Mathematics and Science PYQs

The correct answer is Isoprene.

Key Points

• The monomer (“one-part”) of natural rubber is isoprene.
• The chemical name for natural rubber is polyisoprene.
• Its IUPAC name of isoprene is 2-methyl but-1,3-diene.

Important Points

 

Additional Information

• An addition polymerization joins thousands of isoprene units together to form poly(isoprene) or natural rubber.
• Vulcanization is a manufacturing process discovered by Charles Goodyear in 1839 to convert raw rubber into a tough useful product.
• In this process, about 1 – 3% by weight of sulphur is added to raw rubber and the mixture is heated carefully.

The correct answer is n-Pentane.

Key Points

• The n-pentane melts higher than others because it can assume a rodlike shape in which adjacent molecules can pack closely together.
• The long n-pentane molecules fit rather nearly together, they present the greatest intermolecular attraction.
• The boiling point of n-pentane is 36°C.

Important Points

• The surface area of the simple long chain compound will be more as compared to the branched-chain compound.
• So, less temperature will be required for branched-chain compounds as compared to a simple long chain compound.
• The boiling point of iso-pentane is 28°C, and the boiling point of neo-pentane is 10°C.
• The boiling point of isobutane is -11.7°C.
• **Iso-pentane, neo-pentane and Isobutane are branched-chain compounds.**​

Additional Information

• Hydrocarbons with the same numbers of atoms but different shapes form different compounds called isomers.

The correct answer is 4π + 2.

Key Points

• The pi-electron count is defined by the series of numbers generated from 4n+2 where n = zero or any positive integer (i.e, n = 0, 1, 2, etc.).
• The most common case in six pi electrons (n = 1) is found for example in benzene, pyrrole, furan, and pyridine.
• For a compound to be considered aromatic, it must be flat, cyclic, and conjugated and it must obey Huckel's rule.

 

Important Points

• Huckel's Rule (4n+2 rule): To be aromatic, a molecule must have a certain number of pi electrons (electrons with pi bonds, or lone pairs within p orbitals) within a closed loop of parallel, adjacent p orbitals.
• π bonds are simply the second bond made in a double bond. Any pure double bond is one sigma/σ and one pi/π bond.

The correct answer is the nucleus in the atom.

CONCEPT:

Rutherford Experiment: 

• Rutherford with his co-workers initiated a series of groundbreaking experiments that completely changed the accepted model of the atom.
• The very thin sheets of gold foil were bombarded with fast-moving alpha particles.
• Bombardment of alpha particles on the gold foil showed that a very small percentage of alpha particles were deflected.
• This experiment showed that 'The nuclear model of the atom consists of a small and dense positively charged interior surrounded by a cloud of electrons.'

Figure (A) The experimental setup for Rutherford's gold foil

Figure (B) The plum pudding model (top) says that all of the alpha particles should have been passed through the gold foil with no deflection or very little deflection. Rutherford found that a small percentage of alpha particles were deflected at large angles, which could be explained by an atom with a very small, dense, positively-charged nucleus at its center (bottom).

EXPLANATION:

• Rutherford’s atomic model accounts for the positively charged central core of an atom.
• So that positive charge is termed as the nucleus.

Important Points

• Electron was discovered by J. J. Thomson in 1897 when he was studying the properties of cathode rays.
• Proton was discovered by Ernest Rutherford.
• By his experiment, nitrogen under alpha-particle bombardment ejects what appear to be hydrogen nuclei.
• He had accepted the hydrogen nucleus as an elementary particle, naming its proton.

The correct answer is Air.

• Air is a homogeneous mixture of gaseous substances like nitrogen, oxygen, and smaller amounts of other substances
• Each layer of the Earth's atmosphere has a different density, each layer of air is its own homogeneous mixture.

Important Points

• Whole milk is actually a heterogeneous mixture composed of globules of fat and protein dispersed in water.
• Smoke is a heterogeneous mixture as smoke does not have uniform composition throughout.
• Jelly is a mixture of fruit juice and sugar and it is a heterogeneous mixture.

Key Points

• Mixing two solids, without melting them together, typically results in a heterogeneous mixture.
• Heterogeneous mixtures are not uniform.
• If two samples from different parts of the mixture, they will not have an identical composition.
• Use a mechanical method to separate components of a heterogeneous mixture.
• Examples include sand and sugar, salt and gravel, ice cubes in a drink, a mixture of oil and water.
• Homogeneous mixtures appear uniform to the eye. They consist of a single phase.
• Example: brass, bronze, steel, sterling silver, vodka, vinegar, and dishwashing liquid.

The correct answer is Coke.

Key Points

• Coke is prepared by heating coal in the absence of air (destructive distillation).
• It contains high carbon and few impurities.
• Coke is used as a fuel and a reducing agent in smelting iron ore in a blast furnace.
• Coke is one of the purest forms of carbon when impurities are removed.

 

Important Points

• Petrol (also known as gasoline), is a product of petroleum, produced by distilling and refining petroleum.
• Paraffin wax is a soft colorless solid derived from petroleum.
• Lubricating oil is derived from the refining of petroleum crude oil, processed by solvent extraction or hydrogenation.
• Petroleum is known as Liquid Gold or Black gold.

