Explore topic-wise InterviewSolutions in Current Affairs.

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.

Two point charges `(q_(1)` and `q_(2)`) are placed on x-axis, figure shows graph potential (V) on x-axis. With x-co-ordinate:A. `q_(1)gt0`B. `q_(1)=(x_(1))/(x_(2))q_(2)`C. `|q_(1)|lt|q_(2)|`D. `|q_(1)|gt|q_(2)|`

Answer» Correct Answer - A::C
Potential near `q_(1)` is `+infty` so `q_(1)gt0` potential near `q_(2)` is `-infty` so `q_(2)lt0` potential is zero `x_(2)ltx_(1)` so `|q_(1)|gt|q_(2)|`
2.

Concerning the digital radio graphic image: a. Signal digitization expresses the image as continuous grey-scale values b. The function of the analogue-to-digital converter (ADC) is to digitize the input while maintaining resolution of the information c. A binary system is used for signal digitization d. Eight bits is typically sufficient for most images e. Data compression for storage may lead to loss of data

Answer»

a. False. They are expressed as discrete grey-scale values. It is the analogue input that is continuous. 

b. True.

c. True. 

d. False. Eight bits may be sufficient for images with high noise (e.g. radionuclide imaging), but for imaging requiring greater resolution, 12 bits or more are required. 

e. True. Greater compression can lead to loss of data (using ‘irreversible algorithms’).

3.

Concerning the digital radio graphic image: a. It is represented numerically in digital form b. The image is divided into a matrix consisting of multiple pixels c. The greater the sampling frequency, the greater the spatial resolution d. Decreasing the detector sampling frequency may reduce detector sensitivity e. The image may be read out directly from the detector electronically

Answer»

a. True. 

b. True. A pixel has an assigned value, which represents the signal intensity within the corresponding part of the image. The matrix size refers to the total number of pixels. 

c. True. 

d. False. Increasing the detector sampling frequency will reduce sensitivity. This is due to a decrease in the relative proportion of the pixel that is sensitive to image detection (known as the ‘fill factor’). 

e. True. In solid-state designs of detector, the micro-circuitry may be integrated with the X-ray absorption layer. 

4.

A uniform squre plate of mass m is supported in a horizaontal plane by a vertical pin at B and is attached at A to a spring of constant K. If corner A is given a small displacement and released, determine the period of the resulting motion. A. `2 pi sqrt(2Mb)/(3K)`B. `2pi sqrt(Mb)/(6K)`C. `pi sqrt(Mb)/(3K)`D. none of the above

Answer» Correct Answer - A
`-kbx=(2Mb^(2)alpha)/(3)`
`alpha = - (3)/(2)(kx)/(Mb)`
5.

The voltage-current (V-I) graph of a metallic conductor at two different temperature T1 and T2 is shown in figure. At which temperature is the resistance higher?

Answer»

At T2  the resistance higher.

6.

`Br_(2)` and `I_(2)` are added to a solution containing `1M` each of `Br^(c-)` and `I^(c-)` . What reaction will occur ?

Answer» `Br_(2)+2I^(c-)rarr 2Br^(c-)+I_(2)`,`(EMF=+ve)`
7.

`{:("Column-I","Column-II"),((A)NH_4Cl "in water", (p)"Neutral solution which does not undergo hydrolysis"),((B)CH_3COONa "in water" ,(q)"Cationic hydrolysis"),((C )NH_4CN "in water",(r)"Anionic hydrolysis"),((D)"NaCl in water",(s)"Both Cation and Anionic hydrolysis"):}`

Answer» Correct Answer - A`to` q ; B`to` r ; C`to`s ; D`to`p
`(A)toq` `NH_4Cl` only `NH_4^(o+)` will get hydrolysed
`(B)tor` `CH_3COONa` only `CH_3COO^(-)` will get hydrolysed
`( C)tos` `NH_4CN` Both `NH_4^(o+)` and `CN^(-)` will get hydrolysed
`(D)to p` NaCl Neither `Na^(o+)` nor `Cl^(-)` will get hydrolysed
8.

`[H^(+)]` concentration in 0.01 M `H_2O_2` solution `(K_(a_(1))=3xx10^(-12) and K_(a_2)~~0)` is `xxM`.Fill first two digits of `10^(8)x` as answer.

Answer» Correct Answer - 20
`[H^+]=sqrt(K_aC_0+K_w)=sqrt(3xx10^(-12)xx10^(-2)+10^(-14))=2xx10^(-7)M=x " " therefore 10^8" " x=20`
9.

