Explore topic-wise InterviewSolutions in Current Affairs.

Sum of four ionisation enthalpy for Pt is less than Ni so Pt(IV) is more stable than Ni(IV) where as sum of two ionisation enthalpy for Ni is less than Pt hence Ni(II) is more stable. 

(c) argon

Argon has completely filled configuration. So addition of the electron is not possible and has positive electron gain enthalpy.

This is due to the presence of unpaired electrons in (n-1)d orbitals i.e. d¹ to d⁹ . In presence of ligands, the degeneracy of d orbitals is lost & it splits up into two sets of orbitals t_2g& e_g (according to CF theory). When electrons of lower energy state of d orbitals absorb energy of white light & jump to higher d orbitals, the remaining energy emitted is the colour of the compound. 

Example: If an ion of transition metal absorbs light of wavelength 4500Å(blue), the complementary colour emitted is yellow. [d¹& d¹⁰ are colourless or white].

(i) Zn is more powerful reducing agent in comparison to copper. Zn is also cheaper than Cu.

(ii) Partial roasting of sulphide ore forms some oxide. This oxide then reacts with remaining sulphide ore to give copper i.e. self-reduction occurs.

2Cu₂S + 3O₂ →2Cu₂O+2SO₂, 

2Cu₂O+2Cu₂S→ 6Cu + SO₂. 

(iii) Though carbon is good reducing agent for oxide but it is poor reducing agent for sulphides. The reduction of metal sulphide does not have large negative value.

(i) Telescope

L₂ : objective

L₃ : eyepiece

Reason : Light gathering Power and magnifying power will be larger.

(ii) Microscope

L₃ : objective

L₁ : eyepiece

Reason : Angular magnification is more for short focal length of objective and eyepiece.

4, 7 and 2, x are in inverse proportion . 

Therefore, 4 × 7 = 2 × x 

⇒ x = 4 x 7/2 = 14

(a) The values displayed by Rani are :

(i) high degree of general awareness, and

(ii) helping and caring nature.

(b) In normal adjustment of telescope, the final image is formed at infinity. 

(c) Out of these three lenses, lens of power 0.5 D will be used for objective while lens of power 10 D as eye piece for construction of such telescope.

We have,

f(x) = x³ - 6x² - 19x + 84 and g(x) = x - 7

In order to find whether g (x) = x – 7 is a factor of f (x) or not, it is sufficient to show that f (7) = 0 Now,

f(x) = x³ - 6x² - 19x + 84

f (7) = (7)³ – 6 (7)² – 19 (7) + 84

= 343 – 294 – 133 + 84

= 0

Hence, 

g (x) is a factor of f (x).

If g(x) is a factor of f(x), then the remainder will be zero that is g(x) = 0. 

g(x) = x – 7 = 0, then x = 7

Remainder = f(7)

Now, 

f(7) = (7)³ – 6(7)² – 19 x 7 + 84 

= 343 – 294 – 133 + 84 

= 343 + 84 – 294 – 133 

= 0 

Therefore, g(x) is a factor of f(x).

27x³ − y³ – z³ – 9xyz 

= (3x)³ − y³ – z³ – 3(3xyz)  [a³ + b³ + c³ −3abc = (a + b + c)(a²+b²+c²−ab−bc−ca)] 

Here a = 3x, b = -y and c = -z 

= (3x – y – z){ (3x)² + (- y)² + (– z)² + 3xy – yz + 3xz)} 

= (3x – y – z){ 9x² + y² + z² + 3xy – yz + 3xz)}

1. The angle of incidence at the first surface (i). 

2. The angle of prism (A). 

3. Refractive index of the material.

Yes, we can. Draw a tangent line to the curve, parallel to X – axis, at the lowest point of the graph. The point where this line cuts Y – axis gives the angle of minimum deviation.

(i₁ + i₂ )= A + D 

i + r = A + D

The angle of deviation decreases first and then increases with increase in the angle of incidence.

The angle between the extended incident and emergent rays is called angle of deviation. 

(OR) 

Extend both incident and emergent rays till they meet at a point ‘O’. Measure the angle between these two rays. This is the angle of deviation.

