100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS

1.	It is a legal requirement: a. To report incidences when the wrong patient is given a medical exposureb. To carry out quality assurance measurements on diagnostic X-ray equipment c. To carry out representative patient dosimetry measurements d. To stop working with ionizing radiation if you have received a dose over the dose limit e. For employees to report suspected incidents involving radiation
Answer» a. True. b. True. c. True. d. False. Regulation 26: the employer must ensure that employees do not, during the remainder of the calendar year, receive a dose greater than the proportion of any dose limit that is equal to the fraction of the remaining dose limit period. e. True. Employees should inform either the radiation protection supervisor (RPS) or their employer regarding these incidents.

1.

It is a legal requirement: a. To report incidences when the wrong patient is given a medical exposureb. To carry out quality assurance measurements on diagnostic X-ray equipment c. To carry out representative patient dosimetry measurements d. To stop working with ionizing radiation if you have received a dose over the dose limit e. For employees to report suspected incidents involving radiation

Answer»

a. True.

b. True.

c. True.

d. False. Regulation 26: the employer must ensure that employees do not, during the remainder of the calendar year, receive a dose greater than the proportion of any dose limit that is equal to the fraction of the remaining dose limit period.

e. True. Employees should inform either the radiation protection supervisor (RPS) or their employer regarding these incidents.

Discussion

2.	Events A and B are mutually exclusive and have nonzero probability. Which of the following statement(s) are true? (a) (A∪ B) = (A) + (B) (b) (B) > (A) (c) (A ∩ B) = (A)(B) (d) (B) < (A)
Answer» (a) (A∪ B) = (A) + (B) For mutually exclusive events A and B (A∪ B) = (A) + (B)

2.

Events A and B are mutually exclusive and have nonzero probability. Which of the following statement(s) are true? (a) (A∪ B) = (A) + (B) (b) (B) > (A) (c) (A ∩ B) = (A)(B) (d) (B) < (A)

Answer»

(a) (A∪ B) = (A) + (B)

For mutually exclusive events A and B (A∪ B) = (A) + (B)

Discussion

3.	Then 6 identical cells of no internal resistance are connected in series in the second arycircuit of a potentiometer, the balancing length is `l` if two of them are wrongly connected to balacing length becomesA. `(l)/(4)`B. `(l)/(3)`C. `l`D. `(2l)/(3)`
Answer» Correct Answer - B `(E_(1))/(E_(2))=(6E)/(2E)=(l)/(l_(2))`

Discussion

4.	Which of the following are types of stochastic effects? a. Skin erythema b. Infertility c. Breast cancer d. Cataracts e. Leukaemia
Answer» a. False. This is a deterministic effect. b. False. This is a deterministic effect. c. True. d. False. This is a deterministic effect. e. True.

4.

Which of the following are types of stochastic effects? a. Skin erythema b. Infertility c. Breast cancer d. Cataracts e. Leukaemia

Answer»

a. False. This is a deterministic effect.

b. False. This is a deterministic effect.

c. True.

d. False. This is a deterministic effect.

e. True.

Discussion

5.	Magnetic force on a charged particle is given by `vec F_(m) = q(vec(v) xx vec(B))` and electrostatic force `vec F_(e) = q vec (E)`. A particle having charge q = 1C and mass 1 kg is released from rest at origin. There are electric and magnetic field given by `vec(E) = (10 hat(i)) N//C for x = 1.8 m` and `vec(B) = -(5 hat(k)) T` for `1.8 m le x le 2.4 m` A screen is placed parallel to y-z plane at `x = 3 m`. Neglect gravity forces. Time after which the particle will collide the screen is (in seconds)A. `1/5 (3+(pi)/(6)+(1)/(sqrt(3)))`B. `1/5 (6+(pi)/(3)+(sqrt(3)))`C. `1/3 (5+(pi)/(6)+(1)/(sqrt(3)))`D. `1/5 (6+(pi)/(18)+(sqrt(3)))`
Answer» Correct Answer - A `t_(OA)=sqrt((2s)/(a))=sqrt((2sm)/(qE))=sqrt((2 xx 1.8 xx 1)/((1)(10)))` `=0.6 sec=3//5 sec` `t_(AB)=(30^(@))/(360^(@))T= 1/12 xx (2pi m)/(qB)` `((2pi)(1))/((12)(1)(5))=(pi)/(30)sec` `t_(BC)=(BC)/(v)=(0.6 sec theta)/(v)=(0.6((2)/(sqrt(3)))/(6))=(1)/(5sqrt(3)) sec` Hence `3/5+(pi)/(30)+(1)/(5sqrt(3))=(1)/(5)(3+(pi)/(6)+(1)/(sqrt(3))) sec` Hence choice (a) is correct.

