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. |
Assuming the expression for the pressure exerted by an ideal gas, show that the rms speed of a gas molecule is directly proportional to the square root of its absolute temperature. Calculate the kinetic energy of 10 grams of argon molecules at 127^(@)C. [Atomic weight of argon= 40] |
|
Answer» |
|
| 2. |
The antenna current of an AM transmitter is 8A when only the carrier is sent, but it increases to 8.93A when the carrier is modulated by a single sine wave. Find the percentage modulation |
|
Answer» `60.1`% |
|
| 3. |
What is pole ? |
| Answer» SOLUTION :It is the MIDDLE POINT of the REFLECTING SURFACE. | |
| 4. |
Which of the following is in the increasing order of penetrating power ? |
|
Answer» `ALPHA,BETA,GAMMA` |
|
| 5. |
A stretched rubber has |
|
Answer» INCREASED KINETIC energy |
|
| 6. |
The most commonly employed analog modulation technique in satellite communi-cation is the |
|
Answer» AMPLITUDE modulation |
|
| 7. |
A slit 4.0 cm wide is irradiated with microwaves od wavelengths 2.0 cm. Find the angular spread of the central maximum, assuming incidence normal to the plane of the slit. |
|
Answer» |
|
| 8. |
A dipole is placed in xy plane parallel to the line y = 2x. There exists a uniform electric field along z-axis. Net force acting on the dipole will be zero. But it can experience some torque. We can show that the direction of this torque will be parallel to the line. |
|
Answer» `y = 2x + 1` Torque will be perpendicualr to the line `y=2x` and it should be in `xy` plance, because electric field is in `z-`directio. The lines in options (c) and (d) are perpendicular to `y=2x. |
|
| 9. |
Weightlessness experienced while orbitting the earth in spaceship is the result of : |
|
Answer» acceleration Hence correct CHOICE is (a). |
|
| 10. |
(A): The ratio for time taken for light emssion from an atom to that for release of nuclear energy in fission is 1 : 100 (R): Time taken for the light emission from an atom is of the order of 10^(-8) S. |
|
Answer» Both .A. and .R. are TRUE and .R. is the correct explanation of .A. |
|
| 11. |
A uniform ring of mass m, with the outside radius r_2, is fitted tightly on a shaft of radius r_1. The shaft is rotated about its axis with a constant angular acceleration beta. Find the moment of elastic forces in the ring as a function of the distance r from the rotation axis. |
|
Answer» Solution :Consider an ELEMENTARY RING of width `dr` at a distant `r` from the axis. The part OUTSIDE EXERTS a couple `N+(dN)/(dr)dr` on this ring while the part inside exerts a couple N in the opposite DIRECTION. We have for equilibrium `(dN)/(dr)dr=-dIbeta` where `dI` is the moment of inertia of the elementary ring, `beta` is the angular acceleration and minus sign is needed because the couple `N(r)` decreases, with distance vanshing at the outer radius, `N(r_2)=0`. Now `dI=(m)/(pi(r_2^2-r_1^2))2pirdrr^2` Thus `dN=(2mbeta)/((r_1^2-r_1^2))r^3dr` or, `N=1/2(mbeta)/((r_2^2-r_1^2))(r_2^4-r^4)`, on integration 
|
|
| 12. |
A compound microscope has a eyepiece of focal length 10m and an objective of focal length 4cm. Calculate the magnification if an object is kept at a distanceof 5cm from the objective then final image is formed at the least distance of distinct vision. |
|
Answer» Solution :(c) We know `1/(v_(0))+1/5=1/4` `v_(0)=20cm` and magnification `(M)=(v_(0))/(u_(0))[1+D/(f_(e))]` `M=20/5[1+20/10]=12` |
|
| 13. |
What should be the resistance of an ideal ammeter ? |
|
Answer» very less |
|
| 14. |
Find potential at point A and B due to the small charge -system fixed near origin. (distance between the charges is negligible). |
|
Answer» SOLUTION :Dipole MOMENT of the system is `vec(P) = (qa) + (qa) hat(j)` Potential at POINT A due to the dipole `V_(A) = K((vec(P).vec(r )))/(r^(B)) = (K[(qa) hat(i) +(qa)hat(j)].(4 hat(i) + 3HAT(i)))/(5^(3)) = (k(qa))/(125)(7)` `rArr V_(B) = (K[(qa) hat(i) +(qa)hat(j)](3hat(i) - 4hat(j)))/(5^(3)) = (k(qa))/(125)` |
