Saved Bookmarks
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. |
The values of `wedge_(m)^(oo)` for`NH_(4)Cl,NaOH,` and `NaCl` are, respectively, `149.74,248.1`,and `126.4 ohm^(-1)cm^(2)eq^(-1)`. The value of `wedge_(eq)^(oo)NH_(4)OH` isA. ` 371.44`B. ` 271.44`C. ` 71. 44`D. It cannot be calculated from the data given |
|
Answer» Correct Answer - B `Lambda_(eq)^(infty) ` or ` Lambda_(CH_4OH)^(infty) = Lambda_(eq_(NH_4Cl))^(infty) + Lambda_(eq_(NaOH)^(infty) -Lambda(eq_(NaCl))^(infty)` ` = (149. 74 + 248.1 - 126.4) = 271. 44 "ohm"^(-1) cm^2 eq^(-1)`. |
|
| 2. |
Which among the following solutions is NOT used in determination of the cell constant?A. `10^(-2)` M KClB. `10^(-1)` M KClC. 1 M KClD. Saturated KCl |
|
Answer» Correct Answer - C Saturated KCl. |
|
| 3. |
`AgNO_(3)` solution will not give white precipitate withA. `[Co(NH_(3))_(3)Cl_(3)]`B. `[Co(NH_(3))_(4)Cl_(2)]Cl`C. `[Co(NH_(3))_(5)Cl]Cl_(2)`D. `[Co(NH_(3))_(6)]Cl_(3)` |
|
Answer» Correct Answer - A no free `Cl^(-)` in the ionisation sphere. |
|
| 4. |
Which of the following two reagents form silicates when react with silica `(SiO_(2))`?A. `NaCl` and `Na_(2)S`B. `NaOH` and `Na_(2)CO_(3)`C. `Na_(2)SO_(4)` and `NaNO_(3)`D. `NaBr` and `NaHSO_(4)` |
|
Answer» Correct Answer - B `SiO_(2)+2NaOHtoNa_(2)SiO_(3)+H_(2)O` `SiO_(2)+Na_(2)CO_(3)toNa_(2)SiO_(3)+CO_(2)` |
|
| 5. |
Which among the following is a feature of adiabatic expansion ?A. `DeltaV lt 0`B. `Delta U lt 0`C. `Delta U gt 0`D. `Delta T =0` |
|
Answer» Correct Answer - B A/c to first law of thermodyanimcs `Delta U = q+w` In adiabatic process `q=0 therefore DeltaU =W` In adiabatic expansion `W = -ve therefore -W = - DeltaU` i.e., `DeltU lt O` |
|
| 6. |
Calculate the work done during compression of 2 mol of an ideal gas from a volume of `1m^(3)` to `10 dm^(3)` 300 K against a pressure of 100 KPa .A. `-99 kJ`B. `+99` kJC. `+22.98` kJD. `-22.98` kJ |
|
Answer» Correct Answer - B `n=2 moles, V_(1) = 1m^(3) = 10 + 3 dm^(3), V_(2) = 10 dm^(3), pi =100 k Pa` `W=-P_(ex) (V_(2)-V_(1))` `=-100 kPa (10 dm^(3) -1000 dm^(3))` `=-100 kPa (-990 dm^(3))` `+100 xx 990 J` `(therefore kPa xx dm^(3) =J)` `=+99 kJ` |
|
| 7. |
Molar heat capacity of water in equilibrium with ice at constant pressure is (a) zero (b) ∞(c) 40.45 kJ K–1 mol–1 (d) 75.48 J K–1 mol–1 |
|
Answer» The Correct option is (b) ∞ |
|
| 8. |
Entropy is a measure of (a) disorder (b) internal energy (c) efficiency (d) useful work done by the system |
|
Answer» Entropy is a measure of disorder. |
|
| 9. |
The colour and magnetic nature of manganate ion `(MnO_(4)^(2-))` isA. green, processB. purple, diamagneticC. green, diamagneticD. purple, paramagnetic |
|
Answer» Correct Answer - A `MnO_(4)^(2-)` green and paramagnetic, because it has one unpaired electron. |
|
| 10. |
Which method would you suggest for the separation of the metals in the following mixtures ? (a) Zinc and iron (b) Copper and magnesium (c ) Rare earths Give reasons for your choice. |
|
