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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
What is the angular velocity of a second hand and minute hand of a clock ?A. `(59pi)/(900) rad/s`B. `(59pi)/(1800) rad/s`C. `(59pi)/(2400) rad/s`D. `(59pi)/(3600) rad/s` |
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Answer» Correct Answer - b Angular speed of minute hand `omega_(m)=360^(@)` per hour `=(360x(pi)/(180))` per hour =`(2pi)/(3600)` rad/s angular speed of second hand `omega_(s)=360^(@)` per minute =`(2pi)/(60)` rad / s `therefore` differnce between angular speeds of minute hand and second hand of a clock `=omega_(m)-omega_(s)=(2pi)/(3600)-(2pi)/(60)=-(59pi)/(1800` so difference between `omega_(m)` and `omega_(s)` is `(59pi)/(1800)` rad/s |
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| 2. |
For the reaction `O_(3(g)) +O_((g)) to 2O_(2(g)) ` , if the rate law expression is , rate `=k [O_3][O]` the molecularity and order of the reaction are respectively ______.A. 2 and 2B. 2 and 133C. 2 and 1D. 1 and 2 |
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Answer» Correct Answer - a Given reaction is `O_(3)(g)+O(g) rarr2O_()(g)` also rate law expression is given as `rate = k[O_(3))] [O]` now since molecularity is simply the sum of reactant molecules in a single step of a chemical reaction so molecularity is two for this reaction on the other hand order is the sum of the pwoers raised to the concentration of eractant molecules from eq (i) order is 1+1 =2 `therefore` the molecularity is 2 and order is also 2 |
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| 3. |
A potentiometer wire of length 10 m is connected in series with a battery the emf of a cell balances against 250 cm length of wire if length of potentiometer wire is increased by 1 m then new balancing length of wire will beA. 2.00 mB. 2.25 mC. 2.50 mD. 2.75 m |
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Answer» Correct Answer - a Balancing condition of potentiometer is given by if `I_(1)` is incressed by 1m then new length will `I_(1)=10m+1m=11m` `therefore` new balancing length of wire `(l_(2))` is `(mu^_(1))/(l_(2))=(E_(1))/(E_(2)) rarr (11)/(l_(2))=4 therefore L_(2)=11/4=2.75 m` |
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| 4. |
Which among the following solution is not used in deterination of the cell constant ?A. `10^(-2) MKCI`B. `10^(-1) MKCI`C. 1 MKCID. saturated KCI |
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Answer» Correct Answer - d During the determination of cell constant oly standard solution of KCI is taken here options (a), (b) and (c ) are standard solution while (d) i.e saturated KCI is not a standard solution hence it is not used in cell constant determination |
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| 5. |
Which of the following compounds is most acidic in nature ?A. 4 chlorobutanoic acidB. 3 chlorobutanoic acidC. 2 achlorobutanoic acidD. butanoic acid |
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Answer» Correct Answer - c In options (A), (b) and (c ) CI atom being electron withdrawing group produces more acidic carboxylic acid here 2 chlorobutanioic acid has CI atom very nearer to carboxyl carbon and conjugate base nearer becomes most stabl and its corresponding acid is most acidic in nature therefore 2 chlorobutaniooic acid is the most acidic in nature |
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| 6. |
`[Cr(NH_(3))_(6)]Cr(SCN)_(6)` and `[Cr(NH_(3))_(2)(SCN)_(4)][Cr(NH_(3))_(4)(SCN)_(2)]` are the examples of what type of isomerism ?A. lonisation isomerismB. linkage isomerismC. coordination isomerismD. solvate isomerism |
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Answer» Correct Answer - c Here in given question two coordination spherers in a compund are given which indicates that this is an example of coordination isomerism |
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| 7. |
Molarity is defined asA. the number of moles of solute dissolved in `1 dm^(3)` of the solutionB. the number of moles of solute dissolved in 1 kg of solventC. the number of moles of solute dissolved in `1 dm^(3)` of the solventD. the number of moles of solute dissolved in 100 mL of the solvent |
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Answer» Correct Answer - a We have the defineition for molarity as the number of moles of solute present in one litre of the solution `therefore` molarity (M)=`("Number of moles of solute")/("Volume of solution" (in L))` |
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| 8. |
Which halogen forms an oxyacid that contains the halogen atom in tripositive oxidation state?A. fluorineB. chlorineC. bromineD. lodine |
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Answer» Correct Answer - b The oxyacid of chlorine atom has tripositive oxidation state ofCI e.g `HCIO_(2)` , cI is in +3 oxidation state |
