Explore topic-wise InterviewSolutions in Current Affairs.

• To improve education and skills, government has implemented the following projects.

ICDS, SSA, RMSA, RUSA 

National Skill Development and Monetary Reward Scheme.

• To provide education, there are institutions at various levels in the country. 
• Our country has made education fundamental right and has passed the Right to Education Act (RTE) in 2009.

• All the students are not getting proper education
• Drop out still continues
• Basic facilities are not available
• Quality of education has to be improved.
• The economically and socially backward are denied the opportunity of education.

In India network operates widely at different levels to ensure the people’s health in rural sector. There are different layers of health facilities. Most of the people live in rural areas and they possess sufficient medical institutions which realize the objectives of health. Health indices is developed by attaining basic facilities in health sector. Co-operative and private sectors are the main hospitals.

The multi specialty hospitals operates modem treatment facilities, there are several institutions which provide different streams in medicine like Ayurveda, Yoga, Naturopathy, Unani, Sidha and Homeopathy. The National Rural Health Mission operates in the rural sector. The National Urban Health Mission provides improved health services towards urban slums and township.

• Multi specialty hospitals
• Ayurveda, Yoga, Naturopathy, Unani, Sidha, Homeopathy
• National Rural Health Mission (NRHM)
• National Urban Health Mission (NUHM)

[Hint : U (s) + 3ClF3 (l) -> UF6 (g) + 3ClF (g)]

(c) It is H₂ SO₅ .

Oxidation number of sulphur in  Caro's - Peroxymonosulfuric acid (H₂SO₅ )

2 + x - 10

x= +8

x= oxidation number of Sulphur

But this cannot be true as maximum oxidation number for S cannot exceed + 6. Since S has only 6 electrons in its valence shell.

This exceptional value of oxidattion state for sulphur in H₂SO₅  is due to the peroxy linkage shown by two atoms.

so, Oxidation number of S in H₂SO₅

2 (+1) + x + 3 (-2) + 2 (-1) [for peroxy linkage] = 0

2 + x -6 -2 = 0

x = + 6

so, Oxidation number of S in H₂SO₅ is +6.

• District Hospitals 
• Community Health centers 
• Primary Health Centers 
• Health Sub Centers.

(b) Phosphorus from stable P₄ molecule

[Hint : PCl₃ hydrolyses in the presence of moisture giving fumes of

 HCl. PCl₃ + 3H₂O => H₃PO₃ + 3HCl ↑]

Correct answer is (B) \(-\frac{3}{5}\)

Correct answer is (D) \(\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}\)

Correct answer is (C) (9/10)⁵ 

Correct answer is (A) sin^-1\(\frac{273}{305}\)

Correct answer is (A) 0

Option: (b) 1

(i) Since, x-coordinate of point is 4 which is positive and y-coordinate of point is -1 which is negative.

And we know that if x-coordinate of a point is positive and y-coordinate of that point is negative then that point lies in IV quadrant.

∴ (4, -1) lies is 4^th quadrant.

(ii) (-3, 2) lies in 2^nd quadrant.

Reason :- Abscissa = negative and ordinate = positive

Then point lies in 2^nd quadrant.

(iii) (0, 1/2) lies on positive y-axis.

Reason :- The equation of y-axis is x=0.

i.e. if abscissa is 0 then point lies on y-axis.

(iv) (7/20, -3/4) lies in 4^th quadrant.

Reason :- Abscissa = positive and ordinate = negative

Then point lies in 4^th quadrant.

2(m - 2) -3(m - 3) = 5(m - 5) -4(m - 3)

2m - 4 - 3m + 9 = 5m - 25 - 4m + 12

2m - 3m - 5m + 4m = -25 + 12 + 4 - 9

6m - 3m - 5m = -34 + 16

-2m = -18

m = 18/2

m = 9

Let width of rectangle be W

∴  length would be  L=2W−5

If the length is decreased by 5cm, then new L1=2W−10cm  

and width is increased by 2cm, new width W1=W+2cm 

Given, Perimeter =74cm

=>2(2W−10+W+2)=74

=>3W−8=37

=>3W=45

=>W=15cm

=>L=2W−5=2(15)−5=25cm

(D² -5D + 6)y = x sin3x.

It's auxiliarly equation is,

m² - 5m + 6 = 0

⇒ (m-2)(m-3) = 0

⇒ m = 2 or 3 (real and distinct)

Therefore,

C. F = C₁e^2x + C₂e^3x

P.I = \(\frac{1}{D^2-5D+6}x\,sin\,3x\)

=  \(x\frac{1}{D^2-5D+6}\,sin\,3x\) + {\(\frac{d}{dD}\,\)\(\frac{1}{D^2-5D+6}\)} sin3x

(∵ \(\frac{1}{f(D)}\)xv = x\(\frac{1}{f(D)}\)v + {(\(\frac{d}{dD}\,\) \(\frac{1}{f(D)}\))v} and here v = sin3x)