Discussion

6.	Magnetic force on a charged particle is given by `vec F_(m) = q(vec(v) xx vec(B))` and electrostatic force `vec F_(e) = q vec (E)`. A particle having charge q = 1C and mass 1 kg is released from rest at origin. There are electric and magnetic field given by `vec(E) = (10 hat(i)) N//C for x = 1.8 m` and `vec(B) = -(5 hat(k)) T` for `1.8 m le x le 2.4 m` A screen is placed parallel to y-z plane at `x = 3 m`. Neglect gravity forces. The speed with which the particle will collide the screen isA. `3 m s^(-1)`B. `6 m// s^(-1)`C. `9 m//s^(-1)D. `12 m// s^(-1)`
Answer» Correct Answer - B Velocity at `A v=sqrt(2 as)` `=sqrt(2xx((qE)/(m))s) = sqrt((2 xx 1 xx 10 xx 1.8)/(1))=6 m//s` In magnetic field, speed does not change. Hence particle will collide with speed `6m//s`.

Discussion

7.	In the figure shows below, the electric current flowing through 2R resistor is(a) from left to right(b) from right to left(c) no current flows(d) double to that through any other resistors
Answer» Correct option(b) Explanation: Electric current flows from high potential to low potential. In given circuit B point is connected with ve terminal of battery i.e., at low potential and C point is connected with the terminal of battery i.e., at high potential. Hence current flows from right to left through 2R resistor.

7.

In the figure shows below, the electric current flowing through 2R resistor is(a) from left to right(b) from right to left(c) no current flows(d) double to that through any other resistors

Answer»

Correct option(b)

Explanation:

Electric current flows from high potential to low potential.

In given circuit B point is connected with ve terminal of battery i.e., at low potential and C point is connected with the terminal of battery i.e., at high potential. Hence current flows from right to left through 2R resistor.

Discussion

8.	The spark plug in an automobile engine is an `R-L` circuit as shown in figure. The circuit that provides the spark uses an inductor as the energy source. Initially switch is closed and allows current to build through the inductor. When the switch is open the current decreases rapidly through inductor and a large emf is induced by inductor Given `verepsilon = 12V, L=10mH,Rc=10Omega,R_(p)=7kOmega` If switch must be closed for up to three time constants. Find this timeA. 3 msB. `1.5` msC. 6 msD. `1//3` ms
Answer» Correct Answer - A Time constant `tau_(L)=L/(R_(C))=((1xx10^(-2)))/10=1`m//s `3tau_(L)=3`ms

Discussion

9.	Which of the following unit does not represent the unit of power ?(a) ampere/volt (b) (ampere)2 × ohm (c) joule/second (d) ampere × volt
Answer» (a) ampere/volt Explanation: power = p = voltage × current • unit – volt × ampere power = p = current² × resistance • unit – (ampere)² × ohm power = p = (Energy / time) • unit = (Joule / second)

9.

Which of the following unit does not represent the unit of power ?(a) ampere/volt (b) (ampere)2 × ohm (c) joule/second (d) ampere × volt

Answer»

(a) ampere/volt

Explanation:

power = p = voltage × current • unit – volt × ampere

power = p = current² × resistance • unit – (ampere)² × ohm

power = p = (Energy / time) • unit = (Joule / second)

Discussion

10.	In the figure shown, a circuit contains two identical resistors with resistance R = 5Ω and an inductance with L = 2mH. An ideal battery of 15V is connected in the circuit. What will be the current through the battery long after the switch is closed?(1) 6 A (2) 3 A (3) 7.5 A (4) 5.5 A
Answer» The correct option (1) 6 A Explanation: ∵ R = 5Ω I_{steady state} = 15/R + 15/R = 6A

10.

In the figure shown, a circuit contains two identical resistors with resistance R = 5Ω and an inductance with L = 2mH. An ideal battery of 15V is connected in the circuit. What will be the current through the battery long after the switch is closed?(1) 6 A (2) 3 A (3) 7.5 A (4) 5.5 A

Answer»

The correct option (1) 6 A

Explanation:

∵ R = 5Ω

I_{steady state} = 15/R + 15/R = 6A

Discussion

11.	Write the unit of angular acceleration in the SI system.(a) N.Kg (b) rad/(sec)2 (c) m/sec (d) N/kg
Answer» (b) rad/(sec)² Angular acceleration is measured in (rad / sec²)

11.

