100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS

1.	Statement I: A conducting body is connected to the earth, hence it is electrically neutral. Statement II: The potential of a conducting body connected to the earth is zero.A. Statement I is true, Statement II is True, Statement II is a correct explanation for statement I.B. Statement I is true, Statement II is True, Statement II is Not a correct explanation for statement I.C. Statement I is True, Statement II is False.D. Statement I is False, Statement II is True.
Answer» Correct Answer - D An electrically neutral body means charge on the body is zero. A body connected to earth may possess some charge.

Discussion

2.	Statement I:A uniformly charged disk has a pin hole at its centre. The electric field at the center of the disk is zero. Statement II: Disk can be supposed to be made up of many rings. Also, electric field at the center of a uniformly charged ring is zero.A. Statement I is true, Statement II is True, Statement II is a correct explanation for statement I.B. Statement I is true, Statement II is True, Statement II is Not a correct explanation for statement I.C. Statement I is True, Statement II is False.D. Statement I is False, Statement II is True.
Answer» Correct Answer - A The electric field due to disk is superposition of electric field due to its constituent rings as given in reason. Assertion is true reason is true, reason is a correct explanation for assertion.

Discussion

3.	Free radicals are- *positively chargednegatively chargedelectrically neutralnone of these
Answer» neutral species.

Discussion

4.	A conductor carrying current I is of the type as shown in figure. Find the magnetic field induction at the common centre O of all the three arcs. A. `(5mu_(0)Itheta)/(24pir)`B. `(mu_(0)Itheta)/(24pir)`C. `(11mu_(0)Itheta)/(24pir)`D. zero
Answer» Correct Answer - A `B = B_(1)-B_(2) + B_(3)` `= (mu_(0)Itheta)/(4pir)-(mu_(0)Itheta)/(8pir) + (mu_(0)Itheta)/(12pir)` `= (mu_(0)Itheta)/(4pir) (1-(1)/(2) + (1)/(3))` `= (mu_(0)Itheta)/(4pir)((6-3+2)/(6))` `= (mu_(0)Itheta)/(4pir) xx (5)/(6)`

Discussion

5.	The current growth in two L-R circuits (b) and (c) is as shown in Fig. Let `L_(1), L_(2), R_(1) and R_(2)` be the corresponding values in two circuits. Then A. `R_(1) gt R_(2)`B. `R_(1) = R_(2)`C. `L_(1) gt L_(2)`D. `L_(1) lt L_(2)`
Answer» Correct Answer - B::D Steady state current for both teh circuits is same. Therefore, `(V)/(R_1)=(V)/(R_2) or R_(1)=R_(2)` Further `tau_(L_(1)) lt tau_(L_2)` (`tau_(L) = "time costant"`) `(L_1)/(R_1) lt (LL_2)/(R_2) or L_(1) lt L_(2)` `(R_(1)=R_(2))`.

Discussion

6.	An external magnetic field is decreased to zero, due to which a current is induced in a circular wire loop of radius r and resistance R placed in the field. This current will not become zero at the instant when B stops changingA. At the instant when external magnetic field stops changing (t=0), the current in the loop is `i_(0)`. The current in the loop as a function of time for `t gt 0` is given by `i_(0)e^(-2Rt//mu_(0)pi)`.B. For the same as in option (a), the current in the loop as a function of time `t=0` is given by `(mu_(0)iR)/(2 r)`.C. The time in which current in loop decreases to `10^(-3) i_(0)` (from t=0) for `R=100 Omega` and `r=5 cm` is given by `(3 pi^(2)1n 10)/(10^(10))s`.D. Fro the same as in option (c), the time in which current in loop decreases to `10^(-3) i_(0)` (from t=0) for `R=100 Omega` and `r-5 cm` is given by `(3pi^(2))/(10^(6))s`.
Answer» Correct Answer - A::C Flux linked with loop due to its own magnetic field. `phi=(mu_(0)i)/(2r) (pi r^(2))=(mu_(0)piri)/(2)` emf induced = `-(dphi)/(dt) = e = -(mu_(0)pir)/(2R)(di)/(dt)` `int_(i_0)^(i) (di)/(dt)=-int_(0)^(t) (2R)/(mu_(0)pir)* dt` `i=i_(0)e^(-2RT//mu_(0)pit)` Now, `10^(-3)i_(0)=i_(0)e^(-(2Rt)/(mu_(0)pir))` which gives `t=(3pi^(2)1n10)/(10^(10))s`.

