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.
| 9151. |
Using the equation of state pV = nRT show that at a given temperature, the density of the gas is proportional to the gas pressure p. |
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Answer» According to ideal gas equation : pV = nRT or p = \(\frac{nRT}{V}\) n = \(\frac{Mass\ of\ gas(m)}{Molar\ mass\ of\ gas(M)}\) p = \(\frac{mRT}{MV}\) Now density, d = \(\frac{m}{V}\) ∴ p = \(\frac{dRT}{M}\) Or d ∝ p at constant temperature. |
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| 9152. |
The average kinetic k energy of the gas molecules is directly proportional to the absolute temperature.1. Which theory is related to this assumption? 2. Write the other postulates of this theory. |
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Answer» 1. Kinetic molecular theory of gases. 2.
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| 9153. |
Copper is the most noble of the first transition metals and occurs in small deposits in several countries. Ores of copper include chalcanthite `(CuSO_(4).5H_(2)O)`, atacamite `(Cu_(2)Cl(OH)_(3))`, cuprite `(Cu_(2)O)`, copper glance `(Cu_(2)S)` and malachite `(Cu_(2)(OH)_(2)CO_(3))`. Howerver `80%` of the world copper production comes from the ore chalcopyrites `(CuFeS_(2))` the extraction of copper from chalcopyrites involves partial roasting of iron and self reduction. Q. In self reduction the reducing species is .... Select the correct options for blank.A. `Cu^(+)`B. `O^(-2)`C. `S^(-2)`D. `SiO_(2)` |
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Answer» Correct Answer - C::D `2CuFeS_(2)+4O_(2)overset(Delta)toCu_(2)S+2FeO+3SO_(2)` `Cu_(2)S+FeO+SiO_(2)toFeSiO_(3)` (Fusible slag) `+Cu_(2)S` (matte) |
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| 9154. |
Copper is the most noble of the first row transition elements it occurs in small deposits in several countries. Ores of Copper include chalcanthite `(CuSO_4.5H_2O)`, atacamite `[Cu_2Cl(OH)_3]`, cuprite `(Cu_2O)`, copper glance `(Cu_2S)`, and malachite `[Cu_2(OH)_2CO_3]`. However, `80%` of the world copper production comes from the ore chalcopyrite `(CuFeS_2)`. Extraction of copper from chalcopyrite includes roasting, iron removal, and self-reduction. Q. Partial roasting of chalcopyrite producesA. `Cu_(2)O and FeO`B. `Cu_(2)S and FeO`C. `CuO and Fe_(2)O_(3)`D. `CuO and FeS` |
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Answer» Correct Answer - B Extraction of copper |
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| 9155. |
Partial pressure of a vessel containing Cl2 , CO2 and CO is the sum of the partial pressures of Cl2 , O2 and CO. 1. If so, is it correct to say partial pressure of a vessel containing NH3 and HCl gases is the sum of their partial pressures? Justify. 2. Which law helped you to answer this? 3. State the law. |
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Answer» 1. No. NH3 reacts with HCl to form NH4CI. Since they are not non-interacting gases, their sum of partial pressures may not be equal to the total pressure. 2. Dalton’s law of partial pressures. 3. It states that the total pressure exerted by the mixture of non-reactive gases is equal to the sum of the partial pressures of individual gases. |
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| 9156. |
Copper is the most noble of the first transition metals and occurs in small deposits in several countries. Ores of copper include chalcanthite `(CuSO_(4).5H_(2)O)`, atacamite `(Cu_(2)Cl(OH)_(3))`, cuprite `(Cu_(2)O)`, copper glance `(Cu_(2)S)` and malachite `(Cu_(2)(OH)_(2)CO_(3))`. Howerver `80%` of the world copper production comes from the ore chalcopyrites `(CuFeS_(2))` the extraction of copper from chalcopyrites involves partial roasting of iron and self reduction. Q. Partial roasting of chalcopyrites with silica added to it, produces,A. `FeSiO_(3)`B. `Fe_(2)O_(3)`C. `Cu_(2)S`D. `CuO` |
