100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS

1.	A particle moves in such a way that its position vector at any time t is `vec(r)=that(i)+1/2 t^(2)hat(j)+that(k)`. Find as a function of time: (i) The velocity `((dvec(r))/(dt))` (ii) The speed `(\|(dvec(r))/(dt)\|)` (iii) The acceleration `((dvec(v))/(dt))` (iv) The magnitude of the acceleration (v) The magnitude of the component of acceleration along velocity (called tangential acceleration) (v) The magnitude of the component of acceleration perpendicular to velocity (called normal acceleration).
Answer» Correct Answer - (i) `hat(i)+that(j)+hat(k)`, (ii) `sqrt(t^(2)+2)`, (iii) `hat(j)`, (iv) `t//sqrt(t^(2)+2)`, (vi) `sqrt(2)//sqrt(t^(2)+2)` `vec(r)=that(i)+t^(2)/2hat(j)+that(k)` (i) `vec(v)=(dvec(r))/(dt)=hat(i)+that(j)+hat(k)" "` (iii) speed `\|vec(v)\|=sqrt(t^(2)+2)` (iii) `vec(a)=(dvec(v))/(dt)=hat(j)" "` (iv) `\|vec(a)\|=1` (v) `vec(a)_(T)=(vec(a).hat(v))hat(v)=([hat(j)((hat(i)+that(j)+hat(k)))/(sqrt(t^(2))+2)])((hat(i)+that(j)+hat(k)))/sqrt(t^(2)+2)` `vec(a)_(T)=(t/sqrt(t^(2)+2))hat(v)=(t(hat(i)+that(j)+hat(k)))/((t^(2)+2)), \|vec(a)_(T)\|=t/sqrt(t^(2)+2)` As `a_(N)^(2)+a_(T)^(2)=a^(2)` so `a_(N)=sqrt(a^(2)-a_(T)^(2))=sqrt(2)/sqrt(t^(2)+2)`

Discussion

2.	A ball `A` is thrown up vertically with a speed `u` and at the same instant another ball `B` is released from a height `h`. At time `t`, the speed `A` relative to `B` isA. uB. `u-2gt`C. `sqrt(u^(2)-2gh)`D. `u-gt`
Answer» `a_("rel")=0 U_("rel")-cosnt=U`

Discussion

3.	Identify the pair whose dimensions are equalA. Troque and workB. stress and energyC. force and stressD. Force and work
Answer» Correct Answer - A The dimensions of torque and work are `[ML^(2)T^(-2)]`

Discussion

4.	When two concentric shells are connected by a thin conducting wire, whole of the charge of inner shell transfers to the outer shell and potential difference between them becomes zero. Surface charge densities of two thin concentric spherical shells are `sigma` and `-sigma` respectively. Their radii are R and `2R`. Now they are connected by a thin wire. Suppose electric field at a distance `r=(3R)/2` was `E_(1)` before connecting the two shells and `E_(2)` after connecting the two shells, then `\|E_(2)/E_(1)\|` is :-A. zeroB. 1C. `9//8`D. `8//9`
Answer» Correct Answer - A Electric field at `r=(3R)/2` [Before connecting] `x=(3R)/2` `E_(1)=1/(4pi in_(0))xx(sigma4piR^(2))/((3R)/2)^(2)+0` After connecting `E_(2)=0` Hence `\|E_(2)/E_(1)\|=0`

Discussion

5.	When two concentric shells are connected by a thin conducting wire, whole of the charge of inner shell transfers to the outer shell and potential difference between them becomes zero. Surface charge densities of two thin concentric spherical shells are `sigma` and `-sigma` respectively. Their radii are R and `2R`. Now they are connected by a thin wire. Suppose electric field at a distance `r (gt 2R)` was `E_(1)` before connecting the two shells and `E_(2)` after connecting the two shells, then `\|E_(2)/E_(1)\|` is :-A. zeroB. 1C. 2D. `.^(1)//_(2)`
Answer» Correct Answer - B Electric field `(r gt 2R)` before connecting the shells `E_(1)=(1xxsigmaxx4piR^(2))/(4pi in_(0)xxr^(2))+(1(-sigma)xx4piR^(2)xx4)/(4pi in_(0)xxr^(2))=sigma/in_(0)[R^(@)/r^(2)-(4R^(2))/r^(2)]` After connecting `E_(2)=1/(4pi in_(0))([Q_(1)-Q_(2)])/r^(2)=sigma/in_(0)[R^(2)/r^(2)-(4R^(2))/r^(2)]` `:. E_(1)//E_(2)=1`