Discussion

7.	Which one of the following equation is correct for the circuit shown. The currents in each brach of circuit are indicated and all three cells are ideal. A. `2-I_(1)-2I_(2)=0`B. `2-2I_(1)-2I_(2)-4I_(3)=0`C. `4-I_(1)+4I_(3)=0`D. `-2-I_(1)-2I_(2)=0`
Answer» Correct Answer - D Applying Kirchoff voltage law in left loop. `-2+I_(1)+4+2I_(2)=0 thereofre 2+I_(1)+2I_(2)=0`

Discussion

8.	The deviation caused for red, yellow and violet colours for crown glass prism are `2.84^(@),3.28^(@)` and `3.72^(@)` respectively. The dispersive power of prism material is:A. `(11)/(41)`B. `(92)/(250)`C. `(117)/(250)`D. `(22)/(57)`
Answer» Correct Answer - A Dispersive power is given by `omega=(delta_(v)-delta_(r))/(delta_(y))=(3.72-2.84)/(3.28)=(11)/(41)`.

Discussion

9.	A telescope of diameter `2m` uses light of wavelength `5000 Å` for viewing stars.The minimum angular separation between two stars whose is image just resolved by this telescope isA. `4xx 10^(-4)rad`B. `0.25 xx 10^(-6)rad`C. `0.31 xx 10^(-6)rad`D. `5xx 10^(-3)rad`
Answer» Correct Answer - C `Delta theta = (1.22 lambda)/(a)` `:. Delta theta = (1.22 xx 5000 xx 10^(-10))/(2)` `= 3050 xx 10^(-10) = 3.05 xx 10^(-7)rad`. `~~ 0.31 xx 10^(-6)rad`.

Discussion

10.	a) Explain de-Broglie argument to propose his hypothesis. Show that de-Broglie wavelength of photon equals electromagnetic radiation. b) If, deuterons and alpha particle are accelerated through same potential, find the ratio of the associated de-Broglie wavelengths of two.
Answer» (a) De-Broglie put forward the bold hypothesis that moving particles of matter should display wave-like properties under suitable conditions. If radiation shows dual aspects, so should matter. De-Broglie proposed that the wave length \(\lambda\) associated with a particle of momentum 'P' is given as: \(\lambda\)= \(\frac{h}{P}\) = \(\frac{h}{mv}\) ; where, m = mass of the particle v = particle speed For a photon: P = \(\frac{hv}{c}\) Therefore, \(\frac{h}{p}= \frac{c}{v}=\lambda\) Thus, De-Broglie equation equals the wavelength of em radiation of which the photon is a quantum of energy and momentum. (b) De-broglie wavelength is given by: \(\lambda\) = \(\frac{h}{p}\) \(\lambda\) = \(\frac{h}{\sqrt{2mqv}}\) \(\frac{\lambda_d}{\lambda_{\alpha}}\) = \(\frac{\sqrt{m_{\alpha}q_{\alpha}}}{\sqrt{m_dq_d}}\) = \(\frac{\sqrt{4\times2}}{\sqrt{2\times1}}\) = 2 De-Broglie reasoned out that nature was symmetrical and two basic physical entities –mass and radiation must be symmetrical.If radiation shows shows dual aspect than matter should do so. De-Broglie equation- \(\lambda\)=\(\frac{h}{p}\) For photon – P=\(\frac{hv}{C}\) Therefore,\(\frac{h}{P}=\frac{C}{v}=\lambda\) As \(\lambda=\frac{h}{\sqrt2mk}\) So,alpha particle will be having shortest de-Broglie wavelength compared to deutrons.

