Electronics and Experimental Methods MCQ Quiz in मल्याळम - Objective Question with Answer for Electronics and Experimental Methods - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Resistance in series: When two or more  resistances are connected in series, the total or equivalent resistance ($R_{eq}$) is the sum of the individual resistances, and the uncertainties (standard deviations) are propagated similarly.

$R_{eq}=R_1+R_2+R_3+...$

Where, R₁, R₂, R₃, ... are resistances in series.

Propagation of Uncertainty: The resultant uncertainty (standard deviation) of two or more resistances in series is given by the square root of the sum of the squares of the individual standard deviations.

${\sigma }_R=\sqrt{\Delta R_1^2+\Delta R_2^2+\Delta R_3^2+...}$

Where, Δ R₁ is the uncertainty in R₁.

 Δ R2 is the uncertainty in R₂ and so on...

Explanation:

Given,  R₁ = 100 and  Δ R₁=5Ω,  R₂ = 150 and  Δ R2=15Ω:

$R_{eq}=R_1+R_2=100+150=250\mathrm{\Omega }$

Resultant standard deviations${\sigma }_R=\sqrt{(5\mathrm{\Omega }{)}^2+(15\mathrm{\Omega }{)}^2}=\sqrt{250\mathrm{\Omega }}\approx 15.8\mathrm{\Omega }$

Hence, 

$R_{eq}=250\pm 15.8\mathrm{\Omega }$

The correct answer is Option 3.

Concept:

Reactance of Inductive Coil: The reactance ($X_L$) of an ideal inductor is given by:

$X_L=\omega L$

where, ω = 2πf which is the angular frequency, and (L) is the inductance.

Effect of Stray Capacitance: Stray capacitance in an inductive coil causes the coil to behave as a resonant circuit at certain frequencies, effectively forming a parallel LC circuit.

Effective Reactance ($X_{\text{eff}}$):

1.   Below the resonant frequencies (ω<ω_c ), the inductive reactance ( $X_L$ ) dominates.
2. At the resonant frequency which is (ω=ωc ), the inductive reactance and capacitive reactance cancel each other out, leading to a minimum in the effective reactance.
3. Above the resonant frequency (ω>ωc ), the coil's behaviour is dominated by the capacitive reactance ( $X_C$ ), and the effective reactance starts to decrease.

Hence, the effective reactance of an inductive coil changes due to the presence of stray capacitance as the frequency of the applied signal changes.

Explanation:

Because of the stray capacitance, the effective reactance of an inductive coil: Initially increases with frequency because of the inductive reactance ( $X_L$ ). Decreases at high frequencies because the capacitive reactance ( $X_C$ ) becomes significant and ultimately dominates.

So, the effective reactance of an inductive coil decreases because of stray capacitances as the frequency increases.

The correct answer is Option 2.

Concepts:
1. Force on a Charge in an Electric Field : 
   The force F on a charged particle in a uniform electric field E is given by:
  $F=qE$
   
   where q is the charge of the particle (both proton and electron have the same magnitude of charge e ).

2. Acceleration : 
   Newton’s second law tells us that the acceleration a of a particle due to a force is:
 $a=\frac{F}{m}=\frac{qE}{m}$
   
   where m is the mass of the particle. For the electron, the acceleration is:
  $a_e=\frac{eE}{m_e}$
   and for the proton, the acceleration is:
   $a_p=\frac{eE}{m_p}$
   

3. Kinematic Equation : 
   The distance d moved by the particle under constant acceleration (starting from rest) is given by the equation:
  $d=\frac{1}{2}a{t}^2$
   
   Rearranging for time:
  $t=\sqrt{\frac{2d}{a}}$
   

Explanation:

1. For the electron :
  $t_1=\sqrt{\frac{2d}{a_e}}=\sqrt{\frac{2d}{\frac{eE}{m_e}}}=\sqrt{\frac{2d{m}_e}{eE}}$
   

2. For the proton :
  $t_2=\sqrt{\frac{2d}{a_p}}=\sqrt{\frac{2d}{\frac{eE}{m_p}}}=\sqrt{\frac{2d{m}_p}{eE}}$
   

3. Taking the ratio \frac{t_2}{t_1} :
  $\frac{t_2}{t_1}=\frac{\sqrt{\frac{2d{m}_p}{eE}}}{\sqrt{\frac{2d{m}_e}{eE}}}=\sqrt{\frac{m_p}{m_e}}$
The correct option is (2).

Concept:

1. The least count (LC) of the instrument is 0.1 mm.
2. The number of divisions on the Vernier scale (VSD) is 20.
3. The number of divisions on the main scale (MSD) that coincide with the Vernier scale is 19.

The least count (LC) of a Vernier caliper is given by the formula:

" id="MathJax-Element-28-Frame" role="presentation" style="position: relative;" tabindex="0">$$ $LC=Valueof1MainScaleDivision(MSD)-Valueof1VernierScaleDivision(VSD)$

$\text{Calculation:}$

Here,

20 VSD = 19 MSD

1VSD = $\frac{19}{20}$ MSD

L.C. = 1 MSD – 1 VSD

0.1 mm = 1 MSD - $\frac{19}{20}$ MSD

0.1 = $\frac{1}{20}$ MSD

1 MSD = 2mm

∴ The correct option is 3

Explanation:

This is the tank circuit of the Hartley oscillator. 

