Saved Bookmarks
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 non-conducting sphere of radius R charged uniformly with surface density simga rotates with an angular velocity omega about the axis passing through its centre. Find the magnetic induction at the centre of the sphere. |
| Answer» SOLUTION :` mu_0 OMEGA sigmaR|cos THETA - (cos^3theta)/(3)|_(0)^(pi//2) = 2/3mu_0sigmaR` | |
| 2. |
Which one of the following do not represent travelling Harmonic wave |
|
Answer» `A (ax - BT)` |
|
| 3. |
What is the least amount of work that must be done to freeze ten gram of water at 0^@C by means of a refrigeration machine ? The temperature of the surrounding is 27^@C |
|
Answer» 70 cal |
|
| 4. |
Figure shows an arrangement of four charged particles, with angle theta = 35.0^(@) and distance d = 2.00 cm. Particle 2 has charge q_(2) = +8.00 xx 10^(-19) C, particle 3 and 4 have charges q_(3) = q_(4) = -1.60 xx 10^(-19)C, (a) What is distance D between the origin and particle 2 if the net electrostatic force on particle 1 due to the other particles is zero ? (b) If particles 3 and 4 were moved closer to the x axis but maintained their symmetry about that axis, would the required value of D be greater than, less than, or the same as the part (a) ? |
| Answer» SOLUTION :(a) 2.27 CM, (B) LESS | |
| 5. |
A charged particle can be deflected by electric and magnetic filed both The electric forceoversetrarrF_(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 A charged particle moves in simultaneous electric field oversetrarr(E) and magnetic field oversetrarr(B) If the charged particle passes undeflected , then the value of electric field vectore oversetrarr(E) must be . |
|
Answer» `oversetrarr(E) =oversetrarrv XX oversetrarr(B)` |
|
| 6. |
Curie temperature is the temperature below which it is _____. |
| Answer» SOLUTION :FERROMAGNETIC | |
| 7. |
Compute the magnetic of the magnetic field of a long, straight wire carrying a current of 1 A at distance of 1m from it. Compare it with Earth's magnetic field . |
|
Answer» Solution :Given that I = l A and radius r = l m `B_("straightwire") = (mu_(0)l)/(2pi r) = (4 PI XX 10^(-7) xx l)/(2 pi xx l ) = 2 xx 10^(-7) `T but the EARTH.s magnetic field is `B_("Earth") ~ 10^(-5)` T So, `B_("straightwire")` is one hundred times smaller than `B_("earth")`. |
|
| 8. |
In Fig 5.4 (b), the magnetic needle has magnetic moment 6.7 xx 10^(-2) Am^(2) and moment of inertia g=7.5 xx 10^(-6) Kg m^(-2). It performs 10 complete oscillation in 6.70s. What is the magnitude of the magnetic field? |
|
Answer» SOLUTION :The time period of oscillation is, `T=(6.70)/(10)=0.67s` From Eq. (5.5) `B=(4pi^(2)g)/(m T)` `=(4 xx (3.14)^(2) xx 7.5 xx 10^(-6))/(6.7 xx 10^(-2) xx (0.67)^(2))` =0.01T |
|
| 9. |
An object is placed at a distance of 20 cm from a thin plano convex lens of focal length 15 cm. The plane surface of lens is now silvered. What is the position of image? |
|
Answer» 12 cm to the LEFT of lens SYSTEM |
|
| 10. |
A circular current carrying loop of radius R, carries a current l. The magnetic field at a point on the axis of coll is (1)/(sqrt(8)) times the value of magnetic field at the centre. Distance of point from centre is |
|
Answer» `(R )/(SQRT(2))` If rise due to this force be `x_(1)` then `=(epsilon_(0)bv^(2)(k-1))/(2d)=(DL)x_(1)pg` Then `x_(2)` is the rise due to surface tension `T((1)/(r_(1))+(1)/(r_(2)))=T((1)/(oo)+(1)/(d//2))=(2T)/(d)=(x_(2))pg` `x_(1)+x_(2)=(epsilon_(0)bv^(2)(k-1))/(2d^(2)Lpg)+(2T)/(dpg)=5+2=7mm` |
|
| 11. |
Which one of the following is dimensionally incorrect ? |
|
Answer» Capcitance `C=[M^(-1)L^(-2)T^(4)A^(2)]` `B=(F)/(q.v)=ML^(0)T^(-2)A^(-2)` `L=(EPSILON)/(di//dt)=(Wt)/(QI)=ML^(2)T^(-2)A^(-2)( :.q=It)` `rho=(R.a)/(e)=ML^(3)T^(-3)A^(-2)` `:.L`is wrongly represented. `(c )` is correct choice. HENCE correct choice is `(c )`. |
