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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. |
Temperature of an ideal gas, initially at 27^(@)C, is raised by 6^(@)C. The rms velocity of the gas molecules will |
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Answer» increase by NEARLY 2% `therefore` Percentage increase in velocity `v_(rms)`, `(Deltav_(rms))/(v_(rms))xx100=(1)/(2)(DeltaT)/(T)xx100` `or (Deltav_(rms))/(v_(rms))xx100=(1)/(2)XX(6)/(300)xx100` `(Deltav_(rms))/(v_(rms))xx100=1%` HENCE, the rms velocity increases by 1%. |
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| 2. |
STATEMENT-1 if white light is used in YDSE coloured fringes are obtained STATEMENT-2 For sound waves,air is denser as compare to wall. |
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Answer» STATEMENT-1 is TRUE Statemetnt-2 is True,Statement -2 is a correct explanation for statement -1. |
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| 3. |
Soft ironis prefereed as core of transformer in theformof sheets due to its |
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Answer» high RETENTIVITY LOW coercivity and low hystersis loss |
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| 4. |
For Q. 25 find the minimum constant force to be applied to body of mass m_(1) in order to shift body of mass m_(2). Take m_(1)=m_(2)=10kg, mu=0.5 |
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Answer» Solution :`F.X=mum_(1)gx-1/2kx^(2)=0""...(i)` But `Kx=mu_(2)G.` (for just moving body of mass `m_(2))` Put the value of K in (i) to get F. F.x `-mum_(1)gx-1/2kx^(2)=0` But `kx=mum_(2)g` (for moving body of mass `m_(2)`) `Fx-mum_(1)gx-1/2((mum_(2)g)/(x))x^(2)=0` `F=mum_(1)g+1/2mum_(2)g` `=mu(m_(1)+(m_(2))/(2))g` `=0.5(10+10/2)10` `=75N` |
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| 5. |
To increase magnification of telescope ....... |
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Answer» `f_0`, should be GREATER and `f_e` SMALLER |
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| 6. |
Energy of one photon of radiant energy is 15 keV. Then this radiation belongs to ……. Parts of electromagnetic waves. |
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Answer» ULTRAVIOLET `E = (hc)/(lambda)` `therefore lambda = (hc)/(E )` `= (6.625xx10^(-34)xx3xx10^(8))/((15xx10^(3))(1.6xx10^(-19)))` `=0.8281xx10^(-10)m=0.8281 Å` `rArr` Above wavelength corresponds to X - rays in the electromagnetic SPECTRUM. |
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| 7. |
A block of mass m is resting inside a smooth horizontal tube. If the tube accelerates vertically downward with acceleration a=kt where k is a positive constant, the correct plot showing the variation of normal reaction received by the block as a function of time (considering that upward direction of normal reaction is positive and taken towards positive direction y axis) will be |
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| 8. |
When a current is passed in moving coil galvanometer, the coil gets deflected because, |
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Answer» CURRENT in the COIL produces an electric field |
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| 9. |
A Carnot engine takes 3xx10^6 cal of heat from a reservoir at 627^@C and gives it to sink at 27^@C. What is the work done by. the engine ? |
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Answer» (a) ZERO |
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| 10. |
The equivalent resistance between points A and B of an infinite network of resistance each of 1 Omega connected as shown is |
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Answer» INFINITE |
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| 11. |
A wooden cubical block of mass m and side a is resting on a horizontal surface. A wire carrying current I, is wrapped around it in from of a square of side a. A uniform magnetic fieldvec (B) = B_(0)vec(j)is switched on in the region. Neglect the mass of the wire. (a)At what distance from the x axis does normal force applied by the horizontal surface on the wooden cube act? (b)What is the maximum value of current for which the block will not topple? |
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| 12. |
(A):Two balls of different masses are thrown vertically upward with same speed,they will pass through their point of projection in the downward direction with the same speed . (neglect air resistance) (R ):The maximum height and downward velocity attained at the point of projection are independent of the mass of the ball. |
