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. |
When solving for vectors. Give two reasons why it is good practice to measure all the angles from the positive x-axis and in the counterclockwise direction |
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Answer» Finding the component of a vector we draw the vector add in the triangle legs. Any vector directed at an angle to the horizontal can throught of as having two parts, that lie on the axis. The process of identifying these two components is known as the resolution of the vector. |
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| 2. |
The acceleration of particle at position x is given by a=3-2x. If the particle is rest at x=0, at what value of x is it at rest again ? |
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Answer» Correct Answer - 3 `a=3-2x` `(vdv)/(dx)=3-2x` `(v^(2))/2=3-2x` `(v^(2))/2 =3-2x` `rArr v^(2)=3x-x^(2)+C` `rArr v^(2)=3x-x^(2)=x(3-x)` v=0 at x=0, x=3 |
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| 3. |
The velocity vector of a body is given by `vec(v)=2hati+(3-6t)hatj m//s`. At the initial moment the body is at origin. Find x-coordinate (in m) at the time when its y-coordinate is maximum. |
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Answer» Correct Answer - 1 `vec(v) =2hati+(3-6t)hatj` x=2t+C`" " `[t=0,x=0] `rArr x=2t......(i)` `y=3t-3t^(2)+C` `y=3t-3t^(2)......(ii)` Put `t=x/2` inequation (2) `y=(3x)/2 3(x/2)^(2)` `y=(3x)/2-(3x^(2))/4` `y=3/2(x-(x^(2))/2)` For `y_(max)` `(dy)/(dx)=0` `(dy)/(dx)=3/2 (1-x)=0 " " rArr x=1m` |
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| 4. |
The diagram shows plane wavefronts for sound wave travelling in air towards right Each of these wavefronts represent successive pressure maxima for the pressure wave. Initally the source S, observer O and medium are all at rest. The source is a large plane diaphragm and observer is a detector Wave fronts being considered in column-II have been emitted after the action in column-II has taken place. `{:(,"Column-I",,"Column-II"),("(A)","Source starts moving towards right",,"(P)distance between any two wavelength will increase"),("(B)","Air starts moving towards right",,"(Q)distance between any two wavefronts will decrease"),("(C)",underset("towards left with same speed")"Observer and source both move",,underset("point A to B in space will increase")"(R)the time needed by sound to move"),("(D)",underset("move towards right with same speed")"Source and medium air both",,underset("point A to B in spce will decrease")"(S)time needed by sound to move from"),(,,,"(T)frequency received by observer is increase"):}` |
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Answer» `(DeltaQ)/(Deltat)= (k_(0)[pi(2R)^(2)-piR^(2)])/(3R)(T_(1)-T_(2))=pirk_(0)(T_(1)-T_(2))` (B)`(DeltaQ)/(Deltat)=(4pikR_(1)R_(2))/((R_(2)-R_(1)))(T_(1)-T_(2))=(4pik_(0)3R.R)/((3R-R))(T_(1)-T_(2))=6pik_(0)R(T_(1)-T_(2))` `(C) (DeltaQ)/(Deltat)=(2pik_(0)l)/(l n(R_(2)/R_(1))) (T_(1)-T_(2))=(4pik_(0)R)/(l n2)(T_(1)-T_(2))` (D)`(DeltaQ)/(Deltat)=-piR^(2)k.(dT)/(dx)=-piR^(2)k_(0)(1+x/(3R))(dT)/(dx)` :.`(DeltaQ)/(Deltat)=underset0overset(3R)int (dx)/((1+x/(3R)))=-piR^(2)k_(0) underset(T_(0))overset(T_(2))int dt` `(DeltaQ)/(Deltat)=(l n(1+x/(3R)))/((R//3)):|_(0)^(3R)=piR^(2)k_(0)(T_(1)-T_(2))` `(DeltaQ)/(Deltat)=(pik_(0)R(T_(1)-T_(2)))/(3 l n2)` |
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| 5. |
In each situation of column-I a process A rightarrow B rightarrow C is given for an ideal gas. Match each situation of column-I with correct result in column II |
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Answer» (B) Due to string constraint bob will move in circular path with respect to lift, w.r.t. ground acceleration is not constant hence path will not be parabolic. (C) Coefficient of friction for both objects are not sufficient for pure rolling hence acceleration will be same (a=g `sin theta-mugcostheta)` Also time will be same but work doen by frictional force will not be same |
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| 6. |
Following logic circuit is equivalent to |
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Answer» The correct option is 2 OR gate
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| 7. |
The work function of Cs is 2.14eV.Find (a) threshold frequency for Cs (b) Wavelength of incident light if the photo current is brought to zero by stopping potential of 0.6 V. |
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Answer» Wave function `omega = 2.14eV` (a) Threshold frequency `omega = hnu_(0)` `nu_(0) = (omega)/(h) = (2.14 xx 1.6 xx 10^(-19))/(6.62 xx 10^(-34))` `=5.174 xx 10^(14)H_(2)` (b) `"As "k_(max) = eV_(0)= 0.6eV` = 2.74eV Wave length of photon `lambda = (hc)/(E) = (6.62 xx 10^(-34) xx 3 xx 10^(-8))/(2.74 xx 1.5 xx 10^(-19))` `=4530 Å` |