|
| 12. |
In the vicinity of the triple point the saturated vapour pressure p of carbon dioxide depends on temperature T as log p = 1 - b//T, where a and b are constants. If p is expressed in atomspheres, then for the sublimation processa = 9.05 and b = 1.80 kK, and for the vaporization process a = 6.78 and b = 1.31 kK. Find : (a) temperature and pressure at the triple point , (b) the values of the specific latent heats of sublimation, vaporization, and melting in the vicinity of the triple point. |
|
Answer» Solution :(a) The equations of the transition LINES are `log p = 9.05 - (1800)/(T)` : Solid GAS =`6.78 - (1310)/(T)` : Liquid gas At the triple point they intersect. Thus `2.27 = (490)/(T_(tr))` or `T_(tr) = (490)/(2.27) = 216 K` corresponding `p_(t r)` is 5.14 atmosphere In the formula `log p = a - (b)/(T)`, we compare `b` with the corresponding term in the equation in `2.210`. Then `1n p = 1 XX 2.303 - (2.303 b)/(T)` So, `2.303 = (Mq)/( R)` or `q_("SUBLIMATION") = (2.303 xx 1800 xx 8.31)/(44) = 783 J//gm` `q_("liquid-gas") = (2.303 xx 1310 xx 8.31)/(44) = 570 J//gm` Finally `q_("solid - liquid") = 213 J//gm` on subtraction. |
|
| 13. |
A particle is projected vertically upwards and attains maximum height H. If the ratio of the times to attain heighth on two way journey t (h |
|
Answer» `2/3` |
|
| 14. |
Electric force on an electron, in an electric field is ....... |
|
Answer» in the DIRECTION of electric FIELD |
|
| 15. |
In the following· experiment the rheostat is at a fixed resistance and the resistance R in the ohm box is increased in steps while switching on the key: |
|
Answer» The voltmeter (V) READING will INCREASE and the ammeter (A) reading will DECREASE |
|
| 16. |
What inference was drawn from Division and Germer experiment regarding the nature of electrons ? |
| Answer» SOLUTION :This experiment CONFIRMS the EXISTENCE of ELECTRON waves. | |
| 17. |
A diatomic gas initially occupying a volume 3 litres at 300 K and one atmospheric pressure is adiabatically compressed to 1/3 of the initial volume. It is then isobarically expanded till its temperature becomes 300 K, and finally isothermally expanded to restore it to the initial P-V-T conditions. Find the work done during complete cycle of operations. |
|
Answer» |
|
| 18. |
A block of ice, initially at -20^(@)C, is heated a steady rate until the temperature of the sample reaches 120^(@)C. Which of the following graphs best illustrates the temperature of the sample as a function of time? |
|
Answer»
|
|
| 19. |
It is necessary to use satellites for long distance TV transmission.Why ? |
| Answer» Solution :TV signals are not REFLECTED by ionosphere. HENCE to reflect the signals back to the surface of the earth for long DISTANCE transmission, satellites are USED. | |
| 20. |
If the current gain in CB configuration is 0.95. Then the current gain in CE configuration is |
| Answer» Solution :Here , `BETA = ALPHA/(1- alpha)=(0.95)/(1-0.95)=(0.95)/(0.05)=19` | |
| 21. |
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 |
|
Answer» `E_e = 2E_a` |
|
| 22. |
The current land-speed record was set in the Black Rock Desert of Nevada in 1997 by the jet-powered car Thrust SSC. The car's speed was 1222 km//h in one direction and 1233 km//h in the opposite direction. Both speeds exceeded the speed of sound at that location (1207 km/h). Setting the land-speed record was obviously very dangerous for many reasons. One of them had to do with the car's wheels. Approximate each wheel on the car Thrust SSC as a disk of uniform thickness and mass M = 170 kg and assume smooth rolling. When the car's speed was 1233 km/h, what was the kinetic energy of each wheel? |
|
