Step 1: Mobility \( \mu \) is given by the relation:
\[
\mu = \frac{q \tau}{m^}
\]
where \( q \) is the charge, \( \tau \) is the relaxation time, and \( m^ \) is the effective mass.
Step 2: The mobility ratio between hole and electron is:
\[
\frac{\mu_h}{\mu_e} = \frac{\tau_h \cdot m_e}{\tau_e \cdot m_h}
\]
where \( \tau_h \) and \( \tau_e \) are the relaxation times for hole and electron, and \( m_h \) and \( m_e \) are the effective masses of hole and electron.
Step 3: Substituting the given values:
\[
\frac{\mu_h}{\mu_e} = \frac{1 \times 11}{2 \times 2} = \frac{11}{4}
\]
Thus, the correct answer is (D).
The figure shows an opamp circuit with a 5.1 V Zener diode in the feedback loop. The opamp runs from \( \pm 15 \, {V} \) supplies. If a \( +1 \, {V} \) signal is applied at the input, the output voltage (rounded off to one decimal place) is:
A wheel of mass \( 4M \) and radius \( R \) is made of a thin uniform distribution of mass \( 3M \) at the rim and a point mass \( M \) at the center. The spokes of the wheel are massless. The center of mass of the wheel is connected to a horizontal massless rod of length \( 2R \), with one end fixed at \( O \), as shown in the figure. The wheel rolls without slipping on horizontal ground with angular speed \( \Omega \). If \( \vec{L} \) is the total angular momentum of the wheel about \( O \), then the magnitude \( \left| \frac{d\vec{L}}{dt} \right| = N(MR^2 \Omega^2) \). The value of \( N \) (in integer) is:
In the transistor circuit shown in the figure, \( V_{BE} = 0.7 \, {V} \) and \( \beta_{DC} = 400 \). The value of the base current in \( \mu A \) (rounded off to one decimal place) is: