Explore topic-wise InterviewSolutions in Current Affairs.

1. Consider the reaction,

C(grahite) + O₂ (g) → CO₂ (g); ∆H = x 

CO₂ can also be prepared through the following two steps:

(i) C(graphite)+ ½ O₂ (g) → CO(g); ∆H =110.5 kJ mol^-1

(ii) CO(g) + ½ O₂(g) → CO₂ (g); ∆H =-283.0 kJ mol^{​​​​​​​-1} 

Then by Hess’s law, 

x = (-110.5+-283.0) kJ = -393.5 kJ

2. Certain endothermic process are found to be spontaneous in nature. Hence, spontaneous behaviour of a process cannot be explained only on the basis of energy consideration.

For a spontaneous process ∆S_Total is +ve. 

For a nonspontaneous process ∆S_Total is -ve.

1. Enthalpy is the sum of internal energy and pressure volume energy.

 i.e. H = U + pV

2. ∆H = ∆U + ∆pV

∆H = ∆U + p∆V + V∆p 

At constant pressure, ∆p=0 

∆H = ∆U + p∆V 

But ∆U= q + w 

∆H= q+ w+ p∆V 

w = -p∆V 

i.e; ∆H= q – p∆V + p∆V 

∆H = qp

3. Enthalpy change is heat absorbed or released at constant pressure.

(a) It is the enthalpy change when one mole of a liquid is converted into its vapour at its boiling point. 

(b) It is the enthalpy change when one mole of a solid is converted into its vapour at its transition temperature.

Moles of Al (n) = \(\frac{60g}{27g\ mol^{-1}}\) = 2.22 mol 

Molar heat capacity (C_m) = 24 J mol K 

∆T = 55 °C – 35 C° = 20C° or 20 K 

Now, q = C_m × n × ∆T 

= 24.0 J mol K × 2.22 mol × 20 K 

= 1065.6 J 

= 1.067 kJ

1. The entropy will increase. 

2. The entropy will decrease. 

3. The entropy will increase.

(a) Gives off thermal energy

(a) C (Diamond)

1. ∆G = ∆H -T∆S 

= 26000 – 300 × 8.3 

= 23.510 

Since ∆G is positive, the process is non-spontaneous.

2. ∆G = ∆H -T∆S 

= -393400 – 300 × 6 

= -391600 

Since ∆G is negative, the process is spontaneous.

(c) different intermediate reactions

Any spontaneous process must lead to a net increase in entropy of the universe. TRUE

(c) exothermic 

• In the 18th century the periodical press combined information about current affairs with entertainment.
•  Newspapers and journals carried information about wars and trade as well as the news of development in other places. Similarly, ideas of philosophers now became accessible to common people. 
• Ancient and medieval scientific texts were compiled and published, maps and scientific diagrams were widely printed. 
• When scientists like Issac Newton began to publish their discoveries, they could influence a much wider circle of scientific minded readers. . 
• The writings of thinkers such as Thomas Paine, Voltaire and Rousseau were also widely printed and read. Thus their ideas about science, reason and rationality found way into popular literature.

Plaster of Paris.

In water medium velocity of light is more.

Convex lens. The object is at infinity.

At centre of curvature.

2 principal focus are in lens.

Infinity images can be originate from object kept between two parallel plane mirror.

\(A=\begin{bmatrix}3&4\\-2&5\end{bmatrix}\)

Applying R₁ → R₁ + 3R₂

Then \(A\sim\begin{bmatrix}3+3\times-2&4+3\times5\\-2&5\end{bmatrix}\)

\(=\begin{bmatrix}3-6&4+15\\-2&5\end{bmatrix}\)

\(=\begin{bmatrix}-3&19\\-2&5\end{bmatrix}\)

After applying transformation R₁ → R₁ + 3R₂ the elementary matrix of A is \(\begin{bmatrix}-3&19\\-2&5\end{bmatrix}.\)

Suppose P = (x, y, z).  Now 3PA = 2PB

9PA² = 4PB²

= 9[(x + 2)² + (y - 2)² + (z - 3)²]

= 4[(x - 13)² + (y + 3)² + (z - 13)²]

5(x² + y² + z²) + 140x - 60y + 50z - 1235 = 0

x² + y² + z² + 28x - 12y + 10z - 247 = 0

The correct answer is Pradhan Mantri Kisan Urja Suraksha evam Utthaan Mahabhiyan Yojna.

