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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.
| 17451. |
The reaction `2NO_((g)) + C1_(2(g)) to 2NOC1_((g))` is first order in `C1_(2)` and second order in `NO` Write the rate law for the reaction |
| Answer» Rate law : Rate = `k[NO]^(2) [C1_(2)]` | |
| 17452. |
The reaction `F_(2(g)) + 2C1O_(2(g)) to 2FC1O_(2(g))` if first order in each of the reactants . The rate of the the reaction is `4.88 xx 10^(-4) M//s` when `[F_(2)] =0.015 M` and `[C1O_(2) =0 .025 M` Calculate the rate constant of the reaction . |
| Answer» Rate constant = `k = 1.3 M^(-1) s^(-1)` | |
| 17453. |
Explain the following related to the West Asian problems. |
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Answer» a. The idea of establishing a jewish homeland, b. Establishing a nation for Palestines, c. The agreement mediated by the U.S to recognize Palestine as a free nation. |
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| 17454. |
Which planet besides Neptune is known as the 'Blue Planet'? 1. Earth2. Mars 3. Pluto 4. Jupiter |
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Answer» Correct Answer - Option 1 : Earth The correct answer is Earth.
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| 17455. |
What is the order for the following reactions ? (a) `2NO_(2(g)) + F_(2(g)) to 2NO_(2) F_((g))` rate `= k[NO_(2)][F_(2)]` `(b) CHC1_(3(g)) + C1_(2(g)) to CC1_(4(g)) + HC1_((g))` rate `= k [ CHC1_(3)][C1_(2)]^((1)/(2))` |
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Answer» (a) order of the reaction =2 (b) The order of the reaction `=(3)/(2)` |
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| 17456. |
Complete the table : |
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Answer» a. South Africa, b. Ghana, c. Kenya |
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| 17457. |
Evaluate the causes for the disintegration of Soviet Union. |
Answer»
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| 17458. |
The rate of diffusion of two gases x and y is in the ratio of 1 : 5 and that of y and z in the ratio of 1 : 6 the ratio of the rate of diffusion of z with respect to x is :A. `5 : 6`B. `1 : 30`C. `6 : 5`D. `30 : 1` |
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Answer» Correct Answer - D `(r_(x))/(r_(y))=(1)/(5)" "....(i)" "(r_(g))/(r_(z))=(1)/(6)" "....(ii)` `(i)xx(ii) : (r_(x))/(r_(z))=(1)/(30)implies (r_(z))/(r_(x))=(30)/(1)` |
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| 17459. |
Which of the following does not react with ethers to form coordination complexes ?A. `BF_(2)`B. `RMgX`C. `AICI_(3)`D. `C_2H_5ONa` |
| Answer» Correct Answer - D | |
| 17460. |
What will be normality of `20VH_(2)O_(2)` solution ?A. 1.78 NB. 3.56 NC. 10 ND. 0.28 N |
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Answer» Correct Answer - B `N=(20)/(5.6)=3.56` |
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| 17461. |
Which of the following solution have highest boiling point. (Assume all salts completely dissociates)A. 0.1 M `Al_(2)(SO_(4))_(3)`B. 0.1 `MBaCI_(2)`C. 0.1 M glucoseD. 0.1 `M AlCI_(3)` |
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Answer» Correct Answer - A ic is max.for `Al_(2)(SO_(4))_(3)` |
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| 17462. |
`Cl_(2)` can disproportionate in.A. `NaOH,HF`B. `K_(2)MnO_(4),NaOH`C. `K_(2)MnO_(4),HF`D. `Ca(OH)_(2)H_(2)O` |
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Answer» Correct Answer - D `Cl_(2)+K_(2)MnO_(4)rarrKMnO_(4)+KCl` `Cl_(2)+HFrarr"No reaction"` `Cl_(2)+H_(2)OrarrHCl+HOCl` `Cl_(2)+NaOHrarrNaCl+NaOCl+H_(2)O` |
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| 17463. |
What will be the freezing point `("in"^(@)C)` of solution obtained by dissolving 0.1 g potassium ferricyanide (mol wt = 329) in 100 g water. If `K_(f)` for water is `1.86 K kg mol^(-1)`A. `-2.3xx10^(-2)`B. `-5.7xx10^(-2)`C. `-5.7xx10^(-3)`D. `-1.2xx10^(-2)` |
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Answer» Correct Answer - A `DeltaT_(f)=4xx(0.1)/(329)xx(1000)/(100)xx1.86=2.26xx10^(-2)` `K_(3)[Fe(CN)_(6)]:.i=4` |
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| 17464. |