10.

a) Explain de-Broglie argument to propose his hypothesis. Show that de-Broglie wavelength of photon equals electromagnetic radiation. b) If, deuterons and alpha particle are accelerated through same potential, find the ratio of the associated de-Broglie wavelengths of two.

Answer»

(a) De-Broglie put forward the bold hypothesis that moving particles of matter should display wave-like properties under suitable conditions. If radiation shows dual aspects, so should matter.

De-Broglie proposed that the wave length \(\lambda\) associated with a particle of momentum 'P' is given as:

\(\lambda\)= \(\frac{h}{P}\) = \(\frac{h}{mv}\) ;

where,

m = mass of the particle

v = particle speed

For a photon: P = \(\frac{hv}{c}\)

Therefore, \(\frac{h}{p}= \frac{c}{v}=\lambda\)

Thus, De-Broglie equation equals the wavelength of em radiation of which the photon is a quantum of energy and momentum.

(b) De-broglie wavelength is given by: \(\lambda\) = \(\frac{h}{p}\)

\(\lambda\) = \(\frac{h}{\sqrt{2mqv}}\)

\(\frac{\lambda_d}{\lambda_{\alpha}}\) = \(\frac{\sqrt{m_{\alpha}q_{\alpha}}}{\sqrt{m_dq_d}}\) = \(\frac{\sqrt{4\times2}}{\sqrt{2\times1}}\) = 2

De-Broglie reasoned out that nature was symmetrical and two basic physical entities –mass and radiation must be symmetrical.If radiation shows shows dual aspect than matter should do so.

De-Broglie equation-

\(\lambda\)=\(\frac{h}{p}\)

For photon –

P=\(\frac{hv}{C}\)

Therefore,\(\frac{h}{P}=\frac{C}{v}=\lambda\)

As \(\lambda=\frac{h}{\sqrt2mk}\)

So,alpha particle will be having shortest de-Broglie wavelength compared to deutrons.

Discussion

11.	Find the friction force due to air on a body of mass 1 kg falling with acceleration `8 m//s^(2)` :-A. 2 NB. 4 NC. ZeroD. None
Answer» Correct Answer - A

Discussion

12.	If a force F =500-100t, then impulse as a function of time will be :-A. `500t-50t^(2)`B. ` 50t-10`C. `50-t^(2)`D. `100t^(2)`
Answer» Correct Answer - A

Discussion

13.	If the current constant for a transistor are α & β(a) α β =1(b) β > 1, α < 1(c) α = β(d) β < 1, α > 1
Answer» Correct answer is (b) β > 1, α < 1

Discussion

14.	A prism shaped styrofoam of density `rho_(styrofoam) (ltrho_(water))` is held completely submerged in water. It lies with its base horizontal. The base of foam is at depth `h_(0)` below water surface and atmospheric pressure is `P_(0)` Surface of water is open to atmosphere. Styrofoam prism is held in equilibrium by the string attached symmetrically as shown (Take `rho_(styrofoam)=rho_(f), rho_(water)=rho_(w))`. Magnitude of force on any one of the slant face of styrofoam isA. `(P_(0)+rho_(w)g(h_(0)-sqrt(2)l))Ll`B. `(P_(0)+rho_(w)g(h_(0)-l/sqrt(2))Ll`C. `(P_(0)-rho_(w)g(h_(0)-l/sqrt(2)))Ll`D. `(P_(0)+rho_(w)g(h_(0)-l/(2sqrt(2))))Ll`
Answer» Correct Answer - D Average pressure on slant surface `P_(avg)=((P_(0)+rho_(w)gh_(0))+(P_(0)+rho_(w)gh_(0)-1/sqrt(2)))/2=(P_(0)+rho_(w)gh_(0)-l/(2sqrt(2)))` Force on any one of the slant face `=(P_(0)+rho_(w)g(h_(0)-l/(2sqrt(2)))Ll`