Write the unit of angular acceleration in the SI system.(a) N.Kg (b) rad/(sec)2 (c) m/sec (d) N/kg

Answer»

(b) rad/(sec)²

Angular acceleration is measured in (rad / sec²)

Discussion

12.	Why alum is added for purification of water ?
Answer» The water obtained from natural sources often contains suspended impurities.Alum is added to such water to coagulate the suspended impurities and make water fit for drinking purposes.

Discussion

13.	What happen's when an electric current is passed through colloidal solution ?
Answer» The colloidal particles move towards oppositely charged electrodes, get discharged and precipitated.

Discussion

14.	Write definition of adsorption.
Answer» The accumulation of molecular species at the surface rather than in the bulk of a solid or liquid is termed adsorption. The molecular species or substance, which concentrates or accumulates at the surface is termed adsorbate and the material on the surface of which the adsorption takes place is called adsorbent.

Discussion

15.	State and illustrate Pauli’s exclusion principle.
Answer» No two electrons in an atom can have all same set of four quantum numbers alike.

Discussion

16.	Write definition of chemical adsorption.
Answer» When the forces of attraction existing between adsorbate and adsorbent are strong chemical bonds, the adsorbtion called chemical adsorption. In chemical adsorption, the adsorbate forms product by reaction at the surface of adsorbent.

Discussion

17.	What happen's when a beam of light is passed through the colloidal solution ?
Answer» Tyndall effect is observed .The Tyndall effect is due to the fact that colloidal particles scatter light in all directions in space. This scattering of light illuminates the path of beam in the colloidal dispersion.

Discussion

18.	State Avegadro’s law. How many atoms of Hydrogen are present in 1 mole of water?
Answer» Under the same temperature & pressure equal volume of gases contains equal number of molecules. = 2 × 6.022 × 10²³ atoms

18.

State Avegadro’s law. How many atoms of Hydrogen are present in 1 mole of water?

Answer»

Under the same temperature & pressure equal volume of gases contains equal number of molecules.

= 2 × 6.022 × 10²³ atoms

Discussion

19.	Why the colour of sky appears blue ?
Answer» Dust particles along with water suspended in air scatter blue light which reaches our eyes and the sky looks blue to us.

Discussion

20.	Write oxidation state of nitrogen in nitric acid.
Answer» HNO₃ 1 + x + (–2)3 = 0 x = + 5

20.

Write oxidation state of nitrogen in nitric acid.

Answer»

HNO₃

1 + x + (–2)3 = 0

x = + 5

Discussion

21.	What mass of Calcium carbonate is to be decomposed to obtain 4.4 of CO2 in the following reaction CaCO3 → CaO + CO2 (Molecular mass of CaCO3 = 100)
Answer» 100 x 4.4/44 = 10g or g

Discussion

22.	Expess 9.8 g H2SO4 in mole (Molecular mass of H2SO4 = 98)
Answer» 9.8 g H₂SO₄ = 0.1 mole or 0.1 mole

Discussion

23.	You made a good decision there. A) crow B) call C) catch
Answer» Correct option is B) call

Discussion

24.	What is the mass percentage of Carbon in Methane (CH4) (Molecular mass of CH4 =16)
Answer» 12/16 x 100 = 75% or 75%

Discussion

25.	The spring force is given by `F=-kx`, here k is a constant and x is the deformation of spring. The `F-x` graph isA. B. C. D.
Answer» Correct Answer - C given `F=-10x` The standard equation of straight line is `y=mx+c` If we compare given equation with the standard equation of Straight line, the intercapt is zero and slope is negative. Hence the graph is passing through origin. Hence, option (c ) is correct.

Discussion

26.	A body is attached to a spring whose other end is fixed. If the spring is elongated by x, its potential energy is `U=5x^2`, where x is in metre and U is in joule. U-x graph isA. B. C. D.
Answer» Correct Answer - A `U=5x^2`, it is the equation of parabola with concavity is upward. At `x=0`,`U=0`, it means graph is passing through origin. Hence, option (a) is correct.

Discussion

27.	A Bohr hydrogen atom undergoes a transition `n=5rarrn=4` and emits a photon of frequency v. Frequency of circular motion of electron in n = 4 orbit is `V_(4)`. Find the ratio `v//v_(4).(54)/(25N)`, where N is an integer find N
Answer» Correct Answer - `0003` `0003` `E_(n)=-(mz^(2)e^(4))/(8 epsilon_(0)^(2)n^(2)h^(2))` so , `hv=+(mz^(2)e^(4))/(8 epsilon_(0)^(2)h^(2))[(1)/(16)-(1)/(25)]` `:. v=(mz^(2)e^(4))/(8 epsilon_(0)^(2)h^(3))[(9)/(16xx25)] " " ...(1)` and frequency `v_(4)=(1)/(T)=(v)/(2pir)=((ze^(2))/(2 epsilon_(0)nh))(1)/(2pi)((pi m z e^(2))/(epsilon_(0)h^(2)n^(2)))=(z^(2)e^(4)m)/(4 epsilon_(0)^(2)n^(3)h^(3)) " " ....(2)` `:. v//v_(4)=18//25`

