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The vertices of the triangle formed by the cube roots of unity = 1 , ω, ω² .

the vertices in the complex plane are described by :

1 = (1, 0)        ω = (-1/2 , √3/2 )   ω² = ( -1/2 , -√3/2)

The side of this equilateral  triangle is = √3

Altitude of the triangle is =  √3/2 * √3 = 3/2

So area of triangle = 1/2 * 3/2 * √3  = 3√3 / 4 

Leaching followed by reduction is called hydrometallurgy. It is used in extraction of copper.

Mathematics 
1. IIT Mathematics by M.L. Khanna. 
2. Differential Calculus by A Das Gupta. 
3. Class XI and XII Mathematics by R.D. Sharma. 
4. Trigonometry, Geometry Books by S.L. Loney. 
5. Problems in Calculus of One Variable by I.A. Maron. 
6. Higher Algebra by Hall and Knight.

(a) Current through arm ABC,

I = \(\frac{48}{10+20}\)

= 1.6A

Potential difference across B and C, 

= V_B - V_C

= 1.6 × 20

= 32V 

As,

V_c = 0

So,

V_B = +32V

(b) Current through arm ADC,

I' = \(\frac{48}{5+11}\) 

= 3A

Potential difference across D and C,

= V_D - V_C

= 3 × 11 

= 33V

As,

V_c = 0

So,

V_D = 33V

(c) As V_D > V_c, 

So direction of current will be from D to B.

(d) The bridge will be balanced when : 

\(\frac{10}{R}\) = \(\frac{5}{11}\)

or R = \(\frac{10\times 11}{5}\)  

= 22Ω.

Correct answer is 

(b) AC² = AB² + BC²

Let the equation of the ellipse is x²/a² + y²/b² = 1 (a > b) 

Length of the latus rectum = 2b²/a² = 4 ⇒ b² = 2a. 

Foci are S (ae, 0), S' (– ae, 0) 

Distance between the foci = 2ae = 4√2 

ae = 2√2

b² = a² (1 – e²) = a² – (ae)² 

2a = a² – 8 ⇒ a² – 2a – 8 = 0 

(a – 4) (a + 2) = 0 

a = 4 or – 2 

a > 0 ⇒ a = 4 

b² = 2a = 2. 4 = 8

Equation of the ellipse is 

x²/a² + y²/b² = 1

x²/16 + y²/8 = 1

x² + 2y² = 16

Required equation of circle passing through intersection S = 0 

and L = 0 is S + λL = 0 

(x² + y² – 2x + 4y – 8) + λ(x + y – 3) = 0 

(x² + y² + x(–2 + λ) + y (4 + λ) – 8 – 3λ = 0 —— (i) 

x² + y² + 2gx + 2fy + c = 0 —— (ii) 

Comparing (i) and (ii) we get

g = (-2 + λ)/2, f = (4 + λ)/2

Centre lies on x + y = 3

∴ - ((-2 + λ)/2) - ((4 + λ)/2) = 3

2 – λ – 4 – λ = 6 

–2λ = 8 ⇒ λ = – 4 

Required equation of circle be 

(x² + y² – 2x + 4y – 8) – 4(x + y – 3) = 0 

x² + y² – 6x + 4 = 0

The given differential equation is 

y = c(x – c)² 

dy/dx = 2c(x – c) 

∴ Order of the differential equation is 1.

Equation of the circle is x² + y² – 5x + 8y + 6 = 0 

Polar of P(4, 2) is 

x.4 + y.2 – (5/2) (x + 4) + 4 (y + 2) + 6 = 0 

8x + 4y – 5x – 20 + 8y + 16 + 12 = 0 

3x + 12y + 8 = 0 

P(4, 2), Q(k, –3) are conjugate points 

Polar of P passes through Q 

∴ 3k – 36 + 8 = 0 

3k = 28 ⇒ k = 28/3 .

∫ 2x. sin (x² + 1) dx 

t = x² + 1 

⇒ dt = 2x dx 

∫ 2x. sin (x² + 1) dx 

= ∫ sin t dt = – cos t + C 

= – cos (x² + 1) + C

Equation of the circle is x² + y² + ax + by – 12 = 0 

Centre = (-a/2, -b/2) = (2, 3) 

−a/2 = 2, − b/2 = 3 

a = – 4, b = – 6 

g = –2, f = –3, c = –12 

radius = √(g² + f² - c) = √(4 + 9 + 12) = 5

Equation of the normal is

(x – x₁) (y₁ + f) – (y – y₁) (x₁ + g) = 0 

(x – 3) (– 4 + 1/2) – (y + 4) (3 + 1/2) = 0 

– 7/2 (x –3) – 7/2 (y + 4) = 0 

⇒ (x – 3) + (y + 4) = 0 

x – 3 + y + 4 = 0 

x + y + 1 = 0

S – S' = 0 is radical axis. 

(x² + y² + 4x + 6y – 7) 

– (x² + y² + 2x + 3y - (9/4)) = 0 

2x + 3y – (19/4) = 0 

⇒ 8x + 12y – 19 = 0

Phototherapy is a type of medical treatment that involves exposure to fluorescent light bulbs or other sources of light like halogen lights, sunlight, and light emitting diodes (LEDs) to treat certain medical conditions.

There are different types of phototherapy and the kind, as well as the technique used by your doctor, will depend on the condition you have that is being treated.

Phototherapy is also known as light therapy and heliotherapy.

n-bexane has the highest boiling point.

Murchana is a special pharmaceutical procedure before subjecting the drugs to Snehapaka and is used for Amadoshaharatwa, removal of odor, imparting color and increasing potency of the drug. 

Sildenafil mediates its effects by inhibiting phosphodiesterase 5. Some biochemical and molecular biological aspects of this enzyme are presented here. To achieve satisfactory erection, normal penile innervation is required.

Answer: White precipitate is formed which is soluble in excess of sodium hydroxide.

Silver nitrate and ammonium nitrate.

Correct option is A) One dozen is fine, thank you.

Correct option is C) Yesterday's homework

Diamond non metal which has lustre.

Mercury a liquid metal.

The author was left with his grandmother in the village because his parents had to go to the city for work.

Correct option is D) she cleans her room every day.

The contents of the bag were frequent flyer card, newspaper cuttings, loose papers, tobacco pipe, magazines, passport, English money and film.

Correct option is A) She speaks it very well.

D) How long and how does he get to the place?

Correct option is D) had tidied up / washed

C) When did his dream realize?

Once Mary Kom and her team mates were given chopsticks to eat their food in China. Other friends, asked for spoons and managed. But Mary Kom ended up using both her hands to hold the chopsticks to pick up the food and push it into her mouth. She managed the complex work and satisfied her hunger.

Correct option is B) Borax

Correct option is E) Teachers and pupils do

Correct option is D) doesn’t know

Correct option is D) make

Correct option is B) had left / returned

Correct option is D) finish / shall enter