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| 9. |
Which among the following is a tranquilizer?A. aspirinB. valiumC. penicillinD. suphanilamide |
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Answer» Correct Answer - b Valium is a tranquilizer aspirin is an analgesic penicillin is an antibiotic while suphanilamide is an antimicrobial |
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| 10. |
Name the metl that is purified by plcing the impure metal on sloping hearth of reverberatory furnace and heating that above its melting point in the absence of airA. mercuryB. galliumC. ziroconiumD. copper |
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Answer» Correct Answer - d During the pruification of copper metal an impure metal is placed on sloping hearth of a reverberatory furnace and heated that above its melting point in the absenece of air |
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| 11. |
Which polymer among the following does not soften on heating ?A. bakeliteB. polytheneC. polystyreneD. pvc |
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Answer» Correct Answer - a Bakelite is a highly cross linked polymer which does not soften on heating |
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| 12. |
The relationship between rate constant and half life period of zero order reaction is give byA. `t_(1//2)=[A_(0)]2K`B. `t_(1//2)=(0.693)/(K)`C. `t_(1//2)=[(A_(0))]/(2K)`D. `t_(1//2)=(2[A_(0)])/(K)` |
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Answer» Correct Answer - c The realtionship between rate constant (k) and half life period of zero order reaction is written as `t_(1//2)=([A_(0)])/(2K)` where `[A_(0)]` is the initial concentration and k= rate constant for zero order reaction |
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| 13. |
Half life period of first order reaction `A rarr` product is 6.93 h what is the value of rate constant ?A. 1.596 `h^(-1)`B. 0.1`h^(-1)`C. 4.802 `h^(-1)`D. 10 `h^(-1)` |
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Answer» Correct Answer - b We have the formula for half life of a first order reaction `t_(1//2)=(0.693)/(k)` where k= rate constant `therefore 6.93 h=(0.693)/(k) rarr k=(0.693)/(6.93h)` `k=0.1 h^(-1)` |
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| 14. |
Which among the following functional groups has been given the highest priority while assigning R-S configurationA. `-C_(6)H_(5)`B. `-CN`C. `-C_(2)H_(5)`D. `-CH_(3)` |
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Answer» Correct Answer - b Cn has been given the hightest priority while assigning the R-S configuration The priority order is written for the following as `-CN gt - C_(6)HH_(5) gt -C_(2)H_(5) gt - CH_(3)` |
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| 15. |
Diethyl amine when treated with nitros acid yieldsA. diethyl alcoholB. N nitroso diethyl amineC. N nitroso diethyl amneD. triethyl ammonium nitrite |
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Answer» Correct Answer - c When diethyl amine treated with nitrous acid it yields N nitroso diethyl amine `H_5C_2-undersetundersetH|N-C_2H_5+HONO overset(-H_2O)to underset"N-nitroso diethl amine ( yellow oily liquid )"(H_5-oversetoverset(C_2H_5)(|)N-N=O)` |
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| 16. |
Give R= 8.314 `JK^(-1) mol^(-1)` the work done during combustion of 0.090 kg of ethane (molar mass=30) at 300 K isA. `-18.7 kJ`B. `18.7 kJ`C. `6.234 kJ`D. `-6.234 kJ` |
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Answer» Correct Answer - b Since combustion takes place at contant pressure hence `DeltaH=3W` `DeltaH= n C_(P)dT` `rarr 3W =(0.090)/(30)xx5/2xx8.314xx300` `=18.7 kJ` |
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| 17. |
What is the possible number of monohydroxy derivatives of a hydrocarbon consisting of five carbon atoms with one methyl group as a branch ?A. 2B. 3C. 4D. 5 |
| Answer» Correct Answer - b | |
| 18. |
If M,W and V represent molar mass of solute then mass of solute and volume of solution in litres respecitively which among following equation is true ?A. `pi=(MWR)/(TV)`B. `pi=(TMR)/(WV)`C. `pi=(TWR)/(VM)`D. `pi=(TRV)/(WM)` |
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Answer» Correct Answer - c We have the realtion between osmotic pressure and molecular mass as `pi CRT` where `pi` = osmotic pressure C= coincentration `pi =(n)/(V)` RT `rarr pi = (M)/(V) xxRT rarr pi =(TWR)/(MV)` |
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| 19. |
A hollow spere of mass M and radius R is rotating with angular frequency `omega` it suddenly stops rotating and 75% of kinetic energy is converted to heat if s is the speicific heat of the material in j / kg k then rise in temperature of the spere is (MI of hollow sphere `=2/3MR^(2)`A. `(R omega)/(4s)`B. `(R^(2)omega^(2))/(4s)`C. `(R omega)/(2s)`D. `(R^(2)omega^(2))/(2s)` |