Discussion

6.	Two metal sphere of capacitance `C_(1) andC_(2)` carry some charges. They are put in contact and then separated. If final charges `Q_(1) and Q_(2)` on them, then:A. `(Q_(1))/(Q_(2)) lt (C_(1))/(C_(2))`B. `(Q_(1))/(Q_(2)) =(C_(1))/(C_(2))`C. `(Q_(1))/(Q_(2)) gt (C_(1))/(C_(2))`D. `(Q_(1))/(Q_(2))lt(C_(2))/(C_(1))`
Answer» Correct Answer - B `Q prop C rArr (Q_(1))/(Q_(2))=(C_(1))/(C_(2))`

Discussion

7.	When a piece of polythene is rubbed with wool, a charge of `-2xx10^(-7) C` is developed on polythene. What is the amount of mass, which is transferred to polythene.
Answer» From Q= ne, So, the number of electrons tranferred `n=Q/(e)=(2xx10^(-7))/(1.6xx10^(-19))=1.25xx10^(12)` Now mass of transferred electrons`=n xx` mass of one electron`=1.25xx10^(12)xx9.1xx10^(-31)=11.38xx10^(-19) kg`

Discussion

8.	Two bodies begin to fall freely from the same height but the second falls T second after the first. The time (after which the first body begins to fall) when the distance between the bodies equals L isA. `(T)/(2)`B. `(T)/(2)+(L)/(gT)`C. `(L)/(gT)`D. `T+(2L)/(gT)`
Answer» Correct Answer - D `alpha=(T_(1)-T_(2))/(T_(1))" and "beta=(T_(2))/(T_(1)-T_(2)),alphabeta=(T_(2))/(T_(1))`

Discussion

9.	In solar radiation, the intensity of radiation is maximum around the wavelength λm. If R is the radius of the sun and c is the velocity of light, the mass lost by the sun per unit time is proportional to(a) R2 / λ4C2(b) R2 / λ2C2(c) R3 / λ4C3(d) R3 / λ4C2
Answer» Correct Answer is: (a) R²/ λ⁴C² Let T be the temperature of the surface of the sun. ∴ Wien constant, b = λ_mT. The energy lost by the sun per unit time is (σT⁴) (4πR²) = mc².

9.

In solar radiation, the intensity of radiation is maximum around the wavelength λm. If R is the radius of the sun and c is the velocity of light, the mass lost by the sun per unit time is proportional to(a) R2 / λ4C2(b) R2 / λ2C2(c) R3 / λ4C3(d) R3 / λ4C2

Answer»

Correct Answer is: (a) R²/ λ⁴C²

Let T be the temperature of the surface of the sun.

∴ Wien constant, b = λ_mT.

The energy lost by the sun per unit time is (σT⁴) (4πR²) = mc².

Discussion

10.	Two circular rings A and B each of radius a = 30 cm are placed coaxially with their axis horizontal in a uniform electric field `E = 10^(5) NC^(-1)` directed vertically upward as shown in figure. Distance between centers of the rings A and B `(C_A and C_B)is 40 cm`. Ring A has positive charge `q_A = 10muC` and B has a negative charge `q_B = -20muC`. A particle of mass m and charge `q = 10muC` is released from rest at the center of ring A. If particle moves along `C_AC_B,` then Work done by electric field, when particle moves from `C_A to C_B` isA. `-1.2 J`B. 1.2 JC. `-3.6 J`D. `3.6J`
Answer» Correct Answer - D `W=q(V_(2)-V_(1))=9xx10^(-2)[((10)/(0.3)-(20)/(05))-((10)/(0.5)-(20)/(0.3))]` `=3.6J` Since only conservative forces act on the system, potential energy changes to kinetic energy. `3.6=(1)/(2)mv^(2)` or `v^(2)=72` or `v=6sqrt(2)ms^(-1)`

