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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. |
Upon heating the length of each side of the cube changes by 2%. The volume of the cube would change by : |
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Answer» `1%` `:.(DeltaV)/(V)xx100=3(Deltal)/(l)xx100=3xx2%=6%` Hence the correct CHOICE is `(d)`. |
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| 2. |
Which statement is incorrect regarding for p-n junction. |
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Answer» Donor atoms are depleted of their holes in junction |
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| 3. |
A large flat metal surface has uniform charge density + sigma. An electron of mass m and charge e leaves the surface at point A with speed v, and return to it at point B. The maximum value of AB is |
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Answer» `(VM in_0)/(SIGMA E)` |
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| 4. |
Rain is falling vertically with a speed of 30 m/s. wind starts blowing after some time with a speed of 15 m/s in east to west direction. In which direaction should a boy waiting at a bus hold his umbrella ? |
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Answer» Solution :The direction in which boy should hold his umbrealla is opposite to resultant velocityof rain and wind. Let vertically up bey-axis and WEST to east be x -axis. Now, ` vecv_(R) = - 30hatj , vecv_(w)jj = - 15 hatj` Theresultant velocity of rain and wind is `vecv_(R) + vecv_(w) = -30 hatj - 15 hatj ` From , the figure , ` tan tehta = 15/20 = 1/2` ` Rightarrow THETA = tan^(-1) (1/2)` So, the boy should hold his umbrealla inclined at an angle ` tan^(-1)(1/2)`with vertically ,TOWARDS east. 
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| 5. |
The distace between the first and the sixth minima in the diffraction pattern of a single slit is 0.5 mm. The screen is 0.5 m away from the slit. If wavelength of the light used is 5000 A^(0), then the slit width will be |
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Answer» `5MM` |
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| 6. |
When a particle in U.C.M. performs complete circle on a reference circle, its projection |
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Answer» PERFORMS one to fro motion onhorizontal diameter. |
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| 7. |
Does Lenz's law violate the law of conservation of energy? |
| Answer» SOLUTION :No, Lenz.s LAW is in CONFORMITY with the law of CONSERVATION of ENERGY. | |
| 8. |
A student is given a transister. He is asked to find ot the teminals of p-n-p transistor as emitter, base and collector. He is told that the terminal marked with red dot is emitter. He touches red probe with known terminal as emitter and marks other two lead wires as A and B. He measures resistance between emitter and lead A. Then measured resistance between emitter and leadB and finds that resistance increases. This shows- |
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Answer» A is BASE and B is COLLECTOR |
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| 9. |
A battery consists of a variable number n of identical cells having internal resistance connected in series. The terminals of the battery are short circuited and the current I measured. Which one of the graph below shows the relationship between I and n? |
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Answer»
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| 10. |
In a Young's double slit experiment green light is incident on the two slits. The interference pattern is observed on a screen. Which of the following changes would cause the observed fringes to be more closely spaced ? |
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Answer» Reducing the separation between the slits |
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| 11. |
What does ‘Absolutist’ mean? |
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Answer» A Philosophy |
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| 12. |
A satellite in force-free space sweeps stationary interplanetary dust at a rate of dM//dt = alphav, where M is the mass and v the speed of the satellite, and alpha is a constant. What is the deceleration that satallite experiences? |
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Answer» `ALPHA nu` `a=(upsilonxxaupsilon)/(M)=(aupsilon^(2))/(M)` is the CHOICE. |
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| 13. |
Te modulating wave is given by V_(m)=6sin omegat and the carrier wave is given by V_(c)=12 sin omegat. The percentage of modulation is |
