Explore topic-wise InterviewSolutions in Current Affairs.

Relative Humidity (%): The relative humidity in an air parcel is defined as the ratio of the amount of water vapor actually in the air to the maximum amount of water vapor required for saturation at a particular temperature.

• Relative Humidity (%) = water vapor content / water vapor capacity
• Relative Humidity (%) = U/U_s
• Saturation mixing ratio, U_s: the maximum amount of water vapor that can be in the air
• Mixing ratio, U: the ratio between the actual (measured) mass of water vapor (in the parcel) in grams and mass of dry (non-water vapor) air (in the parcel) in kilograms

Given data

• Temperature = 25⁰C
• Saturation mixing ratio, Us = 20 grams
• Mixing ration, U = 5 grams

Calculation

• Relative Humidity = U/Us
• RH (%) = (5/20)*100 = 100/4 = 25
• Therefore, the correct Relative Humidity (%) is 25%.

The correct answer is Carbon Trade - Montreal Protocol.

• The Montreal Protocol on Substances that Deplete the Ozone Layer is the landmark multilateral environmental agreement that regulates the production and consumption of nearly 100 man-made chemicals referred to as ozone-depleting substances (ODS).
• Montreal Protocol was signed in 1987 and entered into force in 1989. The parties to the Protocol meet once a year to make decisions aimed at ensuring the successful implementation of the agreement.

• First World Climate Conference - 1979 
• First Earth Summit - Agenda 21
• Earth Summit + 5 - 1997

The Correct Answer is Coffee.

• Tea is grown, largely for export, on plantations in Assam, West Bengal, Kerala, and Tamil Nadu, while coffee is grown almost exclusively in southern India, mainly in Karnataka.
• Tobacco is cultivated chiefly in Gujarat and Andhra Pradesh.
• More than 50% of the bamboo species occur in Eastern India- Arunachal Pradesh, Assam, Manipur, Meghalaya, Mizoram, Nagaland, Sikkim, Tripura, and West Bengal.
• Other areas rich in bamboo are the A&N Islands, Chhattisgarh, Madhya Pradesh, and the Western Ghats.
• Cotton is a Kharif crop in the major parts of the country viz. Punjab, Haryana, Rajasthan, Uttar Pradesh, Madhya Pradesh, Gujarat, Maharashtra, and parts of Andhra Pradesh & Karnataka.

The correct answer is ​Regur Soil.

• Regur soil is considered most suitable for sunflower cultivation.

• This soil is formed by lava/ash.
• This soil is found mainly in Gujarat, Madhya Pradesh, Maharashtra and Karnataka in India, so sunflower is cultivated on vast area of Maharashtra and Karnataka.
• This soil is also called black soil and for cotton cultivation this soil is the most appropriate.

[Hint : To provide energy of activation.]

Answer: (1)

Kjeldahl method is not applicable to nitro or diazo groups present in the ring, as nitrogen atom can't be converted to ammonium sulfate under the reaction conditions. 

[Hint :  (a) Finely divided Fe/FeO, MO as promoter.
(b) Pt (Platinised asbestos)]

[Hint : To absorb poisonous gases.]

Answer: (1) Both assertion and reason are correct and reason is correct explanation of assertion. 

Answer is (4) The assertion is true, but the reason is false.

Answer is (4)

(1) Both assertion and reason are correct and reason is correct explanation of assertion. 

Let three digit number is xyz 

x + y + z = 10 ; x ≥ 1, y ≥ 0 z ≥ 0 ..... (1) 

Let T = x – 1 ⇒ x = T + 1 where T ≥ 0 

Put in (1) 

T + y + z = 9 ; 0 ≤ T ≤ 8, 0 ≤ y, z ≤ 9 

No. of non negative integral solution 

= ^9+3–1 C_3 –1 – 1 (when T = 9)

= 55 - 1 

= 54

Answer is (1) 28

N₁λ₁ = N₂λ₂

16 × 700 = N₂ × 400 

⇒ N₂ = 28

(3) a,b,c,d are in G.P. 

