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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.
| 93401. |
Which charge for the `N_2` molecule would give a bond order of 2.5 ?A. `+1`B. `+2`C. `-1`D. `-2` |
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Answer» Correct Answer - AC (A)`N_2=(10-4)/2=3.0 implies N_2^(+)=(9-4)/2=2.5` ( C)`N_2=(10-4)/2=3.0 implies N_2^(-) =(10-5)/2=2.5` |
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| 93402. |
The correct order for triple bond energy in `CO, N_2, CN` and `C-=C` is ,A. `COgtN_(2)gtCNgtC-=C`B. `N_(2)gtCOgtC-=CgtCN`C. `C-=CgtCOgtN_(2)gtCN`D. `CNgtCOgtN_(2)gtC-=C` |
| Answer» Correct Answer - A | |
| 93403. |
The ratio of `t_(0.75)` and `t_(0.5)` for first order reaction :-A. `4:3`B. `3:2`C. `2:1`D. `1:2` |
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Answer» Correct Answer - C `(t_(0.75))/(t_(0.50))=((2.3)/(k)log(A_(0)/(0.25A_(0))))/((2.3)/(k)log(A_(0)/(0.5A_(0))))=(log(4))/(log2)=(2)/(1)` |
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| 93404. |
What is the correct sequence of osmotic pressure of `0.01 M aq.` solution of `:` `(a) Al_(2)(SO_(4))_(3)" "(b) Na_(3)PO_(4)" "(c ) BaCl_(2)" " (d) Gluc ose`A. `pi_(4)gtpi_(2)gtpi_(3)gtpi_(1)`B. `pi_(3)gtpi_(4)gtpi_(2)gtpi_(1)`C. `pi_(3)gtpi_(4)gtpi_(1)gtpi_(2)`D. `pi_(1)gtpi_(2)gtpi_(3)gtpi_(4)` |
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Answer» Correct Answer - D order of I `pi_(1)gtpi_(2)gtpi_3gtpi_(4)` `:.` order of osmotic pressure is also `pi_(1)gtpi_(2)gtpi_(3)gtpi_(4)` |
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| 93405. |
Why should a magnesium ribbon be cleaned before burning in air? |
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Answer» Magnesium ribbon should be cleaned with sandpaper before burning in air due to the following reasons: • Magnesium is very reactive element which rapidly reacts with oxygen in the air to form a white layer of magnesium oxide, which may prevent or slow down the burning of magnesium ribbon. • Unwanted impurities deposited on the magnesium ribbon can be removed and only pure magnesium can be used for the reaction. Magnesium ribbon should be cleaned with sandpaper before burning in air because Magnesium is very reactive element which rapidly reacts with oxygen in the air to form a white layer of magnesium oxide, which may prevent or slow down the burning of magnesium ribbon. |
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| 93406. |
What happens when ferrous sulphate crystals is heated in a dry boiling tube? |
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Answer» On heating, ferrous sulphate crystals lose water of crystallisation (FeSO4.7H2O). So, their colour changes from light green to white due to the formation of anhydrous ferrous sulphate(FeSO4). When FeSO4 is heated, it gives ferric oxide, sulphur dioxide and sulphur trioxide. FeSO4 → Fe2O3 + SO2 + SO3 |
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| 93407. |
List any two observations when Ferrous Sulphate is heated in a dry test tube? |
Answer»
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| 93408. |
Among the metals `Cr,Fe,Mn,Ti,Ba`, and `Mg`, the one that cannot be obtained by reduction of metal oxide by aluminium isA. `Cr`B. `Fe`C. `Mn`D. `Mg` |
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Answer» Correct Answer - D `Mg` reacts with `C` to form Magnesium carbide. |
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| 93409. |
Lysol is solution of cresol inA. Soaphy waterB. Simple waterC. AcidD. Heavy water |
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Answer» Correct Answer - A Soaphy water. |
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| 93410. |
The cell reaction for the given cell is spontaneoous if: `Pt_(Cl_(2))|Cl^(-)(1M)||Cl^(-)(1M)|Pt_(Cl_(2))`A. `P_(1) gt P_(2)`B. `P_(1) lt P_(2)`C. `P_(1) = P_(2)`D. `P_(2) = 1` atm |
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Answer» Correct Answer - b `E_(cell) = E_(OP_(CI))^(@) +E_(RP_(CI))^(@) + (0.059)/(2) log` `([CI^(-)]_(LHS)^(2)P_(2))/([CI^(-)]_(RHS)^(2)P_(1)) = (0.059)/(2)log.(P_(2))/(P_(1))` `E_(cell)` is `+ve` when `P_(2) gt P_(1)` |
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| 93411. |
Which acts as poison for Pb-charcual in Lindle catayat?A. `BaSo_(4)`B. QuinolineC. both (a) and (b)D. None of these |
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Answer» Correct Answer - c Either `BaSO_(4)` or quinoline is used so reduce the activity of pd-chamical .The catayst Pb- chemical `BaSO_(4)` or quinoline in y=used to hydrogen alkene only . `CH + CH + H_(2) overset("Linthe cable")to CH_(2) = CH_(2)` |
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| 93412. |
The vapour presure of a dilute aqueous solution of glucose in `750` atom of fig `373 K` .The mole fraction of solute isA. `(1)/(10)`B. `(1)/(7.6)`C. `(1)/(35)`D. `(1)/(76)` |