Answer» Zinc and iron : Zinc and iron can be separated from the mixture by fractional distillation. The mixture is distilled, when zinc with low boiling point distils over leaving behind iron. (ii) Copper and magnesium : Copper and magenesium metals can be separated from the mixture by electrolytic refining. The mixture metal is converted into a rod and made the anode, while a thin thin plate of Pure copper serves as the cathode. The electrolytic tank contains a solution of copper sulphate acidified with dil. `H_(2)SO_(4)`, which acts as the electrolyte on passing electricity. `Cu^(2+)` and thus these `Cu^(2+)` ions are discharged at the cathode as pure metal. (iii) Rare earths : Rare earths includes lanthanides and actinides. All lanthanide ions are typically trivalent `M^(3+)` and almost identical in size. Their chemical properties. which are determined by the size and charge of their ions, are almost identical. This renders the separation of one metal from another difficult . Different methods employed for their separation given below are based on the light diffenerces in their solubility, stability, and basic properties. Modern methods are based on valence change and ion exchange. (1) Ion exchange method (This is a very effective and rapid method) (2) Complex formation (3) Solvent extraction (4) Fractional crystallisation Valency change method (5) Thermal reaction (6) Precipitation. |
|
| 11. |
Name the main steel plants which are operated by the Steel Authority of India. |
|
Answer» The main steel plants which are operated by Steel Authority of India are (i) Tata and Iron steel Company, Jamshedpur. (ii) Indian Iron and Steel Company, Hirapur(Asansol) (iii) Mysore Iron and Steel works, Bhadravai Three stell plants set up in public sector are located in Ruoukela, Bhilai and Durgapur under the manegement of Hindustan Steel Ltd. |
|
| 12. |
(ii) Carbon reduction is used for the extraction of :A. FeB. KC. AlD. None of these |
|
Answer» Correct Answer - A Carbon reduction process is used for extraction of less electropositive metals like `Pb,Fe,Zn,Sb` etc from their ores. |
|
| 13. |
Enzyme is:(a) Carbohydrate(b) Lipid(c) Proteins(d) None of these |
|
Answer» Answer is (c) Proteins |
|
| 14. |
Natural rubber is a polymer of:(a) Styrene(b) Isoprene(c) Chloroprene(d) Butadiene |
|
Answer» Answer is (b) Isoprene |
|
| 15. |
Antibiotic used for the treatment of typhoid is:(a) Penicillin(b) Chloramphenicol(c) Terramycin(d) Sulphadiazine |
|
Answer» Answer is (a) Penicillin |
|
| 16. |
State and explain faraday's 2nd Law of Electrolysis. |
|
Answer» Faraday's second law of electrolyses: The amount of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights (Atomic mass of metal - Number of electrons required to reduce the metal) |
|
| 17. |
Pink colour of LiCl crystals is due to: (a) Schottky defect (b)Frenkel defect (c) Metal excess defect (d) Metal deficiency defect |
|
Answer» Correct answer is (c) Metal excess defect- It creates F-center in the crystal lattice which is due to a type of crystallographic defect in which an anionic vacancy in a crystal lattice is occupied by one or more unpaired electrons and that is responsible for the color. Answer is (c) Metal excess defect (formation of F centres) |
|
| 18. |
The compound that is least readily nitrated is .......(a) phenol (b) Toluene (c) Ethylbenzene (d) Benzoic acid(e) Xylene |
