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The correct answer is Savitri.

• Gayatri has been revered as Savitur (Sun God).
• Savitur is also called Savitri.
• The popular 'Gayatri Mantra' is given in the third mandala of the Rig Veda, which is dedicated to the god Savitur (Savitri).
• Gayatri Mantra has been composed by Maharishi Vishwamitra.

• The Rigveda is an ancient Indian collection of Vedic Sanskrit hymns.
• It is one of the four sacred canonical texts (śruti) of Hinduism known as the Vedas.
• The Rigveda is the oldest known Vedic Sanskrit text. 

1. In CH₄, NH₃ and H₂O the central atom undergoes sp³ hybridisation. But their bond angles are different due to .the presence of lone pair of electrons. 

2. It can be explained by VSEPR theory. According to this theory, even though the hybridisation is same, the repulsive force between the bond pairs and lone pairs are not same. 

3. Bond pair – Bond pair < Bond pair – Lone pair < Lone pair -Lone pair So due to the varying repulsive force the bond pairs and lone pairs are distorted from regular geometry and organise themselves in such a way that repulsion will be minimum and stability will be maximum. 

4. In case of CH₄ , there are 4 bond pairs and no lone pair of electrons. So it remains in its regular geometry, i.e., tetrahedral with bond angle = 109° 28.'

5. H₂O has 2 bond pairs and 2 lone pairs. There is large repulsion between lp – lp. Again repulsion between lp – bp is more than that of 2 bond pairs. So 2 bonds are more restricted to form inverted V shape (eg;) bent shape molecule with a bond angle of 104° 35. 

6. NH₃ has 3 bond pairs and 1 lone pair. There is repulsion between lp – bp. So 3 bonds are. more restricted to form pyramidal shape with bond angle equal to 107° 18’.

1. Bond order = ½ (N_b -N_a) 

where N_b = No. of electrons occupying bonding orbitals and N_a = No. of electrons occupying antibonding orbitals.

2. As the bond order increases, bond length decreases and bond energy increases, i.e., bond order is directly proportional to bond energy and inversely proportional to bond length.

Carbon 1 → sp³ 

Carbon 2 → sp² 

Carbon 3 → sp² 

Carbon 4 → sp

(i) They are compared to fields of sunlit corn.

(ii) Some bangles are like the flame of a marriage fire.

(iii) Tinkling, luminous, tender, and clear.

(iv) Bangles.

To
The Deputy Superintendent of Police,
Bareilly.
Dated : March 18th, 20….

Sir,

Şubject : Request for use of Loudspeaker Permit. Most respectfully, I beg to say that the marriage ceremony of my younger sister Beena will be performed on Nov. 24th, 20… So I need to use loudspeaker for the function. I, therefore, request you to grant permission for use of loudspeaker from 4 P.M. to 10 P.M. on Nov. 24th, 20….I shall be highly obliged…..

Yours faithfully,
Satya Prakash Gupta
(Address………………)

1. sp³ hybridisation involves mixing up of one – s and three-p orbitals of the valence shell of an atom to form four sp³ hybrid orbitals of equivalent energies and shape. 

2. Each sp³ hybrid orbital has 25% scharacter and 75% p-character. 

3. The angle between the sp³ hybrid orbitals in methane is 109°28′. 

4. Tetrahedron.

The answer is 'heredity'

The phenomenon of genetic transfer of characteristics from one generation to the next, where an offspring  expresses characteristics similar to that of the parent is like begets like. This phenomenon is best understood when the offspring of a mango plant is a mango plant and that of a human is a human and not a monkey or a lion.

Myoglobin is a protein located primarily in the striated muscles of vertebrates. MB is the gene encoding myoglobin in humans. It encodes a single polypeptide chain with one oxygen binding site. Myoglobin contains a heme prosthetic group that can reversibly bind to oxygen.