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Answer» Correct Answer - d `(P^(0) - P_(s))/(P^(0)) = (n)/(n + N)` `(At 373 K V.P of H_(2)O = 760 min Hg)` `(n)/(n +N) = (760 - 750)/(760) = (1)/(76)` |
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| 93413. |
The increasing order of specific charge for the following substances electron (e), proton (p) unipositive helium atom (h), neutron (n) and `alpha`-particle is correctly represented in which of the following options.A. e,n,p,h,`alpha`B. n,h,`alpha`,p,eC. n,`alpha`,h,p,eD. None of the above options |
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Answer» Correct Answer - B `q/m` =specific charge For `eto e/(m_(e)) " " `For `He^(+)toe/(4xx1840 m_(e))` For `P to e/(1840 m_(e))" " ` for `alpha to (2e)/(4xx1840 m_(e))` For `n to 0 ` `n lt h lt alpha lt p lt e` |
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| 93414. |
Find out of compounds which show mutarotation (A) Cellobiose `(4-O-(beta-D-"glucopyranosyl")-D-"glucophranose)` (B) Maltose `(4-O-(alpha-D-"glucopyranosyl")-D-glucopyranose)` (C) Lactose `(4-O-(beta-D-"galactopyranosyl")-D-"glucopyranose")` (D) Gentiobiose `(6-O-(beta-D-"glucopyranosyl")-D-"glucopyranose")` (E) Sucrose `(alpha-D-"glucopyranosyl"-beta-D-"fructofuranoside")` (F) Methyl `alpha-D-"galactopyranoside"`. (G) `alpha-D-` allopyranose (G) `alpha-D-` glucopyranose (I) `alpha-D-` fructofuranose |
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Answer» Correct Answer - 7 `A,B,C,D,G,H,I` show mutarotation. A sugar whose IUPAC name endsd with the suffix ose is a reducing sugar and they do mutarotation |
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| 93415. |
The diameyer of zinc atom is `2.6 A`. Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of `1.6 cm` if the zinc atoms are arranged side by side lengthwise. |
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Answer» a. Radius `=(2.6 Å)/2=1.3xx10^(-10)m=130xx10^(-12)m` `=130 p m` b. Given length `=1.6 cm=1.6xx10^(-2)m` Diameter of one atom `=2.6Å=2.6xx10^(-10)m` `:.` Number of atoms present along the length `=(1.6xx10^(-2))/(2.6xx10^(-10))=6.154xx10^(7)` |
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| 93416. |
A certain particle carries `2.5xx10^(-16) C` of static electric charge. Calculate the number of electrons present in it. |
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Answer» Charge carried by one electron `=1.6022xx10^(-19)C` `:.` Electrons present in particle carrying `2.5xx10^(-16)C` charge `=(2.5xx10^(-16))/(1.6022xx10^(-19))=1560` |
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| 93417. |
When `5mL` of a `1.0M HCl` solution is mixed with `5mL` of a `0.1M NaOH` solution, temperature of solution increases by `t^(@)C`. Which of the following(s) can be predicted from this observationA. If `5mL` of `0.1M HCl` is mixed with `5mL 0.1M NH_(3)` solution, the temperature rise will be less than `t^(@)C`B. If `5mL 0.1M CH_(3)COOH` is mixed with `5mL 0.1 M NaOH`, the temperature rise will be less than `t^(@)C`C. If `10mL` of `1.0M HCl` is mixed with `10mL` of `0.1M NaoH`, the temperature rise will be `2t^(@)C`D. If `10 ML` of `0.1M HCl` is mixed with `10 mL` of `0.1 M NaCl` the temperature rise will be `t^(@)C` |
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Answer» Correct Answer - A::B ab Reaction proceeds by limiting reagent only. Enthalpy of neutralization decreases, when acid or base or both are weak: |
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| 93418. |
If the diameter of a carbon atom is `0.15 nm`, calculate the number of carbon atom which can be placed side by side in a straight line length of scale of length `20 cm` long. |
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Answer» Diameter of carbon atom `=0.15 nm` `=0.15xx10^(-9) m=1.5xx10^(-10) m` Length along which atoms are to be placed `=20 cm` `=20xx10^(-2) m=2xx10^(-1)m` :. Number of C atoms which can be placed along the length `=(2xx10^(-1))/(1.5xx10^(-10))=1.33xx10^(9)` |
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| 93419. |
A stagnation point is the point on the immersed body where the magnitude of velocity is (a) small (b) large (c) zero (d) none of the above |
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Answer» Correct option is (c) zero |
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| 93420. |
The stagnation pressure (ps) and temperature (Ts) are (a) less than their ambient counterparts (b) more than their ambient counterparts (c) the same as in ambient flow (d) none of the above. |
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Answer» (b) more than their ambient counterparts |
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| 93421. |
A standard vapour is compressed to half its volume without changing is temperature. The result is that : (a) All the vapour condenses to liquid (b) Some of the liquid evaporates and the pressure does not change (c) The pressure is double its initial value(d) Some of the vapour condenses and the pressure does not change |
