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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. - Physics

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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

Sum

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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Relation Between Volume of Submerged Part of a Floating Body, the Densities of Liquid and the Body
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Chapter 5: Upthrust and Archimedes’ Principle - Practice Problems 1

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Goyal Brothers Prakashan A New Approach to ICSE Physics Part 1 [English] Class 9
Chapter 5 Upthrust and Archimedes’ Principle
Practice Problems 1 | Q 2
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