A cell contain two hydrogen electrodes. The negatove electrode is in contact with a solution of `pH = 5.5`. The emf of the cell is `0.118 V` at `25^(@)C`. Calculate the `pH` of solution positive electrode. (assume pressure of `H_(2)` in the both electrondes `= 1` bar )A. `3 . 5`B. `7 . 5`C. `4.5`D. `6.5`

Answer» Correct Answer - A
Anode `(-) rArr underset(1 atm) (H_(2))(g) rarr underset(10^(-55))(2H^(+)) (aq)+2e^(-)`
Cathode `(+) rArr underset(xM)(2H^(+)) (aq) + 2e^(-) rarr underset((1 atm)) (H_(2))(g)`
`0.118V = 0-(0.059)/(2)log_(10). (10^(-55))^(2)/((x)^(2)) rArr x=10^(-35)M`
`rArr pH= 3.5`
10.

What is the ratio of moles of `Mg(OH)_(2)` and `Al(OH)_3` present in 1 lit saturated aqueous solution of `Mg(OH)_2` & `Al(OH)_3` (`K_(SP)` of `Mg(OH)_2=4xx10^(-12)` and (`K_(SP)` of `Al(OH)_3=1xx10^(-33)` .Give answer by multiply by `10^(-16)`.

Answer» Correct Answer - 80
`Mg(OH)_2hArrMg^(+2)+2OH^(-)`
`x " " 2x+3y " " K_(SP)` of `Mg(OH)_2 gt K_(SP)"of" Al(OH)_3`
`Al(OH)_3hArrAl^(3+)+3OH^(-)`
`y " " 3y+2x" " so " " xgtgty`
`2x+3y=2x`
`because 4xx10^(-12)=[Mg^(2+)][OH^(-)]^2`
`=x xx (2x)^2`
`x=10^(-4)`
`1xx10^(-33)=[Al^(3+)][OH^(-)]^3`
`1xx10^(-33)=(y)(2x)^3`
`1xx0^(-33)=yxx(10^(-4)xx2)^3`
`y=10^(-21)/8 " " so " " x/y=8xx10^17 " " 8xx10^17xx10^(-16)=80`
11.

When NaOH solution is gradually added to the solution of a weak acid (HA), the pH of the solution is found to be 5.0 at the addition of 10.0 mL of NaOH and 6.0 at the further addition of 10.0 mL of same NaOH.(Total volume of NaOH=20 mL) calculate `pK_a` for `HA[log 2 =0.3]` [Fill your answer in the form of multiple of `10^(-1)` for example if your answer is 2.1 then fill 21 as your answer ]

Answer» Correct Answer - 51
Let initial conc. Of HA & NaOH be `C_1 & C_2`mol/L and initial volume of HA=`V_1mL`
`therefore 5.0=pK_a+"log"(10C_2)/((C_1V_1-10C_2))`...(1)
`6.0=pK_a+"log"(20C_2)/((C_1V_1-20C_2))`...(2) from these we get
`C_1V_1=22.5 C_2`
`:.5.0=pK_a+"log"(10C_2)/(22.5C_2-10C_2)=pK_a+log(0.8)`
`:. pK_a=5.1`
12.

A graph given below deals with the reversible reaction.a. What happened to the forward and backward reactions as time passes? b. In which minute does the system attain equilibrium? c. What change occurs to the equilibrium, when a catalyst is used?

Answer»

a. As time passes the speed of forward reaction decreases and backward reaction increases.

b. 25th-minute 

c. Attain equilibrium fast ie. before 25th minute.

13.

Consider the following statements on Boyle's law.1. It states that the pressure exerted by a gas is directly proportional to the volume occupied by it.2. It states that the volume of an ideal gas is directly proportional to the absolute temperature at constant pressure.Which among the above statements is/are correct.1. 1 only2. 2 only3. Both 1 and 24. Neither 1 nor 2

Answer» Correct Answer - Option 4 : Neither 1 nor 2

The correct answer is Neither 1 nor 2.

  • Boyle's Law
    • It states that the pressure exerted by a gas is Inversely proportional to the volume occupied by it. (statement 1 is incorrect)
    • The pressure and volume of a gas are inversely proportional to each other as long as the temperature and the quantity of gas are kept constant.
    • PV = k where P is the pressure exerted by the gas, V is the volume occupied by it and k is the constant.
  • Charle's Law
    • It states that the volume of an ideal gas is directly proportional to the absolute temperature at constant pressure. (statement 2 is incorrect)
    • Also, When there is a decrease in temperature it will lead to a decrease in volume.
    • Charle’s law is a special case of the ideal gas law. The law is applicable to the ideal gases that are held at constant pressure but the temperature and volume keep changing.
    • VI /T=VF /TF where VI=Initial volume, VF=Final volume, TI= Initial absolute temperature, TF=Final absolute temperature. Please note that the temperature is in kelvin.
14.