Discussion

11.	Look at the following sentence from the story. Suddenly a strong wind began to blow and along with the rain very large hailstones began to fall. ‘Hailstones’ are small balls of ice that fall like rain. A storm in which hailstones fall is a ‘Hailstorm’. You know that a storm is bad weather with strong winds, rain, thunder and lightning. There are different names in different names parts of the world for storms, depending on their nature. Can you match the names in the box with their descriptions below, and fill in the blanks? You may use a dictionary to help you.galewhirlwindcyclonehurricanetornadotyphooni. A violent tropical storm in which strong winds move in a circle: _ _c_ _ _ _ ii. An extremely strong wind: _a_ _ iii. A violent tropical storm with very strong winds: _ _p _ _ _ _ _ _ _______ iv. A violent storm whose centre is a cloud in the shape of a funnel: _______n_______ v. A violent storm with very strong winds, especially in the western Atlantic Ocean: _______r_______ vi. A very strong wind that moves very fast in a spinning movement and causes a lot of damage: _______i_______
Answer» i. Cyclone ii. Gale iii. Typhoon iv. Tornado v. Hurricane vi. Whirlwind

11.

Look at the following sentence from the story. Suddenly a strong wind began to blow and along with the rain very large hailstones began to fall. ‘Hailstones’ are small balls of ice that fall like rain. A storm in which hailstones fall is a ‘Hailstorm’. You know that a storm is bad weather with strong winds, rain, thunder and lightning. There are different names in different names parts of the world for storms, depending on their nature. Can you match the names in the box with their descriptions below, and fill in the blanks? You may use a dictionary to help you.galewhirlwindcyclonehurricanetornadotyphooni. A violent tropical storm in which strong winds move in a circle: _ _c_ _ _ _ ii. An extremely strong wind: _a_ _ iii. A violent tropical storm with very strong winds: _ _p _ _ _ _ _ _ ___ iv. A violent storm whose centre is a cloud in the shape of a funnel: _n_ v. A violent storm with very strong winds, especially in the western Atlantic Ocean: _r_ vi. A very strong wind that moves very fast in a spinning movement and causes a lot of damage: _i_____

Answer»

i. Cyclone

ii. Gale

iii. Typhoon

iv. Tornado

v. Hurricane

vi. Whirlwind

Discussion

12.	What is the S.I. unit of electric dipole moment ?
Answer» Coulomb meter.

Discussion

13.	Give two factors which affect capacitance of a conductor.
Answer» (i) Area of conductor, (ii) Presence of another conductor.

Discussion

14.	Electrical resistivities of silver and nichrome at \( 20^{\circ} C \) are \( 1.6 \times 10^{-6} \) and \( 1.0 \times 10^{-4} \), then explain why silver is used as good conductor and nichrome is used in making filaments ?
Answer» Out of silver and copper, silver has less resistivity, that's why it is a good conductor as compared to copper. The material used in electrical heating devices must have high resistance. So, a material with high resistivity should be used from the table. Nichrome has the highest resistivity, it would have the highest resistance among all the given materials. Hence, it should be used for electrical heating devices.

14.

Electrical resistivities of silver and nichrome at \( 20^{\circ} C \) are \( 1.6 \times 10^{-6} \) and \( 1.0 \times 10^{-4} \), then explain why silver is used as good conductor and nichrome is used in making filaments ?

Answer»

Out of silver and copper, silver has less resistivity, that's why it is a good conductor as compared to copper.

The material used in electrical heating devices must have high resistance. So, a material with high resistivity should be used from the table. Nichrome has the highest resistivity, it would have the highest resistance among all the given materials. Hence, it should be used for electrical heating devices.

Discussion

15.	what causes the potential difference between two terminal of a cell
Answer» The flow of current or charge causes the potential difference between the two terminals of a cell.

