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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. |
An LR circuit with a battery is connected at t=0. Which of the following quantities is not zero just after the connection?A. Current in the circuitB. Magnetic field energy in the inductorC. Power delivered by batteryD. emf induced in the inductor |
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Answer» Correct Answer - A::B::C Initialy current `I=0, so U=1/2 Li^(2)=0` `P=ei=0` But initially `(di)/(dt) != 0`, so `e=L(di)/(dt) != 0`. |
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| 2. |
If a + b + c + d = 2, then the maximum value of (1 + a) (1 + b) (1 + c) (1 + d) is _______.1. \(\frac{{91}}{9}\)2. \(\frac{{81}}{16}\)3. \(\frac{{63}}{22}\)4. \(\frac{{54}}{13}\) |
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Answer» Correct Answer - Option 2 : \(\frac{{81}}{16}\) Given: If a + b + c + d = 2. Calculation: If a + b + c + d = 2 For maximum value a = b = c = d ⇒ 4a = 2 ⇒ a = 2/4 ⇒ a = 1/2 Putting the value of a in (1 + a)(1 + b)(1 + c)(1 + d) ⇒ (1 + 1/2)(1 + 1/2)(1 + 1/2)(1 + 1/2) ⇒ (3/2)(3/2)(3/2)(3/2) ⇒ 81/16 ∴ The maximum value of (1 + a) (1 + b) (1 + c) (1 + d) is 81/16. |
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| 3. |
The sum of the number of male and female students in an institute is 100. If the number of male students is x, then the number of female students becomes x% of the total number of students. Find the number of male students.1. 602. 653. 504. 45 |
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Answer» Correct Answer - Option 3 : 50 Given: Total number of students = 100 Calculation: Number of male students = x So, number of female students = 100 – x According to question ⇒ 100 – x = x/100 × 100 ⇒ 100 – x = x ⇒ 2x = 100 ⇒ x = 50 ∴ Number of male students are 50 |
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| 4. |
A spring of spring constant ‘k’ is used to suspend a mass ‘m’ at its free end while the other end of the spring is rigidly fixed.1. If the mass is slightly depressed and released, then name the motion of the mass.2. Write down the expression for the period of oscillation of the mass.3. If this system is taken into outer space then what happens to its period? Why? |
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Answer» 1. Simple Harmonic Motion 2. T = 2π\(\sqrt{m/k}\) 3. Period of oscillation does not change. |
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| 5. |
A simple pendulum has a bob of mass m is suspended from the ceiling of a lift which is lying at the ground floor of a multistoried building.1. Find the period of oscillation of pendulum when the lift is stationary.2. What is the tension of the string of the pendulum when it is ascending with an acceleration ‘a’?3. What is the period of oscillation of the pendulum while the lift is ascending? |
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Answer» 1. T = 2π\(\sqrt{l/g}\) 2. Tension, T= m (g + a) 3. T = 2π\(\sqrt{\frac{l}{g+a}}\) |
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| 6. |
In the reaction 4HNO3 + P4O10 → 4HPO3 + X , the product X is(a) N2O5 (b) N2O3 (c) NO2 (d) H2O |
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Answer» The product X is (a) N2O5 |
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| 7. |
Ammonia on catalytic oxidation gives an oxide from which nitric acid is obtained. The oxide is :(a) N2O3 (b) NO (c) NO2 (d) N2O5 |
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Answer» (c) [Fe(H2O)5NO]2+ ion is formed |
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| 8. |
N in N2O3 oxidation number |
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Answer» Nitrogen is in the +3 oxidation state in N2O3. |
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| 9. |
The activation energy of a reaction is 155326 J/ mol. The rate constant of the reaction at and 300K as a function of rate constant at 400K, obtained by the Collision theory is ____(a) k1 = 1.5 × 10^-7k2(b) k1 = 1.2 × 10^-6 k2(c) k1 = 1.5 × 10^-6 k2(d) k1 = 1.2 × 10^-7 k2 |
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Answer» Right choice is (a) k1 = 1.5 × 10^-7k2 The explanation: By collision theory, ln((frac{k_2}{k_1}) = -frac{E}{R} (frac{1}{T_2} – frac{1}{T_1}) + 0.5ln(frac{T_2}{T_1}) ) ln((frac{k_2}{k_1}) = -frac{155326}{8.314}(frac{1}{400} – frac{1}{300}) + 0.5ln(frac{400}{300}) ) Hence, k1 = 1.5 × 10^-7k2. |