|
Answer» (d) Benzoic acid |
|
| 19. |
The oimpossible resonating structures of fluorobenzene is/are :A. B. C. D. |
| Answer» Correct Answer - C::D | |
| 20. |
Arrange the following compounds in increasing order of their acid strengths :(CH3)2CHCOOH, CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH |
|
Answer» (CH3)2CHCOOH < CH3CH(Br)CH2COOH < CH3CH2CH(Br)COOH |
|
| 21. |
What is the shortest wavelength in Bracket series of `He^(+)` spectrum ?A. `(100)/(9R_(H))`B. `(16)/(R_(H))`C. `(25)/(R_(H))`D. `(9)/(R_(H))` |
|
Answer» Correct Answer - B Shortest wavelength of Bracket series is correspond to `oo rarr 4` transition `n_(1) = 4, n_(2) = oo` `(1)/(lambda_("min")) = R_(H) xx 2^(2) ((1)/(4^(2)) - (1)/(oo^(2))) rArr lambda_("min") = (4)/(R_(H))` |
|
| 22. |
`40 ml` of `0.5 M K Br` and `60 ml 1 M K Br` are mixed. The solution is then heated to evaporate water until the total volume is `20 ml`. What is the final molarity of `K Br`:A. `1 M`B. `3 M`C. 4 MD. 2 M |
|
Answer» Correct Answer - C `[K Br]_("final") = (n_(K Br))/(V_("final")) = (M_(1)V_(1) + M_(2)V_(2))/(V_("final")) = (0.5 xx 40 + 1 xx 60)/(20)` `= (80)/(20) = 4M` |
|
| 23. |
Consider the equilibrium `SO_(3) (g) hArr SO_(2) (g) + (1)/(2) O_(2) (g) " " K_(C ) = 1` What should be the initial concentration so that at equilibrium `[SO_(3)] = [O_(2)]`A. `(4)/(3) M`B. `(1)/(3)M`C. `(1)/(4)M`D. `(3)/(4)M` |
|
Answer» Correct Answer - D `{:(,SO_(3) (g),hArr,SO_(2) (g),+,(1)/(2) O_(2) (g),K_(C) = 1,),(t = 0,C_(0),,0,,0,,),("conc.",,,,,,,),(t = t_("eqm"),C_(0) - x,,x,,(x)/(2),,),("conc.",,,,,,,):}` Given `[SO_(3)] = (O_(2)]` `C_(0) - x = (x)/(2)` `(3x)/(2) = C_(0) rArr x = (2C_(0))/(3)` at equilibrium `[O_(2)] = (x)/(2) = (C_(0))/(3), [SO_(2)] = x = (2C_(0))/(3), [SO_(3)] = C_(0) - x` `[SO_(3)] = C_(0) - (2C_(0))/(3)` `[SO_(3)] = (C_(0))/(3)` `K_(C) = 1 = ([SO_(2)] [O_(2)]^(1//2))/([SO_(3)]) = ((2C_(0))/(3) xx ((C_(0))/(3))^(1//2))/((C_(0))/(3))` `1 = 4 (C_(0))/(3)` `C_(0) = (3)/(4) M` |
|
| 24. |
Light with a wavelength `310 nm` fell on strontium surface, the electrons were ejected. If maximum kinetic energy of an ejected electron is `1.5 eV`. Then [Given : `lambda_(e) = sqrt((150)/(DeltaV)) Å` where `Delta V=` Voltage difference of battery]A. de-Broglie wavelength of electron is `10 Å`B. Work fuction of strontium is `2.5 eV`C. Threshold wavelength for strontium metal will `496 nm`D. All ejected phot electrons will have kinetic energy `= 1.5 eV` |
|
Answer» Correct Answer - A::B::C `Delta E = (1240)/(310 nm) eV = 4.0 eV` `(KE_("max"))_(e) = 1.5 eV = q Delta V` `Delta V = 1.5 V` (A) `lambda_(e) = sqrt((150)/(Delta V))Å = sqrt((150)/(1.5)) Å = 10 Å` (B) `Delta E = KE_(e) + w.f` `3 = 1.5 + w.f.` wf `= 2.5 eV` (C) `lambda = (1240)/(w.f.) nm = (1240)/(2.5) nm = 496 nm` |
|
| 25. |
A `6.4 gm` sample of methanol `(CH_(3) OH)` was placed in an otherwise empty 1 litre flask and heated to `227^(@)C` to varpoise the methanol. Methanol vapour decomposes by following gasesous compound to effuse out of flask. Measurement shows it contains 32 times as much as `H_(2) (g)` as `CH_(3)OH (g)`. Then Value of `K_(C)` for this reaction is :A. `(16)/(25)`B. `(4)/(5)`C. `((4)/(5))^(3)`D. `((16)/(25))^(2)` |