Discussion

28.	When the voltage applied to an X-ray tube increased from `V_(1)15.5` KV to `V_(2)` = 31 KV, the wavelength interval between the `K_(alpha)` line and the short wavelength cut-off of the continuous X-ray spectrum increases by a factor of 1.3. If the atomic number of the element of the target is 13N, where N is an integer find N (Take : hc = 1240 eV and R=`1xx10^(7)//m`)
Answer» Correct Answer - `0002` `lambda_(th)=(hc)/(eV_(a))` `(1)/(lambda_(K alpha))=R(z-1)^(2)((1)/(1^(2))-(1)/(2^(2)))` `(143)/(10)(lambda_(K alpha)-lambda_(th))=(lambda_(Kalpha)-(lambda_(th))/(2))` `(3)/(10)lambda_(K alpha)=((13)/(10)-(1)/(2))lambda_(th)` `(3)/(10)((4xx10^(-7))/(3(z_(7))^(2)))=((8)/(10))(12.4xx10^(-7))/(15.5xx10^(3))rArr(5000)/(8)=(z-1)^(2)` `625=(z-1)^(2) " " rArr z = 26`

Discussion

29.	Find the sum of the first 13 odd natural numbers.1. 1442. 1213. 1694. 196
Answer» Correct Answer - Option 3 : 169 Given: First 13 odd natural numbers. Formula used: Sum of first 'n' odd natural numbers = n² Calculations: Sum of first '13' odd natural numbers = (13)2 ⇒ 169 ∴ The sum of first '13' odd natural is 169.

29.

Find the sum of the first 13 odd natural numbers.1. 1442. 1213. 1694. 196

Answer» Correct Answer - Option 3 : 169

Given:

First 13 odd natural numbers.

Formula used:

Sum of first 'n' odd natural numbers = n²

Calculations:

Sum of first '13' odd natural numbers = (13)2

⇒ 169

∴ The sum of first '13' odd natural is 169.

Discussion

30.	Find the value of 9.6363… – 7.8….1. 764/992. 173/993. 123/994. 145/99
Answer» Correct Answer - Option 2 : 173/99 Concept Value of 0.6363…. = 63/99 Value of 0.8… = 8/9 Explanation 9 + 0.6363…. – (7 + 0.8… ) ⇒ 9 + (63/99) – [7 + (8/9)] ⇒ 9 + (63/99) – 7 – (8/9) ⇒ 9 – 7 + [(63/99) – (8/9)] ⇒ 2+ [(63 – 88)/99] ⇒ 2 + (-25/99) ⇒ [(198 – 25)/99] ⇒ 173/99

30.

Find the value of 9.6363… – 7.8….1. 764/992. 173/993. 123/994. 145/99

Answer» Correct Answer - Option 2 : 173/99

Concept

Value of 0.6363…. = 63/99

Value of 0.8… = 8/9

Explanation

9 + 0.6363…. – (7 + 0.8… )

⇒ 9 + (63/99) – [7 + (8/9)]

⇒ 9 + (63/99) – 7 – (8/9)

⇒ 9 – 7 + [(63/99) – (8/9)]

⇒ 2+ [(63 – 88)/99]

⇒ 2 + (-25/99)

⇒ [(198 – 25)/99]

⇒ 173/99

Discussion

31.	In a test (+5) marks are given for each correct answer, (-1) marks for each wrong answer and 0 marks for questions not attempted. The student attempts 7 questions right, some questions wrong and rest not answered. If the total marks scored by a student is 30 and the total number of questions is 15, find the no of questions not attempted by the student.1. 32. 53. 24. 7
Answer» Correct Answer - Option 1 : 3 Given : Marks awarded for correct answer = +5 Marks awarded for incorrect answer = -1 Marks awarded for questions not answered = 0 No. of right questions = 7 Total marks he got = 30 Total no. Of questions = 15 Calculations: Let the number of wrongly attempted questions be x According to the question; 7 × (5) + (x) × (-1) = 30 ⇒ x = 5 Now , total question are 15 Que. Answered right + Que. Answered wrong + Que. not answered = 15 ⇒ 7 + 5 + Que. not answered = 15 ⇒ Questions not answered = 3 ∴ The no of questions not attempted by the student is 3.

31.