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Answer» (c) The pressure is double its initial value |
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| 93422. |
Across a normal shock (a) the entropy remains constant (b) the pressure and temperature rise (c) the velocity and pressure decrease (d) the density and temperature decrease. |
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Answer» (b) the pressure and temperature rise |
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| 93423. |
A system of 100 kg mass undergoes a process in which its specific entropy increases from 0.3 kJ/kg-K to 0.4 kJ/kg-K. At the same time, the entropy of the surroundings decreases from 80 kJ/K to 75 kJ/K. The process is (a) Reversible and isothermal (b) Irreversible (c) Reversible (d) Impossible |
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Answer» (b) Irreversible |
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| 93424. |
When the discharge pressure is too high in a refrigeration system, high pressure control is installed to (a) stop the cooling fan (b) stop the water circulating pump (c) regulate the flow of cooling water (d) stop the compressor. |
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Answer» (d) stop the compressor. |
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| 93425. |
A control mass undergoes a process from state 1 to state 2 as shown in During this process, the heat transfer to the system is 200 kJ. If the control mass returned adiabatically from state 2 to state 1 by another process, then the work interaction during the return process (in kN.m) would be(a) – 400 (b) – 200 (c) 200 (d) 400. |
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Answer» Correct option is (c) 200 |
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| 93426. |
When we boil a liquid in a closed vessel and open vessel ,which one take more time to boil .(Pressure cooked is closed vessel) |
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Answer» Liquid can be boiled in closed container like vacuum chamber by controlling pressure but we can not boil liquid just by increasing temperature in a closed vessel for boiling, the vapour pressure of liquid must be equal to external pressure continously increases(volume is constant). There is not a particular temperature of which external pressure become equal to vapour pressure of liquid . Therefore in closed vessels, boiling cannot occurs. but in open vessels boiling occurs. Hence, liquid in closed vessel take more times to boil. |
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| 93427. |
Vehicle will accelerate as long asA. air resistance is greater than thrustB. air resistance is greater than inertiaC. thrust is greater than air resistance and frictionD. friction is greater than thrust |
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Answer» C. thrust is greater than air resistance and friction |
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| 93428. |
1 N is equal toA. 1 kg m s-2B. 10 kg m s-1C. 10 kg m s-2D. 100 kg m s-2 |
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Answer» The Correct option is A. 1 kg m s-2 |
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| 93429. |
At terminal velocity theA. air resistance and weight are equalB. air resistance is less than weightC. weight is more than air resistanceD. air resistance is more than weight |
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Answer» A. air resistance and weight are equal |
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| 93430. |
Density of air isA. 1/8 of waterB. 1/7 of waterC. 1/45 of waterD. 1/800 of water |
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Answer» D. 1⁄800 of water |
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| 93431. |
If there is no net force acting on body, then its acceleration isA. zeroB. constantC. increasingD. decreasing |
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Answer» The Correct option is A. zero |
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| 93432. |
Combinations of base units areA. simple unitsB. derived unitsC. scalarsD. vectors |
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Answer» B. derived units |
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| 93433. |
Rate of falling object in vacuum isA. independent of weightB. dependent on massC. independent of massD. dependent of weight |
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Answer» C. independent of mass |
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| 93434. |
Acceleration of a rocket having mass 5000 kg and resultant force acting on it is 200,000 N isA. 50 m s-2B. 56 m s-2C. 70 m s-2D. 40 m s-2 |
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Answer» The Correct option is D. 40 m s-2 |
