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. |
What is the instrument used to measure the specific gravity? |
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Answer» The instrument used to measure the specific gravity is called Hydrometer |
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| 2. |
What is the difference between circuit breaker and isolator? |
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Answer» Isolator is a disconnecting switch which is not having the making and breaking capacity. Bus coupler - 2000A Feeders - 1600 A |
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| 3. |
What is Rotation tester? |
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Answer» This device is used during the installation of a motor to determine the direction of the motor once it is installed. The shaft is mechanically rotated in the desired direction and the meter indicates if that is the direction in which the motor will rotate. |
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| 4. |
What is Phase sequence indicator? |
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Answer» Phase sequence indicator used to indicate the 3-phase direction- comes in two styles; Lights and meters. In the lighted variety, a sequence of light goes on for the phase sequence being read, while the meter indicates which phase direction it is reading. |
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| 5. |
What is Infrared or Thermal scanner? |
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Answer» Infrared or thermal scanners are used to measure temperature without contact with the equipment. They produce an image of the component showing temperature variations, this is effective in spotting worn or loose connections and components in industrial circuits. |
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| 6. |
What are the parts of the battery? |
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Answer» Parts of the battery are a. Battery container. b. Battery cover. c. Positive plate (Pb o2). d. Negative plate (Pb). e. Cell connector. f. Grid. g. Cell separator (porous material). h. Sediment chamber. i. Positive and negative terminals. j. Vent plugs. k. Dilutes sulphuric acid (electrolyte). |
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| 7. |
How battery capacity is expressed? |
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Answer» Always battery capacity is expressed in Ampere – hour. |
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| 8. |
How you will prepare electrolyte for a lead acid battery? |
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Answer» While preparing electrolyte for lead acid battery sulphuric acid is added to distilled water. |
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| 9. |
Why is internal resistance of a secondary cell low? |
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Answer» As in secondary cell distance between two plates is greater and their sizes are big. There is no polarization in secondary cell. Due to these cause internal resistance of secondary cell is low. Efficiencey of secondary cell is measured in ampere-hour. |
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| 10. |
What are the basic principles of ED? |
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Answer» Basic principles of ED are, a) All the contacts of corresponding relays and contactors are shown in de-energised condition. b) Control circuit gives us idea about ON / OFF selection of motor, fuse rating, forward reverse control, seal in protections etc. c) Power circuits are drawn in thick lines and control circuits are drawn in thin lines. d) When relay or contactor energises normally open contact closes and normally closed contact opens. e) Auxillary contacts acts with main device such as contactors and relay. |
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| 11. |
How you will connect start and stop push button to control the motor from two different places? |
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Answer» Start push button should be connected in parallel and stop push button in series in the circuit. |
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| 12. |
How special current limiting resistance is connected with the seal in contact? |
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Answer» Special current limiting resistance is connected in series with the seal in contact |
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| 13. |
DC relay coil or contactor coils must be connected to which side? |
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Answer» Negative side of the DC supply to avoid galvanic effect on the coil, which will corrode the coil. |