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| 8. |
The magnetic field in a plane em wave is given by : By = 12× 10−8 sin (1.20× 107Z + 3.60 × 1015t) Tesla. Calculate the : (i) energy density associated with the em wave (ii) speed of the wave |
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Answer» (i) Here, B0 =12× 10−8 Tesla E0 = cB0 = 3× 10−8 × 12 × 10−8 = 36 V/m Average energy density of electric field, uE = \(\frac{1}{4}\)∈0\(E^2_0\) = \(\frac{1}{4}\) \(\times\) 8.85 × 10−12 × 36 × 36 = 2867.4 × 10−12 = 2.87× 10−9 J/m3 Average energy density of magnetic field, uB = uE ∴ Energy density associated with the em wave. u = uE+ uB = 2uE = 2× 2.87 × 10−9 = 5.74 × 10−9 J/m3 (ii) \(\frac{1}{c}\) = \(\frac{k}{ω}\) or c = \(\frac{ω}{k}\) = \(\frac{3.6\times 10^{15}}{1.2\times 10^7}\) = 3 × 108 m/s |
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| 9. |
Define 1 watt of power. |
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Answer» Power can be defined as the rate at which work is done i.e. energy converted. The formula for power is P = W/t A body is said to have power of 1 watt if it does work at the rate of 1 joule in 1 s. i.e. 1 W= (1 J)/(1 S)
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| 10. |
Two wires A and B of the same material and having same length, have their cross-sectional area in the ratio 1:6. What would be the ratio of heat produced in these wires when same voltage is applied across each? |
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Answer» H = \(\frac{V^2t}{R}\) = V2t x \(\frac{A}{ρl}\); A = area of cross-section ∴ H∝ A Or, \(\frac{H_1}{H_2}\) = \(\frac{A_1}{A_2}\) = \(\frac{1}{6}\) |
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| 11. |
An `alpha` - particle and a proton are accelerated through same potential difference. Find the ratio `(v_(a)//v_(p))` of velocities acquired by two particles. |
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Answer» Gain in KE of particle = Qv `(1)/(2)"m"_(p)"v"_(p)^(2) = K_(p) = q_(p)V_(p) .............(i)V_(p) = Vprop = V` `(1)/(2)"m"_(prop)"v"_(prop)^(2) = K_(prop) = q_(prop)V_(prop) .............(i)` (ii)/(i) `(m_(prop)v_(prop)^(2))/(m_(p)v_(p)^(2)) = (q_(prop))/(q_(p)) = (2)/(1)` `(v_(prop)^(z))/(v_(p)^(2)) = (m_(p) xx 2)/(m_(alpha) xx 1) = (Zm_(p))/(4m_(p) xx 1) = (1)/(2)` `V_(prop) : V_(p) = 1 : sqrt(2)` |
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| 12. |
A ball of mass0.2 kg is thrown vertically upwards by applying a force by hand . If the hands moves 0.2m while applying the force and the ball goes upto 2m height further, find the magnitude of the forceA) 16NB ) 20NC)22ND)4N |
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Answer» Well explained here |
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| 13. |
Distinguish between any two types of propagation of e.m. waves based on (a) frequency range over which they are applicable and (b) communication system in which they are used. |
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Answer» Radio-waves in frequency range of 88-108 MHz are used for commercial FM radio and radio waves in frequency range of 300 MHz – 3000 MHz for UHF band used in cellular phones communication. |
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| 14. |
(i) In a refracting type of telescope, what is the impact on its magnifying power if the objective and eyepiece lens are interchanged? Explain your answer. (ii) Give one advantage of using an objective lens with a large aperture in a telescope. |
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Answer» (i) In a refracting type of telescope, if we interchange objective and eyepiece lence, then it will behave like a microscope. (ii) Advantage (a) Brighter image (b) Higher diameter mirrors can be easily made higher radiation is obtained. (c) No spherical and no. chromatic abberation. |
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| 15. |
The first nuclear reaction ever observed was by ernest Rutherford in 1919. It was triggered by alpha particles incident on an isotope of nitrogen `._(7)^(14)` N. He observed a proton was emitted along with another element x. Let us assume that `._(7)^(14)` N nucleus was initially stationary. For this reaction to occur, alpha-particle must touch the nitrogen nucleus. The distance between their centres at this moment is d. For this problem, we will neglect the effect of outer electrons in `._(7)^(14)` N. Symbols have their usual meanings. X is an isotope ofA. NitrogenB. OxygenC. FluorineD. Carbon |