Answer» Solution :KEY IDEAS Equation 11-5 gives the kinetic energy of a rolling object, but we need three ideas to use it: (1) When we speak of the speed of a rolling object, we always mean the speed of the center of MASS, so here `v_("com") = 1233 km//h = 342.5 m//s`. (2) Equation 11-5 requires the angular speed `omega` of the rolling object, which we can relate to `v_("com")` with Eq. 11-2, writing `omega= v_("com")/R`, where R is the wheel.s radius. (3) Equation 11-5 also requires the rotational inertia `I_("com")` of the object about its center of mass. From Table 10-2c, we find that, for a uniform disk, `I_("com") (1/2)MR^(2)`. Calculations: Now Eq. 11-5 gives us `K= (1)/(2)I_("com")omega^(2)+ (1)/(2)Mv_("com")^(2)` `((1)/(2))((1)/(2)Mr^(2))((v_("com"))/(R))^(2)+ (1)/(2)Mv_("com")^(2)` `= (3)/(4)(170kg)(342.5m//s)^(2)` `= 1.50 xx 10^(7)J` (Note: The wheel.s radius R cancels out of the calculation.) Reasoning: This ANSWER gives one measure of the dan ger when the land-speed record was set by Thrust SSC: The kinetic energy of each (cast aluminum) wheel on the car was huge, almost as MUCH as the kinetic energy (`2.1 xx 10^(7) J`) of the spinning steel disk that exploded in Sample Problem 10.8. Had a wheel hit any hard obstacle along the car.s path, the wheel would have exploded the WAY the steel disk did, with the car and driver moving faster than sound! |
|
| 23. |
What are eddy currents ? Write any two applications of eddy currents. |
|
Answer» Solution :Whenever magnetic flux linked with a metallic BULK conductor changes, induced CURRENTS are set up in the body of the conductor in the form of closed loops and are known as eddy currents. Thus, eddy currents are the currents induced in a conductor when placed in a CHANGING magnetic field. Alternately eddy currents are induced in a metallic conductor when it is MOVED in a magnetic field and as a result magnetic flux linked with the conductor changes. Electric brakes, induction furnace, SPEEDOMETER, a.c. induction motor etc., are important applicationsof eddy currents. |
|
| 24. |
In R-C circuit when charge on the plate of the capacitor is increasing, the energy obtain from the source is stored in .... |
|
Answer» electric field |
|
| 25. |
How does the huge ball ( 5.4 xx 10^(5) kg ) hanging on the 22^(nd) floor of one of the world's tallest builgind ( chapter opening question mentioned in the introduction) counter the sway of the building? |
| Answer» Solution :The huge ball hangs from four cables and SWINGS like a pendulum when the building WAYS. When the building ways-say, WESTWARD-the ball does also but delayed enough so that ASIF finally swings westward, the building is swaying eastward. So, the motion of the ball is out of step with the BUILIDING motion, and thus counters it. | |
| 26. |
Two equal point charges are fixed at x = a and x = +a on thte axis. Another point charge Q is placed at the origin. The change in the electric potential energy of Q, when it is displaced by a small distance x along the X-axis is approximently proportional to |
| Answer» Answer :B | |
| 27. |
A straight current carrying wire has its one end attached to an infinity conducting sheet (shown as a circle in the figure). The other end of the wire goes to infinity and the wire is perpendicular to the sheet. The current spreads uniformly on the surface of the sheet. Calculate the magnitude of magnetic induction field at a point P at a distance d from the straight wire. Current in the wire is I. |
|
Answer» |
|
| 28. |
Obtain the magnification for the image form at infinity for simple microscope. |
|