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Answer» 20 |
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| 14. |
Try to generalize the result of the previous problem to include the case when one part of the vessel contains k out of n balls (k le n)in conditions when (a) the probabilities of a ball being in the left-hand and the right-hand parts of the vessel are different, (b) the probabilities of a ball being in either part of the vessel are equal. |
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Answer» Solution :(a) Let the probability of finding the given ball in one PART of the vessel bep, then the probability of finding it in the other part of the vessel is `q = 1 - p`. The probability of finding kc particular balls in the first part of the vessel is `p^k`, the probability of finding the rest in the other half is `q^(n-k)`. Consequently the probability of the event when k particular balls are found in the first part of the vessel and n - k are found in the second part of the vessel is `p^k q^(n - k)`. However, we assume all the balls to be identical, and because of that this result may be realized in `C_n^h` WAYS in other words, the thermodynamic prob: ability W of this state is `C_n^h`. To obtain the MATHEMATICAL probability w, one should multiply the number of ways by the probability of a favourable combination, Therefore `w_k = C_n^h p^h q^(n-h) = C_n^h (1 - p)^(n-h) p^h` Such a distribution of probabilities is called binomial distribution.  (b) If the probability of finding the ball in both parts of the vessel is the same, then p = q = 1/2, and we have `w_k = C_n^h (1/2)^h (1/2)^(n-h) = C_n^k (1/2)^n` We have generalized the result of PROBLEM 18.4 for the case of n balls. |
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| 15. |
Find equivalent resistance of the network in Fig. between points (i) A and B and (ii) and C. |
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Answer» Solution :(i) The `100 OMEGA` and `30Omega` resistors are connected in parallel between POINTS A and B. The EQUIVALENT resistance between A and B is `R_(1)=(10xx30)/(10+30)" OHM "=7.5Omega` (ii) The resistance `R_(1)` is connected in series with resistor of `7.5Omega`, hence the equivalent resistance between points A and B is C is, `R_(2)=(R_(1)+7.5)" ohm "=(7.5+7.5)" ohm "=15Omega.` |
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| 16. |
A simple pendulum consists of a ball of mass m suspended from the ceiling using a string of length L. The ball is displaced from its equilibrium position by a small angle theta. What is the magnitude of the restoring force that moves the toward its equilibrium position and produces simple harmonic motion? |
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Answer» kx |
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| 17. |
Answer the question in one sentence each (Alternatives are to be noted): On what condition a convex lens produces real images of an object on a screen at two positions of the lens? Or, Sunnray, sodium light and headlight of an automobile - which of these light are polarised? |
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Answer» SOLUTION :The distance (D) between the object and the screen should be greater than four times the focal length (f) of the CONVEX lens. That is, `D gt 4F` Or, NONE of the three is POLARISED. |
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| 18. |
A thin biprism (see Fig..) of obtuse angle alpha=178^(@) is places at a distance l=20cm from a slit. How many images are formed and what is the separation between them? Refrative index of the material mu=1.6 . |
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Answer» Solution :Two images are formed by the two thin PRISMS - one above the axis and the other below the axis by the same distance. The refracting ABLE of each thin prism is `(PI-alpha)/(2)=(1)/(2)(pi-alpha)` Then, `delta` (deviation of a ray) =`(mu-1)(1)/(2)(pi-alpha)` `d/2=ldelta` `d=2l(mu-1)(1)/(2)(pi-alpha)` `d=(mu-1)l(pi-alpha)` `d=(1.6-1)xx0.20(pi-178xx(pi)/(180))` `= 0.6xx0.20xxpixx(1)/(90)=0.004m=4mm` |
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| 19. |
The plots of intensity versus wavelength for three black bodies at temperatures T_(1),T_(2) and T_(3) respecttively are shown in fig. Their temperatures are such that : |
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Answer» `T_(1)gtT_(2)gtT_(3)` `lamda_(m)PROP(1)/(T)`. Since `lamda_(m_(1))ltlamda_(m_(3))ltlamda_(m_(2))."":.T_(1)gtT_(3)gtT_(2)`. correct choice is (c ). |
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| 20. |
