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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. |
Why a sound heard more intense in carbon dioxide in comparison to air? |
| Answer» SOLUTION :The INTENSITY of sound INCREASES with increase in the density of the MEDIUM. | |
| 2. |
In which option products is not correct. |
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Answer»
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| 3. |
State Ampere's circuital law. |
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Answer» Solution :The LINE INTEGRAL of magnetic field over a CLOSED loop is `mu_(0)` times NET current enclosed by the loop. `oint_(C) vec(B).vec(dl) = mu_(0) I_("enclosed")` |
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| 4. |
If hot air rises, why is it cooler at the top of a mountain than near sea level ? |
| Answer» Solution :We know that the pressure DECREASES with height. When the HOT air rises, it suffers adiabatic expansion. Now according to first law of thermodynamics `DeltaQ=DeltaU+DeltaWor0=DeltaU+DeltaW(thereforeDeltaQ=0)thereforeDeltaU=DeltaW` This CAUSES a decrease in INTERNAL ENERGY. So the temperature falls | |
| 5. |
What is modulation? Explain its necessity. |
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Answer» Solution :Modulation. The process by which some characteristics of a relatively high frequency wave ( called carrier wave ) is varied in accordance with the instantaneous value of a low frequency wave ( called modulating wave ) is called modulation. Reasons for modulation `(i)` For transmission of signal, the height of transmitting antenna should be comparable to wavelength of signal. For audio frequency having range of frequency from `20Hz` to `20,000Hz,` the height of antenna should be from `15km` to `15000km` which is not possible, hence modulation of signal is REQUIRED. `(ii)` The effective power RADIATED by an antenna is inversely PROPORTIONAL to square of LENGTH of antenna height. So for more power to be radiated, the height of the antenna should be small, hence modulation is needed. `(iii)` If large number of signals are to be transmitted in the audio range `(20Hz` to `20KHZ)` these signals get mixed and it becomes almost impossible to distinguish them. To remove this difficulty we communicate at higher frequencies and allotting a band of frequencies to each user. Hence modulation of the signal of low frequency to high frequency is required. |
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| 6. |
If 'b' represents the size of object interacting with light L represents the distance between object and screen X is the wavelength of light then, match the following lists. |
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Answer» |
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| 7. |
Magnetic meridian is a |
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Answer» Point |
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| 8. |
Modern communication systems is based on |
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Answer» analog CIRCUITS |
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| 9. |
The motion of copper plate is damped when it is allowed to oscillate between the two plates of a magnet. What is the cause of this damping? |
| Answer» Solution :The damping is due to eddy currents induced in the copper plate. If slots are CUT in the plate, the eddy CURRENT loops BECOME much shorter-as a result, the net value of the eddy current DECREASES. Then the damping BECOMES less. | |
| 10. |
A circular coil of radius 10 cm, 500 turns and resistance 2Omega is placed with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through 180^@ in 0.5 s. Estimate the magnitudes of the emf and current induced in the coil. Horizontal component of the earth's magnetic field at the place is 3.0 xx 10^(-5) T. |
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Answer» |
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| 11. |
A beam of unpolarised light is incident on a tourmaline crystal C_(1). The intensity of the emergent light is I_(0) and it is incident on another tourmaline crystal C_(2). It is found that no light emerges from C_(2). If now C_(1) is rotated through 45^(@) towards C_(2), the intensity of the light emerging from C_(2) is |
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Answer» ZERO |
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| 12. |
Which of the graphs shown in fig. correctly represents the variation of beta =- (d_(v)// d_(p))//V with P for an ideal gas at constant temperature: |
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Answer»
At constant temperature PV =constant `THEREFORE` Differentiating both side `PDV +vdP =0` `(DV)/(dP) =-(V)/(P)` `rArr -((1)/(V) (dV)/(dP))=(1)/(P)` or `beta =(1)/(P)` THUS, `beta` decrease as P INCREASES Hence, the correct choice is (a). |