Ka for `H_(3)BO_(3) "is" 5.8xx10^(-10)` what buffer component ratio `[NaH_(2)BO_(3)(aq.)]//[H_(3)BO_(3)(aq.)]` would be needed to make a buffer solution of pH=10.A. 10B. 0.18C. 4.3D. 5.8 |
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Answer» Correct Answer - D `pH=pKa+log.([NaH_(2)BO_(3)])/([H_(3)BO_(3)])` `[H^(+)]=Kaxx([H_(3)BO_(3)])/([NaH_(2)BO_(3)])` `10^(-10)=5.8xx10^(-10)xx([H_(3)BO_(3)])/([NaH_(2)BO_(3)])` |
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| 17465. |
The correct sequence of the stability of dihalides of group –14 elements is:(a) SiX2 < SnX2 < GeX2 < PbX2(b) SiX2 > SnX2 >GeX2 > PbX2(c) SiX2 > GeX2 > SnX2 > PbX2(d) SiX2 < GeX2 < SnX2 < PbX2 |
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Answer» Correct option (d) SiX2 < GeX2 < SnX2 < PbX2 Explanation: As due to inert pair effect stability of divalent compound increase ↓ gp |
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| 17466. |
A solution of substance containing `1.05` g per 100 mL was found to be isotonic with `3%` glucose solution . The molecular mass of the substance is :A. 31.5B. 6.3C. 630D. 63 |
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Answer» Correct Answer - D `(1.05)/(Mw)xx10=(3)/(180)xx10` `Mw=1.05xx60=63` |
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| 17467. |
Which is not possible product of heating of ortho boric acid.A. `H_(2)B_(2)O_(6)`B. `H_(2)B_(4)O_(7)`C. `HBO_(2)`D. `B_(2)O_(3)` |
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Answer» Correct Answer - A `H_(3)BO_(3)overset(Delta)(rarr)HBO_(2)overset(Delta)(rarr)H_(2)B_(4)O_(7)overset(Delta)(rarr)B_(2)O_(3)` |
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| 17468. |
Ethers are not distilled to dryness for fear of explosion. This is due to formation ofA. oxidesB. peroxidesC. AlcoholsD. ketones |
| Answer» Correct Answer - B | |
| 17469. |
A solution of substance containing `1.05` g per 100 mL was found to be isotonic with `3%` glucose solution . The molecular mass of the substance is :A. 31.54B. 6.3 uC. 630 uD. 63 u |
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Answer» Correct Answer - D `pi_(x)=pi_("glucose")` `iC_(x)RT=iC_("glucose")RT` `C_(x)=C_("glucose")` `(1.05)/(MM)xx(1000)/(100)=(3g)/(180)xx(1000)/(100)` `MM=63` |
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| 17470. |
In a multielectron atom, which of the following orbitals described by three quantum numbers will have same energy in the absence of magnetic field and electric field : (I) n = 1, l = 0, m = 0(II) n = 2, l = 0, m = 0(III) n = 3, l = 1, m = 1(IV) n = 3, l = 2, m = 1(V) n = 3, l = 2, m = 0(a) I and II(b) II and III(c) III and IV(d) IV and V |
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Answer» Correct option (d) IV and V Explanation: Energies of orbitals are compared by ‘n + l’. In IV and V cases ‘n + l’ values are 5 and n value are 3 |
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| 17471. |
`R-overset(O)overset(||)(C )-O-Hoverset(x)(rarr)R-CH_(2)-OH` X can beA. `LiAlH_(4)//"either",H_(3)O^(+)`B. `B_(2)H_(6),H_(3)O^(+)`C. `NaOH & CaO,Delta`D. Both (A) and (B) |
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Answer» Correct Answer - D `R-overset(O)overset(||)(C )-OHunderset(H_(3)O^(+))overset(B_(2)H_(5))(rarr)RCH_(2)-OH` `R-overset(O)overset(||)(C )-OHoverset(LiAlH_(4))(rarr)RCH_(2)-OH` `R-overset(O)overset(||)(C )-OHunderset(CaO)overset(NaOH)(rarr)R-H` |
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| 17472. |
During reaction of anisole with conc. HI at `375K`, which bond of ether cleaves ?A. `CH_(3)-O`B. `C_6H_5-O`C. `C-C`D. Any of the above |
| Answer» Correct Answer - A | |
| 17473. |
Which salt precipitates as metal hydroxide on reaction with `Na_(2)CO_(3)` solution.A. `FeCl_(3)`B. `CaCl_(2)`C. `AgNO_(3)`D. `(CH_(3)COO)_(2)Pb` |
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Answer» Correct Answer - A `+3` metal cation precipitates in form of hydroxide when react with `Na_(2)CO_(3)` |
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| 17474. |
Isoflurane -an ether is used asA. a flavouring agentB. a solventC. an inhalation anaestheticD. a perfume |