|
Answer» Correct Answer - D `{:(CH_(3)OH (g),hArr,CO (g),+,2H_(2) (g),{"Given " a_(0) = (n_((CH_(3)OH)))_("initial") = (6.4)/(32) = 0.2},),(t = 0 "mole " a_(0) - a_(0) alpha,,a_(0) alpha,,2a_(0) alpha,,):}` `sum n_("eqm") = a_(0) (1 + 2 alpha)` `(r_(H_(2)))/(r_(CH_(3)OH)) = (p_(H_(2)))/(P_(CH_(3)OH)) sqrt(M_(CH_(3)OH)/(M_(H_(2)))) = (2 a_(0) alpha)/(a_(0) (1 - alpha)) sqrt((32)/(2))` `(32)/(1) = (Deltan_(H_(2)))/(Deltan_(CH_(3)OH)) = (2 alpha)/(1 - alpha) xx 4` `4 = (alpha)/(1 - alpha) rArr 4 - 4 alpha = alpha` `alpha = (4)/(5)` `K_(C) = (a_(0) alpha (2a_(0) alpha)^(2))/(a_(0) (1 - alpha)) xx (1)/(V^(2))` `K_(C) = (4 alpha^(3) xx a_(0)^(2))/(1 - alpha) xx (1)/(V^(2)) = 4 xx ((4)/(5))^(3) xx ((0.2)^(2))/(1 - (4)/(5)) = ((16)/(25))^(2)` |
|
| 26. |
Write any two characteristics of Chemisorption |
|
Answer» 1. Chemisorption is highly specific in nature. It occurs only if there is a possibility of chemical bonding between the adsorbent and the adsorbate. |
|
| 27. |
Which of the following d-orbital participates in the hybridization of central atom in the molecule of `IF_(7)` ?A. `d_(xy)`B. `d_(yz)`C. `d_(zx)`D. `d_(z^(2))` |
|
Answer» Correct Answer - A::D `IF_(7)` has `sp^(3) d^(3)` hybridization involving `s, p_(x), p_(y), p_(z), d_(z^(2)), d_(x^(2) - y^(2)), d_(xy)` |
|
| 28. |
A `6.4 gm` sample of methanol `(CH_(3) OH)` was placed in an otherwise empty 1 litre flask and heated to `227^(@)C` to varpoise the methanol. Methanol vapour decomposes by following gasesous compound to effuse out of flask. Measurement shows it contains 32 times as much as `H_(2) (g)` as `CH_(3)OH (g)`. Then Total pressure of mixture at equilibrium.A. `2.08` atmB. `20.8` atmC. `5.2` atmD. `10.4` atm |
|
Answer» Correct Answer - B `K_(C) = (4^(2))/(5^(2)) xx (1)/(5^(2)) = ((4)/(5))^(4)` `n_("total") = a_(0) (1 + 2 alpha) = 0.2 (1 + (8)/(5)) = (2.6)/(5)` `P_("total") = (n_(t))/(V) RT = (2.6)/(5) xx 0.08 xx 500` `P_("total") = 20.8` atm |
|
| 29. |
Instead of principal quantum number (n), azimuthal quantum number `(l)` & magnetic quantum number (m), a set of new quantum number s, t & u was introduced with similar logic but different values as defined below `s = 1, 2, 3,.......oo` all positive integral values. `t = (s^(2) - 1^(2)), (s^(2) - 2^(2)), (s^(2) - 3^(2))`.........No negative value `u = - ((t + 1))/(2) "to" + ((t + 1))/(2)` (including zero, if any) in integral steps. Each orbital can have maximum four electrons. `(s + t)` rule is defined, similar to `(n + l)` rule. Number of electrons that can be accommodated in `s = 2` and `s = 3` shell.A. `14, 38B. 28, 76C. 8, 28D. None of these |
|
Answer» Correct Answer - B `{:(s = 1,t = 0,u = - (1)/(2)","(1)/(2),),(s = 2,t = 0,u = - (1)/(2)","(1)/(2),),(,t = 3,u = - 2"," -1"," 0","+1","+2 ,),(s = 3,t = 0,u = - (1)/(2)","(1)/(2),),(,t = 5,u = -3"," -2"," -1"," 0"," +1"," +2"," +3,),(,t = 8,u = - (9)/(2) "," - (7)/(2)"," - (5)/(2)"," - (3)/(2)"," - (1)/(2)"," (1)/(2)"," (3)/(2)"," (5)/(2)"," (7)/(2)"," (9)/(2),):}` According to `(s + t)` rule increasing order of energy 1s 2s 3s 4s 2f 5s `{:((i),"Number of electron in " s = 2,7 xx 4 = 28,,),(,"Number of electron in " s = 3,19 xx 4 = 76,"Hence (B)",):}` `{:((ii),"For " Z = 24,1s^(8) 2s^(8) 3s^(8) " i.e. number electrons (s) in " s = 2,t = 3 " " [2f^(@)] = 0,),(,"Hence" (C),,,):}` |
