Mathematics and Science PYQs

Concept:

 

Total Internal Reflection (TIR): 

• When a ray of light goes from denser to the rarer medium it bends away from the normal and as the angle of incidence in denser medium increases, the angle of refraction in the rarer medium also increases and at a certain angle, the angle of refraction becomes 90°, this angle of incidence is called critical angle (C).
• When the Angle of incidence exceeds the critical angle then light ray comes back into the same medium after reflection from the interface.
• This phenomenon is called Total internal reflection (TIR).

Explanation:

• TIR happens when a ray of light travels from an optically denser medium to a rarer medium.
• Glass is optically denser than water.
• The speed of light in glass is less than that of water.
• When light travels from glass to water, TIR may happen.
• Glass to water is the correct option.

Considering other options

• Air to water: It is not possible as air is optically rarer than water.
• Air to glass:  air is optically rarer than glass.
• Water to glass: water is optically rarer than glass.

Additional Information

Snell's law:

• The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for the light of a given colour and for the given pair of media.

​\(⇒ \frac{sin, i}{sin, r}=Constant\)

Where I = angle of incidence and r = angle of refraction

Refractive index:

• When light travels from medium 1 to medium 2, then the ratio of the speed of light in medium 2 to the speed of light in medium 1 is called the refractive index of medium 1 with respect to medium 2.

\(⇒ μ_{12}=\frac{v_2}{v_1}\)

• If medium 1 is denser compared to medium 2, then the refractive index of medium 1 with respect to medium 2 will be greater than 1.
• If medium 1 is rarer compared to medium 2, then the refractive index of medium 1 with respect to medium 2 will be less than 1.

Momentum

• Momentum is defined as the product of the mass and velocity of the body.
• It is a vector quantity directed toward velocity.
• It is given as

p = mv      ----(1)

m is mass, v is the velocity, p is momentum

Kinetic Energy

• The energy of a body in motion due to its state of motion is called kinetic energy.
• It is given as

\(K = \frac{1}{2}mv^2\)        ----(2)

K is kinetic energy, m is mass, v is speed.

Relationship between Momentum and Kinetic Energy

If we combine equation (1) and (2) we will get the relationship between momentum and kinetic energy as

\(K = \frac{p^2}{2m}\)

Concept:

Electrical Energy and Joules Heating

The heat generated by a resistive coil or device is given as

\(H = \frac{V^2}{R}t = I^2 Rt\)

H is energy, t is time, R is resistance, V is the potential difference. 

If the conductor is purely resistive, then the whole electrical energy is converted into heat energy. 

Resistivity and Resistance

• The property of a conductor that opposes the flow of electric current through them and is independent of the shape and size of the materials but depends on the nature and temperature of the materials is called resistivity.
• Resistance: The property of any conductor that opposes the flow of electric current through it and depends on the shape and size of the materials, temperature, and nature of the materials is called resistance.
• It is denoted by R and the SI unit is the ohm (Ω).
• The resistance is given by:

\(⇒ R=\frac{ρ l}{A}\)

Calculation:

Given, 

Potential difference V is constant

Resistance is proportional to resistivity e. 

Given

\(H = \frac{V^2}{R}t \)

\(\implies H \propto \frac{1}{{R}}\)

R is proportional to e

\(\implies H \propto \frac{1}{{e}}\)

Additional Information

• The unit for resistivity is the ohm-meter (Ω-m).
• The resistivity of a material depends on its nature and the temperature of the conductor.
• The resistivity of a material doesn't depend on its shape and size (length and area).

CONCEPT:

Logic Gates:

• The logic gates are used in electronic circuits that give a certain output for a given set of Inputs.
• and which is based on a certain logic.
• Input and output are in binary form that is 1 and 0.
• It is based on certain Logic.
• Some basic logic gates are AND gate, OR gate, NOT gate.
• To build digital logic we develop a combination of these gates.

AND Gate

• If both the inputs are high, it produces a high output.
• The Boolean algebra for AND gate is X = A. B
• It means that if any of the input of A and B is zero, the output will be zero.
• If both are one, the output will be one.

 

The correct option is AND Gate. 

Additional Information

NOT gate:  

• It inverts the input. Whatever the input is given, it changes its value at the output.
• The Boolean algebra for NOT gate is X = X̅

Concept:

• A vector quantity has both magnitude and direction.
• The force is a vector that has unit Newton and is defined as the product of mass and acceleration.
• Acceleration is the rate of change of velocity.
• Both acceleration and velocity are vectors.

Calculation:

Given 

Mass = 2 kg

Force F = 10 N

Acceleration a = F / m = 10 N / 2 kg = 5 m s ^{- 2} . 

Initial velocity u = 10 ms ^-1 

Final speed v = ?

