The electrical conductivity of an electrolytic cell depends on various factors, including:

i. Concentration of electrolyte: The conductivity of an electrolytic solution generally increases with an increase in the concentration of the electrolyte. This is because a higher concentration of ions allows for more charge carriers, leading to increased conductivity.

ii. Amount of electrolyte added: Similar to concentration, adding more electrolyte to a solution increases the number of ions present and therefore increases the conductivity.

iii. Temperature: Generally, the conductivity of electrolytic solutions increases with an increase in temperature. This is because higher temperatures lead to increased ion mobility, allowing ions to move more freely and increasing conductivity.

iv. Nature of electrode: The nature of the electrode can affect the conductivity of an electrolytic cell to some extent. For example, certain electrode materials may have a higher resistance, which can affect the overall conductivity of the cell.

Out of the given options, the electrical conductivity of an electrolytic cell does not depend on the nature of the electrode (iv). While the nature of the electrode may have some influence, it is not a primary factor affecting conductivity compared to the other factors listed.

Therefore, the correct answer is (iv) Nature of electrode.

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) \( 7\sqrt{5} \, \text{m/s} \).

Ethanol reacts with Lucas reagent (conc. HCl + anhydrous ZnCl2) to form a cloudy or turbid solution due to the formation of an alkyl chloride.

The reaction occurs as follows:

\[ \text{CH}_3\text{CH}_2\text{OH} + \text{HCl} \xrightarrow{} \text{CH}_3\text{CH}_2\text{Cl} + \text{H}_2\text{O} \]

This reaction is a typical test for the presence of a primary or secondary alcohol. Primary alcohols react slowly with Lucas reagent and may not produce a turbid solution immediately. However, secondary alcohols react more readily and form turbidity faster.

Therefore, the correct answer is likely option (b) After 5 to 7 mins, as this timeframe is typical for the turbidity to become apparent with ethanol.