Problem set 49

1. If the parameters $y$ and $x$ are related by $y=\log{(x)}$, then the circuit that can be used to produce an output voltage $V_0$ varying linearly with $x$ is
1. 
2. 
3. 
4. 
Option (A) is a integrator circuit having output $V_0=-\frac{1}{R*C}\int\limits_0^tV_{in}\:dt$.

Option (B) is a logarithmic circuit having output $V_0=-V_t*\ln{\frac{V_{in}}{I_{d0}R}}$, where $I_{do}$ is reverse saturation current,

Option (C) is a antilog circuit having output $V_0=-I_0*R*e^{ V_{in}/V_t}$, where $V_t$ is the voltage equivalent of temperature

Option (D) is a differentiation circuit having output $V_0=-R*C\frac{dV_{in}}{dt}$.

In our case $V_{in}=\log{(x)}$, hence, clearly, antilog circuit will give desired output

Hence, answer (C) is correct.

2. A hollow metallic sphere of radius $a$, which is kept at a potential $V_0$, has a charge $Q$ at its centre. The potential at a point outside the sphere, at a distance $r$ from the centre, is
1. $V_0$
2. $\frac{Q}{4\pi\epsilon_0r}+\frac{V_0a}{r}$
3. $\frac{Q}{4\pi\epsilon_0r}+\frac{V_0a^2}{r^2}$
4. $\frac{V_0a}{r}$

The potential on metallic sphere is $V_0$. Let $Q_0$ be charge on the sphere. Hence, $$V_0=\frac{Q_0}{4\pi\epsilon_0 a}$$ $$Q_0=V_04\pi\epsilon_0 a$$ Hence, potential at a distance $r$ is $$V=\frac{Q_0}{4\pi\epsilon_0 r}=\frac{V_0a}{r}$$

Hence, answer is (D)

3. In the schematic figure given below, the initial values of 4 bit shift registers A and B are 1011 and 0010, respectively. The values at $SO_A$ and $SO_B$ after the pulse $T_2$ are respectively
   
1. 1110 and 1001
2. 1101 and 1001
3. 1101 and 1100
4. 1110 and 1100

Clock pulses will be applied to shift registers A and B, when the SHIFT at input of AND gate is 1. When SHIFT is 1, two clock pulses will be applied to shift registers A and B. When clock pulses appears, the bits of two registers will shift as shown below



Hence, answer is (D)

4. The anomalous dispersion of light observed in gases is well represented by the Cauchy's formula. The refractive index of the medium in such case is given as $n=1+A+\frac{AB}{\lambda^2}$. ($\lambda$ is wavelength of light in air)

   What does the terms $A$ and $B$ respectively represent?

1. Coefficients of refraction and dispersion
2. Coefficients of dispersion and refraction
3. Coefficients of dispersion and polarization
4. Coefficients of refraction and scattering

(A) Coefficients of refraction and dispersion

5. Two data sets A and B consist of 60 and 10 readings of a voltage measured using voltmeters of resolution of 1 mV and 0.5 mV respectively. The uncertainty in the mean voltage obtained from the data sets $A$ and $B$ are $U_A$ and $U_B$, respectively. If the uncertainty of the mean of the combined data sets is $U_{AB}$, then which of the following statements is correct?
1. $U_{AB} < U_A$ and $U_{AB} > U_B$
2. $U_{AB} < U_A$ and $U_{AB} < U_B$
3. $U_{AB} > U_A$ and $U_{AB} < U_B$
4. $U_{AB} > U_A$ and $U_{AB} > U_B$

For $N$ measurements $x_1,x_2,\cdots x_N$ with uncertainties $\sigma_1,\sigma_2\cdots\sigma_N$ weighted average is given by $$x_{avg}=\frac{\sum\limits_{i=1}^Nw_ix_i}{\sum\limits_{i=1}^Nw_i}\quad--(1)$$ where, $w_i=\frac{1}{\sigma_i}$. The uncertainty on $x_{avg}$ is $$\sigma_{avg}=\frac{1}{\sqrt{\sum\limits_{i=1}^Nw_i}}\quad--(2)$$

For an instrument having resolution $R$ has uncertainty $\sigma=\frac{R}{2}$

$\sigma_A=\frac{R_A}{2}=\frac{1}{2}=0.5 \:mV$

$\sigma_B=\frac{R_B}{2}=\frac{0.5}{2}=0.25 \:mV$

Using equation (2)

$U_A=\frac{\sigma_A}{\sqrt{N}}=\frac{0.5}{\sqrt{60}}=0.065$

$U_B=\frac{\sigma_B}{\sqrt{N}}=\frac{0.25}{\sqrt{10}}=0.08$

$$\begin{align*} U_{AB}&=\frac{1}{\sqrt{\sum\limits_{i=1}^Nw_i}}\\ &=\frac{1}{\sqrt{60\times\frac{1}{0.5}+10\times\frac{1}{0.25}}}\\ &=0.08 \end{align*}$$

Hence, answer is (B)