• PM-KUSUM stands for Pradhan Mantri Kisan Urja Suraksha evam Utthaan Mahabhiyan Yojna.
• The government has approved Pradhan Mantri Kisan Urja Suraksha evam Utthaan Mahabhiyan (PM-KUSUM).
• PM-KUSUM Scheme's objective is to increase farmers’ income, provide reliable sources for irrigation, and de-dieselize the farm sector.

• While presenting the budget for 2020-21, the minister said that farmers would be provided funds to set up grid-connected 15 lakh solar pumps.
• The previous Modi government in February last year had unveiled the PM KUSUM scheme with an outlay of ₹34,422 crores.

• This scheme was launched by the Ministry of New and Renewable Energy (MNRE) for the installation of solar pumps and other renewable power plants across the nation.  

Let numbers are x, y and z

x + y + z = 2

x + z  - 2y = 8

3x + y + z = 4

Matrix form is

\(\begin{bmatrix}1&1&1\\1&1& -2\\3& 1&1\end{bmatrix}\)\(\begin{bmatrix} x\\y\\z\end{bmatrix}\) = \(\begin{bmatrix} 2\\8\\4\end{bmatrix}\) 

∴ \(\begin{bmatrix} x\\y\\z\end{bmatrix}\) = \(\begin{bmatrix}1&1&1\\1&1& -2\\3& 1&1\end{bmatrix}^{-1}\)\(\begin{bmatrix} 2\\8\\4\end{bmatrix}\)

Let A = \(\begin{bmatrix}1&1&1\\1&1& -2\\3& 1&1\end{bmatrix}\)

A = IA

\(\begin{bmatrix}1&1&1\\1&1& -2\\3& 1&1\end{bmatrix}\) = \(\begin{bmatrix}1&0&0\\0&1& 0\\0& 0&1\end{bmatrix}\)A

Applying R₂→R₂ - R₁, R₃ → R₃ - 3R₁

\(\begin{bmatrix}1&1&1\\0&0& -3\\0& -2&-2\end{bmatrix}\) = \(\begin{bmatrix}1&0&0\\-1&1& 0\\-3& 0&1\end{bmatrix}A\)

Applying R₃→R₂

\(\begin{bmatrix}1&1&1\\0&-2& -2\\0& 0&-3\end{bmatrix}\) = \(\begin{bmatrix}1&0&0\\-3&0& 1\\-1& 1&0\end{bmatrix}A\)

Applying R₁→R₂/-2, R₃ → R₃/-3

\(\begin{bmatrix}1&1&1\\0&1& 1\\0& 0&1\end{bmatrix}\) = \(\begin{bmatrix}1&0&0\\3/2&0& -1/2\\1/3& -1/3&0\end{bmatrix}A\)
Applying R₁ → R₁ - R₂, R₂ → R₂ - R₃

\(\begin{bmatrix}1&0&0\\0&1& 0\\0& 0&1\end{bmatrix}\) = \(\begin{bmatrix}-1/2&0&1/2\\7/6&1/3& -1/2\\1/3& -1/3&0\end{bmatrix}\)

∴ A^-1 = \(\begin{bmatrix}-1/2&0&1/2\\7/6&1/3& -1/2\\1/3& -1/3&0\end{bmatrix}\)

∴ x = A^-1b = \(\begin{bmatrix}-1/2&0&1/2\\7/6&1/3& -1/2\\1/3& -1/3&0\end{bmatrix}\)\(\begin{bmatrix} 2\\8\\4\end{bmatrix}\)

\(\begin{bmatrix} x\\y\\z\end{bmatrix}\) = \(\begin{bmatrix} 1\\3\\-2\end{bmatrix}\)

∴ Numbers are 1, 3 and -2

Concept:

Bisection method:

Used to find the root for a function. Root of a function f(x) = a such that f(a)= 0

Property: if a function f(x) is continuous on the interval [a…b] and sign of f(a) ≠  sign of f(b). There is a value c belongs to [a…b] such that f(c) = 0, means c is a root in between [a….b]

Note:

Bisection method cut the interval into 2 halves and check which half contains a root of the equation.

