Explore topic-wise InterviewSolutions in Current Affairs.

x – 2 = 5 

∴ x = 7 

y + 3 = -6 

∴ y = - 9.

1. Yes. 

2. As we move to the top of a mountain atmospheric pressure decreases and hence boiling point decreases. So water boils quickly. 

3. Boiling point of a liquid is the temperature at which the vapour pressure of a liquid is equal to the external pressure or atmospheric pressure. 

4. The boiling point at 1 atm pressure is called normal boiling point. The boiling point at 1 bar pressure is called standard boiling point.

(a) When atmospheric pressure decreases boiling point of the liquid also decreases. So the boiling point of water at sea level is not same as that at the top of a mountain. Atmospheric pressure decreases from sea level as we go high. Hence, the boiling point at the top of the mountain is less than that at sea level.

(b) P₁V₁ = P₂V₂

P₂ = \(\frac{P_1V_1}{V_2}\) = \(\frac{1.2\,bar\times120\,mL}{180\,mL}\) = 0.8 bar

1. The attractive intermolecular forces are known as van der Waals’ forces. 

2. Dispersion forces/London forces, Dipole-Dipole forces, Dipole-Induced dipole forces and Hydrogen bonding.

3. Dispersion forces/London forces < Dipole-Induced dipole forces < Dipole-Dipole forces < Hydrogen bonding.

(c) Number of molecules

The unit of ‘b’ is in Vander Waals equation of state l mol^-1.

1. It is the highest temperature at which a gas can be liquified by applying external pressure.

2. The critical temperature (T_c) of CO₂ is 30.98°C. This is the highest temperature at which liquid CO₂ is observed. Above this temperature it is gas. 

3. Ammonia. This is because, on cooling, critical temperature of ammonia will be reached first. Liquefaction of carbon dioxide will require more cooling.

1. Absolute zero temperature 

2. -273.15 °C 

3. Absolute zero is the lowest hypothetical or imaginary temperature at which gases are supposed to occupy zero volume.

1. Surface tension. 

2. The magnitude of surface tension of a liquid depends on the attractive forces between the molecules. When the attractive forces are large, the surface tension is large. Increase in temperature increases the kinetic energy of the molecules and effectiveness of intermolecular attraction decreases, so surface tension decreases as the temperature is raised. 

3. N m^-1

(b) Decreases

1. Boyle’s law. 

2. It states that at constant temperature, pressure of a fixed amount of gas varies inversely with its volume.

3. Mathematically, Boyle’s law can be written as p ∝ \(\frac{1}{V}\) (at constant T and n) Or p = k × \(\frac{1}{V}\) where k is the proportionality constant which depends upon the amount of the gas, temperature of the gas and the units in which p and V are expressed.

(a) A gas below its critical temperature can be liquified by applying pressure. Under these conditions, it is called vapour of the substance. 

(b) water > ethyl alcohol > carbon tetrachloride > diethyl ether

(c) M = \(\frac{dRT}{p}\) 

\(\frac{2.92\,gL^{-1}\times\,0.0821L\,atm\,K^{-1}\,mol^{-1}\times 300\,K}{2\,atm}\) 

= ^-136g mol

1. Boyle’s law. 

2. According to Boyle’s law volume is inversely proportional to pressure. At the bottom of the pond, pressure is greater. So the volume (size) of the bubble was the least. But on coming up, pressure decreases and hence size of the bubble increases.

Vapour pressure, Boiling point, Viscosity and Surface tension.

1. Compressibility factor (Z) is the ratio of product pV and nRT. Mathematically, Z = \(\frac{pV}{nRT}\)

2. The deviation of real gases from ideal behaviour can be measured in terms of compressibility factor. 

3. (a) Real gas.

\(\Big[p+\frac{an^2}{V^2}\Big]\) [V-nb] = nRT (van der Waals’ equation)

Properties which increase with increase in temperature: 

Evaporation & Vapour pressure

Properties which decrease with increase in temperature: 

Surface tension & Viscosity

1. Yes. 

2. Graham’s law of diffusion. 

3. Rate of diffusion of a gas is inversely proportional to the square root of its molecular mass.

1. At any point in the dome shaped shaded area liquid and gaseous CO₂ exists in equilibrium. 

2. 73 atm. Critical pressure (p_c).

1. Both the graphs are correct. 

2. According to Boyle’s law, v α 1/p or p α 1/v. This is clear from graph 1. Also, according to Boyle’s law, pv is a constant at constant n and T. This is clear from graph 2.