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Answer» Correct Answer - B `._(2)^(4)He+._(7)^(14)Nrightarrow._(1)^(1)p+._(8)^(17)X` |
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| 16. |
The energy-level scheme for the hypothetical one electron element Bansalium is shown in figure. The potential energy is taken to be zero for an electron at an infinite distance from the nucleus. A sample of atoms Bansalium are in all the 3 excited state shown above. What is the possible wavelength that can be emitted by atom in visible range?A. 414 nmB. 620 nmC. 124 nmD. 920 nm |
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Answer» Correct Answer - A `n=4rightarrow 3 E=3eV` `Rightarrow lambda =1242/3=414` nm |
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| 17. |
The energy-level scheme for the hypothetical one electron element Bansalium is shown in figure. The potential energy is taken to be zero for an electron at an infinite distance from the nucleus. If a Bansalium atom is in ground state, which of the following photons cannot excite the atom to a higher state?A. 10 eVB. 15 eVC. 18 eVD. 12 eV |
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Answer» Correct Answer - D `DeltaE` = exactly the energy difference |
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| 18. |
The energy-level scheme for the hypothetical one electron element Bansalium is shown in figure. The potential energy is taken to be zero for an electron at an infinite distance from the nucleus. If photons emitted from Bansalium transitions `n=4 rightarrow n=2 and from n=2 rightarrow n=1` will eject photoelectrons from an unknown metal but the photon emitted from the transition `n=3 rightarrow n=2` will not, what are the limits (maximum and minimum possible values) of the work function of the metal ?A. `8 eVlt philt10eV`B. `5eVltphi lt 10 eV`C. `5 eV lt phi lt 8 eV`D. `5 eV lt phi lt 12 eV` |
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Answer» Correct Answer - C `E_(4rightarrow2)=8eV` `E_(2rightarrow1)=10eV` `philt8eV` `E_(3rightarrow2)=5eV` `phigt5eV` |
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| 19. |
A convex lens is made of 3 layers of glass of 3 different materials as in the figure. A point object is placed on its axis. The number of images of the object areA. `1`B. `2`C. `3`D. `4` |
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Answer» Correct Answer - C Number of images formed by the lens is equal to number of different media. |
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| 20. |
The sensitivity of a galvanometer of resistance `990Omega` is increased by `10` times. The shunt used isA. `100Omega`B. `120Omega`C. `110Omega`D. `50Omega` |
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Answer» Correct Answer - C `S=(G)/(n-1)=(990)/(10-1)=110Omega` |
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| 21. |
The binding energy of 21Sc45 is 368 MeV. What is its atomic mass? Given that : Mass of 1H1 = 1.008143 u Mass of neutron = 1.008986u. |
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Answer» The 21Sc45 atom has 21 protons and 24 neutrons in its nucleus. Mass of 21 protons = 21 × 1.008143 = 21.171003u Mass of 24 neutrons = 24 × 1.008986 = 24.215664u Total mass = 21.171003 + 24.215664 = 45.386667u Mass defect = \(\frac{368}{931}\) = 0.395274u ∴Atomic mass of 21Sc45 = 45.386667 – 0.395274 = 44.991393u |
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| 22. |
A planet is revolving around the sun in an elliptcal orbit. Its `KE` is different for different points and the total energy is negative. Its linear momentum is not conserved the eccentricity decides the shape of the orbit. Linear momentum of the planet isA. different of different points of the orbitB. conservedC. non conservedD. none of these |
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Answer» Correct Answer - A::C At net force `=0` `F=(vecdP)/(dt)` `P=` constant or conserved. |
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| 23. |
A planet is revolving around the sun in an elliptcal orbit. Its `KE` is different for different points and the total energy is negative. Its linear momentum is not conserved the eccentricity decides the shape of the orbit. Net torque on the planet isA. constant at all pointsB. zero at all pointC. maximum at `A`D. minimum at `D` |
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Answer» Correct Answer - B Gravitational force, net torque is zero at all points. `vectau=(dvecL)/(dt)` Since angular momentum is conserved. So `vectau=0`, then `L=` constant and conserved. |
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| 24. |