Given, one of the adjacent angle ∠D = 135°

Let other adjacent angle ∠A be = x°

We know that sum of adjacent angles = 180°

x° + 135° = 180°

x° = 180° – 135°

= 45°

∠A = x° = 45°

We know that measure of opposite angles are equal in a parallelogram.

So, ∠A = ∠C = 45°

And ∠D = ∠B = 135°

Given, one of the adjacent angle ∠A = 130°

Other adjacent angle ∠B be = x°

We know that sum of adjacent angles = 180°

x° + 130° = 180°

x° = 180° – 130°

= 50°

∠B = x° = 110°

We know that measure of opposite angles are equal in a parallelogram.

So, ∠A = ∠C = 130°

And ∠D = ∠B = 50°

In ΔBEC

∠BEC + ∠ECB +∠CBE = 180° [Sum of angles of a triangle is 180°]

90° + 40° + ∠CBE = 180°

∠CBE = 180°-130°

∠CBE = 50°

∠CBE = ∠ADC = 50° (Opposite angles of a parallelogram are equal)

∠B = ∠D = 50° [Opposite angles of a parallelogram are equal]

∠A + ∠B = 180° [Sum of adjacent angles of a triangle is 180°]

∠A + 50° = 180°

∠A = 180°-50°

So, ∠A = 130°

In ΔDFC

∠DFC + ∠FCD +∠CDF = 180° [Sum of angles of a triangle is 180°]

90° + ∠FCD + 50° = 180°

∠FCD = 180°-140°

∠FCD = 40°

∠A = ∠C = 130° [Opposite angles of a parallelogram are equal]

∠C = ∠FCE +∠BCE + ∠FCD

∠FCD + 40° + 40° = 130°

∠FCD = 130° – 80°

∠FCD = 50°

∴ ∠EBC = 50^o, ∠ADC = 50^o and ∠FCD = 50^o

Given, one of the adjacent angle ∠A = 70^o

Other adjacent angle ∠B be = x°

We know that sum of adjacent angles = 180°

x° + 70° = 180°

x° = 180° – 70°

= 110°

∠B = x° = 110°

We know that measure of opposite angles are equal in a parallelogram.

So, ∠A = ∠C = 70°

And ∠D = ∠B = 110°

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

Let each angle be x°

So,

150° + x° + x° + x° = 360°

3x° = 360° – 150°

x°= 210°/3

= 70°

∴ Value of equal angles is 70° each.

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

Let each angle be x°

So,

65° + 65° + x° + x° = 360°

2x° = 360° – 130°

x° = 230°/2

= 115°

∴ Value of two angles is 115° each.

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

Let each angle be x°

So,

x° + x° + x° + x° = 360°

x° = 360°/4

= 90°

∴ Value of angle is 90° each.

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

So,

80° + 80° + 80° + x° = 360°

x° = 360° – 240°

x° = 120°

∴ Value of fourth angle is 120°

(i) Name a pair of adjacent sides.

Adjacent sides are: AB, BC or BC, CD or CD, DA or AD, AB

(ii) Name a pair of opposite sides.

opposite sides are: AB, CD or BC, DA

(iii) How many pairs of adjacent sides are there?

Four pairs of adjacent sides i.e. AB BC, BC CD, CD DA and DA AB

(iv) How many pairs of opposite sides are there?

Two pairs of opposite sides. AB, DC and DA, BC

(v) Name a pair of adjacent angles.

Four pairs of Adjacent angles are: ∠DAB ∠ABC, ∠ABC ∠BCA, ∠BCA ∠CDA or ∠CDA ∠DAB

(vi) Name a pair of opposite angles.

Four pair of opposite angles are: ∠DAB ∠BCA and ∠ABC ∠CDA

(vii) How many pairs of adjacent angles are there?

Four pairs of adjacent angles. ∠DAB ∠ABC, ∠ABC ∠BCA, ∠BCA ∠CDA and ∠CDA ∠DAB

(viii) How many pairs of opposite angles are there?

Two pairs of opposite angles. ∠DAB ∠BCA and ∠ABC ∠CDA

The measure of interior angle A of a polygon of n sides is given by A = [(n-2) ×180^o]/n

Angle of quadrilateral is 160°

150^o = [(n-2) ×180^o]/n

150^on = (n-2) ×180^o

150^on = 180^on – 360^o

180^on – 150^o = 360^o

30^on = 360^o

n = 360^o/30

= 12

∴ Number of sides are 12.

