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25501.	The process by which a gas through a small hole into vacuum is called effusion. The rate of change of pressure (p) of a gas at constant temperature due to effusion of gas from a vessel of constant volume can be related to rate of change of number of molecules by the expression: (dp)/(dt)=(kT)/(V)((dN)/(dt)) where Rate ofchange of number of molecules rArr""-(dN)/(dt)=(pA_(0))/((2pimkR)^(1//2)) where, k= Boltzmann constant N_(A) = Avogadro's number T= temperature (in K) rArr""-(dN)/(dt)=(pA_(0))/((2pimkR)^(1//2)) N_(A) = Avogadro's number T=temperature (in K) V=volume of vessel N=1 no.of molecules A_(0)= area of aperture m=mass of single molecule gamma=(V)/(A_(0))sqrt((2pim)/(kT)) If under any conditions gamma=100 sec then select the correct graph of pressure of gas at any time (P_(t)) us Time (t). [Given : In 2=0.7]
Answer» ANSWER :B

25502.	The process by which vegetable ghee is manufactured known as:
Answer» SAPONIFICATION HYDROGENATION esterification hydrolysis Answer :B

25503.	The process by which a gas through a small hole into vacuum is called effusion. The rate of change of pressure (p) of a gas at constant temperature due to effusion of gas from a vessel of constant volume can be related to rate of change of number of molecules by the expression: (dp)/(dt)=(kT)/(V)((dN)/(dt)) where Rate ofchange of number of molecules rArr""-(dN)/(dt)=(pA_(0))/((2pimkR)^(1//2)) where, k= Boltzmann constant N_(A) = Avogadro's number T= temperature (in K) rArr""-(dN)/(dt)=(pA_(0))/((2pimkR)^(1//2)) N_(A) = Avogadro's number T=temperature (in K) V=volume of vessel N=1 no.of molecules A_(0)= area of aperture m=mass of single molecule gamma=(V)/(A_(0))sqrt((2pim)/(kT)) If initial pressure is P_(0) then pressure of gas at any time t in the container at constant V and T is given by:
Answer» <P>`P_(0)E^(-t//y)` `P_(0)e^(t//gamma)` `P_(0)(1-e^(-t//gamma))` `P_(0)(1-e^(t//gamma))` ANSWER :A

25504.	The process by which synthesis of protein takes place based on the genetic information present in m-RNA is called
Answer» TRANSLATION transcription replication messenger hypothesis Solution :Translation.

25505.	The process associated with the manufacture sodium carbonate is known as:
Answer» Chamber Haber Le-Blanc Castner Answer :C

25506.	The process of zone refining is used for
Answer» Sillicon GERMANIUM GALLIUM all THEABOVE ANSWER :D

25507.	The process in which ore is heated in air below its melting point is known as :
Answer» ROASTING Calcination Reduction Distillation ANSWER :A

25508.	The probable formula for prussian blue is :
Answer» `Fe_3[Fe(CN)_6]_2` `Fe_2[Fe(CN)_6]_3` `Fe_4[Fe(CN)_6]_3` `Fe_2[Fe(CN_6]_4` Answer :C

25509.	The probability of finding the electron in P_(x) orbital is
Answer» maximum on two OPPOSITE sides of the NUCLEUS along X-axis zero at the nucleus they produces heating effect they can affect PHOTOGRAPHIC plate. Answer :A::B::D

25510.	The prinicipal constituent of superphosphate of lime is
Answer» `Ca_3(PO_4)_2` `CaHPO_4` `CA(H_2PO_4)_2` CaO Answer :C