Discussion

15.	In a thermally insulated tube of cross sectional area `4cm^(2)` a liquid of thermal expansion coefficeint `10^(3)K^(-1)` is flowing. Its velocity at the entrance is `0.1m//s`. At the middle of the tube a heater of a power of 10 kW is heating the liquid. The specific heat capacity of the liquid is 1.5 kJ/(kg,K), and its density is `1500kg//m^(3)` at the entrance. Q. How much bigger is the volume rate of flow at the end of the tube than at the entrance in cubic meters?A. `9xx10^(-5)`B. `1/3xx10^(-5)`C. `4/9xx10^(-5)`D. None of these
Answer» Correct Answer - C

Discussion

16.	Consider PT graph of cyclic process shown in the figure. Maximum pressure during the cycle is twice the minimum pressure. The heat received by the gas in the process 1-2 is equal to the heat received in the process 3-4. The process is done on one mole of monoatomic gas. Correct PV diagram for the process is-A. B. C. D.
Answer» Correct Answer - D

Discussion

17.	A metal cylinder of mass `0.5` kg is heated electrically by a 12 W heater in a room at `15^(@)` C. The cylinder temperature rises uniformly to `25^(@)` C in 5 min and finally becomes constant at `45^(@)` C Assuming that the rate of heat losss is proportional to the excees temperature over the surroundings,A. the rate of loss of heat of the cylinder to surrounding at `20^(@)` C is 2WB. the rate of loss of heat of the cylinder to surrounding at `45^(@)` C is 12WC. the rate of loss of heat of the cylinder to surrounding at `20^(@)` C is 5WD. the rate of loss of heat of the cylinder to surrounding at `45^(@)` C is 30W.
Answer» Correct Answer - A::B Rate of heat loss `((dQ)/(dt))_(loss)=k(theta-theta_(0))` At `45^(@)` C Rate of heat loss = Rate of heat supplied =12W `k(45-15)=12` `Rightarrow k=2//5 W//^(@)C` `((dQ)/(dt))_(loss)=k(20-15)=2W`

Discussion

18.	`y=x+x^(2)+(1)/(x)+(1)/(x^(3))`. Find `(dy)/(dx)`
Answer» Correct Answer - `(dy)/(dx)=1+2x-(1)/(x^(2))-(3)/(x^(4)).` `y=x+x^(2)+(1)/(x)+(1)/(x^(3)),(dy)/(dx)=1+2x-(1)/(x^(2))-(3)/(x^(4))`

Discussion

19.	`y=x^(2)+(1)/(x^(2))`. Find `(dy)/(dx)`
Answer» Correct Answer - `(dy)/(dx)=2x-(2)/(x^(3))` `y=x^(2)+(1)/(x^(2)),(dy)/(dx)=2x-(2)/(x^(3))`

Discussion

20.	A plane wavefront `A_(1)B_(1)` is incident at a boundary `A_(1)B_(2)` as shown. It takes time tau for the wavefront to travel from `B_(1)` to `B_(2)` Speeds of propagation of light in medium 1 and 2 are `v_(1) & v_(2)` respectively and `v_(2) gt v_(1)`. For the total internal reflection of wavefront :- A. `v_(1)taugtA_(1)B_(2)`B. `v_(2)taugtA_(1)B_(2)`C. `v_(1)taultA_(1)B_(2)`D. `v_(2)tau lt A_(1)B_(2)`
Answer» Correct Answer - B If `v_(2)1gtA_(1)B_(2)`, then refrected wavefront will be absent. (Not possible to draw wavefront)

Discussion

21.	A quantity of 2 mole of helium gas undergoes a thermodynamic process, in which molar specific heat capacity of the gas depends on absolute temperature` T` , according to relation: `C=(3RT)/(4T_(0)` where `T_(0)` is initial temperature of gas. It is observed that when temperature is increased. volume of gas first decrease then increase. The total work done on the gas until it reaches minimum volume is :-A. `(3)/(2) RT_(0)`B. `(3)/(4)RT_(0)`C. `(3)/(8)RT_(0)`D. `(3)/(10)RT_(0)`
Answer» Correct Answer - B When it reaches minimum volume, the process can be treated isochoric. `(3RT)/(4T_(0)) = (3R)/(2)` ` T = 2T_(0)` ` DeltaQ = DeltaU + W` `int_(T_(0))^(2T_(0))(2)(3RT)/(4T_(0))dT = (2)(3R)/(2)(2T_(0)-T_(0))+W` ` (6R)/(4T_(0))[T^(2)/(2)]_(T_(0))^(2T_(0))= 3RT _(0)+W` ` W=(9RT_(0))/(4)-3RT_(0)` ` W=-(3RT_(0))/(4)` So work done on the gas `(3RT_(0))/(4)`

Discussion

22.