A planet is revolving around the sun in an elliptcal orbit. Its `KE` is different for different points and the total energy is negative. Its linear momentum is not conserved the eccentricity decides the shape of the orbit. Velocity of the planet is minimum atA. `C`B. `D`C. `A`D. `B` |
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Answer» Correct Answer - B Angular momentum is conserved i.e., `mvr=`cosntant |
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| 25. |
Several ____ friends attended last night’s concert.A) them B) my C) of my D) of them |
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Answer» Correct option is C) of my |
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| 26. |
From a balloon rising vertically upwards at 5m/s a stone thrown up at 10m/s relative to the balloon . Its velocity with respect to ground after 2s is |
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Answer» We have given: Speed of balloon w.r.t ground=5 m/s (Upwards) Speed of stone w.r.t balloon=10 m/s So, speed of stone w.r.t ground=10+5=15 m/s u=15 m/s (upwards) Using relation v=u+gt =15-10 x 2=-5 Thus the velocity of stone is 5 m/s (downwards) w.r.t to ground after 2 second. |
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| 27. |
In the given velocity `v//s` position graph find acceleration at x = 2m A. `16m//s^(2)`B. `4m//s^(2)`C. `8m//s^(2)`D. `2m//s^(2)` |
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Answer» Correct Answer - A `a=v(dv)/(dx)=8((4)/(2))=16m//s^(2)` |
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| 28. |
If `x = (4t^(2)+2)` any y = 2t, then find `((dx)/(dy))` at t = 2 sec. (x in meter and t is in second)A. 16B. 8C. 4D. 2 |
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Answer» Correct Answer - B `((dx)/(dt))=8(t),((dy)/(dt))=2` `((dx)/(dy))=((8t)/(2))=(4t)` at `t = 2 sec` `((dx)/(dy))=8` |
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| 29. |
If side length of cube changing at `0.1mm//sec`. Then what will be rate of change of volume, when side length is 2 meterA. `8xx10^(-4)m^(3)//sec`B. `4xx10^(-3)m^(3)//sec`C. `12xx10^(-4)m^(3)//sec`D. `12xx10^(-3)m^(3)//sec` |
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Answer» Correct Answer - C `v=l^(3)` `(dv)/(dt)=3l^(2)((dl)/(dt))` `((dv)/(dt))=3(2)^(2)(10^(-4)m^(3)//sec)` `((dv)/(dt))=12xx10^(-4)m//sec` |
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| 30. |
During one dimensional motion on x-axis Velocity time graph for particle is given. Find total displacement during (t = 0 sec) to (t = 6 sec)A. 6 mB. 10 mC. 8 mD. 2 m |
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Answer» Correct Answer - A Net area under curve `=(A_(1)-A_(2))=6m` |
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| 31. |
If `V = sqrt(gammap)/(rho)`, then the dimensions of `gamma` will be (where p is pressure, `rho` is density and V is velocity) :A. `M^(0)L^(0)T^(0)`B. `M^(0)L^(0)T^(1)`C. `M^(1) L^(0) T^(0)`D. `M^(0) L^(1) T^(0)` |
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Answer» Correct Answer - A `gamma = v^(2) (rho)/(p)` `[gamma] = M^(0)L^(0)T^(0)` |
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| 32. |
During one dimensional motion on x-axis Velocity time graph for particle is given. Find total distance during (t = 0 sec) to (t = 6 sec)A. 6 mB. 10 mC. 8 mD. 2 m |
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Answer» Correct Answer - B Area `=A_(1)+A_(2)=10m` |
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| 33. |
During one dimensional motion on x-axis Velocity time graph for particle is given. Average acceleration between (t = 0 sec) to (t = 6 sec)A. `(5)/(3)m//s^(2)`B. `(1)/(2)m//s^(2)`C. `1m//s^(2)`D. zero |
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Answer» Correct Answer - D `(:vec(a):)=(0-0)/(6)=0` |
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| 34. |
A body P starts from rest with an acceleration `a_(1)`. After 2 seconds, another body Q starts from rest with an acceleration `a_(2)` from the same point. If they travel equal distances in the fifth second after starting of the motion of the P, then the ratio `a_(1) : a_(2)` is equal toA. `5 : 9`B. `5 : 7`C. `9 : 5`D. `9 : 7` |
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Answer» Correct Answer - A As given in the questions, `(a_(1))/(2) (2 xx 5 - 1) = (a_(2))/(2) (2 xx 3 - 1)` `rArr (a_(1))/(a_(2)) = (5)/(9)` |
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| 35. |
Hyena-Mammal, Crocodile-______? |
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Answer» Correct answer is Reptile |
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| 36. |