25511.	The principle on conductometric titration is based in the fact that during the titration, one of the ions is replaced by the otherand invariably thses two ions differ in the ionicconductivity with the result that the conductivity of the solution varies during the course of the titration.Take,for example,the titration between strong acid, say HCl. and a strong base, say NaOH.Before NaOH is added the conuctance of HCl solution has a high value due ti the persence if highly hydrogen ions.As NaOH is added,H^(+) ions are the replaced by relatively slower moving Na^(+) ions.Consequently, the conductance of the solution decrease and this continues right upto the enuivalence point where the solution contains only NaCL.Beyond the equivalence point if more of NaOH is added, then the solution contains an excess of the fast moving OH^(-) with the result that its as more of NaoH added, we plot the conductance value vs the amount of NaOH added we get curve of the type shown in figure.The descending portion AB represents the conductance before the equivalence point (solution cintains a mixture if acid HCl and the salt NaCL and the ascending portion CD represents the conductances after the equivalence point (solution contains the salt NaCL and the excess of NaOH). The point E which repersents only the minimum conductance is alkali and thus repersents the equivalence point.This point can, however, be obtained by the extrapoltion of the lines AB and CD and therefore one is not very particular in location this point experimentally as it is in the case of ordinary acid -base involving the acid base indicators LET us take the specific exaanple of acetic acid being titration against NaOH.Before the addition the of alkali, the solution shows poor conductance due to feeble inization of acetic acid.Initially the addtion of alkali causes not only the replacement of H^(+) by Na^(+) but also suppresses the thus the conductance of the solution decrease in the begnning .But very soon the conductance starts increasing as addetion NaOH thus causing neutralizes the undissociated HAc to HAc with strong condutance eletroyte Na^(+)Ac^(-).The increase in conducting OH^(-) ions gtthe graph near the highly equivalence point actual equivalence point can as usual be ontained by the extrapoiation method In all thses graphs it been assumed that the vloume change due additionof solution from burette in negligible heane vloume changeof the sloution in beaker the conuctance of which is measured is almost constant throughout the measurement If a 100 ml solution of 0.1 MHBr is titration using a very concentrated solution of NaOH then the conductivity (specific conductance0 of this solution at the equivalence wil be (assume volume change is negligible due addition of NaOH). Report your answer after mulitiplyingit with 100 inSm^(-1) [Given : lambda_((Na^(+)))^(@)=8xx10^(-3)Sm^(2)mol^(-1), lambda_((Br^(-)))^(@) = 4 xx 10^(-3) Sm^(2)mol^(-1)]
Answer» 6 12 15 24 Answer :B

25512.	The principle on conductometric titration is based in the fact that during the titration, one of the ions is replaced by the otherand invariably thses two ions differ in the ionicconductivity with the result that the conductivity of the solution varies during the course of the titration.Take,for example,the titration between strong acid, say HCl. and a strong base, say NaOH.Before NaOH is added the conuctance of HCl solution has a high value due ti the persence if highly hydrogen ions.As NaOH is added,H^(+) ions are the replaced by relatively slower moving Na^(+) ions.Consequently, the conductance of the solution decrease and this continues right upto the enuivalence point where the solution contains only NaCL.Beyond the equivalence point if more of NaOH is added, then the solution contains an excess of the fast moving OH^(-) with the result that its as more of NaoH added, we plot the conductance value vs the amount of NaOH added we get curve of the type shown in figure.The descending portion AB represents the conductance before the equivalence point (solution cintains a mixture if acid HCl and the salt NaCL and the ascending portion CD represents the conductances after the equivalence point (solution contains the salt NaCL and the excess of NaOH). The point E which repersents only the minimum conductance is alkali and thus repersents the equivalence point.This point can, however, be obtained by the extrapoltion of the lines AB and CD and therefore one is not very particular in location this point experimentally as it is in the case of ordinary acid -base involving the acid base indicators LET us take the specific exaanple of acetic acid being titration against NaOH.Before the addition the of alkali, the solution shows poor conductance due to feeble inization of acetic acid.Initially the addtion of alkali causes not only the replacement of H^(+) by Na^(+) but also suppresses the thus the conductance of the solution decrease in the begnning .But very soon the conductance starts increasing as addetion NaOH thus causing neutralizes the undissociated HAc to HAc with strong condutance eletroyte Na^(+)Ac^(-).The increase in conducting OH^(-) ions gtthe graph near the highly equivalence point actual equivalence point can as usual be ontained by the extrapoiation method In all thses graphs it been assumed that the vloume change due additionof solution from burette in negligible heane vloume changeof the sloution in beaker the conuctance of which is measured is almost constant throughout the measurement Themost appropriate titration curve ibtained when a mixture of a strong acid (say HCl) and a weak acid (sayCH_(3)COOH0 is titration with (say NaOH) will be
Answer» ANSWER :C

