A simple pendulum with a solid metal bob has a period `T`. What will be the period of the same pendulum if it is made to oscillate in a non - viscous liquid of density one - tenth of the of the metal of the bob ?

Explanation

To solve the problem of finding the period of a simple pendulum with a solid metal bob oscillating in a non-viscous liquid, we can follow these steps:

### Step 1: Understand the formula for the period of a simple pendulum
The period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where:
- \( L \) is the length of the pendulum,
- \( g \) is the acceleration due to gravity.

### Step 2: Determine the effective acceleration due to gravity in the liquid
When the pendulum bob is submerged in a liquid, the effective acceleration due to gravity \( g' \) is modified by the buoyant force acting on the bob. The effective gravitational acceleration can be calculated as:
\[ g' = g - \frac{F_b}{m} \]
where:
- \( F_b \) is the buoyant force,
- \( m \) is the mass of the bob.

### Step 3: Calculate the buoyant force
The buoyant force \( F_b \) can be calculated using Archimedes' principle:
\[ F_b = V \cdot \rho_{liquid} \cdot g \]
where:
- \( V \) is the volume of the bob,
- \( \rho_{liquid} \) is the density of the liquid.

Given that the density of the liquid is one-tenth of the density of the metal bob (\( \rho_{bob} \)), we have:
\[ \rho_{liquid} = \frac{1}{10} \rho_{bob} \]

### Step 4: Substitute the values into the equation for \( g' \)
Substituting the expression for \( F_b \) into the equation for \( g' \):
\[ g' = g - \frac{V \cdot \left(\frac{1}{10} \rho_{bob}\right) \cdot g}{m} \]
Since the mass \( m \) of the bob is given by \( m = \rho_{bob} \cdot V \), we can simplify:
\[ g' = g - \frac{V \cdot \left(\frac{1}{10} \rho_{bob}\right) \cdot g}{\rho_{bob} \cdot V} = g - \frac{1}{10}g = \frac{9}{10}g \]

### Step 5: Substitute \( g' \) back into the period formula
Now we can substitute \( g' \) back into the period formula:
\[ T' = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{\frac{9}{10}g}} = 2\pi \sqrt{\frac{10L}{9g}} \]

### Step 6: Express \( T' \) in terms of \( T \)
Since \( T = 2\pi \sqrt{\frac{L}{g}} \), we can express \( T' \) as:
\[ T' = \sqrt{\frac{10}{9}} T \]

### Final Answer
Thus, the period of the pendulum when oscillating in the non-viscous liquid is:
\[ T' = \frac{\sqrt{10}}{3} T \]