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| 13. |
Derive the equations of stored energy for series or parallel connection of many capacitors. |
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Answer» Solution :For series connection : Charge REMAINS same Q on each capacitor `:.` TOTAL energy stored, `U =(Q^(2))/(2).(1)/(C)` `=(Q^(2))/(2)[(1)/(C_(1))+(1)/(C_(2))+CDOTS+(1)/(C_(n))]` `=(Q^(2))/(2C_(1))+(Q^(2))/(2C_(2))+cdots+(Q^(2))/(2C_(n))` `:. U = U_(1)+U_(2)+cdots+U_(n)` For parallel connection : Potential difference V remains same for each capacitor `:.` Total energy stored , `U=(1)/(2) CV^(2)` `=(1)/(2)[C_(1)+C_(2)+cdots+C_(n)]V^(2)` `=(1)/(2) C_(1)V^(2)+(1)/(2)C_(2)V^(2)+cdots+(1)/(2)C_(n)V^(2)` `:. U_(1)+U_(2)+cdots+U_(n)` So the total energy stored in both the connection is the sum of energy stored in each capacitor. |
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| 14. |
Consider the following reaction The major products formed in the reaction are |
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Answer» `(CH_(3))_(2) CHCH_(2)OH and CH_(3)CH_(2)I` 
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| 15. |
A hydrogen target is bombarded by pions. Calculate the threshold values of kinetic energies of these poins making possible the following reactions: (a) pi^(-)+prarr K^(+)+Sigma^(-), (b)pi^(0)+prarr K^(+)^^ ^(0) |
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Answer» Solution :(a) Here `T_(th)=((m_(k)+m_(Sigma))^(2)-(m_(x)+m_(p))^(2))/(2m_(p))C^(2)` SUBSTITUTION GIVES `T_(th)=0.904GeV` (b) `T_(th)=((m_(k)^(+)+m_(^^))^(2)-(m_(PI^(0))m_(p))^(2))/(2m_(p))c^(2)` Substitution gives `T_(th)= 0.77GeV` |
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| 16. |
If the threshold wavelength for a certain metal is 2000Å, then the work function of the metal is |
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Answer» `6.2`J |
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| 17. |
Temperature of a conductor increases by 5^(@)C passing electric current for same time. The increase in its temperature when double current is passed through 1he same conductor the same time is ..... ""^(@)C. |
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Answer» 10 ` H = (I^(2) Rt)/(J)` MS `Delta theta = (I^(2) Rt)/(J) ` `therefore Delta theta prop I^(2)` `therefore Delta theta_(1)prop I_(1)^(2)` `therefore Delta theta_(2) prop I_(2)^(2) ` ` therefore (Delta theta_(2))/(Delta theta_(1))= ((I_(2))/(I_(1)) )^(2) ` `therefore Delta theta_(2) = ((I_(2))/(I_(1)))^(2) Delta theta_(1)` Now `I_(2)= 2I_(1)` `therefore Delta theta_(2) = ((2I_(1))/(I_(1)))^(2) XX 5 ` `therafore Delta theta_(2) = 20""^(@)C` |
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| 18. |
Two closed organ pipes, A and B have the same length. A is wider than B. They resonante in the fundamental mode at frequencies n_(A) and n_(B) respectively. |
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Answer» `n_(A)=n_(B)` |
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| 19. |
How can you represent experimentally that (i) there are two types of charges and (ii) there is repulsion between two like charges and attraction between two unlike charges ? |
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Answer» Solution :Only two types of charges are there on different bodies that is proved by following two experiments. Experiment 1 If two glass rods rubbed with wool or SILK cloth are brought CLOSE to each other, they repel each other as shown in figure (a).  The two strands of wool or two pieces of silk cloth, with which the rods were rubbed, also repel each other. However, the glass rod and wool attracted each other as shown in figure (b) but attracted the fur. On the other hand, the plastic rod attracts the glass rod as shown in figure (c) and repel the silk or wool with which the glass rod is rubbed. If a plastic rod rubbed with fur is made to touc two small pith balls (polystyrene balls suspended by silk or nylon thread, then the bal repel each other as shown in figure (d) and at also repelled by the rod. Experiment 2 A similar effect is found if the pith balls ar touched with a glass rod rubbed with silk a shown in figure (e). A dramatic observation is that a pith bal touched with glass rod attracts another pith bal touched with plastic rod as shown in figure (f) It is concluded after many careful studies b, different scientists, that there were only twi kinds of an entity which is called the electric charge. There are two types of charges : Positive anc: negative like charges repel and UNLIKE charge attract each other. The experiments also demonstrated that the charges are transferred from the rods to the pith balls on contact. It is said that the pith balls are charged by contact. The property which differentiates the two kinds of charges is called the polarity of charge. DUE to internal property, there is attractive oi repulsive force between balls when they are charged. They are ELECTRICALLY neutral when they have no charge. |