In a test (+5) marks are given for each correct answer, (-1) marks for each wrong answer and 0 marks for questions not attempted. The student attempts 7 questions right, some questions wrong and rest not answered. If the total marks scored by a student is 30 and the total number of questions is 15, find the no of questions not attempted by the student.1. 32. 53. 24. 7

Answer» Correct Answer - Option 1 : 3

Given :

Marks awarded for correct answer = +5

Marks awarded for incorrect answer = -1

Marks awarded for questions not answered = 0

No. of right questions = 7

Total marks he got = 30

Total no. Of questions = 15

Calculations:

Let the number of wrongly attempted questions be x

According to the question;

7 × (5) + (x) × (-1) = 30

⇒ x = 5

Now , total question are 15

Que. Answered right + Que. Answered wrong + Que. not answered = 15

⇒ 7 + 5 + Que. not answered = 15

⇒ Questions not answered = 3

∴ The no of questions not attempted by the student is 3.

Discussion

32.	Assertion (A) : Due to Schottky defect, there is no effect on the density of a solid.Reason (R) : Equal number of cations and anions are missing from their normal sites in Schottky defect.(A) Both A and R are true and R is the correct explanation of A.(B) Both A and R are true but R is not the correct explanation of A.(C) A is true but R is false.(D) A is false but R is false.
Answer» Option : (D) A is false but R is false. Schottky defect arieses when equal Number of cations and anions are missing from their normal sites, due to this, the density of solid decreases.

32.

Assertion (A) : Due to Schottky defect, there is no effect on the density of a solid.Reason (R) : Equal number of cations and anions are missing from their normal sites in Schottky defect.(A) Both A and R are true and R is the correct explanation of A.(B) Both A and R are true but R is not the correct explanation of A.(C) A is true but R is false.(D) A is false but R is false.

Answer»

Option : (D) A is false but R is false.

Schottky defect arieses when equal Number of cations and anions are missing from their normal sites, due to this, the density of solid decreases.

Discussion

33.	We’ll get along just fine _______ he minds his own business. A) as B) therefore C) as long as D) so that
Answer» Correct option is C) as long as

Discussion

34.	Apparently Betty is jealous ______ I get better grades than she does. A) in spite of B) in case C) moreover D) because
Answer» Correct option is D) because

Discussion

35.	Blush-on is also called___
Answer» Blush-on is also called BLUSHER.

Discussion

36.	EXERCISE 4.21. If \( B + C =12, B - C =4 \), and \( AB +7 C =108 \), then find the values of \( A , B \), and \( C \).
Answer» A=10 B=8 C=4

Discussion

37.	Write Bohr’s postulates for the hydrogen atom model.
Answer» First Postulate: Electron revolves round the nucleus in discrete circular orbits called stationary orbits without emission of radiant energy. These orbits are called stable orbits or non-radiating orbits. Second Postulate: Electrons revolve around nucleus only in such orbits for which the angular momentum is some integral multiple of (h/2π). Third Postulate: When an electron makes a transition from one of its non-radiating orbits of another of lower energy, a photon is emitted having energy equal to the energy difference between the two state. The frequency of the emitted photon is the given by, h_v = E_i – E_f

37.

Write Bohr’s postulates for the hydrogen atom model.

Answer»

First Postulate: Electron revolves round the nucleus in discrete circular orbits called stationary orbits without emission of radiant energy. These orbits are called stable orbits or non-radiating orbits.

Second Postulate: Electrons revolve around nucleus only in such orbits for which the angular momentum is some integral multiple of (h/2π).

Third Postulate: When an electron makes a transition from one of its non-radiating orbits of another of lower energy, a photon is emitted having energy equal to the energy difference between the two state. The frequency of the emitted photon is the given by, h_v = E_i – E_f

Discussion

38.	Calculate moles of C formed if 40 moles each of A and B are taken. `A + 2B overset(40%)(rarr) D` `2D overset(100%)(rarr) C`
Answer» Correct Answer - 4 `A + 2B overset(40%)(rarr) D` `2D overset(100%)(rarr) C` B is LB `n_(c ) = (1)/(2) xx 40 xx (40)/(100) xx (1)/(2) xx (100)/(100)` `= (1)/(4) xx 40 xx (40)/(100) = 4` moles