 

In vector form 

\(Initial \ velocity\ u = 10\ \widehat{i}\)

\(Acceleration \ a= 5\ \widehat{j}\)

Now acceleration is applied along the vertical direction and velocity is in the horizontal direction. 

In horizontal direction 

Initial velocity u_x = 10 

Acceleration a_x = 0

Final speed speed = u_x (As no acceleration is along this direction of Force) 

In vertical direction 

Initial velocity u_y = 0

Acceleration a_y = 5 

time t =  2 s

Final speed v_y = u_y + at = 0 + 5 × 2 = 10 m / s

Now Net velocity 

\(v = \widehat{v_x} + \widehat{v_y}\)

\(\overrightarrow{v} = v_x\widehat{i} + v_y\widehat{j}\)

\(\implies \overrightarrow{v} = 10\widehat{i} + 10\widehat{j}\)

The magnitude of the speed 

\(v = \sqrt{(v_x)^2 +(v_y)^2} = \sqrt{(10)^2 +(10)^2} = 10\sqrt{2} \ ms ^{-1}\)

Angle \(θ = tan^ {-1} \left ({\frac{v_y}{v_x}} \right ) = tan^ {-1} \left ({\frac{10}{10}} \right ) \)

θ = tan ^{- 1} (1) = 45 ° 

The angle is made to the horizontal axis along North-East direction. 

10√2 ms-1 at 45° North of East is the correct option.

Explanation:

Hydrogenation of unsaturated oils

• Unsaturated vegetable oils are liquid at room temperature.
• They can be hardened by reacting them with hydrogen in the presence of a Nickel(Ni) catalyst at about 60o C.
• This is called hydrogenation.
• The hydrogen adds to the double-bonded carbons.

The catalyst used during the Hydrogenation of Oils is Nickel.

Concept:

• 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.
• Tollens' reagent is basically an ammoniacal silver nitrate solution.
• It can also be used to identify C2H2 (Acetylene)

Explanation:

If Ethyne has acidic hydrogen which when passed through ammonium silver nitrate solution we get a white precipitate of  disilver acetate and the reaction can be given as 

\(\rm \underset{ethyne}{HC\equiv CH} + \underset{Tollen's \ reagent}{2[Ag(NH_3)_2]OH}\xrightarrow{ \ \ \ \ \ \ \ \ \ \ \ } \rm \underset{diSilver \ acetylide}{Ag \ C \equiv CAg} + 4 \ NH_3+2H_2O\)

Calculation:

The molar mass of SO₂ 

The molecular mass of Sulphur + 2 (molecular mass of oxygen) 

32 gm  + 2 (16 gm) = 64 gm

Number of moles in 64 gm = 1 mole

Number of moles in 128 gm = \(\frac{1}{64} × 128 = 2\)

Important Points 

1 mole of a substance is the atomic mass of it expressed in grams. 

In 1 mole 6.022 × 10 ²³ Particles are present

Oxidation state: oxidation state is the total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom.

The oxidation state of Iron in Fe(CO)5

Let Oxidation state Fe be X,

then

As CO has 0 oxidation state,

X + 0(5) = 0

Therefore oxidation state of Fe in Fe(CO)5 is 0.

The hardness of water: The hardness of water is that characteristic that prevents the formation of sufficient leather or foam.

• It occurs due to the presence of carbonates, bicarbonates, sulphates, chlorides of calcium and magnesium(Ca and Mg).
• Hardness of water is due to the presence of dissolved minerals of Ca²⁺, Mg²⁺, Al³⁺, Iron and other heavy elements. Specifically due to Calcium Sulphate, Calcium Bicarbonate, Calcium chloride, Magnesium Sulphate, Magnesium Bicarbonate and Magnesium chloride.​

Key Points

Hardness is of following two types-

Temporary hardness (Carbonate hardness):

• Temporary hardness is caused by the presence of dissolved bicarbonates of Calcium (Ca(HCO3)2 **), Magnesium (**Mg(HCO3)2 ), Iron and other heavy elements.
• It can be eliminated by mere boiling.

Ca(HCO₃)₂ → CaCO₃↓ + H₂O + CO₂ ↑

Mg(HCO₃)₂ → MgCO3 + H2O + CO2 ↑

MgCO₃ → Mg(OH)₂↓ + CO₂ ↑

Permanent Hardness (non-carbonate hardness):

• Permanent Hardness is due to the presence of dissolved chlorides and sulphates of Calcium, Magnesium, Iron and other heavy elements.
• It cannot be eliminated by boiling.

Hence, we can conclude that Mg(HCO3)2 is responsible for temporary hardness of water.