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| 13. |
The time taken by a photoelectron to come out after photon strikes is approximately |
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Answer» `10^(-14)` s |
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| 14. |
A : When tiny circular obstacle is placed in the path of light from some distance, a bright spot is seen at the centre of the shadow of the obstacle. R : Destructive interference occurs at the centre of the shadow |
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Answer» Both A and R are TRUE and R is the CORRECT explanation of A |
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| 15. |
Statement I: Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion. Statement II : For heavy nuclei, binding energy per nucleon increases with increasing Z while for light nuclei it decreases with increasing Z. |
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Answer» Statement I is TRUE, statement II is false. |
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| 16. |
A small block of mass m and a concave mirror of radius R fitted with a stand, lie on a smooth horizontal table with a separation d between them. The mirror together with its stand has a mass m. The block is pushed at t = 0 towards the the mirror so that it starts moving towards the mirror at a constant speed V and collides with it. The collision is perfectly elastic. Find the velocity of the image (a) at a time t < dV (b) at a time t > d/V. |
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Answer» `(-R^2 V)/([2(d-VT)-R]^2) . V (1+(R^2)/([2(Vt - d) -R]^2) )` |
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| 17. |
If instantaneous current is given by i=4 cos (omegat+phi)ampere, than the rms value of current is ...... ampere. |
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Answer» 4 `therefore i_(RMS)=i_m/sqrt2 = 4/sqrt2 =2sqrt2A` |
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| 18. |
A magnetic needle is free to rotate in a vertical plane which makes an angle of 60^(@) with the magnetic meridian. If the needle stays in a direction making an angle of tan^(-1)(2)/(sqrt(3)) with the horizontal, what would be the dip at that place? |
| Answer» SOLUTION :`30^(@)` | |
| 19. |
Q - factor for resonance curve of L-C-R series circuit is defined as .... |
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Answer» `omega_0 Deltaomega` |
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| 20. |
A direct current flows in a solenoid of length L and radius R,(Kgt gtR) , producing a magnetic field of magnetic B_(0) inside the solenoid . |
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Answer»
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| 21. |
The radius of a cupper wie is 4 mm . What force is required to stretch the wire by 20% of its length, assuming that the elastic limit is not exceeded ? (Y = 1.2 xx 10^11 N/m^2) |
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Answer» a)`13 xx 10^5 N` |
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| 22. |
Refractive index of glass with respect to air is 1.8. So refractive index of air with respect to glass = .......... |
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Answer» 1 |
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| 23. |
A small air bubble is situated at a distance of 3 cm from the center of a glass sphere of radius 9 cm. When viewed from the nearest side, the bubble appears to be at a distance of 5 cm from the surface. Its apparent distance when viewed from the farthest side is nxx5 cm where n is? |
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Answer» |
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| 24. |
What is the equivalent conductivity of rods A & B – |
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Answer» `50 w//m -^@C` |
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| 25. |
In an a.c. circuit with voltage V and current I, the power dissipated is |
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Answer» VI |
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| 26. |
The quantities A and B are related by the relation A//B=m, where m is the linear density and A is force, the dimensions of B will be : |
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Answer» same as that of pressure `:.B=[M^(0)L^(2)T^(-2)]` `Latent heat=("Heatenergy")/("mass")` `=(ML^(2)T^(-2))/(M)=[M^(0)L^(2)T^(-2)]` Thus `B` has same dimensions as that of latent heat. So cannot choice is `(d)`. |
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| 27. |
Assertion: An applied electric field will polarize the polar electric material Reason:In polar dielectrices, each molecule has a permanent dipole moment but these are randomly oriented in the absence of an externally applied electric field |