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| 14. |
What is the operating voltage of 3C? |
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Answer» The operating voltage of 3C is 48V DC. |
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| 15. |
What is anti pumping? |
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Answer» When a breaker is closed on fault condition there will be continuous tripping and closing of the breaker because 3C is energized. Anti pumping in circuit avoids frequent tripping and closing of circuit breaker when the breaker is closed in fault condition. |
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| 16. |
In a coaxial cable, a thin straight conductor is fixed along the axis of an outer hollow conductor. The two carry equal currents flowing in opposite directions. Let X denote the region between the two conductors and Y denote the region outside the outer conductor. The magnetic field is zero in (a) Y but not in X (b) X but not in Y (c) both X and Y (d) neither X nor Y |
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Answer» Correct Answer is: (a) Y but not in X |
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| 17. |
When x amount of heat is given to a gas at constant pressure, it performs x/3 amount of work. The average number of degrees of freedom per molecule of the gas is (a) 3 (b) 4 (c) 5 (d) 6 |
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Answer» Correct Answer is: (b) 4 |
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| 18. |
The total energy of a hydrogen atom in its ground state is -13.6 eV. If the potential energy in the first excited state is taken as zero then the total energy in the ground state will be (a) -3.4 eV (b) 3.4 eV (c) -6.8 eV (d) 6.8 eV |
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Answer» Correct Answer is: (c) -6.8 eV The total energy in the first excited state is - 13.6/4 eV = -3.4 eV. This consists of the kinetic energy of 3.4 eV and the potential energy of -6.8 eV. In order to take the PE here as zero, we add 6.8 eV to all energy levels. The total energy in the ground state then becomes (- 13.6 + 6.8) eV = - 6.8 eV. |
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| 19. |
The radius of gyration of a square plate of side length l about a diagonal is(a) l/3(b) l/3√2(c) l/2√3(d) l/6 |
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Answer» Correct Answer is: (c) l/2√3 |
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| 20. |
A spaceship orbits the earth at a constant speed along a circular path. When an astronaut inside the spaceship releases an object, it does not move away from him. Which of the following is the most accurate reason for this? (a) The astronaut and the object move along the same circular path due to the earth’s gravitational pull. (b) An object moving in a circular path round the earth experiences no gravitational pull. (c) The gravitational forces on the object due to the spaceship exactly balance the gravitational pull on it due to the earth. (d) The gravitational pull on the object due to the earth is very weak at a large distance from the earth. |
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Answer» Correct Answer is: (a) |
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| 21. |
Define nutrition? Why nutrition is important for all the organism |
| Answer» Nutrition is a nourishment or energy that is obtained from food consumed or the process of consuming the proper amount of nourishment and energy | |
| 22. |
Carius method of halogens |
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Answer» The Carius halogen method in analytical chemistry is a method for the quantitative determination of halogens in chemical substances. A known mass of an organic compound is heated with fuming nitric acid in the presence of silver nitrate contained in a hard glass tube known as carius tube, in a furnace. |
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| 23. |
Give an example each for the mixture having the following characteristics. Suggest a suitable method to separate the components of these mixtures (a) A volatile and a non-volatile component. (b) Two volatile components with appreciable difference in boiling points. (c) Two immiscible liquids. (d) One of the components changes directly from solid to gaseous state. (e) Two or more coloured constituents soluble in some solvent. |
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Answer» (a) Mixture of acetone and water. It can be separated by distillation. (b) Mixture of petrol and kerosene. Can be separated by distillation. (c) Mixture of oil and water. Can be separated by fractional distillation. (d) Mixture of naphthalene and ammonium chloride. Separating naphthalene by filtration and then separation of ammonium chloride from water by evaporation. (e) Mixture of pigments from a flower petal extract. Can be separated by chromatography. |
