To find the velocity of sound using the given problem, we'll utilize the principles of sound waves in organ pipes. Since both pipes are in fundamental modes, we'll use the fundamental frequency formulas for open and closed pipes.
The calculated velocity of sound \(v = 294 \, \text{m/s}\) is within the specified range (294, 294).
For a closed pipe of length \(L = 150 \, \text{cm} = 1.5 \, \text{m}\):
The fundamental frequency is given by:
\(f_c = \frac{v}{4L}.\)
For an open pipe of length \(L = 350 \, \text{cm} = 3.5 \, \text{m}\):
The fundamental frequency is given by:
\(f_o = \frac{v}{2L}.\)
Given that the beat frequency is:
\(|f_c - f_o| = 7 \, \text{Hz}.\)
Substituting:
\(\left| \frac{v}{4 \cdot 1.5} - \frac{v}{2 \cdot 3.5} \right| = 7.\)
Simplifying:
\(\left| \frac{v}{6} - \frac{v}{7} \right| = 7.\)
Solving for \(v\):
\(\frac{v}{42} = 7 \implies v = 42 \cdot 7 = 294 \, \text{m/s}.\)
The Correct answer is: 294
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 : \(A_2 \rightleftharpoons 2A\)