|
| 15. |
How many of the following has hydrogen bonding (a) NH_(3) , (b) CH_(4) , (c) H_(2)O (d) HI , (e) HF , (f) HCOOH (g) B(OH)_(3) , (h) CH_(3)COOH , (i) HCO_(3^(-))ion |
|
Answer» |
|
| 16. |
The relation between surface tension and surface energy |
|
Answer» W = TA |
|
| 17. |
What is the effective capacitance between point A and B in the given figure? |
|
Answer» `1 MUF` |
|
| 18. |
The maximum amplitude of the modulated wave is 16 v and minimum amplitude is 4 V. The percentage modulation is |
| Answer» ANSWER :C | |
| 19. |
What do you understand by active and passive satellites ? |
|
Answer» Solution :REMOTE SENSING is a technique of obtaining information about an object/area from a distance without having any physical contact. Two types of satellites are used for remote sensing. `(i)` Passive satellites. A passive satellite is not equipped with electronic devices to process or AMPLIFY the signal. The signal is simply reflected back to earth, hence a very powerful transmitter is required. `(ii)` Active satellite. Active satellites have high resolution cameras, infrared scanners etc. These devices can record the information and them BEAMED down to earth in digital form which is them converted into maps and pictures by the COMPUTERS. Applications `(i)` Remote sensing is used for geological surveys, water resource surveys, urban land use survey and natural disaster surveys. `(ii)` Remote sensing data is used in monitoring climate changes. `(iii)` Remote sensing can be used to assess the damage caused by floods. |
|
| 20. |
A ray of light incident at an angle theta on a refracting face of a prism emerges from the other face normally. If the angle of the prism is 5^@ and the prism is made of a material of refractive index 1.5, the angle of incidence is |
|
Answer» `7.5^@` `delta=(mu-1)A=(1.5-1)xx5^@` `THEREFORE delta=0.5xx5^@=2.5^@` Here angle of refraction is `5^@` at the first SURFACE and so from the geometry of figure,  `delta=theta-r` `therefore theta=delta+r` `2.5^@+5.0^@` `7.5^@` |
|
| 21. |
At what depth in a salt water lake is the absolute pressure equal to three times atmospheric pressure ? Density of a sea water = 1.3 xx 10^3 (kg)/m^3. Atmospheric pressure =1.013 xx 10^5 N/m^2.g = 9.8 m/s^2. |
|
Answer» SOLUTION :ABSOLUTE PRESSURE = gauge pressure + atmospheric pressure, From the condition of the problem THEREFORE we have gauge pressure =` 2 xx` atmospheric pressure it h be the required depth hpg = `2 xx` atmospheric pressure h =` (2 xx 1.013 xx 10^5)/(1.03 xx 10^3 xx 9.8`) = 20 m. |
|
| 22. |
The ratio of length of two wires of same mass arc made up of same material is 1 : 2 Therefore ratio of their resistance is ........ |
|
Answer» `1: 1` `R = (rho L)/(A)= (rho l^(2))/(V) = (rho l^(2) d)/(m)` ` "(DENSITY" d = (m)/(V)rArr V = (m)/(d) ` `R PROP l^(2) ` (other terms are EQUAL. `therefore (R_(1))/(R_(2)) = ((l_(1))/(l_(2)))^(2) = ((1)/(2))^(2)= (1)/(4) ` |
|
| 23. |
Column-I gives some incomplete nuclear reaction match them with the reaction in Column-II. {:(,"Column I",,"Column II"),((A),.^(237)Nprarr.^(233)Pa+.......,,(P)beta^(+)"decay"),((B),._(228)Rararr......+.........+bar(v)("Anti neutrino"),,(Q)beta^(-)"decay"),((C),.^(41)._(20)Cararr......+.........+v("Neutrino"),,(R)alpha " decay"),((D),.^(197)._(80)Hg^(+)rarr.^(197)._(60)Hg+......,,(S)gamma" decay"),(,,,(T)"None"):} |
|
Answer» |
|
| 24. |
When a p-type semiconductor is brought into a close contact with n-type semiconductor, we get a p-n junction with a barrier potential 0.4V and the width ofdepletion region is 4.0 xx 10^(-7)m. this p-n junction is forwaed biased with a battery of voltage 3V adn negligible internal resistance, in series with a resistance of resistance R, ideal milliammeter and key K as shown in Fig. Wattage of diode is |
|
Answer» 0.060 W `=0.4xx20xx10^(-3)=0.008 W` |
|
| 25. |
Assertion In front of a concave mirror, a point object is placed between focus and centre of curvature. If a glass slab is placed between object and mirror, then image from mirror may become virtual. Reason Glass slab always makes a virtual image of a real object. |
