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(b) Acceptable Quality Level (AQL) 

(a) It is paramagnetic 

(b) its bond order is two 

(c) In liquid state it is blue coloured 

(d) It has two unqaired electrons 

(b) Both statement (I) and Statement (II) are individually true but Statement (II) is NOT the correct explanation of Statement (I) 

Statement I is condition for fanno line and statement II is condition for Rayleigh line but statement II not explaining the statement I 

(a) hygroscopic agent 

(b) oxidizing agent 

(c) sulphonating agent 

Nitrogen does not possess 2d - subshell and it cannot excite its 2s paired electron to get unpaired whereas phosphorus does so on account of availability of 3d - subshell. 

The high boiling point and viscosity of sulphuric acid are due to the presence of hydrogen bonding which binds a number of simple molecules into clusters.  

3MnO₄ ^–2 + 2H₂O → MnO₂ + 2MnO₄ ^– + 4OH^–

c).  Assertion is true but reason is false

(a) equal to the sonic speed 

c). F- center

The correct order of increasing nucleophilicity is Cl^- < Br^- < I^-

a). H₂O > H₂S > H₂Se > H₂Te

a). [Cr(NH₃)₆]³⁺

c). entropy increases

b). Methyl cyanide

α -D glucose and β -D glucose are epimers

A blood cell retain their shape in solution which are isotonic to blood

rt - s² > 0 and r > 0 is the condition of point of minima for function z = f(x, y) at stationary point.

Calculation:

The series follows following pattern

⇒ 2 × 0.5 = 1

⇒ 1 × 1 = 1

⇒ 1 × 1.5 = 1.5

⇒ 1.5 × 2 = 3

⇒ 3 × 2.5 = 7.5

∴ The value of ? is 7.5

Given:

p : p²  :: p² : 1000

p² + 2pn + n² = 21

Calculation:

From statement 1

p/p² = p²/1000

⇒ 1/p = p²/1000

⇒ p³ = 1000

⇒ p = 10

From statement 1, we can find the value of p

From statement 2

We can't find the value of p because we can't get the value of two variable from one equation

∴ Only 1 is sufficient

Total no. of families considered = 2400 

(i) P(a family earning Rs 10000 – 13000 per month and owning exactly 2 vehicles) 

= No. of families earning Rs 10000 – 13000 per month and owning 2 vehicles /Total no. of families = 29 / 2400

(ii) P (a family earning Rs 16000 or more per month and owning exactly 1 vehicle)

= No. of families earning Rs 16000 or more per month and owning 1 vehicle/ Total no. of families = 579/ 193 = 2400/ 800

(iii) P(a family earning less than Rs 7000 per month and does not own any vehicle) 

= No. of families earning less than Rs 7000 per month and does not own any vehicle /Total no. of families = 10 / 2400 = 1 / 240

(iv) P(a family earning Rs 13000 – 16000 per month and owing more than 2 vehicles) 

= No. of families earning Rs 13000 – 16000 per month and owning more than 2 vehicles / Total no. of families 

= 25/ 2400 = 1 /96

(v) P (a family owning 0 vehicle or 1 vehicle) = P (a family not owning more than 1 vehicle)

= (10 +0+ 1+ 2+ 1+ 160+ 305+ 535+ 469+ 579) / 2400  

= 2062 /  2400 = 1031/ 1200

Given:

A is 20% more than B, B is 25% more than C, C is 50% less than D and D is 10% more than E.

Concept used:

Ratio and proportion

Calculation:

A is 20% more than B 

⇒ A : B = 6 : 5

B is 25% more than C

⇒ B : C = 5 : 4

C is 50% less than D

⇒ C : D = 1 : 2

D is 10% more than E

⇒ D : E = 11 : 10

Combining all the ratios

A : B : C : D : E = 330 : 275 : 220 : 440 : 400

Now going through all the options

A : E = 330 : 400 

A is less than E by 70

Percentage = \(\frac{{70}}{{400}} \times 100 = 17.5\% \)

Option A is true.

Given

7 : 15 :: 350 : x

Concept

Product of extremes terms = Product of middle terms

Calculations

7 × x = 15 × 350

x = (15 × 350)/7

x = 750

Concept used:

Mean proportion of 'a' and 'b' = √ab 

Calculations:

Mean proportion of 6.25 and 6.4 = √(6.25 × 0.64)

⇒ 2.5 × 0.8 = 2

∴ Mean proportion is 2 

Given:

The ratio between the present ages of A and B is 3 : 5.

The ratio of their ages five years after becomes 13 : 20.

Calculation:

Let the proportional ratio be x

According to the question,

(3x + 5)/(5x + 5) = 13/20

⇒ 60x + 100 = 65x + 65

⇒ 5x = 35

⇒ x = 7

The present age of B = 5x = 5 × 7 = 35

∴ The present age of B is 35 years.

Given:

B is twice as old as A today

In 10 years, the ratio of ages of A and B will be 2 : 3

Calculation:

B is twice as old as A today

⇒ B = 2A

⇒ B/A = 2/1

Let the present age of A be ‘x’

⇒ B = 2x, A = x

In 10 years ratio of A : B becomes 2 : 3

⇒ (x + 10)/(2x + 10) = 2/3

⇒ 3x + 30 = 4x + 20

⇒ x = 10

Age of B

⇒ 2 × 10

⇒ 20 years

The correct option is : (D) 5:4

The correct option is : (C) 36°

The correct option is : (D) 3 : 1

Diagonal of square = 20 cm.

Let side of square = x

∴ x² + x² = 20² ......(By Pythagoras theorem)

∴ 2x² = 400

∴ x² = 200

∴ x = 10 √2 cm. 

Perimeter of square = 4 × 10 √2 = 40 √2 

(i) Side of square = 10 √2 cm. 

(ii) Perimeter of square = 40 √2 cm. 

In figure, PQ = 12, PR = 8

PQ² = PR × PS .......(Tangent secant theorem)

∴ 12² = 8 × PS 

∴ 144 = 8 × PS 

∴ PS = 144 / 8 

∴ PS = 18

Correct option is (1) 4

9¹⁰¹¹ = (10 - 1)¹⁰¹¹ = 10λ - 1 = 5\(\mu\) - 1

⇒ remainder = 4

The current can be written as i = 10√2 sin (314 t − 25° + 90°) = 10√2 sin (314 t + 65°). 

It is seen that applied voltage leads by 20° and current leads by 65° with regards to the reference quantity, their mutual phase difference is = 65° −(20°) = 45°. 

Hence, p.f. = cos 45° = 1/ √2 (lead). 

Now, V_m = 200√2 and  I_m =  10√2

∴ Z = V_m/I_m = 200 √2/10√2 = 20 Ω 

R = Z cos φ = 20/√2Ω = 14.1 Ω ; X_c = Z sin φ = 20√2 = 14.1 Ω 

Now, f = 314/2π = 50 Hz. Also, X_c = 1/2π fC 

∴ C = 1/2π × 50 × 14.1 = 226μF 

Hence, the given circuit is an R-C circuit.

Given f(z) = xy + iy

u = xy, v = y

x and y are continuous functions ,therefore u and v are also continuous.

But u = xy, v = y

u_x = y, v_x = 0

u_y = x, v_y = 1

u_x ≠ v_y and v_x - u_y

C-R equations are not satisfied. 

Hence f(z) is not differentiable anywhere though it is continuous everywhere.

Let f(z) = z/(z - 2)

z = 2 lies outside c. 

∴ f(z) is analytic inside and on c . 

f'(z) is continuous inside c 

Hence by Cauchy’s theorem R c f(z)dz = 0