Which of the following complex have maximum value of crystal filed splitting energy.(1) [Co(CN)6]3-(2) [Cu(NH3)4]2+(3) [Co(H2O)6]2+(4) [Ti(H2O)6]3+

Answer»

(2) [Cu(NH₃)₄]²⁺

Complex	Hybridisation
(1) [Co(CN)₆]^3-	d²sp³
(2) [Cu(NH₃)₄]²⁺	dsp²
(3) [Co(H₂O)₆]²⁺	sp³d²
(4) [Ti(H₂O)₆]³⁺	d²sp³

Δ_sp= 1.3 Δo

Discussion

23.	What is meant by the statement that the rating of fuse in a circuit is 5A?
Answer» It means maximum current of 5A can pass through the fuse without melting it.

Discussion

24.	Calculate the spin only magentic moment of `M^(2+)` ion `(Z=27)`.
Answer» In a medium (like Water) `d^3` is more stable as compared to `d^5`. `M(Z=27)implies3d^74s^2,M^(2+)implies3d^74s^0` Thus it has 3 unpaired electrons and hence. `mu=sqrt(3(3+2))=sqrt(15)BM=3.87BM`.

Discussion

25.	Which is a stronger reducing agent `Cr^(2+)` or `Fe^(2+)` and why?
Answer» (i). `Cr^(2+)(Z=24)=3d^54s^1,Cr^(2+)=3d^4,Cr^(3+)=3d^3` `Cr^(2+)toCr^(2+)+e^(ɵ)` (ii). `Fe(Z=36)=3d^64s^2,Fe^(2+)=3d^6,Fe^(3+)=3d^5` `Fe^(2+)toFe^(3+)+e^(ɵ)`. (Oxidation) (In a medium (like water) `d^3` is more stable as compared to `d^5`.

Discussion

26.	`overset((i)CH_(3)CO_(3)H)rarr` product is:-A. B. C. D.
Answer» Correct Answer - C

Discussion

27.	Mathc the column (I) and column (II) :- A. a-pB. d-sC. b-rD. c-q
Answer» Correct Answer - A

Discussion

28.	Which of the following is natural polymer ?A. BakeliteB. PolytheneC. Buna-SD. Protein
Answer» Correct Answer - D

Discussion

29.	Correct set of quantum number for last electron of Pd.A. `n=5,l=2,m=0,s=-(1)/(2)`B. n=4,l=2,m=0,s=`-(1)/(2)`C. n=4,l=0,m=0,s=`-(1)/(2)`D. n=6,l=0,m=0,s=`+(1)/(4)`
Answer» `Pd_(46)=[Kr]_(36)5s^(0)4d^(10)` `n=4, l=2,m=0,s=-(1)/(2)`

Discussion

30.	Pure sulphur was burnt. the gaseous products are `SO_(2)=60%` (mol), `SO_(3)=20%` (mol) and `O_(2)=20%` (mol). If initially 50 moles of sulphur was taken then how many moles of `O_(2)` should be taken.A. 110B. 68.75C. 55D. 50
Answer» Assume that mole of product mixture =x mole `n_(a)=50=xxx0.6+x xx0.2-0.8x` `x=(50)/(0.8)=(500)/(8)=62.5` `n_(o_(2))` remain=62.5xx0.2-12.5 `n_(o)=2xxn_(SO_(2))+3xxn_(SO_(3))=2xx x xx0.6+3xx x xx0.2` `n_(o)=1.2x+0.6x=1.8x=1.8xx62.5` `n_(o_(2))=(1.8xx65)/(2)=0.9xx65=56.25` `n_(o_(2))` total require `=56.25+12.5=68.75`