The correct answer is Cu.

• Copper (Cu) does not liberate H₂ when it reacts with acid.
• Cu is least reactive than Hydrogen. So it cannot displace hydrogen from acid.
• When Cu reacts with oxygen, it gives a black colour product, that is a Copper oxide.

2Cu + O₂ ⇒ 2CuO

Important Points

• Sodium (Na) metal dissolves readily in dilute sulphuric acid to form solutions containing the aquated Na⁺ ion with hydrogen gas.
• 2Na(s) + H₂SO₄(aq) → 2Na⁺(aq) + SO₄^2-(aq) + H₂(g)
• Magnesium (Mg) metal dissolves readily in dilute sulphuric acid to form solutions containing the aquated Mg²⁺ ion with hydrogen gas.
• Mg(s) + H₂SO₄(aq) → Mg²⁺(aq) + SO₄^2-(aq) + H₂(g)
• Calcium (Ca) metal dissolves readily in dilute or concentrated hydrochloric acid to form solutions containing the aquated Ca²⁺ ion with hydrogen gas.
• Ca(s) + 2HCl(aq) → Ca²⁺(aq) + 2Cl^-(aq) + H₂(g)

Key Points

• Sodium metal reacts rapidly with water to form a colorless solution of sodium hydroxide (NaOH) and hydrogen gas (H₂).
• 2Na(s) + 2H₂O → 2NaOH(aq) + H₂(g)

The correct answer is 16 g of CH₄

• Find the mole number of each:

N₂ = 16/28 = 0.5714

CO₂ = 16/44 = 0.3636

CH₄ = 16/16 = 1

O₂ = 16/36 = 0.5

So, mol number of CH₄ is maximum, so 16 g of CH₄ has maximum number of atoms.

Important Points

• One mole of a substance is equal to 6.022 × 10²³ units of that substance (such as atoms, molecules, or ions).
• The number 6.022 × 10²³ is known as Avogadro's number or Avogadro's constant.
• The concept of the mole can be used to convert between mass and number of particles.

The correct answer is Tollens reagent.

• In the laboratory, ethylene and acetylene can be distinguished by Tollens reagent.
• Tollens' reagent is a chemical reagent used to determine the presence of aldehyde, aromatic aldehyde, and alpha-hydroxy ketone functional groups.
• The reagent consists of a solution of silver nitrate and ammonia.

Important Points

• Fehling's solution is a deep blue alkaline solution that is used to identify the presence of aldehydes, groups that contain an aldehyde functional group or a ketone.
• Fehling’s A containing blue color copper(II) sulfate.
• Fehling’s B is a clear liquid containing potassium sodium tartrate (Rochelle salt) and a strong alkali (sodium hydroxide).

Ammonical silver nitrate solution(Tollen's reagent) is used to distinguish between ethene and ethyne.

Ethyne + tollens reagent --------> silver acetylide (white ppt) + NH₃

Ethene + tollens reagent ---------> no reaction

• Br₂/H₂O or Bromine water, also called Bromide bromate solution.
• It is used to identify the alkene or alkane functional groups present in the compound.
• Alkene groups react with bromine water in the dark condition and undergo an addition reaction, to give a decolourized solution.
• Alkane doesn’t react with bromine water and the colour of the bromine water remains the same.
• KMnO₄/OH⊖/H₂O or Potassium permanganate (purple color compound) is a strong oxidizing agent.
• It is used for medication for dermatitis, for cleaning wounds, and general disinfection.

Key Points

• The main difference between ethylene (H₂C=CH₂) and acetylene (C₂H₂) is that ethylene is an alkene whereas acetylene is an alkyne.

The correct answer is OF₂.

Key Points

• Oxygen shows +ve oxidation states only in compounds with fluorine because fluorine is the only element more electronegative than oxygen.
• In OF₂, the oxidation state of O is +2.
• OF₂ pulls the bonded electrons towards itself and **leaving oxygen with a positive charge.**​

Important Points

• The oxidation number is also called the oxidation state, the total number of electrons that an atom either gains or loses to form a chemical bond with another atom.
• In Na₂O oxidation state of Na is +1.
• The oxidation number of a free element is always 0.
• The oxidation number of a monatomic ion equals the charge of the ion.

Additional Information

• The oxidation number of H is +1, but it is -1 when combined with less electronegative elements.
• The oxidation number of O in compounds is usually -2, but it is -1 in peroxides.
• The oxidation number of a Group 17 element in a binary compound is -1.
• The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

Given:

The total price of shares = Rs. 2500000

Price of one share = Rs. 100

Concept used:

Number of shares = Total price of shares/Price of one share

Calculation:

The total price of shares = Rs. 2500000

Price of one share = Rs. 100

Number of shares = Total price of shares/Price of one share

⇒ Number of shares = 2500000/100 = Rs. 25000

∴ The number of shares circulated by the company is 25000.