Discussion

16.	What is dry ice? Why is it so called?
Answer» Dry ice is solid `CO_(2)`. When it is kept in air under `1 atm` pressure, it sublimes. In other words, unlike ordinary ice, it does not melt and hence does not wet the surface on which it is kept. Therefore, it is called dry ice.

Discussion

17.	The first transition series is called __________transition series.
Answer» Correct Answer - `3d`

Discussion

18.	Second period ends with __________.
Answer» Correct Answer - Neon

Discussion

19.	What is the correct structural formula of borax?
Answer» `Na_(2)[B_(4)(OH)_(4)O_(5)].8H_(2)O`

Discussion

20.	Lother Meyer atttempts to classify the elements by plotting graph between ________and ________.
Answer» Atomic weight and atomic volume

Discussion

21.	The formula for fluoride of carbon is ____________.
Answer» Correct Answer - `CF_(4)`

Discussion

22.	Out of `C Cl_(4)` and `SiCl_(4)`, which one react with water.
Answer» Due to the presence of vacant `d`-orbitals in `Si,SiCl_(4)` reacts with water. But due to the absence of `d`-orbitals in `C, C Cl_(4)` does not react with water.

Discussion

23.	The solubility of alkali metal carbonate_________as one goes down the group.
Answer» Correct Answer - Increases

Discussion

24.	The electronic configuration of `Re^(3+)` is `(Xe)4f^(14)5d^(4)`, the number of unpaired electrons in this ion is ____________.
Answer» Correct Answer - Four

Discussion

25.	Explain why silicon shows a higher covalency than carbon.
Answer» Silicon because of the presence of vacant `d`-orbitals can show a covalency up to `6`, whereas carbon due to the absence of `d-`orbitals cannot have a covalency of more than `4`.

Discussion

26.	Bernoulli equation is applicable for a _________state process.(a) Steady(b) Un-steady(c) Equilibrium(d) None of the mentioned
Answer» Right answer is (d) None of the mentioned To explain: Bernoulli equation is applicable for a steady state process.

Discussion

27.	What is indicator? Give its types.
Answer» An indicator is a substance which changes its colour at the end point or neutral point of the acid-base titration i.e. the substance which is used to indicate neutral point of acid-base titration are called indicators. At end point N₁V₁ = N₂V₂ Indicators are of two types : (i) Acidic (ii) Basic

27.

What is indicator? Give its types.

Answer»

An indicator is a substance which changes its colour at the end point or neutral point of the acid-base titration i.e. the substance which is used to indicate neutral point of acid-base titration are called indicators. At end point N₁V₁ = N₂V₂

Indicators are of two types :

(i) Acidic

(ii) Basic

Discussion

28.	The major product of mono-ntitration of phenyl ethonate isA. B. C. D. Both (A) and (B)
Answer» Correct Answer - D `-OCOCH_3` is an o^(-),p-directing group.

Discussion

29.	Adsorption when interaction between the solid and the condensed molecules is relatively strong as contrasted with physical adsorption.(a) Absorption(b) Adsorption(c) Chemisorption(d) None of the mentioned
Answer» Correct answer is (b) Adsorption Easiest explanation: “Chemisorption is the adsorption when interaction between the solid and the condensed molecules is relatively strong as contrasted with physical adsorption.

Discussion

30.	Permissible value of orbit angular momentum of electron in a H-like atom will beA. `1.5((h)/(pi))`B. `(h)/(4pi)`C. `(h)/(1.5pi)`D. `(h)/(3pi)`
Answer» Correct Answer - A Permissible value of orbit angular momentum of `e^(-)=n((h)/(2pi))` n = 1, 2, 3………… for n = 3, orbit angular momentum = `1.5 ((h)/(pi))`

Discussion

31.	0.01M acetic acid 5% ionization occurs. So find its dissociation immediately.
Answer» We have given, Concentration of acetic acid (c) = 0.01 M Percent ionization = 5% CH₃COOH(l) + H₂O(l) ⇌ CH₃COO^⊖(aq) + H₃O^⊕(aq) We know that, K_a = α²c and percentionization = α x 100 where, k_a = dissociation constant α = degree of dissociation ∴ degree of dissociation α = 5/100 ∴ k_a = (5/100)² x 0.01 k_a = 2.5 x 10^-5 Hence, dissociation constant k_a = 2.5 x 10^-5.