Discussion

39.	Unit cell of the iron crystal has edge length of 288 pm and density of 7.86 g `cm^(-3)`. Determinte the type of crystal lattice. Atomic mass of Fe= 56 g `mol^(-1)`
Answer» Correct Answer - Type of crystal lattice is bcc. Given : Edge length =a= 288 pm `=2.88 xx 10^(-8)` cm Density of crystal = d= 7.86 g `cm^(-1)` A vogadro number = 6.022 `xx10^(-23) mol^(-1)` Atomic mass of Fe= 56 g `mol^(-1)` Type of crystal lattice = ? Mass of one Fe atom = `(56)/(6.022 xx 10^(23)) = 9 .3 xx 10^(-23) g` If there are z atoms in the unit cell , then Mass of unit cell= mass of z atoms = z `xx 9.3 xx 10^(-23) g` Volume of unit cell `= a^(3) = (2.88 xx 10^(-8) )^(3)` `=23 .88 xx 10^(-24) cm^(3)` Density of unit cell `d= ("mass of unit cell")/("Volume of unit cell")` `7.86 = (z xx 9.3 xx 10^(-23))/(23. 88 xx 10^(-24))` `:. z =(7.86 xx 23.88 xx 10^(-24))/(9.3xx 10^(-23)) = 2.01 =2` Since the number of atoms in the unit cell is 2, the crystal lattice must be of bcc type.

Discussion

40.	Copper crystallises into a fcc structure and the unit cell has length of edge `3.61 xx 10^(-8)` cm. Calculate the density of the copper Atomic mass of copper is 63.5 g `mol^(-1)`
Answer» Correct Answer - Density of Cu = 8. 97 g `cm^(-3)` Given : Crystalline structure of Cu is fcc Edge length =a= `3.61 xx 10^(-8) cm` Atomic mass of Cu= 63.5 g `mol^(-1)` Avogadro number `=6.022 xx 10^(23) mol^(-1)` Density = d= ? In fcc structure there are 8 Cu atoms at 8 corners and 6 Cu atoms at 6 face centres . `:. ` Total number of Cu atoms `=(1)/(8) xx 8 + (1)/(2) xx 6 = 1 + 3 =4` Mass of Cu atoms `=4 xx 1.054 xx 10^(-22)= 4.216 xx 10^(-22) g` Mass of unit cell = Mass of 4 Cu atoms `= 4.216 xx 10^(-22) g` Volume of unit cell =`a^(3) (3.61 xx 10^(-8))^(3) = 4.7 xx 10^(-23) cm^(3)` Density of unit cell `= ("mass of unit cell" )/( "volume of unit cell" ) ` `:. d = (4.216 xx 10^(-23))/( 4.7 xx 10^(-23)) = 8.97 g cm^(-3)`

Discussion

41.	An experiment is done as shown in diagram then final pressure after stop cock is opened. [Assume reaction : `NH_(3) (g) + HCl (g) rarr NH_(4)Cl (s)` ] A. 3 atmB. `(1)/(3)` atmC. 1 atmD. `1.5` atm
Answer» Correct Answer - B `NH_(3) (g)` `n = (PV)/(RT) = (1 xx 22.4)/(0.0821 xx 273 xx 2) = (1)/(2)` mole `HCl (g)` `n = (PV)/(RT) = (1 xx 44.8)/(0.0821 xx 54.6) = 1` mole `{:(NH_(3) (g),+,HCl (g),rarr,NH_(4)Cl(s),),((1)/(2),,1,,0,),(0,,(1)/(2),,(1)/(2),):}` `n_(HCl) = (1)/(2)` mole `P = (0.5 xx 0.0821 xx 546)/(67.2)` `P = (1)/(3)` atn

Discussion

42.	A solution can be expressed as 25% w/w as well as 20% w/v then mass of 20 ml of such solution is :A. 25 gmB. 8 gmC. 16 gmD. 10 gm
Answer» Correct Answer - C `{:(25% w//w,rArr,25 gm " solute in 100 gm solution" ,),(20% w//v,rArr,"20 gm solute in 100 L solution",),(,,"25 gm solute in x ml solution",),(,,x = (25 xx 100)/(20) = 125 ml,):}` `125 ml rarr 100 gm` (For solution) `20 ml rarr x gm` `x = (2000)/(125) = (400)/(25) = (80)/(5) = 16 gm`

Discussion

43.	To form one molecule of `Al_(2)O_(3)` the toal number of electron transferred from metal to non metal isA. `6 N_(A)`B. `3 N_(A)`C. 3D. 6
Answer» Correct Answer - D In one molecule of `Al_(2)O_(3) . 2Al^(+3)` are present `:.` tota number of electron transferred `= 3 xx 2 = 6`

Discussion

44.	The correct set of quantum numbers for the unpaired electron of Xenon (single positive) ion isA. `{:(n,l,m,),(6,1,0,):}`B. `{:(n,l,m,),(4,1,1,):}`C. `{:(n,l,m,),(5,1,1,):}`D. `{:(n,l,m,),(3,0,0,):}`
Answer» Correct Answer - C `[Kr] 5s^(2) 4d^(10) 5p^(5)` n = 5 l = 1 `m = -1 // 0// + 1` `s = pm (1)/(2)`

