The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.
Mathematically, the frequency of the vibration of a string can be expressed as

Where,
L = Vibrating length string
T = Tension in the string
Linear mass density
At the same time we have the expression for the number of beats described as

Where
= First frequency
= Second frequency
From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:

If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well


Replacing
for n and 202Hz for 



The frequency of the tightened is 205Hz
Answer:

Explanation:
Given that,
Radius of a spherical shell, r = 0.7 m
Torque acting on the shell, 
Angular acceleration of the shell, 
We need to find the rotational inertia of the shell about the axis of rotation. The relation between the torque and the angular acceleration is given by :

I is the rotational inertia of the shell

So, the rotational inertia of the shell is
.
Explanation:
A worker picks up the bag of gravel. We need to find the speed of the bucket after it has descended 2.30 m from rest. It is case of conservation of energy. So,

h = 2.3 m

So, the speed of the bucket after it has descended 2.30 m from rest is 6.71 m/s.
I believe the answer is the second one.