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Answer» Correct Answer - C Isoflurance is used as an inhalation anaesthetic. It is a better anaesthetic than diethyl ehter. `CI-overset(CI)overset(|)CH-C-O-CHF_2` |
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| 17475. |
When aqueous solution of `NH_(4)Cl` reacts with `NaNO_(2)` two gases are formed, one is in large amount and other is in small amount, these gases are respectivelyA. `N_(2),NO`B. `N_(2),N_(2)O`C. `N_(2)O,NO`D. `NO,N_(2)` |
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Answer» Correct Answer - A `NH_(4)Cl+NaNO_(3)rarrNaCl+N_(2)uarr+H_(2)O` in small amount `HNO_(2)` is also formed `HNO_(2)rarrHNO_(3)+NOuarr` |
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| 17476. |
Which of the following cell can produce maximum electrical work? `["Given" : P_(H_(2))=1 atm, K_(a)(CH_(3)COOH)=K_(b)(NH_(4)OH)=10^(-5)]`A. `Pt|H_(2)(g)|0.1MNH_(4)Cl(aq.)||0.1MCH_(3)COOH(aq.)|H_(2)(g)|Pt`B. `Pt|H_(2)(g)|0.1MHNO_(3)(aq.)||0.1MKOH(aq)|H_(2)(g)|Pt`C. `Pt|H_(2)(g)|0.1MHCl(aq.)||0.1MCH_(3)COOK(aq.)|H_(2)(g)|Pt`D. `Pt|H_(2)(g)|0.1MCH_(3)COONa(aq.)||0.1MHCl(aq.)|H_(2)(g)|Pt` |
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Answer» Correct Answer - D For maximum electrical work `E_("cell")` = Maximum `E_("cell") = 0.0591"log" ([H^(+)]_(c))/([H^(+)]_(a))" "[H^(+)]_(c)gt[H^(+)]_(a)` `therefore` Anodic compartment should be more basic and cathodic compartment should be more acidic. |
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| 17477. |
Write Vander Walls equation for one mole of a gas and name any two terms in it. |
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Answer» (P + a/V2)(V - b) = RT |
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| 17478. |
Vanilin is a ring substituted anisole withA. a phenolic groupB. an aldehydic groupC. both (A) and (B)D. an ester group |
| Answer» Correct Answer - C | |
| 17479. |
Why does the following reaction occur? `XeO_(6)^(4-)(aq)+2F^(Θ)(aq)+6H^(o+)(aq) rarr XeO_(3)(g)+F_(2)(g)+3H_(2)O(l)` What conclusion about the compound `Na_(4)XeO_(6)` (of which `XeO_(6)^(4-)` is a part) can be drawn from the reaction? |
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Answer» `overset(+8)(Xe)O_(6)^(4-)(aq)+2F^(Θ)(aq)+6H^(o+)(aq)rarr overset(+6)(Xe)O_(3)(s)+overset(0)(F_(2))(g)+3H_(2)O(l)` Here, oxidation number of `Xe` decreases from `+8` in `XeO_(6)^(4-)` to `+6` in `XeO_(3)`, while that of F increases from `-1` in `F^(Θ)` to `0` in `F_(2)`. Therefore, `XeO_(6)^(4-)` is reduced while `F^(Θ)` is oxidised. This reaction occure because `Na_(2)XeO_(6) (or XeO_(6)^(4-))` is a stronger oxidising agent than `F_(2)`. |
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| 17480. |
In the circuit shown the cell will deliver maximum power to the network if the value of R isA. `(2)/(3)Omega`B. `(4)/(3)Omega`C. `(1)/(3)Omega`D. `(5)/(3)Omega` |
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Answer» Correct Answer - A The cell will deliver maximum power to the network if `R_(eq)=r` `implies(3R)/(2)=r` `thereforeR=(2r)/(3)=(2)/(3)Omega` |
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| 17481. |
Two wires each of radius of cross section `r` but of different materials are connected together end to end (in series). If the densities of charge carries in the two wires are in the ratio `1:4`, the drift velocity of electrons in the two wires will be in the ratio:A. `1 : 2`B. `2 : 1`C. `4 : 1`D. `1 : 4` |
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Answer» Correct Answer - C Two wires carry same current `I=n_(1)eA V_(d_(1)) =n_(2)eA Vd_(d_(2))` so `n_(1)V_(d_(1))=n_(2)V_(d_(2))` `(V_(d_(1)))/(V_(d_(2)))=(n_(2))/(n_(1))=4 rArr V_(d_(1)):V_(d_(2))=4:1`. |
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| 17482. |
Write the basic features of photon pictures of e.m. radiation on which Einstein’s photoelectric equation is based. |
