Explore topic-wise InterviewSolutions in Current Affairs.

2x + y = 7

For x = 0,

2(0) + y = 7

⇒ y = 7

Therefore, (0, 7) is a solution of this equation.

For x = 1,

2(1) + y = 7

⇒ y = 5

Therefore, (1, 5) is a solution of this equation.

For x = −1,

2(−1) + y = 7

⇒ y = 9

Therefore, (−1, 9) is a solution of this equation.

For x = 2,

2(2) + y = 7

⇒y = 3

Therefore, (2, 3) is a solution of this equation.

In parallelogram ABCD

∠C = ∠A = 55° [In a parallelogram opposite angles are equal in a parallelogram]

In parallelogram AEFG

∠A = ∠F = 55° [In a parallelogram opposite angles are equal in a parallelogram]

∴ Measure of ∠F = 55°

We know that Sum of angles of a quadrilateral is = 360°

In the quadrilateral ABCD

Given, ∠C =100° and ∠D = 50°

∠A + ∠B + ∠C + ∠D = 360°

∠A + ∠B + 100° + 50° = 360°

∠A + ∠B = 360° – 150°

∠A + ∠B = 210° ……. (Equation 1)

Now in Δ APB

½ ∠A + ½ ∠B + ∠APB = 180° (since, sum of triangle is 180°)

∠APB = 180° – ½ (∠A + ∠B)………. (Equation 2)

On substituting value of ∠A + ∠B = 210 from equation (1) in equation (2)

∠APB = 180° – ½ (210^o)

= 180° – 105°

= 75°

∴ The measure of ∠APB is 75°

In parallelogram RISK

∠RKS + ∠KSI = 180° [sum of the adjacent angles is equal to 180° in a parallelogram]

120° + ∠KSI = 180°

∠KSI = 180° – 120°

∠z = 60°

In parallelogram CLUE

∠CEU = ∠z = 70° [opposite angles are equal in a parallelogram]

In ΔEOS

70° + ∠x + 60° = 180° [Sum of angles of a triangles is 180°]

∠x = 180° – 130°

∠x = 50°

∴ x = 50°

(i) No, opposite angles are equal in a parallelogram.

(ii) Yes, opposite sides are equal and parallel in a parallelogram.

(iii) No, diagonals bisect each other in a parallelogram.

(i) It has two pairs of parallel sides.

True, Rhombus is a parallelogram having two pairs of parallel sides.

(ii) It has two pairs of equal sides.

True, Rhombus has two pairs of equal sides .

(iii) It has only two pairs of equal sides.

False, Rhombus has all four sides equal.

(iv) Two of its angles are at right angles.

False, rhombus has no angle of right angle.

(v) Its diagonals bisect each other at right angles.

True, in rhombus diagonals bisect each other at right angles.

(vi) Its diagonals are equal and perpendicular.

False, in rhombus diagonals are of unequal length.

(vii) It has all its sides of equal lengths.

True, Rhombus has all four sides equal.

(viii) It is a parallelogram.

True, Rhombus is a parallelogram since opposite sides equal and parallel.

(ix) It is a quadrilateral.

True, Rhombus is a quadrilateral since it has four sides.

(x) It can be a square.

True, Rhombus becomes square when any one angle is 90°.

(xi) It is a square.

False, Rhombus is never a square. Since in a square each angle is 90°.

(i) 3y – 1 = 26 [opposite sides are of equal length in a parallelogram]

3y = 26 + 1

y = 27/3

y = 9

3x = 18 [opposite sides are of equal length in a parallelogram]

x = 18/3

x = 6

∴ x = 6 and y = 9

(ii) y – 7 = 20 [diagonals bisect each other in a parallelogram]

y = 20 + 7

y = 27

x – y = 16 [diagonals bisect each other in a parallelogram]

x -27 = 16

x = 16 + 27

= 43

∴ x = 43 and y = 27

(i) AD = BC. Because, diagonals bisect each other in a parallelogram.

(ii) ∠DCB = ∠BAD. Because, alternate interior angles are equal.

(iii) OC = OA. Because, diagonals bisect each other in a parallelogram.