Discussion

31.	Which of the following is correct statement about energy of an orbital in multielectronic species?A. `4sgt3d`B. `5plt4d`C. `4fgt6s`D. `4s=3d`
Answer» `(n+l)_(4f)=4+3=7` `(n+1)_(6s)=6+0=6` `4fgt6s("Energy order")`

Discussion

32.	Identify structure of Adenine :-A. B. C. D.
Answer» Correct Answer - A

Discussion

33.	Which of the following has maximum number of paired electrons.A. `Cu^(+)`B. `Fe^(3+)`C. `Zr^(+)`D. `Se^(+)`
Answer» (1) `Cu_(29)^(+)-[Ar]_(18)3d^(10)` (2) `Fe_(26)^(+2)=[Ar]_(18)3d^(6)` (3) `Zr_(40)^(+)=[Kr]_(36)5s^(1)3d^(2)` (4) `Se_(21)^(+)=[Ar]_(18)4s^(1)3d^(1)`

Discussion

34.	Among the following polymer, identify compolymer ?A. Nylon-6B. StarchC. PVCD. Protein
Answer» Correct Answer - D

Discussion

35.	Among the following reactions, which form salicylic acid (after acidification):-A. B. C. D.
Answer» Correct Answer - C

Discussion

36.	What is known as White Vitriol?
Answer» Answer: Zinc Sulphate

Discussion

37.	What could be the product for the following reaction ? A. B. C. D.
Answer» Correct Answer - D

Discussion

38.	What could be the product for the following reaction ? A. B. C. D.
Answer» Correct Answer - D

Discussion

39.	In the given reaction :- The final product (Y) is :A. B. C. D.
Answer» Correct Answer - D

Discussion

40.	During electrolysis of conc. \( H _{2} SO _{4} \), perdisulphuric acid \( \left( H _{2} S _{2} O _{8}\right) \), and \( O _{2} \) form in equimolar amount. The amount of \( H _{2} \) that will form simultaneously will be : (A) Thrice that of \( O _{2} \) in moles. (B). Twice that of \( O _{2} \) in moles. (C). Equal to that of \( O _{2} \) in moles. (D). Half of that of \( O _{2} \) in moles.
Answer» Correct option is (A) Thrice that of O₂in moles At anode - 2H₂SO₄ \(\longrightarrow\) H₂S₂O₈ + 2H⁺+ 2e^- 2H₂O \(\longrightarrow\) O₂ + 4H⁺ + 4e^- At Cathode- (2H₂O \(\longrightarrow\) H₂ + 2OH^- - 2e^- x 3 Net reaction- 2H₂SO₄ + 8H₂O \(\longrightarrow\) H₂S₂O₈ + O₂ + 3H₂ + 6H⁺ + 6OH^- Hence, ratio of moles of O₂ and H₂ is 1 : 3 Hence, The amount of H₂ that will form simultaneously will be Thrice that of O₂ in moles.

40.

During electrolysis of conc. \( H _{2} SO _{4} \), perdisulphuric acid \( \left( H _{2} S _{2} O _{8}\right) \), and \( O _{2} \) form in equimolar amount. The amount of \( H _{2} \) that will form simultaneously will be : (A) Thrice that of \( O _{2} \) in moles. (B). Twice that of \( O _{2} \) in moles. (C). Equal to that of \( O _{2} \) in moles. (D). Half of that of \( O _{2} \) in moles.

Answer»

Correct option is (A) Thrice that of O₂in moles

At anode - 2H₂SO₄ \(\longrightarrow\) H₂S₂O₈ + 2H⁺+ 2e^-

2H₂O \(\longrightarrow\) O₂ + 4H⁺ + 4e^-

At Cathode- (2H₂O \(\longrightarrow\) H₂ + 2OH^- - 2e^- x 3

Net reaction- 2H₂SO₄ + 8H₂O \(\longrightarrow\) H₂S₂O₈ + O₂ + 3H₂ + 6H⁺ + 6OH^-

Hence, ratio of moles of O₂ and H₂ is 1 : 3

Hence, The amount of H₂ that will form simultaneously will be Thrice that of O₂ in moles.