|
| 30. |
Instead of principal quantum number (n), azimuthal quantum number `(l)` & magnetic quantum number (m), a set of new quantum number s, t & u was introduced with similar logic but different values as defined below `s = 1, 2, 3,.......oo` all positive integral values. `t = (s^(2) - 1^(2)), (s^(2) - 2^(2)), (s^(2) - 3^(2))`.........No negative value `u = - ((t + 1))/(2) "to" + ((t + 1))/(2)` (including zero, if any) in integral steps. Each orbital can have maximum four electrons. `(s + t)` rule is defined, similar to `(n + l)` rule. Number of electrons foe which `s = 2, t = 3` for an element with atomic number 24A. 8B. 4C. 0D. None of these |
|
Answer» Correct Answer - C `{:(s = 1,t = 0,u = - (1)/(2)","(1)/(2),),(s = 2,t = 0,u = - (1)/(2)","(1)/(2),),(,t = 3,u = - 2"," -1"," 0","+1","+2 ,),(s = 3,t = 0,u = - (1)/(2)","(1)/(2),),(,t = 5,u = -3"," -2"," -1"," 0"," +1"," +2"," +3,),(,t = 8,u = - (9)/(2) "," - (7)/(2)"," - (5)/(2)"," - (3)/(2)"," - (1)/(2)"," (1)/(2)"," (3)/(2)"," (5)/(2)"," (7)/(2)"," (9)/(2),):}` According to `(s + t)` rule increasing order of energy 1s 2s 3s 4s 2f 5s `{:((i),"Number of electron in " s = 2,7 xx 4 = 28,,),(,"Number of electron in " s = 3,19 xx 4 = 76,"Hence (B)",):}` `{:((ii),"For " Z = 24,1s^(8) 2s^(8) 3s^(8) " i.e. number electrons (s) in " s = 2,t = 3 " " [2f^(@)] = 0,),(,"Hence" (C),,,):}` |
|
| 31. |
Select the correct statement (s):A. Average translational kinetic energy of `O_(2)` is more than He if both are taken at same temperature.B. At room temperature He shows positive deviation.C. At constant pressure average translational kinetic energy depends on volume.D. If we increase temperature in a rigid closed vessel then mean free path will remain unchanged. |
|
Answer» Correct Answer - B::C::D (A) `KE = (3)/(2) nRT` KE depend on number of moles (B) For He, `Z gt 1` (C) `KE = (3)/(2) nRT = (3)/(2) PV` At constant pressure average translational kinetic energy depends on volume. (D) `lambda = (1)/(sqrt2 pi sigma^(2)N^(**))` `N^(**) = (N)/(V)` at constant volume (rigid closed vessel) `N^(**)` remains constant |
|
| 32. |
Match the following :(a) Pure nitrogen(i) Chlorine(b) Haber process(ii) Sulphuric acid(c) Contact process (iii) Ammonia(d) Deacon’s process(iv) Sodium azide or Barium azide Which of the following is the correct option? (a) (b) (c) (d)(1) (i) (ii) (iii) (iv)(2) (ii) (iv) (i) (iii)(3) (iii) (iv) (ii) (i)(4) (iv) (iii) (ii) (i) |
||||||||
|
Answer» Correct option (4) (iv) (iii) (ii) (i) Explanation:
|
|||||||||
| 33. |
Mass if `CO_(2)` Produced on heating 20g of 40% pure limestone :-A. 8gmB. 8.8gmC. 3.52gmD. none of these |
|
Answer» Correct Answer - C `{:(CaCO_(3),to, CaO,+CO_(2),),( 100g , ,56g , 44g,):}` wt of pure limestone ` = (20xx40)/(100)=8gm` 100 gm pure `CaCO _(3)` produce ` to 44g CO_(2)` 8 g pure `Caco_(3)` produce to `3.52 g CO_(2)` |
|
| 34. |
Calculate the amount of 50% `H_(2)SO_(4)` required to decompose 25g calcium carbonate :-A. 98gmB. 49gmC. 24.5gmD. 196gm |
