1. Fundamental frequency of a closed pipe

For an air column closed at one end, the fundamental frequency is

$f = \frac{v}{4L}$

where:

• $v$ = speed of sound
• $L$ = length of air column

2. Frequencies of the two columns

Given:

• $L_1 = 100\text{ cm} = 1.0\text{ m}$
• $L_2 = 120\text{ cm} = 1.2\text{ m}$

So their fundamental frequencies are

$f_1 = \frac{v}{4\times 1.0} = \frac{v}{4}$

$f_2 = \frac{v}{4\times 1.2} = \frac{v}{4.8}$

3. Use beat frequency condition

Beat frequency is the difference of frequencies:

$|f_1-f_2| = 15$

Thus,

$\left|\frac{v}{4} - \frac{v}{4.8}\right| = 15$

Take the difference:

$v\left(\frac{1}{4} - \frac{1}{4.8}\right)=15$

Now,

$\frac{1}{4} = 0.25, \qquad \frac{1}{4.8} = 0.20833$

So,

$v(0.25 - 0.20833)=15$

$v(0.04167)=15$

$v=\frac{15}{0.04167}\approx 360\text{ m/s}$

4. Match with options

$v = 360\text{ m/s}$

So the correct option is D.