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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. |
A vessel is partitioned in two equal halves by a fixed diathermic separator. Two different ideal gases ae filled in left (L) and right (R) halves the rms speed of the molecules in L part is equal to the mean speed of moleucles in the R equal to the ratio of the mass of a molecules in L part to that of a molecules in R part is A. `sqrt((3)/(2))`B. `sqrt(pi//4)`C. ` sqrt(2//3)`D. ` 3pi//8` |
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Answer» Correct Answer - D Root means square velocity of molecule in left part `V_(n ns) =sqrt((3KT)/(m_(L)))` Mean or average speed of moleucles in right part `V_(av)=sqrt((8KT)/(pim_(k)))` According to problem `sqrt((3KT)/(m_(L)))=sqrt((8)/(pi)(KT)/(m_(R)))` ` rArr " " (3)/(m_(L))=(8)/(pim_(R))` `rArr" " (m_(L))/(m_(R))=(3pi)/(8)` |
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| 2. |
Identify the option which is used to mute the selected key. a) tangent menu b) keys menu c) mute key d) dope sheet outliner |
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Answer» c) mute key mute key is used to mute the selected key. |
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| 3. |
What is UV-VIS spectrometry? |
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Answer» The UV-VIS spectrometry is one of the oldest instrumental techniques of analysis and is the basis for a number of ideal methods for the determination of micro and semimicro quantities of analytes in a sample. It concerns with the measurement of the consequences of interaction of Electromagnetic radiations in the UV and/or visible region with the absorbing species like, atoms, molecules or ions. |
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| 4. |
Which are commonly used bacteriophages as cloning vectors? |
| Answer» M13 and lambda phage are commonly used bacteriophages. | |
| 5. |
Name any two natural cloning vectors. Give reasons that make them act as cloning vectors. Write the two characteristics the engineered vectors are made to possess |
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Answer» The two natural cloning vectors are Plasmids and bacteriophages. They have ability to replicate within bacterial cell and high copy number within the bacterial cell. Characteristics of engineered Vectors:
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| 6. |
What is the natural ageing of a lake called? How is it caused? Write the impact of uncontrolled human activities on it? |
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Answer» Natural ageing of a lake is called Eutrophication. It is caused due to nutrients such as nitrates and phosphates being drained by streams into lakes. Effluent from industries and home and agriculture wastes have resulted in cultural or Accelerate Eutrophication. These uncontrolled human activities increase the flow of both inorganic nutrients and organic substances into the ecosystem. |
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| 7. |
“A very small sample of tissue or even a drop of blood can help determine paternity”. Provide a scientific explanation to substantiate ho it is possible |
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Answer» DNA from all cells of an organism like tissue or even a drop of blood show the same degree of polymorphism, which is inheritable. Thus DNA finger printing of the given sample can be useful for determining paternity |
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| 8. |
If a Hydra is cut into two pieces it will result in (a) growth of mouth and disc according to their position in the parent. (b) growth without mouth and basal disc. (c) no regeneration. (d) growth of mouth and disc at any end. |
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Answer» (a) growth of mouth and disc according to their position in the parent. |
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| 9. |
Asexual life cycle of plasmodium is called a. Schizogony b. Sporogony c. Both a and b d. None of all |
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Answer» a. Schizogony |
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| 10. |
If flowers ______ enough water, they die. A) don’t get B) get C) didn’t get D) doesn’t get |
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Answer» Correct option is A) don’t get |
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| 11. |
What is apomixis and what is its importance? |
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Answer» Apomixis is the mechanism of seed production without involving the process of meiosis and syngamy. It plays an important role in hybrid seed production. The method of producing hybrid seeds by cultivation is very expensive for farmers. Also, by sowing hybrid seeds, it is difficult to maintain hybrid characters as characters segregate during meiosis. Apomixis prevents the loss of specific characters in the hybrid. Also, it is a cost-effective method for producing seeds. |