1) Suppose interval [a…b] .

2) Cut interval in the middle to find m : \(m =\frac{{a+b}}{{2}}\)

3) sign of f(m) not matches with f(a) proceed the search in the new interval.

Calculation:

The bisection method is applied to a given problem with [1, 9]

After 1 iteration

\({x_1} = \frac{{1\; + \;9}}{2} = 5\)

Now since f(x₁) > 0, x₂ replaces x₁

Now, x₀ = 1 and x₁ = 5

And after 2^nd iteration 

\({x_2} = \frac{{1\; + \;5}}{2} = 3\)

Now since f(x₁) f(x₂) > 0, x₂ replaces x₁ and x₀ = 1 and x₁ = 3 and after 3^rd iteration

\({x_2} = \frac{{1\; + \;3}}{2} = 2\)

Now, f(x₂) = f(2) = 2⁴ – 2³ – 2² – 4 = 0

So the method converges exactly to the root in 3 iterations.

\(A=\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]\)

\(A = 0\begin{bmatrix}1&0\\0&0\end{bmatrix} - 0\begin{bmatrix}0&0\\1&0\end{bmatrix} + 1\begin{bmatrix}0&1\\1&0\end{bmatrix}\)

\(A= 0(1 - 0) - 0(0 - 1) + 1(0 - 1)\)

\(​​A = -1\)

\(⇒ A^2 = 1\)

\(⇒ A^6 = 1\)

 (a) f(-x) = \(([\frac{-x}{\pi}]+\frac{1}{2})sin(-x)\)

\(=-([\frac{-4}{\pi}]+\frac{1}{2})sinx\)

\(≠\) f(x) or \(≠\) -f(x)

∴ f(x) is neither even nor odd function

(b) f(-x) = \(\frac{(a^{-x}+1)^5}{a^{-x}}=\frac{(1+a^x)^5}{a^{5x}a^{-x}}\)

\(=\frac{(1+a^x)^5}{a^{4x}}≠ f(x)\) or \(≠-f(x)\)

∴ f(x) is nether even nor odd function

(c) f(-x) = \(\frac{-x}{e^{-x}-1}-\frac{x}{2}+1\)

\(=-\frac{xe^x}{1-e^x}-\frac{x}{2}+1\)

\(≠ \) f(x) or \(≠\) -f (x)

∴ f(x) is neither even nor odd function

(d) f(-x) \(=\frac{g(-x)-g(x)}{5}=-(\frac{g(x)-g(-x)}{5})\)

= -f (x)

∴ f(x) B odd function

(6a²- a + 1) - (4 -10a - 3a²)

= 6a²- a +1 - 4 +10a +3a²

= 9a² + 9a - 3

= 9 (a² + a - \(\frac{1}{3}\))

= 9 (a² + a + \(\frac{1}{4}\) - \(\frac{1}{4}\) - \(\frac{1}{3}\))

= 9 ((a + \(\frac{1}{2}\))² - (\((\sqrt{\frac{7}{12}})^2\))

= 9 (a + \(\frac{1}{2}\) - \(\frac{\sqrt{21}}{6}\)) (a + \(\frac{1}{2}\) + \(\frac{\sqrt{21}}{6}\))

(∵ a² - b² = (a+b)(a-b) & \(\sqrt{\frac{7}{12}}\) = \(\frac{1}{2}\)\(\sqrt{\frac{7}{3}}\) = \(\frac{\sqrt 7 \times \sqrt 3}{2\sqrt 3\times \sqrt 3}\) = \(\frac{\sqrt{21}}{6}\))

Given series is 

sin(p + x) sin(p + q) + x²/2! sin(p + 2q) + ....

We know that e^iα = cos α + i sin α.

We develop new series

cos p + x cos(p + q) + x²/2! cos(p + 2q) + ... + i(sin p + x sin(p + q) + x²/2! sin(p + 2q)+....)

