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(a) Three resistances are connected in star. Each resistance is of 100 ohms and 254 – V appears across it. Hence, a current of 2.54 A flows through the resistors and the concerned power-factor is unity. Due to star-connection,

 Line-current = Phase-current = 2.54A 

(b) Three inductive reactance are delta connected. 

Line-Voltage = Phase – Voltage = 440 V 

Phase Current = 440/100 = 4.4 A 

Line current = 1.732 × 4.4 = 7.62 A

The current has a zero lagging power-factor.

Total Line Current = 2.54 - j 7.62 A

= 8.032 A, in each of the lines. 

Power factor = 2.54/8.032 = 0.32 Lag.

Voltage per phase = 1100/1.732 = 635 V 

Impedance = 635/80 = 7.94 ohms. 

Due to the leading current, a capacitor exists. 

Resistance R can be evaluated from current and power consumed 

3I²R = 100 × 1000, giving R = 5.21 ohms 

X_c = (7.942 – 5.212)^0.5 = 6 ohms 

At 50 Hz, C = 1/(314 × 6) = 531 micro farads.

Load kVA = 200/0.8 = 250 

Load kVAR = 250 × 0.6 = 150 (lag) [cos φ = 0.8 sin φ = 0.6] 

Total combined load = 50 + 200 = 250 kW 

kVA of combined load = 250/0.9 = 277.8 

Combined kVAR = 277.8 × 0.4356 = 121 (inductive) (combined cos φ = 0.9, sin φ = 0.4356) 

Hence, leading kVAR supplied by synch, motor = 150 – 121 = 29 (capacitive)

kVA of motor alone = √(kW² + kVAR²) = √(50² + 29²) = 57.8

p.f. of motor = kW/kVA = 50/57.8 = 0.865 (leading)

tanφ = (√3(W₁ - W₂))/(W₁ + W₂) ....(i)

Now, cosφ = 0.4; φ cos^-1(0.4) = 66.6°; tan 66.6° = 2.311

W₁ + W₂ = 30  ....(ii)

Substituting these values in equation (i) above, we get

2.311 = (√3(W₁ - W₂))/30

∴ W₁ - W₂ = 40 .... (iii)

From Eq. (ii) and (iii), we have W₁ = 45 kW and W₂ = – 5 kW 

Since W₂ comes out to be negative, second watt meter reads ‘down scale’. Even otherwise it is obvious that p.f. being less than 0.5, W₂ must be negative.

We are given that W₁ :W₂ = 2:1.

Hence, W₁/W₂ = r = 1/2 = 0.5.

cosφ = (1 + r)/(2√(1 - r + r²)) = (1 + 0.5)/(2√(1 - 0.5 + 0.5²)) = 0.866 lag

As given in, when W₂ reads negative, then we have

tanφ = √3(W₁ + W₂)/(W₁ - W₂).

Substituting numerical values of W₁ and W₂, we get

tanφ = √3(10.4 + 3.4)/(10.4 - 3.4) = 1.97; φ = tan^-1(1.97) = 63.1°

(a) p.f. = cos cos . φ= ° 631 = 0.45 (lag)

(b) W = 10.4 – 3.4 = 7 KW = 7,000 W

7000 = √3I_L x 400 x 0.45; I_L = 22.4A

(a) Recombination of charge carriers

(d) n

P = mv = \(\frac{nh}{2a}\); V ∝ n

(c) The change in the crystal structure

Option : (B)

Since p–n junction is forward biased, 

Its resistance can be taken to be zero. 

Thus,

I = \(\frac{v}{R}\) 

= \(\frac{5v-2v}{3000Ω}\) 

= 10 mA

So, 

Both assertion and reason are true but the reason is not the correct explanation of assertion.

When a p-n junction is reverse biased, 

A very small current (of the order of few µA) flows due to drifting of minority charge carriers whose number density remains constant upto the critical voltage.

As a result of it,

The current under a reverse bias is almost independent of the applied potential upto critical voltage.

In case of n-type semi- conductors, n>p, 

Where n = majority carrier density 

p = minority carrier density 

On illumination of semi-conductor, there will be production of equal number of electrons and holes. If Δn and Δp be the increase in majority carrier density and minority density carrier density due to illumination of semiconductor, then

fractional change in majority carrier = \(\frac{Δn}{n}\), and 

fractional change in minority carrier = \(\frac{Δp}{p}\) 

Since, 

n>>p,

So,

\(\frac{Δn}{n}\) < \(\frac{Δp}{p}\),

It means, due to photo-effects the fractional change due to minority carriers dominates. As a result of it, the fractional change in the reverse bias current is more easily measurable then the fractional change in the forward bias current. It is due to this reason, photodiodes are preferably used in the reverse bias condition for measuring light intensity.

When an intrinsic semi- conductor is doped with the impurity atoms of valency three like indium or boron, some allowed energy levels are produced, situated in the energy gap slightly above the valence band. These levels are called acceptor energy levels.

Given relation R is defined on L, where L is the set of all lines in a plane and defined by 

R = {(L₁, L₂ ) ∶ L₁ is perpendicular to L₂ }. 

Now, let L₁, L₂ ∈ L such that (L₁, L₂ ) ∈ R. 

⇒ L₁ is perpendicular to L₂. 

