Explore topic-wise InterviewSolutions in Current Affairs.

Importance of Business Planning :

(a) Estimating the money required to be spent 

(b) Estimating quantity of material required 

(c) Figure out how they can make the customer see the business as standing out uniquely when compared with competitors. 

(d) Setting realistic goals for motivating entrepreneurs

Customer needs can be categorized into four types of needs. 

(a) Served Needs: These are needs that customers know and are fulfilled by different businesses or the government. 

(b) Partially-served Needs: These are needs which are served through different products or services, but the customer is not completely satisfied and still faces problems while using.

(c) Unserved and Known Needs: These needs are known by the customers, but not fulfilled by anyone in the market. 

(d) Unknown Needs: These are needs that people have, but are not aware or do not expect for it to get solved by a business.

The stakeholders are :  

•  Government 
•  The private agencies 
•  The people

Correct option(b) 15 Hz

Explanation:

The multiple of fundamental frequency gives 135, 165 Hz 

Hence taking LEAST COMMON MULTIPLE OF 135, 165 

135 = 5 × 3 × 3 × 3 

165 = 5 × 3 × 11 

SO 15 is least common

HENCE 

15 Hz is the fundamental frequency

Given 

m = 25 Kg

v = 3 m/s

P = mv 

p = 25 x 3

p = 75 kg m/s

m = 8 kg

v = 7 m/s

\(\Delta\)E = 1 / 2 mv² - 0

\(\Delta\)E = 1 / 2 x 8 x (7)²

\(\Delta\) E = 4 x 49

\(\Delta\) E = 196 J 

The resistance of the group of 3 resistors is equal to

R = 30 V / 7.5 A = 
4 ohm.

Let R₃ be the resistance of the third resistor, then

1/R = 
1/10 + 1/12 + 1/R₃
so,

1/4 = 1/10 + 1/12 + 1/R₃

1/R₃ = 1/4 
- 1/10 - 1/12 
= 15/60 - 6/60 - 5/60 
= 4/60
Hence 
R₃ = 60/4 = 15 ohm

Given P = 1500 WV = 250 V

i) Current drawn I = \(\frac{P}{V}=\frac{1500}{250}\) = 6A

ii) Electric energy consumed by gyser in 50 hours = Power × time
= 1500 × 50 = 75,000 Wh

= 75 kWh

Since 1 kWh = 1 unit

75 kWh = 75 units.

∴ Energy consumed = 75 units

iii) Cost of one unit = 6.00 Cost

75 units = 75 × 6.00

Rs. 450.00

Solution:

Let B’s share = x

A’s share = 2x/9

Ratio of shares of A and B is 2 : 9.

⇒ Sum of ratio terms = 2 + 9 = 11

Therefore A’s share = 2/11 × 1650 = Rs. 300

(a) The symbol ‘b’ represents impact parametter and ‘θ’ represents the scattering angle. 

(b) When θ= 0, the impart parameter will be maximum and represent the atomic size. 

(c) When θ =π radians, the impact parameter ‘b’ will be minimum and represent the nuclear size.

Cinnabar is HgS

Cassiterite is an ore of Sn

The process employed for the concentration of sulphide ore is Froth floatation

Correct answer is (D) Al

Co-ordination number of sodium ion Na⁺ in NaCl is  5

Glass is non-crystalline or amorphous.

It can be inferred from the given table that the ratio of voltage with current is a constant, which is equal to 19.7. Hence, manganin is an ohmic conductor i.e., the alloy obeys Ohm’s law. According to Ohm’s law, the ratio of voltage with current is the resistance of the conductor. Hence, the resistance of manganin is 19.7 Ω.

(a) Alloys of metals usually have greater resistivity than that of their constituent metals.
(b) Alloys usually have lower temperature coefficients of resistance than pure metals.
(c) The resistivity of the alloy, manganin, is nearly independent of increase of temperature.
(d) The resistivity of a typical insulator is greater than that of a metal by a factor of the order of 10^22.