(a) According to Boyle’s law,

p = \(k\times\frac{1}{V}\) OR  V = \(\frac{k}{p}\) 

Density, d = \(\frac{m}{V}\) 

Substituting the value of V,

d = \(\Big(\frac{m}{k}\Big)p\) 

i.e., d = k'p  OR  d ∝ p

(b) d = \(\frac{pM}{RT}\)

1. Surface tension.

2. This is because the surface area of thin film of water in moist soil is reduced due to surface tension.

1. The pressure exerted by saturated water vapour at a given temperature is called aqueous tension at that temperature. 

2. Pressure of dry gas can be calculated by subtracting aqueous tension from the total pressure of the moist gas. 

P_{dry gas} = P_Total - Aqueous tension

Surface energy is defined as the energy required to rise the surface area of the liquid by one unit. The SI unit of surface energy is J m^-2.

1. Charles’ law 

2. According to Charles’ law, volume a Temperature. Therefore, the volume increases when temperature increases, When the volume of the gas inside the balloon expanded more than that the balloon could afford, it bursted.

(1) Irreversible adiabatic

Bursting of balloon will result in rapid expansion of gas.

Curves are 2y² = x³ ...(1) 

and y = 32x ...(2)

4y(dy/dx) = 3x²

(dy/dx) = 3x²/4y ...(3)

and (dy/dx) = 32 ...(4)

Angles of the curves at Point (0,0)

From (4), (dy/dx) at (0,0) = 32

From (3), (dy/dx) th (0,0) = 0

So, angle of both the curves at (0,0)

= |90° - 0°| = 90° proved

\(\frac{cos\,2A\,cos\,3A-cos\,2A\,cos\,7A+cos\,A\,cos\,10A}{sin\,4A\,sin\,3A-sin\,2A\,sin\,5A+sin\,4A\,sin\,7A}\)

\(=\frac{2\,cos\,2A\,cos\,3A-2\,cos\,2A\,cos\,7A+2\,cos\,A\,cos\,10A}{2\,sin\,4A\,sin\,3A-2\,sin\,2A\,sin\,5A+2\,sin\,4A\,sin\,7A}\)

\(=\frac{cos(2A+3A)+cos(2A-3A)-(cos(2A+7A)\\\;\;\;\;\;\;+cos(2A-7A))+cos(A+10A)+cos(A-10A)}{cos(4A-3A)-cos(4A+3A)-(cos(2A-5A)\\\;\;\;\;\;\;-cos(2A+5A))+cos(4A-7A)-cos(4A+7A)}\)

(∵ 2 cos A cos B = cos (A+B) + cos (A-B) & 2 sin A sin B = cos (A-B) - cos (A+B))

\(=\frac{cos\,5A+cos(-A)-cos\,9A-cos(-5A)+cos(11A)+cos(-9A)}{cos(-A)-cos\,7A-cos(-3A)+cos\,7A+cos(-3A)-cos\,11A}\)

\(=\frac{cos\,5A+cos\,A-cos\,9A-cos\,5A+cos\,11A+cos\,9A}{cos\,A-cos\,7A-cos\,3A+cos\,7A+cos\,3A-cos\,11A}\)

(∵ cos (-θ) = cos θ)

\(=\frac{cos\,A+cos\,11A}{cos\,A-cos\,11A}\)

\(=\cfrac{2\,cos\,\frac{12A}A\,cos\,(\frac{-10A}2)}{2\,sin\,\frac{12A}A\,sin\,\frac{10A}2}\)

(∵ cos C + cos D = 2 cos (C+D/2) cos (C-D/2) and cos C - cos D = 2 sin (C+D/2) sin (D-C/2))

\(=\frac{cos\,6A\,cos\,5A}{sin\,6A\,sin\,5A}\) (∵ cos (-θ) = cos θ)

= cot 6A cot 5A

Hence proved.

Correct answer is: (b) p = r = - 2

Sum of zeroes = 2 + \(\cfrac{1}{2}\)= -5/p

i.e. 5/2 = -5/p . So p = -2

Product of zeroes = 2x 1/2 = r/p

i.e. r/p = 1 or r = p = -2

Correct option is: (b) 1/2

Correct answer is (c) 2

(D²-2D+1)y = e^xx²

It's auxiliary equation is m²-2m+1 = 0

⇒ (m-1)² = 0

⇒ m = 1, 1

Therefore, C.F. = (C₁+C₂x)e^x

P.F. \(=\frac1{D^2-2D+1}\) e^xx² \(=\frac1{(D-1)^2}\) e^xx²

= e^x \(\frac1{(D+1-1)^2}x^2\) (∵ \(\frac1{f(D)}\) e^axV = e^ax \(\frac1{f(D+a)}V)\)

= e^x 1/D² x²

= e^x [∫(∫x² dx)dx] (∵ 1/D V = ∫V dx)

= e^x(∫x³/3 dx) (∵ ∫xⁿ dx \(=\frac{x^{n+1}}{n+1})\)

\(=\frac{e^xx^4}{12}\)

Hence, complete solution of given differential equation is y = C.F.+P.I.

