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Step-by-step Solution
Step 1: Recall the Time Period of a Simple Pendulum in Air
For a simple pendulum of length L executing small oscillations in air (or vacuum), the time period T is given by
T = 2\pi \sqrt{\frac{L}{g}},
where g is the acceleration due to gravity.
In the question, the given time period of the pendulum in air is 10\,\text{s} . Hence,
10 = 2\pi \sqrt{\frac{L}{g}}.
From this, we find
\sqrt{\frac{L}{g}} = \frac{10}{2\pi} = \frac{5}{\pi}.
Step 2: Determine the Effective Acceleration When Bob is Immersed in Water
The bob has a relative density (specific gravity) of 5 , which means its density is 5 times that of water. When the bob is immersed in water, it experiences a buoyant force
F_\text{b} = \rho_{\text{water}}\,V\,g,
where V is the volume of the bob.
Since the bob’s mass m = \rho_{\text{bob}} \, V = 5\,\rho_{\text{water}}\;V, its weight is
mg = 5\,\rho_{\text{water}}\,V\,g.
The buoyant force is
\rho_{\text{water}}\,V\,g.
Therefore, the net downward force on the bob in water is
mg - F_\text{b} = 5\rho_{\text{water}}\,V\,g - \rho_{\text{water}}\,V\,g = 4\rho_{\text{water}}\,V\,g.
Hence, the effective acceleration due to gravity (sometimes called the apparent gravitational acceleration) is
g_\text{eff} = \frac{(4\rho_{\text{water}}\,V\,g)}{(5\rho_{\text{water}}\,V)} = \frac{4g}{5}.
Step 3: Write the Time Period of the Pendulum in Water
When the bob is fully immersed in water, the time period T' is that of a simple pendulum with the effective acceleration g_\text{eff} = \frac{4g}{5} . So,
T' = 2\pi \sqrt{\frac{L}{g_\text{eff}}}
= 2\pi \sqrt{\frac{L}{\frac{4g}{5}}}
= 2\pi \sqrt{\frac{5L}{4g}}.
We already know from Step 1 that
\sqrt{\frac{L}{g}} = \frac{5}{\pi}.
Thus,
T' = 2\pi \sqrt{\frac{5}{4}}\;\sqrt{\frac{L}{g}}
= 2\pi \cdot \sqrt{\frac{5}{4}} \cdot \frac{5}{\pi}
= 10 \cdot \sqrt{\frac{5}{4}}
= 10 \cdot \frac{\sqrt{5}}{2}
= 5\sqrt{5}.
Step 4: Equate the Given Expression and Find x
According to the problem statement, the new time period is given as 5\sqrt{x}\,\text{s} . We have found
T' = 5\sqrt{5}.
Equating these:
5\sqrt{x} = 5\sqrt{5}.
Dividing both sides by 5, we get
\sqrt{x} = \sqrt{5} \quad \Longrightarrow \quad x = 5.
Answer
The value of x is \boxed{5}.