(a² + b² + c²)p² + 2(ab + bc + cd)p + b² + c² + d² = 0 

(a²p² + 2abp + b²) + (b²p² + 2bcp + c²) + (c²p² + 2cdp + d²) = 0 

(ab + b)² + (bp + c)² + (cp + d)² = 0 

This is possible only when 

ap + b = 0 and bp + c = 0 and cp + d = 0 

p = -b/a = -c/b = -d/c

or b/a = c/b = d/c

∴ a,b,c,d are in G.P.

Concept:

If two vectors \(\rm \vec a = {a_1}\hat i + {a_2}\hat j+{a_3}\hat k\) and \(\rm \vec b = {b_1}\hat i + {b_2}\hat j+{b_3}\hat k\) are equal, then a₁ = b₁, a₂ = b₂ and c₁ = c₂.

Calculation:

We have \(\rm \vec c = m\vec a + n\vec b\).

⇒ 2î + 3ĵ = m(-4î + 2ĵ) + n(2î + ĵ)

⇒ 2î + 3ĵ = (-4m + 2n)î + (2m + n)ĵ

Equating the scalar coefficients, we get:

-4m + 2n = 2               ... (1)

2m + n = 3               ... (2)

Multiplying equation (2) by 2 and adding to equation (1), we get:

4n = 8

⇒ n = 2

Using either of the equations above, we also get:

m = \(\frac12\)

∴ m + n = 2 + \(\frac12\) = \(\frac{5}{2}\).

Concept:

• Cross Product: For two vectors \(\rm \vec A\) and \(\rm \vec B\) at an angle θ to each other, the cross product is defined as:

  \(\rm \vec A\times \vec B=\vec n|\vec A||\vec B|\sin \theta\), where \(\rm \vec n\) is the unit vector perpendicular to the plane containing the vectors \(\rm \vec A\) and \(\rm \vec B\).

• If \(\rm \vec A = a_1\hat i +a_2\hat j+ a_3\hat k\) and \(\rm \vec B = b_1\hat i +b_2\hat j+b_3 \hat k\), then their cross product is:

  \(\rm \vec A\times\vec B=\begin{vmatrix} \rm \hat i & \rm \hat j & \rm \hat k \\ \rm a_1 & \rm a_2 & \rm a_3 \\ \rm b_1 & \rm b_2 & \rm b_3\end{vmatrix}\).

• The unit vector \(\rm \vec u\) in the direction of a vector \(\rm \vec A\), can be calculated as:

  \(\rm \vec u = \frac{\vec A}{\left| \vec A\right|}\), where \(\rm \left| \vec A\right|\) is the magnitude (length) of the vector \(\rm \vec A\).

Calculation:

We have \(\rm \vec A = 4\hat i +3\hat j+ \hat k\) and \(\rm \vec B = 2\hat i -\hat j+2 \hat k\). Therefore, their cross product will be:

\(\rm \vec A\times\vec B=\begin{vmatrix} \rm \hat i & \ \ \ \rm \hat j & \rm \hat k \\ 4& \ \ \ 3&1\\2&-1&2\end{vmatrix}\)

Expanding along R₁, we get:

= (6 + 1)î + (2 - 8)ĵ + (-4 - 6)k̂

= 7î - 6ĵ - 10k̂

The magnitude of \(\rm \vec A\times \vec B\) is:

\(\rm \left| \vec A\times\vec B\right|=\sqrt{7^2+(-6)^2+(-10)^2}\)

= \(\rm \sqrt{49+36+100}\)

= \(\rm \sqrt{185}\)

The unit vector N̂ along \(\rm \vec A\times\vec B\) will be:

N̂ = \(\rm \frac{\vec A\times\vec B}{\left| \vec A\times\vec B\right|}\)

⇒ N̂ = \(\rm \frac{1}{\sqrt{185}}\left(7\hat{i}-6\hat{j}-10\hat{k}\right)\).

Concept:

If G is the geometric mean of the numbers a and b and is given by  ⇔ G = \(\rm ​​\sqrt {ab}\)

Calculation:

Given, \(\rm \hat i + b\hat j + \hat k,\hat i + b\hat j, a\hat i + c^2\hat j + c\hat k\) are coplanar

∴\(\rm \left | \begin{array}{ccc} 1 & b & 1 \\ 1 & b & 0 \\ a & c^2 & c \end{array} \right | =0\)

⇒ 1(bc) - b (c) + 1(c² - ab) = 0

⇒ c² = ab

⇒ c = ​\(\rm \sqrt {ab}\)​

Hence, option (2) is correct. 