| Answer» Correct Answer - A::C | |
| 9157. |
Assume that ‘A’, ‘B’ and ‘C’ are three nonreacting gases kept in a vessel at a constant temperature.Then, PTotal = PA + PB + PC 1. Name the related law. 2. How can you explain the above law on the basis of kinetic molecular theory of gases? |
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Answer» 1. Dalton’s law of partial pressures. 2. In the absence of attractive forces, the particles of the gas behave independent of one another. The same is true even if there are more than one type of molecules. Thus, the number of molecules colloding the unit area of the wall per second at a given temperature, fora fixed amount of the gas issame.lt implies that the partial pressure of the gas will be unaffected by the presence of the molecules of other gases. But, the total pressure exerted is duet the impact of molecules of all the gases. Hence, the total pressure would be the sum of the partial pressures of the gases. |
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| 9158. |
Real gases behave ideally at:A. low pressure and low temperatureB. high pressure and low temperatureC. low pressure and high temperatureD. high pressure and high temperature |
| Answer» Correct Answer - C | |
| 9159. |
Real gases behave ideally at low temperature and high pressure. 1. Is the above statement correct or not?2. Justify. 3. Write the van der Waals’ equation for 1 mole of a real gas. |
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Answer» 1. The statement is wrong. 2. This is because real gases behave ideally at high temperature and low pressure, when the gas molecules are far apart. 3. \(\Big(p+\frac{a}{V^2}\Big)\) (v-b) = RT |
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| 9160. |
What correction is applied to obtain pressure of dry gas with total pressure and aqueous tension? |
| Answer» P(dry gas) = P(Total) - Aqueous tension. | |
| 9161. |
Which of the following compound is violating octet Rule.A. `AlF_(3)`B. `PCl_(3)`C. `NH_(4)^(+)`D. `SF_(4)` |
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Answer» Correct Answer - B `CH_(3)COOH + NaOH rarrCH_(3)COONa+H_(2)O` `{:(0.2. 2L,0.1 M. 3L,,),(0.4,0.3 ,,),(0.1,-,0.3,):}` `pH = pKa+log.(S)/(A)` `=3.6 + 0.48` =4.08 pOH = 13 - 4.08 = 8.92 |
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| 9162. |
The dissociation equilibrium of a gas `AB_(2)` can be represented as, `2AB_(2)(g)iff2AB(g)+B_(2)(g)`. The degree of dissociation is x and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant `k_(p)` and total pressure P is :A. `(k_(p))/(P)`B. `(2k_(p))/(P)`C. `((2k_(p))/(P))^(1//3)`D. `((2k_(p))/(P))^(1//2)` |
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Answer» Correct Answer - D `BeCl_(2) lt AlCl_(3) lt SiCl_(4)` `C Cl_(4)ltPCl_(5)ltPF_(5)` `Na^(+) lt Mg^(2+) lt Al^(3+)` |
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| 9163. |
Calculate E.A.N. (Effective atomic number) of Fe in this complex salt. |
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Answer» Let the oxidation no. of Fe = x 1 x 3 + X x 1 (–1) x 6 = 0 x = +3 EAN = No. of electron in Fe3+ + No. of electron in 6 CN- ion = (26 – 3) + 12 = 35 :. EAN = 35 |
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| 9164. |
Calculate E.A.N. (Effective Atomic Number) of Fe in this complex salt. |
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Answer» EAN of Fe in K3[Fe(CN)6] = 26 - 2 + 2 x 6 = 36 [Ar] |
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| 9165. |
Explain Why H2O is in liquid state while H2S is in gaseous state at room condition. |
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Answer» At room temp, H2O form intermoleculer H-bond while H2S dues not; Because Electronegativity of 0-atom is greater while size is smaller then s-atom Hence H2O is liquid while H2S 10 gas. |
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| 9166. |