Cat-Kitten, Pig-______? |
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Answer» Correct answer is Piglet |
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| 37. |
A facultative anaerobic is |
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Answer» Facultative anaerobes are bacteria that can grow in both the presence or absence of oxygen. In addition to oxygen concentration, the oxygen reduction potential of the growth medium influences bacterial growth. |
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| 38. |
Al-Idrisi was an cartographer? |
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Answer» Muhammad al-Idrisi was a Muslim cartographer, geographer, traveler and Egyptologist. |
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| 39. |
Which type of receptors can be activated by angiotensinamide: a) Adrenergic receptors b) Cholinergic receptors c) Dopaminergic receptors d) Angiotensin’s receptors |
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Answer» d) Angiotensin’s receptors |
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| 40. |
Indicate the vasoconstrictor of endogenous origin: a) Ephedrine b) Phenylephrine c) Xylomethazoline d) Angiotensinamide |
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Answer» d) Angiotensinamide |
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| 41. |
Choose the group of antihypertensive drugs which diminishes the metabolism of bradykinin: a) Ganglioblockers b) Alfa-adrenoblockers c) Angiotensin-converting enzyme inhibitors d) Diuretics |
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Answer» c) Angiotensin-converting enzyme inhibitors |
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| 42. |
This drug reduces blood pressure by acting on vasomotor centers in the CNS: a) Labetalol b) Clonidine c) Enalapril d) Nifedipine |
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Answer» b) Clonidine |
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| 43. |
Pick out the sympatholythic drug: a) Labetalol b) Prazosin c) Guanethidine d) Clonidine |
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Answer» c) Guanethidine |
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| 44. |
Hydralazine (a vasodilator) can produce: a) Seizures, extrapyramidal disturbances b) Tachycardia, lupus erhythromatosis c) Acute hepatitis d) Aplastic anemia |
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Answer» b) Tachycardia, lupus erhythromatosis |
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| 45. |
An endogenous vasoconstrictor that can stimulate aldosterone release from suprarenal glands: a) Angiotensinogen b) Angiotensin I c) Angiotensin II d) Angiotensin-converting enzyme |
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Answer» c) Angiotensin II |
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| 46. |
Think and answer:In my situation, when anyone comes my home specially stranger and male, then my mother says that you go in a room. Why? Why do we hide? Why we feel shy when a male comes home? |
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Answer» The mother tells us to go in our room because she thinks that any kind of harm to her child can be caused by the stranger. Also, she thinks that her child might not be comfortable with the stranger style. We hide because we feel a little tense, nervous and anxious seeing a stranger. We feel shy, when males come home because we aren't sure how to act, don't know how others will react, or when attention is on them. |
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| 47. |
Tick the drug with nonselective beta-adreno blocking activity: a) Atenolol b) Propranolol c) Metoprolol d) Nebivolol |
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Answer» b) Propranolol |
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| 48. |
Pick out the diuretic agent for hypertension treatment: a) Losartan b) Dichlothiazide c) Captopril d) Prazosin |
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Answer» b) Dichlothiazide |
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| 49. |
A ganglio blocking drug for hypertension treatment is: a) Hydralazine b) Tubocurarine c) Trimethaphan d) Metoprolol |
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Answer» A ganglio blocking drug for hypertension treatment is Trimethaphan. |
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| 50. |
Choose the vasodilator which releases NO: a) Nifedipine b) Hydralazine c) Minoxidil d) Sodium nitroprusside |
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Answer» d) Sodium nitroprusside |
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