Concept:

If vectors a and b are orthogonal, then \(\rm \vec a.\vec b\) = 0

Equation of circle: x² + y² = r², where r = radius

Calculation:

\(\begin{aligned} \rm &\rm (- \hat i-2 x \hat{j}-3 y \hat{k})(\rm \hat i-3 x \rm\hat j-2 y \rm\hat{k})=0\\ &\Rightarrow(-1)(1)+(-2\rm x)(-3\rm x)+(-3\rm y)(-2 \rm y)=0\\ &\Rightarrow 6 \rm x^{2}+6\rm y^{2}=1\\ &\Rightarrow \rm x^{2}+y^{2}=(1 / \sqrt{6})^{2}\\ &\text { Locus of }\rm(x, y) \text { is a circle } \end{aligned}\)

Hence, option (3) is correct.

Concept:

Area of triangle when two vectors are given:

\(\rm \frac12 \times |\vec {AB} \times \vec {AC}| \)

Cross product:

\(\begin{array}{l} \rm \vec{a}=x_{1} \hat{1}+y_{1} \hat{j}+z_{1} \hat{k}\\ \rm \vec{b}=x_{2} \hat{1}+y_{2} \hat{j}+z_{2} \hat{k} \\ \rm \vec{a} \times \vec{b}=\left|\begin{array}{ccc} \rm \hat{i} &\rm j &\rm \hat{k} \\ \rm x_{1} &\rm y_{1} & \rm z_{1} \\ \rm x_{2} & \rm y_{2} &\rm z_{2} \rm \end{array}\right| \end{array}\)

Calculation:

Here, Let A = (0,2,2), B = (2,0,-1) and C =(3,4,0) 

AB = (2-0, 0-2, -1-2) = (2, -2, -3) and

AC = (3-0, 4-2, 0-2) = (3, 2, -2)

Area of triangle = \(\rm \frac12 \times |\vec {AB} \times \vec {AC}| \)

 \(\begin{array}{l} =\rm \frac{1}{2}\left|\begin{array}{ccc} \rm \hat i &\rm \hat j & \rm \hat k \\ 2 & -2 & -3 \\ 3 & 2 & -2 \end{array}\right| \\ =\frac{1}{2}|[\rm \hat i(4+6)+ \rm \hat j(-4+9)+ \rm \hat k(4+6)]| \end{array}\)

=1/2(10 i + 5 j + 10 k)

= 1/2 √(100 + 25 + 100)

= 15/2 sq unit

Hence, option (1) is correct.

Concept:

\(\text { Let } \overrightarrow{\mathrm{a}}=\mathrm{a}_{1} \overrightarrow{\mathrm{i}}+\mathrm{b}_{1} \overrightarrow{\mathrm{j}}+\mathrm{c}_{1} \overrightarrow{\mathrm{k}}, \overrightarrow{\mathrm{b}}=\mathrm{a}_{2} \overrightarrow{\mathrm{i}}+\mathrm{b}_{2} \overrightarrow{\mathrm{j}}+\mathrm{c}_{2} \overrightarrow{\mathrm{k}} \text { and } \overrightarrow{\mathrm{c}}=\mathrm{a}_{3} \overrightarrow{\mathrm{i}}+\mathrm{b}_{3} \overrightarrow{\mathrm{j}}+\mathrm{c}_{3} \overrightarrow{\mathrm{k}} \text { be the three vectors }\)

Condition for coplanarity:

\( \overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=\left|\begin{array}{lll} \rm a_{1} & \mathrm{b}_{1} & \mathrm{c}_{1} \\ \mathrm{a}_{2} & \mathrm{b}_{2} & \mathrm{c}_{2} \\ \mathrm{a}_{3} & \mathrm{b}_{3} & \mathrm{c}_{3} \end{array}\right|=0 \)

​Calculation:

Here, \(\rm \hat i-\hat j+\hat k, 2\hat i+\hat j-\hat k, \hat iλ-\hat j+\hat k λ \) are coplanar

\(\begin{array}{l} \Rightarrow \left|\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ λ & -1 & λ \end{array}\right|=0\end{array}\)

1(λ - 1) + 1(2λ + λ) + 1(-2 - λ) = 0

λ - 1 + 2λ + λ + -2 - λ = 0

3λ - 3 = 0

λ = 0

Hence, option (4) is correct.

