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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. |
How are infrared waves produced ? Why are these referred to as .heat waves. ? Write their one important use. |
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Answer» Solution :Infrared waves are produced by hot bodies and MOLECULES on account of vibrations of atoms and molecules. Infrared waves are referred to as heat waves because water molecules present in most materials readily absorb infrared waves. After absorption, their thermal motion increases, that is, they heat up and heat their SURROUNDINGS. Infrared radiation play an important ROLE in maintaining the earth.s warmth or average temperature through the greenhouse effect. Infrared detectors are used in earth satellites for remote sensing PURPOSES. |
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| 2. |
The number of decibels present in 10 bel is |
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Answer» 100 dB |
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| 3. |
A Carnot engine whose sink is at 300 K has an efficiency of 40 %. By how much should the temperature of source be increased so as to increase its efficiency by 50% of original efficiency? |
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Answer» `380 K` or, `T_(2)/T_(1) = 1 - eta = 1 - 40/100 = 3/5` ` :. T_(1) = 5/3 xx T_(2) = 5/3 xx 300 = 500 K`. Increase in efficiency = `50% " of " 40% = 20%` New efficiency, `eta. = 40% + 20% = 60%` ` :. T_(2)/T_(1) = 1 - 60/100 = 2/5` ` rArr T_(1). = 5/2 xx T_(2) = 5/2 xx 300 = 750 K` ` :. ` Increse in temperature of SOURCE ` = T_(1). - T_(1)` ` = 750 - 500 = 250 K` |
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| 4. |
A convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens. Then its focal length will |
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Answer» become infinite. |
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| 5. |
The equation of a stationary wave is given by y=0.05cos(pix/3)sin40pit where y nad x are given in meter and time in sec, then the amplitude of each wave is, |
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Answer» 5cm |
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| 6. |
What is the value of resistance of the following resistor? |
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Answer» `100K Omega` |
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| 7. |
Energy generation in stars is mainly due to : |
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Answer» NUCLEAR fusion |
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| 8. |
An electric capacitor consists of this round parallel plates. eachof radius R, separated by a distancel (l lt lt R) and uniformlycharged with surfacedensitiessigma and -sigma . Find the potentialof the electircfield and the magnitude of tis strengthvector at the axesof thecapacitor as functionsof a distanace x fromthe platesif x gt gt l. Investigateand obtainedexpressions at x gt gt R. |
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Answer» Solution :For `X gt 0` we can use the result as given above and write `varphi = (sigma l)/(2 epsilon_(0)) (1 - (|x|)/((R^(2) + x^(2))^(1//2)))` for the solution that solution that vanishes at `alpha` . There is a discontiunity in potentail for `|x| = 0`. The solution for NEGATIVE `x` is obtained by `sigma rarr -sigma`. THUS `varphi = (sigma//x)/(2 epsilon_(0) (R + x^(2))^(1//2))` + constant Hence ignorning the jump `E = - (DEL varphi)/(del x) = (sigma l R^(2))/(2epsilon_(0) (R^(2) + x^(2))^(3//2))` for large `|x| varphi ~~ +- (p)/(2pi epsilon_(0) x^(2))` and`E ~~ (p)/(2pi epsilon_(0)|x|^(3))` (where `p = pi R^(2) sigma l`) |
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| 9. |
A narrow beam of light after reflection by a plane mirror falls on a scale at a distance 100cm from the mirror. When the mirror is rotated a little, the light spot moves through 2 cm. The angle through which the mirror is rotated is |
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Answer» 0.02 RAD |
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| 10. |
Explain AC circuit for circuit with only resistor. |