= e^ip + xe^{i(p + q)} + x²/2! e^{i(p + 2q)}+.....

= e^ip(1 + x^iq + x²/2! e^i2q + ....)

= e^ip(1 + (x e^iq) + \(\frac{(xe^{i2q})^2}{2!}\)+....)

= e^ip. e^{\(xe^{iq}\)} (\(\because\) e^x = 1 + x + x²/2! + .......)

= e^ip + xe^iq

= e^{ip + x(cos q + i sin q)} (\(\therefore \) e^iα = cos α + i sin α)

= e^{xcosq + i (p + x sin q)}

 = e^xcosq. e^{i(p +x sin q)} (\(\because\) e^{a + b} = e^a. e^b)

= e^xcosq (cos(p +sin q) + i sin (p + x sin q))

By comparing imaginary part we get

sin p + x sin (p + q) + x²/2! sin(p + 2q) = e^xcosq sin(p + x sin q).

Hence, option (A) is correct.

Given series is 

sin(p + x) sin(p + q) + x²/2! sin(p + 2q) + ....

We know that e^iα = cos α + i sin α.

We develop new series

cos p + x cos(p + q) + x²/2! cos(p + 2q) + ... + i(sin p + x sin(p + q) + x²/2! sin(p + 2q)+....)

= e^ip + xe^{i(p + q)} + x²/2! e^{i(p + 2q)}+.....

= e^ip(1 + x^iq + x²/2! e^i2q + ....)

= e^ip(1 + (x e^iq) + \(\frac{(xe^{i2q})^2}{2!}\)+....)

= e^ip. e^{\(xe^{iq}\)} (\(\because\) e^x = 1 + x + x²/2! + .......)

= e^ip + xe^iq

= e^{ip + x(cos q + i sin q)} (\(\therefore \) e^iα = cos α + i sin α)

= e^{xcosq + i (p + x sin q)}

 = e^xcosq. e^{i(p +x sin q)} (\(\because\) e^{a + b} = e^a. e^b)

= e^xcosq (cos(p +sin q) + i sin (p + x sin q))

By comparing imaginary part we get

sin p + x sin (p + q) + x²/2! sin(p + 2q) = e^xcosq sin(p + x sin q).

Hence, option (A) is correct.

Given:

Distance travelled by boat in downstream = 16 km

Time boat takes to complete the distance in downstream = 40 min

Speed of stream = (3/5) × the speed of boat in still water

Formula used:

Downstream speed = Speed of boat + Speed of stream

Upstream speed = Speed of boat – Speed of stream

Speed = Distance/Time

Calculation:

Let speed of boat in still water be u

Let speed of stream be v

v = (3/5)u

Let downstream speed be a

Let upstream speed be b

a = 16/40 km/min

⇒ a = (16 × 60)/40 km/h

⇒ a = 24 km/h

Downstream speed = Speed of boat + Speed of stream

⇒ a = u + v

⇒ 24 = u + (3/5)u

⇒ 24 = (5u + 3u)/5

⇒ 120 = (5u + 3u)

⇒ 8u = 120

⇒ u = 15 km/h

v = (3/5)u

⇒ v = 9 km/h

Upstream speed = Speed of boat – Speed of stream

⇒ b = u – v

⇒ b = 15 – 9 = 6 km/h

Upstream distance covered in 45 min 

Distance = 6 × (45/60) = 4.5 km

∴ The upstream distance covered by boat in 45 min is 4.5 km.

Foundational Challenge:

Different countries face different kinds of challenges. At least one fourth of the globe is still not under democratic government. The challenge for democracy in these parts of the world is very stark. These countries face the foundational challenge of making the transition to democracy and then instituting democratic government. This involves bringing down the existing non-democratic regime, keeping military away from controlling government and establishing a sovereign and functional state. For example, Nepal was under Monarchy till recent times. Now Nepal has changed to a democratic system. Certain mindsets and systems will take years to change as they have taken years to develop. Nepal is a very good example of foundational challenge of democracy.