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Answer» Correct Answer - b Intial rotational kinetic energy of the hollow sphere `k_(i)=1/2xxIomega^(2)=1/2 xx2/3mr^(2)xx(v^(2))/(r^(2))` `=1/3 mv^(2)=1/3m xx omega^(2)R^(2)` according to the question `rarr 3/4xx1/3 omega^(2)R^(2)=S Delta theta` `therefore` Rise in temperature of the sphere `Delta theat =1/4 (omega^(2)R^(2))/(S)` |
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| 20. |
How is ore of aluminium concentrated ?A. roastingB. leachingC. froth floatationD. using wilfley table |
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Answer» Correct Answer - b Leaching is the method of extraction of aluminum form its ore (bauxite) by the treatment of powdered ore with a concentrated solution of NaOH resulting in dissolution of `AI_(2)O_(3)` leaving impurities behind as `AI_(2)O_(33)(s)+2OH^(-)(aq)+3H_(2)Orarr 2[aI(OH)_(4)]^(-)(aq)` This Process is called bayer process |
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| 21. |
For the same angle of incidence the angles fo refreaction in media P,Q ,R and S are `50^(@) ,40^(@),30^(@),20^(@)` respectively the speed of lights is minimum in mediumA. PB. QC. RD. S |
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Answer» Correct Answer - c The speed of light reduces when it travels from rarer to densere medium when light travels from rarer to denser medium it bendds towards athe normal thus speed of light is minimum when it bends maximum towards the normal for same angle of incidence i.e andgle of refraction is least so angle of refraction is least i.e `20^(@)` in medium s |
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| 22. |
Same current is flowing in two alternating circuits. The first circuit contains only inductances and the other contains only a capacitor, if the frequency of the e.m.f of AC is increased, the effect on the value of the current will beA. increase in first circuit and decrease in secondB. increase in both circuitsC. decrease in both circuitsD. decrease in first circuit and increase in second |
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Answer» Correct Answer - c Current corresponding to inductive circuit `I=(V)/(Z)=(V)/(omegaL) rarr I_("inductive") prop (1)/(omega)` similary for capocitive circuit wgeb frequency of AC is increased from Eq (i) `I_("inductive")` decrease abd from Eq (ii) `I_("capactive")` increases |
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| 23. |
The differeence in the effective capacity of two similar capacitor when joined in series and then in paralllel is 6 `mu` F the capacity of each capacitor isA. `2 mu F`B. `4 mu F`C. `8 mu F`D. `16 mu F` |
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Answer» Correct Answer - a When two similar capactors are joined in series then effective capacity is `(1)/(C_(s))=(1)/(c )+(1)/(C )+(1)/(C )=(2)/(C )` `therefore C_(s)=(C )/(2)` when capacitors are joined in parallel then `C_(p)=C+C=2C` since `C_(p)-C_(s)=6 mu f` `therefore 2c-(C )/(2)=6 rarr 4C-c =12` or C=4 muF` |
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| 24. |
A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration `a_(c)` is varying with time t as `a_(c) = k^(2)rt^(2)`, where k is a constant. The power delivered to the particle by the forces acting on it is :A. `m^(2)k^(2)r^(2)t^(2)`B. `mk^(2)r^(2)t`C. `mk^(2)rt^(2)`D. `mkr^(2)t` |
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Answer» Correct Answer - b The centripetal accertation `a_(c)=K^(r)rt^(r ) ` or `V^(2)/ r = k^(2)rt^(2)` v= krt power =`f_(t)v cos 0^(@)=(ma_(1))(krt)` `(mkr)(krt)=mk^(2)r^(2)t` |
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| 25. |
An open and closed organ pipe have the same length the ratio pth mode of frequency of vibration of air in two pipe isA. `P(2p+1)`B. `(2p)/(2p-1)`C. pD. 1 |
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Answer» Correct Answer - c The frequency of note produced in the pth mode of viration in open pipe is `v_("open") =(2p-1)(v)/(4L)` `therefore` Ratio of pth mode of frequency of vibration of air in two pipes is `(V_("open"))/(v_("closed"))=(pv)/(2L)/(29-1)(v)/(4L)=(2p)/(2p-1)` |
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| 26. |
A capacitor `C_(1)=4mu` F is connected is series with another capacitor `C_(2)=1mu`F the combination is connected across DC source of 200 V the ratio of potential across `C_(2)` to `C_(1)` isA. `2:1`B. `4:1`C. `8:1`D. `16:1` |
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Answer» Correct Answer - a In series combination `(1)/(C_(s))+(1)/(C_(2))=1/4+1/1=5/4` `C_(s)=4/5 mu F` now total charge in series combination is given by `q=c_(s)xxV=4/5 xx200 =600 V` `therefore V_(1)` potential across `R_(1)=(160)/(C_(1))=160/4=40 V` so `(V_(2))/(V_(1))=160/40=4/1 rarr V_(2) L: V_(1)=4:1` |
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