx10` `=10N` and the normal force R of the floor on the block. Net force = 0, According to first law `rArr R=10N` directed upwards (b) The system (block + cylinder) accelerates DOWNWARD with `0.1ms^(-2)`. The FBD shows weight of hte system `(=(20+1)xx10=210N)` and normal force R' by the floor. FBD does not show internal forces between the block and the cylinder. `210-R'=21xx0.1 ""` (Applying second law to the system) `R'=207.9N` Action of the system on the floor = 207.9 N vertically downwards Action-reaction pairs For (a) : (i) Action : Force of gravity (10N) on the block by the earth. Reaction: Force of gravity on the earth by the block. (ii) Action: Force on the floor by the block. Reaction: Force on the block by the floor. For (b): (i) Action: The force of gravity (210 N) on the system by the earth. Reaction: The force of gravity (210 N) on the earth by the system. (ii) Action: Force on the floor by the system. Reaction: Force on the system by the floor. Remember that an action-reaction pair consists of mutual forces which are always equal and opposite between two bodies. The forces on the same BODY CONNOT constitute an action-reaction pair. For example, the force of gravity on the block in (a) or (b) and the normal reaction on the block by the floor are not action - reaction pairs. These forces are equal and opposite for (a) as the block is at rest. For(b), they are notsame. Weight = 210 N, while R' = 207.9 N. |
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| 31. |
A particle is projected along the line of greatest slope up a rough inclined plane at an angle of 45^(@) with the horizontal. If the coefficient of friction is (1)/(2) then the retardation is : |
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Answer» `(G)/(2)` |
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| 32. |
The Bohr atom model is derived with the assumption that the nucleus of the atom is stationary and only electrons revolve around the nucleus. Suppose the nucleus is also in motion, then calculate the energy of this new system. |
Answer»  Let V be the velocity of the nuclear motion and v be the velocity of electron motion. Since the total linear momentum of the system is zero, `(-)MV + Mv = 0 or MV = mv = p` `vec P_(e) + vec P_(n) = vec 0 or |vec P_(e)| = |vec P_(n)| = P ` Hence , the kinetic energy of the sytem is `KE = (p_(n)^(2))/(2M) + (p_(e)^(2))/(2m) = (p^(2))/(2) ((1)/(M) + (1)/(m))` Let `(1)/(M) + (1)/(m) + (1)/(mu_(m))`. Here the reduced mass is `mu_(m) = (mM)/(M+ m)` Therefore , the kinetic energy of the system now is `KE = (p^(2))/(2mu_(m))` Since the potential energy of the system is same, the total energy of the hydrogen can be EXPRESSED by replacing mass by reduced mass, which is `E_(n) = -(mu_(m)e^(4))/(8 epsilon_(0)^(2) H^(2)) (1)/(n^(2))` Since the nucleus is very heavy compared to the electron , the reduced mass is CLOSER to the mass of the electron. |
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| 33. |
The poem draws a contrast between.........and............... |
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Answer» ANIMALS, HUMAN beings |
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| 34. |
Oscillatory disturbance travelling through the medium is |
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Answer» energy |
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| 35. |
Energy required for the electron excitation in Li^(++) from the first to the third Bohr orbit is |
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Answer» `12.1` eV `Delta E = E_(3) - E_(2) = 13.6(3)^(2)[(1)/(1^(2)) - (1)/(3^(2))]` ` = (13.6 xx 9 xx 8)/(9) = 108.8 eV` |
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| 36. |
A body cools from 60^(@)C,50^(@)C in 10 minutes. If the room temperature is 25^(@)C and assuming Newton's law of cooling to hold good, the temperature of the body at the end of the next 10 minutes will be : |
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Answer» `40^(@)C` `(theta_(1)-theta_(2))/(t)=K[(theta_(1)+theta_(2))/(2)-theta_(0)]` `:.""(60-50)/(10)=K[(60+50)/(2)-25]`. . . (i) In SECOND case, if T be the final temperature after 10 MINUTES. Then `(50-theta)/(10)=K[(50+theta)/(3)-25]` . . . (ii) Dividing (i) by (ii) `(10)/(50-7)=(30)/(theta/2)rArr10theta=60xx50-60T`. `70theta=60xx50` `theta=(60xx50)/(70)=(300)/(7)=42*85^(@)C` Correct CHOICE is (d). |
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| 37. |
Explain series connection of resistors. Derive equation of equivalent resistance (R_(S)). |
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Answer» Solution :When more than one resistors are CONNECTED between two POINT in such a way that current FLOWING through each resistor is equal (same) then such connection of resistors is called series connection. In series connection sum of p.d. (voltage) across each resistor is equal to total voltage supplied (emf of cell). `RARR` As shown in figure two resistor `R_(1) and R_(2)` are connected between point A and B ,  ` rArr` Current flowing through both resistors is same (I) . `therefore `p.d. across resistor `R_(1)`. p.d. across `R_(2)`, `V_(2) = IR_(2) ""` ....(2) V = terminal voltage of battery ` V = V_(1) + V_(2)` = `IR_(1) + IR_(2)` `V = I (R_(1) + R_(2))` `I = I_(1) + I_(2)` ` = (V)/(R_(1)) + (V)/(R_(2))` = I = V`((1)/(R_(1)) + (1)/(R_(2) ) )` `(I)/(V) = (1)/(R_(1)) + (1)/(R_(2))` `(V)/(I) = R_(eq)` `(I)/(V) = (1)/(R_(eq))` `(1)/(R_(eq)) = (1)/(R_(1)) + (1)/(R_(2))` `rArr` In parallel connection RECIPROCAL equivalent resistance is equal to sum of reciprocal of individual resistors. `rArr` In parallel connection of resistors equivalent resistance is smaller than smallest value of resistance. `rArr R_(1), R_(2), .... , R_(n)` resistors are connected inparallel then `R_(eq) = R_(p)`, `(1)/(R_(eq))= (1)/(R_(p)) = (1)/(R_(1)) + (1)/(R_(2)) + .... + (1)/(R_(n))` `rArr " If " R_(1) = R_(2) ...., R_(n) = R,` then equivalent resistance, `(1)/(R_(eq)) = (1)/(R_(p)) = (1)/(R) + (1)/(R) + .... + ` n times. `rArr` In such connection sum of current flowing through each resistor is equal to total current in the circuit.  `rArr " Let " R_(1) and R_(2)` resistance are connected between a and b and battery of terminal voltage V is connected. `rArr` Let current in the circuit is I. It is divided into two branch `I_(1) and I_(2)` at point a. `rArr` Let current in `R_(1) " be" I_(1)` and current in `R_(2) " be " I_(2)` at point a, `I = I_(1) + I_(2) ` .... (1) `rArr` By Ohm.s law , p.d . across `R_(1)`, V = `I_(1) R_(1)` `I_(1) = (V)/(R_(1)) "" `... (2) `rArr`p.d. across `R_(2)` , `V = I_(2) R_(2)` `I_(2) = (V)/(R_(2)) `...(3) `(1)/(R_(eq)) = (n)/(R) ` `R_(eq) = (R)/(n)` |