Discussion

41.	Statement-1:But-1-ene gives 2-Bromobutane with HBr/Peroxide Statement-2:The reaction involves formation of more stable free radical the product obtained by anti Markovnikof rule.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True.
Answer» Correct Answer - D But-1-ene gives 1-Bromobutane with HBr/Peroxide because this is governed by anti Markovnikof rule .

Discussion

42.	Statement-1 : A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True.
Answer» Correct Answer - A 1,2-addition is major product at low temperature but 1,4- addition is major product at high temperature

Discussion

43.	List down the reasons that helped Gandhiji to gain the trust and recognition of the common people.His priests in south Africa made him famous.
Answer» Gandhiji became familiar to the Indians after his involvement in addressing the problems of the Indian expatriates in South Africa. His protests in South Africa made him famous. Gandhiji was very popular because he lived among the common people and his food and clothing was similar to that of the common man in India and he spoke in their language. Common people found in Mahatma Gandhi a savior who could solve their problems.

Discussion

44.

What is the formula of Ozone?1. O32. O3. O24. O4

Answer» Correct Answer - Option 1 : O₃

The correct answer is O3.

A chemical formula is an alphanumeric notation assigned to chemical compounds for their instant and swift representation and understanding.
Explanation:
- Ozone is a protective layer that envelopes the earth's atmosphere.
- Ozone is an allotrope of oxygen.
- It is also represented and identified by the chemical formula O3.
- It is composed of three oxygen atoms.
- Ozone is responsible for shielding the earth from the harmful ultraviolet (UV) radiation emitted by the sun.

Symbol	Name of compound
O	Atomic oxygen
O2	Oxygen molecule
O3	Ozone
O4	Oxozone

Ozone is formed in the atmosphere by the combination of atomic oxygen with an oxygen molecule.
- Dissociation of an oxygen molecule in presence of sunlight,
- Combination of atomic oxygen and oxygen molecule,
Ozone can also be synthesized in a laboratory by using the electric discharge method.

Discussion

45.	The face diagonal length of `f.c.c.` cubic cell is `660 sqrt(2) p m`. If the radius of the cation is `110 p m`, the radius of the anion isA. `249 p m`B. `220 p m`C. `608 p m`D. `176 p m`
Answer» Correct Answer - 2 `sqrt(2)a=660sqrt(2)p m` so `a=660 p m` Now if tetrahedral void is occupied by cations than `(sqrt(3))/(4)a=(r_(+)+r_(-))` `r_(-)((sqrt(3)xx600)/(4)-110)=110[(3)/(2)sqrt(3)-1]=1.598xx110. ` So, `(r_(+))/(r_(-))=(1)/(1.598)=(1)/(1.6)=(10)/(16)=0.625` but `(r_(+))/(r_(-))gt0.414` so it must not be occupying tetrahedral void then `a=2(r_(+)+r_(-))" "implies" "330=r_(+)+r_(-)` `r_(-)=220p m` `{(r_(+))/(r_(-))=0.5` it can occupy octahedral void `}`

Discussion

46.	Complete the table.Early struggle of Gandhiji is IndiaRegion Year......a.....Bihar.....b.....Ahammadabad Cotton Mill Strikes.....c..........d..........e..........f.....1918
Answer» a. Farmers in Champaran, b. 1917, c. Gujarat, d. 1918, e. Peasant struggle in Kheda, f. Gujarat.

46.

Complete the table.Early struggle of Gandhiji is IndiaRegion Year......a.....Bihar.....b.....Ahammadabad Cotton Mill Strikes.....c..........d..........e..........f.....1918

Answer»

a. Farmers in Champaran,

b. 1917,

c. Gujarat,

d. 1918,

e. Peasant struggle in Kheda,

f. Gujarat.