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| 21. |
Three metal rods are located relative to each other shown in figure - 6.101, where L_(1) + L_(2) = L_(3) Values of density and Young's modulus of the three materials are: (rho_(1)) = 2.7 xx 10^(3)kg//m^(3)y_(1) = 7 xx 10^(10)Pa, (rho_(1)) = 11.3 xx 10^(3)kg//m^(3)Y_(2) = 1.6 xx 10^(10) pa, (rho_(1)) = 8.8 xx 10^(3)kg//m^(3)Y_(3) = 11 xx 10^(10) pa If L_(3) = 1.5m What musht the ratio L_(1)//L_(_(2)) be if a sound wave is to travel the length of rods 1 and 2 in the same time as required to travel the length of rod 3 ? |
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| 22. |
Use Lenz's law to determine the direction of induced current in the situations described by Fig. 6.04 : (a). A wire of irregular shape turning into a circular shape, (b) A circular loop being deformed into a narrow straight wire. |
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Answer» Solution :(a) When a wire of irregular shape turns into a circular shape, the area enclosed by the loop increases and, consequently, the MAGNETIC FLUX linked with the loop.increases. In accordance with Lenz.s law the induced CURRENT tends to oppose the increase in flux and it is possible only when induced current flows along adcba (i.e., in an anticlockwise direction). (b) When a circular loop is being deformed into a narrow straight wire type rectangular loop as shown in Fig. 6.04(b), SURFACE area and hence outward magnetic flux decreases. Consequently, in accordance with Lenz.s law, the induced currents want to add to the magnetic flux. It is possible if induced current is anticlockwise i.e., along a.d.c.b. so that the flux DUE to induced current is in same direction as flux due to existing magnetic field. |
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| 23. |
A carbon resistor has three bands as Brown Black and Green in order. What will be that range of resistance it offers. |
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Answer» `7 XX 10^(5) Omega - 13 xx 10^(5) Omega` resistance R = `10 xx 10^(5) pm 20 ` % `therefore 20 % ` of 1000000 = 200000 `Omega` `therefore ` Resistance R = ` (1000000 pm 2000000) Omega` = `8 xx 10^(5) Omega " OR " 12 xx 10^(5) Omega` |
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| 24. |
A constant current is maintained in a resistive solenoid having iron rod as its core. Which of the following quantities will change when iron rod is removed from solenoid? |
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Answer» Rate of HEAT dissipation |
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| 25. |
In Balmer series of hydrogen spectrum line of 486.1 nm wavelength is called ...... |
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Answer» `H_(alpha)` |
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| 26. |
The magnetic needle has magnetic moment 3.14 xx 10^(-2) "Am"^(2) and moment of inertia 1=2 xx 10^(-6) kg m^(2). It performs 314 complete oscillations in 100 s. What is the magnitude of the magnetic field ? |
| Answer» SOLUTION :`8 XX 10^(-4)` T | |
| 27. |
A converging lens of focal length 15 cm and a converging mirror of focal length 20 cm are placed with their principal axes coinciding. A point source S is placed on the principal axis at a distance of 12 cm from the lens, as shown in figure. It is found that the final beam comes out parallel to the principal axis. Let the separation between the mirror and the lens be 10 xx k. Find k. |
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| 28. |
यदि n(A)=12,n(B)=8,n(AnnB)=4 तो n(AuuB) का मान ज्ञात कीजिए । |
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Answer» 24 |
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| 29. |
Is the magnetic field same at all points which are at the same distance from the element? |
| Answer» Solution :No. It depends on SINE of the angle between current element and radius vector. (i.e., `DB PROP sin THETA )` | |
| 30. |
Two identical circular wires P and Q each of radius R and carrying current 'I' are kept in perpendicular planes such that they have a common centre as shown in fig. Find the magnitude and direction of the net magnetic field at the common centre of the two coils. |