Give one test from which methyl alcohol and ethyl alcohol are distinguished. |
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Answer» Iodoform test: On heating ethyl alcohol, with I, & NaOH gives yellow PPt of iodoform which has foul smell. C2H5OH + 3I2 + 4NaOH → HCl3↓ + HCOONa Yellow PPt (Iodoform) + 3HI + 3H2O While Methyl alcohol does not give this test |
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| 9167. |
Explain why H2O is neutral whereas H2S is acid. |
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Answer» The size of central atom H2S is larger than that in H2O2S — H bond is larger and weaker than O-H bond H2O, hence H2S is more acidic than H2O. |
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| 9168. |
Why Gd3+ is colourless? |
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Answer» Gd – Electronic Configuration : [Xe] 4f75d1 6s2 Gd3+ Electronic Configuration : [Xe] 4f7 In Gd3+ , no electrons are there in outer d-orbitals. d-d transition is not possible. So it is colourless. |
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| 9169. |
Mention the uses of the potash alum. |
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Answer» The uses of the potash alum:
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| 9170. |
What is the decreasing order of basicity?(a) NH3 > C2H5NH2 > (C2H5)2NH > (C2H5)3N(b) (C2H5)3N > (C2H5)2NH > C2H5NH2 > NH3(c) (C2H5)2NH > C2H5NH2 > (C2H5)3N > NH3(d) (C2H5)2NH > C2H5NH2 > NH3 > (C2H5)3N |
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Answer» Answer is (b) (C2H5)3N > (C2H5)2NH > C2H5NH2 > NH3 |
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| 9171. |
How will you prepare potash alum? |
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Answer» The alunite the alum stone is the naturally occurring form and it is K2SO4 . Al2(SO4) .4Al(OH)3 . When alum stone is treated with excess of sulphuric acid, the aluminium hydroxide is converted to aluminium sulphate. A calculated quality of – potassium sulphate is added and the solution is crystallised to generate potash alum. It is purified by recrystallisation. K2SO4.Al2(SO4)3 .4Al(OH)3 + 6H2SO4 ⟶ K2SO4 + Al2(SO4)3 + 12H2O K2SO4 + Al2(SO4)3 +24H2O ⟶K2SO4.Al2(SO4)3.24H2O |
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| 9172. |
Eye lense are manufactured by(a) PVC(b) Teflon(c) Buna-N(d) PMMA |
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Answer» Eye lense are manufactured by PMMA |
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| 9173. |
Which of the following undergoes Cannizaro's reaction?(a) CH3CHO(b) CH3CH2CHO(c) (CH3)2CHCHO(d) HCHO |
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Answer» Answer is (a) CH3CHO |
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| 9174. |
Which compound does not perform Cannizaro’s reaction among ? |
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Answer» CH3 – CHO compound does not perform Cannizaro’s reaction among . |
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| 9175. |
Chlorine gas is passed through dilute NaOH solution. The compounds formed are-(a) NaCl(b) NaOCl(c) NaCl & NaOCl (d) NaCl and NaClO3 |
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Answer» The compounds formed are NaCl & NaOCl |
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| 9176. |
Which compound does not give colourless gas?(a) Cu(NO3)2 (b) NaNO3(c) Ca(NO3)2(d) Al(NO3)2 |
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Answer» NaNO3 compound does not give colourless gas. |
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| 9177. |
Ammonia gas is passed through aqueous CuSO4 solution produces deep blue colouration. The molecular formula of formed deep blue colouration is(a) CuSO4 .NH3(b) CuSO4 .4NH3(c) [Cu(NH3)4] SO4(d) [Cu(NH3)6] SO4 |
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Answer» The molecular formula of formed deep blue colouration is [Cu(NH3)4] SO4 |
