Question #4ad26

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Question #4ad26

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Answer 1 · 2017-08-24T04:57:24

Let the volume of the air bubble at the bottom of the lake be $V_{b}$ $V_b$ . So its volume at the surface $V_{s} = 2 V_{b}$ $V_s=2V_b$

Let the depth of the lake be $h c m$ $hcm$

So pressure at the bottom of the lake

$P_{b} = pressure of water of height h+ atmospheric prssure$ $P_b="pressure of water of height h+ atmospheric prssure"$

$P_{b} = h \times d_{w} \times g + 75 \times d_{H g} \times g$ $P_b= hxxd_wxxg+ 75xxd_(Hg)xxg$

Pressure at the surface

$P_{s} = 75 \times d_{H g} \times g$ $P_s=75xxd_(Hg)xxg$

Now applying Boyle's law we get

$P_{b} \times V_{b} = P_{s} \times V_{s}$ $P_bxxV_b=P_sxxV_s$

$\Rightarrow (h \times d_{w} \times g + 75 \times d_{H g} \times g) V_{b} = (75 \times d_{H g} \times g) \times 2 V_{b}$ $=>(hxxd_wxxg+ 75xxd_(Hg)xxg)V_b=(75xxd_(Hg)xxg)xx2V_b$

$\Rightarrow h + 75 \times \frac{d_{H g}}{d_{w}} = 75 \times \frac{d_{H g}}{d_{w}} \times 2$ $=>h+ 75xxd_(Hg)/d_w=75xxd_(Hg)/d_wxx2$

$\Rightarrow h = 75 \times \frac{d_{H g}}{d_{w}}$ $=>h=75xxd_(Hg)/d_w$

$\Rightarrow h = 75 \times \frac{40}{3} = 1000 c m = 10 m$ $=>h=75xx40/3=1000cm=10m$

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Question #4ad26

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