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प्रश्न
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.
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उत्तर
(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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