31.

0.01M acetic acid 5% ionization occurs. So find its dissociation immediately.

Answer»

We have given,

Concentration of acetic acid (c) = 0.01 M

Percent ionization = 5%

CH₃COOH(l) + H₂O(l) ⇌ CH₃COO^⊖(aq) + H₃O^⊕(aq)

We know that,

K_a = α²c and percentionization = α x 100

where, k_a = dissociation constant

α = degree of dissociation

∴ degree of dissociation α = 5/100

∴ k_a = (5/100)² x 0.01

k_a = 2.5 x 10^-5

Hence, dissociation constant k_a = 2.5 x 10^-5.

Discussion

32.	Identify the miller indices for the corresponding direction given below.(a) [1 1 0] , [2 0 1] , [1 2 2] , [1 0 1 ](b) [1 1 1] , [2 2 1] , [1 1 2] , [1 0 1](c) [1 0 0] , [1 1 1] , [2 0 1] , [1 12](d) [1 2 1] , [2 0 1] , [1 1 2] , [1 1 0]
Answer» (c) [1 0 0] , [1 1 1] , [2 0 1] , [1 12]

Discussion

33.	The angle between [111] and [112] directions in cubic unit cell is(a) 0°(b) 45°(c) 90°(d) 180°
Answer» The angle between [111] and [112] directions in cubic unit cell is 90°

Discussion

34.	Wilkinson catalyst is : (1) [(Et3P)3 IrCl] (2) [(Ph3P)3 IrCl] (3) [(Et3P)3RhCl] (4) [(Ph3P)3RhCl] (Et = C2H5)
Answer» Correct option: (4) [(Ph₃P)₃RhCl] (Et = C₂H₅) Explanation: Wilkinson catalyst [Rh(PPh₃) 3Cl] is used for the hydrogenation of alkenes.

34.

Wilkinson catalyst is : (1) [(Et3P)3 IrCl] (2) [(Ph3P)3 IrCl] (3) [(Et3P)3RhCl] (4) [(Ph3P)3RhCl] (Et = C2H5)

Answer»

Correct option: (4) [(Ph₃P)₃RhCl] (Et = C₂H₅)

Explanation:

Wilkinson catalyst [Rh(PPh₃) 3Cl] is used for the hydrogenation of alkenes.

Discussion

35.	The stacking arrangement of FCC crystal structure is(a) ABCABCABC…(b) ABABAB…(c) ABABCABA(d) ABCABABC……
Answer» (a) ABCABCABC…

Discussion

36.	1. For the reaction PCL ⇌ PCl3 + Clc (i) Write the expression of Kc . (ii) What happens if pressure is increased? 2. Write the conjugate acid and base of the following species: (i) H2O (ii) HCO3. Name the phenomenon involved in the preparation of soap by adding NaCI.
Answer» 1. (i) K_c = \(\frac{[PC_3][Cl_2]}{[PCl_5]}\) (ii) If we increase the pressure the system will try to decrease the pressure. For this system will proceed in the direction where there is minimum number of moles, i.e., rate of backward reaction increases by decreasing the pressure. 2.(i) Conjugate acid of H₂O is H₃O⁺ Conjugate base of H₂O is OH” (ii) Conjugate acid of HCO,” is H₂CO₃ Conjugate base of HCO^-₃ is CO^2-₃ 3. Common ion effect.

36.

1. For the reaction PCL ⇌ PCl3 + Clc (i) Write the expression of Kc . (ii) What happens if pressure is increased? 2. Write the conjugate acid and base of the following species: (i) H2O (ii) HCO3. Name the phenomenon involved in the preparation of soap by adding NaCI.

Answer»

1. (i) K_c = \(\frac{[PC_3][Cl_2]}{[PCl_5]}\)

(ii) If we increase the pressure the system will try to decrease the pressure. For this system will proceed in the direction where there is minimum number of moles, i.e., rate of backward reaction increases by decreasing the pressure.

