Question

Air column of 20 cm length in a resonance tube resonates with a certain tuning fork when sounded at its upper open end. The lower end of the tube is closed and adjustable by changing the quantity of mercury filled inside the tube. The temperature of the air is 27°C. The change in length of the air column required, if the temperature falls to 7°C and the same tuning fork is again sounded at the upper open end is nearly
(a) 1mm
(b) 7mm
(c) 5mm
(d) 13mm

Answer with proper explaination​

Answer 1

Given:

• Length of the air column(L₁) = 20 cm
• Initial temperature(T₁) = 27°C = 300 K
• Final temperature(T₂)  = 7°C = 280 K

To find:

• Change in length of the air column

Explanation:

As per the question, the upper end of the tube is open and the lower end is closed.

We know that the fundamental frequency of an open organ pipe is,

$\to\boxed{ \sf{n = \dfrac{v}{4L}}}$

As the same tuning fork is used for both cases the frequency of the tuning fork remains constant.

Hence,

$\to \sf{ v \propto L ....(1)}$

Velocity of the sound in air is given by,

$\to\boxed{ \sf{v = \sqrt{\dfrac{\gamma RT}{M}}}}$

here,

• γ = adiabatic cnst
• R = universal gas cnst
• T = temperature
• M = molecular mass of the gas

As γ, R, M are constant we can say that,

$\to \sf{ v \propto \sqrt{T}....(2)}$

From eq (1) and (2) we can conclude that,

$\to\sf{\dfrac{L_1}{L_2} =\sqrt{ \dfrac{T_1}{T_2}}}$

$\to\sf{\dfrac{L_1}{L_2} =\sqrt{ \dfrac{300}{280}}}$

$\to\sf{\dfrac{L_1}{L_2} =\sqrt{1.0714}}$

$\to\sf{\dfrac{L_1}{L_2} =1.035}$

$\to\sf{L_2= \dfrac{L_1}{1.035} }$

$\to\sf{L_2= \dfrac{20}{1.035} }$

$\to\sf{L_2= 19.32\: cm }$

Change in length of the air column

= 20 - 19.3

= 0.7 cm

= 7 mm

The change in length of the air column is 7mm.

The answer is option (b).

Answer 2

Answer:

Hello 'K'

$n (oop) = \frac{v}{2l}$

refer to attachment !

answers are taken in approx. value!