Explore topic-wise InterviewSolutions in Current Affairs.

(1) The inflated waist of the dog showed that the dog ate a lot those days.

(2) The dog kept the same hours as humans for going to bed, i.e., at ten o’clock at night.

(3) The dog tried a walk before bedtime. He tried a hot drink. He tried a full meal. He also tried no meal to promote sleep.

(4) The sure remedy recommended by doctors for sleeplessness is counting imaginary sheep.

(5) This passage is about a doctor examining a patient.

Some of the changes in the international trade in the last fifteen years are as follows:
a). Bulk imports as a group registered a growth accounting for 39.09 per cent of total imports. This group includes fertilizers (67.01 per cent), cereals (25.23 per cent), edible oils (7.94 per cent) and newsprint (5.51 per cent).
b). International trade has under gone a sea change in the last fifteen years. Exchange of commodities and goods have been superseded by the exchange of information and knowledge.
c). India has emerged as a software giant at the international level and it is earning large foreign exchange through the export of information technology.

M.L. Jaisimha

There are eleven players who actually play the game of cricket and five are substitutes.

Correct option  (4) None of the above statement is correct

Explanation:

Resistivity of a wire is independent of dimensions of wire.

Correct option  (4) Drift velocity will decrease

Explanation:

I = same

Area changes

V_d decreases

J = 1/A(change) 

Let,

S=1+w+w²+w³+w⁴+.....

As we know that 1+w+w²=0 then,

S=1+w+w²+w(1+w+w²)+w²(1+w+w²)+...

S=0+w×0+w²×0+w³×0+....

S=0+0+0+0+0+...

S=0

Hence final answer is zero.

A + B + C = 3π/2 ---------(1) 

L.H.S = cos 2A + cos 2B + cos 2C 

= 2 cos (A + B). cos (A – B) + 1 – 2 sin² C 

= 2 cos (270° – C). cos (A – B) – 2 sin² C 

= 1 – 2 sin C [cos (A – B) –2 sin² C 

= 1 – 2 sin C[cos (A – B) + sin C] 

= 1 – 2 sin C [cos (A – B) + sin (270° - bar(A + B))]

= 1 – 2 sin C [cos (A – B) – cos (A + B)]

= 1 – 2 sin C [2 sin A sin B] 

= 1 – 4 sin A sin B sin C

Equation of the chord of contact is S₁ = 0 

i.e., xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0 

Equation of the circle is 

S = x² + y² – 5x + 4y – 2 = 0 

Equation of the chord of contact is 

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

Multiplying with 2 

10y – 5x + 4y + 20 – 4 = 0 

– 5x + 14y + 16 = 0 

or 5x – 14y – 16 = 0

We have given,

Weight of element M = 35.3g

Weight of compound M₃N₂ = 43.5g

3M + N2 → M₃N₂ (balance equation)

From above balance equation, use can clearly see that 3 mole of element M react with one mole of Nitrogen to form one mole M₃N₂.

Let say. molar mass of element M = x g.

∴ Number of moles of element M = \(\frac {35.3}{x}\)mole and 

Number of moles of M₃N₂ =  \(\frac {43.5}{28 + 3x}\) mole.

∴ 3 mole of element produce = 1 mole M₃N₂

∴ 1 mole of element M produce = \(\frac 13\) mole of M₃N₂

∴ \(\frac {35.3}{x}\) mole of element M produce = \(\frac {35.3}{3x}\) mole of  M₃N₂ 

but \(\frac {43.5}{28 + 3x}\) mole of M₃N₂ formed

Therefore - \(\frac {43.5}{28 + 3x}\) = \(\frac {35.3}{3x}\)

= 130.5x = 988.4 + 105.9 x

= 24.6x = 988.4

x = 40.18g.

i) Hence, the molar mass of element M will be 40.18g

ii) We know that, change on Nitrogen is -3.

∴ Oxidation state of M will be

= Let say - M has oxidation state  y then,

3y + 2 (-3) = 0

= 3y = 6

= y = +2

it means element has oxidation state (valency) +2 and molar mass 40.17g.

We know that Ca has valency +2 and it has molar mass approx 40.078

Therefor compound will be Ca₃N₂ 

And the name of compound is Calcium nitride.

A 35.3 g of element M is reacted with nitrogen to produce 43.5 g of compound M₃N₂.

Generally, the equation for the Chemical Reaction  is mathematically given as

3M+N₂=M₃N₂

Therefore

moles of N₂ is given as

m= mass of N₂/molar mass

Where

mass of N₂=43.5-35.3

mass of N₂=8.2

Therefore

m =8.2/32

m=0.25625mole

Since 2 moles of N₂ reacts with 3mols of M

Hence, 0.25625 mole of N₂ reacts with 3mole of M

x=3/2*0.256

x=0.384mol

In conclusion, the molar mass of M

M = mass of x/mole of x

M = 35.3/0.384

M=91.927g/mol

\(\frac{a+b\omega+c\omega^2}{c+a\omega+b\omega^2}\) = \(\frac{a+b/\omega^2+c\omega^2}{c+a/\omega^2+b\omega^2}\) (\(\because\omega^3=1, \omega=1/\omega^2\))

= \(\frac{a\omega^2+b+c\omega^4}{c\omega^2+a+b\omega^4}\)\(=\frac{b+\omega^2(a+c\omega^2)}{a+\omega(c+b\omega^2)}\)

Tell your question properly.

Do'nt mess with others

Given:

Denominations 50, 100, 200, 500 and 2,000

Calculation:

Let each denomination is single

Total denominations = 5

According to the question:

Taking at least three denominations at a time

⇒ 3 denominations, 4 denominations, 5 denominations

⇒ 3 denominations out of 5 in ⁵C₃ = 10 ways

⇒ 4 denominations out of 5 in 5C₄ = 5 ways

⇒ 5 denominations out of 5 in 5C₅ = 1 way

Total ways = 10 + 5 + 1 = 16

∴ The answer is 16.