• There are two inductors in the series connection. The equivalent inductance is : $L=12\mu H+8\mu H=20\mu H$
• The capacitance is given as : $C=0.05\mu f$
• The formula for the frequency is given as : 

$f=\frac{1}{2\pi \sqrt{LC}}$

$f=\frac{1}{2\pi \sqrt{(20\mu H\times 0.05\mu f)}}$

$f=\frac{1}{2\pi \mu }Hz$

$f=\frac{{10}^6}{2\times 3.14}=159235Hz=159KHz$

Explanation:

The circuit appears to be a full wave rectifier with a 4-stage voltage multiplier, often known as a ladder network or Cockcroft–Walton generator. This kind of circuit is used to generate a high DC voltage from an AC power source. It is made by a combination of diodes and capacitors, allowing for increasing the voltage value.

• Given that the knee voltage for the silicon diodes is 0.7 volts, this will have to be subtracted for every diode as we move through the voltage multiplier.
• The output voltage is approximately equal to $4V_{in}-4V_{knee}=4V_{in}-4(0.7V)=4V_{in}-2.8V$, where V_in is the peak voltage of the AC source, and V_knee is the knee voltage of a diode. The factor of 4 in front of the V_knee originates from the fact that there are 4 diodes in the current path from the source to the output, and each diode subtracts its knee voltage from the overall voltage.

Explanation:

• A lock-in amplifier primarily recognizes the component of the signal that is in phase with the reference signal (B = B₀sinΩt in this case) and measures its amplitude and phase, not just the amplitude.
• When $sin\Omega t$ is at its peak amplitude, the Hall voltage $V_H$ is also at its peak: $V_{H_{max}}=\frac{I{R}_{H}B₀}{d}+RI$ .
• However, the DC offset term RI will get removed by the lock-in amplifier. What remains is the AC part of the signal, which follows the sinusoidal modulation.
• The observed amplitude of this sinusoidal signal, having accounted for our removal of the DC offset, is the root-mean-square (RMS) value.
• The RMS value of a sinusoidal function A sin(x) is $\frac{A}{\surd 2}$.
• Recognising that, and applying it to the signal we're talking about, we get: $V_H^{\prime }=\frac{I{R}_{H}B₀}{(\surd 2d)}$

Concept:

Every measurement has an air of uncertainty in it, so error is measured by standard deviation ${\sigma }_i$ of the measurement.

${\sigma }_x^2={\left({\displaystyle \frac{\delta x}{\delta a}}\right)}^2{\sigma }_a^2+{\left({\displaystyle \frac{\delta x}{\delta b}}\right)}^2{\sigma }_b^2+{\left({\displaystyle \frac{\delta x}{\delta c}}\right)}^2{\sigma }_c^2$

Explanation:

Given, efficiency of the motor is the ratio between the work done by the motor and the energy delivered to it.

So, $e=\frac{W}{E}=\frac{mgh}{VIt}$

Given, $M=2.00\pm 0.02kg,h=1.00\pm 0.01m,V=10.0\pm 0.1V,I=2.00\pm 0.02A,t=300\pm 15s$

Using propagation of errors, we get,

• $\frac{\delta e}{e}=\sqrt{{\left({\displaystyle \frac{\delta m}{m}}\right)}^2+{\left({\displaystyle \frac{\delta h}{h}}\right)}^2+{\left({\displaystyle \frac{\delta V}{V}}\right)}^2+{\left({\displaystyle \frac{\delta I}{I}}\right)}^2+{\left({\displaystyle \frac{\delta t}{t}}\right)}^2}$    

• $\frac{\delta e}{e}=\sqrt{{\left({\displaystyle \frac{0.02}{2}}\right)}^2+{\left({\displaystyle \frac{0.01}{1}}\right)}^2+{\left({\displaystyle \frac{0.01}{1}}\right)}^2+{\left({\displaystyle \frac{0.02}{1}}\right)}^2+{\left({\displaystyle \frac{15}{300}}\right)}^2}$
• $\frac{\delta e}{e}=\sqrt{(0.01{)}^2\times 4+(0.05{)}^2}$
• $\frac{\delta e}{e}=\sqrt{2}9\times {10}^{-2}$
• $\frac{\delta e}{e}\approx 0.05$

So, the correct answer is $\frac{\delta e}{e}\approx 0.05$

Explanation:

Case-1-When diode is in forward biasing

Given, $V_{rms}=1V$

Now, peak voltage $V_p=\sqrt{2}{V}_{rms}=\sqrt{2}\times 1=1.414$

• Current in $20\mathrm{\Omega }=\frac{voltagedifference}{resistance}=\frac{1.414-0.7}{20}\approx 0.03A$

This is low current

• Current in $10\mathrm{\Omega }=\frac{voltagedifference}{resistance}=\frac{0.7-0}{10}=0.07A$

This is high current as comparison to $20\mathrm{\Omega }$ resistor in forward biasing 

So, Forward biasing is not possible in this circuit.

Case-2-When the circuit is in Reverse biasing-

This circuit is possible in reverse biasing.

In Reverse biasing the current in the diode should be zero.

So, the correct answer is $0A$

Concept:

 An ammeter will be always connected in series because it has a low resistance and the Voltmeter will be always connected in parallel because it will be having a high resistance.

Explanation:

• Option-1 is incorrect as voltmeter is connected in series and wheat stone bridge principle is not applied. 
• Option-4 is incorrect as circuit does not follow wheat stone bridge principle. 
• Option-3 is incorrect as ammeter is connected in parallel. 
• Option-2 is correct as it follows the wheat stone bridge principle and the voltmeter is connected in parallel.

So, the correct answer is Option-(2).