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| 9178. |
Heat exchanged in a chemical reaction at constant temperature and pressure is called(a) entropy(b) enthalpy (c) internal energy (d) free energy |
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Answer» (b) enthalpy |
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| 9179. |
Calculate the e.m.f. of the following cell at `298K :` `Pt(s)|Br_(2)(l)|Br^(-)(0.010M)||H^(+)(0.030M)|H_(2)(g)(1"bar")|Pt(s)` Given `: E_((1)/(2)Br_(2)//Br^(-))^(@)=+1.08V`.A. `-1.288V`B. `1.288V`C. `0.128V`D. `-128V` |
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Answer» Correct Answer - 2 Cell reaction `:` `2Br^(-)+2H^(+)rarrBr_(2)+H_(2)` Nernst eqn. `E_(cell)=E_(cell)^(@)-(0.0591)/(2)log.(1)/([Br^(-2)][H^(+)]^(2))` `:. " "E_(cell)=(0-1.08)-(0.0591)/(2)log.(1)/((0.01)^(2)(0.03)^(2))=-1.08-(0.0591)/(2)log(1.111xx10^(7))` `=-1.08-(0.0591)/(2)(7.0457)=-1.08-0.208=-1.288V` Thus, oxidation will occur at the hydrogen electrode and reduction on the `Br_(2)` electrode `E_(cell)=1.288V`. |
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| 9180. |
The colligative properties of a dilute solution depend upon – (A) the nature of the solute (B) the nature of solvent (C) the number of particles of solute (D) the number of particles of solvent |
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Answer» Correct answer is (C) the number of particles of solute |
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| 9181. |
The osmotic pressure of a solution is expressed by the relation –(A) π = CR/T(B) π/C = RT(C) π = CT/R(D) π = RT/C |
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Answer» Correct answer is (B) π/C = RT |
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| 9182. |
The amount of electricity required to deposit one mole of Copper from Copper sulphate solution is – (A) 1 Faraday (B) 2.33 Faraday (C) 2 Faraday (D) 1.33 Faraday |
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Answer» Correct answer is (C) 2 Faraday |
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| 9183. |
If two waves having amplitudes a and b superimposes on each other and resultant amplitude is A = √(a2 +b2 +2ab cosϕ), where ϕ = phase constant.Intensity (I) is directly proportional to square of amplitudeI ∝ A2I ∝(a2 +b2 + 2ab cosϕ)In constructive interference Imax = (a +b)2In destructive interference lmin = (a -b)2Light waves from two coherent sources producing interference have intensity ratio 81:1.Using these information choose correct option from the following :The ratio of maxima and minima in the interference pattern is(a) 9: 1(b) 81 : 1(c) 25 : 16(d) 16 : 25 |
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Answer» Correct answer is (b) 81 : 1 |
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| 9184. |
If two waves having amplitudes a and b superimposes on each other and resultant amplitude is A = √(a2 +b2 +2ab cosϕ), where ϕ = phase constant.Intensity (I) is directly proportional to square of amplitudeI ∝ A2I ∝(a2 +b2 + 2ab cosϕ)In constructive interference Imax = (a +b)2In destructive interference lmin = (a -b)2Light waves from two coherent sources producing interference have intensity ratio 81:1.Using these information choose correct option from the following :(i) The ratio of amplitude of light waves from two sources is(a) 9:1 (b) 81 : 1(c) 1: 9 (d) 1 : 81(ii) For maximum intensity the value of cosϕ is(a) -1(b) +1(c) ∞(d) 0 |
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Answer» Correct answer is (i)(a) (ii) b |
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| 9185. |
Define critical angle and write its conditions. |
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Answer» The angle of incidene corresponding to which angle of refraction becomes 90º is called critical angle. Conditions: (i) Light should travell from denser medium to rarer medium (ii) For total internal reflection angle of incidence must be greater than critical angle. |