= (C₁+C₂x)e^x \(+\frac{e^xx^4}{12}.\)

L.H.S. = (7+3)(7-3) = 10 x 4 = 40

R.H.S. = (5+4)(5-4) = 9 x 1 = 9 ≠ 40

Hence, L.H.S. ≠ R.H.S.

Thus, (7+3)(7-3) ≠ (5+4)(5-4).

The oxidising flame is the flame produced with an excessive amount of oxygen. When the amount of oxygen increases, the flame shortens, its colour darkens. All of the carbon that gets used can be turned into CO, so there's actually no soot. This means that there is complete oxidation and due to increasing darkening, it doesn't produce much light and becomes non-luminous. Due to this reason, non-luminous flame is also called oxidising flame. 

In luminous flame, there is partial oxidation due to lesser availability of oxygen. A luminous flame is a bright, light emitting, usually yellow due to unburnt carbon or hydrocarbons which reduce the oxygen contained in the materials processed with the flame. Due to this reason, luminous flame is also called as reducing flame.

\((\overline{A \cup B}) \cup C= (\overline A \cap \overline B) \cup C\)

\(= (\overline A \cup C) \cap (\overline B \cup C)\)

Answer is Lead oxide

Given:

4515 ÷ 17.5 = 258 We have to find the value of 45.15 ÷ 1.75

Concept Used:

Concept of fractional division.

\({{{a} \over {b}} \div {{c} \over {d}}} = {a \over b} \times {d \over c}\)

Calculation:

4515 ÷ 17.5 = 258

⇒ 4515 ÷ \( {175} \over {10}\) = 258

⇒ 4515 × \({10} \over {175}\) = 258

 \(⇒ {{4515} \over {10}} \times {{100} \over {175}} = 258\) 

\(⇒ {{4515} \over {100}} \times {{100} \over {175}} = {{258} \over {10}}\)

\(⇒ {{4515} \over {100}} ÷ {{175} \over {100}} = {{258} \over {10}}\)

⇒ 45.15 ÷ 1.75 = 25.8

∴ The required value of 45.15 ÷ 1.75 = 25.8.

We've all been taught that water freezes at 32 degrees Fahrenheit, 0 degrees Celsius, 273.15 Kelvin. When liquid water is cooled below -42 degrees F, it crystallizes into ice too quickly for scientists to measure the temperature of the liquid.

Concept:

If α and β are the two roots of the quadratic polynomial Ax² + Bx + C, then:

• Ax² + Bx + C = (x - α)(x - β) = x² - (α + β)x + αβ = 0
• α + β = \(\rm -\frac{B}{A}\) and αβ = \(\rm \frac{C}{A}\).

Calculation:

Comparing the given equation 9x2 - 3x + (m - 4) = 0 with the general equation Ax2 + Bx + C = 0, we can say that:

A = 9, B = -3, C = (m - 4)

It is given that:

α - β = 1              ... (1)

Using α + β = \(\rm -\frac{B}{A}\), we get:

α + β = \(\rm -\left(\frac{-3}{9}\right)=\frac13\)              ... (2)

Adding equations (1) and (2), we get:

\(\rm 2\alpha = 1+\frac13=\frac43\)

⇒ \(\rm \alpha = \frac23\)

Using equation (2), we get:

\(\rm \beta = \frac13-\frac23=-\frac13\)

Now, the product of the roots is: αβ = \(\rm \frac{C}{A}\).

⇒ \(\rm \left(\frac23\right)\times\left(-\frac13\right)=\frac{m-4}{9}\)

⇒ m - 4 = -2

⇒ m = 2.

Concept used:-

Trick to multiply two numbers ending in 5 with difference of 10 between them.

Method:-

• Write 75 at the end.
• Add 1 to ten’s place of the bigger number.
• Now multiply the increased digit with the ten’s digit of the smaller number
• Write the product so obtained on the left side of 75

Calculation:-

Let the answer be A75.

A = (8 + 1) × 7 = 63 (using step 2 and 3)

Hence, 6375 is our answer.

Given that,

2x = Sec A and 

2/x = tanA

Therefore,

sec²A = 4x² and

tan²A = \(\frac{4}{x^2}\) 

We know that,

sec²A – tan²A = 1.

Therefore,

4x² − \(\frac{4}{x^2}\) 

= 1 

(By putting values of sec²A and tan²A)

⇒ 4(x² - \(\frac{1}{x^2}\)) = 1

⇒  x² - \(\frac{1}{x^2}\) = \(\frac{1}{4}\).

Hence,

The value of  x² - \(\frac{1}{x^2}\) is \(\frac{1}{4}\).

Given:

Even numbers between 100 and 250.