|
Answer» Correct Answer - B `{:(CaCO_(3)+,H_(2)SO_(4),to,CaSo_(4),+H_(2)O+CO_(2),),(100g,98g,,,,):}` `100g CaCO_(3) require 98 gm H_(2)SO_(4)` ` 25 g CaCO _(3)require (98)/(100)xx25 gm pure H_(2) SO_(4)` ` = 24.5g pure H_(2)SO_(4)` Hence, wt 50% pure `H_(2)SO_(4)` will be 49.5 gm |
|
| 35. |
Explain the terms : (i) CMC, (ii) Kraft temperature (Tk). |
|
Answer» [Hint : CMC : Concentration above which micelle formation took place. Kraft Temperature : It is the temperature above the micelle formation took place.] |
|
| 36. |
Number of electron having `l+m` value equal to zero in `._(26)Fe` may beA. 13B. 14C. 7D. 12 |
|
Answer» Correct Answer - A,B `._26Feto1s^2,2s^2,2p^6,3s^23p^6,3d^6,4s^2` `{:(l+m=0implies,l=0,m=0,i.e.s-"subshell"),(,l=1,m=-1,i.e."one orbital of p"),(,l=2,m=-2,i.e. "one of d-orbitals"):}` hence there are 13 or 14 electron as in d-orbital it may be one or two electron having m=-2 |
|
| 37. |
The ration of the `e//m` (specific charge) values of an electron and an `alpha-"particle"` isA. `2:1`B. `1:1`C. `1:1`D. None of these |
|
Answer» Correct Answer - D `(e/m)_e=e/m_e(e/m)_alpha=(2e)/m_alpha=(4e)/(4m_p)=(2e)/(4xx1840m_e)implies ((e//m)_e)/((e//m)_alpha)=3680` |
|
| 38. |
Which of the following statement is correct for `3d_(xy)` orbitals ? A. The orbitals drawn has two nodal planes , xz and yz.B. The minimum probability point lie along `theta=45^@`C. `+ve` and -ve signs represent sign of amplitude of electron waveD. It is a non-axial orbital |
|
Answer» Correct Answer - A,C,D These are the fact. |
|
| 39. |
Electrons are revolving around the nucleus in `n_(1^(th))` orbit of an atom, have atomic number `Z_1`, and in the `n_2` orbit of other atom, have atomic number `Z_2`, then [Where P= Linear momentum, L=Angular momentum, f=frequency of revolution and K.E. =kinetic energy]A. `L_1/L_2=n_1/n_2`B. `P_1/P_2=(Z_1n_2)/(Z_2n_1)`C. `f_1/f_2 = (Z_2/Z_1)^2 (n_1/n_2)^3`D. `((K.E)_1)/((K.E)_2)=(Z_1/Z_2 . n_2/n_1)^2` |
|
Answer» Correct Answer - A,B,D `L_1/L_2=(mv_1r_1)/(mv_2r_2)=(Z_1/n_1xxn_1^2/Z_1)/(Z_2/n_2xxn_2^2/Z_2)=n_1/n_2implies P_1/P_2(mv_1)/(mv_2)=(Z_1/n_1)/(Z_2/n_2)=(Z_1n_2)/(Z_2n_1)` `(f_1)/(f_2) =(v_1/(2pir_1))/(v_2/(2pir_2))=(Z_1/n_1xxZ_1/n_1^2)/(Z_2/n_2xxZ_2/n_2^2)=(Z_1/Z_2)^2.(n_2/n_1)^3implies (K.E._1)/(K.E._2)=(1/2mv_1^2)/(1/2mv_2^2)=(Z_1/Z_2)^2xx(n_2/n_1)^2=((Z_1n_2)/(Z_2n_1))^2` `(K.E_1)/(K.E_2)=(1/2(KZ_1e^2)/r_1)/(1/2(KZ_2e^2)/r_2)=(Z_1/n_1^2.Z_1)/(Z_2.Z_2/n_2^2)=(Z_1/Z_2.n_2/n_1)^2` |
|
| 40. |
In a hydrogen like sample two different types of photons A and B are produced by electronic transition. Photon B has its wavelength in infrared region. If photon A has more energy than B, then the photon may belong to the region:A. ultrovioletB. visibleC. infraredD. None |
|
Answer» Correct Answer - A,B,C Since B is in infrared region and A has more energy than B hence it will have lesser wavelength i.e. ultra violet, visible and infrared region. |
|
| 41. |
Select the correct statementA. The value of spin only magnetic moment of `Co^(3+)` ion (in BM)=`sqrt(24)`B. The number of radial nodes in a 3p-orbital =1C. The number of electrons with (m=0) in `Mn^(2+)` ion =11D. The orbital angular momentum for the unpaired electron in `V^(4+)=(sqrt6h)/(4pi)` |