State modern periodic law.

Answer»

Properties of elements are periodic functions of their atomic numbers.

15.

What is electron gain enthalpy?

Answer»

The enthalpy change occurs when an electron is added to isolated gaseous atom to convert into anion.

16.

What is the effect of a Catalyst on the equilibrium of a reversible reaction?

Answer»

A catalyst has no effect on the position of equilibrium but it helps the reaction to attain equilibrium quickly.

17.

State Markownikoff’s rule.

Answer»

When an asymmetric reagent adds up to an unsymmetrical alkene, the negative part of the adding molecule goes to the carbon atom with lesser number of hydrogen atoms while the positive part to the other carbon atom.

18.

Which orbital is specified by l = 2 and n = 3?

Answer»

3d orbital is specified by l = 2 and n = 3.

19.

Name the process in which hydrogen is obtained as a by product?

Answer»

Electrolysis of brine solution.

20.

What is effect of catalyst on the equilibrium of a reversible reaction.

Answer»

A catalyst has no effect on the position of equilibrium but it helps the reaction to attain equilibrium quickly.

21.

Write the IUPAC name of

Answer»

3, 4- dimethyl hexane.

22.

Define functional group.

Answer»

Atom or group of atom which determine the characteristic properties of organic compound.

23.

How many significant figures in 6.022 × 1023?

Answer»

Significant figures in 6.022 × 1023 :

4 or Four.

24.

Which alkali metal is the strongest reducing agent?

Answer»

Lithium or Li.

25.

Chemical equation is dynamic. Give reason.

Answer»

Both forward and backward reactions are occurring at the same rate.

The concentration of the reactants and products remains constant.

26.

What is resonance effect? Name one group each showing + R and – R effect.

Answer»

The resonance effect is defined as the polarity produced in the molecule by the interaction of two π bonds or between a π -bond and lone pair of electrons present in conjugated molecular.

+Reffect: -NH2, – NHCOCH3, – CH3

-Reffect: -NO2, -CN, -CHO, -COOH

27.

Write the three postulates of Bohr’s atomic model.

Answer»

1. An electron in an atom could revolve around the nucleus only in certain stable orbits without the emission of radiant energy. 

2. An electron revolves around the nucleus only in those orbits for which the angular momentum is an integral multiple of (h/2π), where h is the Planck’s constant. 

3. An electron can jump from higher energy orbit to lower energy orbit. In this process a photon is emitted whose energy is equal to the energy difference between the initial and final states.

28.

Write any three postulates of Bohr’s atomic model.

Answer»

Bohr’s Model of an atom, the postulates are 

1. Electrons revolve around the nucleus of an atom in a certain definite path called Orbit or stationary state of shell. 

2. The shells are having different energy levels denoted as K, L, M, N, …… 

3. As long as the electron remains in an orbit, they neither absorb nor emit energy. 

4. The electron can move only in that orbit in which angular momentum is quantized, i.e., the angular momentum of the electron is an integral multiple of h/2π.

29.

Horizontal and vertical components of earth's magnetic field at a place are equal. the angle of dip at that place is

Answer»

The angle of dip is given by δ=tan−1(Bv/Bh

where Bv ​ = vertical component of earth's magnetic field and Bh ​ = horizontal components of earth's magnetic field. 

Here, Bv ​ =Bh ​ 

So, δ=tan−1 (1)=45o

30.

Which of the following is correct?(1) An orbital is represented by only n, l..(2) Hydrogen spectrum helps to prove Bohr's model.(3) Electron revolve in circular orbit proposed by Bohr's.(4) An atomic orbital is the wave function \(\Psi\) for an electron in an atom.

Answer»

Correct option is (4) An atomic orbital is the wave function \(\Psi\) for an electron in an atom.

An orbital is represented by only n, l, m

Bohr model fails for explanation of the details of hydrogen atom spectrum.

31.

Horizontal and vertical components of earth’s magnetic field at a place are equal. The angle of dip at that place is ________.

Answer» `45^(@)` or vertical
32.

Where on the Earth’s surface is the magnetic dip zero?

Answer»

At equator, dip is zero.

33.

State Curie’s law in magnetism.