|
Answer» If both Assertion and Reason are TRUE and Reason is the correct explanation of Assertion. |
|
| 26. |
When a p-type semiconductor is brought into a close contact with n-type semiconductor, we get a p-n junction with a barrier potential 0.4V and the width ofdepletion region is 4.0 xx 10^(-7)m. this p-n junction is forwaed biased with a battery of voltage 3V adn negligible internal resistance, in series with a resistance of resistance R, ideal milliammeter and key K as shown in Fig. If an electron with speed 4.0xx10^(5) ms^(-1) approaches the p-n junction from the n-side, the speed with which it will enter the p-side is |
|
Answer» `1.39xx10^(5) ms^(-1)` K.E. of the incident electron =workdone against potential barrier +K.E. of the emerging electron. i.e. `1/2mv_(1)^(2)=eV_+1/2mv_(2)^(2)` or `1/2xx(9.1xx10^(-31))xx(4xx10^(5))^(2)` `=(1.6xx10^(-19)xx(0.4)+1/2xx9.1xx10^(-31)xxv_(2)^(2)` on solving, we GET, `v_(2)=1.39xx10^(5) ms^(-1)` |
|
| 27. |
Statement -I : During thunderstoms light is seen much eartear than the hearing of sound. Statement-II : Speed of light is much more thant speed of sound . |
|
Answer» Statement -I is true, Statement -II is true and |
|
| 28. |
Sketch the output Y from a NAND gate having inputs A and B given below . |
|
Answer» Solution :For `t LE t_(1), "" A = 1 , B = 1 ,` Hence Y = 0 For `t_(1)` to `t_(2), "" A = 0 , B = 0 ,` Hence Y = 1 For `t_(2)` to `t_(3), "" A = 0 , B = 1` , Hence Y = 1 For `t_(3)` to `t_(4) , "" A = 1 , B = 0,`Hence Y = 1 For `t_(4)` to `t_(5), "" A = 1 , B = 1`,Hence Y = 0 For `t_(5)` to `t_(6), "" A = 0 , B = 0` , Hence Y = 1 For `t gt t_(6) , "" A = 0 , B = 1` , Hence Y = 1 |
|
| 29. |
What are de-Brogli Waves ? How does the de-Broglie wavelength vary with momentum of moving particle ? |
|
Answer» Solution :WAVES ASSOCIATED with material particles in motion are called matter waves. de Broglieequation is `lambda=h/(mv)`where `lambda`is the de Broglie WAVELENGTH, m is the mass and v is the velocity of the particle. de-Broglie wavelength `lambda` isinversely proportional to the linear momentum of the particle. i.e., `lambdaalpha 1/(MV)` where linear mmentum p=mv |
|
| 30. |
A body starting with a velocity 'v' returns to its initial position after 't' second with the same speed ,along the same line.Acceleration of the particle is |
|
Answer» `(-2V)/(t)` |
|
| 31. |
Maximum magnetization of a paramagnetic and ferromagnetic sample |
|
Answer» is of the same order |
|
| 32. |
Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon). |
| Answer» SOLUTION :`lambda=h//p=h//(hv//c)=c//v` | |
| 33. |
If the wavelength of an electromagnetic radiation is doubled, what will happen to the energy of photons ? |
|
Answer» Solution :Energy of a photon E = hv = `(hv)/LAMBDA` i.e. , E `prop 1/lambda` clearly, ENRERGY of a photon reduces to one HALF when the RADIATION is DOUBLED. |
|
| 34. |
The ratio of S.I. unit to C.G.S. unit of a Young's modulus |
| Answer» Answer :C | |
| 35. |
A circular coil of 100 turns has an effective radiusof 0.05 m and current of 0.1 A how much work is done to trun it through 180^(@) in auniform fieldof 1.5 wbm^(2) if the planeof the coil is initally perpendicularto the magnetic field |
|
Answer» 0.5523 j =MB `(cos 0^(@)-cos 180^(@))` `=2xx100xx0.1xx3.14xx(0.05)^(2)xx1.5=0.2355 j` |
|
| 36. |
A boy releases a hall from the top of a building. It will clear a window 2 m high at a distance 10 m below the top in nearly : |
|
Answer» 1 s `/_\=((2xx12)/(10))^(1//2) -((2XX10)/(10))^(1//2) =0.135 SEC`. |
|
| 37. |
Absolute zero temperature is not the temperature of zero energy, explain. |
| Answer» Solution :At absolute zero,4he ENERGY of translatory motion of molecules CEASES (being a FUNCTION of temperature) but the other forms of energy such as inter-molecular, POTENTIAL, energy of moleclar motion ETC. do not become zero. Thus absolute zero temperature is not the temperature of zero energy | |