25513.	The principle on conductometric titration is based in the fact that during the titration, one of the ions is replaced by the otherand invariably thses two ions differ in the ionicconductivity with the result that the conductivity of the solution varies during the course of the titration.Take,for example,the titration between strong acid, say HCl. and a strong base, say NaOH.Before NaOH is added the conuctance of HCl solution has a high value due ti the persence if highly hydrogen ions.As NaOH is added,H^(+) ions are the replaced by relatively slower moving Na^(+) ions.Consequently, the conductance of the solution decrease and this continues right upto the enuivalence point where the solution contains only NaCL.Beyond the equivalence point if more of NaOH is added, then the solution contains an excess of the fast moving OH^(-) with the result that its as more of NaoH added, we plot the conductance value vs the amount of NaOH added we get curve of the type shown in figure.The descending portion AB represents the conductance before the equivalence point (solution cintains a mixture if acid HCl and the salt NaCL and the ascending portion CD represents the conductances after the equivalence point (solution contains the salt NaCL and the excess of NaOH). The point E which repersents only the minimum conductance is alkali and thus repersents the equivalence point.This point can, however, be obtained by the extrapoltion of the lines AB and CD and therefore one is not very particular in location this point experimentally as it is in the case of ordinary acid -base involving the acid base indicators LET us take the specific exaanple of acetic acid being titration against NaOH.Before the addition the of alkali, the solution shows poor conductance due to feeble inization of acetic acid.Initially the addtion of alkali causes not only the replacement of H^(+) by Na^(+) but also suppresses the thus the conductance of the solution decrease in the begnning .But very soon the conductance starts increasing as addetion NaOH thus causing neutralizes the undissociated HAc to HAc with strong condutance eletroyte Na^(+)Ac^(-).The increase in conducting OH^(-) ions gtthe graph near the highly equivalence point actual equivalence point can as usual be ontained by the extrapoiation method In all thses graphs it been assumed that the vloume change due additionof solution from burette in negligible heane vloume changeof the sloution in beaker the conuctance of which is measured is almost constant throughout the measurement the nature of curve obtained for the titration between weak acid vs strong base as described in the above will be
Answer» ANSWER :A

25514.	The principle which gives a way to fill the electrons in the available energy level is:
Answer» HUND's rule Pauli's EXCLUSION principle Aufbau principle None of these Answer :C

25515.	The principle of cleaning by soap is
Answer» surface tension floatation viscosity elasticity SOLUTION :Soaps form surface films, REDUCE surface tension of solution and help in REMOVING dirt and DUST by emulsifying GREASE.

25516.	The principle involved in zone refining method is that _______.
Answer» IMPURITIES of low boiling metals can be seperated by distillation impurities are more SOLUBLE in molten metal than in solid metal different components of a mixture are DIFFERENTLY adsorbed on an adsorbent vapours of VOLATILE compound can be decomposed into PURE metal Answer :B

25517.	The principle behind the acidity order Cl_3C COOH > Cl_2CHCOOH > CH_3COOH is the withdrawal of electrons and liberation of protons is due to which effect?
Answer» ANSWER :A

25518.	The principal source of helium is:
Answer» Air Monazite sand Radium All Answer :B

25519.	The principal reducing agent in the metallurgy of iron is ___________.
Answer» ALUMINIUM CARBON Carbon monoxide Carbon dioxide SOLUTION :`DeltaG` is negative

25520.	The principal reaction in the zone of fusion of blast furnace employed in the metallurgy of iron is
Answer» `C+O_(2) to CO_(2)` `2C + O_(2) to 2CO` `CO_(2) + C to CO` `Fe_(2) O_(3) + 3CO to 2FE + 3CO_(2)` Solution :`ZnO + C to ZN + CO`, MIXTURE of ZnO + C to form briquettes.

25521.	The principal reaction in the zone of heat absorption of blast furnace employed in the metallurgy of iron is
Answer» `2C + O_(2) to CO` `2CO + O_(2) to 2CO_(2)` `CaCO_(3) UNDERSET(-CO_(2))to CaO underset(+SiO_(2))to CaSiO_(3)` `C+O_(2) to CO_(2)` Answer :C

25522.	The principal products obtained on heating iodine with concentrated caustic soda solution are:
Answer» NAOI + NAI `NaIO_3 + NaI` `NaOI + NaIO_3 + NaI` `NaIO_4 + NaI` ANSWER :B

25523.	The principal organic compound formed in the reaction CH_2=CH(CH_2)_8COOH+HBroverset("Peroxide")rarr............ is
Answer» `(CH_3)_8COOH` `CH_2=CH(CH_2)_8COBr` `UNDERSET(Br)underset(\|)CH_2-CH_2(CH_2)_8COOH` `CH_2=CH(CH_3)_7-underset(Br)underset(\|)CH-COOH` Solution :`:.CH_2=CH(CH_2)_8COOHoverset(HBR)underset("Peroxide")rarrunderset(Br)underset(\|)CH_2-CH_2(CH_2)_8COOH` HENCE, (C) is the correct answer.