Challenge of Expansion:

Most of the established democracies face the challenge of expansion. This involves applying the basic principle of democratic government across all the regions, different social groups and various institutions. Ensuring greater power to local governments, extension of federal principle to all the units of the federation, inclusion of women and minority groups, etc., falls under this challenge. This also means that less and less decisions should remain outside the arena of democratic control. Most countries including India and other democracies like the US face this challenge. In India certain socio-economically backward groups have yet to get the political powers. Additionally, some north-eastern states have not yet fully integrated with the mainstream India. These are examples of challenge of expansion of democracy.

Deepening of Democracy

The third challenge of deepening of democracy is faced by every democracy in one form or another. This involves strengthening of the institutions and practices of democracy. This should happen in such a way that people can realise their expectations of democracy. But ordinary people have different expectations from democracy in different societies. In the early nineties, when T N Sheshan became the Chief Election Commissioner, he brought about a revolutionary change of disciplining political parties. This in turn ensured fairer elections. This is an example of strengthening a particular political institution.

From above descriptions it is clear that different country face different kinds of challenges to democracy. These challenges depend on what stage of social development the country is. Solutions to these challenges also depend on a particular country’s unique situation. It can be said that there is no pre-prescribed method to tackle the challenge faced by a democracy.

Solution:
Since x and y are odd positive integers, we have
x = 2m + 1 and y = 2n + 1
=> x² + y² = (2m + 1)² + (2n + 1)²
= 4m² + 4m + 1 + 4n² + 4n + 1
= 4(m² + n²) + 4(m + n) + 2
Hence, x² + y² is an even number but not divisible by 4.

Following are the duties as consumers if we visit a shopping complex :

Alertness : it is the duty for us to be more alert and we should question about the price and quality of goods and services that we intend to use. 

Asserting : We should assert and act to ensure that we get a fair deal. as long as we remain passive, we will continue to be exploited. 

Association of consumer : Consumer should organise themselves to develop the strength and influence to promote and protect their interest. 

Asking for warrantee : We should insist for warrantee card. 

Cash memo : We should get cash memo from sellers 

Complaint Against Exploitation : A consumer should make complaints against a seller if he feels exploited.

Relation between mean, median and mode is 

Mode = 3 median – 2mean. … (1)

Given that median and mode of frequency distribution are 26 and 29 respectively. 

i.e., median = 26 and mode = 29. 

Now, putting the values of median and mode in equation (1), we get 

29 = 3 × 26 – 2 Mean 

⇒ 2 Mean = 78 – 29 = 49 

⇒ Mean = \(\frac{49}2\) = 24.5.

Hence, the mean of the frequency distribution is 24.5.

α + β = \(\frac{k}{3}\)

3 = \(\frac{k}{3}\)

K = 9

Four decimal places

(LCM)(3) = 180

LCM = 60

N = 51

⇒ \(\frac{N}{2}\) = \(\frac{51}{2}\) = 25.5 

As 29 is just greater than 25.5, therefore median class is 145-150. 

Median = I + \(\frac{(\frac{N}{2}-C)}{f}\) X h

Here, l = lower limit of median class = 145 

C = C.F. of the class preceding the median class = 11 

h = higher limit - lower limit = 150 − 145 = 5 

f = frequency of median class = 18 

∴ median = 145 + \(\frac{(25.5-11)}{18}\) X 5

=149.03 

Mean by direct method

Mean = \(\frac{∑fx}{N}\)

= \(\frac{7637.5}{51}\)

= 149.75

Nayapura
Kota
29 July 2020
The Post Master
Kota

Sir,

I have been informed by my elder brother Shri Ravinder Seth that on July 2, he sent a money-order to me for Rs. 200/- from Sojati Gate, Jodhpur post-office vide receipt No. 195. I am sorry to inform you that the same has not yet been received by me. Please make an enquiry into the matter and arrange to pay me the amount of the money-order at an early date.

Thanking you

Yours faithfully

XYZ

Radha told Sheenu (a) that her watch was exclusive. She (b) further requested her to tell her where she had bought it from. Sheenu replied that (c)  she had bought it  from a show-room which was situated in Maharana Nagar. Radha asked Sheenu (d)if she would tell her how much it had cost her Sheenu replied that (e) it had cost her only eight hundred rupees.