Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

An air column, closed at one end and open at the other, resonates with a tuning fork when the smallest length of the column is $50 \text{ cm}$ . The next larger length of the column resonating with the same tuning fork is

$100 \text{ cm}$

$150 \text{ cm}$

$200 \text{ cm}$

$66.7 \text{ cm}$

Step-by-Step Solution

Identify the Type of Air Column: An air column closed at one end and open at the other acts as a closed organ pipe.
Resonance Conditions for Closed Pipe: For a closed organ pipe, resonance occurs at odd multiples of $\lambda/4$ . The resonating lengths are given by $L = (2n-1)\frac{\lambda}{4}$ , where $n = 1, 2, 3, \dots$
Find the Smallest Length: The smallest length (fundamental mode, $n=1$ ) is given as $L_1 = \frac{\lambda}{4} = 50 \text{ cm}$ .
Calculate the Next Larger Length: The next larger length (first overtone, $n=2$ ) will be $L_2 = \frac{3\lambda}{4}$ . Since $\frac{\lambda}{4} = 50 \text{ cm}$ , we substitute this value into the equation for $L_2$ : $L_2 = 3 \times 50 \text{ cm} = 150 \text{ cm}$

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started