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physics ncert solutions class 11th Physics Class 11th Kinetic Theory

13.11 A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?

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Payal Gupta | Contributor-Level 10

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13.11 Length of the narrow bore, L = 1 m = 100 cm

Length of the mercury thread, l = 76 cm

Length of the air column between mercury and the closed end, $l_{o}$ = 15 cm

Since the bore is held vertically in air with the open end at the bottom, the mercury length that occupies the air space is 100 – (76 + 15) = 9 cm

Hence, total length of the air column = 15 + 9 = 24 cm

Let h cm of mercury flow out as a result of atmospheric pressure.

Length of the air column in the bore = 24 + h cm

Length of the mercury column = 76 – h cm

Initial pressure, $P_{1}$ = 76 cm of mercury

Initial volume, $V_{1}$ = 15 ${c m}^{3}$

Final pressure, $P_{2}$ = 76 – (76 – h) = h cm

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13.11 Length of the narrow bore, L = 1 m = 100 cm

Length of the mercury thread, l = 76 cm

Length of the air column between mercury and the closed end, $l_{o}$ = 15 cm

Since the bore is held vertically in air with the open end at the bottom, the mercury length that occupies the air space is 100 – (76 + 15) = 9 cm

Hence, total length of the air column = 15 + 9 = 24 cm

Let h cm of mercury flow out as a result of atmospheric pressure.

Length of the air column in the bore = 24 + h cm

Length of the mercury column = 76 – h cm

Initial pressure, $P_{1}$ = 76 cm of mercury

Initial volume, $V_{1}$ = 15 ${c m}^{3}$

Final pressure, $P_{2}$ = 76 – (76 – h) = h cm of mercury

Final volume, $V_{2}$ = (24 + h) ${c m}^{3}$

Since temperature remained constant throughout the process,

$P_{1} V_{1}$ = $P_{2} V_{2}$ or 76 $* 15$ = h $* (24 + h)$

Or, $h^{2}$ + 24h - 76 $* 15$ =0

h = $\frac{- 24 \pm \sqrt{24^{2} + 4 * 76 * 15}}{2}$ = $\frac{- 24 \pm 71.67}{2}$

So h = 23.84 cm or -47.84 cm

Since height cannot be negative, so h = 23.84 cm

23.84 cm of mercury will flow out from the bore and (76-23.84) 52.16 cm of mercury will remain in the bore. The length of the air column will be 24 + 23.84 = 47.84 cm

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13.11 Length of the narrow bore, L = 1 m = 100 cmLength of the mercury thread, l = 76 cmLength of the air column between mercury and the closed end, <math><msub><mrow><mrow><mi>l</mi></mrow></mrow><mrow><mrow><mi>o</mi></mrow></mrow></msub></math>  = 15 cmSince the bore is held vertically in air with the open end at the bottom, the mercury length that occupies the air space is 100 – (76 + 15) = 9 cmHence, total length of the air column = 15 + 9 = 24 cmLet h cm of mercury flow out as a result of atmospheric pressure.Length of the air column in the bore = 24 + h cmLength of the mercury column = 76 – h cmInitial pressure,  <math><msub><mrow><mrow><mi>P</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math> = 76 cm of mercuryInitial volume, <math><msub><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math>= 15 <math><msup><mrow><mrow><mi>c</mi><mi>m</mi></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup></math>Final pressure, <math><msub><mrow><mrow><mi>P</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math> = 76 – (76 – h) = h cm of mercuryFinal volume, <math><msub><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math> = (24 + h) <math><msup><mrow><mrow><mi>c</mi><mi>m</mi></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup></math>Since temperature remained constant throughout the process,  <math><msub><mrow><mrow><mi>P</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub><msub><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math>=<math><msub><mrow><mrow><mi>P</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub><msub><mrow><mrow><mi>V</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math> or 76 <math><mo>*</mo><mn>15</mn></math> = h <math><mo>*</mo><mo> (</mo><mn>24</mn><mo>+</mo><mi>h</mi><mo>)</mo></math><math><mi></mi></math>Or, <math><msup><mrow><mrow><mi>h</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> + 24h - 76 <math><mo>*</mo><mn>15</mn></math> =0h = <math><mfrac><mrow><mrow><mo>-</mo><mn>24</mn><mo>±</mo><msqrt><mrow><msup><mrow><mrow><mn>24</mn></mrow></mrow><mrow><mrow><mn>2</mn><mi></mi></mrow></mrow></msup><mo>+</mo><mi></mi><mn>4</mn><mo>*</mo><mn>76</mn><mo>*</mo><mn>15</mn></mrow></msqrt></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac></math>=<math><mfrac><mrow><mrow><mo>-</mo><mn>24</mn><mo>±</mo><mn>71.67</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac></math>So h = 23.84 cm or -47.84 cmSince height cannot be negative, so h = 23.84 cm23.84 cm of mercury will flow out from the bore and (76-23.84) 52.16 cm of mercury will remain in the bore. The length of the air column will be 24 + 23.84 = 47.84 cm

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13.11 A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?

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