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Answer» Photon picture of em radiations : (i). In interaction of radiation with matter, radiation behaves as if it is made of particles like photons. (ii). Each photon has energy E(= h) and momentum p (=\(\frac{h}{\lambda}\)) (iii). Irrespective of the intensity of radiation, all the photons of a particular frequency and wavelength have the same energy and same momentum. (iv). Photon energy is independent of the intensity of radiations. (v). Photons emitted from a source of radiations travel with the speed of light. (vi). The rest mass of a photon is zero. (vii). Photons are electrically neutral (viii). In a photon particle collision, the energy and momentum are conserved. However, Number of photons may not be conserved, photon may be absorbed or new photon may be created. |
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| 17483. |
I.` 8/sqrtX+6/sqrtX = sqrtX` II.` y^(2) -(14^(5//2)//Y^(1//2)) = 0`A. ` x gt y`B. ` x ge y`C. ` x lt y`D. x = y or a relationship between x and y cannot be established. |
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Answer» Correct Answer - D From I , `(8)/(sqrt(x))+(6)/(sqrt(x))=sqrt(x)` `rArrX=14` From II , `y^(2)(14^(5//2)//y^(1//2))=0` `rArry^(5//2)=14^(5//2` `rArry=14` So , x = y |
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| 17484. |
(i) ` 2x^(3) = sqrt(256)` II.` y^(2) -9y+14 = 0`A. x = y or no relation can be established between x & y.B. ` x lt y`C. ` x le y`D. ` x ge y` |
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Answer» Correct Answer - C (i) `2x^(3)=16` `x^(3)=8` `x=2` (ii) `y^(2)-9y+14=0` `y^(2)-2y-7y+14=0` `y(y-2)-7(y-2)=0` `(y-7)(y-2)=0` `y=2,7` `xley` |
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| 17485. |
What is that instrument which can detect the presence of electric current in a circuit.a) Galvanometerb) Motorc) Generatord) None of above |
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Answer» a) Galvanometer |
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| 17486. |
In the circuit in figure, the current flowing `5 Omega` resistance is : A. 0.5 AB. 0.6 AC. 0.9 AD. 1.5 A |
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Answer» Correct Answer - B `I_(2)=(R_(1))/(R_(1)+R_(2))I=(10)/(35)xx2.1 =(21)/(35)=(3)/(5)` `I_(2) =0.6A`. |
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| 17487. |
Imagine a short dipole at the center of a spherical surface. If the magnitude of the electric field at a certain point on the surface of he sphere is `10NC^(_1)` then which of the following cannot be the magnitude of the electric field anywhere on the surface of the sphere?A. `4 N//C`B. `8 N//C`C. `16 N//C`D. `20 N//C` |
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Answer» Correct Answer - A If `E_("axial") = 10 N//C` then `E_(eq) = 5N//C` If `E_(eq) = 10 N//C` then `E_("axial") = 20 N//C` So magnitude of electric field at any point `5N//C le E le 20 N//C` |
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| 17488. |
A car is moving on circular path of radius 100 m such that its speed is increasing at the rate of `5(m)/(s^(2))` at `t=0` it starts from rest. The radial acceleration of car at the instant it makes one compelete round trip, will be equal to `10pix(m)/(s^(2))` then find x. |
| Answer» Correct Answer - 2 | |
| 17489. |
Which of these is/are representing electrostatic field? (`x` and `y` are space coordinates)A. `yhat(i)+xhat(j)`B. `sinxhat(i)`C. `xhat(i)+yhat(j)`D. `x^(2)hat(i)+yhat(j)` |
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Answer» Correct Answer - A::C::D For consevative `(delEx)/(dely)=(delEy)/(delx)` |
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| 17490. |
A string fixed at both the ends of length 2m. Vibrating in its 7th overtone. Equation of the standing wave is given by `y=Asinkxcos(omegat+(pi)/(3))`. All the symbols have their usual meaning. Mass per unit length of the string is `0.5(gm)/(cm)`. Given that `A=1` cm and `omega=100pi(rad)/(sec)` Answer the following 2 questions based on information given (use `mu^(2)=10)` Q. Starting from `t=0`, energy of vibration is completely kinetic at time t`t`, where t is:A. `(1)/(600)sec`B. `(5)/(600)sec`C. `(19)/(600)sec`D. `(25)/(600)sec` |