25524.	The principaloxidation state of lanthanides is :
Answer» `+3` `+2` `+4` 0 Answer :A

25525.	The principal hormone responsible for development of female characteristics in controlling of menstrual cycle is…………………….
Answer» PROGESTERONE TESTOSTERONE ESTRADIOL CHOLESTEROL SOLUTION :Estradiol

25526.	The principal gas evolved from sludge digestion tank is :
Answer» CO `CO_2` `CH_4` `N_2` ANSWER :C

25527.	The primary valency of iron in K_(4)[Fe(CN)_(6)] is :
Answer» 2 3 4 5 Answer :A

25528.	The principal buffer present in the blood is:
Answer» `CH_3COONH_4` `CH_3COOH//CH_3COONa` `H_2CO_3//HCO_3^-` `NaH_2PO_4Na_2HP_4` ANSWER :C

25529.	The principal buffer present in human blood
Answer» `NaH_(2)PO_(4) + Na_(2)HPO_(4)` `H_(3)PO_(4) + NaH_(2)PO_(4)` `CH_(3)COOH + CH_(3)COONa` `H_(2)CO_(3) + HCO_(3)^(-)` Solution :`H_(2)CO_(3) + HCO_(3)^(-)` buffer is present in human BLOOD, which maintain PH of blood.

25530.	The primary valency of the metal ion in the co-ordination compound, K_2[Ni(CN)_4] is
Answer» Four Zero Two Six Answer :C

25531.	The primary valency of Iron in K_(4)[Fe(CN)_(6)] is satisfied by
Answer» Six `CN^(-)` IONS TWO `CN^(-)` ion FOUR `K^(+)` ions Two `K^(+)` ions Answer :B

25532.	The primary valency of the central transition metal ion in a complex compound (Cr(NH_(3))_(4)Cl_(2)]Cl
Answer» 3 2 1 0 Answer :A

25533.	The primary valency of 'Fe' in the complex K_(4)[Fe(CN)_(6)] is
Answer» 2 3 6 4 Answer :A

25534.	The primary test to check the sol is colloid or not is…
Answer» by the motion of PARTICLES. by ultrafiltration. TYNDAL EFFECT by the VOL. of particles. SOLUTION :tyndal effect.

25535.	The primary valency of Fe in K_3[Fe(CN)_6] is
Answer» 3 2 1 Zero Answer :C

25536.	The primary structure of a protein tells about
Answer» 3D ARRANGEMENT of all atoms shape of poly PEPTIDE chain SPECIFIC SEQUENCE of amino acids 3D arrangement of oligo peptide chains ANSWER :C

25537.	The primary standard solution for estimation of Na_(2)S_(2)O_(3) is
Answer» `I_(2)` solution `KMnO_(4)` `K_(2)Cr_(2)O_(7)` OXALIC acid Answer :A

25538.	The primary structure of a protein refers to
Answer» `ALPHA `-helical backbone hydrophobic interaction sequence of `alpha`-amino acids fixed CONFIGURATION of the polypeptide backbone Solution :Primary STRUCTURE INVOLVES sequence of `alpha`-amino acid in polypeptide chain. Secondary structure involves `alpha`-helical and `beta`-PLEATED sheet like structures.

25539.	The primary products of photosynthesis in green plants . It contains the element
Answer» Fructose Glucose Maltose Cellulose Answer :B

25540.	The primary pollutant that leads to photochemical smog is :
Answer» nitrogen oxides sulphur dioxide OZONE acrolein ANSWER :B

25541.	The primary and secondary valencies of chromium in the complex ion, dichloro-dioxalatochromium (III) are respectively
Answer» 3, 4 4, 3 3, 6 6, 3 Answer :C

25542.	The primary and secondary valencies of the central metal ion in the complex [Co(NH_(3))_(6)]Cl_(3) respectively are
Answer» 6 and 3 6 and 6 3 and 3 3 and 6 Answer :D

25543.	The primary bateries are ……………. .
Answer» RECHARGEABLE non-rechargeable REVERSIBLE RENEWABLE ANSWER :B

25544.	The pricipal oxidation state of lanthanide is
Answer» `+2` `+3` `+4` `+5` Answer :B

25545.	The pri., sec. and ter.amines can be distinguished by :
Answer» Hinsberg's REAGENT Grignard's reagent Fehling's SOLUTION TOLLEN's reagent ANSWER :A