For an HCI molecule find the rotational quantum numbers of two neighbouring levels whose energies differ by 7.86 meV. Thenuclei of the molecule are separated by distacne of 127.5p m. |
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Answer» SOLUTION :The axis of rotation passes through the centre of mass of the `HCl` molecule.The distance of the TWO atoms from the centre of mass are `d_(H)=(m_(cl))/(M_(Hcl))d, =(m_(H))/(m_(Hcl))d` Thus `I=` moment of interita about the axis `(4)/(2) m_(H)d_(H)^(2)+m_(cl)d_(cl)^(2)=(m_(H)M_(cl))/(m_(H)+m_(cl))d^(2)` The energy difference between two NEIGHBOURING levels WHOSE quantum numbers are `J&J-1` is `( ħ^(2))/(2I).2J=( Jħ^(2))/(I)= 7.86 meV` Hence `J=3` and the levels have quantum numbers `2 & 3`. |
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| 21. |
Circuit shown here uses an air filled parallel plate capacitor. A mica sheet is now introduced between the plates of capacitor. Explain with reason the effect on brightness of the bulb B. |
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Answer» Solution :Brightness of bulb depends on current. `P prop I^2` and `I= (V)/(Z)` where `Z = sqrt(X_( c)^(2) + R^2) and X_(C) = (1)/(omega C) = (1)/(2pi v C)` `X_(C ) prop (1)/(C )`, when MICA SHEET is introduced capacitance C increases `(C = (K in_(0) A)/(d))`, `X_(C)` decreases, current increases and THEREFORE brightness increases. |
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| 22. |
A particle is found to be at rest when seen from a frame S_(1) and moving with a constant velocity when seen from another frame S_(2)(a)Both the frames are inertial(b) Both the frames are non inertial (c ) S_(1) is inertial and S_(2) is non inertial(d)S_(1) is non inertial and S_(2) is inertial |
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Answer» a, B are TRUE |
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| 23. |
Derive a relation between electric field and potential |
Answer» Solution :CONSIDER two equipotential surfaces A and B with the potential difference dV between them as shown in figure. Let dl be the perpendicular DISTANCE between them and `vec(E )` be the electric field normal to these surfaces  The WORK done to MOVE a unit positive charge from B to A against the field `vec(E )` through a DISPLACEMENT `vec(dl)` is `dW= vec(E ).vec(dl)= E dl cos pi= - E dl` This is equal to the potential difference, therefore `dV= dW rArr dV= - E dl` `E= - (dV)/(dl)` |
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| 24. |
In viewing through a magnifying glass , one usually positions one' s eyes very close to the lens . Does angular magnification change if the eye is moved back? |
| Answer» SOLUTION :Yes, there is a very small decrease in the angular MAGNIFICATION . Because , the angle subtended by the OBJECT at the EYE becomes somewhat less thanthe angle subtended at the lens . HOWEVER , if the object is at a large distance , then the decreases is negligible. | |
| 25. |
In Young's experiment , the distance of the screen from the two slits is 1.0 m . When a light of wavelength 6000 Å is allowed to fall on the slits , the width of the fringes obtained on the screen is 2.0 mm . Determine (a) the distance between the two slits and (b) the width of fringe if the wavelength of the incident light is 4800 Å. |
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Answer» |
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| 26. |
In shot putting, many athletes elect to launch the shot at an angle that is smaller than the theoretical one (about 42^(@)) at which the distance of a projected ball at the same speed and height is greatest. One reason has to do with the speed the athlete can give the shot during the acceleration phase of the throw. Assume that a 7.260 kg shot is acclerated along a straight path of length 1.650 m by a constant applied force of magnitude 380.0 N, starting with an initial speed of 2.500 m/s(due to the athlete's preliminary motion). What is the shot's speed at the end of the acceleration phase if the angle between the path and the horizontal is (a) 30.00^(@) and (b) 42.00^(@) ? (c) By what percentage is the launch speed decreased if the athlete increases the angle from 30.00^(@) to 42.00^(@) ? |
| Answer» SOLUTION :(a) 47.44 m/s, (B) 12.54 m/s, ( C ) 1.69% | |
| 27. |
A galvanometer has coil of resistance 50Omega and shows full deflection at 100 μA . The resistance to be added for the galvanometer to work as an ammeter of range 10mA is nearly |
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Answer» `0.5Omega` in SERIES |
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| 28. |