Discussion

47.	Which Indian State has the highest percentage of children suffering from malnutrition ?1. Madhya Pradesh2. Rajasthan3. Jharkhand 4. Bihar
Answer» Correct Answer - Option 3 : Jharkhand The correct answer is Jharkhand. Jharkhand State has the highest percentage of children suffering from malnutrition. Jharkhand is a state in eastern India. It's known for its waterfalls, the elegant Jain temples of Parasnath Hill and the elephants and tigers of Betla National Park. It features the 17th-century Jagannath Temple, a Hindu shrine and the Jharkhand War Memorial. Tagore Hill is a monument honouring Nobel Prize-winning author Rabindranath Tagore. Capital: Ranchi Chief minister: Hemant Soren Population: 3.19 crores (2012) Jharkhand state animal: Elephant State tree: Sal tree

47.

Which Indian State has the highest percentage of children suffering from malnutrition ?1. Madhya Pradesh2. Rajasthan3. Jharkhand 4. Bihar

Answer» Correct Answer - Option 3 : Jharkhand

The correct answer is Jharkhand.

Jharkhand State has the highest percentage of children suffering from malnutrition.

Jharkhand is a state in eastern India.
It's known for its waterfalls, the elegant Jain temples of Parasnath Hill and the elephants and tigers of Betla National Park.
It features the 17th-century Jagannath Temple, a Hindu shrine and the Jharkhand War Memorial.
Tagore Hill is a monument honouring Nobel Prize-winning author Rabindranath Tagore.
Capital: Ranchi
Chief minister: Hemant Soren
Population: 3.19 crores (2012)
Jharkhand state animal: Elephant
State tree: Sal tree

Discussion

48.	Complete the following table.OrganisationLeadersYear in which formed.....a..........b.....1923Hindutan Socia list Republican Association.....c..........d..........e.....jai Parkash Narayan, Aruna Asif Ali.....f.....
Answer» a. Swaraj Party. b CR Das and Mothilal Nehru. c. Bhagat Singh Chandra Sekhar Azad, Guru and Sukh Dev. d. 1928. e. Congress Socialist Party. f. 1934.

48.

Complete the following table.OrganisationLeadersYear in which formed.....a..........b.....1923Hindutan Socia list Republican Association.....c..........d..........e.....jai Parkash Narayan, Aruna Asif Ali.....f.....

Answer»

a. Swaraj Party.

b CR Das and Mothilal Nehru.

c. Bhagat Singh Chandra Sekhar Azad, Guru and Sukh Dev.

d. 1928.

e. Congress Socialist Party.

f. 1934.

Discussion

49.

Complete the table. Revolutionary MovementsLeaders......a.......N D SarvakarAnuseelan Samathi......b............c......Lala HardayalIndian Republican......d......

Answer»

a. Abhinar Bharat Society.

b. Bareender Kumar Ghose, Pulin Bihari Das.

c. Ghadar party.

d. Suryasen.

Discussion

50.

Which of the following leader was awarded the Bharat Ratna in 2015?1. Dr. C. N. R. Rao2. Sachin Tendulkar3. P. Bhimsen Joshi4. Atal Bihari Vajpayee

Answer» Correct Answer - Option 4 : Atal Bihari Vajpayee

The correct answer is Atal Bihari Vajpayee.

The awardees of Bharat Ratna 2015 are - Madan Mohan Malaviya and Atal Bihari Vajpayee.
Atal Bihari Vajpayee was an Indian statesman who served three terms as the Prime Minister of India, first for a term of 13 days in 1996, then for a period of 13 months from 1998 to 1999, followed by a full term from 1999 to 2004.
Vajpayee was one of the co-founders and a senior leader of the Bharatiya Janata Party (BJP).
He was a member of the Rashtriya Swayamsevak Sangh, a Hindu nationalist volunteer organization.

The Government has awarded Bharatiya Jana Sangh leader, late Nanaji Deshmukh, singer and musician, late Bhupen Hazarika, and former President of India Pranab Mukherjee with the Bharat Ratna Award.
The Bharat Ratna is the highest civilian award of the Republic of India.
It was instituted on 2 January 1954, the award is conferred in recognition of exceptional service/performance of the highest order, without distinction of race, occupation, position, or sex.

Discussion

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