25546.	The pressure volume work for an ideal gas can be calculated by using the expression W=-int_(V_(i))^(V_(f))p_(ext)dV. The work can also be calculated from the pV-plot by using the area under the curve within the specified limits. When an ideal gas is compressed (a) reversibly or (b) irreversibly from volume V_(i) to V_(f). choose the correct option.
Answer» `W_(("reversible")) = W_(("IRREVERSIBLE"))` `W_(("reversible")) lt W_(("irreversible"))` `W_(("reversible")) gt W_(("Irreversible"))` `W_(("reversible"))=W_(("irreversible"))+p_(ext)*DELTAV` Solution :`W_(("reversible")) lt W_(("irreversible"))` Area under the curve is ALWAYS more in irreversible compression as can be seen from FIGURE. (a) and (b)

25547.	The pressure of two pure liquid A and B which form an ideal solutions are 400 mm Hg and 800 mm Hg respectively at temperature T.A liquid containing 3:1 molar composition pressure can be varied.The solutions is slowly vapourized at temperature T by decreasing the applied pressure starting with a pressure of 760 mm Hg.A pressure gauge (in mm) Hg is connected which give the reading of pressure applied. The reading of pressure Gauge at bubble point is
Answer» <P>500 600 700 None Solution :`X_A=0.75 " " X_B=0.05` above 500 mm HG only liquid phase exists. `P_("bubble") "POINT"=X_A P_A^@+X_B^@ P_B^@` `P_("cqycqfyr fcUnq") =X_A P_A^@+X_B^@ P_B^@` =0.75+400+0.25x800=500 mm `y_A=0.75 , " " y_B=0.5` At DEW point `1/P_T=y_A/P_A^@+y_B/P_B^@implies1/P_T=0.75/400+0.25/800=(1.5+0.25)/800` `implies P_T=800/1.75=457.14` mm Hg Below dew point only vapour phase exists

25548.	The pressure of two pure liquid A and B which form an ideal solutions are 400 mm Hg and 800 mm Hg respectively at temperature T.A liquid containing 3:1 molar composition pressure can be varied.The solutions is slowly vapourized at temperature T by decreasing the applied pressure starting with a pressure of 760 mm Hg.A pressure gauge (in mm) Hg is connected which give the reading of pressure applied. The reading of pressure Gauge at which only vapour phase exists is
Answer» 501 457.14 425 525 Solution :`X_A=0.75 " " X_B=0.05` above 500 MM Hg only liquid PHASE EXISTS. `P_("BUBBLE") "point"=X_A P_A^@+X_B^@ P_B^@` `P_("cqycqfyr fcUnq") =X_A P_A^@+X_B^@ P_B^@` =0.75+400+0.25x800=500 mm `y_A=0.75 , " " y_B=0.5` At DEW point `1/P_T=y_A/P_A^@+y_B/P_B^@implies1/P_T=0.75/400+0.25/800=(1.5+0.25)/800` `implies P_T=800/1.75=457.14` mm Hg Below dew point only vapour phase exists

25549.	The pressure under which liquid and its vapour can coexist in equilibrium is known as:
Answer» NORMAL VAPOUR pressure Saturated vapour pressure Read vapour pressure Limiting vapour pressure Answer :B

25550.	The pressure of two pure liquid A and B which form an ideal solutions are 400 mm Hg and 800 mm Hg respectively at temperature T.A liquid containing 3:1 molar composition pressure can be varied.The solutions is slowly vapourized at temperature T by decreasing the applied pressure starting with a pressure of 760 mm Hg.A pressure gauge (in mm) Hg is connected which give the reading of pressure applied. The reading of pressure Gauge at which only liquid phase exists.
Answer» <P>499 399 299 None Solution :`X_A=0.75 " " X_B=0.05` above 500 mm HG only liquid PHASE exists. `P_("bubble") "point"=X_A P_A^@+X_B^@ P_B^@` `P_("cqycqfyr fcUnq") =X_A P_A^@+X_B^@ P_B^@` =0.75+400+0.25x800=500 mm `y_A=0.75 , " " y_B=0.5` At dew point `1/P_T=y_A/P_A^@+y_B/P_B^@implies1/P_T=0.75/400+0.25/800=(1.5+0.25)/800` `IMPLIES P_T=800/1.75=457.14` mm Hg Below dew point only VAPOUR phase exists

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