1. Increasing magnitude of hydrogen bonding among NH₃, H₂O and HF is 

HF > H₂O > NH₃ 

2. The extent of hydrogen bonding depends upon electronegativity and the number of hydrogne atoms available for bonding. 

3. Among N, F and O the increasing order of their electronegativities are

N < O < F 

4. Hence the expected order of the extent of hydrogen bonding is 

HF > H₂O > NH₃

(II) Conc. H₂SO₄ cannot be used because it acts as an oxidizing agent also and gets reduced to SO₂ . 

Zn + 2H₂SO₄ (Conc) → ZnSO₄ + 2H₂O + SO₂ Pure Zn is not used because it is non-porous and reaction will be slow. The impurities in Zn help in constitute of electrochemical couple and speed up reaction.

HF, because Fluorine is more electronegative.

Strength of H-bond depends upon the atomic size and electronegativity of the other atom to which H- atom is covalently bonded. Smaller size and higher electronegativity favour H-bonding. Among N, F and O atoms, F is the smallest and its electronegativity is highest. Hence, HF will have highest magnitude of H-bonding.

Thermodynamically most stable allotrope of carbon is Graphite.

Correct answer is b. You will apply moist-heat on it. 

ΔG = ΔH – TΔS 

ΔG = -1648000 + 163721.2 

= -1484278.8 J/mol 

= -1484.2788 kJ/mol 

The reaction is spontaneous.

Correct answer is 4

We can not calculate I.E. of lithium atom.

Initial moles of H₂SO4 (in/Lit.) = 0.02

In 50% solution moles of H₂SO₄ = 0.01

Added moles of H₂SO₄ = 0.01

Total moles of H₂SO₄ in resulting solution = 0.02 

= 20 x 10^-3 moles

= 20 milimoles

\(n_{H_2SO_4}\) in Solⁿ A = 50% of original solution

 = 0.01 m mol

\(n_{H_2SO_4}\) in Final solution = 0.01 + 0.01

 = 0.02 mmol

 = 0.00002 x 10³ mmol

The answer is 0

The Vaccine used for whooping cough in children is D.P.T. 

232.79 kJ/mol

t₂ = 2t₁ because for 3/4th of the reaction to complete time required is equal to two
half lives.

Given:

Provider marks up 50% above the CP

Discount % = 20%

GST = 18%

Concept used:

Loss = CP – SP

Loss% = {(CP – SP)/CP} × 100

CP = cost price

SP = selling price

MP = marked price

Calculation:

Let CP of internet plan be 2x

MP = 2x × 150% = 3x

SP = 3x × 80% = 12x/5

According to the question,

The provider has paid GST of 18% on SP

So,

Actual amount got by the provider = 12x/5 × 82/100

⇒ 492x/250

Loss = {2x – (492x/250)}

⇒ (500x – 492x)/250

⇒ 8x/250

Loss percentage = {(8x/250)/2x} × 100

⇒ (8x/500x) × 100

⇒ 8/5

⇒ \(1\frac{3}{5}\%\)

∴ Required loss percentage is \(1\frac{3}{5}\%\)

GIVEN :

The shop owner gave 40% discount instead of 30% during Diwali but he also bumped up the MP by 20%.

CONCEPT :

This is a profit and loss problem revolving around MP and discount. Here, we don’t need to calculate the CP.

ASSUMPTION :

Let’s take the initial MP = 100x

CALCULATION :

So, MP during Diwali = 100x × 120/100 = 120x

SP in Non-Diwali days = 100x × 70/100 = 70x

SP in Diwali days = 120x × 60/100 = 72x

Percentage change in SP = \(\frac{{\left( {72x - 70x} \right)}}{{70x}} \times 100 = \frac{{20}}{7} = 2.86\% \)

Concept used:

Successive discount

Calculation:

Let the cost price of the article be x

For 1^st type =  \({\rm{x}} \times \frac{{90}}{{100}} \times \frac{{80}}{{100}} \times \frac{{110}}{{100}} = {\rm{\;}}0.792{\rm{x}}\)

