Certainly! Let's go through this problem step-by-step.
The periodic time [tex]\( T \)[/tex] of a pendulum is given to vary as the square root of its length [tex]\( l \)[/tex].
This means we can write:
[tex]\[ T \propto \sqrt{l} \][/tex]
To express this as an equation, we introduce a constant of proportionality [tex]\( k \)[/tex]:
[tex]\[ T = k \sqrt{l} \][/tex]
Given that:
[tex]\[ T = 6 \quad \text{when} \quad l = 9 \][/tex]
We can substitute these values into the equation to find [tex]\( k \)[/tex]:
[tex]\[ 6 = k \sqrt{9} \][/tex]
Since [tex]\( \sqrt{9} = 3 \)[/tex], we have:
[tex]\[ 6 = k \cdot 3 \][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{6}{3} = 2 \][/tex]
Now, we need to find [tex]\( T \)[/tex] when [tex]\( l = 25 \)[/tex]. Using the proportionality with the constant [tex]\( k \)[/tex] we just found:
[tex]\[ T = k \sqrt{25} \][/tex]
Since [tex]\( k = 2 \)[/tex] and [tex]\( \sqrt{25} = 5 \)[/tex]:
[tex]\[ T = 2 \cdot 5 = 10 \][/tex]
Therefore, the periodic time [tex]\( T \)[/tex] when [tex]\( l = 25 \)[/tex] is:
[tex]\[ T = 10 \][/tex]