Answer» SOLUTION :Suppose the object has a height h the MAXIMUM angle it can subtend and be clearly visible. (without a lens) is when it is at the near POINT i.e., a distance D. `tan theta_0=h/D` For small angle, `tan theta_0 ~~ theta_0` `therefore theta_0=h/D` ...(1)  Now if the angle subtended at the eye by the image when the object is at u `theta_i`, then `tan theta_i=h/f` For small angle, `tan theta_i~~theta_i` `therefore theta_i=h/f` ...(2)  From the figure angular magnification, `m=(theta_i)/(theta_0)` From EQUATION (1) and (2), `m=(h//f)/(h//D)` `therefore m=D/f` This is one less than the magnification when the image is at the near point but the viewing is more comfortable and the difference in magnification is usually small. The maximum limited magnification is (`le 9`) for real focal lengths for simple microscope. |
|
| 29. |
A flexible steel cable of total length 'L' and mass per unit length H hangs vertically (under it's own weight) from a support at upper end. If transverse pulse starts to move down in the wire from its support. The ratio of acceleration of the pulse at distance L/4 from the support end to the acceleration of the of the pulse at distance L/2 from the support end K:1 then the value of K is |
|
Answer» |
|
| 30. |
An image of an object obtained by a convex mirror is n times smaller than the object. If the focal length of lens is f, the object distance would be ........... |
|
Answer» `F/n` here, `m=1/n` `THEREFORE 1/n=v/u` `therefore v =u/n` now,`1/f=1/u+1/n` `therefore1/f =(1)/(-u)+n/u` `therefore 1/f=(-1+n)/(u)` `therefore u=(n-1)f` |
|
| 31. |
A projectile is thrown at an angle of 30^@ with a velocity of 10m/s. the change in velocity during the time interval in which it reaches the highest point is |
| Answer» Answer :B | |
| 32. |
When is the torque acting on an electric dipole placed in a uniform electric field maximum? |
| Answer» SOLUTION :When electric dipole is placed PERPENDICULAR to the direction of UNIFORM electric FIELD. | |
| 33. |
(a) What is the energy E of the hydrogen-atom electron whose probability density is represented by the dot plot of Fig. 38-16? (b) What minimum energy is needed to remove this electron from the atom? |
| Answer» SOLUTION :(a) `-3.4eV`, (B) `.34eV` | |
| 34. |
A radioactive substance X decays into another radioactive substance Y. Initially only X was present.lambda_x and lambda_y are the disintegration constants of X and Y. N_x and N_y are the number of nuclei of X and Y at any time t. Number of nuclei N_y will be maximum when |
|
Answer» `N_y/(N_x-N_y)=lambda_y/(lambda_x-lambda_y)` |
|
| 35. |
K_(alpha) radiation of Mo (Z = 42) has a wavelength of 0.71Å. Calculate wavelength of the corresponding radiation of Cu (Z=29) |
|
Answer» `1.45Å` `V=a(Z-SIGMA)^(2)` for `K_(alpha)` radiationscreening CONSTANT `sigma =1 and v=(c)/(lambda)` `:. (1)/(lambda) prop (Z-1)^(2)` `(lambda_(Cu))/(lambda_(MO))=((Z_(Mo)-1)^(2))/((Z_(Cu)-1)^(1))=(41^(2))/(28^(2))` `:. lambda_(Mo)=0*71Å` Then `lambda_(Cu)=(0.71)Åxx(41^(2))/(28^(2))=1*52Å` |
|
| 36. |
The variation of potential energy of a particle of mass m = 5kg to its x-coordinate . The particle moves under the effect of this conservative force only along the x-axis. Then, choose the correct option. |
|
Answer» If the particle is released at the origin, then it will move towards x-axis. |
|
| 37. |
A converging lens has a focal length of 0.2 m. The image fromed will be virtual, when the object is at a distance of |
|
Answer» `0.25 m` |
|
| 38. |
A satellite revolving close to earth has a K.E., E. What energyshould be given to it so make it escape the gravitational pull of earth? |
|
Answer» `(E )/(2)` `E.=(1)/(2)mv_(e )^(2)=(1)/(2)m(sqrt(2)v_(0))^(2)=2((1)/(2)mv_(0)^(2))` `therefore E.=2E` `therefore ` Additional energy to be given `=E.-E=2E-E=E` Thus correct choice is (b). |
|
| 39. |
Which one of the following relations is correct |
|
Answer» `1N//C = 10^(8)` Volt/m |
|
| 40. |