For 2^nd type = \({\rm{x}} \times \frac{{80}}{{100}} \times \frac{{90}}{{100}} \times \frac{{110}}{{100}} = {\rm{\;}}0.792{\rm{x}}\)

For 3^rd type = \({\rm{x}} \times \frac{{110}}{{100}} \times \frac{{80}}{{100}} \times \frac{{90}}{{100}} = {\rm{\;}}0.792{\rm{x}}\)

For every types effective discount is same

∴ Option 4 is the correct answer

Given∶

Ronaldo, Messi and Neymar got Rs. 6000, 2000 and 6000 to start a business after 8 months, there was a gain of = Rs. 7500

Formula used∶

Basic knowledge of Ratio.

Calculation∶

Ratio of Ronaldo, Messi and Neymar = (6000 × 6) ∶ (2000 × 8) × (6000 × 8)

= 36000 ∶ 16000 ∶ 48000

= 36 ∶ 16 ∶ 48 = 9 ∶ 4 ∶ 12

∴ Messi’s share = Rs. (7500 × 4/25) = Rs. 1200

∴ After 8 months, Messi got Rs. 1200 out of Rs. 7500. 

Given:

A grocery shop owner usually makes a profit of 20% in a certain transaction.

Formula used:

SP = CP × (100 + P%)/100

Profit% = (P/CP) × 100

SP = MP × (100 - D)/100

Where,

CP  → cost price

MP  → marked price

P → Profit 

D  → Discount

SP  → selling price

Calculations:

Let the cost of 1 kg of goods be Rs. 100 but shopkeeper charges Rs. 120 to the customer.

But now shopkeeper is giving only 900 grams to a customer that cost him at Rs. 90.

But he gives 20% discount on his previous selling price = (80/100) × 120 = Rs. 96.

Profit = 96 - 90 = 6 

Profit % = (6/90)×100 = (20/3)%

∴ His actual profit percentage is (20/3)%.

Given:

A fruit seller buys avocados at Rs.100, Rs. 80 and Rs. 60 per kilogram.

Formula used:

Selling price (SP) = CP × (100 + P%)/100

Where, 

CP →  cost price 

P → Profit

Calculations:

Let the weight of avocados be 2x, 4x, and 9x.

So total weight = 2x + 4x + 9x = 15x kg

Cost of 15x kg of avocados = Rs. (100 × 2x) + (80 × 4x) + (60 × 9x) = Rs. 1060x

Thus selling price = CP × (100 + P%)/100 = 1060x × (130/100) = 1378x

SP of fruits per kg = 1378x/15x = 91.86 ≈ Rs. 92/kg

∴ The approximate price/kg he sells the dry fruit is Rs. 92.

Given:

Cost price(CP) of milk = Rs. 40 per litre

Total quantity of milk purchased = 50 litre

Water added to the milk = 10 litre

Selling price(CP) of milk = Rs. 50 per litre

Formula used: 

Profit = SP - CP

Profit% = (Profit/CP) × 100

Calculation:

Total CP = 40 × 50 = 2000

Total SP = 50 × 60 = 3000

Profit = 3000 - 2000 = 1000

Profit% = (1000/2000) × 100 = 50

∴ Profit = 50%

The answer is increases.

Correct option is (4) P^o = \(\frac{P}{1-S}\) 

As we know, relative lowering of vapour pressure is given by--

\(\frac{P^o-P_1}{P^o}=x\)----(1)

where,

P^o - P₁ = lowering in vapour pressure

P^o = Vapour pressure of pure solvent

x = mole fraction of solute

we have given,

lowering of vapour pressure = P^o - P₁ = P

mole fraction of solvent  = 'S'

∴ mole fraction of solute = 1 - S

putting these values in equation (1)

\(\frac{P}{P^o}=1-S\) 

P^o = \(\frac{P}{1-S}\)

P^o = \(\frac{P}{1-S}\) 

Therefore, vapour pressure of pure solvent will be

P^o = \(\frac{P}{1-S}\)