Question

Two masses one of 60 g and the other of 90 g balance each other when
suspended in water from the arms of a tarik balance. If the density of the
& mass is 8 g/ce, find that of the other,

Answer 1

The density of the body is 3 g/cm^3.

Explanation:

Correct statement:

Two bodies are equilibrium when suspended in water from the arm of a balance.the mass of one body is 36 g and its density is 9 g/cm^3 . If the mass of the other is 48 g.what is its density ?

Solution:

• The volume of body 1, V1 = m1 / ρ1 =36 / 9 = 4 cm^3
• The weight of body 1 in the water "W1" = m1g − V1ρwg
• The weight of the body 2 in water "W2" = m2g − V2ρwg
• As the two weights balances each other: Therefore W1 = W2

⇒ m1g − V1ρwg = m2g − V2ρwg

⇒ m1−V1ρw = m2 − V2ρw

⇒ 36 − 4 = 48 − V2

⇒ V2 = 48 − 32 = 16 cm^3

Therefore, the density of body 2 "ρ2" =m2 / V2 = 48 / 16 = 3 g/cm^3

Thus the density of the body is 3 g/cm^3.

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