Given:

Total number of cards = 52

Formula used:

Probability = Number of favorable outcome/ Total number of outcome

Calculation:

Total number of the spade in a standard deck of card = 13

Total number of aces in a standard deck of card = 4

Number of an ace in spade = 1

Number of favorable outcome = 13 + 4 – 1 = 16

Total number of outcome = 52

Probability = Number of favorable outcome/ total number of outcome

⇒ Probability = 16/52 = 4/13

∴ The probability that the card is a spade or an ace is 4/13.

Given:

ABCD is a cyclic quadrilateral

∠ADC = 4y + 20°

∠ABC = 4x°

∠BAD = 3y – 5°

∠BCD = 7x + 5°

Concept used:

Sum of the opposite angle of a cyclic quadrilateral is 180°.

Calculation:

ABCD is a cyclic quadrilateral

∴ ∠ADC + ∠ABC = 180°

⇒ 4y + 20° + 4x° = 180°

⇒ x + y + 5° = 45°

⇒ x + y = 40°        ……(1)

And,

∠BAD + ∠BCD = 180°

⇒ 3y – 5° + 7x + 5° = 180°

⇒ 7x + 3y = 180°     …….(2)

Multiply 3 in equation (1) and subtract it in equation (2).

7x + 3y – 3x – 3y = 180° - 120°

⇒ 4x = 60°

⇒ x = 15°

Put value of x in equation (1)

x + y = 40°

⇒ 15° + y = 40°

⇒ y = 25°

∴ The ordered pair (x, y) is (15°, 25°).

Given:

25, 26, 27, 23, 27, 26, 24, x, 27, 26, 25, 25

Concept used:

The mode is the value that appears most often in a set of data values.

Calculation:

Following data is arranged in ascending order:

23, 24, 25, 25, 25, 26, 26, 26, 27, 27, 27, x.

The number of repetitions in data.

23 and 24 - 1 time

25 and 26 - 3 times

27 - 3 times

If Mode = 27

i.e., 27 appears most often (4 times) in given sets of data values.

∴ x = 27

Given:

x = 2, y = 1, z = -3

Formula used:

x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)

Calculation:

x + y + z = 2 + 1 + (-3) = 3 – 3 = 0

If x + y + z = 0,

Then x3 + y3 + z3 – 3xyz = 0

∴ x3 + y3 + z3 – 3xyz = 0.

Given:

Jatin buys an article for 10% less than its value

sells it for 10% more than its value

Formula used:

Profit% = [(S.P – C.P)/C.P] × 100

Calculation:

Let original value be Rs. 10

He bought it at = Rs. 10 × (90/100) = Rs. 90

He sells it at = Rs. 10 × (110/100) = Rs. 110

Gain = Rs. 110 – Rs. 90 = Rs. 20

Gain% = (20/90) × 100%

⇒ 200/9% = \(22\frac{2}{9}\)%

∴ The gain percentage is \(22\frac{2}{9}\)%

Basic quadratic equation is ax² + bx + c  where sum of roots =-b/a and product of roots = c/a

⇒ Here two roots be α and β which are 2x and 3x.

⇒ α × β = c/a  

⇒ α × β = 54

⇒ 6x² = 54

⇒ x = 3 so α = 6 and β = 9

⇒ Now sum of roots = α + β = -b/a 

⇒ p = 15

Given:  

The ratio of the heights of the two right circular cones and their diameters is given.

Formula used:

The volume of cone = \(\frac{1}{3}\pi {r^2}h\)

Calculation:

Here we have two cones with two different radius and height.

Let the radius and the height of the first cone be r₁ and h₁. Similarly the radius and the height of the second cone be r₂ and h₂.

Volume of the first cone be \(\frac{1}{3}\pi {r_1}^2{h_1}\) and the volume of the second cone be \(\frac{1}{3}\pi {r_2}^2{h_2}\)

⇒ r₁ = 2 and r₂ = 2.5, h₁ = 25 and h₂ = 64

⇒ placing the above value in the equation \(\frac{{\frac{1}{3}\pi {r_1}^2{h_1}}}{{\frac{1}{3}\pi {r_2}^2{h_2}}}\)

⇒ \(\frac{{\pi {r_1}^2{h_1}}}{{\pi {r_2}^2{h_2}}}\) = \(\frac{{4 \times 25}}{{6.25 \times 64}}\) = \(\frac{1}{4}\)

∴ Required ratio = 1: 4

Given: 

A = {BELOW}, B = {WOOL}

Concept used:

If a set has “n” elements, then the number of subset of the given set is 2ⁿ

Calculation:

⇒ C = A – B

⇒ C = {B, E} taking out the common letters.

The subsets of C = 2² = 4

∴ The number of subsets of C is 4

Given:

Diagonals BD = 24 and AC = 10.

Concept used: 

Rhombus has all four sides equal and their diagonals bisect each other at 90° 

Let both the diagonals meet at O,

OD = OB = 12 and AO = OC = 5

In Δ OCD use the Pythagoras theorem.

In Δ OCD, OD² + OC² = DC² ,

12² + 5² = DC²

⇒ DC = 13

The perimeter of the Rhombus is 4a where a is the sides of the rhombus.

∴ Perimeter = 4 × 13 = 52 cm.