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| 12. |
In the figure shown when the massless spring is in relaxed state, its free end is at point B. A very small block is pressed against the spring by a distance `delta` and then released from rest. Except the portion BC where coefficient of kinetic is `mu`, track is smooth everywhere. Determine the spring compression `delta` so that the block enters a small hole at E. Consider all value shown in the figure.If `delta` is given by `delta=x[(mgd(5+mu))/(2K)]^((1)/(2))`, find x. Block never leaves contact with surface |
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Answer» Correct Answer - 9 Mg is parallel to centre line and normal reaction passes through it so Torque about centre line is zero. So angular momentum is conserved about centre line. `mv_(0)r_(0)=m(vcostheta)r` `vcostheta=(v_(0)r_(0))/(r )=(v_(0)r_(0))/(sqrt(r_(0)^(2)-h^(2)))=v_(0)sqrt(2)` Energy conservation `(1)/(2)mv^(2)=(1)/(2)mv_(0)^(2)+mgh` `v^(2)=v_(0)^(2)+2gh` `=(10)^(2)+2(10)(15)=400` `V=20` `costheta=(10sqrt(2))/(20)=(1)/(sqrt(2))` |
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| 13. |
In the figure shown when the massless spring is in released state its free end is at point `B`. A very small block is ressed against the spring by a distance `delta` and the released from rest. Except the portion `BC` where coefficient of kinetic friction is `mu_(k)`, track is smooth everywhere. Determine the spring compression `delta` so that the block enters a small hole at `E`. Consider all values shown in the figure. |
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Answer» We determine the speed required at `D` so that the block falls into the hole at `E` by following a parbolic trajectory. `v_(d)=`speed at `D` `v_(d)=sqrt(8gd)` Applying conservation of energy for motion from `A` to `D` `(1)/(2)kdelta^(2)=mu_(k)mgd+mgd+(1)/(2)m(8gd) rArr delta=sqrt((2mgd)/(k)(mu_(K)+5))` |
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| 14. |
A small ball is rolled with speed `u` from piont A along a smooth circular track as shown in figure. If `x=3R`, then Determine the required speed u so that the ball returns to A, the point of projection after passing through C, the highest point. |
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Answer» We determine the velocity at `C` so that it reaches `A` along a parbolic trajectory, then applying the law of conservation of energy to the motion `ABC` (as non-conservative force are absent) `v=` velocity at point `C=(3)/(2)sqrt(gR)` applying conservation of energy from `A` to `C` `(1)/(2)mu^(2)=2mgR+(1)/(2)mv^(2)` `rArr u=(5)/(2)sqrt(gR)` |
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| 15. |
A ball strikes a horizontal floor at `45^(@)`. 25% of its kinetic energy is lost in collision. Find the cofficient of resultationA. `(1)/(2)`B. `(1)/sqrt(2)~C. `(1)/(2sqrt(2))`D. `(1)/(4)` |
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Answer» `((1)/(2)mU^(2)-(1)/(2)mV^(2))/((1)/(2)mU^(2))=(1)/(4)rArrV=Usqrt((3)/(4))` `V=sqrt((3)/(4))U=[((U)/sqrt(2))^(2)+((eU)/sqrt(2))^(2)]^(1//2)` `e=(1)/sqrt(2)` |
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| 16. |
Directions : Each of these questions contains two statements, Assertion and Reason. Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select one of the codes (a), (b), (c) and (d) given below.(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion(c) Assertion is correct, reason is incorrect(d) Assertion is incorrect, reason is correct.Assertion : The optical instruments are used to increase the size of the image of the object.Reason : The optical instruments are used to increase the visual angle. |
| Answer» (d) Assertion is incorrect, reason is correct. | |
| 17. |
The correct sign convention for the following figure where the object is at ‘c’ will be(a) u = –ve; v = +ve; H = +ve; h = –ve(b) u = +ve; v = –ve; H= –ve; h = +ve(c) u = –ve; v = –ve; H = +ve; h = –ve(d) u = +ve; v = +ve; H = –ve; h = +ve |
| Answer» (c) According to sign-convention, distance left or below pole is – (ve) and right or above pole is +(ve) | |
| 18. |
The graph between angle of deviation (d) and angle of incidence (i) for a triangular prism is represented by |
| Answer» (c) For the prism as the angle of incidence (i) increases, the angle of deviation δ first decreases goes to minimum value and then increases. | |
| 19. |
The equation n2/v -n1/u= n2 - n1 /R holds true for(a) only concave mirror(b) only convex mirror(c) both concave and convex mirror(d) any type of reflecting surface |