The measure of interior angle A of a polygon of n sides is given by A = [(n-2) ×180^o]/n

Angle of quadrilateral is 175°

175^o = [(n-2) ×180^o]/n

175^on = (n-2) ×180^o

175^on = 180^on – 360^o

180^on – 175^o = 360^o

5^on = 360^o

n = 360^o/5

= 72

∴ Number of sides are 72

In parallelogram BDEF

BD = EF and BF = DE [opposite sides are equal in a parallelogram]

In parallelogram DCEF

DC = EF and DF = CE [opposite sides are equal in a parallelogram]

In parallelogram AFDE

AF = DE and DF = AE [opposite sides are equal in a parallelogram]

So, DE = AF = BF

Similarly: DF = CE = AE

Given, DE = DF

Since, DF = DF

AF + BF = CE + AE

AB = AC

∴ ΔABC is an isosceles triangle.

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

So,

110° + 50° + 40° + x° = 360°

x° = 360° – 200°

x° = 160^o

∴ Value of fourth angle is 160^o

297/6 sq. units

It is given that

Length (l) of room = 5 m

Breadth (b) of room = 4 m

Height (h) of room = 3 m

It can be observed that four walls and the ceiling of the room are to be white-washed. The floor of the room is not to be white-washed.

Area to be white-washed = Area of walls + Area of ceiling of room

= 2lh + 2bh + lb

= [2 × 5 × 3 + 2 × 4 × 3 + 5 × 4] m²

= (30 + 24 + 20) m²

= 74 m²

Cost of white-washing per m2 area = Rs 7.50

Cost of white-washing 74 m2 area = Rs (74 × 7.50)

= Rs 555

Let length, breadth, and height of the rectangular hall be l m, b m, and h m respectively.

Area of four walls = 2lh + 2bh

= 2(l + b) h

Perimeter of the floor of hall = 2(l + b)

= 250 m

∴ Area of four walls = 2(l + b) h = 250h m²

∴ 15000 = 2500h h = 6

Therefore, the height of the hall is 6 m.

Height (h) of cylindrical pipe = Length of cylindrical pipe = 28 m

Radius (r) of circular end of pipe = 5/2 = 2.5 cm = 0.025m

CSA of cylindrical pipe = 2πrh

= ( 2 x 22/7 x 0.025 x 28) m²

= 4.4 m²

The area of the radiating surface of the system is 4.4 m².

In a cross between AABB x aabb, the ratio of F, genotypes between AABB, AaBB, Aabb and aabb would be 1 : 2 : 2 : 1. Genotype is the genetic make up of an individual. This may refer to just one trait or it may refer to the combination of many or all of the traits of the individuals.

AABB : AaBB : Aabb : aabb 

 2ⁿ where n= number of heterozygotes 

2⁰ : 2¹ : 2¹ : 2⁰

1 : 2 : 2 : 1 are the ratio of above genotypes.

The ability of a substance to flow is called fluidity. The substances which show fluidity are called fluids. Liquid and gas are examples of fluids. The particles of liquid and gas are able to move freely because there is a large space between them.

Usually acids generate hydrogen gas on reacting with metals.

Test: 

When a burning splinder is brought near to the collected gas (H₂), it puts off with a pop sound. This test proves that the gas is H₂

The conditions which affect the decomposition of nitric acid are: 

1. Presence of sunlight 

2. Higher temperature.

Potassium nitrate and cone, sulphuric acid.

Excess of air carries the reactions in forward direction as oxygen is needed in all the three reactions, leading to the formation of nitric acid.

 The exothermicity of catalytic reaction helps in stopping external heating, there by saving on energy. 

The first step in the manufacture of HNO₃ is the catalytic oxidation of NH₃ Platinum is used.

Cu + 4HNO₃ (cone.) → Cu (NO₃ )₂ + 2H₂O + 2NO₂

Low temperature (less than 50°C). 

This salt gives nitrogen dioxide on heating. Pb(NO₃ )₂

Aspergillus flavus infest dried fidoos and produ.