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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 3201. |
A boy and a block both of same mass, are suspended at the same horizontal level from each end of a light string that moves over a frictionless pulley as shown. The boy starts moving upwards with an acceleration of 2.5 m/s2 relative to the rope. If the block is to travel a total distance 10 m before reaching at the pulley, the time taken by the block in doing so is equal to : |
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Answer» Applying newton's 2nd law of motion, you will get the answer as time = 4s |
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| 3202. |
Element X crystallizes in a 12 coordination FCC lattice. On applying high temperature it changes to 8 coordination BCC lattice. Find the ratio of the density of the crystal lattice before and after applying high temperature ? |
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Answer» Correct answer is option (D) |
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| 3203. |
A substance whose relative permeability is more than the permeability of free space is?(a) Diamagnetic(b) Paramagnetic(c) Ferromagnetic(d) Both paramagnetic and ferromagnetic |
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Answer» The correct option is (d) Both paramagnetic and ferromagnetic The explanation: Relative permeability=1+Magnetic susceptibility Since both paramagnetic and ferromagnetic materials have positive susceptibility, their relative permeability is greater than unity i.e. their permeability is more than the permeability of free space. |
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| 3204. |
Ionic solid `B^+A^(-)` crystallizes in rock salt type of structure.1.296 gm ionic solid salt `B^+A^(-)` is dissolved in water to make one litre solution.The pH of the solution is measured to be 6.0.if the value of face diagonal in the unit cell of `B^+A^(-)` be `600sqrt2` pm.Calculate the density of ionic solid in gm/cc.`(T=298 K, K_b "for" BOH "is" 10^(-6)`,(Avogadro Number =`6.0xx10^(23)`)) |
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Answer» Correct Answer - 4 g/cc `pH=1/2{pK_w -pK_b- log C} implies square6=7.0-5/2-1/2"log C" " " implies " " C=0.01 M` `:. C=1.296/("Molar mass")=1/100 " " implies ` molar mass of salt =129.6 For rock salt, `600sqrt2=sqrt2a` a=600 pm=`600xx10^(-10) cm` effective number of formula units z=4 Now density `=Z/N_A (M/a^3)=4/(6.0xx10^23)[129.6/((600)^3xx10^(-30))]=4` gm/cc. |
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| 3205. |
A substance whose relative permeability is less than the permeability of free space is?(a) Diamagnetic(b) Paramagnetic(c) Ferromagnetic(d) Not a magnetic substance |
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Answer» Right choice is (a) Diamagnetic The explanation is: A diamagnetic material creates a magnetic field opposing that of the external magnetic field and it repels the external magnetic field. Hence its relative permeability is less than that of the free space. |
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| 3206. |
The relation between the direction of induced emf and the direction of motion of the conductor is?(a) Parallel(b) Equal(c) Not related(d) Perpendicular |
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Answer» Correct option is (d) Perpendicular Easy explanation: According to Fleming’s right hand rule, the induced emf, the motion of the conductor and the magnetic flux are mutually perpendicular. |
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| 3207. |
An ionic compound MCl is simultaneously doped with `10^(-6)` mole % `NCl_2` and `10^(-7)` mole % `RCl_3`.Calculate the concentration of cationic vacancies x. Express your answer as `x/9xx10^14`. (take N & R =divalent & trivalent cations) |
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Answer» Correct Answer - 8 Conc.of cationic vacancies `=10^(-6)xx6xx10^(23)+10^(-9)xx6xx10^(23)xx2=6xx10^(15)+12xx10^(14)=(60+12)xx10^14=72xx10^(14)` |
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| 3208. |
The development of lateral surfaces of a pentagonal pyramid is ________(a) Five rectangles(b) Five squares(c) Five triangles(d) Five circles |
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Answer» Right option is (c) Five triangles To explain: A pentagonal pyramid is a pyramid with a pentagonal base upon which are erected five triangular faces that meet at a point (the vertex). Like any pyramid, it is self-dual. The regular pentagonal pyramid has a base that is a regular pentagon and lateral faces that are equilateral triangles. |