(iv) ∠DAB+ ∠CDA = 180°. Because sum of adjacent angles in a parallelogram is 180°.

(a) Mr. Sharma experienced the shock because due to the electricity leakage. Which occurred due to bad wirings or there might be a bad insulation of those electric wire. When last night rain occurred, the basement got flooded with water. And that water got in contact with those badly insulated wire or open wire. And due to the fact that water is a good conductor of electricity, Ms. Sharma experienced the shock. 

(b) There are no of precaution which Mr. Sharma should take. Some of them are: 

•Monthly checking of all the wirings of the house. So that he can find out about open wires. 

•At the times of rain/monsoon, we must check the water before going into it. 

•Always use appropriate insulated rubber gloves and goggles while working on any branch circuit or any other electrical circuit. 

•Always wear rubber slippers/shoes while handling electricity. 

(c) Mr. Sharma's father is a very intelligent man. He quickly off the main switch which shows that he is very responsible man. Moreover we can also say that he has very caring attitude towards his son.

A number is divisible by 3 if the sum of its digits is divisible by 3, So, 261, 111 are divisible by 3.

A number having its unit digit 2,4,6,8 or 0 is divisible by 2, So, Number 192, 1660 are divisible by 2.

At 273.15K and 1atm pressure every gas has molar volume = 22.4L.

Thermal energy is the energy of a body arising from motion of its atoms or molecules. It is directly proportional to the temperature of the substance.

These mixtures have two or more than two substance or constituents mixed. We may require only one or two separate constituents of a mixture for one use. So tey need to be separated the process which is used for it is based on the various physical properties of the constituents like boling point, melting point, volatility density etc.

Commonly used process which are used to separate the constituents of mixture are

• Sublimation

• Filtration

• Centrifugation

• Evaporation

• Crystallization

• Chromatography

• Distillation

• Fractional Distillation

• Separating funnel

In order to learn the separation of mixtures, we will consider the following three cases:

1. Mixture of two solids

2. Mixture of a solid and a liquid

3. Mixture of two liquids.

Blue ink consists of two or more colour. Separation of individual colour components can be done through the use of chromatography. In this technique, the component that is more soluble with the solvent moves faster as compared with the less soluble ones and as such gets separated.

The residue left on the watch glass is the solute, dye. Due to heating of the ink the water, solvent in the ink got evaporated leaving behind the dye, the solute.

On heating the blue ink which is a mixture of water (volatile) and dye (non-volatile), the solvent water got evaporated from the watch glass leaving the solute, dye. The process involved is evaporation of water.

Nitrogen is used as fertilizers, oxygen is used in hospitals and argon is used in bulbs.

Mixture of two solids are separated by the following method

a. By using a suitable solvent (sugar and sand mixture)

b. By the process of sublimation (Ammonium chloride and common salt)

c. By using a magnet (mixture of iron filling and sulphur power)

Colloid: A colloid is a heterogeneous mixture whose size is too small to be individually seen by naked eyes.

Pure substance: A pure substance consists of single type of particles.

Another French chemist, Joseph Proust, stated this law as 'A chemical compound always contains the same elements combined together in the same proportion by mass'.

The value of pH is inversely related with the concentration of hydrogen ions. Hence, if the concentration of hydrogen ions then pH decreases.

Saturated solution: At any particular temperature a solution that has dissolved as much solute as it is capable of dissolving is said to be a saturated solution.

Chemical properties of acids :

1) Active metals react with acids and liberate hydrogen gas

Zn + HCl → ZnCl₂ + H₂ ↑

2) Acids react with bases to form salt and water. 

HCl_(aq) + NaOH_(aq) → NaCl_(aq) + H₂O_(I) 

3) Acids react with metallic oxides to form salt and water. 

MgO_(s) + 2HCl_(aq) → MgCl_2(aq) + H₂O_(I) 

4) Acids react with carbonates and hydrogen carbonates and release carbon dioxide gas. 