2.(i) Conjugate acid of H₂O is H₃O⁺

Conjugate base of H₂O is OH”

(ii) Conjugate acid of HCO,” is H₂CO₃ Conjugate base of HCO^-₃ is CO^2-₃

3. Common ion effect.

Discussion

37.	The effect of lanthanoid contraction in the lanthanoid series of elements by and large means : (1) increase in both atomic and ionic radii (2) decrease in both atomic and ionic radii (3) decrease in atomic radii and increase in ionic radii (4) increase in atomic radii and decrease in ionic radii
Answer» Correct option: (2) decrease in both atomic and ionic radii Explanation: The effect of lanthanoid contraction is overall decrease in atomic & ionic radii from lanthanum to lutetium.

37.

The effect of lanthanoid contraction in the lanthanoid series of elements by and large means : (1) increase in both atomic and ionic radii (2) decrease in both atomic and ionic radii (3) decrease in atomic radii and increase in ionic radii (4) increase in atomic radii and decrease in ionic radii

Answer»

Correct option: (2) decrease in both atomic and ionic radii

Explanation:

The effect of lanthanoid contraction is overall decrease in atomic & ionic radii from lanthanum to lutetium.

Discussion

38.	Math the following items in column-I with the corresponding, items in column-II.Column-IColumn-II(i) Na2CO3 x 10H2O(A) Portland cement ingredient(ii) Mg(HCO3)2(B) Castner-Kellner process (iii) NaOH(C) Solvay process(IV) Ca3Al2O6(D) Temporary hardness(1) (i) → (C); (ii) → (D); (iii) → (B); (iv) → (A)(2) (i) → (D); (ii) → (A); (iii) →(B); (iv) → (C)(3) (i) → (C); (ii) → (B); (iii) →(D); (iv) →(A)(4) (i) → (B); (ii) → (C); (iii) →(A); (iv) →(D)
Answer» Correct option (1) (i) → (C); (ii) → (D); (iii) → (B); (iv) → (A) Explanation: Na₂CO₃ . 10H₂O → Solvay process Mg(HCO₃)₂ → Temporary hardness NaOH → Castner – Kellner process Ca₃Al₂O₆ →Portland cement ingredient

38.

Math the following items in column-I with the corresponding, items in column-II.Column-IColumn-II(i) Na2CO3 x 10H2O(A) Portland cement ingredient(ii) Mg(HCO3)2(B) Castner-Kellner process (iii) NaOH(C) Solvay process(IV) Ca3Al2O6(D) Temporary hardness(1) (i) → (C); (ii) → (D); (iii) → (B); (iv) → (A)(2) (i) → (D); (ii) → (A); (iii) →(B); (iv) → (C)(3) (i) → (C); (ii) → (B); (iii) →(D); (iv) →(A)(4) (i) → (B); (ii) → (C); (iii) →(A); (iv) →(D)

Answer»

Correct option (1) (i) → (C); (ii) → (D); (iii) → (B); (iv) → (A)

Explanation:

Na₂CO₃ . 10H₂O → Solvay process

Mg(HCO₃)₂ → Temporary hardness

NaOH → Castner – Kellner process

Ca₃Al₂O₆ →Portland cement ingredient

Discussion

39.	What happens to the entropy, when liquid is converted into vapours?
Answer» Entropy increases.

Discussion

40.	What type of graph will you get when PV is plotted against P at constant temperature?
Answer» A straight line parallel to pressure axis.

Discussion

41.	Which of the following represents the correct order of electron affinities?A. `F gt Cl gt Br gt I`B. `C lt N lt O lt F`C. `N lt C lt O lt F`D. `C lt Si lt P lt N`
Answer» Correct Answer - C

Discussion

42.	An atom with high `EA` generally hasA. Tendency to form `+ve` ionsB. High ionisation energyC. Large atomic sizeD. Low electron affinity
Answer» Correct Answer - B

Discussion

43.	Why vegetables are cooked with difficulty at hill station?
Answer» At hill station the atmospheric pressure is less and so boiling point decreases.

