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.
| 7351. |
If train B passes a platform in 90 sec, length of train 150 m and speed of train 20 kmph. Find the speed of train B if he passes to the platform in 180 sec and the ratio between the length of the platform to train B is 7 : 5.1. 10 kmph2. 12 kmph3. 15 kmph4. 18 kmph5. None |
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Answer» Correct Answer - Option 2 : 12 kmph Given: Speed of train A = 20 kmph Calculation: Let be assume the length of the platform is y ⇒ (y + 150)/{20 × (5/18)} = 90 ⇒ y + 150 = 100 × 5 ⇒ y = 350 Length of train B = (5/7) × 350 = 250 Speed of train B = (350 + 250)/180 = 600/180 mps ⇒ (600/180) × (18/5) ⇒ 12 kmph ∴ The result will be 12 kmph. |
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| 7352. |
Statement-1: Low activation energy means the reaction will be faster. Statement-2 A thermodynamically stable product is always formed easily.A. If both the statement are TRUE and Statement -2 is the correct explanation of Statement-1:B. If both the statement are TRUE but Statement-2 is not the correct explanation of Statement-1C. If statement-1 is TRUE and Statement-2 is FALSED. If statement -1 is FALSE and Statement-2 is TRUE |
| Answer» A thermodynamically stable product with high activation energy is not formed esily, particularly at low temperature. | |
| 7353. |
The reaction of dimerisation of `NO_(2)` in `N_(2)O_(4)` is `2NO_(2)hArrN_(2)O_(4)`. The reaction is carried out by taking `1` mole each of `NO_(2)` and `N_(2)O_(4)` in a closed vessel of `1` litre at `400 K`. The equilibrium pressure was found to be `77 atm`. Which statements is correct for given values of teh reactions.?A. Dissociation of `N_(2)O_(4)` occurs with degree of dissociation of `N_(2)O_(4) 0.35`B. Formation of `NO_(2)` occurs and total moles of `NO_(2)` at equilibrium `1.35`C. Dissociation of `N_(2)O_(4)` occurs leaving `0.35` moles at equilibriumD. Formation of `NO_(2)` occurs with total moles at equilibrium `2.70` |
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Answer» `underset(1)(2NO_(2)) hArr underset(N_(2)O_(4))` Total moles taken `=2`, `:. P=2xx0.0821xx400=65.7 atm` Given equilibrium pressure is `77` atm, thus reaction should show the change where total number of moles increases i.e., dissociation of `N_(2)O_(4)`. `{:(N_(2)O_(4),hArr,2NO_(2)),(1,,1),(1-x,,1+2x):}` Total moles at eq.=`2+x` `=(PV)/(RT)=(77xx1)/(0.0821xx400)=2.35` `:. x=0.35` |
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| 7354. |
The minimum value of `y=5x^(2)-2x+1` isA. `(1)/(5)`B. `(2)/(5)`C. `(4)/(5)`D. `(3)/(5)` |
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Answer» Correct Answer - C The minimum/minimum value `(dy)/(dx)=0rArr5(2x)-2(1)+0=0=0rArrx=(1)/(5)` Now at `x=(1)/(5)m(d_(y^2))/(dx^(2))=10 x=(1)/(5)m(d_(y^2))/(dx^(2))=10` which is positive so minima at `x=(1)/(5)`. Therefore `y_("min")=5((1)/(5))^(2)-2((1)/(5))+1=(4)/(5)` |
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| 7355. |
Evaluate `int (x^(2)-cos x+(1)/(x))dx` |
| Answer» `=intx^(2)dx-intcos xdx+int(1)/(x)dx=(x^(2+1))/(2+1)-sin x +"log"_(e)x+c` | |
| 7356. |
The reaction of dimerisation of `NO_(2)` in `N_(2)O_(4)` is `2NO_(2)hArrN_(2)O_(4)`. The reaction is carried out by taking `1` mole each of `NO_(2)` and `N_(2)O_(4)` in a closed vessel of `1` litre at `400 K`. The equilibrium pressure was found to be `77 atm`. The equilibrium pressure at which dissociation of `N_(2)O_(4)` will show degree of dissociation of `N_(2)O_(4)` to be `0.50` in the above case:A. `82.1 atm`B. `65.7 atm`C. `72.0 atm`D. `70.0 atm` |
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Answer» `{:(N_(2)O_(4),hArr,2NO_(2)),(1,,1),(0.5,,1+0.5xx2):}` Moles at equilibrium `=2.5` `:. P_(at eq.)=2.5xx0.0821xx400` `=82.1 atm` |
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| 7357. |
Intergrate w.r.t. x.: (i) `x^(11//2)` (ii) `x^(-7)` (iii) `x^(p//q)` |
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Answer» (i) `intx^(11//2)dx=(x^(11/2+1))/((11)/(2)+1)+c=(2)/(13)xx^(13//2)+c` (ii) `intx^(-7)dx=(x^(-7+1))/(-7+1)+c=-(1)/(6)x^(-6)+c` (iii) `intx^((p)/(q))dx=(x^((p)/(q)+1))/((P)/(q)+1)+c=(q)/(p+q)x^((p+q)//q)+c` |
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| 7358. |
