
When a narrow resonance column apparatus is filled with ozone gas $(y = \dfrac{4}{3})$, it resonates in $1st$ resonance with a tuning fork. When it is filled with nitrogen gas ($N2$), it resonates with the same tuning fork in $1st$ resonance when the water column is shifted down by $10$cm. (Take $V5 = 2.25$)
A. wavelength of sound in ozone is approximately $1$m.
B. wavelength of sound in Nitrogen is approximately $1.6$m.
C. For second resonance in ozone gas, the water column should be shifted down by $60$cm.
D. For second resonance in Nitrogen gas, the water column should be shifted by $80$cm.
Answer
524.7k+ views
Hint: Here we will use the formula for the speed of sound, the standard formula and accordingly will find the ratios of two different speeds. Then will check for the given multiple choices are applicable or not.
Complete step by step answer:
Speed of Sound can be given by
Speed $ = \sqrt {\dfrac{{rRT}}{M}} $
When there are two different speeds, the ratio of both the speed can be given as-
$\dfrac{{{v_1}}}{{{v_2}}} = \sqrt {\dfrac{{{r_1}{M_2}}}{{{r_2}{M_1}}}} $
Place the given values in the above equation-
$\dfrac{{{v_1}}}{{{v_2}}} = \sqrt {\dfrac{{4 \times 5 \times 28}}{{3 \times 7 \times 40}}} $
Simplify the above expression-
$\dfrac{{{v_1}}}{{{v_2}}} = \sqrt {\dfrac{2}{3}} $
Now, the ratios of the wavelength can be given as –
$\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{{v_1}}}{{{v_2}}} = \sqrt {\dfrac{2}{3}} $
Also, we have ${\lambda _2} = {\lambda _1} + 10$
Place in the above equation –
$\dfrac{{{\lambda _1}}}{{{\lambda _1} + 10}} = \sqrt {\dfrac{2}{3}} $
Simplify the above equation –
$\dfrac{{{\lambda _1}}}{{{\lambda _1} + 10}} = 0.8$
Make the required unknown the subject –
$
0.2{\lambda _1} = 8 \\
{\lambda _1} = 40cm \\
$
This is the required solution.
So, the correct answer is Option B,CD.
Note: Always remember that the numerator’s denominator goes in denominator and the denominator’s denominator goes to the numerator of the fraction. Be good in simplification of basic mathematical concepts. Always remember when we take a ratio of two speeds, then the constant terms of both the speeds are removed since they cancel each other.
Complete step by step answer:
Speed of Sound can be given by
Speed $ = \sqrt {\dfrac{{rRT}}{M}} $
When there are two different speeds, the ratio of both the speed can be given as-
$\dfrac{{{v_1}}}{{{v_2}}} = \sqrt {\dfrac{{{r_1}{M_2}}}{{{r_2}{M_1}}}} $
Place the given values in the above equation-
$\dfrac{{{v_1}}}{{{v_2}}} = \sqrt {\dfrac{{4 \times 5 \times 28}}{{3 \times 7 \times 40}}} $
Simplify the above expression-
$\dfrac{{{v_1}}}{{{v_2}}} = \sqrt {\dfrac{2}{3}} $
Now, the ratios of the wavelength can be given as –
$\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \dfrac{{{v_1}}}{{{v_2}}} = \sqrt {\dfrac{2}{3}} $
Also, we have ${\lambda _2} = {\lambda _1} + 10$
Place in the above equation –
$\dfrac{{{\lambda _1}}}{{{\lambda _1} + 10}} = \sqrt {\dfrac{2}{3}} $
Simplify the above equation –
$\dfrac{{{\lambda _1}}}{{{\lambda _1} + 10}} = 0.8$
Make the required unknown the subject –
$
0.2{\lambda _1} = 8 \\
{\lambda _1} = 40cm \\
$
This is the required solution.
So, the correct answer is Option B,CD.
Note: Always remember that the numerator’s denominator goes in denominator and the denominator’s denominator goes to the numerator of the fraction. Be good in simplification of basic mathematical concepts. Always remember when we take a ratio of two speeds, then the constant terms of both the speeds are removed since they cancel each other.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

