Explore topic-wise InterviewSolutions in Current Affairs.

A. homogeneous

C. up thrust

The Correct option is D. nano

The Correct option is D. drag

C. smaller the acceleration

Phase margin is that amount of additional phase lag at the gain crossover frequency required to bring the system to the verge of instability.

Gain Crossover Frequency: 

Gain crossover frequency is that frequency at which, [G(jω)], the magnitude of the open loop transfer function is unity. 

The Phase margin PM is 180° plus the phase angle φ of the open loop transfer function at the gain crossover frequency.

PM = (180°(+φ)at GCF)

Phase Crossover Frequency:

Phase crossover frequency is that frequency at which, ∠G(jω), the phase angle of the open loop transfer function equals ‐180°.   

Thus, Gain Margin,  

GM = (1/jG(jω))_{at PCF}

In terms of decibels,

GM dB = 20 logIG(jω))_{at PCF}

(a) Fusion takes place when two low-mass isotopes, typically isotopes of hydrogen, unite under conditions of extreme pressure and temperature whereas fission is the splitting of a heavy, unstable nucleus into two lighter nuclei.

(b) \(^{56}_{26}Fe\) → 2 \(^{28}_{13}Al\)

Q = Δm x 931 Mev

= \(\big(\)m[\(^{56}_{26}Fe\)] - 2m[ \(^{28}_{13}Al\)]\(\big)\) x 931 Mev

= (55.93494 - 2 x 27.98191) x 931 Mev

= -0.02886 x 931 Mev

= -26.88 Mev < 0

The fission is not possible energetically.

One difference between nuclear fission and nuclear fusion 

Calculating Q=((m) Fe-2(m) Al)C² =26.90MeV 

Justification not possible

The electric field at the point P at a distance r from the point charge q is the force experienced by a unit charge and is given by

\(\vec E\) = \(\frac{\vec F}{q_0}\)

The electric field is a vector quantity and its Sl unit is Newton per Coulomb (NC^-1).

The basic principle of a potentiometer is that when a constant current flows through a wire of uniform cross-sectional area and composition, the potential drop across any length of the wire is directly proportional to that length.

Given:

focal length,f = 150 cm (or) f= 1.5 m 

Equation for power of lens is, P = 1/f 

Substituting the values, 

P = \(\frac{1}{1.5}\) 0.067 diopter 

As the power is positive, it is a converging lens.

⁶C₃=\(\frac{6!}{3!3!}=\frac{6\times5\times4\times3!}{3!\times3!}\)

\(\frac{120}{6}=20\)

The circuit used to feedback a portion of the output to the input is called the feedback network. If the portion of the output fed to the input is in phase with the input, then the magnitude of the input signal increases. It is necessary for sustained oscillations.

I₃ is zero as the diode in that branch is reverse bised. Resistance in the branch AB and EF are each (125 + 25)Ω = 150Ω.

As AB and EF are identical parallel branches, their effective resistance is 150/2=75Ω

Net resistance in the circuit = (75 + 25) Ω = 100Ω .

Current I₁=5/100=0.05A.

As resistances of AB and EF are equal, and I₁ = I₂ + I₃ + I₄, I₃ = 0

I₂=I₄=0.05/2=0.025A

Enzymes are biological catalyst they decrease the activation energy of reaction and speed up the rate of reaction. they increases the rate of biological reaction upto 10⁷ to 10²⁰ times.

Cryoscopic constant is the depression of the freezing point of the solvent produced on dissolving are mole of substance in 1000g of it. it is also called molal depression constant. it is denoted by K_f.

As we know,

depression in freezing point

△T_f = K_f m

where,

m = molality 

K_f = Cryoscopic constant

△T_f = depression in freezing point

we have given

△T_f = 177.5 - 147

= 30.5 °C

Number of moles of  Naphthalene = \(\frac1{128}\) mole

molality = \(\cfrac{\frac1{128}}{10}\times1000\)

= 0.78 m

Therefore,

K_f = \(\frac{30.5}{0.78}\)

K_f = 39.1 °C kg mol^-1

Hence, the cryosopic constant will be 39.1 °C kg mol^-1

(c) affinity of a substance for another substance 

(b) the chart is for a given pressure 

(c) square root of temperature 

(b) zone of silence 

(a) less than the critical pressure 

K_a = α²/C = (0.018)²/0.05

= 6.48 x 10^-3