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| 3209. |
The figure below represents the ________ view of the pentagonal base joined to a circular top.(a) side(b) front(c) top(d) bottom |
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Answer» Correct option is (c) top Easiest explanation: Figure 1 shows the top view and pictorial view of two transition pieces: (a) the pentagonal base joined to a circular top and (b) circular base connected to a square top. The lateral surface of the transition piece must be divided in to curved and non-curved triangles as shown in figure 1.Divide the curved cross section in to a number of equal parts equal to the number of sides of non-curved cross-section. Division points on the curved cross section are obtained by drawing bisectors of each side of the non-curved cross section. The division points thus obtained when connected to the ends of the respective sides of the non-curved cross-section produces plane triangles. In between two plane triangles there lies a curved triangle. After dividing in to a number of triangles, the development is drawn by triangulation method. |
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| 3210. |
Partnership form of organisation was developed due to the inherent limitations of sole proprietorship. One of them is limited capital, identify the other two? a. Limited managerial ability, limited continuity b. Limited continuity, unlimited liability c. Limited managerial ability, unlimited liability d. Lack of secrecy, limited continuity |
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Answer» a. Limited managerial ability, limited continuity |
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| 3211. |
How many types of two-dimensional lattice exist? Why pentagonal latticis not possible? |
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Answer» Five types, namely (a) hexagonal, (b) square, (c ) rectangular, (d) rhombic, and € parallelogram. Pentagonal lattice is not possible because the interior angle of a regular pentagon is `108^(@)` which is not an integral factor of `360^(@)`. |
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| 3212. |
A beaker containing an ideal fluid executes plane `SHM` in a horizontal plane according to the equation `x=(sqrt(3)g)/(omega^(2))sinomegat`, `O` being the mean position. A bob is suspended at `S` through a string of length `L` as shown in the figure. The line `SO` is vertical. Assuming `L gtgt (sqrt(3)g)/(omega^(2))`. The magnitude of maximum buoyant force and the time with it occurs for the second time are, respectively A. `2g`, `pi//2omega`B. `g`, `3pi//omega`C. `2g`, `3pi//2`D. `g`, `pi//omega` |
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Answer» (`C`) `2rhoVg`, `3pi//2omega` Magnitude of component of Buoyant force `=rhova` Where a is acceleration of beaker, `a=-sqrt(3) g sin omega t` Net buoyant force `=sqrt(rho^(2)v^(2)g^(2)+rho^(2)v^(2)a^(2))` `=rhovgsqrt(1+3sin^(2)omega t)` Magnitude of maximum buoyant force `=2rhovg` `sin^(2)omegat=1` at `t=(pi)/(2omega)`, `(3pi)/(2omega)` `:. 2rhovg`, `(3pi)/(2omega)` |
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| 3213. |
Horizontal analysis is done by analyzing ____________. A. Quarterly statement B. Half yearly statement C. Financial statements of several years D. Financial statements of a particular year |
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Answer» D. Financial statements of a particular year |
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| 3214. |
Two lighter nuclei combine to form a comparatively heavier nucleus by the relation given below: \(_1^2X + _1^2X = \,_2^4Y\)The binding energies per nucleon \(_1^2X\) and \(_2^4Y\) are 1.1 MeV and 7.6 MeV respectively. The energy released in this process is _____ MeV. |
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Answer» Energy released in the given process = Binding energy of product – Binding energy of reactants = 7.6 × 4 – (1.1 × 2) × 2 = 30.4 – 4.4 = 26 MeV |
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| 3215. |
Commerical `11.2` volume `H_(2)O_(2)` solution has a molarity ofA. `1.0`B. `0.5`C. `11.2`D. `1.12` |
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Answer» Correct Answer - A `V = M xx 11.2` `:. M = (V)/(11.2) =(11.2)/(11.2) = 1` |
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| 3216. |
The number of protons, neutrons and electrons in \(^{175}_{71}\)Lu , respectively, are : (1) 175, 104 and 71 (2) 71, 104 and 71 (3) 104, 71 and 71 (4) 71, 71 and 104 |
