both questions of gaseous state
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9. The force will be equal to hAρ = 0.1 x (π x 0.4²) x 13600 = 13.6 x 16 x π = 144.16π
10. [tex]h_{1}\rho_{1} = h_{2}\rho_{2} \\ \\ h_{1} = h_{2} \times 13.6 \\ \\ h_{1}-h_{2} = h_{2}\times 12.6 = 2 \implies h_{2} = 6.3 cm \\ \\
h_{1} =[/tex]Height of water column =
10. [tex]h_{1}\rho_{1} = h_{2}\rho_{2} \\ \\ h_{1} = h_{2} \times 13.6 \\ \\ h_{1}-h_{2} = h_{2}\times 12.6 = 2 \implies h_{2} = 6.3 cm \\ \\
h_{1} =[/tex]Height of water column =
smritisaraswat:
both sol. are wrong the answer of first one is 67 N and second one is 27.2 . but still concept was correct it helped me .
Is this question from allen's book?
It shouldn't be wrong
nice guess .
but the answers are diff. still thanku for helping.
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9.
Force = mgh = V ρ g = AL ρg = π * (0.04)^2 * 0.1 * 13560 * 9.8 ≈ 67N
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10.
Let h1 and ρ1 be the height of mercury and it’s density respectively
h2 and ρ2 be the height and density of water respectively
Specific gravity = ρ1/ρ2 = 13.6
h1 ρ1 = h2 ρ2
h1/h2 = ρ2/ρ1
h1/h2 = 1/13.6
h2 = 13.6h1
Given that
h2 - h1 = 2
13.6h1 - h1 = 2
h1 * (13.6 - 1) = 2
h1 = 2/12.6
h1 = 0.158
If h1 is 0.158cm then h2 will be 2 + 0.158 = 2.158 cm
Force = mgh = V ρ g = AL ρg = π * (0.04)^2 * 0.1 * 13560 * 9.8 ≈ 67N
______________________________
10.
Let h1 and ρ1 be the height of mercury and it’s density respectively
h2 and ρ2 be the height and density of water respectively
Specific gravity = ρ1/ρ2 = 13.6
h1 ρ1 = h2 ρ2
h1/h2 = ρ2/ρ1
h1/h2 = 1/13.6
h2 = 13.6h1
Given that
h2 - h1 = 2
13.6h1 - h1 = 2
h1 * (13.6 - 1) = 2
h1 = 2/12.6
h1 = 0.158
If h1 is 0.158cm then h2 will be 2 + 0.158 = 2.158 cm
thanks for it . but the second one not right. its ans is 27 cm
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