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Answer» (c) both concave and convex mirror |
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| 20. |
Which of the following (referred to a spherical mirror) depends on whether the rays are paraxial or not?(a) Pole (b) Focus(c) Radius of curvature (d) Principal axis |
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Answer» The correct answer is (b) Focus |
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| 21. |
An equilateral prism is placed on a horizontal surface. A ray PQ is incident onto it. For minimum deviation(a) PQ is horizontal(b) QR is horizontal(c) RS is horizontal(d) Any one will be horizontal |
| Answer» (b) For minimum deviation, incident angle is equal to emerging angle. QR is horizontal | |
| 22. |
Which of the following statements is/are incorrect?I. At sunset or sunrise, the sun’s rays have to pass through a small distance in the atmosphere.II. At sunset or sunrise, the sun’s rays have to pass through a larger distance in the atmosphere.III. Rayleigh scattering which is proportional to (1/λ)2.(a) I and II (b) I and III(b) II and III (d) I, II and III |
| Answer» (b) At sunset or sunrise the sun’s rays have to pass through a larger distance in the atmosphere and most of the blue or other shorter wavelengths are removed by scattering. | |
| 23. |
Figure shows two rays A and B being reflected by a mirror and going as A' and B'. The mirror(a) is plane(b) is convex(c) is concave(d) may be any spherical mirror |
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Answer» The mirror (a) is plane |
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| 24. |
Energy of molecules of any substance is known as A. kinetic energy B. potential energy C. internal energy D. chemical energy |
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Answer» C. internal energy |
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| 25. |
Measure of average kinetic energy of molecules is A. temperature B. energy C. internal energy D. enthalpy |
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Answer» A. temperature |
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| 26. |
Average kinetic energy of gas molecules is proportional to A. internal energy B. thermodynamic temperature C. enthalpy D. condensation point |
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Answer» B. thermodynamic temperature |
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| 27. |
On compression, the gat gets hotter due to A. increase in kinetic energy B. decrease in kinetic energy C. increase in potential energy D. increase in atomic collisions |
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Answer» A. increase in kinetic energy |
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| 28. |
Two vessel A and B in volume ratio 2 : 1 maintained at temperature 300 K and 600 K contain equal mass of `CH_(4)` and `O_(2)` respectively. The ratio of toal translational kinetic energy of gas in flask A to that of flask B is -A. 1B. 2C. 4D. 0.25 |
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Answer» Correct Answer - A `(KE_(A))/(KE_(B)) = ((3)/(2) xx n_(CH_(4)) RT_(A))/((3)/(2) xx n_(O_(2))RT_(B))` `((w)/(16) xx 300)/((w)/(32) xx 600) = (1)/(1)` |
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| 29. |
Vessel A contains an ideal gas at a pressure `5xx10^(5)` Pa and is connected with a heat source which maintains its temperature at `300 K`. Another vessel B which has four time greater inner volume contains the same gas at a pressure `10^(5)` Pa and is connected to a heat source which maintains its temperature at `400 K`. What will be the pressure of entire system if two vessels are connected by a narrow tube tap:-A. `10^(5)Pa `B. `2xx10^(5)Pa`C. `4xx10^(5)Pa`D. `5xx10^(5)Pa` |
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Answer» Correct Answer - B `(P_(A)V_(A))/(RT_(A))+(4P_(B)V_(A))/(RT_(B)) = (pV_(A))/(RT_(A))+(4pV_(A))/(RT_(B))` `(p_(A))/(T_(A))+(4p_(B))/(T_(B))=(p)/(T_(A))+(4p)/(T_(B))` `(5xx10^(5))/(300)+(4xx10^(5))/(400) = (p)/(300)+(4p)/(400)` `((20+12)xx10^(5))/(1200) = (4p+12p)/(1200)` `32xx10^(5) = 16p` `p = 2xx10^(5) N//m^(2)` |
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| 30. |
Which of the following is correct for pure liquid :A. Molality is independent of density of liquid.B. Molarity depends on density of liquid.C. Molarity = density `xx` molality (where density is in gm/lit)D. All are correct |
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Answer» Correct Answer - D (1) `m = ("moles of solute")/("mass of solvent in kg")` Independent of density as no volume is present (2) `M = ("moles of solute")/("volume of solution in Lt")` For pure liquid solvent = solution so, m depends on density (3) `M = (n_("solute"))/((m_("sovent"))/(d)) = (n_("solute"))/(m_("solvent")) xx d = m xx d` `"Molarity" = "Molality" xx d` |