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Answer» Solution :AC current through a resistor is in phase with voltage. But this is not so in the case of an inductor, a capicator or a combination of these circuit elements. In ORDER to show phase relationship between voltage and current by using notion of phasors. A phasor is a vector which rotates about the origin with angular speed `omega ` as shown in figure .  The vertical components of phasors V an dI are `V_(m) SIN omega t ` and `I_(m) sin omega t ` where `V_(m)` and `I_(m)` are oscillating quantities or INDICATES peak values. According to figure a resistor is CONNECTED with ac SOURCE.  Figure (a) showsthe voltage and current phasors and their relationship at time `t_(1)` corresponding to the circuit. The projection of V and I phasors on vertical axis are `V_(m) sin omega t_(1)` and `I_(m) sin omega t ` respectively represent the value of voltage and current at that instant. The rotating vectors with frequency `omega`is shown in figure (b) are generated. Phasors V and I for the case of resistor are in the same direction. This means that the phase angle between voltage and the current is zero. |
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| 11. |
A ferromagnetic material is heated above its curie temperature. Which one is a correct statement ? |
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Answer» FERROMAGNETIC domains are perfectly arranged. |
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| 12. |
A drops of mercury, 2mm in a diameter is broken up into 1000 droplets, all of the same size. Calculate the work done in the prosses, if surface tension of mercury is 460 xx 10^-3 N/m. |
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Answer» `5.2 XX 10^-5 J` |
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| 14. |
Two particles P and Q describe coplanar concentric circles of radii a and a' with angular velocities omega andomega', (in the same sense ) respectively. If the angular velocity of P with respect to Q is x, then fill (49x)/(12) in OMR sheet. Take theta = (2pi)/(3), a = 2m, a' = 1m, omega = 2 "rad/s&" omega' = 1 "rad/s". |
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Answer» <P> `V_(P) = a_(w)` and `V_(Q) = a' omega'` respectively Now the angluar velocity of P relative to Q is `omega_(r ) = (a omega cos d + a' omega' cos beta)/(d)` `= (a omega [(a^(2) + d^(2) - a^('2))/(2AD)] + a' omega' [(a^('2) + d^(2) - a^(2))/(2a'd)])/(d)`  `= ((omega + omega'))/(2) + ((omega + omega') (a^(2) - a^('2)))/(2 (s^(2) + a^('2) - 2 a a' cos theta))` If `omega_(r ) = 0` then`cos theta = (a^(2) omega + a^('2) omega')/(2a a' (omega + omega))` |
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| 15. |
Statement-1: The magnetic field at the centre of a circular loop carrying current zero Statement -2: Due to symmetry the field produced by the the field produced by the diametrically opposite points add to each other. |
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Answer» STATEMENT-1 is TURE, Statement-2 is Ture, Statement-2 is CORRECT EXPLANATION for Statement-1 |
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| 16. |
Do all materials have a net magnetic moment in the normal state ? |
| Answer» Solution :N. For an unmagnetised SUBSTANCE, under NORMAL state , the net magnetic MOMENT is ZERO. | |
| 17. |
Two masses m_(1) and m_(2) concute a high spring of natural length l_(0) is compressed completely and tied by a string. This system while conving with a velocity v_(0) along +ve x-axis pass thorugh the origin at t = 0, at this position the string sanps, Position of mass m_(1) at time t is given by the equation x_(1)t = v_(0)(A//1-cosomegat). Calculate (i) position of the particle m_(2) as a funcation of time, (ii) l_(0) in terms of A. |