Answer»

The magnetic susceptibility of a para magnetic substance varies inversely to its absolute temperature(T). 

i.e x α\(\frac{C}{T}\) (x = \(\frac{C}{T}\))

34.

State and explain curie’s Law in magnetism.

Answer»

The magnetic susceptibility (X) of a para magnetic substance is inversely proportional to the absolute temperature (T). 

X = \(C\frac{\mu_0}{T}\)

C – Curie constant.

35.

Give any two application of ultraviolet radiations.

Answer»

1. used in LASIK eye surgery 

2. UV lamps are used to kill germs in water purifiers 

3. Disinfection for virus and bacteria 

4. To produce photo electric current in burglar alarm.

36.

What is polarisation of light? Name any one method of producing plane polarised light.

Answer»

The phenomenon of confining the vibrations of light in a single plane is called polarisation. Reflection/scattering.

37.

Assertion `:-` In projectile motion a particle is projected at some angle from horizontal. At highest point of its path radius of curvature is least. At highest point of path normal acceleration of projectile is equal to acceleration due to gravity.A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertiion.C. If Assertion is True but the Reason is False.D. If both Assertion & Reason are False

Answer» Correct Answer - A
38.

A particle is projected from the ground at an angle of `60^(@)` with horizontal at speed `u = 20 m//s.` The radius of curvature of the path of the particle, when its velocity. makes an angle of `30^(@)` with horizontal is : `(g=10 m//s^(2)`A. 10.6 mB. 12.8 mC. 15.4 mD. 24.2 m

Answer» Correct Answer - A
`bar(V)_(B//P)=(3hat(i)+hat(j))-(7hat(i)-2hat(j))`
`vec(f)_(k)=-hatv_(B//P)=(4)/(5)hat(i)-(3)/(5)hat(j)`
friction will be opposite to relative velocity.
39.

A block of `4kg` mass starts at rest and slides a distance `d` down a friction less incline ( angle `30^@`) where it runs into a spring of negligible mass. The block slides an additional `25cm` before it is brought to rest momentarily by compressing the spring. The force constant of the spring is `400Nm^-1`. The value of `d` is (take `g=10ms^-2`) A. 25 cmB. 37.5 cmC. 62.5 cmD. None of the above

Answer» Correct Answer - C
`h=(1)/(2)g t^(2)=(1)/(2)xx10xx4=20m`. The height of the ball from ground at `t = 2 s " is "45-20=25m`
40.

Figure shown an arrangement of three points charges. The total potential energy of this arrangement zero. The ratio `(q)/(Q)` is :A. `4 : 1`B. `2 : 1`C. `1 : 1`D. `1: 2`

Answer» Correct Answer - A
`U_(sys)=(1)/(4 pi epsilon_(0))[(-qQ)/(r)+((+q)(+q))/(2r)+(Q(-q))/(r)]=0`
`-Q+(q)/(2)-Q=0`
or `2Q=(q)/(2)or (q)/(Q)=(4)/(1)`
41.

use de Moivre's theorem to express tan 5x in term of power of tan x

Answer»

 a water molecule in flowing stream can be located using a vector A = yzi - yj + xz2k find the divergence of the molecule

42.

Write three properties of ferromagnetic materials.

Answer»

1. Ferromagnetic substance is strongly attracted by a magnet.

2. The relative permeability of a ferromagnetic substance is very large i.e. Mr >> 1. 

3. The magnetic susceptibility of a ferromagnetic substance is high +ve value.

43.

Consider the following statements and arrange in the order of true/false as given in the codes. `S_1:Na_2O_2ltMgOltZnOltP_4O_(10)`:Acidic property. `S_2:NaltSigtMgltAl`:First ionisation energy. `S_3:FgtClgtBr`:Electron affinity.A. TTTB. TTFC. TFFD. TFT

Answer» Correct Answer - C
`S_1:Na_2O_2ltMgOltZnOlt P_4O_10`: as non-metallic character increases the acidic character increases `MgO,Na_2O` are basic: ZnO amphoteric and `P_4O_10` acidic.
`S_2:NaltSigtAlltMg:Mg` has higher than Na due to small size and higher nuclear charge .Mg has highest than Al because of `ns^2` configuration (has extra stability and high electron penetration power of s-subshell electrons) and Si has highest than Al because of higher nuclear charge and small size.`IE_1 :Na=496,Al=577,Mg=737 and Si=786` kJ/mole
`S_3:`There is more interelectronic repulsion in 2p-subshell of fluorine than chlorine (3p), So extra electron will be added easily in 3p-subshell of chlorine as compared to 2p-subshell of fluorine.Down the group electron affinity values generally decreases with increasing atomic number due to increase in atomic size.
So`ClgtFgtBr`
44.