Discussion

45.	For the reaction : `2 NH_(3) (g) rarr N_(2) (g) + 3H_(2) (g)`. What is the % of `NH_(3)` converted if the mixture diffuses twice as fast as that `SO_(2)` under similar conditions.A. 3.125B. 6.25C. 12.5D. None of these
Answer» Correct Answer - B `{:(,2NH_(3) (g),rarr,N_(2) (g),+,3H_(2) (g),),("Initially",1,,,,,),("After",1 - 2x,,x,,3x,):}` decomposition `(r_("mix"))/(r_(SO_(2))) = 2 = sqrt((M_(SO_(2)))/(M_("Mix")))` `M_("mix") = (64)/(4) = 16 = M_("avg")` `M_(avg) = (1 xx 17)/(1 + 2x) = 16` `17 = 16 + 32 x` 1 = 32 x `x = (1)/(32)` `% NH_(3) = (2x)/(1) xx 100` `= (1)/(16) xx 100 = 6.25 %`

Discussion

46.	Calculate the molecular weight of a gas `X` which diffuses four times as fast as another gas `Y`, which in turn diffuses twice as fast as another `Z`. Molecular weight of the gas `Z` is `128`.
Answer» Correct Answer - 2

Discussion

47.	The equilibrium constants for two reactions are given. In which case the yield of product will be the maximum? For first reaction: K = 3.2 × 10-6For second reaction: K = 7.4 × 10-6
Answer» Higher the value of K, greater will be the yield of product. So maximum yield will be in the second case.

Discussion

48.	Write an expression for equilibrium constant, Kc for the ‘ reaction,4NH3(g) + 5O2(g) ⇌ 4NO(g) + 6H2O(g)
Answer» K_c = \(\frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^5}\)

Discussion

49.	What is the equilibrium constant (K) in the following cases? 1. Reaction is reversed.2. Reaction is divide by 2. 3. Reaction is multiplied by 2. 4. Reaction is splitted into two.
Answer» 1. 1/K 2. √K 3. K² 4. K₁ K₂

49.

What is the equilibrium constant (K) in the following cases? 1. Reaction is reversed.2. Reaction is divide by 2. 3. Reaction is multiplied by 2. 4. Reaction is splitted into two.

Answer»

1. 1/K

2. √K

3. K²

4. K₁ K₂

Discussion

50.	1. What is homogeneous equilibrium?2. Suggest an example for this.
Answer» 1. The equilibrium in which the reactants and products are in the same phase, 2. N₂(g) + 3H₂(g) ⇌ 2NH₃(g) In this equilibrium, the reactants and products are in the gaseous phase. In this equilibrium, the reactants and products are in the gaseous phase.

50.

1. What is homogeneous equilibrium?2. Suggest an example for this.

Answer»

1. The equilibrium in which the reactants and products are in the same phase,

2. N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

In this equilibrium, the reactants and products are in the gaseous phase.

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Events A and B are mutually exclusive and have nonzero probability. Which of the following statement(s) are true? (a) (A∪ B) = (A) + (B) (b) (B) &gt; (A) (c) (A ∩ B) = (A)(B) (d) (B) &lt; (A)

Then 6 identical cells of no internal resistance are connected in series in the second arycircuit of a potentiometer, the balancing length is `l` if two of them are wrongly connected to balacing length becomesA. `(l)/(4)`B. `(l)/(3)`C. `l`D. `(2l)/(3)`

Which of the following are types of stochastic effects? a. Skin erythema b. Infertility c. Breast cancer d. Cataracts e. Leukaemia

In the figure shows below, the electric current flowing through 2R resistor is(a) from left to right(b) from right to left(c) no current flows(d) double to that through any other resistors

Which of the following unit does not represent the unit of power ?(a) ampere/volt (b) (ampere)2 × ohm (c) joule/second (d) ampere × volt

In the figure shown, a circuit contains two identical resistors with resistance R = 5Ω and an inductance with L = 2mH. An ideal battery of 15V is connected in the circuit. What will be the current through the battery long after the switch is closed?(1) 6 A (2) 3 A (3) 7.5 A (4) 5.5 A

Write the unit of angular acceleration in the SI system.(a) N.Kg (b) rad/(sec)2 (c) m/sec (d) N/kg

Why alum is added for purification of water ?

What happen's when an electric current is passed through colloidal solution ?

Write definition of adsorption.

State and illustrate Pauli’s exclusion principle.

Write definition of chemical adsorption.