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| 24. |
Based on solute solvent interaction, arrange the solubility of benzene in toluene, bromobenzene, potassium hydroxide. |
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Answer» Benzene is a non polar compound and non-polar solide dissolve in non-polar solvent, and polar so solution dissolve in polar solvent. It means as the polarity of solvent increases the solubility of non-polar solute decreases. The order of polarity of toluene, bromo benzene and potassium Hydroxide is Potassium Hydroxide > bromobenzene > tolulene therefore, the order of solubility of benzene will be--- tolune > bromo benzene > potassium Hydroxide. |
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| 25. |
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased. a. what is the initial effect of the change on vapour pressure? b. How do rates of evaporation and condensation change initially? c. What happens when equilibrium is restored finally and what will be the final vapour pressure? |
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Answer» a. Initially, the vapour pressure will decrease. b. The rate of evaporation remains constant at constant temperature in a closed vessel. However, the rate of condensation will be low initially because there are fewer molecules per unit volume in the vapour phase and hence the number of collisions per unit time with the liquid surface decrease. c. When equilibrium is restored rate of evaporation =rate of condensation. the final vapour pressure will be same as it was originally, |
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| 26. |
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.(a) What is the initial effect of the change on vapour pressure? (b) How do rates of evaporation and condensation change initially? (c) What happens when equilibrium is restored finally and what will be the final vapour pressure? |
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Answer» (a) Vapour pressure decreases due to increase in volume. (b) Rate of evaporation remains same and rate of condensation decreases. (c) Finally the same vapour pressure is restored and the rate of evaporation becomes equal to the rate. |
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| 27. |
1. What is pH? What is its significance?2. The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10-3. What is its pH? |
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Answer» 1. pH is a logarithmic scale used to express the hydronium ion concenration in molarity more conveniently. The pH of a solution is defined as negative logarithm to the base 10 of the activity of hydrogen ion. pH = — log aH+ = —log[H+] 2. [H+] = 3.8 × 10-3 pH = -log[H+] = -log [3.8 × 10-3] = -(-2.42) = 2.42 |
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| 28. |
The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10-3 M. What is its pH? |
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Answer» pH = -log[H+] =-log(3.8 x10-3) = 2.42 |
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| 29. |
The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ions in it. |
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Answer» pH= -log[H+] = 3.76 log[H+] = – 3.76 [H+] = antilog of (- 3.76) = 1.738 × 10-4 mol L-1 |
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| 30. |
The ionization constants of HF, HCOOH and HCN at 298 K are 6.8 × 10-4 , 1.8 × 10-4 and 4.8 × 10-9 respectively. Calculate the ionization constants of the corresponding conjugate base. |
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Answer» The relation between ionization constant of an acid and that of its conjugate base is Ka x Kb= Kw Ka x Kb = Kw ∴ Kb = \(\frac{K_w}{K_a}\) The conjugate base of HF is F- , ∴ Kb(F-) = \(\frac{K_w}{K_{a(HF)}}\) = \(\frac{1\times10^{-14}}{6.8\times10^{-4}}\) = 1.5 x 10-11 The conjugate base of HCOOH is HCOO-. ∴ Kb(HCOO-) = \(\frac{K_w}{K_{a(HCOOH)}}\) = \(\frac{1\times10^{-14}}{1.8\times10^{-4}}\) = 5.6 x 10-11 The conjugate base of HCN is CN-. ∴ Kb(CN-) = \(\frac{K_w}{K_{a(HCN)}}\) = \(\frac{1\times10^{-14}}{4.8\times10^{-9}}\) = 2.1 x 10-9 |
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| 31. |
If you set up the seventh harmonic on a string fixed at both ends, how many nodes and antinodes are set up in it -A. 8,7B. 7,7C. 8,9D. 9,8 |
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Answer» Correct Answer - A 8,7 |
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| 32. |
The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it. |
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Answer» H= -log [H+] or log [H+] = -3.76 = -4.24 [H+] = antilog (-3.76) = 1.74 × 10-4 M |
