A tuning fork vibrating with a frequency of 512Hz is kept close to the open end of a tube filled with water (Fig. 15.4). The water level in the tube is gradually lowered. When the water level is 17cm below the open end, maximum intensity of sound is heard. If the room temperature is 20° C, calculate

(a) Speed of sound in air at room temperature

(b) Speed of sound in air at 0° C

(c) If the water in the tube is replaced with mercury, will there be any difference in your observations?

This is a short answer type question as classified in NCERT Exemplar

consider the frequency of tuning fork= 512Hz

(a) L= $\frac{\lambda }{4}$

$\lambda =4L$

V= $v\lambda$ = 512 $\times 4\times 17\times {10}^{-2}$

= 348.16m/s

(b) we know that v $\propto \sqrt[]{T}$

$\frac{V_{20}}{V_0}$ = $\sqrt[]{\frac{273+20}{273}}=\sqrt[]{\frac{293}{273}}$

$\frac{V_{20}}{V_0}=\sqrt[]{1.073}=1.03$

V_o= $\frac{V_{20}}{1.03}=\frac{348.16}{1.03}=338m/s$

(c) Resonance will be observed at 17cm length of air column only intensity of sound heard may be greater due to more complete reflection of the sound waves at the mercury surface because mercury is more denser than water.

<p><span data-teams="true">This is a short answer type question as classified in NCERT Exemplar</span></p><p>consider the frequency of tuning fork= 512Hz</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745488164phplOW5Sq_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745488164phplOW5Sq.jpeg" alt="" width="152" height="140"></picture></div><div><p><strong> (a)</strong> L=<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </mfrac> </math> </span></p><p><span contenteditable="false"> <math> <mi>λ</mi> <mo>=</mo> <mn>4</mn> <mi>L</mi> </math> </span></p><p>V=<span contenteditable="false"> <math> <mi>v</mi> <mi>λ</mi> </math> </span>= 512<span contenteditable="false"> <math> <mo>×</mo> <mn>4</mn> <mo>×</mo> <mn>17</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </mrow> </msup> </math> </span></p><p>= 348.16m/s</p><p><strong> (b)</strong> we know that v <span contenteditable="false"> <math> <mo>∝</mo> <mroot> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow></mrow> </mroot> </math> </span></p><p><span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mn>20</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </math> </span>=<span contenteditable="false"> <math> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mn>273</mn> <mo>+</mo> <mn>20</mn> </mrow> </mrow> <mrow> <mrow> <mn>273</mn> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> <mo>=</mo> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mn>293</mn> </mrow> </mrow> <mrow> <mrow> <mn>273</mn> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> </math> </span></p><p><span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mn>20</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> <mo>=</mo> <mroot> <mrow> <mrow> <mn>1.073</mn> </mrow> </mrow> <mrow></mrow> </mroot> <mo>=</mo> <mn>1.03</mn> </math> </span></p><p>V_o=<span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mn>20</mn> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <mn>1.03</mn> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>348.16</mn> </mrow> </mrow> <mrow> <mrow> <mn>1.03</mn> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>338</mn> <mi>m</mi> <mo>/</mo> <mi>s</mi> </math> </span></p><p><strong> (c)</strong> Resonance will be observed at 17cm length of air column only intensity of sound heard may be greater due to more complete reflection of the sound waves at the mercury surface because mercury is more denser than water.</p></div></div>

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