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Answer» (2) 71, 104 and 71 \(^{175}_{71}\)Lu p+ = 71 n0 = 175 – 71 = 104 e– = 71 |
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| 3217. |
The freezing point depression constant (Kf) of benzene is 5.12 K kg mol–1. The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is (rounded off upto two decimal places) : (1) 0.60 K (2) 0.20 K (3) 0.80 K (4) 0.40 K |
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Answer» (4) 0.40K ΔTf = Kf × m = 5.12 × 0.078 ΔTf = 0.40 K |
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| 3218. |
`CH_(3)-overset(O)overset(||)(C)-OHoverset(ND_(3))underset(Delta)(to)(A)overset(Br_(2))underset(KOH)(to)(B)` Product (B) is -A. `CH_(3)-ND_(2)`B. `CH_(3)-NH_(2)`C. D. |
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Answer» Correct Answer - B `CH_(3)-NH_(2)` |
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| 3219. |
, `underset((ii)H^(+)//Delta)overset((i)Br_(2)//KOH)rarr` Product isA. B. C. D. |
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Answer» Correct Answer - D Molar mass due to `CMxx(12)/(100)=12" "M=100` Molar mass due to `OMxx(4)/(100)=16" "M=400` Molar mass due to `SMxx(16)/(100)=32" "M=200` Minimum molar mass should be 400 Number of C-atom =`(400xx12)/(12xx100)=4-"atom"` |
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| 3220. |
Which of the following species is more soluble in water ?A. `M(OH)_(3)`,`(k_(sp)=10^(-35))`B. `M(OH)_(2)`,`(k_(sp)=10^(-30))`C. `MOH`,`(k_(sp)=10^(-28))`D. `MOH`,`(k_(sp)=10^(-26))` |
| Answer» Correct Answer - A | |
| 3221. |
Which of the following is/are correact for the formation of `pi` bond ?A. overlapping of `p_(x)-p_(x)`(`Z` internuclear axis)B. overlapping of `p_(z)-p_(z)` orbitals(`Z` internuclear axis)C. overlapping of `p_(y)-p_(y)` orbitals(`Z` internuclear axis)D. overlapping of `s-p_(z)` orbitals (`Z` internuclear axis) |
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Answer» Correct Answer - A::C (A) overlapping of `p_(x)-p_(x)` (`Z` internuclear axis) (C) overlapping of `p_(y)-p_(y)` orbitals (`Z` internuclear axis) |
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| 3222. |
Which of the following will result in zero overlap if molecular axis is `x-`axis ?A. `1s - 2p_(x)`B. `2s - 2p_(z)`C. `2p_(x) - 2p_(x)`D. `1s - 2p_(y)` |
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Answer» Correct Answer - B::D If direction of approching orbital is not proper then it is known as zero overtapping. In `2s - 2p_(z)` and `1s - 2p_(y)` direction of approching orbital is not proper. |
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| 3223. |
How many kg of Na2SO4 will be produced using 80 kg of NaCl? Assume CaSO4 in excess.(a) 122 kg(b) 142 kg(c) 162 kg(d) 182 kg |
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Answer» The correct answer is (b) 142 kg Explanation: 2 mole of NaCl = 1 mole of Na2SO4. |
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| 3224. |
How many pounds of NaCl required producing 2.5 pound moles of Na2SO4? Assume CaSO4 in excess.(a) 200(b) 400(c) 90800(d) 181600 |
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Answer» Right answer is (a) 200 For explanation: 2 pound mole of NaCl = 1 pound mole of Na2SO4. |
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| 3225. |
2 moles of an ideal monoatomic gas undergo a reversible process for which `PV^(2)=` constant. The gas sample is made to expand from initial volume of 1 litre of final volume of 3 litre starting from initial temperature of 300 K. find the value of `DeltaS_("sys")` for the above process. Report your answer as `Y` where `DeltaS_("sys")=-YRIn3` |
| Answer» Correct Answer - 1 | |
| 3226. |
Taj Mahal is being slowly disfigured and discoloured. This is primarily due to :(1) global warming (2) acid rain(3) water pollution(4) soil pollution |
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Answer» Correct option (2) acid rain Explanation: Acid rain reacts with marble, CaCO3 of Taj Mahal causing damage to monument and Monument is being slowly disfigured and the marble is getting discoloured and lustreless. CaCO3 + H2SO4 → CaSO4 + H2O + CO2 |
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| 3227. |
Vector product of three vectors is given by `vec(A)xx(vec(B)xxvec(C))=vec(B)(vec(A).vec(C))-vec(C)(vec(A).vec(B))` The plane of vector `vec(A)xx(vec(A)xxvec(B))` lies in the plane ofA. `vec(A)`B. `vec(B)`C. `vec(A)xxvec(B)`D. `vec(A)` & `vec(B)` |
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Answer» Correct Answer - D `vec(A)xx(vec(A)xxvec(B))=vec(A)(vec(A).vec(B))-vec(B)(vec(A).vec(A))rArr` this vector lies in plane of `vec(A)` & `vec(B)` |