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Answer» Correct Answer - A::C::D For energy to be completely potential `cos(omegat+(pi)/(3))=+-1impliesomegat+(pi)/(3)=npi` `t=((3n-1)/(3))(pi)/(omega)=((3n-1)/(300))sec`. For energy to be completely kinetic `cos(omegat+(pi)/(3))=0` `omegat+(pi)/(3)=(2n-1)(pi)/(2)` `t=((6n-5)/(600))sec` |
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| 17491. |
A string fixed at both the ends of length 2m. Vibrating in its 7th overtone. Equation of the standing wave is given by `y=Asinkxcos(omegat+(pi)/(3))`. All the symbols have their usual meaning. Mass per unit length of the string is `0.5(gm)/(cm)`. Given that `A=1` cm and `omega=100pi(rad)/(sec)` Answer the following 2 questions based on information given (use `mu^(2)=10)` Q. Starting from `t=0`, energy of vibration is completely potential at time `t`, where t isA. `(1)/(150)sec`B. `(1)/(60)sec`C. `(3)/(100)sec`D. `(11)/(300)sec` |
| Answer» Correct Answer - A::B::D | |
| 17492. |
A potential difference of `10^(3)` V is applied across an X-ray tube. The ratio of the de-Broglie wavelength of the incident electrons to the shortest wavelength of X-rays products is - `(e//m=1.8xx10^(14)C//kg`for an electron)A. `1//20`B. `1//100`C. 1D. None of these |
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Answer» Correct Answer - C For incident electron `lambda_(1) =(h)/(p) =(h)/sqrt(2Km)=(h)/(sqrt(2m Ve))` and shortest wavelength of X- rays is `lambda_(2)=(hc)/(Ve)` `(lambda_(1))/(lambda_(2))=1/c sqrt(((V)/(2))((e)/(m)))` Substituting the value we get `(lambda_(1))/(lambda_(2)) =1` |
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| 17493. |
Rs 9000 becomes Rs 18000 at SI in 8 years . Find the rate of interest . |
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Answer» Principal = Rs. 9000 Amount = Rs. 18000 Time = 8 years S.I = A - P = 18000 - 9000 = 9000 S.I = (P × R × T)/100 9000 = (9000 x R x 8)/ 100 9000 x 100 = 72000 x R 900000 = 72000 x R 900000/ 72000 = R 25/2 = R R = 12.5 % |
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| 17494. |
Antibodies are produced by? |
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Answer» Antibodies are host proteins that are produced by the immune system in response to foreign molecules that enter the body. These foreign molecules are called antigens, and their molecular recognition by the immune system results in selective production of antibodies that are able to bind the specific antigen. |
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| 17495. |
The respiratory centre is present in the |
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Answer» The medulla oblongata is the primary respiratory control center. Its main function is to send signals to the muscles that control respiration to cause breathing to occur. |
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| 17496. |
The ratio of static enthalpy rise in the rotor to the static enthalpy rise in the stage of an axial flow compressor is defined as: (a) Power input factor(b) Flow coefficient (c) Temperature coefficient (d) Degree of reaction |
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Answer» (d) Degree of reaction Degree of reaction = \(\frac{(\Delta h)_{rotar}}{(\Delta h)_{stage}}\) |
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| 17497. |
Hybrid vigour is due to -A. ChiasmaB. LinkageC. Crossing overD. Heterozygosity |
| Answer» Correct Answer - D | |
| 17498. |
In nephorgenic diabetes insipidus one should be |
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Answer» All adults and children with nephrogenic diabetes insipidus should take frequent bathroom breaks. This helps to avoid over-distending the bladder, which can cause long-term problems, though rarely. The most important treatment for nephrogenic diabetes insipidus is to ensure constant access to lots of water. |
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| 17499. |
Progeny of a cross made between two pure parents show increased vigour and productivity. This is due to:A. SelectionB. HybridisationC. Hybrid vigourD. Mutation |
| Answer» Correct Answer - C | |
| 17500. |
Methods of selection in plants showing vegetative propagation is -A. Clonal selectionB. Mass selectionC. Pureline selectionD. Pedigree selection |
| Answer» Correct Answer - A | |