To solve this problem, we need to understand the relation between the frequencies of closed and open organ pipes.
Therefore, the length of the closed pipe is 15 cm.
Hence, the correct answer is 15 cm.
The relationship for the fundamental frequency of a closed organ pipe (length \( L_1 \)) and the first overtone of an open organ pipe (length \( L_2 \)) is given by:
\[ \frac{\lambda}{4} = L_1 \quad \text{and} \quad 2 \left( \frac{\lambda}{2} \right) = \lambda \]
The velocity of sound is \( v \), thus:
\[ v = f \lambda \]
For the closed pipe:
\[ v = f_1 (4L_1) \]
For the open pipe:
\[ f_2 = \frac{v}{2L_2} \]
Equating the fundamental frequency of the closed pipe to the first overtone of the open pipe:
\[ f_1 = f_2 \] \[ \frac{v}{4L_1} = \frac{v}{2L_2} \] \[ L_2 = 4L_1 \]
Given \( L_2 = 60 \, \text{cm} \):
\[ 60 = 4 \times L_1 \] \[ L_1 = 15 \, \text{cm} \]
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)