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Answer» Velocity of centre of mass `= v_(0)` `:.` Location/x -coordinate of centre of mass of TIME `t = v_(0)t` `:. BARV = (m_(1)x_(1) + m_(2)x_(2))/(m_(1) + m_(2))` `rArr v_(0)t = (m_(1)[v_(0)t - A(1-cosomegat)]+m_(2)x_(2))/(m_(1) + m_(2))` `rArr (m_(1)+m_(2))v_(0)t=m_(1)[v_(0)t-A(1-cosomegat)]+m_(2)x_(2))` `rArr m_(1)v_(0)t + m_(2)v_(0)t = m_(1)v_(.0)t - m_(1)A(1-cosomegat)] + m_(2)x_(2)` `rArr m_(2)x_(2) = m_(2)v_(0)t + m_(1)A(1-cosomegat)` `rArr x_(2)=v_(0)t + (m_(1)A)/(m_(2))(1-cosomegat)"......"(i)` To express `l_(0)` in terms of A. `:. x_(1) = v_(0)t - A(1-cosomegat) :. (dx_(1))/(dt^(2)) = -Aomega^(2) sinomegat` `:. (d^(2)x^(2))/(dt^(2)) = - Aomega^(2) cosomegat "........"(ii)` `x_(1)` is displacement of `m_(1)` at time t. `:. (d^(2)x_(1))/(dt^(2)) =` acceleration of `m_(1)` at time t. When the spring attains its natural length `l_(0)`, then acceleration is zero and `(x_(2) - x_(1)) = l_(0))` `:. x_(2) x_(1) = l_(0)` , Put `x_(2)` from (i) `rArr [v_(0)t + (m_(1)A)/(m_(2)) (1-cosomegat)] - [v_(0)t - A(1-cosomegat)] = l_(0)` `rArr l_(0) = ((m_(1))/(m_(2)) + 1)A(1-cosomegat)` When `(d^(2)x_(1))/(dt^(2)) = 0, cosomegat = 0` from (ii). `:. l_(0) = ((m_(1))/(m_(2)) + 1)A`. |
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| 18. |
A rock is 1.5xx10^(9) year old. The rock contains ""^(238)U which disigntegrates to form ""^(206)U. Assming that there is no ""^(206)Pb in the rock initially at t=0) and its is the only stable product formed in the decay Consider (T_(1//2))_(v)=4.5xx10^(9) years and 2^(1//3)=2^(1//3)=1.259. The ratio of the number of nuclei of ""^(218)U and""^(206)Pb in the rock is nearly ______ (round off to the nearcst integer) |
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Answer» 1 |
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| 19. |
Convert the following binary numbers into decimal numbers. |
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Answer» 110011 |
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| 20. |
Domain formation is the necessary feature of... |
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Answer» ferromagnetism |
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| 21. |
Four metallic plates are placed as shown in the figure .Plate 2 is given a cgharge Q whereas all other plates are uncharged . Plates 1 and are joined together . The area of each plate is same . . The potential difference beetween plates 1 and 2 is - |
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Answer» ` 3/2 (QD)/ (varepsilon_0A)` `V_1 V_2 = (3Q)/(4 int_0 ) D/A`. |
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| 22. |
Four metallic plates are placed as shown in the figure .Plate 2 is given a cgharge Q whereas all other plates are uncharged . Plates 1 andare joined together . The area of each plate is same . . The charge appearing on right side of plate 4 is - |
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Answer» ` zero` ` 2 Q' - Q_2 - Q_1 =0` ` 2Q' = Q_1 +q_2` ` 2 Q' = Q` ` Q' = Q/2`. |
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| 23. |
Two identical balls each having a density p are suspended from a common point by two insulating strings of equal length. Both the balls have equal mass and charge. In equilibrium each string makes an angle theta with the vertical. Now both the balls are immersed in a liquid. As a resultant the angle theta does not change. The density of liquid is sigma. Find the dielectric constant of the liquid. |
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Answer» `(RHO)/(rho - SIGMA)` |
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| 24. |
A mass of 1kg is suspended by a thread. It is 1. lifted up with an accleration 4.9m//s^(2). 2. lowered with an acceleration 4.9 m//s^(2). The ratio of the tensions is |
| Answer» ANSWER :A | |
| 25. |
Obtain an expression for the total energy of an electron in the n^(th) orbit of hydrogen atom in terms of absolute constants. |
Answer» Solution : Consider an electron of mass m and charge-e revolving round the nucleus of an atom of atomic number Z in the stationary orbit of radius .r.. Let v be the velocity of the electron. The electron posses potential energy because it is in the electrostatic field of the nucleus it also possess KINETIC energy by virtue of its motion. For stationary orbit, total energy E = KE + PE...(1) From Rutherford.s atom model, For stationary orbit, Centripetal force = Electrostatci force. `(mv^(2))/( r )=(1)/(4pi epsilon_(0))(ZE)/(r^(2))(-e)` (`because` Electrostatic force, ACCORDING to Coulomb.s law) On dividing by 2 on both sides, we get `(mv^(2))/(2)=(Ze^(2))/(8pi epsilon_(0)r)` `KE=(Ze^(2))/(8pi epsilon_(0)r)- - - - (2) "" [because KE=(1)/(2)mv^(2)]` We have PE = Electric potential at a distance are due to `+Ze xx(-e)` `PE=(1)/(4pi epsilon_(0))(Ze)/( r )(-e)""[because V=(1)/(4pi epsilon_(0))(q)/( r )," Hence "q=+Ze]` `PE=-(Ze^(2))/(4pi epsilon_(0)r)_ _ _ _ to(3)` Equation (2) and (3) in (1), we get `E=(Ze^(2))/(4pi epsilon_(0)r)[(1)/(2)-1]` `E=-(Ze^(2))/(8pi epsilon_(0)r)` |