Additional Information

Degree of Hardness

• Degree of hardness is extent of hardness.
• It is expressed as weight in milligrams of CaCO₃ equivalent to all hardness causing substance in one million milligram of water. That is parts per million (ppm). Or it can be expressed as weight in milligrams of CaCO₃ equivalent to all hardness causing substance in one litre of water.​

CaCO₃ as reference for Hardness

• a. CaCO₃ is stable, non-hygroscopic and is obtained in pure form. Therefore a standard hard water solution can be prepared by dissolving accurately weighed CaCO₃ in dilute HCl and make up to a known volume.
• b. CaCO₃ is insoluble in water. Therefore it can be easily precipitated in water treatments.
• c. Molecular weight of CaCO₃ is 100, so mathematical calculations are easy

Shortcut Trick

Solve inner exponents: 8^1/3 = 2 and 27^1/3 = 3

Evaluate sum and power: (2 + 3)³ = 5³

Combine with outer factor: 5 × 5³ = 5⁴

Final root: (5⁴)^1/4 = 5

∴ The correct answer is 5.

Alternate Method

Given expression: [5(8^1/3 + 27^1/3)³]^1/4

Formula Used: (a^m)ⁿ = a^m×n and a^m × aⁿ = a^m+n

⇒ [5((2³)^1/3 + (3³)^1/3)³]^1/4

⇒ [5(2^3×1/3 + 3^3×1/3)³]^1/4

⇒ [5(2¹ + 3¹)³]^1/4

⇒ [5(5)³]^1/4

⇒ [5¹⁺³]^1/4

⇒ [5⁴]^1/4

⇒ 5^4×1/4 = 5¹ = 5

∴ The correct answer is 5.

Additional Information

Laws of Indices: Product Rule

When multiplying expressions with the same base, the exponents are added: a^m × aⁿ = a^m+n.

Laws of Indices: Power of a Power

When a base with an exponent is raised to another power, the exponents are multiplied: (a^m)ⁿ = a^m×n.

Fractional Exponents

An exponent of 1/n represents the n^th root of the number: a^1/n = √ⁿa. For example, 27^1/3 = 3.

Concept used:

The product of consumption and price is equal to total expenditure.

Calculation:

Let the initial price of petrol be 100x and initial travel be 10y.

⇒ Initial expenditure = 100x × 10y = 1000xy      ----(1)

When the price of petrol is increased by 25%,

The new price of the petrol = 100x + 25% of 100x = 125x

Now, let new travel be T.

As A/Q expenses cannot change

⇒ 125x × T = 1000xy      [From (1)]

⇒ T = 1000xy/125x = 8y

⇒ New travel is 8y

Reduction in travel = 10y - 8y = 2y

Reduction in travel in % = (2y/10y) × 100 = 20%

∴ one must reduced travel by 20%.

Given:

In a parallelogram ABCD, diagonals AC and BD bisect at O.

⇒ ∠DAC = 32°, ∠AOB = 70°

Concept used:

If AB and CD are two straight lines parallel to each other, and transverse by a line, 

Then, ∠1 = ∠2 and ∠3 = ∠4 as, both are alternate angles.

Calculation:

Here, AB is parallel to CD,

⇒ ∠DAC = ∠ACB = 32°    [ as both are alternate angles]

As ∠AOB and ∠BOC are linear pair angles,

⇒ ∠AOB + ∠BOC = 180°

⇒ 70° + ∠BOC = 180°

⇒ ∠BOC = 180° - 70° = 110°

Now, in ΔBOC,

⇒ ∠BOC + ∠OBC + ∠OCB = 180°

⇒ 110° + ∠OBC + 32° = 180°

⇒ ∠OBC = 180° - 142° = 38° = ∠DBC

∴ The ∠DBC equals to 38°.

Given:

The diameters of parks are 16 m and 12m.

Formula used:

Area of circle = πr², where 'r' is the radius of the circle.

Calculation:

As diameters of parks are 16m and 12m

⇒ The radius of the two parks is 16/2 = 8 m and 12/2 = 6m

⇒ Area of the new circular park = sum of the areas of two circular parks

Let the radius of the new circular park be 'R'

⇒ πR² = π(8)² + π(6)²

⇒ πR² = π(64 + 36)

⇒ πR² = π × 100

⇒ R² = 100

⇒ R = √100 = 10 m

∴ The radius of the new park is 10m.

Given:

The edge of a cube is r cm.