Cause: Damage to liver cells leads to increase in bilirubin resulting in jaundice.

Prevention: Healthy diet and exercise.

Control: generous intake of water is necessary, Clear liquid including fruit juices, dal or rice water.

Microbes are tiny living things that are found all around us and are too small to be seen by the naked eye. They live in water, soil, and in the air. The human body is home to millions of these microbes too, also called microorganisms. Some microbes make us sick, others are important for our health.

Let E be the event that the number picked from 1 to 20 is a prime and S be the sample space.

∴ n(S) = ²⁰C₁ = 20

E = {2, 3, 5, 7, 11, 13, 17, 19}

n(E) = 8

P(E) = n(E)/n(S)  = 8/20  = 2/5

Let y = (x² + x + 1)/(x² - x + 1)

⇒ x²y – xy + y = x² + x + 1

⇒ x²y – xy + y – x² – x – 1 = 0

⇒ x²(y – 1) – x(y + 1) + (y – 1) = 0

x is real ⇒ b² – 4ac ≥ 0

⇒ (y + 1)² – 4(y – 1)² ≥ 0

⇒ (y + 1)² – (2y – 2)² ≥ 0

⇒ (y + 1 + 2y – 2) (y + 1 – 2y + 2) ≥ 0

⇒ (3y – 1) (–y + 3) ≥ 0

⇒ –(3y – 1) (y – 3) ≥ 0

a = coeff. of y² = –3 < 0.,But

The expression ≥ 0

⇒ y lies between 1/3 and 3

∴ The range of (x² + x + 1)/(x² - x + 1) is [1/3, 3]

CONCEPT:

If P represents the nonzero complex number z = x + iy.

Here \(r = \sqrt {{x^2} + {y^2}} = \left| z \right|\) is called modulus of given complex number.

CALCULATION:

Given:  \(tanθ = \frac{3}{4}\:\)

As we know that, if z = x + iy then x = r cos θ and y = r sin θ

⇒ \(tan\ θ = \frac{y}{x}\:\)

⇒ \(tanθ = \frac{3}{4}\:= \frac{y}{x}\)

\(\therefore r = \sqrt {{{\left( x \right)}^2} + {{\left( y \right)}^2}} = \sqrt {{{\left( 4 \right)}^2} + {{\left( 3 \right)}^2}} = 5\)

\( ⇒ cosθ = \frac{x}{r} = \frac{4}{5};\;sinθ = \frac{y}{r} = \frac{3}{5}\)

\(\therefore z = r\left( {cosθ + i\;sinθ } \right) = 5\left( {\frac{4}{5} + \frac{3}{5}i} \right)\)

⇒ z = 4 + 3i

Concept:

Let z = x + iy be a complex number.

Modulus of z = \(\left| {\rm{z}} \right| = \sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} = \sqrt {{\rm{Re}}{{\left( {\rm{z}} \right)}^2} + {\rm{\;Im\;}}{{\left( {\rm{z}} \right)}^2}} \)

\(\left| {\frac{{{{\rm{z}}_1}}}{{{{\rm{z}}_2}}}} \right| = {\rm{\;}}\frac{{\left| {{{\rm{z}}_1}} \right|}}{{\left| {{{\rm{z}}_2}} \right|}},{\rm{\;}}\left| {{{\rm{z}}_2}} \right| \ne 0\)

Equation of a circle

• The equation of a circle whose centre is at a point having affix z₀ and radius r is |z - z₀| = r
• If the centre of the circle is at origin and radius r, then its equation is |z| = r

Calculation:

Given: |z| = 1

Let z' = \(\rm \frac 2 z\)

Taking the mod of both sides, we get

⇒ |z'| = \(\rm \left| \frac 2 z \right|\)

⇒ |z'| = \(\rm \frac {|2|}{| z|}\)

⇒ |z'| = \(\frac 21\)            (∵  |z| = 1)

⇒ |z'| = 2

it will be a circle of radius 2 units and with centre as origin.