|
Answer» Correct Answer - A,B,C (a)`Co^(3+):1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 :. 4` unpaired electrons `:. Mu=sqrt(4(4+2))=sqrt24=4.9 BM` (b)Number of radial nodes=n-l-1 Number of radial nodes in 3p orbital =3-1-1=1 ( c)Number of electrons with (m=0) in `Mn^(2+) (1s^2 2s^2 2p^6 3s^2 3p^6 3d^5)` ion=`1s(2)+2s(2)+2p(2)+3s(2)+3p(2)+3d(1)=11` (d)Orbital angular momentum for the unpaired electron in `V^(4+)` lies in 3d orbital `:. l=2` `:.` Orbital angular momentum=`sqrt(l(l+1))h/(2pi)=(sqrt6h)/(2pi)` |
|
| 42. |
The ejection of the photoelectron from the silver metal in the photonelectric effect exeriment can be stopped by applying the voltage of `0.35 V` when the radiation `256.7 nm` is used. Calculate the work function for silver metal. |
|
Answer» Energy of incident radiation = Work function + Kinetic energy of photoelectron Energy of incident raidation `(E)` `=hv=h c/lambda` `=((6.626xx10^(-34)J s)(3.0xx10^(8) m s^(-1)))/((256.7xx10^(-9)m))` `=7.74xx10^(-19)J` `=4.83 eV(1 eV=1.602xx10^(-19)J)` The potential applied gives the kinetic energy to the electron. Hence, kinetic energy of the elecyron`=0.35 eV` :. Work function `4.83 eV-0.35 eV=4.48 eV` |
|
| 43. |
1)flowering plants, 2)hydra, 3)salmonella, 4)green algae |
|
Answer» In bacteria both reproduction and growth are synonymous so according to options it is salmonella typhymurium |
|
| 44. |
Which velocity is known as “If unequal displacements in equal intervals of time is moving” A. average velocity B. instantaneous velocity C. non-uniform velocity D. none of the above |
|
Answer» C. non-uniform velocity |
|
| 45. |
A wire of resistance R is cut into n equal parts. These parts are then connected in parallel. The equivalent resistance of the combination will be (a) nR(b) R/n(c) n/R(d) R/n2 |
|
Answer» Correct option(d) Explanation: Resistance of wire = R If the wire is cut into n equal parts, the resistance of each wire is R/n and connected in parallel. Req = R/n2 |
|
| 46. |
Which velocity is known as “If unequal displacements in equal intervals of time is moving” A. instantaneous velocity B. uniform velocity C. non-uniform velocity D. average velocity |
|
Answer» B. uniform velocity |
|
| 47. |
Which type of motion is “a train moving on a track “ ? A. Circular B. rectilinear C. Periodic D. none of the above |
|
Answer» Rectilinear. “Rectilinear motion is the motion of an object that moves in a straight line. |
|
| 48. |
Which of the following is a type of motion ? A. Circular B. Rectilinear C. Periodic D. All the above |
|
Answer» D. All the above |
|
| 49. |
If an object moves along a straight path it is said to be ………………… motion A. linear B. one-dimensional C. Both A and B D. two-dimensional. |
|
Answer» C. Both A and B |
|
| 50. |
Which type of motion of an object that moves in a straight line ? A. Rectilinear motion B. Periodic motion C. Circular motion D. none of the above |
|
Answer» Rectilinear motion. (a train moving on a track, a parade, coins tossed in the air are all in rectilinear motion.) |
|