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| 33. |
The solubility product of silver sulphate is 2.3 × 10-5. Calculate the concentration of Ag+ in the saturated solution of salt containing 1.8 × 10-2 moles/dm3 of SO42- ions. |
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Answer» Ksp = 2.3 x 10-5 [SO4-] = 1.8 x 10-2 Ag2SO4 \(\rightleftharpoons\) 2Ag+ + SO4-2 Ksp = [Ag+]2[SO4-2] \(\therefore\) [Ag+]2 = \(\frac{K_{sp}}{[SO_4^{-2}]}\) [Ag+]2 = \(\frac{2.3\times10^{-5}}{1.8\times10^{-2}}\) = 1.28 x 10-3 or [Ag+]2 = 12.8 x 10-4 [Ag+] = \(\sqrt{12.8\times10^{-4}}\) = 3.58 x 10-2 M or [Ag+] = 0.0358 M |
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| 34. |
The initial and final temperatures of water in a container are observed as `16 +-0.6^(@)C` and `56+-0.3^(@)C`. What is the rise in the temperature of water. |
| Answer» Rise in temperature `=[(56+-0.6)]=[(56-16)+-(0.3+0.6)]=40+-0.9^(@)C` | |
| 35. |
The mass of a cube has error of 2% and the length of its edge has error of 1%. What will be the maximum percentage error in calculation of its density. |
| Answer» Density of cube `(p)=(M)/(V)=(M)/L^(3)) therefore((Deltap)/(P))_("max")xx100=((DeltaM)/(M)xx100)+3(DeltaL)/(L)xx100)=2%+3(1%)=5%` | |
| 36. |
Area of a square is `(100+-2)m^(2)` Determine its side. |
| Answer» Let side of square be a then area `=a^(2)=(100+-2)m^(2) because` error in `a^(2)` is `2%` so error in a will be 1% Hence `a=(10+-0.1)m` | |
| 37. |
A sphere has mass of `(20+-0.4)kg` and radius of `(10+-0.1)` m. Find the maximum percentage error in the measurement of density. |
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Answer» Density `(p)=("Mass")/("Volume") therefore p=(M)/(4//3piR^(3))therefore(Deltap)/(p)xx100=(DeltaM)/(M)xx100+3(DeltaR)/(R)xx100` `rArr "maximum error" =(0.4)/(20)xx100+3xx(0.1)/(10xx100)=2%+3%=5%` |
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| 38. |
The maximum percentage errors in the measurement of mass (M), radius (R) and angular velocity `(omega)` of a ring are 2%,1% and 1% respectively, then find the maximum percentage error in the measurement of its- (a) Moment of inertia `(1=(1)/(2)MR^(2))` about its geometric axis. (b) Rotational kinetic energy `(K=(1)/(2)Iomega^(2))` and (c) Angular momentum `(J=I omega)` about geometrical axis. |
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Answer» (a) Moment of inertia `(I)=(1)/(2)MR^(2)` `therefore (DeltaI)/(I)xx100=(DeltaM)/(M)xx100+2(DeltaR)/(R)xx100` `=2%+(2xx1%)=4%` (b) Rotational kinetic energy `(K)=(1)/(2) I omega^(2)=(1)/(2)MR^(2)omega^(2)` `therefore (DeltaK)/(K)xx100=(DeltaM)/(M)xx100+2(DeltaR)/(R)xx100+2(Deltaomega)/(omega)xx100` `=2%+(2xx1%)=6%` (c) Angular momentum `(J)=I omega=(1)/(2)MR^(2)omega` `therefore (DeltaJ)/(J)xx100=(DeltaM)/(M)xx100+(Deltaomega)/(omega)xx100` `=2%+(2xx1%)+1%=5%` |
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| 39. |
Integrate the following function (a) `int_(o)^(2) 2t dt` (b) `int _(pi//6)^(pi//3) sin x dx`(c) `int _(4)^(10)(dx)/(x)`(d) `int _(o)^(pi) cos x dx` (e) `int _(1) ^(2)(2t -4) dt` |
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Answer» (a) `vec0` (b) 1 (c) 0 (d) `(1)/(x)` (e) `-(1)/(2)` (f) `int_(0)^(pi//4)sin 2x dx=[-(cos 2x)/(2)]_(0)^(pi//4)=-(1)/(2)(cos(2pi)/(4)-cos 0) =-(1)/(2)(0-1)=(1)/(2)` (g) `(d^(2))/(dx^(2))(e^(3x))=(d)/(dx)(e^(3x)(d)/(dx)(3x))=(d)/(dx)(3e^(3x))=3e^(3x).3=9e^(3x)` (h) `|2hati-hatj-5hatk|=sqrt(2^(2)+(-1)^(2)+(-5)^(2))=sqrt(30)` (i) `(d^(2))/(dx^(2))(4x^(3)-3x^(2)+2x+1)=(d)/(dx)(12x^(2)-6x+2)=24x-6` (k) `int_(-1)(+1)t^(-3dt)=[(t^(-2))/(-2)]_(-1)^(+1)=[-(1)/(2t^(2))]_(-1)^(+1)=(-(1)/(2(1)^(2)))-(-(1)/(2(-1)^(2)))=-(1)/(2)+(1)/(2)=0` (l) `hati.(hatjxxhatk)+hatj.(hatkxxhati)+hatk.(hatixxhatj).=hati.hati+hatj.hatj+hatk.hatk=1+1+1=3` |
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| 40. |
Find the directional consines of vector `(5hati+2hatj+6hatk)`. Also write the value of sum of squares of directional cosines of this vector. |
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Answer» Let `alpha, beta & lambda` are the angles of vector `(5hati+2hatj+6hatk)` from x,y `&` z-axis respectively. then `cos alpha=(A_(x))/(A)=(6)/(|5hati+2hatj+6hatk|)=(6)/sqrt(65)` `cos gamma=(A_(z))/(A)=(6)/(|55hati+2hatj+6hatk|)=(6)/sqrt(65)` The sum of squares of directional cosines of this vector `cos^(2)alpha+cos ^(2) beta+cos ^(2) gamma=(5^(2)+2^(2)+6^(2))/(65)=(65)/(65)=1` |
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| 41. |
If `vecA=4veci-2hatj+4hatk` and `vecB=-4hati+2hatj+alphahatk` are perpendicular to each other then find value of `(alpha)` ? |
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Answer» `because vecA & vecB` are perpendicular to each other `therefore vecA. vecB=0rArr (4hati-2hatj+4hatk).(-4hati+2hatj+alpha hatk)=0` `rArr-16-4+4alpha=0rArr alpha=5` |