The rate of a reaction quadruples when the temperature changes from `293K` to `313K`. Calculate the energy of activation of the reaction assuming that it does not change with temperature. |
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Answer» Correct Answer - `E_(a)=52.8 mol^(-1)` Let `k` at `293K=k_(1),k at 313K=k_(2)` `:.k_(2)=4k_(1) or (k_(2))/(k_(1))=4` Using Arrhenius equation, `log``(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))` `log4=(E_(a))/(2.303xx8.314J K^(-1)mol^(-1))xx((313-293))/(313xx293)` `0.606=(E_(a))/(2.303xx8.314)xx(10)/(313xx293)` Solvie for `E_(a):` `:. E_(a)=52860J mol^(-1)=52.86mol^(-1)` |
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| 7359. |
The approx value of `K` for a reaction `AhArrB` is `:` Given `Delta_(r)H_(298K)^(@)=-54.07KJmol^(-1)`,`Delta_(r)S_(298K)^(@)=10Jk^(-1)mol^(-1)` and `R=8.314JK^(-1)mol^(-1)`.A. `10^(+10)`B. `10^(-100)`C. `10^(-10)`D. `10^(+100)` |
| Answer» Correct Answer - A | |
| 7360. |
Which of the following option is correct of the above reaction?A. `x=PhCOOH y=C_(2)H_(5)OH`B. `M=PhPhCOONa N=CH_(3)OCH_(3)`C. `x=PhOH y=CH_(3)OCH_(3)`D. `M=C_(2)H_(5)ONa N=CH_(3)OCH_(3)` |
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Answer» Correct Answer - D `PhCOOH` is more acidic than `H_(2)CO_(3)` while `PhOH` is less acidic than `H_(2)CO_(3)` |
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| 7361. |
`6ml` of `0.1M CH_(3)COOH` when mixed with `x` ml of `0.1 M NaOH` the `pH` of the resulting buffer solution was found to be `5.04`. The value of `x` is `(log2=0.3)pK_(a(CH_(3)COOH))=4.74` |
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Answer» Correct Answer - 4 `underset(0.6-0.x) underset(6xx0.1=0.6)(CH_(3)COOH)+underset(x xx 0.1)(NaOH)tounderset(0.x)underset(0)(CH_(3)COONa)+H_(2)O` The resulting solution is a buffer solution. `pH=pK_(a)= "log"([CH_(3)COO^(-)])/([CH_(3)COOH]) 0.3 = "log" (0.x)/(0.-0.x)` `pH=pK_(a)+"log"(0.x)/(0.6-0.x) log 2="log" (0.x)/(0.6-0.x)` `5.04=4.74+"log"(0.x)/(0.6-0.x) (0.x)/(0.6-0.x)=2` `x=4` |
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| 7362. |
How many of the following species are paramagnetic? `N_(2)^(+), O_(2)^(-), B_(2), C_(2), CN^(-), NO^(+), CO, H_(2)^(-), He_(2)^(+)` |
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Answer» Correct Answer - 5 The following species are paramagnetic `N_(2)^(+),O_(2)^(-), B_(2), H_(2)^(-),He_(2)^(+)` |
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| 7363. |
Consider the reactions:2S2O32-(aq) + I2(s) ⟶ S4O62-(aq) + 2I-(aq)S2O32-(aq) + 2Br2(I) ⟶ 5H2O(I) ⟶ 2SO42-(aq) + 4Br-(aq) + 10H+ (aq)Why does the same reductant, thiosulphate react differently with iodine and bromine? |
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Answer» The average O.N. of S in S2O32- is + 2 while in S4O62- it is + 2.5. The O.N. of S in SO42- is+6. Since Br2 is a stronger oxidising agent that l2 , it oxidises S of S2O32- to a higher oxidation state of + 6 and hence forms SO42- ion. l2 , however, being a weaker oxidising agent oxidises S of S2O32- ion to a lower oxidation of + 2.5 in S4O62- ion. |
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| 7364. |
When phenol reactes with `CHCl_(3)` and NaOH followed by acidification, salicyladehyde is obtained. Which of the following species are involed in the above-mentioned reaction as intermediates ?A. B. C. D. |
| Answer» Correct Answer - A::D | |
| 7365. |
Which oxide of nitrogen is obtained on heating ammonium nitrate at 250ºC ?(a) Nitric oxide (b) Nitrous oxide (c) Nitrogen dioxide (d) Dinitrogen tetraoxide |
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Answer» (b) Nitrous oxide |
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| 7366. |
On heating ammonium dichromate and barium azide separately we get(a) N2 in both cases(b) N2 with ammonium dichromate and NO with barium azide(c) N2O with ammonium dichromate and N2 with barium azide(d) N2O with ammonium dichromate and NO2 with barium azide |
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Answer» (a) N2 in both cases |
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| 7367. |
CF2 Cl2 is used as a/an– (a) Antiseptic(b) Insecticide(c) Analgesic(d) Refrigerant |