Concept:

 Let z  = x + iy, then \( {\bar z} \) is the conjugate of a complex number given by,

\(\rm {\bar z} = x - iy\)

Calculations:

Let z  = x + iy lies on |z| = 2

⇒ \(\rm \sqrt {x^2+y^2} = 2\)

⇒ \(\rm {x^2+y^2} = 4\), which is a circle with centre at (0, 0) and radius is 2.

 Let z  = x + iy, then \( {\bar z} \) is the conjugate of a complex number given by,

\(\rm {\bar z} = x - iy\)

Now, \(\rm \dfrac 4 {\bar z} = \dfrac {4}{x -iy}\)

 \(⇒ \rm \dfrac 4 {\bar z} = \dfrac {4}{x -iy} \times \dfrac {x+iy}{x+iy}\)

\(⇒ \rm \dfrac 4 {\bar z} = \dfrac {4(x+iy)}{x^2+y^2}\)
\(⇒ \rm \dfrac 4 {\bar z} = \dfrac {4z}{4}\)          (∵ \(\rm {x^2+y^2} = 4\))
\(⇒ \rm \dfrac 4 {\bar z} = z\)

Taking mod both sides, we get

\(⇒ \rm \left|\dfrac 4 {\bar z} \right|= |z|\)

Let \( \rm \dfrac 4 {\bar z} = z'\)

Hence, |z'| = 2            (∵  |z| = 2)

|z'| lies on the circle.

Concept:

The cube-root of unity 'ω' follows:

• ω3 = 1
• 1 + ω + ω2 = 0
• ω3n = 1
• ω3n + 1 = ω
• ω3n + 2 = ω2

Calculation:

S = ω¹⁷

S = ω^{3 × 5} .ω²

S = 1 × ω² = ω2

Concept:

Identities:

• i² = -1
• i³ = -i
• i⁴ = 1
• i⁴ⁿ = 1
• i4n+1 = i
• i4n+2 = -1
• i4n+3 = -i

Calculation:

Let S = i^-1245 = \(\rm 1\over i^{1245}\)

S = \(\rm 1\over i^{1244 +1}\)

S = \(\rm 1\over i^{4(311)+1}\)

S = \(\rm 1\over i\)

Multiply i both numerator and denominator, we get

S = \(\rm i\over i^{2}\)

S = -i

Concept:

1, ω and ω2 are cube root of unity

Where ω = \( \rm \frac{-1+\sqrt3i}{2}\) , ω² = \( \rm \frac{-1-\sqrt3i}{2}\) , ω³ = 1

Calculation:

\(\left(\dfrac{\sqrt{3}+i}{\sqrt{3}-i}\right)^6\)

\(= \rm (\dfrac{\sqrt{3}+i}{\sqrt{3}-i}\times\dfrac{\sqrt{3}+i}{\sqrt{3}+i} )^6\)

\(= \rm (\dfrac{(\sqrt{3}+i)^2}{3+1})^6\)

\(= \rm (\dfrac{({3}-1+2\sqrt3i)}{4})^6\)

\(= \rm (\dfrac{2(1+\sqrt3i)}{4})^6\)

\(= \rm (\dfrac{1+\sqrt3i}{2})^6\)

=  (-ω²)⁶                                    (∵ -ω² = \( \rm \dfrac{1+\sqrt3i}{2}\) )

=  (ω³)⁴

= 1

Concept:

Properties of cube root of unity:

ω3 = 1 and (ω2 + ω + 1) = 0

Calculations:

Consider, ω10 + ω-10

= ω10 + \(\dfrac {1}{ω^{10}}\)

= \(\dfrac {ω^{20}+1}{ω^{10}}\)

= \(\dfrac {ω^{18}.ω^{2}+1}{ω^{9.}ω}\)

= \(\dfrac {{(ω^3)^6}.ω^{2}+1}{(ω^3)^3ω}\)

Hence, if ω is a complex cube root of unity, then  ω10 + ω-10 equal to -1.

