Problem set 78

1. The experimentally measured spin $g$ factors of a proton and a neutron indicate that
1. both proton and neutron are elementary point particles
2. both proton and neutron are not elementary point particles
3. while proton is an elementary point particle, neutron is not
4. while neutron is an elementary point pat1icle, proton is not

both proton and neutron are not elementary point particles

Hence, answer is (B)

2. The tank circuit of a Hartley oscillator is shown in the figure. If $M$ is the mutual inductance between the inductors, the oscillation frequency is
   
1. $\frac{1}{2\pi\sqrt{(L_1+L_2+2M)C}}$
2. $\frac{1}{2\pi\sqrt{(L_1+L_2-2M)C}}$
3. $\frac{1}{2\pi\sqrt{(L_1+L_2+M)C}}$
4. $\frac{1}{2\pi\sqrt{(L_1+L_2-M)C}}$

Frequency of Hartley oscillator is given by $$f=\frac{1}{2\pi\sqrt{LC}}$$ where $L=L_1+L_2$ if the coils are assumed to be winded on different cores and $L_1+L_2+2M$ if they are winded on a single core.

Hence, answer is (A)

3. In the given digital logic circuit, $A$ and $B$ form the input. The output $Y$ is
   
1. $Y=\bar A$
2. $Y=A\bar B$
3. $Y=A\oplus B$
4. $Y=\bar B$

\begin{align*} Y&=\overline{\left(\bar A+B\right) \left( A+B\right)}\\ &=\overline{\left(\bar A+B\right)}+ \overline{\left( A+B\right)}\\ &=\overline{\left(\bar A\right)}\bar B+\bar A\bar B\\ &=A\bar B+\bar A\bar B\\ &=\left(A+\bar A\right)\bar B\\ &=\bar B\quad \text{since, }(A+\bar A=1) \end{align*}

Hence, answer is (D)

4. The largest analog output voltage from a 6-bit digital to analog converter (DAC) which produces 1.0 V output for a digital input of 010100, is
1. 1.6 V
2. 2.9 V
3. 3.15 V
4. 5.0 V

For a n-bit DAC $$V_{out}={\textstyle V_R\left(\frac{b_{n-1}}{2^1}+\frac{b_{n-2}}{2^2}+\cdots+\frac{b_{0}}{2^n}\right)}$$ For a 6-bit DAC $$V_{out}=\!\!{\textstyle V_R\!\!\left(\frac{b_{5}}{2^1}\!+\!\frac{b_{4}}{2^2}\!+\!\frac{b_{3}}{2^3}\!+\!\frac{b_{2}}{2^4}\!+\!\frac{b_{1}}{2^5}\!+\!\frac{b_{0}}{2^6}\right)}$$ For 1.0 V output of a digital input of 010100 $$1=\!{\textstyle V_R\!\!\left(\frac{0}{2^1}+\frac{1}{2^2}+\frac{0}{2^3}+\frac{1}{2^4}+\frac{0}{2^5}+\frac{0}{2^6}\right)}$$ $$V_R=\frac{16}{5}$$ $$V_{max}\!=\!{\textstyle \frac{16}{5}\!\left(\frac{1}{2^1}\!+\!\frac{1}{2^2}\!+\!\frac{1}{2^3}\!+\!\frac{1}{2^4}+\frac{1}{2^5}\!+\!\frac{1}{2^6}\right)}$$ $$V_{max}=3.15 V$$

Hence, answer is (C)

5. The low-pass active filter shown in the figure has a cut-off frequency of 2 kHz and a pass band gain of 1.5. The values of the resistors are
   
1. $R_1 = 10\: k\Omega$; $R_2 = 1.3 \Omega$
2. $R_1 = 30\: k\Omega$; $R_2 = 1.3 \Omega$
3. $R_1 = 10\: k\Omega$; $R_2 = 1.7 k\Omega$
4. $R_1 = 30\: k\Omega$; $R_2 = 1.7 k\Omega$

Using $$Gain=1+\frac{15k}{R_1}$$ $$R_1=30\:k\Omega$$ Using $$f_c=\frac{1}{2\pi RC}$$ $$R=\frac{1}{2\pi f_cC}=1.7\:k\Omega$$

Hence, answer is (D)

Problem set 77

1. The energy $E(\vec k)$ of electrons of wavevector $\vec k$ in a solid is given by $E(\vec k) = Ak^2 + Bk^4$, where $A$ and $B$care constants. The effective mass of the electron at $|\vec k| = k_0$ is
1. $Ak_0^2$
2. $\frac{\hbar^2}{2A}$
3. $\frac{\hbar^2}{2A+12Bk_0^2}$
4. $\frac{\hbar^2}{2A+12Bk_0^2}$

Effective mass is given by $$m*=\frac{\hbar^2}{\frac{d^2E}{dk^2}}$$ $$m*=\frac{\hbar^2}{2A+12Bk_0^2}$$

Hence, answer is (C)

2. Which one of the following statements is NOT correct about the Brillouin zones (BZ) of a square lattice with lattice constant a?
1. The first BZ is a square of side $2\pi/a$ in $k_x$-$k_y$ plane
2. The areas of the first BZ and third BZ are the same
3. The $k$-points are equidistant in $k_x$ as well as in $k_y$ directions
4. The area of the second BZ is twice that of the first BZ

The area of the second BZ is twice that of the first BZ is not correct

Hence, answer is (D)

3. In a crystal of $N$ primitive cells, each cell contains two monovalent atoms. The highest occupied energy band of the crystal is
1. one-fourth filled
2. one-third filled
3. half filled
4. completely filled

Each band can be occupied by $2N$ electrons, if there are $N$ primitive cells in a crystal. Hence, if each primitive cell contains 2 electrons, the highest occupied energy band of the crystal is completely filled.

Hence, answer is (D)

4. If the number density of a free electron gas changes from $10^{28}$ to $10^{26}$ electrons/$m^3$, the value of plasma frequency (in Hz) changes from $5.7\times 10^{15}$ to
1. $5.7\times 10^{13}$
2. $5.7\times 10^{14}$
3. $5.7\times 10^{16}$
4. $5.7\times 10^{17}$

Plasma frequency is given by $$\omega_p=2\pi f_p=\sqrt{\frac{ne^2}{m\epsilon_0}}$$ $$f_{p_1}\propto\sqrt{n_1}$$ $$f_{p_2}\propto\sqrt{n_2}$$ $$\frac{f_{p_2}}{f_{p_1}}=\sqrt{\frac{n_2}{n_1}}$$ $$f_{p_2}=f_{p_1}\sqrt{\frac{n_2}{n_1}}$$ $$f_{p_2}=5.7\times 10^{15}\times\sqrt{\frac{10^{26}}{10^{28}}}$$ $$f_{p_2}=5.7\times 10^{14}$$

Hence, answer is (B)

5. Which one of the following statements about superconductors is NOT true?
1. A type I superconductor is completely diamagnetic
2. A type II superconductor exhibits Meissner effect upto the second critical magnetic field ($H_{c_2}$)
3. A type II superconductor exhibits zero resistance upto the second critical magnetic field
4. Both type I and type II superconductors exhibit sharp fall in resistance at the superconducting transition temperature

In type II superconductor the transition from superconducting to normal state occurs after going through a broad "mixed state" region.

Hence, answer is (D)