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| 38. |
In a medium of refractive index n_(1), a monochromatic light of wavelength lamda_(1) is travelling. When it enters in a denser medium of refractive index n_(2), the wavelength of the light in the second medium is |
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Answer» `lamda_(1)((n_(1))/(n_(2)))` `lamda_(1)=c/(vn_(1)),lamda_(2)=c/(vn_(2))` or `lamda_(1)n_(1)=lamda_(2)n_(2)` `LAMDA=lamda_(1)((n_(1))/(n_(2)))` |
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| 39. |
A metallic ring is attached with the wall of a room. When the north pole of a magnet is brought near it, the induced current in the ring will be |
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Answer» FIRST CLOCKWISE then ANTICLOCKWISE. |
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| 40. |
Give the average value of sin^(2)omegat(ltsin^(2)omegatgt). |
| Answer» SOLUTION :SINCE `ltsin^(2)omegatgt =(1)/(2)(1- ltcos2omegatgt)andltcos2omegatgt =0,ltsin^(2)omegatgt =1//2` | |
| 41. |
A circle of radius a has charge density given by lambda=lambda_(0)cos^(2)theta on its circumference, where lambda_(0) is a positive constant and theta is the angular position of a point on the circle with respect to some reference line. The potential at the centre of the circle is |
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Answer» `(lambda_(0))/(4epsilon_(0))` |
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| 42. |
How do the two lenses of an astronomical telescope differ from each other? |
| Answer» Solution :FOCAL LENGTH of the objective is GREATER than focal length of eye PIECE. | |
| 43. |
In a vertical smooth hollow thin tube, a block of same mass as that of tube is released as shown. When it is slightly disturbed it moves towards right. By the time the block reaches the right end of the tube, the displacement of the tube will be (where ‘R’ is the mean radius of tube, assume that the tube remains in vertical plane). |
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Answer» `(2R)/(pi)` |
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| 44. |
(A): When the base region of a transistor has large width, the collector current decreases (R): Electron hole combination in base results in increase of base current |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the correct EXPLANATION of 'A' |
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| 45. |
What are the applications of surface tension? |
| Answer» Solution :(i) Determination of ATOMIC diameter. (ii) Manufacture of lead shots. (iii) manufacture of waterproof material. (iv) Surface tension of URINE is USED for diagnosis of JAUNDICE in PATHOLOGICAL test. | |
| 46. |
If A is the atomic mass number of an element, N is the Avogadro number and a is the lattice parameter, then the density of the element, if it has crystal structure, is : |
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Answer» `(A)/(Na^(3))` |
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| 47. |
The overall gain of a multistage amplifier is 100. When negative feedback is applied, the gain reduces to v 10. Find the fraction of the output that is feedback to the M input. |
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Answer» Solution :We KNOW that `A_f = (A )/(1 + A BETA)` Here, `A_f= 10 and A= 100` ` therefore1= (100)/(1 +100 beta) or10 + 1000beta= 100` ` beta = (100-10)/( 1000) = (90)/( 1000 ) = (9)/(100) = 0.09` |
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| 48. |
A parallel plate capacitor of capacitance C is charged to a potential V by a battery . Without disconnecting the battery , the distance between the plates is tripled and a dielectric medium of dielectric constant 10 is introduced between the plates of the capacitor . Explain giving reasons , how will the following be affected : capacitance of the capacitor |
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Answer» Solution :Initially capacitance of given parallel capacitor, `C = (in_(0) A)/(d)` and it has been charged to a potential DIFFERENCE V . Hence charge on capacitor Q = CV and electric field between the plates of capacitor E `= (V)/(d)` and so the energy density `u = 1/2 in_(0) E^2` As per question the new distance between the plate `d. = 3d` and a dielectric medium of dielectric constant K = 10 is INTRODUCED between the plates of capacitor . Moreover , as battery REMAINS connected , the potential difference V remains unchanged . `therefore ` New capacitance `C. = (k in_(0) A)/(d.) = (10 xx in_(0) A)/(3d) = (10)/(3) C` |
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| 49. |
Half life period of a radio active element A is 10 hours. In certain time 2g of A has become 0.25g. In the same time 4g of B reduced to 0.5g. Half life period of B is |
| Answer» ANSWER :A | |
| 50. |
NAND and NOR gates are called universal gates primarily because they : |
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Answer» are AVAILABLE universally |
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