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| 10. |
From the following date , mark the opation(s) where `Delta H` is correctly written for the given reaction . Given: `H+(aq) +OH-(aq) to H_(2)O(l),` `DeltaH=-57.3 kJ` `DeltaH_(solution) HA(g)=-70.7 kJ"mole"` `DeltaH_(solution) BOH(g)=- 20 kJ"mole"` `DeltaH_("ionzatoin") ` of `HA=15 kJ//"mole"` and BOH is a strong base. A. `{:("Reaction",DeltaH_r(kJ//"mol")),((A)HA(aq)+BOH(aq)toBA(aq)+H_2O,-42.3):}`B. `{:("Reaction",DeltaH_r(kJ//"mol")),((B)HA(g)+BOH(g)toBA(aq)+H_2O,-93):}`C. `{:("Reaction",DeltaH_r(kJ//"mol")),((C )HA(g)toH^+(aq)+A^(-)(aq),-55.7):}`D. `{:("Reaction",DeltaH_r(kJ//"mol")),((D )B^+(aq)+OH^(-) ("aq"),-20):}` |
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Answer» Correct Answer - ABC (D)BOH(g)+aqueous `to B^+ (aq)+OH^(-)(aq) " " DeltaH=-20` kJ/mole |
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| 11. |
Amongs V2O3, V2O4, V2O5, determine value of spin only magnetic moment for most basic oxide in BM (Bohr's megnatone): |
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Answer» Correct answer is (3) V2O3(V+3) is most basic oxide amongs the following Total no. of unpaired electrons in v+3 is n = 2 23V+3 = [Ar]3d2 → \(\sqrt{n(n+2)}\) or √8 = - 2.87 BM |
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| 12. |
Select the incorrect option regarding test of NO2- radical:(1) A nitrosyl complex is formed. (2) A brown ring is obtained in brown ring test at the junction of two liquids.(3) Formula of brown ring is [Fe(H2O)5(NO)]SO4.(4) When concentrated H2SO4 is added to nitrate salt, very light brown fumes appears. |
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Answer» Correct option is (4) When concentrated H2SO4 is added to nitrate salt, very light brown fumes appears. \(NO_3-+H_2SO_4(conc.)\rightarrow\)NO2↑(reddish brown) + O2 + SO42- + H2O |
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| 13. |
Which oxoacid of phosphorous has the highest number of oxygen atoms present in its chemical formula?(A) Pyrophosphorous acid(B) Hypophosphoric acid(C) Phosphoric acid(D) Pyrophosphoric acid |
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Answer» Correct option is (D) Pyrophosphoric acid Pyrophosphorous acid → H4P2O5. Hypophosphoric acid → H4P2O6. Phosphoric acid → H3PO4. Pyrophosphoric acid → H4P2O7. |
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| 14. |
if the ratio of Cp/Cv for a gas is 1.3 and it's atomic mass is M, the molar mass of the gas will be |
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Answer» We know that, Cpm - Cvm = R where, Cpm = molar heat capacity at constant pressure Cvm = molar heat capacity at constant volume As we know, Cv.m = M.Cv where, m = molar mass of gas ∴ MCp - MCv = R Cp - Cv = R/M Cp/Cv - 1 = \(\frac R{C_v.M}\) ⇒ 1.3 - 1 = \(\frac R{C_v.M}\) ⇒ 0.3 = \(\frac R{C_v.M}\) ⇒ M = \(\frac R{C_v\times0.3}\) |
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| 15. |
When I peel onions. I can’t stop my eyes ________. A) spilling B) watering C) leaking D) dripping E) dropping |
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Answer» Correct option is B) watering |
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| 16. |
An electric fan is switched on in a closed room will the air of the room be cooled? |
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Answer» No. Infact speed of air molecules will increase and this results in increase in temperature |
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| 17. |
The temperature drop through a two layer furnace wall is `600^(@)C`. Each layer is of equal area of cross section. Which of the following actions will result in lowering the temperature `theta` of the interface ? A. By increasing the thermal conductivity of inner layer.B. By increasig the thermal conductivity of outer layerC. By increasing thickness of water layerD. By decreasing thickness of inner layer |
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Answer» Correct Answer - A::B::D In steady state, `(K_(1) A (1000 - theta))/(L_(1))` `= (K_(2) A (theta - 400))/(L_(2))` `1000 (K_(1))/(L_(1)) - (K_(1))/(L_(1)) thetya = (K_(2) theta)/(L_(2)) - 400 (K_(2))/(L_(2))` `theta = (1000(K_(1))/(L_(1)) + 400 (K_(2))/(L_(2)))/((K_(1))/(L_(1)) + (K_(2))/(L_(2)))` |