A (hypothetical) large slingshot is stretched 2.30 m to launched a 170 g projectile with speed sufficient to escape from Earth (11.2 km//s). Assume the elastic bands of the slingshot obey Hooke's law. What is the spring constant of the device if all the elastic potential energy is converted to kinetic energy? |
|
Answer» `4.03 XX 10^(6) N//m` |
|
| 41. |
The minimum distance of the reflector from the listener to hear the echo of monosyllabe sound is nearly equal to : |
|
Answer» 16.5 m `therefore ` time to speak 1 syllable = `(1)/(5) `sec Now,If d is the distance of REFLECTOR ` (2 d)/(v)= t` `d = (vt)/(2) = (v)/(2) xx (1)/(5) = (300)/(10)` ` d = 33 m ` Hence the CORRET choice is (b). |
|
| 42. |
Characol pieces of tree is found from an archeological site. The carbon - 14 content of this characol is only 17.5% that of equivalent sample of carbon from a living tree. What is the age of tree? |
|
Answer» SOLUTION :According to radioactive law `R = R_0e^(-lambdat)` `e^(lambdat) = R_0/R` Taking LOG on both sides `t= 1/lambda In(R_0/R)` HALF LIFE of CARBON, `T_(1//2) = 5730` years `t = (T_(1//2))/(0.6931) In (1/(0.175))` `=(5.730" years")/(0.6931) xx 1.74297` `= 14409.49` years `t = 1.44 xx 10^4` years `R_0 = 100%` `R = 17.5%` `lambda = (0.6931)/(T_1//2)` `T_(1//2) = 5730` years |
|
| 43. |
A sounding body of negligible dimension emitting a frequency of 150 Hz is dropped from a height. During its fall under gravity it passes near a balloon moving up with a constant velocity of 2m/s one second after it started to fall.The difference in the frequency observed by the man in balloon justbefore and just after crossing the body will be : (Given that -velocity of sound = 300m/s, g = 10 m//s^2) |
|
Answer» 12 |
|
| 44. |
What is meant by the term doping of an intrinsic semiconductor ? How does it affect the conductivity of a semiconductor ? |
|
Answer» Solution :Doping of an intrinsic semiconductor means DELIBERATELY introducing a controlled quantity (generally 1 part in `10^(6)` parts or 0.0001%) of impurity element which is either pentavalent or trivalent so as to obtain -type and p-type semiconductors, RESPECTIVELY. The electrical conductivity of an extrinsic semiconductor formed as a result of doping is many fold increased as COMPARED to that of undoped intrinsic semiconductor. |
|
| 45. |
A boat having a length 3 m and breadth 2 m is floating on a lake. The boat sinks by 1 cm when a man gets on it. The mass of the man is : |
|
Answer» 60 Kg WEIGHT of man = wt. of extra water displaced by the boat mg=V.pg `thereforemg=(3xx2xx1/100)xx10^(3)xxg` `rArrm=60kg` Hence the CORRECT choice is (a). |
|
| 46. |
A concave and convex lens have the same focal length of 20 cm and are put into contact to form a lens combination. The combination is used to view an object of 5 cm length kept at 20 cm from the lens combination. As compared to the object, the image will be |
|
Answer» MAGNIFIED and inverted |
|
| 47. |
Current amplification factor of a common base configuration is 0.88. Find the value of base current when the emitter current is 1 mA. |
|
Answer» SOLUTION :In a common-base arrangement, the CURRENT amplification FACTOR. `alpha=((DeltaI_(C))/(DeltaI_(E)))_(V_(CB))=(I_(C))/(I_(B))," Given "alpha=0.88, I_(E)=1mA` `therefore`Collector current `I_(C)=alphaI_(E)=0.88xx1=0.88mA`. Now since `I_(E)=I_(B)+I_(C)` `therefore` Base current `I_(B)=I_(E)-I_(C)=1-0.88=0.12mA`. |
|
| 48. |
two heading coils ofresistances 10Omega and 20 Omega are connected in parallel and connected to a battery of emf 12 V and internal resistance1Omega Thele power consumed by the n are in the ratio |
|
Answer» `1:4` |
|
| 49. |
A closed organ pipe of length 1.2 m vibrates in its first overtone mode. The pressure variation is maximum at: |
|
Answer» 0.8 m from the OPEN end |
|