CaCO_3(s) + 2HCl_(aq) → CaCl_2(aq) + H₂O_(I) + CO_2(g) ↑ 

Ca(HCO₃)_2(I) + 2HCl_(aq) → CaCl_2(aq) + 2H20((| + 2CO_2(g) ↑

(a) Zn + H₂SO₄ → ZnSO₄+H₂

(b) HCl+NaHCO₃ → NaCl + H₂O + CO₂

(c) HCl+NaOH → NaCl + H₂O

H atom is responsible for the common properties of acids.

General properties of Acids:

• Tastes sour

• Reacts with metals such as zinc, magnesium etc. liberating hydrogen gas.

• Changes the colour of litmus from blue to red.

• Conducts electricity.

When fats and oils are left open, they get oxidized in the presence of air, their smell and taste change. This means the oxidation is the main cause for rancidity in fats and oil. 

The substances which prevent oxidation of the food items are called antioxidants. Example, Nitrogen is an inert gas and does not oxidize the food. Thus, it is used as antioxidants by chips manufactures.

• A suspension is of heterogeneous nature- There are two phase. The solid particles represent one phase while the liquid in which these are suspended or distributed forms the other phase

• The particle size in a suspension is more than 100 mn (or 10^-7 m).

• The particles in a suspension can be seen with naked eyes and also under a microscope.

• The solid particles present in the suspension can be easily separated by ordinary filter papers. No special filter papers are needed for the purpose.

• The particles in a suspension are unstable. They settle down after sometime when the suspension is kept undisturbed. This is known as precipitate.

• It should be noted that suspension-and precipitate are actually the same. The particles of the solid in the suspended form represent suspension. When they settle, a precipitate result.

A solution is said to be saturated if it has maximum amount of the solute dissolved in it at a given temperature and no solute can be dissolve further.

The substances which are insoluble in water form suspension. “ It is a heterogeneous mixture in which the small particles of a solid are spread throughout a liquid without dissolving in it’. 

Example: Chalk water mixture, muddy water milk of magnesia, fluorine water etc.

A solution is said to be unsaturated if the more amount of solute can be dissolved in it at a given temperature.

d) Kharif crop

Aluminium and zinc.

Correct Option (c)

Explanation: 

From ideal gas equation PV = RT  .....(i)

or    PΔV = RΔT                   ......(ii)

Dividing equation (ii) by (i) we get ΔV/V = ΔT/T = ΔV/VΔT = 1/T  = δ

:.   δ = 1/t. So the graph between δ and T will be rectangular hyperbola.

(a) \(\sqrt{\frac{2eV}{m}}\)

K.E. gained by an electron when accelerated through a potential difference V, 

\(\frac{1}{2}\)mv = eV or v =\(\sqrt{\frac{2eV}{m}}\)

Correct option: (C) Lutetium

Correct option: (C) Mn²⁺

Correct option: (B) TiCl₄

Correct option is b. –1, 0

Correct option is b. HCOOH and CH₃COOH

Conditions for Hydrogen Bonding In case of inter molecular hydrogen bonding, the most important condition is that the molecules must contain one hydrogen atom linked to one highly electronegative atom.

In case of intra–molecular hydrogen bonding, the following conditions are favourable for hydrogen bonding .

(i) The molecule should contain two groups such that one group contains H–atom linked to a highly electro–negative atom and the other group should also contain a highly electronegative atom linked to a lesser electronegative atom.

(ii) The molecule should be planar.

Correct option is a. CO < CO₂ < CO₃^2-

Correct option is a. H₂S < NH₃ < SiH₄ < BF₃

After studying this unit you will be able to • 

• Explain the three states of matter with respect to intermolecular forces and thermal energy between the particles. 
• Define the different gas laws which explains the behavior of ideal gases. • Differentiate between ideal gas and real gas. 
• Explain the behavior of real gases. 
• Understand the liquefaction phenomenon of gases. 
• Explain the equilibrium between gaseous and liquid state and the continuity between the two states. 
• Describe some properties of liquids and explain some natural phenomenon exhibited by them. 
• Solve problems based on gas laws, gaseous behavior, liquefaction of gases and also on properties of liquids.

The deflection was due to the enormous repulsive force.

The number of electrons in a shell is 2n²