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| 42. |
the midpoints of two small magnetic dipoles of length D in end on positions are separated by |
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Answer» In magnetic dipole Force ∝1/r4 In the given question, Force ∝x−n Hence, n = 4 |
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| 43. |
There are two vectors `vecA=3hati+hatj` and `vecB=hatj+2hatk`. For these two vectors- (a) Find the component of `vecA` along `vecB` in vector form. (b) If `vecA & vecB` are the adjacent sides of a parallalogram then find the magnitude of its area. (c) Find a unit vector which is perpendicular to both `vecA & vecB`. |
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Answer» (a) Component of `vecA` along `vecB=((vecA.vecB)/(B))hatB=((vecA.vecB)/(B))(vecB)/(B)=[((3hati+hatj).(hatj+2hatk))/sqrt(5)]((hatj_2hatk))/sqrt(5)=(1)/(5)(hatj+2hatk)` (b) Area of the parallelogram `=|vecAxxvecB|=|{:(hati,hatj,hatk),(3,1,0),(0,1,2):}|=|2hati-6hatj+3hatk|=sqrt(2^(2)+(-6)^(2)+3^(2))=7 "units"` (c) Unit vector perpendicular to both `vecA & vecB hatn=(vecAxxvecB)/(|vecAxxvecB|)=(2hati-6hatj+3hatk)/(7)=(2)/(7)hati-(6)/(7)hatj+(3)/(7)hatk` |
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| 44. |
If vector `(hata+ 2hatb)` is perpendicular to vector `(5hata-4hatb)`, then find the angle between `hata and hatb`. |
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Answer» Correct Answer - `60^(@)` `because (hata+ 2hatb) `is perpendicular to`(5hata-4hatb)` `therefore (hata+ 2hatb) *( 5hata - 4hatb) =0` `rArr 5hata*hata- 4hata*hatb+ 10 hatb*hata- 8hatb*hatb=0` `rArr 5+ 6hata*hatb-8=0` `rArr 6hata* hatb=3` `rArr hata*hatb= (1)/(2)` `rArr cos theta = (1)/(2)` `rArr theta = 60^(@)` |
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| 45. |
There are two vectors `vecA= 3hati+hatj and vecB=hatj+2hatk`. For these two vectors - (a) If `vecA & vecB` are the adjacent sides of a parallelogram then find the magnitude of its area. (b) Find a unit vector which is perpendicular to both `vecA & vecB`. |
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Answer» Correct Answer - (a) 7 units (b) `(2)/(7)hati- (6)/(7)hatj+ (3)/(7)hatk` (a) Area of the parallelogram `= |vecA xx vecB| = |{:(hati,,hatj,,hatk),(3,,1,,0),(0,,1,,2):}|` `" "= |2hati-6hatj+3hatk|= sqrt(2^(2)+ (-6)^(2)+ 3^(2))` = 7 units (b) Unit vector perpendicular to both `vecA & vecB` `hatn = (vecAxx vecB)/(|vecAxx vecB|)= (2hati-6hatj+3hatk)/(7)` `= (2)/(7)hati- (6)/(7)hatj+ (3)/(7) hatk` |
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| 46. |
The side of a square is increasing at the rate of 0.2 cm/s. The rate of increase of perimeter w.r.t time is :A. 0.2 cm/sB. 0.4 cm/sC. 0.6 cm/sD. 0.8 cm/s |
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Answer» Correct Answer - 4 If side =a, then rate of increase of perpendicular w.r.t time `= 4((da)/(dt)) = 4(0.2) = 0.8` cm/s |
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| 47. |
A block of mass 2 kg is kept at rest on a big box moving with velocity 2i and having acceleration -3i+ 4 j m/s2find the value of pseudo force acting on Block with respect to the box |
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Answer» Mass of the block m= 2kg Acceleration of the system a= -3i+4j So pseudo force on the block = F = -m*a=-2(-3i+4j)=6i-8j N |
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| 48. |
The slope of graph as shown in figure at points 1, 2 and 3 is `m_1, m_2 and m_3` respectively then A. `m_1 gt m_2 gt m_3`B. `m_1 lt m_2 lt m_3`C. `m_1 = m_2 = m_3`D. `m_1 = m_3 gt m_2` |
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Answer» Correct Answer - 2 From point 1 to point 3. `theta ` is acute and increases. `therefore (tan theta)` will be increased as slope (m) = `tantheta` therefore sope will increase from point 1 to point 3. |
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| 49. |
The physical quantities not having same dimensions aretorque and workmomentum and Planck's constantstress and Young's modulusspeed and (µoε0)–1/2. |
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Answer» Correct answer is option (b) : [Momentum] = [MLT–1] [Planck's constant] = [ML2T–1] Momentum and Planck's constant do not have same dimensions. |
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| 50. |
The minimum value of `y=5x^(2)-2x+1` isA. `1/5`B. `2/5`C. `4/5`D. `3/5` |
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Answer» Correct Answer - C For maximum/ minimum value `(dy)/(dx)=0rArr 5(2x)-2(1)+0=0rArr x=1/5` Now at `x=1/5, (d^(2)y)/(dx^(2))=10` which is positive so minima at `x=1/5` Therefore `y_("min")=5(1/5)^(2)-2(1/5)+1=4/5` |
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