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| 31. |
At what temperature, the mean kinetic energy of O2 will be the same for H2 molecules at – 73°C (a) 127°C (b) 527°C (c) – 73°C (d) – 173°C |
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Answer» Correct option (c) – 73°C Explanation: Mean kinetic energy of molecule depends upon temperature only. For O2 it is same as that of H2 at the same temperature of −73°C. |
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| 32. |
The % by volume of `C_(4)H_(10)` in a gaseous mixture of `C_(4)H_(10), CH_(4)` and `CO_(2)` is 40. When 200 ml of this mixture is burnt completely in excess of `O_(2)` and then pass through aq. KOH then volume contraction will be -A. 220 mlB. 340 mlC. 440 mlD. 560 ml |
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Answer» Correct Answer - C `ubrace(C_(4)H_(10), CH_(4),CO_(2))_("200 mL")` `C_(4)H_(10) = 0.4 xx 200 = 80 ml` `CH_(4) = x ml` `CO_(2) = (120 - x) ml` `C_(4)H_(10) + (13)/(2) O_(2) rarr 4CO_(2) + 5H_(2)O` `{:(80 ml,," 320 ml",),(CH_(4) + 2O_(2),rarr,CO_(2) + 2H_(2)O,),(x ml,,x ml,):}` Total `CO_(2) = 320 + x + 120 - x` `= 440 ml` |
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| 33. |
If PD vs P (where P si pressure in atm and D is density in gm/lit). Is plotted for `H_(2)` ideal gas at a particular temperature. If `[(d(PD))/(dP)]_(P = 8.21 "atm") = 10`, then the temperature will beA. 40 KB. 400 KC. 20 KD. 200 K |
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Answer» Correct Answer - A PD v/s P `D = (PM)/(RT)` `PD = (P^(2)M)/(RT)` `(d (PD))/(dP) = (2PM)/(RT) = 10` `(2 xx 8.21 xx 2)/(R xx T) = 10` `T = (4 xx 8.21)/(0.0821 xx 10) = 40 K` |
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| 34. |
If a closed container of volume 200 lit. of O_{2} gas (ideal gas) at 1 atm & 200 K is taken to planet. Find the pressure of oxygen gas at the planet at 821 K in same container(A)10/(e ^ 100)(B)20/(e ^ 50)(C) 1 atm (D) 2 atmhansion #1 (0 No. 8 to 10) |
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Answer» We have given, Volume of container = 200 litre initial pressure of g oxygen gas = 1 atm initial temperature = 200 K Temperature of planet (final Temperature) = 821 K ∵ Oxygen gas in the same container, it means. Volume of gas constant. According to Gay-Lussac law — At constant volume, As pressure increases, temperature also increases. P ∝ T or P1/T1 = P2/T2 where, P1 = initial pressure T1 = initial Temperature P2 = final pressure T2 = final Temperature Therefore, \(\frac{1\,atm}{200\,K}=\frac{P_2}{821\,K}\) P2 = \(\frac{821\,K\times1\,atm}{200\,K}\) P2 = 4.105 atm. |
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| 35. |
In a face centered cubic arrangement of A and B atoms whose A atoms are at the corner of the unit cell and B atoms at the face centers. One of the B atoms missing from one of the face in unit cell. The simplest formula of compounding is:A. `AB_(3)`B. `A_(8)B_(5)`C. `A_(2)B_(5)`D. `AB_(2//5)` |
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Answer» Correct Answer - C `Ato"corners"=8xx(1)/(8)=1` `Bto"at 5 face centers"=5xx(1)/(2)=(5)/(2)` `A : B=1:(5)/(2) = 2:5` ` :. A_(2)B_(5)` |
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| 36. |
The reptile with a four chambered heart is (a) crocodile (b) turtle (c) lizards (d) snakes |
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Answer» (a) A crocodile is any species belonging to the family Crocodylidae sometimes classified instead as the subfamily Crocodylinae. Its a reptile with four chambered heart. |
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| 37. |
The girth of the stem or root increases due to (a) apical meristem (b) intercalary meristem (c) lateral meristem (d) None |
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Answer» (b) The intercalary meristem increases the girth of the stem or root. The intercalary meristems occur only in monocot stems between mature tissues. They are cylindrical meristems located around the nodes and are an adaptation to grazing herbivores and landmowers. |
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| 38. |
Which among the following has specialised tissue for conduction of water? (i) Thallophyta (ii) Bryophyta (iii) Pteridophyta (iv) Gymnosperms (a) (i) and (ii) (b) (ii) and (iii) (c) (iii) and (iv) (d) (i) and (iv) |
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Answer» (c) (iii) and (iv) |
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| 39. |
The process of formation of permanent tissue in plants is called (a) scalarification (b) differentiation (c) cell thickening (d) specialisation |
