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प्रश्न
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.
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उत्तर
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`
ρlead = 10.6 × 1 = 10.6 gcm-3
Mass of lead = m = V × ρlead
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`
ρsugar = 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
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