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| 3228. |
Vector product of three vectors is given by `vec(A)xx(vec(B)xxvec(C))=vec(B)(vec(A).vec(C))-vec(C)(vec(A).vec(B))` The value of `hat(i)xx(hat(i)xxhat(j))+hat(j)xx(hat(j)xxhat(k))+hat(k)xx(hat(k)xxhat(i))` isA. `hat(i)+hat(j)+hat(k)`B. `-hat(i)-hat(j)-hat(k)`C. `vec(0)`D. `-3hat(i)-3hat(j)-3hat(k)` |
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Answer» Correct Answer - B `Sigmahat(i)xx(hat(i)xxhat(j))=Sigmahat(i)(hat(i).hat(j))-hat(j)(hat(i).hat(i))=-Sigmahat(j)=-(hat(i)+hat(j)+hat(k))` |
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| 3229. |
Ohm’s law fails in case of Option(a) Thyristor (b) pn junction crystal (c) Electrolytes (d) All of these |
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Answer» Correct option(d) Explanation: Ohm’s Law fails in case of nonohmic conductors or semiconductors and in case of elctrolytes. |
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| 3230. |
If temperature is decreased, then relaxation of electrons in metals will(a) Increase(b) Decrease(c) Fluctuate(d) Remain constant |
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Answer» Correct option(a) Explanation: When the temperature decreases K.E of electrons decreases, therefore average free path between two successive collisions increases, Hence relaxation time of electrons in metal will increase. |
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| 3231. |
The resistivity of three metals namely iron, silver and mercury are `10xx10^(-8) Omegam, 1.6xx10^(-8) Omega` and `94xx10^(-8) Omega`m respectively. Answer the following using given data (a) Which among iron, silver and mercury is the best conductor ? (b) If the length of the iron conductor is doubled, what will happen to its resistivity ? Given reason for your answers. |
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Answer» (a) Silver is the best conductor, as its resistivity is the least among these metals. (b) The resistivity of iron will remain the same, as resistivity of the material is a constant value for that material. |
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| 3232. |
Name the process by which unsaturated fats are changed to saturated fats |
| Answer» Hydrogenation | |
| 3233. |
Saponification is: a) Alkaline hydrolysis of lipids b) Alkaline hydrolysis of glycerol c) Esterification d) Reduction |
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Answer» Answer: a) Alkaline hydrolysis of lipids |
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| 3234. |
Concerning grids: a. They may be focused or unfocused b. They may produce artefacts on the final image c. Stationary grids are commonly employed d. The higher the grid factor, the greater the exposure required to produce a film of the same radio graphic density e. The effect of grids on scatter depends mainly on the grid ratio |
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Answer» a. True. This describes the orientation of the lead strips in relation to the path of the X-ray beam. As X-rays are divergent from their source, focused grids have their strips tilted to allow more of the primary radiation to pass through the grid. b. True. Shadows from the lead strips may be superimposed on the final image. c. False. As described above, shadows may be projected onto the final image. This effect may be reduced by using moving grids, which oscillate during the exposure. These are more commonly used than stationary grids. d. True. This is a ratio of the exposures required with and without a grid. e. True. The grid ratio describes the effect of the grid in reducing scatter. The higher the grid ratio, the more scatter will be removed. |
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| 3235. |
Why soap is not suitable for washing clothes with hard water? |
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Answer» Soap reacts with calcium and magnesium ions in water to form insoluble precipitates, which is known as scum. Scum sticks to clothes and Interferes with cleaning ability of soap. RCOONa+ Ca2+ or Mg2 →(RCOO)2 Mg ↓or (RCOO)2 Ca ↓ |
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| 3236. |
Along the period `(rarr)` atomic`//`ionic radii and metallic character decreases while `IE`, `EN` , non-metallic character and oxidising power increases. Down the group `(darr)`, atomic`//`ionic radii, metallic character and reducing character increase while `IE` and `EN` decrease. However, `Delta_(eg)H^(ɵ)` becomes less negative down a group but more negative along a period. Correct order of `IE_(2)` of the following isA. `F gt O gt N gt C`B. `O gt N gt F gt C`C. `O gt F gt N gt C`D. `C gt N gt O gt F`. |