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Answer» CF2 Cl2 is used as a Refrigerant |
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| 7368. |
The antiseptic action of CHI3 is due to (a) CHI3 (b) Liberation of free iodine(c) Iodide ions(d) Iodine and CHI3 both |
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Answer» The antiseptic action of CHI3 is due to Liberation of free iodine. |
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| 7369. |
Which of the following is a secondary alkyl halide ? (a) Isobutyl chloride(b) Isopentyl chloride(c) Neopentyl chloride(d) Isopropyl chloride |
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Answer» Isopropyl chloride is a secondary alkyl halide. |
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| 7370. |
Write the chemical formulae of the following ores :(a) Haematite (b) Magnetite(c) Limonite (d) Siderite |
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Answer» (a) Fe2O3 (b) Fe3O4 |
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| 7371. |
Give the principles involved in the manufacture of sulphuric acid by contact process with equations: |
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Answer» Production of sulphur dioxide: It is carried out by burning powdered sulphur or roasting of sulphur rich ores. S8 + 8O2 → 8SO2 4FeS2 + 11O2 → 2Fe2O3 + 8SO2 |
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| 7372. |
Mention one use each of Paracetamol |
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Answer» The use of Paracetamole is to the fever as antipyretic. |
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| 7373. |
Mention one use each of Tincture of iodine. |
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Answer» The use of Tinture iodine as Antiseptic for clerminl wounds. |
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| 7374. |
The IUPAC name of C3H4. |
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Answer» The answer is propene HH-C--C--HH Cyclopropene Cyclopropene is an organic compound with the formula C3H4. It is the simplest cycloalkene. Because the ring is highly strained, cyclopropene is difficult to prepare and highly reactive. This colorless gas has been the subject for many fundamental studies of bonding and reactivity. |
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| 7375. |
CO + H2 + X Catalyst/300°c/300atm → CH3OH, the catalyst X is(1) Fe (2) Cr2O3/ZnO (3) V2O5 (4) Al2O3 |
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Answer» CO + H2 + X Catalyst/300°c/300atm → CH3OH, the catalyst X is Cr2O3/ZnO |
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| 7376. |
Some chemicals are given below. Sodium chloride, Ammonium hydroxide, Nitric acid, cone. Sulphuric, Sodium hydroxide. a. Which are the substances needed to produce hydrogen chloride? b. Suggest a method to identify hydrogen chloride gas. |
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Answer» a. Sodium chloride b. Show a glass rod dipped in ammonia, just above the jar in which hydrogen chloride gas is taken. A white smoke of NH34Cl is produced. |
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| 7377. |
Take some ammonium chloride (NH4Cl) in a boiling tube and heat it. Don’t you sense a peculiar smell? |
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Answer» Yes. Pungent smell |
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| 7378. |
Take a little ammonium chloride (NH4Cl) in a watch glass and add a little calcium hydroxide (Ca(OH)2) to it. Stir well. Can you sense any smell? |
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Answer» Irritating smell is experienced. |
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| 7379. |
the vapour of which alcohol if passed over heated CU at 573K will produce an alkane? |
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Answer» When the vapours of tertiary butyl alcohol are passed through heated copper of 573K when the vapours of an alcohol are passed over copper catalyst heated at 573K. it undergoes dehydrogenation (loss of hydrogen or oxidation). The product formed depends on the alcohol and hence this reaction is also used to distinguish the three classes of alcohols. Tertiary alcohol when reacted with copper catalyst at 573K are not dehydrogenated but undergoes dehydration to give the corresponding alkene. |
