Let’s denote the initial potential energy of the bob as \( PE_i \) and the final potential energy at the lowermost position as \( PE_f \). Similarly, let’s denote the initial kinetic energy as \( KE_i \) and the final kinetic energy at the lowermost position as \( KE_f \).

Since the pendulum bob is initially at rest when the string is horizontal, its initial kinetic energy (\( KE_i \)) is zero.

The potential energy of the bob at the initial position (\( PE_i \)) is fully converted to kinetic energy at the lowermost position, assuming no energy loss. Therefore, at the lowermost position, the kinetic energy (\( KE_f \)) equals the potential energy at the initial position (\( PE_i \)).

Given that 10% of the energy is lost, the final potential energy (\( PE_f \)) is 90% of the initial potential energy (\( PE_i \)).

Using the conservation of mechanical energy, we have:

\[ PE_i – PE_f = KE_f – KE_i \]

Since \( KE_i = 0 \), this simplifies to:

\[ PE_i – PE_f = KE_f \]

\[ PE_i – 0.9PE_i = KE_f \]

\[ 0.1PE_i = KE_f \]

Now, the potential energy of a pendulum bob at a height \( h \) is given by:

\[ PE = mgh \]

Given that the length of the pendulum is \( 10 \) m, at the lowermost position, \( h = 10 \) m. Therefore, the initial potential energy (\( PE_i \)) is:

\[ PE_i = mgh = mg(10) \]

Since the mass (\( m \)) of the pendulum bob cancels out when calculating the speed, we can ignore it.

Now, we can calculate \( KE_f \):

\[ KE_f = 0.1PE_i = 0.1 \times mg \times 10 \]

\[ KE_f = 1 \times g \times 10 \]

\[ KE_f = 10g \]

Now, at the lowermost position, all of the potential energy is converted into kinetic energy, so:

\[ KE_f = \frac{1}{2}mv^2 \]

\[ 10g = \frac{1}{2}v^2 \]

\[ v^2 = 20g \]

\[ v = \sqrt{20g} \]

Given that \( g = 9.8 \, \text{m/s}^2 \) (approximately), we can calculate \( v \):

\[ v = \sqrt{20 \times 9.8} \]

\[ v \approx \sqrt{196} \]

\[ v \approx 14 \, \text{m/s} \]

However, this speed represents the magnitude of the velocity vector. Since the bob is moving vertically downward, the speed is positive.

Therefore, the correct option is (C) \( 7\sqrt{5} \, \text{m/s} \).