| 38. |
Range of resistivity for metals is |
|
Answer» `10^(-6)Omega"m to"10^(-4)OMEGAM` |
|
| 39. |
The bob of a pendulum of mass 'm' suspended by an inextensible string of length L as shown in the figure carries a small charge 'q' . An infinite horizontal plane conductor with uniform surface charge density sigma1 is placed below it . What will be the time period of the pendulum for small amplitude oscillations ? |
|
Answer» `2pisqrt((L)/((g-(MG)/(epsilon_(0)sigma))))` `F_(E)=Qe` or acceleration `a=(qE)/(m)` Now effective value of g is `g_("effective")=g-ag-(qE)/(m)` `thereforeT=2pisqrt((L)/(g-(qE)/(m)))=2pisqrt((L)/(g-(qsigma)/(epsilon_(0)m)))("using"E=(sigma)/(epsilon_(0)))` |
|
| 40. |
State the condition under which a large magnification can be achieved in an astronomical telescope. |
| Answer» Solution :A large magnification can be ACHIEVED in an ASTRONOMICAL TELESCOPE by taking an objective lens of highest possible FOCAL length and an eyepiece of LEAST possible focal length. | |
| 41. |
In young.s double slit is illummated by light of wavelength 6000 A^(@). The slits are 0.1 cm apart and the screen is placed 1 m away .Then the angularposition of 10^(th) maxima is |
|
Answer» 3 |
|
| 42. |
A golfer standing on the ground hits a ball with a velocity of 52 mis at an angle Q above the 5 horizontal if tan theta = 5/12find the time for which the ball is LB at least 15m above the ground? (g = m//s^(2)) |
Answer» Solution : `v_(y) = sqrt(u_(y)^(2) -2gy)` `=sqrt(52 xx 52 xx (5 xx 5)/(13 xx 13) -2 xx 10 xx 15)` `= sqrt(16 xx 25 -300) = 10` `DELTAT =(2u_(y))/10 =(2 xx 10)/10 = 2s` |
|
| 43. |
Consider two coherent monochromatic (wavelength lambda) sources S_(1) and S_(2) separated by a distance d. The ratio of intensity of S_(1) and that of S_(2) at point P is 4. The distance of P from S_(1) if the resultant intensity at point P is equal to (9)/(4), times the intensity due to S_(1) is : (n is a positive integer ) |
|
Answer» `(d^(2)-N^(2)LAMBDA^(2))/(2nlambda)` |
|
| 44. |
A particle of mass m strikes a wall normally with a velocity v and then its velocity reversed. The change in momentum is : |
|
Answer» <P>m`UPSILON` Hence correct choice is (d). |
|
| 45. |
Two plane mirrors P and Q are aligned parallel to each other, as shown in the figure. A light ray is incident at an angle of q at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. The maximum number of times the ray undergoes reflections (including the first one) before it emerges out is: |
|
Answer» `1/(dtantheta)` x = Horizontal distance TRAVELLED by light ray in one reflection. So nx=lalso `tantheta=x/d` 
|
|
| 46. |
Microwaves and visible light have different wavelegnth but travel in vaccum with same speed. |
|
Answer» |
|
| 47. |
A potassium surface is placed 75 cm away from a 100 W bulb. It is found that energy radiated by the bulb is 5% of the input power. Consider each potassium atom as a circular disc of diameter 1Å and determine the time required for each atom to absorb an amount of energy equal to its work function of 2:0 eV: |
|
Answer» 76.5 sec `I=("POWER of bulb")/("Areaof sphere")=(100xx(5)/(100))/(4pixx(0*75)^(2))` `=0*707 wm^(-2)` The power of incident on each potassium atom is `P_(i)`= intensity `xx` area per atom `=0*707xx(pixx(10^(-10))^(2))/(4)=5*56xx10^(21-)W` `:. "Time" =("Energy")/("Power")=(W)/(P_(i))=(2xx1*6xx10^(-19))/(5*56xx10^(-21))` `=57*6SEC` |
|
| 48. |
A coil of 200 turns carries a current of o.4 A. If the magnetic flux of 4 m Wb is linked with the coil, find the inductance of the coil. |
|
Answer» Solution :Number of turns, N= 200, Current, I = 0।4 A Magnetic FLUX LINKED with coil, `phi = 4 mWb = 4 X 10^(-3) `Wb Induction of the coil,`L=phi/I=200xx4xx10^(-3)/(0।4)=800xx10^(-3)/(0।4)=2H` |
|
| 49. |
Value of stopping potential depends on……. Of incident light. |
| Answer» Answer :C | |
| 50. |
Foliose lichens are attached to the substratum with the help of |
|
Answer» BRANCHED MULTICELLULAR rhizoids |
|