Explore topic-wise InterviewSolutions in Current Affairs.

y = sin(x² + 5)

dy/dx = cos(x² + 5)(2x)

किसी भी दिशा में वस्तु द्वारा एकांक समय में तय की गई दूरी को चाल होती हैं। एक निश्चित दिशा में वस्तु द्वारा एकांक समय में तय की गई दूरी को वेग होता हैं। चाल एक अदिश राशि है। वेग एक सदिश राशि है।

(c) Negative feedback

(b) Reduces distortion

(b) Emitter feedback resistance

(b) Current source

(a) Diff amp

(d) AC collector resistance

(d) BIFET op amp

Correct Answer is: (d) zero

For x > X₀, V is constant. Hence, force = 0 and therefore acceleration = 0.

(a) First stage

Correct Answer is: (b) 1/A √(m/α)

V = αx⁴,  ∴ [α] = ML^-2 T^-2.

The only expression which has the dimension M⁰ L⁰ T is (b).

(c) Voltage-to-current converter

The correct answer is: (b) Beta dc

(d) Tolerance of resistors

Correct Answer is: (c) V₀ > E > 0 

For an oscillating system. The kinetic energy must be zero periodically, for a finite value of x. Also, E < V₀ .

Correct Answer is: (a, c)

Error in measuring, T = 2s/40 = 0.05 s = ΔT

T ∝ 1/√g Δg/g = 2ΔT/T = 1/20 Δg/g x 100% = 5%.

Correct Answer is: (a) |Q_1| > |Q₂|, (d) at a finite distance to the right of Q₂ the electric field is zero

Q₁ is positive, Q₂ is negative. Also,|Q₁| > |Q₂|.

Lines of force start from a positive charge and end on a negative charge, and are denser near a larger charge.

Correct Answer is: (a) Internal energies at A and B are the same, (b) Work done by the gas in process AB is P₀ V₀ ln 4

Internal energy ∝ absolute temperature.

∴ U_A = U_B .

In the isothermal process AB,

W = nRT In ( V_f / V_i ) = p₀ V₀ In( V_B / V_A )

(a) bar (A .B) = γ

Correct answer is (d) I₀ = I_rms/√2

(b) Concave Lens

Correct answer is (C) 56 km

Correct answer is (c) 56 km

Eddy current- Eddy currents are loop of electrical current induced with in conductors by a changing magnetic field in the conductor due to Faraday's law of induction. Eddy current flow in closed loop with in conductors in plane perpendicular to the magnetic field. They can be induced in transformer

Use of eddy currents arc following-

(i) Eddy current use the drag force created by eddy current as a brake to slow or stop moving objects. 

(ii) In indentification of metals.

Let vertical component of Earth = B_v

and horizontal component of Earth = B_H

According to Question B_H = B_v√3

So, Angle of dip = 30°

(a) increase of temperature

(a) We can't calculate the value of resistance of galvanometer by this method

Resistance from half deflection method in given by

R_g = RS/R-S

if we do 1/3^rd of deflection then R_g = 2 RS/R-2S

which is not related of actual resistance of galvanometer.

Correct option (a) 1.22 and 0.1

Explanation:

Return loss (dB) = - 20log₁₀|ρ|

Where ρ is the reflection coefficient.

For |ρ| = 1 full reflection

Return loss = 0 dB

If |ρ| = 0.1

R. Loss (dB) = - 20log₁₀(0.1)

= - 20 x (- 1)

= 20dB

VSWR = (1 + |ρ|)/(1 - |ρ|)

= (1 + 0.1)/(1 - 0.1) = 1.1/0.9 = 1.22

Correct option (b) 25%

Explanation:

VSWR = (1 + |Γ|)/(1 - |Γ|)

3 = (1 + |Γ|)/(1 - |Γ|)

Γ = 0.5

P_r/P_i = Γ² = (0.5)²

= 0.25

25% of incident power is reflected.

Correct option (d)1 and ∞

Explanation:

 VSWR = (1 + |ρ|)/(1 - |ρ|)

Where ρ is reflection coefficient ρ can take values between 0 and 1 when ρ = 0, VSWR = 1

ρ = 1, VSWR =  ∞

Correct option (d)0.002

Explanation:

Attenuation constant α to be independent of frequency for distortion less transmission α = √(RG)

For distortion less transmission:

L/C = R/G

z₀ = √(L/C) = √(R/G)

α = √(RG) = √R(√R/z₀)

= R/z₀

= 0.1/50

= 0.002Np/m

Correct answer is (12)

 \(l=tsin\, i\left[1-\frac{cos i}{\sqrt{\mu^2-sin^2i}}\right] \)

\(\Rightarrow 4\sqrt3=t sin 60^{\circ}\left[1-\cfrac{cos60^{\circ}}{\sqrt{3-\frac34}}\right] \)

\(4\sqrt 3 = t \times \frac{\sqrt 3}{2}\)\( \left[1 - \cfrac{\frac 1 2}{\sqrt{3 - \frac 3 4}}\right] \)

\(4\sqrt 3 = t \times \frac{\sqrt 3}{2}\)\( \left[1 - \cfrac{\frac 1 2}{\frac 3 2}\right] \)

\(4\sqrt 3 = t \times \frac{\sqrt 3}{2} \times \frac 2 3\)

\(4\sqrt 3 = t \times \frac{\sqrt 3}{2} \times \frac{2}{\sqrt 3 \times \sqrt 3}\)

\(t = 4 \times \sqrt 3 \times \sqrt 3\)

t = 12

Concave mirror. (when the object is placed between focus and pole).

 (b) \(\begin{bmatrix}0& 1 \\[0.3em]1 &0 \\[0.3em]\end{bmatrix}\)

The line is terminated with 50 ohms at the ends, so matched on both the sides thus S₁₁= 0,S₂₂ = 0 and S₁₂ = S₂₁ = 1

(a) Adding an inductance in series with Z

Point P ( Z = R + jx) on the Smith chart as shown in figure is the intersection of constant resistance circle r = 0.5 and constant reactance circle X = -1, Normalized impedance Z = 0.5 − j1 

The movement from point P along constant resistance circle of 0.5 by 450 in clockwise direction, resistance 0.5 is not changed but positive reactance is added. This is equivalent to adding inductance in series with Z.

Correct option (a) Adding an inductance in series with Z

Explanation:

Point P(z = R + jX) on the Smith chart as shown in figure is the intersection of constant resistance circle r = 0. 5 and constant reactance circle X = −, Normalized impedance Z = . − j 

The movement from point P along constant resistance circle of 0.5 by 450 in clockwise direction, resistance 0.5 is not changed but positive reactance is added. This is equivalent to adding inductance in series with Z.

Correct option (D) 11.11

Explanation:

VSWR = 2 

Reflection coefficient is given by,

|Γ| = (s - 1)/(s + 1) = (2 - 1)/(2 + 1) = 1/3

 The ratio of reflected power to incident power is given by,

P_r/P_i = |Γ|²

where, Γ = reflection coefficient at load

P_r/P_i = (1/3)² = 1/9 = 0.11

In percentage, P_r/P_i = 11.11%