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| 18. |
A proton in a large nucleus: A. attracts all other protons B. repels all other protons C. repels all neutrons D. attracts some protons and repels others E. attracts some neutrons and repels others |
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Answer» D. attracts some protons and repels others |
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| 19. |
Two protons are about 10−10 m apart. Their relative motion is chiefly determined by: A. gravitational forces B. electrical forces C. nuclear forces D. magnetic forcesE. torque due to electric dipole moments |
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Answer» B. electrical forces |
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| 20. |
Figure shows two capacitor connected in series and joined to a battery. The graph shows the variation in potential as one moves from left to right on the branch containing the capacitors. |
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Answer» Correct Answer: C Explanation : According to graph we can say that potential difference across the capacitor C1 is more than that across C2. Since charge Q is same i.e., Q=C1V1=C2V2 => C1/C2=V2/V1 => C1<C2 (V1>V2). |
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| 21. |
A rocket is going upwards with an accelerated motion. A man sitting in it feels his weight increased 5 times of his own weight. If the mass of the rocket including that of the man is 1.0*104 N, how much force is being applied by rocket engine? |
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Answer» if 104 is weight of system than F-- WEIGHT = (MASS OF SYSTEM )(5G) (MASS)(G) = WEIGHT THEREFORE F = 6 (WEIGHT) HENCE F = 6*104 N |
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| 22. |
An electron is projected as shown with speed `u`. If electron just fails to strike the upper plate, find range. (distance from starting point at which electron strike the lower plate again). Consider only electric force. |
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Answer» Correct Answer - 8 `ucosthetaxxt=R` .(1) `darra=((qE)/(m))` `((u)/(sqrt(2)))=((qE)/(m))(t)/(2)impliest=(2mu)/(sqrt(2)qE)` .(2) `((u)/(sqrt(2)))^(2)=2((qE)/(m))dimplies(qE)/(m^(2))=(u^(2))/(4d)` ..(3) `impliest=(2u)/(sqrt(2)u^(2))(4d)=(4sqrt(2)d)/(u)` put in (1) `R=((u)/(sqrt(2)))((4sqrt(2)d)/(u))=4d` |
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| 23. |
Define pollination. Explain the different types of pollination. List two agents of pollination? How does suitable pollination lead to fertilization? |
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Answer» Pollination is defined as the transfer of pollen grains from another to the stigma of a flower of a plant. There are two types of pollination: (i) Self Pollination:- It is the transfer of pollen grains from anther to the stigma of the same or genetically similar flower. (ii) Cross Pollination:- It is the transfer of pollen grain from anther of one flower to the stigma of another flower borne on a different plant of the same species. Two agents of pollination are insects and wind. By the process of pollination, pollens grain reaches to the stigma of flower. After that pollen tube develops from pollen grain. Pollen tube develops from pollen grain. Pollen tube contains two male gametes. One male gamete fuses with the egg to form diploid zygote and other male gametes fuses with polar nuclei to form triploid nucleus, which develops into endosperm. Thus, suitable pollination leads to fertilization. |
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| 24. |
List four characteristics of the image formed by a concave mirror of focal length 40 cm when the object is placed in front of it at a distance of 20 cm from its pole. |
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Answer» Characteristics of the image formed by the concave mirror are - (i) The image is formed behind the mirror at a distance of 40 cm from the pole. (ii) It is enlarged. (iii) It is virtual. (iv) It is erect. |
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| 25. |
Identify the given diagrams. Name the parts 1 to 5 |
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Answer» 1. Fallopian tube 2. Ovary 3. Uterus 4. Cervix 5.Vagina |