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Answer» (b) Differentiated plant cells generally cannot divide or produce cells of a different type. Therefore, cell division in the meristem is required to provide new cells for expansion and differentiation of tissues and initiation of new organs, providing the basic structure of the plant body. |
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| 40. |
The vascular tissues of the plant function in (a) support (b) support and transport of materials (c) secretion of plant hormones (d) All of the above |
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Answer» (b) support and transport of materials |
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| 41. |
Tissues that have long and narrow cells are called (a) cuboidal epithelium (b) squamous epithelium (c) germinal epithelium (d) columnar epithelium |
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Answer» (d) columnar epithelium |
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| 42. |
Hydrocolloid carrageen is obtained from:(1) Phaeophyceae and Rhodophyceae(2) Rhodophyceae only(3) Phaeophyceae only(4) Chlorophyceae and Phaeophyceae |
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Answer» Correct option is (2) Rhodophyceae only |
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| 43. |
Why are cleistogamous flowers invariably autogamous? |
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Answer» Cleistogamous flowers are the bisexual flowers which do not open all even at maturity; the anthers and stigma lie close to each other and self-pollination is affected easily. |
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| 44. |
Given below are two statements:Statement I:Cleistogamous flowers are invariably autogamousStatement II:Cleistogamy is disadvantageous as there is no chance for cross pollination.In the light of the above statements, choose the correct answer from the options given below:(1) Both Statement I and Statement II are incorrect(2) Statement I is correct but Statement II is incorrect(3) Statement I is incorrect but Statement II correct is(4) Both Statement I and correct Statement II are correct |
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Answer» (4) Both Statement I and correct Statement II are correct |
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| 45. |
The thickening of the walls of the sclerenchyma tissues is due to (a) suberin (b) magnesium (c) lignin (d) calcium |
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Answer» (c) Sclerenchyma is a supporting tissue. Two groups of sclerenchyma cells exist: fibres and sclereids. Their walls consist of cellulose and/or lignin. Sclerenchyma fibres are of great economical importance, since they constitute the source material for many fabrics flax, hemp, jute, ramie. |
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| 46. |
The waxy, water resistant layer in the xerophytic plants is secreted by (a) endodermis (b) cortex (c) phloem (d) epidermis |
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Answer» (d) A xerophyte or xerophytic organism is an organism, which is able to survive in an ecosystem with little to no water or moisture; usually in environments where heat accounts for a similar amount of water loss to that of transpiration. |
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| 47. |
Which among the following is incorrectly matched pair?(a) Silicified epidermal layer—Equisetum(b) Waxy deposition—Banana leaves(c) Epidermal hairs of seed coat—Cotton fibres(d) None of the above |
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Answer» Answer is : (d) None of the above |
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| 48. |
When limestone is heated strongly, it gives off `CO_(2)`. In matallurgy this process is known asA. CalcinationB. RoastingC. SmeltingD. Ore dressing |
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Answer» Correct Answer - A `CaCO_(3) rarr CaO+CO_(2)` Heating the ore in absence of air is calcination. |
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| 49. |
When limestone is heated strongly, it gives off `CO_(2)`. In matallurgy this process is known asA. CalcinationB. RoastingC. SmeltingD. Ore-dressing |
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Answer» Correct Answer - A `CaCO_(3) overset(Delta) rarr CaO + CO_(2) uarr`. |
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| 50. |
The nitrogenous excrtory products are formed from the catabolism of amino acids byA. Calvin cycleB. Nitrogen cycleC. Ornithine cycleD. Krebs cycle |
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Answer» Correct Answer - C The nitrogenous excretory products are formed from the catabolism of amino acids by ornithine cycle. |
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