A uniform cylindrical hydrometer of mass 20g and cross sectional area 0.54cm2 floats upright in a liquid. If 25cm of its length is submerged, calculate the relative density of the liquid. (Density of water = 1 gcm-3)

a

1.54

b

1.48

c

1.25

d

0.80

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b

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VPETERS2002
5 years ago

Answer: 1.48

Relative density= density of liquid ÷ density of water

density of water=1gcm^-3

density of liquid
=mass/volume
=mass/Area*height
=20 / (0.54*25)
=20/13.5
=1.48gcm^-3

Therefore;

Relative density
=1.48gcm^-3/1gcm^-3
=1.48 ( Note Relative Density has no unit)

Johnanamboi
5 years ago

answer below

Johnanamboi
5 years ago

Mh = 20g
CSA= 0.54cm3
h of hygrometer in H²O = 25cm
Dw = 1g/cm3
Df = ?
Dr = ?
but Dr = Df/Dw
Df = Mf/Vf
but according to Archimedes principle, a floating body displaces its own mass of fluid.
=> Mf = Mh=20g
and also, according to the law of flotation, a floating body displaces its own volume of fluid.
vf = Vh
Vh = CSA × h
Vh = 0.54× 25 =13.5cm3
Vf = Vh = 13.5cm3
Df = Mf/Vf
=> Df = 20g/13.5cm3 = 1.48g/cm3
Dr = Df/Dw
Dr = (1.48g/cm3)/(1.00g/cm3)
Dr = 1.48
Note that it has no unit because it is a ratio of two similar quantities.
[D = density
r, f, w, h = relative, fluid, water and heigt respectively
V = volume
]

Achisons
3 weeks ago

Ans
1.48

Josephwale
5 years ago

from p=m/v, m=pv, m=pahg from v=area * height. therefore, p=m/ahg

EvaElens
6 years ago

A uniform cylindrical hydrometer of mass 20g and cross sectional area 0.54cm2 floats upright in a liquid. If 25cm of its length is submerged, calculate the relative density of the liquid. (Density of water = 1 gcm-3)

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