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Answer» Both Assertion and Reason are TRUE and Reason is the CORRECT EXPLANATION of Assertion |
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| 28. |
In nuclear fusion, |
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Answer» a heavy nucleus BREAKS into two intermediate nucleiand few high PARTICLES |
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| 29. |
On a cold winter day (5^(@)C), the foundation block or a statue is filled with 2.0m^(3) of concrete. By how much will the concrete's volume increase on a very warm summer day (35^(@)) if its coefficient of volume expansion is 4.0xx10^(-5)//""^(@)C ? |
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Answer» `160cm^(3)` `DeltaV=betaV_(i)DeltaT=(4.0xx10^(_5))/(""^(@)C)(2.0m^(3))(35^(@)C-5^(@)C)=0.0024m^(3)xx((10CM)/(1m))^(3)=2,400cm^(3)`. |
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| 30. |
A resistance R draws power P when connected to an AC source. If an inductance is now placed in series with the resistance, such that the impedance of the circuit becomes Z, the power drawn will be |
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Answer» <P>`P((R )/(Z))^(2)` |
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| 31. |
The logic gate is an electric circuit which |
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Answer» makes LOGIC decisions |
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| 32. |
The magnectic potential at a point at a distance 10 cm on a line inclibned at 60^@ with axis of short magnetic dipole moment 0.8 Am^2 is : |
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Answer» 4 `XX 10^-6 ` J/Am |
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| 33. |
Suppose we consider a large number of containers each containing initially 10000 atoms of a radioactive material with a half life of 1 yr. After 1 yr. |
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Answer» all the containers will have 5000 atoms of the MATERIAL |
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| 34. |
What is the masss of pion plus (pi^(+))? |
| Answer» SOLUTION :MASS of a PION plus is 273 times the mass of an electron. | |
| 35. |
Choose the correct option: A rectangular coil of copper wires is rotated in a magnetic field. The direction of the induced current changes once in each: |
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Answer» ONE REVOLUTION |
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| 36. |
A circular coil of radius 10 cm, 500 turns and resistance 2Omegais placed with its plane perpendicular to the horizontal component of the earth's magnetic field. It is rotated about its vertical diameter through 180^@ in 0.25 s. Estimate the magnitudes of the emf and current induced in the coil. Horizontal component of the earth's magnetic field at the place is 3.0xx10^(-5) T. |
Answer» SOLUTION :INITIALLY:  `Phi_1=NBA_1costheta_1` `=NBAcos 0^@ "" (because A_1=A)` =NBA  `Phi_2=NBA_2costheta_2` `=NBAcos180^@ (because A_2=A)` =-NBA Average induced emf in the coil, `(epsilon)=-(DeltaPhi)/(Deltat)` `therefore (epsilon)=-((Phi_2-Phi_1)/(Deltat))` `=-((-NBA-NBA)/(Deltat))` `=(2NBA)/(Deltat)` `=(2NB(pir^2))/(Deltat)` `=((2)(500)(3XX10^(-5))(3.14)(0.1)^2)/(0.25)` `=3.768xx10^(-3)` V Now, average induced CURRENT , `(I)=(epsilon)/R` `therefore (I)=(3.768xx10^(-3))/2` `thereforelt I gt = 1.884xx10^(-3)` A |
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| 37. |
Three basic elements viz transmitter, channel and receiver in a communication Process comprise ? |
| Answer» SOLUTION :COMMUNICATION SYSTEM | |
| 38. |
A cell can be balanced against 100 cm and 110 cm of a potentiometer wire respectively with and without being short circuited through a resistance of 10 Omega . Its internal resistance is |
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Answer» 0 |
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| 39. |
What is the first overtone in terms of harmonics present in a vibrating air column of a pipe (a) open a both ends (b) closed at one end ? |
| Answer» SOLUTION :For a pipe at both ENDS, the first overtone is the second harmonic. For pipe closed at ONE end, the first overtone is the third harmonic. | |
| 40. |
A certain wire has a resistance R. The resistance of another wire identical with the first except having twice it's diameter is: |
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Answer» 2R |
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| 41. |
Resistance of an ammeter having range 0.5A is 1.8Omega.it is shunted by a 0.2Omegasmall resister. The effective current when its pointer indicates 2A,is: |
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Answer» 20A |
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| 42. |
A motion which repeats itself after a definite interval of time is called. |