Concept:

The polar form of the complex number z = (x + iy) is 

⇒ z = r (\(\rm \cos\;\theta + i sin\;\theta\)) , where r = \(\sqrt {x^2+y^2}\) and \(\rm \theta = tan^{-1}(\dfrac y x)\)

De - Moivers thereom:

Let z = r (\(\rm \cos\;\theta + i sin\;\theta\)) be the polar form of the complex number z = x + iy.

then the value of zⁿ = [r (\(\rm \cos\;\theta + i sin\;\theta\))]ⁿ is rⁿ(\(\rm \cos\;n\theta + i sin\;n\theta\))

Calculations:

Given, complex number is  \((-1+i\sqrt{3})^{48} \)

Consider z = (x + iy) =\((-1+i\sqrt{3}).\) 

⇒ x = 1 and y = \(\sqrt 3\)

The polar form of the complex number is 

⇒ z = r (\(\rm \cos\;\theta + i sin\;\theta\)) , where r = \(\sqrt {x^2+y^2}\) and \(\rm \theta = tan^{-1}(\dfrac y x)\)

Here r = \(\sqrt {1^2+(\sqrt3)^2}\) = 2 and \(\rm \theta = tan^{-1}(\dfrac {\sqrt3}{1})\) = \(\dfrac{\pi}{3}\)

⇒ z = 2 (\(\rm \cos\;\dfrac {\pi}{3} + i sin\;\dfrac{\pi}{3}\))

⇒ z⁴⁸ = [2 (\(\rm \cos\;\dfrac {\pi}{3} + i sin\;\dfrac{\pi}{3}\))]⁴⁸

⇒ z48 = 2⁴⁸ (\(\rm \cos\;(48\times\dfrac {\pi}{3}) + i sin\;(48\times\dfrac{\pi}{3})\)

⇒ z48 = 248 [\(\rm \cos\;(16 {\pi}) + i sin\;(16{\pi})\)]

⇒ z48 = 248 [\(\rm [1 + i(0)]\)]

⇒ z48 = 248

Hence, the value of \((-1+i\sqrt{3})^{48} \) = 248

Concept:

Sum of infinite terms in geometric progression is given by, \(\rm S_\infty= \frac{a}{1-r}\), where, a = first term and r = common ratio

i = \(\sqrt {-1}\), i² = -1

Calculation: 

Given series is (1 + 2i2 + 4i4 + 8i6+.....) 

Clearly it is a GP, with a = 1 and r = 2i²

Now sum = \(\rm S_\infty= \frac{a}{1-r}\)

 \(\rm = \frac{1}{1-2i^2} \\=\frac{1}{1+2}\)

= \(\rm \frac1 3\)

Hence, option (2) is correct.

Equation of parabola is x² = 4ay 

Here parabola passing through(5, 2) 

∴ 5² = 4a(2) 4a = 25/2

∴ Equation is x² = (25/2)y

3x² = -8y ⇒ x² = \(\frac{8}{3}\)y 

Length of L.R = 4a = \(\frac{8}{3}\)

Given Directrix As x = -4 & Focus = (4,0) ⇒ a = 4 

The standard equation is y² = 4ax 

⇒ y² = 4 × 4 x X = 16x 

∴ Equation of the parabola = 16x

4k = 8 ⇒ k = 2

It is given that vertex of the parabola V(0,0)and symmetric about y – axis so its equation is x² = 4ay or x² = – 4ay, 

since the parabola passess through the point \(p(\frac{1}{2}, 2)\) so it lies in the 1st quadrant. 

∴ It is an upward parabola x² = 4ay 

∴ It passess though the point \(p(\frac{1}{2}, 2)\) we have

\((\frac{1}{2})^2 = 4 \times a\times2\) ⇒ a = \(\frac{1}{32}\)

∴ Required parabola is x² = 4 ⇒ x² = \(\frac{1}{8}y\)