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| 7380. |
pH of ammonium chloride (NH4Cl) or copper sulphate (CuSO4) solution beA) 7B) >7C) <7D) 0 |
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Answer» The answer is C) <7 |
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| 7381. |
The conductivity of a solution of complex with formula CoCl3(NH3)4 corresponds to 1 : 1 electrolyte, then the primary valency of central metal ion is ____. |
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Answer» [Co(NH3)4Cl2]Cl Primary valency = oxidation no. = +3 |
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| 7382. |
The elevation of boiling point of 0.10 m aqueous CrCl3 .xNH3 solution is two times that of 0.05m aqueous CaCl2 solution. The value of x is____. [Assume 100% ionisation of the complex and CaCl2 , coordination number of Cr as 6, and that all NH 3 molecules are present inside the coordination sphere] |
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Answer» Correct answer is 5.00 2Tb=i1×kb×0.10 -(1)Tb=i2×kb×0.05 -(2) Solving (1) and (2) , We get i1=i2 Now i2 (for cacl2) = 3 So, i1=3 The country point hence formed is Cr[(cl) (nh3) 5]cl2. |
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| 7383. |
The number of six membered carbon rings in the structure of Buckminsterfullerene (i.e. `C_60`)is . |
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Answer» Correct Answer - 20 It contains twenty six - membered rings and twelve five-membered rings. |
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| 7384. |
How carbon monoxide, emitted by automobiles, prevents transport of oxygen in the body tissues?(a) by forming a stable compound with haemoglobin(b) by obstructing the reaction of oxygen with haemoglobin(c) by changing oxygen into carbon dioxide(d) by destroying the haemoglobin. |
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Answer» (a) by forming a stable compound with haemoglobin |
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| 7385. |
`CH_(3)CH=CH_(2)+BrC Cl_(3)overset("Peroxide")to[A]` |
| Answer» `underset("Propene")(CH_(3)CH=CH_(2)+BrC Cl_(3))overset("Peroxide")tounderset([A])(CH_(3)CH(Br)-CH_(2)C Cl_(3))` | |
| 7386. |
find the total numbers of compounds which contain S-S linkage. `H_(2)S_(2)O_(3),H_(2)S_(2)O_(5),H_(2)S_(4)O_(6),H_(2)S_(2)O_(7),H_(2)SO_(2)O_(8),H_(2)SO_(2)O_(6)` |
| Answer» Correct Answer - 4 | |
| 7387. |
find the total number of compounds containing X-O-X linkage. (where X is the central atom) `N_(2)O_(4),N_(2)O_(5),P_(4)O_(10),H_(2)S_(2)O_(8),H_(4)P_(2)O_(8),C_(3)O_(2),Cr_(2)O_(7)^(2-),Cl_(2)O_(7),(HPO_(3))_(3),(SO_(3))_(3)` |
| Answer» Correct Answer - 6 | |
| 7388. |
IUPAC nme of Isooctane isA. 2-methyl heptaneB. 2,3,4-Trimethyl pentaneC. 2,2,4-Trimethyl pentaneD. 2,3,3-Trimethly pentane |
| Answer» Correct Answer - 3 | |
| 7389. |
which of the following options consits of only those orbitals which have number opf radial nodes exactly same as their angular nodes ?A. 1s,2p,3dB. 1s,2s,3sC. 1s,3p,5dD. 4f,5g,6h |
| Answer» Correct Answer - 3 | |
| 7390. |
Which of the following is correct radial probability distributaion curve for various orbitals?A. B. C. D. |
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Answer» Correct Answer - A `mvr = (nh)/(2pi)` n = 4 `vec(v) = (1)/(lambda)=R(t)^(2)[(1)/((2)^(2))-(1)/((4)^(2))]=(3R)/(16)` |
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| 7391. |
Correct order of the heats of combustion of above compounds is :A. `(i) gt (ii) gt (iii)`B. `(i) gt (iii) gt (ii)`C. `(ii) gt (i) gt (iii)`D. `(ii) gt (iii) gt (i)` |
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Answer» Correct Answer - D `Deltax.Deltap=(h)/(4pi) " Given "Deltax=10^(-13)c.m.` `Deltap=mDeltav " "1 " Meter "= 100 c.m.` `Deltax.m.Deltav=(h)/(4pi) " "1 c.m. = 10^(-2) " meter"` `Deltav=(6.625xx10^(-34))/(4xx3.14xx10^(-11)xx9.1xx10^(-31))=0.0579xx10^(8)=5.79xx10^(6)m//sec` |
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| 7392. |
