Alright, let's solve this problem step-by-step.
We are given the following information:
1. The period of oscillation for a mass "M" suspended from a spring is 25 seconds (T₁ = 25 s).
2. We need to determine the period of oscillation when the mass is increased to 16 times the original mass (16M).
The period of oscillation [tex]\( T \)[/tex] of a mass-spring system is related to the mass [tex]\( M \)[/tex] by the formula:
[tex]\[ T \propto \sqrt{M} \][/tex]
This means that the period [tex]\( T \)[/tex] is proportional to the square root of the mass [tex]\( M \)[/tex].
To find the new period [tex]\( T_2 \)[/tex], we use the relationship between the periods and the masses:
[tex]\[ \frac{T_2}{T_1} = \sqrt{\frac{M_2}{M_1}} \][/tex]
Here, [tex]\( M_1 \)[/tex] is the original mass (M), and [tex]\( M_2 \)[/tex] is the new mass (16M). Substituting these into the equation, we get:
[tex]\[ \frac{T_2}{25} = \sqrt{\frac{16M}{M}} \][/tex]
Simplifying inside the square root:
[tex]\[ \frac{T_2}{25} = \sqrt{16} \][/tex]
[tex]\[ \sqrt{16} = 4 \][/tex]
Thus, we have:
[tex]\[ \frac{T_2}{25} = 4 \][/tex]
To find [tex]\( T_2 \)[/tex], we multiply both sides by 25:
[tex]\[ T_2 = 25 \times 4 \][/tex]
[tex]\[ T_2 = 100 \][/tex]
So, the period of oscillation when the mass is 16M is 100 seconds.
This means the correct answer is not listed in the provided options. However, the calculated period for mass 16M is indeed 100 seconds.
