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Question
A cork cut in the form of a cylinder floats in alcohol of density 0.8 gcm-3, such that 3/4 of its length is outside alcohol. If the total length of the cylinder is 35 cm and area of cross-section 25 cm2, calculate:
(1) density of cork
(2) wt. of cork
(3) extra force required to submerge it in alcohol
Solution
(1) Density of alcohol = `rho_"alcohol"` = 0.8 g cm-3
Total length of cork cylinder = `"h"_"cork" = 35` cm
Area of cross-section of cork cylinder = A = 25 cm2
∵ `3/7` of the length of cork cylinder is outside the alcohol
∴ Length of cork cylinder inside alcohol = `(1 - 3/7)l`
`"h"_"alcohol" = (7-3)/7 xx 35`
`"h"_"alcohol" = 4 xx 5 = 20` cm
Density of water = `rho_"water"` = 1 g cm-3
By a law of floatation:
`"h"_"cork" xx rho_"cork" = "h"_"alcohol" xx rho_"alcohol"`
`35 xx rho_"cork" = 20 xx 0.8`
`rho_"cork" = (20xx0.8)/35 = 0.457` g cm-3
(2) Volume of cork = V = `"A" "h"_"cork"`
V = 25 × 35 = 875 cm3
Mass of cork = m = `"V" xx rho_"cork"`
m = 875 × 0.457
m = 399.9 g ≅ 400 g
Wt. of cork = mg = 400 × g = 400 gf
(3) Total upthrust when cork is completely immersed in water
`= "V" rho_"alcohol" xx "g"`
= 875 × 0.8 × g = 700 g = 700 gf
Extra force required to submerge complete the cork in alcohol = Upthrust - down thrust
= 700 - 400 = 300 gf
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