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A hollow cylinder of copper of length 25 cm and area of cross-section 15 cm2, floats in water with 3/5 of its length inside water. Calculate: - Physics

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Question

A hollow cylinder of copper of length 25 cm and area of cross-section 15 cm2, floats in water with 3/5 of its length inside water. Calculate:
(1) apparent density of a hollow copper cylinder.
(2) wt. of cylinder.
(3) extra force required to completely submerge it in water.

Sum

Solution

(1) Length of hollow cylinder of copper = hcu= l= 25 m
Length of hollow cylinder of copper inside water

`"h"_"water"= 3/5"l"`

`"h"_"water" = 3/5 xx 25`

`"h"_"water" = 15 "cm"`

Area of cross - section of hollow cylinder of copper = 15 cm2

Density of water = `rho_"water" = 1 "g cm"^-3`

Apparent density of hollow copper cylinder = `rho_"cu" = ?`

`"h"_"cu" xx rho_"cu" = "h"_"water" xx rho_"water"`

`25 xx rho_"cu" = 15 xx 1`

`rho_"cu" = 15/25 = 0.6` g cm-3 

(2) Volume of cylinder = V = `"A" xx "h"_"cu"`

= 15 × 25 = 375 cm3

Mass of hollow cylinder of copper = `"V" xx rho_"cu"`

m = `"Ah"_"cu" xx rho_"cu"`

m = 15 × 25 × 0.6

m = 225 g

Weight of hollow cylinder of copper = mg

= 225 × g = 225 gf

(3) Total upthrust when hollow copper cylinder is completely immersed in water = `"V" rho_"water" "g"`

= 375 × 1 × g = 375 gf

Extra force required to submerge complete the cylinder in water = Upthrust - down thrust

= 375 - 225 = 150 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 1
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