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| 26. |
A loop ABCDEFA of straight edges has a six corner points A(0, 0, 0), B(5, 0, 0), C(5, 5, 0), D(0, 5, 0), E(0, 5, 5), F(0, 0, 5). The magnetic field in this region is vecB=(3hati + 4hatk)T. The quantity of the flux through the loop ABCDEFA (in Wb) is ...... |
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Answer» 350  Magnetic FLUX `PHI = vecA.VECB` `therefore phi = (vecA_(x) + vecA_(z))(vecB_(x)+vecB_(z))` `=(25hati + 25hatk).(3hati + 4HATK)` `therefore phi = 175 Wb` |
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| 27. |
The charring of sugar takes p,ace when treated with concentrated H_2SO_4. Wjat is the type of reaction involved in it? |
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Answer» Dehydration reaction `C_12H_22O_11+[H_2SO_4]rarr12C+[H_2O+H_2SO_4]_3` |
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| 28. |
The U-tube shown has a uniform cross-section. A liquid is filled in the two arms up to heights h_(1) and h_(2), and then the liquid is allowed to move. Neglect viscosity and surface tension. When the levels equalize in the two arms, the liquid will |
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Answer» be at rest |
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| 29. |
Crystalline structures can be studied using …. . |
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Answer» UV RAYS |
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| 30. |
Find the maximum value of current when a coil of inductance 2H is connected to 150V, 50 cycles/sec supply. |
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Answer» SOLUTION :Here `L=2H, E_("rms")=150V, V=50Hz` `X_(L)=Lomega=Lxx2piv=2xx2xx3.14xx50="628 ohm"` RMS value of current through the INDUCTOR, `I_("rms")=(E_("rms"))/(X_(L))=(150)/(628)=0.24A` Maximum value *or PEAK value) of current is given by `I_("rms")=(I_(0))/(sqrt2)` or `I_(0)=sqrt2""I_("rms")=1.414xx0.24=0.339A` |
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| 31. |
Any two protons repel each other, then how is this possible for them to remain together in a nucleus. |
| Answer» Solution :Nuclear FORCE between two protons is 100 TIMES stronger than the electrostatic force. | |
| 32. |
A mixture of light , consisting of wavelength 590 nm an unknown ,illuminates Young.sdouble slit and gives rise to two ovelapping interference patterns on the screen. Thecentral maximum of both lights coincides .Further ,it is observed that the third bright fring of known light coincides with the 4th bright fringe of the unknown light . From this data, the wavelength of the unknown light is |
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Answer» `885.0` NM |
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| 33. |
Radio waves with frequencies higher than television signals are |
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Answer» ultrasonic waves |
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| 34. |
Two ice skaters A and B approach each other at right angles. A has mass 30 kg and velocity 1 mis and B has a mass of 20 kg and velocity 2 m/s. They meet and stick together. The final velocity of couple is : |
Answer» Solution :Let v = COMPOSITE velocity after COLLISION, Q = ANGLE with the direction of 30 kg mass Then from momentum conservation,  `5vcos theta = 30 xx 1` `50V sin theta =20 xx 2` From equations (1) and (2) `sqrt(50^(2).v^(2)) = sqrt(900 + 1600)` 50v= 50, v= 1m/s |
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| 35. |
A sample of radioactive material has mass m, decay constant 1, and molecular weight M. Avogadro constant = N_AThe activity of the sample after time twill be |
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Answer» `((mN_A)/M)E^(-lamdat)` |
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| 36. |
In a single slit diffraction experiment, the width of the slit is made double its original width. Then the central maximum of the diffraction pattern will become a) narrower b) fainter c) broader d) brighter |