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| 26. |
In the experimental set up to show that "CO2 is given out during respiration". Name the substance taken in the small test tube kept in the conical flask. State its function and the consequence of its use. |
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Answer» The substance taken in test tube is KOH (Potassium hydroxide). KOH absorbs the CO2 released by the germinating seeds and a partial vaccum is created because of the loss of CO2 in the flask. This vaccum causes the water in the delivery tube to rise up. |
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| 27. |
Define current. Give its S.I. unit. |
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Answer» Electric current is expressed by the amount of charge flowing through a particular area in unit time. Its S.I. Unit is Ampere (A). |
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| 28. |
Why is vegetative propagation practised for growing some types of plants? |
| Answer» Some plants which do not produce viable seeds or produce seeds with prolonged period of dormancy are cultivated through the mode of vegetative propagation. Thus this method is used for growing of plants bearing superior traits, because they produce genically identical plants. | |
| 29. |
What is contraception? List three advantages of adopting contraceptive measures. |
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Answer» The deliberate use of artificial methods or other techniques to prevent pregnancy as a consequence of sexual intercourse is called contraception. Three advantages are: (i) Unwanted pregnancy can be prevented (ii) Birth rate can be controlled. (iii) Transfer of sexually transmitted disease can be prevented. |
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| 30. |
(a) Identify the diagram. Name the parts 1 to 5. (b) What is contraception? List three advantages of adopting contraceptive measures. |
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Answer» (a) The above given diagram is the Human female Reproductive System `{:("the parts are: 1. Oviduct or Fallopin tube", " 2. Ovary"," 3. Uterus"),(" 4. Cervix"," 5. Vagia",""):}` (b) The methods used to prevent pregnancy as a consequence of sexual intercoures is called contrraception. Advantages of adopting contraceptives measures are: (i) It helps in controlling the size of the family which can improve the standard of living. (ii) The general health of the female can be improved as reproduction demands high pressure on the body and mind of the female. (iii) It helps in preventing sexually trancmitted diseases. (iv) These methods check the population growth of a country by controlling child birth rate. (v) Frequent and unwanted pregnancies can be avoided by using contraception. |
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| 31. |
Write four sequentaial step of the procedure of experiment "Preparing a temporary" mount of a leaf peel to show stomata." |
| Answer» Procedure for preparing a temporoary mount of a leaf peel to show stomata: | |
| 32. |
When a tall pea plant was self pollinated, one-fourth of the progeny were dwarf. Give the genotype of the parent and dwarf progenies. |
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Answer» Tt –Parent, tt – dwarf progeny |
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| 33. |
A car of mass `500kg` moving with a speed `36kmh` in a straight road unidirectionally doubles its speed in `1min`. Find the power delivered by the engine.A. 1250 WB. 1250 HPC. 625 WD. 625 HP |
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Answer» Correct Answer - A Its initial speed `v_(1)=(36000)/(3600) =10ms^(-1)` If the car doubles its speed, finally its speed becomes `v^(2) =20ms^(-1)` Change in KE of the car `=Delta KE =(1//2)` `mv_(2)^(2) -(1/2)mv_(1)^(2)` During Dt = 1 min = 60 s the power delivered by the engine, `P=("Work done")/("Time taken") =(|Delta KE|)/(Delta t)` `=((1)/(2) xx m [v_(2)^(2) -v_(1)^(2)])/(Delta t) =(1//2 xx 500 [20^(2)-10^(2)])/(60)` `=1250 W` |
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| 34. |
In two different system of unit an acceleration is represented by the same number, whilst a velocity is represented by numbers in the ratio `1:3`. The ratio of unit of length and time areA. `1/3, 1/9`B. `1/9, 1/3`C. `1, 1`D. None of these |