Two carsof unequa massesuse similartyres .If they are moving at thesame intial speed , the minimum stoppng distance |
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Answer» is smallerfor the HEAVIER car |
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| 29. |
The motion of which charge is the direction of conventional current. |
| Answer» SOLUTION :+ ve CHARGE | |
| 30. |
Starting from the expression for the force on a current carrying conductor in a magnetic field, arrive at the expression for Lorentz magnetic force. |
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Answer» SOLUTION :Force `=BL = B xx (Q)/(t) l But (l)/(t)=v` `therefore F =Bqv or VEC F=q(vec v xx vec B)` - te Lorentz magnetic force. |
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| 31. |
The efficiency of Carnot's heat engine is 0.5 when the temperature of the source is T_(1) and that of sink is T_(2). The eficiency of another Carnot's heat engine is also 0.5. The temperatures of source and sink of the second engine are respectively |
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Answer» `2T_(1),2T_(2)` Efficiency remains same when both `T_(1) and T_(2)` are INCREASED by same factor. So, correct choice is (a). |
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| 32. |
Two discs one of density 7.2 g cm^(-3) and other of density 8.9 gcm^(-3) are of same mass and thickness. Their M.I. are in the ratio of : |
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Answer» `(7.2)/(8.9)` As `piR_(1)^(2)drho_(1)=piR_(2)^(2)drho_(2)` implies `(R_(1)^(2))/(R_(2)^(2))=(rho_(2))/(rho_(1))` |
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| 33. |
An engine develops 10 kW power. How much time will it take to lift a mass of 200 kg to a height of 40 m ? |
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Answer» 10 s h=40 m `:.` Work DONE, W=Mgh =`200xx10xx40=80,000` J Now `p=W/timplies t=W/p=(80,000)/(10,000)=8 sec`. |
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| 34. |
Interference differs from diffraction in that: |
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Answer» it cannot be OBSERVER with white light |
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| 35. |
Consider the following statement A and B and identify the correct answer A) Electric vector of electromagnetic wave is the light vector that affects the retina of the eye B) In a polarized light the sum of all the components of the vibrations in one direction is eaqul to the sum of all the components of the vibrations perpendicular to that direction |
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Answer» A is false but B is true |
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| 36. |
The time taken by a photoelectron to come out after the photon strikes is nearly : |
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Answer» `10^(-1)S` |
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| 37. |
A particle is projected with 10ms^(-1) at angle theta above horizontal from a horizontal ground. Find the value of theta so that area under the parabolic path of the projectile will be maximum (g=10ms^(-2)). |
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Answer» `60^(@)` `=tan theta((R^(2))/(2)-(R^(3))/(3R))` `(DA)/(dtheta)=0=2 sin^(2)thetacos^(2)theta-sin^(3)thetasintheta=0` `=tan theta(R^(2))/(6)=(tantheta)/(6)((u^(2)sin theta cos theta)/(G))^(2)` `sin^(2)theta(3 cos^(2)theta-sin^(2)theta)=0` `=(10^(4)xxu^(2))/(36xx10xx10cos theta)sin^(2)thetacos^(2)theta` `3=tan^(2)theta` `=(200)/(3)sin^(3)theta cos theta` `tan theta=sqrt(3)` `theta=60^(@)` |
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| 38. |
भाज्य a और भाजक b के लिए में a=bq+r में r के लिए कौन-सा सत्य है ? |
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Answer» `0 lerleb` |
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| 39. |
In the above question if the sheets were thick and codnucting .value of E in the space betwee the two sheets wouldbe |
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Answer» `2sigma//epsilon_(0)` |
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| 40. |
What is the meaning of 'reconciled'? |
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Answer» It MEANS that TRUE equality doesn't exist |
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| 41. |
X and Y are two equal size containers. X contains a gas A at a temperature 75^@ C and Y contains a gas B at a temperature 40^@ C. Each gas behaves as an ideal gas and specific heat at constant pressure for both gases has the same value.From the given information, we can conclude that |