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| 93435. |
Until a force acts on a body, it's velocity isA. zeroB. constantC. increasingD. decreasing |
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Answer» B. constant. |
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| 93436. |
Acceleration due to gravity on moon isA. 9.9 m s-2B. 9.5 m s-2C. 6.1 m s-2D. 1.6 m s-2 |
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Answer» The Correct option is D. 1.6 m s-2 |
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| 93437. |
A force similar to friction isA. forward forceB. pulling forceC. drag forceD. contact force |
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Answer» C. drag force |
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| 93438. |
500 MW can be written in powers of 10 asA. 500 × 106B. 500 × 103C. 500 × 10-6D. 500 × 109 |
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Answer» The Correct option is A. 500 × 106 |
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| 93439. |
Point where entire weight of an object acts isA. edgeB. center of gravityC. central pointD. can be anywhere in body |
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Answer» B. center of gravity |
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| 93440. |
Contact force always acts atA. acute angles to the surface producing itB. right angles to the surface producing itC. obtuse angle to the surface producing itD. parallel to the surface producing it |
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Answer» B. right angles to the surface producing it |
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| 93441. |
Forces acting on an object are balanced if resultant force on object isA. constant B. zero C. increasing D. decreasing |
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Answer» The Correct option is B. zero |
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| 93442. |
Two forces which make up Newton's third law canA. act on same objectsB. act on different objectsC. not act at same timeD. not act oppositely |
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Answer» B. act on different objects |
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| 93443. |
When two objects are in contact, they exert forces inA. opposite directionB. same directionsC. can be both A and BD. perpendicular direction |
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Answer» A. opposite direction |
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| 93444. |
Activity is proportional to number ofA. daughter nucleiB. decayed nucleiC. undecayed nucleiD. father nuclei |
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Answer» C. undecayed nuclei |
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| 93445. |
Four cards are drawn successively without replacement from a well shuffled deck of 52 playing cards what is the probability that only 2 cards are spade? |
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Answer» Probability of gating only 2 spade card i.e., P(only 2 are spades)=4Cr(1/4)r(3/4)n-r Here, putting r=2, we get P(only 2 are spades)=4C2(1/4)2(3/4)2 =27/128 |
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| 93446. |
If θ = 90°, then the value of sinθ/2 – cosθ/2 is(A) 1 (B) -1 (C) 2 (D) 0 |
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Answer» Correct option is: (D) 0 |
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| 93447. |
Two unbiased coins are tossed then the probability of getting at least one head is –(A) 1/2(B) 1/4(C) 1/3(D) 3/4 |
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Answer» Correct answer is (D) 3/4 |
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| 93448. |
A card is drawn at random from a well shuffled deck of playing cards, then the probability of getting a red card is –(A) 1/2(B) 2/13(C) 3/13(D) None of these |
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Answer» Correct answer is (A) 1/2 |
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| 93449. |
Describe Galesburg, as it existed in the year 1894? |
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Answer» Galesburg was a wonderful town with big old frame houses, huge lawns, and tremendous trees whose branches met overhead and roofed over the streets. In 1894, summer evenings were twice as long, and people sat on their lawns, the men smoking. |
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| 93450. |
Who had sent the first day cover and what was written on it? |
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Answer» Sam had sent the first day cover. Sam wrote that he had discovered the Third level and had reached Galesburg .He found Galesburg peaceful and friendly .he advised Charley to keep looking for third level and reach Galesburg. |
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