⇒ L₂ is perpendicular to L₁. 

⇒ (L₂, L₁ ) ∈ R ∀ L₁, L₂ ∈ I

Hence, relation R is symmetric.

Which is easily available and burns in air at a moderate rate, producing large amount of heat energy, without leaving behind any undesirable residue is called a fuel. 

(i) It should have high calorific value.

(ii) It should have a proper ignition temperature so that it may burn easily. 

(iii) It should leave no residue (or very small amount of residue) or ash after burning.

Given θ = 90°

(a) Compton effect scattered wavelength incident wavelength compton shift

Δλ \(=\frac h{m_ec}\) (1 - cos θ)

Δλ \(=\frac h{m_ec}\) (1 - cos 90°)

Δλ \(=\frac h{m_ec}\) (1 - 0)

 Δλ \(=\frac h{m_ec}\) 

where h = plank constant

c = speed of light

m_e = mass of electron

λ_f - λ_i \(=\frac h{m_ec}\) (1 - cos θ)  (θ = 90°, cos 90° = 0)

λ_f - 0.2 x 10^-9 \(=\frac h{m_ec}\)

λ_f - 0.2 x 10^-9 \(=\frac{6.63\times10^{-34}}{9.1\times10^{-31}\times3\times10^8}\)

λ_f = 0.2 x 10^-9 + 0.24 x 10^-11

λ_f = 0.20 x 10^-10 + 0.024 x 10^-10

λ_f = 0.224 x 10^-10

λ_f = 0.0224 x 10^-9 m

Compton shift Δλ = λ_f - λ_i

Δλ = 0.0224 - 0.2

Δλ = -0.1776 nm

(b) Kinetic energy imparted to the recoiling electron.

E_k \(=\frac{h^2}{2m\lambda^2}\)

E_k \(=\frac{(6.64\times10^{-34})^2}{2\times9.1\times10^{-31}\times(0.2\times10^{-9})^2}\)

E_k \(=\frac{44.08\times10^{-68}}{0.728\times10^{-18}\times10^{-31}}\)

E_k = 60.54 x 10^-19 J

K.E = 37.83 eV

The correct option (2) (~ p) ∧ (~ q)

Explanation:

(ii) (~ p) ∧ (~ q)

n(A) = 25

n(B) = 7

n(A ∩ B) = 3

n(A ∪ B) = 25 + 7 - 3 = 29

n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

= 25 + 7 – 3 = 29

Option (1), (2), (3) are incorrect for

ƒ(x) = constant and option (4) is incorrect

(f(1) - f(c))/(1 - c) = f'(a) where c < a < 1 (use LMVT)

Also for f(x) = x²  option (4) is incorrect.

The correct option is (3) |f(c) + f(1)| < (1 + c)| f'(c)|

Use LMVT theorem & check option.

n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

= 25 + 7 - 3

= 29

Answer is (2)

f(g(x)) = x

⇒ f'(g(x)).g'(x) = 1

Put x = b

⇒ f'(g(b)) g'(b) = 1

⇒ f'(a) x 5 = 1

⇒ f'(a) = 1/5

Answer is (3)

y = mx + 4 ……(i)

y² = 4x tangent y = mx + a/m

⇒ y = mx + 1/m .....(ii)

from (i) and (ii) 

4 = 1/m ⇒ m = 1/4

So line y = 1/4x + 4 s also tangent to parabola

x² = 2by, so solve and D = 0

x² = 2b(x + 16/4) 

⇒ 2x² – bx – 16b = 0 ⇒ D = 0 

⇒ b² – 4 × 2 × (–16b) = 0

⇒ b² + 32 × 4b = 0 

b = –128, b = 0 (not possible)

Answer is (3) 1/5

f(g(x)) = x

f'(g(x)) g'(x) = 1

put x = a

f'(b) g'(a) = 1

f'(b) = 1/5

Answer is (3) 1/5

f(g(x)) = x

f'(g(x)).g'(x) = 1

x = a

f'(g(a)) - g'(a) = 1

f'(b) = 1/5

Answer is (4)

Let (1/2, - 2) is (2t², 4t) ⇒ t = -1/2

Parameter of other end of focal chord is 2

⇒  point is (8, 8)

⇒ equation of tangent is 8y – 4(x + 8) = 0

⇒ 2y – x = 8 

Answer is  (1) 64/5

L = √S₁ = √16 = 4

R = √(16 + 4 - 16) = 2

Length of Chord of contact = 2LR/√(L² + R²) = √(2 x 4 x 2)/√(16 + 4) = 16/√20

square of length of chord of contac = 64/5

Answer is (2) x(dy/dx)² = 2y(dy/dx) + x

2x = 4by' ⇒ b = x/2y'

So. differential equation is x² = 2x/y, . y + (x/y)²

⇒ x(dy/dx)² = 2y(dy/dx) + x

Answer is (4) (1, 1, –1)

d.r of normal to the plane

10/3, 10/3, 10/3

1, 1, 1

midpoint of P and Q is (-2/3, 1/3, 4/3)

equation of plane x + y + z = 1 

Answer is (4) 0

Using L’Hospital

lim_(x →0) (xsin(10x))/1 = 0

(b) \(\int_0^{a}\)f(a - x) dx

(c) \(\int\limits_0^{2a}\)f(x) dx