The velocity of sound \( v \) in an open organ pipe is given by the formula: \[ v = f \cdot \lambda \] where \( f \) is the frequency, and \( \lambda \) is the wavelength of the wave. The fundamental frequency of an open organ pipe is given by: \[ \lambda_1 = 2L \] where \( L \) is the length of the tube. For the second harmonic, the wavelength \( \lambda_2 \) is: \[ \lambda_2 = L \] Using the relationship \( v = f \cdot \lambda \), we find that the second harmonic corresponds to the frequency of 1.1 kHz.
Therefore, the correct harmonic is the second harmonic. Hence, the correct answer is (d).
Match List-I with List-II.
Choose the correct answer from the options given below :}
There are three co-centric conducting spherical shells $A$, $B$ and $C$ of radii $a$, $b$ and $c$ respectively $(c>b>a)$ and they are charged with charges $q_1$, $q_2$ and $q_3$ respectively. The potentials of the spheres $A$, $B$ and $C$ respectively are:
Two resistors $2\,\Omega$ and $3\,\Omega$ are connected in the gaps of a bridge as shown in the figure. The null point is obtained with the contact of jockey at some point on wire $XY$. When an unknown resistor is connected in parallel with $3\,\Omega$ resistor, the null point is shifted by $22.5\,\text{cm}$ towards $Y$. The resistance of unknown resistor is ___ $\Omega$. 