An open organ pipe is closed at one end such that the third overtone of the closed pipe is found to be higher in frequency by $200 \ Hz$ than the second overtone of the original open pipe. The fundamental frequency of the open pipe is (Neglect end correction). (in $Hz$)

An open tube is in resonance with a string (frequency of vibration of the tube is $n_0$). If the tube is dipped in water so that $75\%$ of the length of the tube is inside the water,then the ratio of the frequency of the tube to the string now will be

Two closed organ pipes,when sounded simultaneously,give $4$ beats per second. If the longer pipe has a length of $1 \ m$,then the length of the shorter pipe will be,... $cm$ $(v = 300 \ m/s)$.

$A$ closed pipe is suddenly opened and changed to an open pipe of the same length. The fundamental frequency of the resulting open pipe is less than that of the $3^{rd}$ harmonic of the earlier closed pipe by $55 \,Hz$. Then, the value of the fundamental frequency of the closed pipe is: (in $\,Hz$)

The closed and open organ pipe have same length and when they are vibrating simultaneously in first overtone produce $3$ beats. The length of open pipe is made $(\frac{1}{3})^{rd}$ and closed pipe is made $3$ times the original, the number of beats produced will be {neglect end correction}

$A$ tuning fork vibrates with frequency $256\, Hz$ and gives one beat per second with the third normal mode of vibration of an open pipe. What is the length of the pipe in $cm$? (Speed of sound in air is $340\, m/s$)