Concept used:

Even numbers between n and (n + 150) = (n + 150)/2 - (n/2) - 1

Calculations:

Even numbers upto 100 including 100 = 100/2

⇒ 50

Even numbers upto 250 excluding 250 = (250/2) - 1

⇒ 124

Difference = (124 - 50)

⇒ 74

∴ The even numbers between 100 and 250 are 74.

Given:

0.0043/? = 0.43

Calculation:

Let ? = x

0.0043/x = 0.43

⇒ 43/10000x = 43/100

⇒ 1/10000x = 1/100

⇒ x = 1/100 = 0.01

∴ 0.01

We have,

sinθ = cos(θ – 45°), 

Where θ is acute angle.

⇒ cos(90° − θ) = cos(θ − 45°) 

(∵ cos(90° − θ) = sinθ) 

⇒ 90° − θ = θ − 45° 

⇒ 2θ = 90° + 45° = 135°

⇒ θ = \(\frac{135°}{2}\) = 62.5°.

Hence,

The value of θ is 62.5°.

Given: 

The number is (5317 - 1)

Concept: 

When is n is odd, then (aⁿ - bⁿ) is divisible by (a - b)

When n is even, then (aⁿ - bⁿ) is divisible by (a - b) and (a + b)

Calculation: 

The number is (5317 - 1)

Also, we can write this

⇒ (5317 - 1¹⁷) 

Here, n is 17(odd)

Then, the number is must be divisible by

⇒ 53 - 1

⇒ 52

Then, the number is also divisible by the factor of 52

Factor 52 is 1, 2, 4, 13, 26, and 52.

∴ The required answer is 13.

Rational Numbers: A number which is written in the form of \(p\over q\) where p. q belongs to integers and q is not equal to zero.

Prime Numbers: A natural number which is divisible by 1 and itself only called a prime number, i.e.2, 3, 5, 7, .....

Integer: = {.......-3, -2, -1, 0, 1, 2, 3, .......}

Irrational number: A number that cannot be expressed in the \(p\over q\) form is called an irrational number or not a rational number, for Example: √2, √3, √5, etc.

Hence, Option 4 is correct.

We have\(\begin{bmatrix} 2 & -1\\ 1 & 0\\ -3 & 4\end{bmatrix}\). A = \(\begin{bmatrix} -1 & -8 & -10\\ 1 & -2 & -5\\ 9 & 22& 15\end{bmatrix}\)

This matrix multiplication is possible only when the order of matrix is 2 × 3. Let matrix A = \(\begin{bmatrix} a & b & c\\ d & e & f\\ \end{bmatrix}\).

Therefore, \(\begin{bmatrix} 2 & -1\\ 1 & 0\\ -3 & 4\end{bmatrix}\)\(\begin{bmatrix} a & b & c\\ d & e & f\\ \end{bmatrix}\) = \(\begin{bmatrix} -1 & -8 & -10\\ 1 & -2 & -5\\ 9 & 22& 15\end{bmatrix}\)

By equality of two matrices, comparing corresponding components, we get 

a = 1, b = −2, c = −5. (By comparing corresponding components of 2^nd row of both matrices. ) 

2a − d = −1, 2b − e = −8, 2c − f = −10. 

d = 2a + 1 = 2 + 1 = 3, e = 2 + 8 = −4 + 8 = 4, f = 2 + 10 = −10 + 10 = 0. 

(By comparing corresponding components of 1^st row of both matrices and putting values of a, b and c. ) We clearly see that values of a,b,c,d,e and f satisfies by corresponding components of 3^rd row of both matrices. 

Therefore, matrix = \(\begin{bmatrix} a & b & c\\ d & e & f\\ \end{bmatrix}\) = \(\begin{bmatrix} 1 & -2 & -5\\ 3 & 4 & 0\\ \end{bmatrix}\)

Given:

First 7 even natural numbers.

Formula used:

Sum of first 'n' even natural numbers = n (n + 1)

Calculations:

Sum of first 'n' even natural numbers = n (n + 1)

⇒ 7 (7 + 1)

⇒ 56

⇒ Unit digit = 6

∴ The unit digit of the sum of first 7 even natural numbers is 6

Given: 

The number is (218 + 5) and the divisor is 9.

Concept used:

When a number is divided by another number then there are two remainder one is negative and other is positive 

Ex:- 8/9 

Negative remainder = (8 - 9) = -1

And positive remainder = 8

Calculation: 

The number is divided by 9

⇒ (218 + 5) ÷ 9

⇒ ((2³)⁶ + 5)/9

⇒ (8⁶ + 5)9

⇒ ((-1)⁶ + 5)/9

⇒ (1 + 5)/9

⇒ 6/9

Remainder = 6

∴ The required remainder is 6.