Let the d.r.’s of the normal to the plane passing through the point (–1, 3, 2) be A, B, C.

So, eq. of plane is : A(x + 1) + B(y - 3) + C(z - 2) = 0 ...(i)

As (i) is perpendicular to the planes x + 2y + 3z = 5 and 3x + 3y + z = 0

So, A + 2B + 3C = 0 ...(ii) and, 3A + 3B + 3C = 0 ...(iii)

Solving (ii) and (iii), we get :

A/(2 - 9) = B/(9 - 1) = C/(3 - 6) i.e.,

A/-7 = B/8 = C/-3 i.e., the d.r.’s are –7, 8, –3. 

By (i), -7(x + 1) + 8(y - 3) - 3(z - 2) = 0 i.e., 7x - 8y + 3z + 25 = 0, 

which is Cartesian equation. Also the vector equation of plane is, vector r.(7i - 8j + 3k) + 25 = 0

(A) always radially outwards. 

Initially bead is applying radially inward normal force. 

During motion at an instant, N = 0, after that N will act radially outward.

A spreadsheet is a file that exists of cells in rows and columns and can help arrange, calculate and sort data. Data in a spreadsheet can be numeric values, as well as text, formulas, references and functions.

Insert or delete a column

1. Select any cell within the column, then go to Home > Insert > Insert Sheet Columns or Delete Sheet Columns. 

2. Alternatively, right-click the top of the column, and then select Insert or Delete.

Insert or delete a row

1. Select any cell within the row, then go to Home > Insert > Insert Sheet Rows or Delete Sheet Rows. 

2. Alternatively, right-click the row number, and then select Insert or Delete.

(b) A new page

'Insert -> Page Break' is used to insert A new page.

(a) Landscape

When the text is printed or typed lengthwise, it is called Landscape.

Correct option: D. Cell

(b) Cell

The intersection of a row and a column in a spreadsheet is known as Cell.

Grandma Turns Eighty

 [Naren, XI A]

Sixteenth August was a grand occasion for our family. My grandmother had turned eighty that day. We organised a family get-together. Messages had been sent to all my uncles, aunts and cousins. The ancestral home was decorated with flowers. A puja was performed in the temple. Then the main function began in the sitting room. It was a very cheerful occasion. All my uncles, aunts and cousins gathered under one roof. She was seated in a high arm chair. My uncle honoured her with a beautiful shawl.

Then my parents presented her an almond coloured silk saree. Then came the turn of youngsters.

She appreciated all the gifts presented to her and blessed us. She is still smart, witty and energetic. Words of wit and wisdom dropped like honey from her lips. Dressed in her usual orange coloured dress, she appeared like a divine personality. Since I was the youngest member of the family, I received love and . affection from everyone. Sometimes I felt it was my birthday.

The intersection point between a row and column is called Cell.

b) File Extension

The most common way to discover the type of a file, is to look at the File Extension.

Correct option: a) Mail Merge

Correct option is A) All my friends will come

Concept:

• The ways of arranging n different things = n!
• The ways of arranging n things, having r same things and rest all are different = \(\rm n!\over r!\)
• The no. of ways of arranging the n arranged thing and m arranged things together = n! × m!
• The number of ways for selecting r from a group of n (n > r) = nCr 

Calculation:

Selection of 2 boys out of 5 = 5C2 = 10

Selection of 3 boys out of 5 = 5C3 = 10

Selection of 3 girls out of 4 = 4C3 = 4

Selection of 4 girls out of 4 = 4C4 = 1

Now the selection can be done in the ways = 2 boys and 4 girls + 3 boys and 3 girls

Hence number ways N = 5C2 × 4C4 + 5C3 × 4C3 

Total number of ways = 10 × 1 + 10 × 4 = 50 ways

Concept:

Arrangements of n different objects around a circle, then the number of arrangements is (n – 1)!

Calculation:

Given: All boys are to sit together

So, All boys can be considered as a single group.