Consider the following statements and arrange in order of true/false as given in the codes. `S_1`: Vapour pressure is a colligative property. `S_2`:Freezing point of a solution is always lower than of the pure solvent `S_3`: Acetic acid undergoes association in benzene The molar mass of acetic acid, determined by elevation of boiling point is always higher than its normal molar mass. `S_4`: Osomotic pressure measurements can be used for determination of molar mass of polymers.A. FFFFB. FTTTC. TTTTD. TFTF

Answer» Correct Answer - B
`S_1:(DeltaP)/P` is a colligative property.
`S_2:`Entropy of solution is more than entropy of pure solvent
`S_3`:Due to association molar mass increases
`S_4`:Larger value of osmotic pressure is easily measurable
45.

From the following statements, concerning ideal gas at any given temperature T, select the correct one(s) (1) The coefficient of volume expansion at constant pressure is same for all ideal gases (2) In a gaseous mixture, the average translational kinetic energy of the molecules of each component is same (3) The mean free path of molecules increases with the decrease in pressure (4) The average translational kinetic energy per molecule of oxygen gas is 3KT (K being Boltzmann constant)A. The coefficient of volume expansion at constant pressure is same for all ideal gasB. The average translational kinetic energy per molecule of oxygen is `3kT`.C. In a gaseous mixture, the average translational kinetic energy of the molecules of each component is same.D. The mean free path of molecules increases with decrease in pressure.

Answer» See the assumptions of kinetic theory of gases.
46.

The temperature of an isotropic cubical solid of length `L`, density `d` and coefficient of expansion `alpha` is raised by `10^(@)C`. To a good approximation, at final tempertureA. length is `L(1+10alpha)`B. surface area of any face of the cube is `L^(2)(1+20alpha)`C. density is `d(1+30alpha)`D. density is `d//(1+30alpha)`

Answer» Use `L=L_(0)(I+alpha t).A=A_(0)(1+2alpha t)` and `d=(d_(0))/(1+3 alpha t)`
47.

what should be the velocity of an electron so that its momentum becomes equal to that of a photon of wavelength `5200 Å`A. 700m/sB. 1000 m/sC. 1400 m/sD. 2800 m/s

Answer» Correct Answer - C
`P_(e)=O_(ph)`
`mv=(hc)/(lambda)`
`" " v=(hc)/(lambdaxxm)`
`" " =(6.67xx10^(-34)xx3xx10^(8))/(5200xx10^(-10)xx9.1xx10^(-31))`
`~~ 1400 m//s`
48.

A Photo sensitive material would emit electrons if excited by photons beyond a threshold. To overcome the threshold, one would increases -A. Voltage applied to the light sourceB. Intensity of lightC. Wavelength of lightD. The frequency of light

Answer» Correct Answer - D
By theory
for photon electric effect
`vgtv_(0)`
49.

N molecules, each of mass m of gas `A and 2 N` molecules each of mass `2m` of gas `B` are containted in the same vessel which is maintained at temperature T. The mean square velocity of molecules of B type is denoted by `v^(2)` and the mean square velocity of A type is denoted by `(omega)^(2)`. the `omega^(2)//v^(2)` is:

Answer» Correct Answer - 2
For gas `A` `V_(A)^(2)=(3KT)/(m)`
since `V_(X)^(2)=V_(y)^(2)=V_(z)^(2)` for molecule
hence `V_(A)^(2)=V_(X)^(2)+V_(y)^(2)=3V_(X)^(2)`
`V_(A)^(2)=3w^(2)=(3KT)/(m)`
for gas `B` `V_(B)^(2)=V^(2)=(3KT)/(2m)`
divide `=(V_(A)^(2))/(V_(B)^(2))=(w^(2))/(V^(2))=(2)/(3)`
50.

The minimum (threshold ) KE of the proton to initiate the nulear reaction `" " p+^(7)Lirarr ""^(7)Be+n` Given `m_(p)=1.0073` amu,`m_(1) =7.0144` amu, `m_(Be)=7.0147` amu, `m_(0)=1.0087` amu.A. `2xx10^(-15) J`B. `4xx10^(-14)J`C. `2.5xx10^(-13)J`D. `8xx10^(-6)J`

Answer» Correct Answer - C
Required energy `=931.5xx10^(6)xx1.6xx10^(-19)`
`[m_(Be)+m_(a)-m_(p)-m_(Li)]=2.5xx10^(-13)` Joule