Q105 / 150 · MHT CET 2024 (03 May Shift 2) · Physics

The end correction of resonance tube is 1 cm . If the shortest length resonating with a tuning fork is 15 cm , the next resonating length will be

A
35 cm
B
40 cm
Correct answer
C
47 cm
D
64 cm

Solution

Let the shortest length be denoted as $l_1=15.0 \mathrm{~cm}$
The end correction, $\mathrm{e}=1.0 \mathrm{~cm}$ $l_1+\mathrm{e}=\frac{\lambda}{4}$
Now, $l_2+\mathrm{e}=\frac{3 \lambda}{4}$ $\begin{array}{ll} \therefore & \frac{l_1+\mathrm{e}}{l_2+\mathrm{e}}=\frac{1}{3} \\ \therefore & \frac{15+1}{l_2+1}=\frac{1}{3} \\ \therefore & l_2+1=48 \\ \therefore & l_2=47 \mathrm{~cm} \end{array}$

Hence, the correct answer is option: 'C'.

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