Sitar maestro Ravi Shankar is playing sitar on its strings, and you, as a physicist (unfortunately without musical ears!), observed the following oddities.
I. The greater the length of a vibrating string, the smaller its frequency.
II. The greater the tension in the string, the greater is the frequency.
III. The heavier the mass of the string, the smaller the frequency.
IV. The thinner the wire, the higher its frequency.
The maestro signalled the following combination as the correct one.
A. II, III and IV
B. I, II and IV
C. I, II and III
D. I, II, III and IV

Hint: When string is fixed from both ends nodes will be formed at both ends. When a string is only fixed from one end and set free at the other end node is formed at fixed end and antinode is formed at the free end. Since tension and linear mass density are related to velocity of standing waves in the string and wavelength is related to length of string(l). We can find frequency from this.

Formula used:
$v = \sqrt {\dfrac{T}{\mu }} $
$v = f\lambda $

Complete step by step answer:
When string is fixed at both ends initially one loop will be formed and hence wavelength of that loop be ${\lambda _1}$
For one loop distance is $\dfrac{{{\lambda _1}}}{2}$
So initially $\dfrac{{{\lambda _1}}}{2} = l$, ${f_1} = \dfrac{v}{{{\lambda _1}}} = \dfrac{v}{{2l}}$ … eq1
Then next it forms two loops this time wave length be ${\lambda _2}$
So distance would be ${\lambda _2} = l$, ${f_2} = \dfrac{v}{{{\lambda _2}}} = \dfrac{v}{l} = \dfrac{{2v}}{{2l}}$ … eq2
Then it forms three loops now wavelength is ${\lambda _3}$
So distance would be $\dfrac{{3{\lambda _3}}}{2} = l$, ${f_3} = \dfrac{v}{{{\lambda _3}}} = \dfrac{v}{{\dfrac{{2l}}{3}}} = \dfrac{{3v}}{{2l}}$ … eq3
From equation 1 and 2 and 3 we get
${f_n} = \dfrac{{nv}}{{2l}}$
Velocity of wave would be $\sqrt {\dfrac{T}{\mu }} {\text{ }}$
Where T is tension and $\mu $ is linear mass density
So whatever the harmonic is we can consider ‘n’ as a constant and deduce the below given relations.
$\eqalign{
  & {f_n} = \dfrac{{n\sqrt {\dfrac{T}{\mu }} {\text{ }}}}{{2l}} \cr
  & \Rightarrow f\alpha \sqrt T \cr
  & \Rightarrow f\alpha \dfrac{1}{{\sqrt \mu }} \cr
  & \Rightarrow f\alpha \dfrac{1}{l} \cr
  & \therefore f\alpha \dfrac{1}{{\sqrt \mu }}\alpha \dfrac{1}{{\sqrt {\dfrac{m}{l}} }}\alpha \dfrac{1}{{\sqrt {\dfrac{{\rho Al}}{l}} }}\alpha \dfrac{1}{{\sqrt A }} \cr} $
Thickness is proportional to area. Hence as area increases thickness increases and frequency decreases.
Hence all the four options are correct.

So, the correct answer is “Option D”.

Note:
The reason why we applied the both ends fixed condition is because when the string is vibrating in the guitar, obviously both ends of the strings will be fixed on the guitar so nodes will be formed at the extreme ends of the guitar. If one end is free then antinode will form there but that’s not the case here.