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Answer» Solution :Magnetic field `B_p` due to current carrying coil `P` = Field BQ due to current carrying coil `Q = (mu_0 I)/(2R)`. Their directions are as shown in fig. As the two field are in mutually PERPENDICULAR directions, HENCE net magnetic field at the common centre O is `B = sqrt(B_(P)^(2) + B_(Q)^(2)) = (mu_0 I)/(sqrt(2)R)` The net field SUBTEND an angle `beta` from `vec(BQ)` such that `tan beta = (BP)/(BQ) = 1` `implies beta = 45^@` 
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| 31. |
If an electron enters into a space between the plates of a parallel plate capacitor at an angle alpha with the plates and leaves at angle betato the plates. The ratio of its kinetic energy while entering the capacitor to that while leaving will be : |
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Answer» `[(COS alpha)/(cos beta)]^(2)` |
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| 32. |
There are two concentric metallic spherical shells of radii a and b such that a < b. An ideal cell of emf is connected across the two shells. The medium between the sheets is filled with a dielectric of dielectric constant and resistivity p. For a point P at a distance from the common centre C where a < r< b. Now choose INCORRECT option. |
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Answer» RATE of fall of potential is `EPSILON/(b-a)` |
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| 33. |
A charged particle can be deflected by electric and magnetic filed both The electric force oversetrarrF_(e) =qoversetrarrE and magnetic force oversetrarr(F_(m)) = oversetrarr(qv)xxoversetrarr(B) q = charge in coulomb, oversetrarrV is velocity in m/s oversetrarrE is electric filed strengh in N//C and B is magnetic field strength in tesla (T) If a moving charged particle enters simultaneous electric and magnetic field, the net force on the charged particle is oversetrarr(F) =q(oversetrarrE +oversetrarr(v) xx oversetrarrB) If oversetrarr(F) =0 the charged particle will move undeflected in simultaneous fields If a charged particle moving along x-axis enters the simultaneous electric and magnetic fields at right angles Electric field is along y-axis and magnetic field along z-axis and vgt(E)/(B) the particle will . |
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Answer» PASS undeflected |
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| 34. |
A charged particle can be deflected by electric and magnetic filed both The electric force oversetrarrF_(e) =qoversetrarrE and magnetic force oversetrarr(F_(m)) = oversetrarr(qv)xxoversetrarr(B) q = charge in coulomb, oversetrarrV is velocity in m/s oversetrarrE is electric filed strengh in N//C and B is magnetic field strength in tesla (T) If a moving charged particle enters simultaneous electric and magnetic field, the net force on the charged particle is oversetrarr(F) =q(oversetrarrE +oversetrarr(v) xx oversetrarrB) If oversetrarr(F) =0 the charged particle will move undeflected in simultaneous fields If electric field is along Y axis and charged particle moves along X axis the magnetic field must be . |
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Answer» `(E)/(v)` ALONG negative z-axis |
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| 35. |
A choke coil is connected in series with a lamp and the combination across a dc supply. What happens to the intensity of light if an iron core is used in the coil? |
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Answer» Solution :INTENSITY of the BULB REMAINS the same. Note : Although `Lpropmu_(r),X_(L)=2pifL=0" for "DC(f=0)`. |
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| 36. |
The objective lenses of two telescopes have the same aperture but their focal lengths are in the ratio 1:2. compare the resolving powers of the two telescopes. |
| Answer» Solution :RESOLVING POWERS of two telescope is equal because APERTURE of their objectives LENSES is same and resolving power does not depend on focal LENGTH of lenses. | |
| 37. |
The p.d across the wire is doubled |
| Answer» SOLUTION :`v_d=(EE)/(m)tau`=`(Ve)/(ml)tau` When V is DOUBLED,DRIFT velocity also gets doubled. | |
| 38. |
A 4.0 kg block is suspended from the ceiling of an elevator through a string having a linear mass density of 19.2 xx10^(-3)kg//m. Find the speed (with respect to the string) with which a wave pulse can proceed on the string if the elevator accelerates up at the rate of 2.0 m//s^2. Take g = 10 m//s^2 |