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Answer» Correct Answer - B Length `l=v^(2)/a`, time `t=v/a` `rArr` ratio of unit of length `=(1/3)^(2)=1/9` and ratio of unit of time `=1//3` |
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| 35. |
A bullet of mass 0.012kg and horizontal speed 70ms-1 strikes a block of wood of mass 0.4kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block. |
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Answer» If V be the velocity of the block after collision, the using law of conservation of momentum, we get 0.012 × 70 + 0 = (0.012 + 0.4)V or V = \(\frac{0.012×70}{0.412}\)ms-1 = 2.04ms-1 If h be the height through which block rises, then (M + m) gh = 1/2 (M + m)V2 or h = v2/2g or h = \(\frac{2.04×2.04}{2×9.8}\)m = 0.212m = 21.2 cm Amount of heat produced in the block = loss of K.E. = \(\frac{1}{2}\) × 0.012 × 70 × 70 – \(\frac{1}{2}\) × 0.412 × 2.04 × 2.04 = 29.4J – 0.857J = 28.543J. |
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| 36. |
A uniform metre rule balance on A knife edge at 55cm mark when a 40g is hung from the 95cm mark |
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Answer» The given answer is below https://bit.ly/2MSFzVB |
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| 37. |
Name the element having highest value for its Electron gain enthiaply. |
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Answer» Chlorine is highest value for its Electron gain enthiaply. |
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| 38. |
Calculate the binding Energy of an alpha (α) particle in Mev from the following data. Mass of Helium Nucleus = 4.00260 u Mass of neutron = 1.008662 u Mass of proton = 1.007825 u. |
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Answer» Alpha particle: \(_2^4He\)Z = 2 and A = 4 Mass defect: Δm = [Zmp + (A-Z)mn] – M Δm = 2 × 1.007825 + 2 × 1.008662 – 4.00260 Δ m = 2.01565 + 2.017324 – 4.00260 Δ m = 0.030374 u Binding energy: Eb = Δm × 931 MeV = 0.030374 × 931 MeV Eb = 28.2782 MeV |
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| 39. |
A student is recuired to measure emf of a cell the should use .A. PotentiometerB. VoltmeterC. ammeterD. either (1) or (2) |
| Answer» Correct Answer - A | |
| 40. |
A potentiometer is an ideal device of measuring potential difference becauseA. it uses a sensitive galvanometerB. it does not disturb the potential difference it measuresC. it is an elaborate arrangementD. it has a long wire hence heat developed quickly radiated |
| Answer» Correct Answer - B | |
| 41. |
A direct current of 5A is superimposed on an alternating current I = 10 `sin omega t` flowing through a wire . The effective value of the resulting current will beA. `((15)/(2))A`B. `5 sqrt(3)`C. `5sqrt(5)A`D. `15 A` |
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Answer» Correct Answer - B Given ` l = 5 + 10 sin omega t` `I_(eff) = [(int_(0)^(T) I^(2)dt)/(int_(0)^(T) dt)]^(1//2) = [ (1)/(T) int_(0)^(T) ( 5 + 10 sin omegat)^(2)dt]^(1//2)` ` = [(1)/(T) int_(0)^(T) (25+ 100 sin omegat dt =0 and (1)/(T)int_(0)^(T) sin^(2) omegat dt =(1)/(2)` So `I_("eff")=[25+1/2xx100]^(1//2) = 5sqrt(3)A` |
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| 42. |
The position vector of a particle is given by `vec(r)=1.2 t hat(i)+0.9 t^(2) hat(j)-0.6(t^(3)-1)hat(k)` where t is the time in seconds from the start of motion and where `vec(r)` is expressed in meters. For the condition when `t=4` second, determine the power `(P=vec(F).vec(v))` in watts produced by the force `vec(F)=(60hat(i)-25hat(j)-40hat(k)) N` which is acting on the particle. |
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Answer» Correct Answer - 1044 `vec(v)=(dvec(r))/(dt)=1.2 hat(i)+1.8that(j)-1.8t^(2)hat(k)` at `t=4s, vec(v)=1.2 hat(i)+7.2 hat(j)-28.8 hat(k)` `P=vec(F).vec(v)=(60hat(i)-25hat(j)-40hat(k)).(1.2hat(i)+7.2hat(j)-28.8hat(k))` `=1044W` |
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| 43. |
The velocity corresponding to a gas "X" (MW=100) corresponding to the maxima of the curve given below at a given temperature is 200 m/s.Then kinetic energy of 300 gram of has "X" will be A. 9000 kJB. 9 kJC. 90 kJD. 810 kJ |