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Answer» a, d only |
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| 37. |
An-type semiconductor is |
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Answer» negatively charged |
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| 38. |
A frog can jump higher than nmormal in a magnetic field because the tissues of a frog are |
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Answer» PARAMAGNETIC |
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| 39. |
A dielectric slab fills the space between the square plates of a parallel-plate capacitor. The side of each plate of the capacitor is L. The magnitude of the bound charges on the slab is 75% of the magnitude of the free charge on the plates. The capacitance is 480muF and the maximum charge that can be stored on the capacitor is 240epsilon_(0)L^(2)E_("max"), where E_("max") is the breakdown strength of the medium: |
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Answer» The dielectric constant for the dielectric SLAB is 4 |
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| 41. |
Where do you find the null points in the combined field due to a bar magnet and the earth , when the N-pole of the magnet is kept towards the north of the earth ? |
| Answer» SOLUTION :On the EQUATORIAL LINE of the MAGNET | |
| 42. |
The ammeter (A) and voltmeter (V) in the given circuit are ideal but the galvanometer has a resistance of 2 Omega In steady state |
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Answer» no current flows in (G) |
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| 43. |
Statement A: If the antenna is vertical the vertically polarized em wave is radiated Statement B : The vertically polarized em wave has electrical variations in the vertical plane |
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Answer» A is TRUE but B is FALSE |
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| 44. |
A: In a forced oscillator the energy transferred from driving force to damped oscillator is maximum in resonance state. R: The amplitude of forced oscillator depends on the frequency of external force. |
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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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| 45. |
परिभुणपोष (पेरीस्पर्म ) अवशेष है |
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Answer» बाह्रय अध्यावरण का |
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| 46. |
In a Young's experiment, the width of one of the two slits is double the other, then interference |
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Answer» INCREASES the intensity of bright and dark fringes. `I_("max")=(E_(1)+E_(2))^(2)` means increase and the intensity of light at point due to distractive interference `I_("min")=(E_(2)-E_(1))^(2) "" "If" E_(1) lt E_(2)` `=(2E_(1)-E_(1))^(2)` means intensity of fringes increases. |
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| 47. |
When did Saalumarada Thimmakka win Padma Shri Award? |
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Answer» 2019 |
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| 48. |
A light ray travelling in glass medium is incident on glass - air interface at an angle of incidence theta. The reflected ( R) and transmitted (T) intensities, both as function of theta, are plotted. The correct sketch is : |
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Answer»
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| 49. |
The diode used in the circuit shown in figure has a constant voltage drop at 0.5 V at all currents and a maximum power rating of 100 mW. What should be the value of the resistance R, connected in series with diode for obtaining maximum current ? |
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Answer» `5OMEGA` Now `P=VI:.I=P/V=(100xx10^(-3))/1=1/10A`. But `IR = 0.5 :. R=(0.5)/I=(0.5)/(1//10)=5Omega` |
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| 50. |
What we call the phenomenon of splitting of polychromatic beam of white light into constituent colors ? |
| Answer» SOLUTION :DISPERSION | |