∴ Total no of students = 4Girls + 1Group = 5

The number of ways of arranging 5 students in a round table is (5 - 1)! = 4!

Now, no of the ways of arranging 5 boys is 5!

Hence, The total number of ways = 4!5!

Calculation:

Given: 4 boys and 4 girls can be arranged in a row so that no two girls and no two boys are together.

It means they can sit alternately

Case-I: 1st person in the row is a boy.

B₁G₁B₂G₂B₃G₃B₄G₄

The no. of ways in which the boys can be rearranged among themselves is 4!. 
Similarly, The no. of ways in which the girls can be rearranged among themselves is 4!.
So, the total number of ways = 4! × 4! = (4!)²

Case-II:1st person in the row is a  girl.

G1B₁G2B2G₃B₃G₄B₄

The no. of ways in which the girls can be rearranged among themselves is 4!. 
Similarly, The no. of ways in which the boys can be rearranged among themselves is 4!.
So, the total number of ways = 4! × 4! = (4!)²

Hence, Total Number of ways =  (4!)2 + (4!)2 = 2(4!)2

Correct option is (4) 160

Keshav can invite exactly 3 girls but he can invite any no. of ways out for 4.

∴ Total no. of ways = \(5_{C_3} (4_{C_0} + 4_{C_1} + 4_{C_2}+ 4_{C_3}+ 4_{C_4})\)

\(= \frac{5\times 4 \times 3!}{2!\times 3!}\times 2^4\)

\(= 10 \times 16\)

\(= 160\)

Hence, Keshav can infinite there friends in 160 ways if there have to be exactly 3 girls in the invitees.

The cumulative frequency curve is called ‘ogive’.

If in a class, the lower or upper limits are not specified are called open classes.

Let x denote the number of bombs hit the target is a Binomial variate with the parameters

n = 5, P = 1/4 = 0.25 and q= 1 – p= 1 – 0.25 = 0:75.

The p.m.f is:-

p(x) = ⁿc_x p^x q^n-x : x = 0,1,2 n

p(x)= ⁵c_x (0.25)^x (0.75)^5-x x = 0,1,2……..5

(i) p(3 bombs hit the target) =p(x = 3) 

= ⁵c₃ (0.25)³ (0.75)^5-3 = 10 × 0.015625 × 0.5625

= 0.08789.

(ii) p(at most two bombs hit the target) = p(x ≤ 2)

= p(x = 0) + p[x = 1) + p(x = 2)

= ⁵c₀ (0.25)° x (0.75)^5-0 + ⁵c₁ (0.25)1 (0.75)^5-1 + ⁵c₂ (0.25)₂ (0.75)^5-2

= 1 × 1 × 0.2373 + 5 × 0.25 × 0.3164 + 10 × 0.0625 × 0.4218

= 0.2373 + 0.3955 + 0.2637 = 0.8965.

Modal(z) size is 8.

γ = ± 1 or -1 ≤ γ≤ 1.

Standard deviation = σ = √variance = √4

σ = 2 cm.

First we arranged the data in ascending order:
15, 18, 20, 24, 27, 28, 30, 47

Now computing the quartile deviation (QD) and the formula is:
QD = (Quartile 3 - Quartile 1) / 2

Computing quartile 1 and quartile 3 as:
As there is even number of data (N = 8, which is even), So,

Quartile 1 = 1/ 4 (n + 1)^th term

= 1/4 ( 8 + 1)

= 1/4 × 9

= 2.25th term

Quartile 3 = 3/4 (n+1 )^th term

= 3/4 (8+ 1)

= 3/4 × 9

= 6.75

Computing the QD as:
2nd term is 18 and adding to this 0.25 × (20 - 18), which is 0.5, the result is 18.5

6th term is 28 and adding to this 0.75 × (30 - 28), which is 1.5. the result is 29.5

Using the QD formula as:
QD = (29.5 - 18.5) / 2

= 11 / 2

= 5.5

2. Computing the coefficient of quartile deviation as:

Coefficient of quartile deviation = (Quartile 3 - Quartile 1) /  (Quartile 3 + Quartile 1)

Putting the values of the Quartile 3 and Quartile 1 in the above formula:
Coefficient of quartile deviation = (29.5 - 18.5) / (29.5 + 18.5)

= 11/ 48

= 0.229