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Answer» NON PERIODIC motion |
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| 43. |
An isolated particle of mass m is moving in horizontal plane along the x - axis at a certain height above the ground. It suddenly explodes into two framents of masses (m)/(4) and (3m)/(4). At an instant the smaller frament is at y=+15 cm. The larger fragment, at this instant is at |
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Answer» <P> Solution :As `vec(F)_(y)=0 "" Delta vec(P)_(y)=0 vec(P)_(y1)=vec(P)_(YF)` `vec(P)_(YI)=0 THEREFORE m_(1)vec(V)_(y1)+m_(2)vec(V)_(y2)=0, m_(1)vec(y)_(1)=-m_(2)vec(y)_(2)` `y_(2)=-((cancel(m))/(4)xx15)/((3cancel(m))/(4))y_(2)=-5cm` |
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| 44. |
The torque of a force vec F = -3hati + hatj + 5hatk acting at a point is hat (tau) . If the position vector of the point vector of the point is 7 hati+3 hatj+hatk , then vec (tau) is : |
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Answer» `14 HATI-hatj+3 HATK` |
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| 45. |
A bob of a simple pendulum of mass 40 mg with a positive charge 4 xx 10^(-6)Cis oscilliating with time period "T_(1). An electric field of intensity 3.6 xx 10^(4) N/c is applied vertically upwards now time period is T_(2). The value of (T_(2))/(T_(1)) is (g = 10 m/s^(2)) |
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Answer» 0.16 |
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| 46. |
What is coordination number in 2-D square close packing |
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Answer» 2 |
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| 47. |
We can see object due to : |
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Answer» REFLECTION of light from it |
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| 48. |
(i) A radioactive nucleus ‘A’ undergoes a series of decays as given below : Aoverset(alpha)(to)A_(1)overset(beta)(to)A_(2)overset(alpha)(to)A_(3)overset(gamma)(to)A_(4) The mass number and atomic number of A_(2) are 176 and 71 respectively. Determine the mass and atomic numbers of A_(4) and A. (ii) Write the basic nuclear processes underlying beta^(+) and beta^(-) decays. |
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Answer» Solution :(i) As per DECAY series, mass number and atomic numbers of various nuclei are : `" "_(72)^(180)Aoverset(alpha)(to)" "_(70)^(176)A_(1)overset(BETA^(-))(to)" "_(71)^(176)A_(2)overset(alpha)(to)" "_(69)^(172)A_(3)overset(GAMMA)(to)" "_(69)^(172)A_(4)^(*)` Therefore, mass number of `A_(4)` is 172 and its atomic number is 69. Similarly, mass number of A is 180 and its atomic number is 72. (ii) At the time of `beta^(-)` emission, a neutron inside a nucleus decays as `" "_(0)^(1)n to " "_(+1)^(1)p +" "_(-1)^(0)e + bar(NU)` (antineutrino) At the time of `beta(+)` decay, a proton inside nucleus decays as : `" "_(+1)^(1)pto " "_(0)^(1)n + " "_(+1)^(0)e + nu` (neutrino) Neutrino as well as antineutrino are neutral particles with very little or no mass. |
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| 49. |
Three identical small spheres each of mass 0.1 g are suspended by three silk threads each of length 20 cm from a certain point. How much charge should be given to each sphere so that each thread will make an angle of 30^(@) with the vertical? You may suppose that each sphere has equal charge. |
Answer» SOLUTION :The sphere are identical and each sphere has equal charge. They will repel each otherw with equal force. So in the positon of equilibrium they will form an equilateral triangle ABC in the horizontal plane. Here, LENGTH of the THREAD `OA=OB=OC=20cm,` mass of each sphere, `m=0.1` g angle of inclination of the thread with the vertical, `theta=30^(@)`  SUPPOSE, charge of each sphere is q. Clearly, the vertical line OG passes through the centre of gravity of the triangle ABC. Suppose , length of each side of the triangle `=a` Median `AD=sqrt(a^(2)-(a^(2)/4)=(sqrt(3))/2a` Now `AG=2/3AD=2/3 . (sqrt(3))/2a=a/(sqrt(3))` Again `sin 30^(@)=(AG)/(OA)` or `AG=OA sin 30^(@)=20xx1/2=10cm` `:.a/(sqrt(3))=10` or `a=10sqrt(3)`cm Force of repulsion between any two balls `F=(q^(2))/(a^(2))` On the sphere A two equal forces of repulsion F act due to the spheres B and C. Suppose, the resultant of these two equal forces is R. `:.R^(2)=F^(2)+F^(2)+2F.Fcos 60^(@) [ :' /_BAC=60^(2),` `:.` internal angle between F and `F=60^(@)]` `=F^(2)+F^(2)+F^(2)=3F^(2)` or `R=sqrt(3)F=sqrt(3) . (q^(2))/(a^(2))` Let the TENSION along `AO=T` `:.` At equilibrium `T sin 30^(@)=R` and `T cos 30^(@)=mg` `:.tan 30^(@)=R/(mg)=(sqrt(3)q^(2))/(a^(2).mg)` or `1/(sqrt(3))=(sqrt(3)q^(2))/(a^(2)mg)` or `q^(2)=(a^(2)mg)/3` or `q=asqrt((mg)/3)=10sqrt(3) . sqrt((0.1xx980)/3)=99` esu (approx) |
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