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| 9186. |
Write two differences between electromagnet and permanent magnet. |
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Answer»
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| 9187. |
What is doping ? |
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Answer» The process of adding desirable impurities in the intrinsic semiconductor is called doping. |
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| 9188. |
The atomic mass number of Radium is `A=226`, its half life is `1622` years. What is the activity of `1g` Radium ?A. `3.6xx10^(10)Bq`B. `0.6xx10^(10)Bq`C. `1.6xx10^(10)Bq`D. `2.6xx10^(10)Bq` |
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Answer» Acitivity `=(dN)/(dt)=lambdaN` `lambda=(0.693)/(t_(1//2))` `N=` number of radium atoms |
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| 9189. |
Why is core of transformer laminated ? |
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Answer» The core of a transformer is laminated to insulate the core due to which the current induced in the iron core due to alternating magnetic flux can be minimised. |
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| 9190. |
What is polaroid? Write its uses. |
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Answer» Polaroid is a device used to produce the plane polarised light. Use of polaroid (i) It is used as glass windows in train and aeroplanes. (ii) lt is used in sun glasses (goggles) to block the glare. (iii) It is used to record and reproduce three dimenional moving pictures. (iv) It is used to elimirate the daggle from the head lights of cars, buses etc. |
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| 9191. |
A manufacturing plant, working in 2 shifts of 8 hrs each, produces 30,000 switch boards using a set of 7 workstations in series. The cycle time, in seconds, is _______ |
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Answer» Concept : Net Production Time = Cycle Time \(\times\) Number of Units Produced Calculation: Given: Net Production Time = \(2\times8 \times3600\) = 57600 sec Number of Units Produced = 30000 \(Cycle\; time = \frac{Net \;Production\; time}{Number\;of\;Units\;Produced}\) \(Cycle\; time = \frac{57600}{30000} =1.9\) ∴ Cycle time = 1.9 sec |
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| 9192. |
Abhilasha is working in an organisation. After every quarter, the manager shifts her from one department to another, so that she may gain a broader understanding of all the departments. Name and explain the method of training she is undergoing. |
| Answer» Job rotation Under this method of training, an individual is transferred from one job to another, one department to another or one shift to another. This enables the trainee to gain a broader understanding of all parts of a business. The trainee gets a chance to test her own aptitude and ability. It also allows trainees to interact with other employees facilitating future cooperation among departments. | |
| 9193. |
Two blocks `A` and `B` of masses `m` and `2m` respectively are connected together by a light spring of stiffness `k` and then placed on a smooth horizontal surface. The blocks are pushed towards each other such that spring gets compressed by a length `x_(0)` and then released from rest. Find the work done on the block `A` by the spring, by the time the spring acquires its natural length. |
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Answer» from `COM` `mv_(A)=2mv_(B)` `rArr v_(B)=(v_(A))/(2)` from `COE` `(1)/(2)kx_(0)^(2)=(1)/(2)mv_(A)^(2)+(1)/(2)xx2m(v_(A)^(2))/(4)=(1)/(2)mv_(A)^(2)(1+(1)/(2))` `rArr(1)/(2)kx_(0)^(2)=(1)/(2)mv_(A)^(2)xx(3)/(2)rArr(1)/(2)mv_(A)^(2)=(kx_(0)^(2))/(3)` |
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| 9194. |