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Answer» the molecules in CONTAINER X are moving as fast as those in container Y. |
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| 42. |
A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m. If a force F is applied at one end of the rope, the force which the rope exerts on the block is : |
| Answer» Answer :C | |
| 43. |
Match Column I (Phenomenon) with Column II (Principle) and select the correct answer using the codes given below the Column. |
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Answer» |
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| 44. |
(A): A beam of charged particles is employed in the treatment of cancer. (R) : Charged particles on passing through a material medium lose their energy by causing ionization of the atoms along their path. |
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Answer» Both .A. and .R. are true and .R. is the CORRECT explanation of .A. |
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| 45. |
Grandfather portrait looks like |
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Answer» WEARING TURBAN FITTING clothes |
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| 46. |
पराग मातृ कोशिका में गुणसूत्रों की कितनी संख्य होती है ? |
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Answer» अगुणित |
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| 47. |
Let two cells of e.m.f.'s E_1 and E_2 be connected in parallel in a circuit . Let r_1 and r_2 be the internal resistances of the cells . Find the value of the current |
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Answer» Solution :Resultant p.d. due to `E_1 and E_2 = E_1 - E_2` SUM of INTERNAL RESISTANCE = `r_1 + r_2` `therefore ` current `I =(E_1 - E_2)/((r_1 +r_2))` |
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| 48. |
A particle is kept at rest at the origin. A constant force rarr F starts acting on it at t =0 . Find the speed of the particle at time t. |
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Answer» SOLUTION :The equation of motion is . `rarr dp / dt = rarr F` As the particle starts from rest and the force is always in the same direction, the motion will be along this direction, THUS, we can write `dp/ dt = F ` or, `(int_(0)^(P) dp = int_(0)^(t) F dt)` ` p = Ft` or, ` m_0 V / (sqrt 1-v^(2) / c^(2)) = Ft ` or, `m_(0) ^(2) V^(2) = F^(2) t^(2) - F^(2) t^(2) / C^(2)V^(2)` V^(2) (m_(0)^(2) + F^(2) t^(2)/ C^(2)) = F^(2) t^(2)` or, `V = Ftc / (sqrt m_(0) ^(2) C^(2) + F^(2) t^(2)) ` Note from example (47.4)that however large t MAY be , V can never EXCEED c, No matter how long you apply a force , the speed of a particl will be less than the speed c. |
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| 49. |
(a) Two stable isotopes of lithium " "_(3)^(6)Liand " "_(3)^(7)Li have respective abundances of 7.5% and 92.5%. These isotopes have masses 6.01512u and 7.01600u, respectively. Find the atomic mass of lithium. (b) Boron has two stable isotopes, " "_(5)^(10)B and " "_(5)^(11)B. Their respective massis are 10.01294u and 11.00931u, and the atomic mass of boron is 10.811 u. Find the abundances of " "_(5)^(10)B and " "_(5)^(11)B. |
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Answer» Solution :(a) Average atomic mass of lithium is the weighted average of MASSES of two isotopes `" "_(3)^(6)Li` and `" "_(3)^(7)Li`, =`(7.5xx6.01512+92.5xx7.0160)/100`u=6.9409u`APPROX`6.941 u. (B) It is given here that mass of `" "_(5)^(10)B` nuclide is 10.01294 u and that of `" "_(5)^(11)B` nuclide is 11.00931 u. Let abundance of `" "_(5)^(10)B` in BORON be x% and abundance of `" "_(5)^(11)B` be (100-x)%. Then Atomic mass of boron 10.811=`(10.01294xxx+11.00931xx(100-x))/100 implies x=19.9%` Hence, abundance of `" "_(5)^(10)B` nuclide in boron is 19.9% and abundance of `" "_(5)^(11)B` isotopes is (100-19.9%)=80.1%. |
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| 50. |
Water of volume 2 liter in a container is heated with a coil of 1 kWat 27^(@)C.The lid of the container is open and energydissipates at therate of160 J s^(-1) . In howmuch time,temperaturewill rise from27^(@) C "to"77^(@) C? |
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Answer» 8 MIN 20 s = Rate at whichenergysupplied - Rateat which energyis lost ` = 1000 j s^(-1)- 160 J s^(-1)= 840 J s^(-1) ` Heat REQUIREDTO raisethe temperatureof water from ` 27^(@) C"to "77^(@) C " is ms " Delta T ` Hence, the required time ` t = ( ms Delta T )/( " Rate by which energy is gainedby water ")` ` ( 2kg xxs 4.2 xx 10^(3) J kg^(-1@)C^(-1) xx (77^(@)C - 27^(@)C))/( 840 J s^(-1)` ` 500 s = 8 min 20 s ` |
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