A simple pendulum with a brass bob has a period `T`. The bob is now immersed in a nonviscous liquid and oscillated. If the density of the liquid is `1//8th` of brass, the time period of the same pendulum will be

Options

A. `sqrt((8)/(7))T`
B. `(8)/(7)T`
C. `(64)/(49)T`
D. `T`

Explanation

To solve the problem of finding the new time period of a simple pendulum with a brass bob immersed in a non-viscous liquid, we can follow these steps:

### Step 1: Understand the Initial Conditions
The initial time period \( T \) of a simple pendulum in air is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

### Step 2: Identify the Forces Acting on the Pendulum in the Liquid
When the pendulum bob is immersed in a liquid, two main forces act on it:
- The gravitational force \( Mg \) acting downwards.
- The buoyant force \( F_b \) acting upwards, which is given by:
\[ F_b = \rho_{liquid} \cdot V \cdot g \]
where \( \rho_{liquid} \) is the density of the liquid and \( V \) is the volume of the bob.

### Step 3: Calculate the Effective Weight of the Bob in the Liquid
The net force acting on the bob when it is in the liquid is:
\[ F_{net} = Mg - F_b = Mg - \rho_{liquid} \cdot V \cdot g \]
Substituting the volume \( V \) of the bob, we have:
\[ F_{net} = Mg - \rho_{liquid} \cdot \frac{M}{\rho_{solid}} \cdot g \]
Here, \( \rho_{solid} \) is the density of the brass bob.

### Step 4: Substitute the Density Values
Given that the density of the liquid is \( \frac{1}{8} \) of the density of brass:
\[ \rho_{liquid} = \frac{1}{8} \rho_{solid} \]
Substituting this into the equation gives:
\[ F_{net} = Mg - \left(\frac{1}{8} \rho_{solid}\right) \cdot \frac{M}{\rho_{solid}} \cdot g = Mg - \frac{Mg}{8} = Mg \left(1 - \frac{1}{8}\right) = Mg \cdot \frac{7}{8} \]

### Step 5: Determine the New Effective Acceleration
The effective weight of the bob in the liquid is \( \frac{7Mg}{8} \). The new effective acceleration \( g' \) can be defined as:
\[ g' = \frac{7g}{8} \]

### Step 6: Calculate the New Time Period
Now, we can calculate the new time period \( T' \) using the modified acceleration:
\[ T' = 2\pi \sqrt{\frac{L}{g'}} \]
Substituting \( g' \):
\[ T' = 2\pi \sqrt{\frac{L}{\frac{7g}{8}}} = 2\pi \sqrt{\frac{8L}{7g}} = \sqrt{\frac{8}{7}} \cdot 2\pi \sqrt{\frac{L}{g}} = \sqrt{\frac{8}{7}} \cdot T \]

### Final Result
Thus, the new time period \( T' \) of the pendulum bob immersed in the liquid is:
\[ T' = T \sqrt{\frac{8}{7}} \]

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