Q106 / 150 · MHT CET 2026 (15 April Shift 1) · Physics
A Pipe open at one end has length $0.8$ m. At the open end of the tube a string $0.5$ m long is vibrating in its first overtone and resonates with fundamental frequency of pipe. If tension in the string is $50$ N, the mass of string is (Neglect end correction) (Speed of sound $= 320$ m/s)
- A$2$ gram
- B$5$ gram
- Correct answerC$10$ gram
- D$20$ gram
Solution
The fundamental frequency of a pipe open at one end (closed pipe) is given by $f_p = \dfrac{v}{4L_p}$.
Substituting the given values, $f_p = \dfrac{320}{4 \times 0.8} = 100$ Hz.
The string is vibrating in its first overtone, which corresponds to the second harmonic. The frequency of the string is $f_s = \dfrac{2}{2L_s} \sqrt{\dfrac{T}{\mu}} = \dfrac{1}{L_s} \sqrt{\dfrac{T}{\mu}}$.
Since the string resonates with the pipe, $f_s = f_p$.
$\dfrac{1}{0.5} \sqrt{\dfrac{50}{\mu}} = 100$
$2 \sqrt{\dfrac{50}{\mu}} = 100 \Rightarrow \sqrt{\dfrac{50}{\mu}} = 50$
Squaring both sides, we get $\dfrac{50}{\mu} = 2500 \Rightarrow \mu = \dfrac{50}{2500} = 0.02$ kg/m.
The mass of the string is $m = \mu L_s = 0.02 \times 0.5 = 0.01$ kg.
Converting to grams, $m = 10$ g.
Answer: $10$ gram
Substituting the given values, $f_p = \dfrac{320}{4 \times 0.8} = 100$ Hz.
The string is vibrating in its first overtone, which corresponds to the second harmonic. The frequency of the string is $f_s = \dfrac{2}{2L_s} \sqrt{\dfrac{T}{\mu}} = \dfrac{1}{L_s} \sqrt{\dfrac{T}{\mu}}$.
Since the string resonates with the pipe, $f_s = f_p$.
$\dfrac{1}{0.5} \sqrt{\dfrac{50}{\mu}} = 100$
$2 \sqrt{\dfrac{50}{\mu}} = 100 \Rightarrow \sqrt{\dfrac{50}{\mu}} = 50$
Squaring both sides, we get $\dfrac{50}{\mu} = 2500 \Rightarrow \mu = \dfrac{50}{2500} = 0.02$ kg/m.
The mass of the string is $m = \mu L_s = 0.02 \times 0.5 = 0.01$ kg.
Converting to grams, $m = 10$ g.
Answer: $10$ gram
Hence, the correct answer is option: 'C'.