| Answer» SOLUTION :`50 m//s` | |
| 39. |
What is Thomas Young's double slit experiment ? |
| Answer» Solution :Bt using a narrow SLIT ( 1 MM wide) illuminated by MONOCHROMATIC light and two slits (0.5 m apart it was CONDUCTED). | |
| 40. |
What are good conductors? Give examples. |
| Answer» SOLUTION :Materials which have LOW VALUES of resistivity are called good CONDUCTORS. Eg: Silver, Copper, ALUMINIUM. | |
| 41. |
Magnetic susceptibility X, magnetic intensity H and magnetisation vector M are correlated as |
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Answer» `X = HM` |
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| 42. |
In the conditions of the previous problem find the kinetic, the potential and the total energy of the oscillator. |
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Answer» |
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| 43. |
The resistance of an ideal diode in forward biased condition is |
| Answer» ANSWER :A | |
| 44. |
The applied input a.c. power to a half-wave rectifier is 100 watts. The d.c. output power obtained is 40 watts. What happens to remaining 60 watts ? |
| Answer» SOLUTION :`40%` EFFICIENCY of rectification does not MEAN that `60%` of power is lost in the rectifier circuit.`thereforepower efficiency=40/50xx100=80%` | |
| 45. |
The applied input a.c. power to a half-wave rectifier is 100 watts. The d.c. output power obtained is 40 watts. What is the rectification efficiency ? |
| Answer» SOLUTION :RECTIFICATION EFFICIENCY `=(d.c.output POWER)/(a.c."INPUT"power)=40/100=0.4=40%` | |
| 46. |
Calculate the radius of_(79)^(197) Au. |
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Answer» Solution :ACCORDING to the EQUATION, `R = 1.2 XX 10^(-15) xx (197)^(1/3) = 6.97 xx 10^(-15) m (or) R = 6.97 F`. |
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| 47. |
A customer sits in an amusement park ride in which the compartment is to be pulled downward in the negative direction of a y axis with an acceleration magnitude of 1.24 g, with g=9.80" m"//"s"^(2). A 0.567 g coin rests on the customer's knee. Once the motion begins and in unit-vector notation, (a) What is the coin's acceleration relative to the ground? |
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Answer» Solution :The customer sitting and the coin on his knee both are in the noninertial frame of reference as the compartment is accelerating downward. Thus the concept of pseudo force comes into picture and we use this concept here. Calculation: The coin undergoes free fall. THEREFORE, with respect to GROUND, its acceleration ISS `vec(a)_("coin")=vec(g)=(-9.8" m"//"s"^(2))hatj`. (b) What is the coin.s acceleration relative to the customer? Calculation: Since the customer is being pulled down with an acceleration of `vec(a)_("customer")=1.24vec(g)=(-12.15" m"//"s"^(2))hatj`, the acceleration of the coin with respect to the customer is `vec(a)_("rel")=vec(a)_("customer")` `=(-9.8" m"//"s"^(2))hatj-(-12.15"m"//"s"^(2))hatj` `=(+2.35" m"//"s"^(2))hatj`. (c) How long does the coin take to reach the compartment ceiling, 2.20 m above the knee? Calculation: The time it takes for the coin to reach the ceiling is `t=sqrt((2h)/(a_("rel")))=sqrt((2(2.20" m"))/(2.35" m"//"s"^(2)))=1.37" s."` (d) In unit-vector notation, what is the actual force on the coin? Calculation: Since gravity is the only force acting on the coin, the actual force on the coin is `vec(F)_("coin")=m vec(a)_("coin")=m vec(g)` `=(0.567xx10^(-3)" kg")(-9.8" m"//"s"^(2))hatj` `=(-5.56xx10^(-3)" N")hatj`. (e) In unit-vector notation, what is the apparent force according to the customer.s measure of the coin.s acceleration? Calculation: In the customer.s frame, the coin travelsupward at a CONSTANT acceleration. Therefore, the apparent force on the coin is `vec(F)_("app")=m vec(a)_("rel")` `=(0.567xx10^(-3)" kg")(+2.35" m"//"s"^(2))hatj` `=(+1.33xx10^(-3)" N")hatj`. |
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| 48. |
A hollow insulated conducting sphere is given a positive charge of 10 muC . What will be the electric field at the centre of the sphere is its radius is 2 meters ? |
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| 50. |
Statement-I : Energy of a particle excuting simple harmonic motion is entirely potenital energy at the extreme possition. Statement-II : Particle at extreme position is at rest. |
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Answer» Statement-I is TRUE, Statement-II is true and So correct choice is (a). |
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