Problem set 54

1. The high input impedance of field effect transistor (FET) amplifier is due to
1. the pinch-off voltage
2. its very low gate current
3. the source and drain being far apart
4. the geometry of the FET

The pinch-off voltage

Hence, answer is (A)

2. The circuit shown in the figure functions as
   
1. an OR gate
2. an AND gate
3. a NOR gate
4. a NAND gate

When either or both inputs A and B will be high, the respective transistors will be made ON and output will be high, hence circuit functions as OR gate

Hence, answer is (A)

3. A linear transformation $T$, defined as $T\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}=\begin{pmatrix}x_1+x_2\\x_2-x_3\end{pmatrix}$, transforms a vector $\vec x$ from a three-dimensional real space to a two-dimensional real space. The transformation matrix $T$ is
1. $\begin{pmatrix}1&1&0\\0&1&-1\end{pmatrix}$
2. $\begin{pmatrix}1&0&0\\0&1&0\end{pmatrix}$
3. $\begin{pmatrix}1&1&1\\-1&1&1\end{pmatrix}$
4. $\begin{pmatrix}1&0&0\\0&0&1\end{pmatrix}$

\begin{align*} T\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}&=\begin{pmatrix}1&1&0\\0&1&-1\end{pmatrix}\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}\\ &=\begin{pmatrix}x_1+x_2\\x_2-x_3\end{pmatrix} \end{align*}

Hence, answer is (A)

4. Three point charges $q$, $q$ and $-2q$ are located at $(0,-a,a)$, $(O,a,a)$ and $(0,0,-a)$, respectively. The net dipole moment of this charge distribution is
1. $4qa\hat k$
2. $2qa\hat k$
3. $-4qa\hat i$
4. $-2qa\hat j$

If the system of charges is electrically neutral as whole, then the net dipole moment of a system of charges $q_1,q_2,\dots, q_n$ with position vectors $r_1,r_2,\dots, r_n$ is given by $$\vec P=\sum\limits_{i=1}^{n}q_i\vec r_i$$ \begin{align*} \vec P&=q\vec r_1+q\vec r_2-2q\vec r_3\\ &=q\left(-a\hat j+a\hat k\right)+q\left(a\hat j+a\hat k\right)-2q\left(-a\hat k\right)\\ &=4qa\hat k \end{align*}

Hence, answer is (A)

5. The vector potential in a region is given as $\vec A(x, y, z) = -y\hat i + 2x\hat j$. The associated magnetic induction $\vec B$ is
1. $\hat i+\hat k$
2. $3\hat k$
3. $-\hat i+2\hat j$
4. $-\hat i+\hat j+\hat k$

\begin{align*} \vec B&=\vec\nabla\times\vec A\\ &=\begin{vmatrix}\hat i&\hat j&\hat k\\\frac{\partial}{\partial x} &\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ -y&2x&0\end{vmatrix}\\ &=3\hat k \end{align*}

Hence, answer is (B)