A body initially rest and sliding along a frictionless trick from a height `h` (as shown in the figure) just completes a vertical circle of diameter `AB = D`. The height `h` is equal to A. `(7)/(5)D`B. d=1kmC. `(3)/(2)D`D. `(5)/(4)D` |
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Answer» Correct Answer - D |
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| 7393. |
Transition metals (with the exception of Zn, Cd and Hg) are hard and have high melting and boiling points. |
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Answer» Transition metals contain unpaired electrons. Hence, they form covalent bonds which results in hardness. Transition metals have close packed structures in which transition metals are held together by strong metallic bonds which have notable covalent character. Hence, transition metals have high melting and boiling points. For Zn, Cd and Hg the d orbitals are completely filled and no covalent bonds are formed. |
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| 7394. |
In general the melting and boiling points of transition metalsA. Increase gradually across the period from left to right.B. Decrease gradually across the period from left to right.C. First increase till the middle of the period and then increase towards the end.D. First decrease regularly till the middle of the period and then increase towards the end. |
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Answer» Correct Answer - C Along the period , the number of unpaired electrons increases and then decreases due to the pairing to electrons so inter-atomic forces (i.e., metallic bond) increases upto middle of the series and then decreases .Boiling and melting points are directly proportionate to strength of metallic bond. |
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| 7395. |
The infinity resistance plug in a post office box (resistance box) has(a) An air gap only(b) A resistance coil of infinite resistance(c) Largest resistance value in box(d) Resistance of value 100 ohm |
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Answer» (a) An air gap only |
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| 7396. |
Consider the following statements regarding the network shown in the figurei) The equivalent resistance of the network between A and B is independent of the value of R’ ii) The equivalent resistance of the network between points A and B is 4 R/3 iii) The current flowing through R’ is zero Which of the above statement through (s) is/are correct? (a) (i) alone(b) (ii) alone(c) (ii) and (iii)(d) (i), (ii) and (iii) |
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Answer» Correct option(d) Explanation: As circuit is balanced Req = 4R/3 , which is independent of R' Current through R' = 0 |
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| 7397. |
The top of a water tank is open to air and its water level is maintained. It is giving out 0.74m3 water per minute through a circular opening of 2cm radius in its wall. The depth of the centre of the opening from the level of water in the tank is close to: (1) 9.6m (2) 4.8m (3) 2.9m (4) 6.0m |
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Answer» The correct option (2) 4.8m Explanation: 0.74/60 = π x 4 x 10-4 x √(2gh) => h = 4.8m |
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| 7398. |
The electron in the first excited state `(n_(1)=2)` of `H`- atom absorbs a photon and is further excited to `n^(th)` shell. The de-Broglie wavelength of the electron in this excited state is 1340 pm. Find the value of `n`. |
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Answer» Correct Answer - 4 `n=(lamda)/(2pir_(1))=1340/(2xx3.14xx53)=4` |
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| 7399. |
The rate law for a reaction is Rate = K [A] [B]3/2 Can the reaction be an elementary process ? Explain. |
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Answer» No, an elementary process would have a rate law with orders equal to its molecularities and therefore must be in integral form. |
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| 7400. |
Identify the order of a reaction from the following rate constant : k = 2.3 x 10-5 L mol-1 s-1 |
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Answer» . Second order |
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