According to Maxwell`-`distribution law, the probability function representing the ratio of molecules at a particular velocity to the total number of molecules is given by `f(v)=k_(1)sqrt(((m)/(2piKT^(2))))4piv^(2)e^(-(mv^(2))/(2KT))` Where `m` is the mass of the molecule, `v` is the velocity of the molecule, `T` is the temperature `k` and `k_(1)` are constant. The dimensional formulae of `k_(1)` isA. `L^(2)T^(-2)`B. `L^(1)T^(-1)K^(-3//2)`C. `L^(1)T^(-1)K^(+3//2)`D. `L^(2)T^(-1)K^(+3//2)` |
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Answer» Correct Answer - C Here `f(v)` is dimensionless and `[mv^(2)]=[kt]` `[k_(i)]=[((2pikT^(2))/(m))^( 1.5)(1)/(4piv^(2))]=[vT^(1.5)]=[L^(1)T^(-1)K^(1.5)]` |
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| 9195. |
N similar slabs of cubical shape of edge b are lying on ground. Density of material of slab is p. Work done ot arrange them one over the other is - A. `(N^(2)-1)b^(3)pg`B. `(N-1)b^(4)pg`C. `1//2(N^(2)-N)b^(4)pg`D. `(N^(2)-N)b^(4)pg` |
| Answer» Correct Answer - A | |
| 9196. |
A block of mass m=1/3 kg is kept on a rough horizontal plane. Friction coefficient is `mu=0.75`. The work done by minimum force required to drag to the block along the plane by a distance 5m is :-A. 8 JB. 4 JC. 2 JD. 6 J |
| Answer» Correct Answer - A | |
| 9197. |
A ball collides elastically with a massive wall moving towards it with a velocity of `v` as shown. The collision occurs at a height of `h` above ground level and the velocity of the ball just before collision is `2v` in horizontal direction. The distance between the foot of the wall and the point on the ground where the ball lands, at the instant the ball lands, will be `:` A. `vsqrt((2h)/(g))`B. `2vsqrt((2h)/(g))`C. `4vsqrt((2h)/(g))`D. `3vsqrt((2h)/(g))` |
| Answer» Correct Answer - A | |
| 9198. |
Consider the equation `d/(dt)(int vec(F). dvec(S))=A(vec(F).vec(p))` where `vec(F) equiv` force, `vec(s) equiv` displacement, t `equiv` time and `vec(p)`= momentum. The dimensional formula of A will beA. `M^(0)L^(0)T^(0)`B. `M^(1)L^(0)T^(0)`C. `M^(-1)L^(0)T^(0)`D. `M^(0)L^(0)T^(-1)` |
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Answer» Correct Answer - C Appying dimensional analysis `([MLT^(-2)][L])/([MLT^(-1)][MLT^(-2)]) = A` `=[M^(-1)L^(0)T^(0)] = A` |
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| 9199. |
An ant starts from origin and moves to the mid- point of CD edge and stops there. It then moves to the mid-point of GF edge and again stops there. If P and Q are the midpoint fo CD and GF respectively, then find `bar(AB) + bar(PQ)`. The figure is a cube of side `1 m`. A. `hat(i) + hat(j) + (hat(k))/(2)`B. `hat(i) - hat(j) + (hat(k))/(2)`C. `- hat(i) - hat(j) - (hat(k))/(2)`D. `hat(i) + hat(j) - (hat(k))/(2)` |
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Answer» Correct Answer - A `bar(AP) + bar(PQ) = bar(AQ)` `A (0, 0, 0)` `Q (1, 1, (1)/(2))` So `bar(AQ) = hat(i) + hat(j) + (hat(k))/(2)` |
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| 9200. |
A ball collides elastically with a massive wall moving towards it with a velocity of `v` as shown. The collision occurs at a height of `h` above ground level and the velocity of the ball just before collision is `2v` in horizontal direction. The distance between the foot of the wall and the point on the ground where the ball lands, at the instant the ball lands, will be `:` A. `vsqrt((2h)/(g))`B. `2vsqrt((2h)/(g))`C. `3v sqrt((2h)/(g))`D. `4v sqrt((2h)/(g))` |
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Answer» Correct Answer - C Solve in the reference frame fixed to the wall. Before collision, velocity of ball `=3v` towards it . `:.` After elastic collision of ball`=3v` away from it Time of flight `=sqrt((2h)/(g))` `:.` distance between wall and ball `=3v. sqrt((2h)/(g)).` `(` Here no pseudo force is applied since the wall keeps on moving with constant velocity `w.r.t.` ground, it being very heavy `)`. |
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