Formula used:

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

Calculation:

When the largest possible cone is cut out from the cube,

Then, the edge of the cube becomes the diameter of the cone, and also the height of the cone equals to edge of the cube

⇒ r/2 = Radius of cone and height of cone = r

Now, the volume of cone = (1/3)πr2h

⇒ the volume of cone = \(\frac{1}{3}\pi × (\frac{r}{2})^2 × r\)

⇒ \(\frac{1}{12}\pi r^3\)

∴ The volume of the cone is \(\frac{1}{12}\pi r^3\) cm³

Important PointsProperties of cone:

• A cone has only one face, which is the circular base but no edges
• A cone has only one apex or vertex point.
• The volume of the cone is ⅓ πr2h.
• The total surface area of the cone is πr(l + r)
• The slant height of the cone is √(r2+h2)

Concept: 

When (p – 1)n is divided by ‘p’, the remainder will be '1' only if 'n' is an even number.

Calculation:

When (p – 1)n is divided by ‘p’, the remainder will be '1' only if 'n' is an even number.

∵ 13 can be written as (14 – 1)

1350 is divided by 14

⇒ (14 – 1)50/14

50 is an even number.

∴ Remainder = 1

Given:

 x = 3 + √8      ----(1)

Formula used:

a³ + b³ = (a + b)(a² - ab + b²)

Calculation:

As,  x = 3 + √8

⇒ x² = (3 + √8)² = 9 + 8 + 2 × 3 × √8

⇒ x² = 17 + 6√8      ----(2)

⇒ \({1\over x} = {1 \over 3+√{8}}\) 

⇒ \({1\over x} = {1 \over 3+√{8}} × {3 - √ 8\over 3 - √{8}}\) 

⇒ \({1\over x} = {3 - √ 8\over (3)^2 - (√{8})^2}\)      ----(3)

⇒ \({1\over x} = {3 - √ 8\over (9 - 8)} = 3 - √ 8\)

⇒ \({1\over x^2} = (3 - √ 8)^2 = 9 + 8 - 2 × 3 × √ 8 = 17 - 6√ 8 \)      ----(4)

From equations (1) and (3)

⇒ x + 1/x = 3 + √8 + 3 - √8 = 6      ----(5)

From equations (2) and (4),

⇒ x² + 1/x² =  17 + 6√8 +  17 - 6√8 = 34      ----(6)

Now, \(x^3+\frac{1}{x^3}\) = \((x+\frac{1}{x})(x^2 - x × \frac{1}{x} +\frac{1}{x^2})\)

⇒ \(x^3+\frac{1}{x^3}\) = (6) × (34 - 1)

⇒ 6 × 33 = 198

∴ The value of \(x^3+\frac{1}{x^3}\) is 198.

Calculation:

⇒ x2 + 3√2x + 4    

⇒ x2 + √2x + 2√2x + 4     

⇒  x(x + √2) + 2√2(x + √2)

⇒ (x + 2√2)(x +√2)

∴ The factors of  x2 + 3√2x + 4 are (x + 2√2)(x + √2​).

Given:

Adjacent sides of a parallelogram are 74 cm and 40 cm

One diagonal of parallelogram is 102 cm.

Formula used:

Area of parallelogram = 2 × area of a triangle that is made by a diagonal.

If a triangle has sides a, b and c respectively.

Then, By Heron's formula 

Area of triangle = \({\sqrt{s(s - a)(s - b)(s - c)}}\) where 's' is the semi-perimeter of the triangle

⇒ s = (a + b + c)/2

Calculation:

Let ABCD be the given parallelogram.

Here, AB = CD = 74 cm and AD = BC = 40 cm

And, the diagonal AC = 102 cm

Now, In ΔACD

⇒ Semi-perimeter = s = (74 + 40 + 102)/2

⇒ s = 216/2 = 108

Now, area of ΔACD

⇒ \({\sqrt{108(108 - 74)(108 - 40)(108 - 102)}}\)

⇒ \({\sqrt{108 × 34 × 68 × 6}}\)

⇒ \({\sqrt{36 × 3 × 34 × 34 × 2 × 3 × 2}}\) = 6 × 3 × 34 × 2

⇒ 1224 sq. cm

Now, area of parallelogram = 2 × area of Δ ACD

⇒ 2 × 1224 = 2448 sq. cm

∴ The area of the parallelogram is 2448 sq. cm.

Important PointsProperties of parallelogram are:

• Number of sides = 4
• Number of vertices = 4
• Mutually Parallel sides = 2 (in pair)
• Area = Base x Height
• Perimeter = 2 × (Sum of adjacent sides length)
• Type of polygon = Quadrilateral

Given:

Each exterior angle of a polygon is 18°.

Concept used:

The sum of the exterior angles of a polygon is 360°.

Formula used:

Number of sides of polygon = 360°/each exterior angle

Calculation:

As each exterior angle of a regular polygon is 18°.

⇒ Number of sides of regular polygon = 360°/18° = 20

∴ The number of sides of the polygon is 20 when each exterior angle is 18°.