What happen's when a beam of light is passed through the colloidal solution ?

State Avegadro’s law. How many atoms of Hydrogen are present in 1 mole of water?

Why the colour of sky appears blue ?

Write oxidation state of nitrogen in nitric acid.

What mass of Calcium carbonate is to be decomposed to obtain 4.4 of CO2 in the following reaction CaCO3 → CaO + CO2 (Molecular mass of CaCO3 = 100)

Expess 9.8 g H2SO4 in mole (Molecular mass of H2SO4 = 98)

You made a good decision there. A) crow B) call C) catch

What is the mass percentage of Carbon in Methane (CH4) (Molecular mass of CH4 =16)

The spring force is given by `F=-kx`, here k is a constant and x is the deformation of spring. The `F-x` graph isA. B. C. D.

A body is attached to a spring whose other end is fixed. If the spring is elongated by x, its potential energy is `U=5x^2`, where x is in metre and U is in joule. U-x graph isA. B. C. D.

A Bohr hydrogen atom undergoes a transition `n=5rarrn=4` and emits a photon of frequency v. Frequency of circular motion of electron in n = 4 orbit is `V_(4)`. Find the ratio `v//v_(4).(54)/(25N)`, where N is an integer find N

Find the sum of the first 13 odd natural numbers.1. 1442. 1213. 1694. 196

Find the value of 9.6363… – 7.8….1. 764/992. 173/993. 123/994. 145/99

We’ll get along just fine _______ he minds his own business. A) as B) therefore C) as long as D) so that

Apparently Betty is jealous ______ I get better grades than she does. A) in spite of B) in case C) moreover D) because

Blush-on is also called___

EXERCISE 4.21. If \( B + C =12, B - C =4 \), and \( AB +7 C =108 \), then find the values of \( A , B \), and \( C \).

Write Bohr’s postulates for the hydrogen atom model.

Calculate moles of C formed if 40 moles each of A and B are taken. `A + 2B overset(40%)(rarr) D` `2D overset(100%)(rarr) C`

Unit cell of the iron crystal has edge length of 288 pm and density of 7.86 g `cm^(-3)`. Determinte the type of crystal lattice. Atomic mass of Fe= 56 g `mol^(-1)`

Copper crystallises into a fcc structure and the unit cell has length of edge `3.61 xx 10^(-8)` cm. Calculate the density of the copper Atomic mass of copper is 63.5 g `mol^(-1)`

An experiment is done as shown in diagram then final pressure after stop cock is opened. [Assume reaction : `NH_(3) (g) + HCl (g) rarr NH_(4)Cl (s)` ] A. 3 atmB. `(1)/(3)` atmC. 1 atmD. `1.5` atm

A solution can be expressed as 25% w/w as well as 20% w/v then mass of 20 ml of such solution is :A. 25 gmB. 8 gmC. 16 gmD. 10 gm

To form one molecule of `Al_(2)O_(3)` the toal number of electron transferred from metal to non metal isA. `6 N_(A)`B. `3 N_(A)`C. 3D. 6

The correct set of quantum numbers for the unpaired electron of Xenon (single positive) ion isA. `{:(n,l,m,),(6,1,0,):}`B. `{:(n,l,m,),(4,1,1,):}`C. `{:(n,l,m,),(5,1,1,):}`D. `{:(n,l,m,),(3,0,0,):}`

For the reaction : `2 NH_(3) (g) rarr N_(2) (g) + 3H_(2) (g)`. What is the % of `NH_(3)` converted if the mixture diffuses twice as fast as that `SO_(2)` under similar conditions.A. 3.125B. 6.25C. 12.5D. None of these

Calculate the molecular weight of a gas `X` which diffuses four times as fast as another gas `Y`, which in turn diffuses twice as fast as another `Z`. Molecular weight of the gas `Z` is `128`.

The equilibrium constants for two reactions are given. In which case the yield of product will be the maximum? For first reaction: K = 3.2 × 10-6For second reaction: K = 7.4 × 10-6

Write an expression for equilibrium constant, Kc for the ‘ reaction,4NH3(g) + 5O2(g) ⇌ 4NO(g) + 6H2O(g)

What is the equilibrium constant (K) in the following cases? 1. Reaction is reversed.2. Reaction is divide by 2. 3. Reaction is multiplied by 2. 4. Reaction is splitted into two.

1. What is homogeneous equilibrium?2. Suggest an example for this.

Events A and B are mutually exclusive and have nonzero probability. Which of the following statement(s) are true? (a) (A∪ B) = (A) + (B) (b) (B) > (A) (c) (A ∩ B) = (A)(B) (d) (B) < (A)