Discussion

44.	The elements of group 1 describe, more clearly than any other group of elements, the effects of increasing the size of atoms or ions on the physical and chemical properties.The chemical and physical properties the elements are closely related to their electronic structures and sizes.These metals are highly electropositive and thus form very strong bases, and have quite stable oxo-salts, In the manufacturing of sodium hydroxide chlorine and sodium carbonate, the sodium chloride is used as starting material. Which of the following acts as an oxidising as well as reducing agent ?A. `Na_2O`B. `NaO_3`C. `NaNO`D. `NaNO_3`
Answer» Correct Answer - D `NaNO_3`(sodium nitrite) acts both as oxidising agent and reducing agent because the nitrogen atom in it is in +3 oxidation state (+3 is intermediate oxidation state for nitrogen) Oxidising property : `2NaNO_2 + 2Kl + 2H_2SO_4 to NaSO_4 +K_2SO_4 +2NO +2H_2O + I_2` Reducing property : `NaNO_2 + H_2O_2 to NaNO_3 + H_2O`

Discussion

45.	The first `(Delta_(i)H_(1))` and second `(Delta_(i)H_(2))` ionisation enthalpies `("in" kJ mol^(-1))` and the `(Delta_(eg)H^(ө))` electron gain enthalpy `("in" kJ mol^(-1))` of a few elements are given below: `{:("Elements",(Delta_(i)H_(1)),(Delta_(i)H_(2)),Delta_(eg)H^(ө)),(I,520,7300,-60),(II,419,3051,-48),(III,1681,3374,-328),(IV,1008,1846,-295),(V,2372,5251,+48),(VI,738,1451,-40):}` Which of the above elements is likely to be a. the least reactive element. b. the most reactive metal. c. the most reactive non-metal. d. the least reactive non-metal. e. the metal which can form a stable binary halide of the formula MX2 (X=halogen). f. the metal which can form a predominantly stable covalent halide of the formula MX(X=halogen).A. PB. QC. SD. U
Answer» Correct Answer - B Metal having low `IE_1` and higher `IE_2` corresponds to alkali metal. Hence, Q is most reactive metal (easily forms cation)

Discussion

46.	(I) State the Newland’s law of octaves? (II) What are the two exceptions of block division in the periodic table?
Answer» (I) The Law of octaves states that, “when elements are arranged in the order of increasing atomic weights, the properties of the eighth element are a repetition of the properties of the first element”. (II) 1. Helium has two electrons. Its electronic configuration is 1s². As per the configuration, it is supposed to be placed in ‘s’ block, but actually placed in 18th group which belongs to ‘p’ block. Because it has a completely filled valence shell as the other elements present in 18th group. It also resembles with 18th group elements in other properties. Hence helium is placed with other noble gases. 2. The other exception is hydrogen. It has only one s-electron and hence can be placed in group 1. It can also gain an electron to achieve a noble gas arrangement and hence it can behave as halogens (17^th group elements). Because of these assumptions, position of hydrogen becomes a special case. Finally, it is placed separately at the top of the periodic table.

46.

(I) State the Newland’s law of octaves? (II) What are the two exceptions of block division in the periodic table?

Answer»

(I) The Law of octaves states that, “when elements are arranged in the order of increasing atomic weights, the properties of the eighth element are a repetition of the properties of the first element”.

(II)

1. Helium has two electrons. Its electronic configuration is 1s². As per the configuration, it is supposed to be placed in ‘s’ block, but actually placed in 18th group which belongs to ‘p’ block. Because it has a completely filled valence shell as the other elements present in 18th group. It also resembles with 18th group elements in other properties. Hence helium is placed with other noble gases.

2. The other exception is hydrogen. It has only one s-electron and hence can be placed in group 1. It can also gain an electron to achieve a noble gas arrangement and hence it can behave as halogens (17^th group elements). Because of these assumptions, position of hydrogen becomes a special case. Finally, it is placed separately at the top of the periodic table.

