Question

A cube of the lead of side 8 cm and R.D. 10.6 is suspended from the hook of a spring balance. Find the reading of spring balance. The cube is now completely immersed in sugar solution of R.D. 1.4. Calculate the new reading of spring balance.

Answer 1

Length of side of cube = l = 8 cm
Volume of cube =β= (8)³ = 512 cm³
V = 512 cm³
Relative density of lead cube = R.D. = 10.6 Relative density of sugar solution = R.D. 1.4
Density of water = 1 g cm^-3 

R.D. of lead = `"Density of lead"/"Density of water"`

`10.6 = rho_"lead"/1`

ρ_lead = 10.6 × 1 = 10.6 gcm^-3 

Mass of lead = m = V × ρ_lead 

m = 512 × 10.6 = 5427.2 g

Wt. of led cube = mg

= 5427.2 × g = 5427.2 gf

Volume of sugar solution displaced = Volume of lead cube

V = 512 cm³ 

R.D. of sugar solution = `"Density of sugar solution"/"Density of water"`

`1.4 = rho_"sugar"/1`

ρ_sugar = 1.4 × 1 = 1.4 g cm^-3 

Upthrust due to sugar solution = `"V" xx rho_"sugar" xx "g"`

= 512 × 1.4 × g

= 716.8 g = 716.8 gf

Now reding of spring balance = Actual weight - Upthrust

= 5427.2 - 716.8 = 4710.4 gf