= x\(\frac{1}{-9-5D+6}\) sin3x - \(\frac{2D-5}{(D^2-5D+6)^2}\) sin3x

(∵ \(\frac{d}{dx}\) \(\frac{1}{x}\) = \(\frac{-1}{x^2}\) and by chain rule Also we use\(\frac{1}{f(D^2)}\)sinax = \(\frac{1}{f(-a^2)}\)sinax)

= - x\(\frac{1}{3+5D}\)sin3x - \(\frac{2D-5}{(-9-5D+6)^2}\)sin3x

= -x \(\times\) \(\frac{3-5D}{9-25D^2}\)sin3x - \(\frac{2D-5}{(3+5D)^2}\)sin3x

= \(\frac{-x(3-5D)sin3x}{9-25\times -9}\) - \(\frac{2D-5}{25D^2+30D+9}\)sin3x

= \(\frac{x}{234}\)(3 sin3x - 5D sin3x) - \(\frac{2D-5}{25\times{(-9)}+30D+9}\)sin3x

= \(\frac{x}{234}\)(3 sin3x - 15 cos3x) - \(\frac{2D-5}{30D-216}\)sin3x

(∵ D sin3x = \(\frac{d}{dx}\) sin3x = 3cos 3x)

= \(\frac{3x}{234}\)(sin3x - 5cos3x) - \(\frac{1}{6}\times\)\(\frac{(2D-5)(5D+36)}{(5D-36)(5D+36)}\)sin3x

= 78x (sin3x - 5cos3x) - \(\frac{1}{6}\times\)\(\frac{(10D^2+47D-180)}{25D^2-1296}\)sin3x

=  78x (sin3x - 5cos3x) - \(\frac{1}{6}\times\) \(\frac{10D^2\,sin3x + 47D\,sin3x-180\,sin3x}{25\times -9-1296}\)

=  78x (sin3x - 5cos3x) - \(\frac{1}{6}\times\) \(\frac{30D\,cos3x +141\,cos3x-180\,sin3x}{1521}\)

=  78x (sin3x - 5cos3x) + \(\frac{1}{9126}\) (-90 sin3x + 141cos3x - 180sin3x)

=  78x (sin3x - 5cos3x) + \(\frac{1}{9126}\)(141 cos3x - 270 sin3x)

Hence,

complete solution given diff.equation is 

I = C.F + P.I

= C₁e^2x + C₂e^3x + 78x(sin3x - 5cos 3x) + \(\frac{1}{9126}\)(141cos3x - 270 sin3x).

Correct option is C) breathing down

Correct option is E) brainstorm

Correct option is C) do

Correct option is C) dig

Correct option is B) stir

Enzymes play a major role in DNA replication because they catalyze several important stages of the entire process. DNA replication is one of the most essential mechanisms of a cell's function and therefore intensive research has been done to understand its processes.Enzymes play a major role in DNA replication because they catalyze several important stages of the entire process. DNA replication is one of the most essential mechanisms of a cell's function and therefore intensive research has been done to understand its processes.

Work = Dot product of Force and Displacement

W = F.D

F₁ = 2i + 3j , W₁ = 8J , D₁ = ?

W₁ = F₁.D₁

Let the displacement vector D₁ have components xi + yj

8 = (2i + 3j).(xi + yj)

8 = (2i.xi) + (2i.yj) + (3j.xi) + (3j.yj)

8 = 2x + 3y

F₂ = 3i + 5j , W₂ = -4 J , D₂ = ?

W₂ = F₂.D₂

Let the components of D₂ = ai + bj

-4 = (3i + 5j).(ai + bj)

-4 = (3i.ai) + (3i.bj) + (5j.ai) + (5j.bj)

-4 = 3a + 5b

5{a² -(a - (a - 2))} = 5(a² - a + a - 2)

= 5(a² - 2)

= 5a² - 10

For its sedimentation. Here dirt particles, sands etc settle down and water is free from any of these impurities.

Light is just one form of electromagnetic radiation, or electromagnetic waves. These waves are all around us and come in many sizes. The largest electromagnetic waves, with wavelengths from a few centimeters to over 100 meters are called radio waves. The smallest electromagnetic waves, with wavelengths the size of an atomic nucleus are called gamma rays. In between are microwaves, infrared light, visible light, ultraviolet light and X-rays.

f (x) = \(\frac {x^2}{1+x^2}\) 

Since, \(x^2 \geq 0\)

\(\Rightarrow\) \(1+ x^2 \geq 1 > 0\)

\(\therefore\) Domain of function f (x) is R.

Also, \(x^2 + 1 > x^2\)

\(\Rightarrow\) \(\frac {x^2}{x^2+1} < 1\) 

Also \(x^2 \geq 0 \) & \(x^2 + 1 \geq 0\) 

\(\therefore\) \(\frac {x^2}{x^2+1}\geq0\) 

Hence, 0 \(\leq \frac {x^2}{x^2+1} < 1\)

\(\therefore\) Range of function f (x) is (0,1)

\(\because\) f (-1) = \(\frac {(-1)^2}{(-1)^2 +1} = \frac 12\)  (\(\because\) f (-1) = f(1))

f (1) = \(\frac 1{1+1} = \frac 12\) (\(\because\) f is many one function)