Discussion

47.	A certain sum amounts to Rs.7350 in 2 years and to Rs.8575 in 3 years.find the sum and rate percent.
Answer» S.I on Rs.7350 for 1 year =Rs.(8575-7350) = Rs.1225. Rate=(1001225/73501)%=16 2/3% Let the sum be Rs.x.then, X(1+50/3100)²= 7350 X7/67/6=7350 X=(735036/49)=5400. Sum = Rs.5400.

47.

A certain sum amounts to Rs.7350 in 2 years and to Rs.8575 in 3 years.find the sum and rate percent.

Answer»

S.I on Rs.7350 for 1 year =Rs.(8575-7350) = Rs.1225.

Rate=(100*1225/7350*1)%=16 2/3%

Let the sum be Rs.x.then,

X(1+50/3*100)²= 7350

X*7/6*7/6=7350

X=(7350*36/49)=5400.

Sum = Rs.5400.

Discussion

48.	Two numbers A and B are such that A > B and their G.M. is 40% lower than their A.M. Find the ratio between the numbers. (a) 4 : 3 (b) 9 : 1 (c) : 1 (d) 3 : 1
Answer» Correct option (b) 9 : 1 Explanation: Trial and error gives us that for option (b): With the ratio 9:1, the numbers can be taken as 9x and 1x. Their AM would be 5x and their GM would be 3x. The GM can be seen to be 40% lower than the AM.

48.

Two numbers A and B are such that A > B and their G.M. is 40% lower than their A.M. Find the ratio between the numbers. (a) 4 : 3 (b) 9 : 1 (c) : 1 (d) 3 : 1

Answer»

Correct option (b) 9 : 1

Explanation:

Trial and error gives us that for option (b): With the ratio 9:1, the numbers can be taken as 9x and 1x. Their AM would be 5x and their GM would be 3x. The GM can be seen to be 40% lower than the AM.

Discussion

49.	X and Y are two numbers whose A.M. is 41 and G.M. is 9. Which of the following may be a value of X? (a) 125 (b) 49 (c) 81 (d) 25
Answer» Correct option (c) 81 Explanation: AM = 41 means that their sum is 82 and GM = 9 means their product is 81. The numbers can only be 81 and 1.

49.

X and Y are two numbers whose A.M. is 41 and G.M. is 9. Which of the following may be a value of X? (a) 125 (b) 49 (c) 81 (d) 25

Answer»

Correct option (c) 81

Explanation:

AM = 41 means that their sum is 82 and GM = 9 means their product is 81. The numbers can only be 81 and 1.

Discussion

50.	Prove that \(\frac{sinA-cosA+1}{sinA+cosA-1}=\frac{1}{secA-tanA}\)
Answer» \(\frac{sinA-cosA+1}{sinA+cosA-1}=\frac{1}{secA-tanA}\) L.H.S. divide numerator and denominator by cos A \(\frac{tanA-1+secA}{tanA+1-secA}\) \(\frac{tanA-1+secA}{1-secA+tanA}\) We know that 1 + tan² A = sec²A Or 1 = sec² A - tan² A = (sec A + tan A)(sec A – tan A) = \(\frac{secA+tanA-1}{(secA+tanA)(secA-tanA)-(secA-tanA)}\) = \(\frac{secA+tanA-1}{(secA-tanA)(secA+tanA-1}\) = \(\frac{1}{secA-tanA}\) , proved.

50.

Prove that \(\frac{sinA-cosA+1}{sinA+cosA-1}=\frac{1}{secA-tanA}\)

Answer»

\(\frac{sinA-cosA+1}{sinA+cosA-1}=\frac{1}{secA-tanA}\)

L.H.S. divide numerator and denominator by cos A

\(\frac{tanA-1+secA}{tanA+1-secA}\)

\(\frac{tanA-1+secA}{1-secA+tanA}\)

We know that 1 + tan² A = sec²A

Or 1 = sec² A - tan² A = (sec A + tan A)(sec A – tan A)

= \(\frac{secA+tanA-1}{(secA+tanA)(secA-tanA)-(secA-tanA)}\)

= \(\frac{secA+tanA-1}{(secA-tanA)(secA+tanA-1}\)

= \(\frac{1}{secA-tanA}\) , proved.

Discussion

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