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Answer» Correct Answer - B The velocity corresponding to the maxima is the most probable velocity which is given by the expression `C_(np)=sqrt((2RT)/M)` `RT=(C_(mp)^2M)/2=((200)^2xx100xx10^(-3))/2=40000xx50xx10^(-3) J mol^(-1) = 40xx50 J mol^(-1)` No of moles =`300/100=3` `E=3/2nRT=3/2xx3xx40xx50=9xx10^(3) J =9 kJ` |
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| 44. |
`2CaSO_4 (s) hArr 2CaO(s)+2SO_(2)(g)+O_2(g), DeltaHgt0` Above equilibrium is establised by taking some amount of `CaSO_4(s)` in a closed container at 1600 K. Then which of the following may be correct options ?A. Moles of CaO(s) will increase with the increase in temperatureB. if the volume of the container is doubled at equilibrium then partial pressure of `SO_2(g)` will change at new equilibriumC. If the volume of the container is halved partial pressure of `O_2`(g) at new equilibrium will remain sameD. If two moles of the He gas is added at constant pressure then the moles of CaO(s) will increase |
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Answer» Correct Answer - A,C,D (A)As reaction is endothermic therefore with increases in temperature it will go in the forward direction hence moles of CaO will increases. (B)With the increase of decrease of volume partial pressure of the gases will remain same. (C )Due to the addition of inert gases at constant pressure reaction will proceed in the direct in which more number of gaseous moles are formed. |
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| 45. |
P-V plots for three gases (assuming ideal behaviour and similar condition) for reversible adiabatic compression are given in the figure below : Plots X,Y and Z should correspond to respectively :A. `CO_2,Cl_2` and NeB. `SO_2, N_2O` and HeC. He, `N_2 and O_3`D. `NH_3,H_2S and Ar` |
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Answer» Correct Answer - B `SO_2(gamma =1.33), N_2O(gamma=1.4),He (gamma=1.67)` As `gamma` increases, for compression graph rises up. |
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| 46. |
30 ml of 0.06 M solution of the protonated form of an anion of acid methonine `(H_2A^(+))` is treated with 0.09 M NaOH. Calculate pH after addition of 20 ml of base `pKa_1=2.28 and pKa_2=9.2`A. 5.5B. 5.74C. 9.5D. None |
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Answer» Correct Answer - B `pH=(pK_(a_1)+pK_(a_2))/2 =(2.28+9.2)/2=5.74` |
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| 47. |
A 0.0200 gm sample containing copper (II) was analysed iodometrically, copper (II) is reduced to copper (I) by iodine, `2Cu^(2+)+4I^(-)to2 CuI+I_2` If 20.0 mL of 0.10 M `Na_2S_2O_3` is required for titration of the liberated iodine then the percentage of copper in the sample will be (Cu=63.5 g/mole)A. `31.75%`B. `63.5%`C. `53%`D. `37%` |
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Answer» Correct Answer - B 2 moles of `Cu^(2+)`=1 moles of `I_2` =2 moles of hypo. so moles of hypo used =`20xx10^(-3)xx0.1=2` milli moles = milli moles of copper hence percentage of copper =`(2xx10^(-3)xx63.5)/0.2xx10%=63.5%` |
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| 48. |
A table is a set of data elements that is organized using a model of vertical _______ and horizontal ________. a. Rows, Tables b. Columns, Rows c. Rows, Columns d. Forms, Reports |
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Answer» Correct answer is b. Columns, Rows Columns, Rows |
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| 49. |
Will a precipitate of Mg(OH)2 be formed in a 0.002 M solution of Mg(NO3)2, if the pH of solution is adjusted to 9 ? Ksp of Mg(OH)2 = 8.9 × 10–12. |
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Answer» pH = 9 ∴ [H+ ] = 10–9 M or [OH– ] = 10–5 M Now if Mg(NO3)2 is present in a solution of [OH–] = 10–5 M, then, Product of ionic conc. = [Mg+2] [OH– ]2 = [0.002] [[10–5]2 = 2 × 10–13 lesser than Ksp of Mg(OH)2 i.e, 8.9 × 10–12 ∴ Mg(OH)2 will not precipitate. |
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| 50. |
The drug used as an oral anticoagulant |
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Answer» Warfarin continues to be the most widely used oral anticoagulant but the use of the newer oral anticoagulants (dabigatran etexilate, rivaroxaban, edoxaban and apixaban) is increasing. |
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