| Answer» Correct Answer - C | |
| 3237. |
Analysis show that nickel oxide consists of nickel ion with `96%` ions having `d^(8)` configuration and `4%` having `d^(7)` configuration. Which amongst the following best represents the formula of the oxide?A. `Ni_(1.02)O_(1.00)`B. `Ni_(0.96)O_(1.00)`C. `Ni_(0.98)O_(0.98)`D. `Ni_(0.98)O_(1.00)` |
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Answer» Correct Answer - D `d^(8)toNi^(2+) " " d^(7)toNi^(3+)` Total charge of nickel. `(0.96xx2)+(0.04xx3)=2.04` No of `O^(2-)` ion`=2.04/2=1.02` Formula of solid`=NiO_(1.02)=Ni_(0.98)O` |
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| 3238. |
The power bandwidth increases when a. Frequency decreases b. Peak value decreases c. Initial slope decreases d. Voltage gain increases |
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Answer» (b) Peak value decreases |
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| 3239. |
A 741C cannot work without a. Discrete resistors b. Passive loading c. Dc return paths on the two bases d. A small coupling capacitor |
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Answer» (c) Dc return paths on the two bases |
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| 3240. |
A 741C uses a. Discrete resistors b. Inductors c. Active-load resistors d. A large coupling capacitor |
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Answer» (c) Active-load resistors |
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| 3241. |
The 741C has a unity-gain frequency of a. 10 Hz b. 20 kHz c. 1 MHz d. 15 MHz |
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Answer» The correct answer is: (c) 1 MHz |
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| 3242. |
The open-loop cutoff frequency of a 741C is controlled by a. A coupling capacitor b. The output short circuit current c. The power bandwidth d. A compensating capacitor |
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Answer» (d) A compensating capacitor |
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| 3243. |
A 741C has a. A voltage gain of 100,000 b. An input impedance of 2 Mohm c. An output impedance of 75 ohm d. All of the above |
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Answer» (d) All of the above |
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| 3244. |
The elliptic approximation has a a. Slow roll off rate compared to the Cauer b. Rippled stopband c. Maximally-flat passband d. Monotonic stopband |
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Answer» (b) Rippled stopband |
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| 3245. |
Above the cutoff frequency, the voltage gain of a 741C decreases approximately a. 10 dB per decade b. 20 dB per octave c. 10 dB per octave d. 20 dB per decade |
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Answer» (d) 20 dB per decade |
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| 3246. |
The all-pass filter is used when a. High rolloff rates are needed b. Phase shift is important c. A maximally-flat passband is needed d. A rippled stopband is important |
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Answer» (b) Phase shift is important |
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| 3247. |
The all-pass filter has a. No passband b. One stopband c. the same gain at all frequencies d. a fast rolloff above cutoff |
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Answer» (c) the same gain at all frequencies |
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| 3248. |
If n = 10, the approximation with the fastest rolloff in the transition region is a. Butterworth b. Chebyshev c. Inverse Chebyshev d. Elliptic |
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Answer» The answer is: (d) Elliptic |
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| 3249. |
If a Butterworth filter has 9 second-order stages, its rolloff rate is a. 20 dB per decade b. 40 dB per decade c. 180 dB per decade d. 360 dB per decade |
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Answer» (d) 360 dB per decade |
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| 3250. |
The center frequency of a bandpass filter is always equal to a. The bandwidth b. Geometric average of the cutoff frequencies c. Bandwidth divided by Q d. 3-dB frequency |
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Answer» (b) Geometric average of the cutoff frequencies |
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