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Vitek1552 [10]
2 years ago
9

There is a small hole of radius r in a hollow sphere, which is immersed in a liquid. Upto what maximum depth it can be immersed

so that liquid may not enter in it? Density of liquid is d and surface tension is T.
Physics
1 answer:
densk [106]2 years ago
4 0

Answer:

h = 4T/dgr

Explanation:

Using Laplace law for spherical bubble, the pressure difference P'- P = 4T/r where T = surface tension and r = radius of sphere.

Now, the pressure difference on the hollow sphere P' - P = dgh where d = density of liquid, g = acceleration due to gravity and h = maximum depth to which sphere must be immersed.  

So dgh = 4T/r

h = 4T/dgr

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Answer:

\displaystyle \rho=1.25\ g/ml

Explanation:

<u>Density </u>

The density of a substance is the mass per unit volume. The density varies with temperature and pressure.

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\displaystyle \rho=\frac{m}{V}

The cube has a mass of m=3.75 g and a volume of V=3 ml, thus the density is:

\displaystyle \rho=\frac{3.75\ g}{3\ ml}

\boxed{\displaystyle \rho=1.25\ g/ml}

Since 1 kg=1000 mg and 1 lt = 1000 ml, the density has the same value but with different units:

\displaystyle \rho=1.25\ kg/l

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rusak2 [61]

Answer : The volume of a sample of 4.00 mol of copper is 28.5cm^3

Explanation :

First we have to calculate the mass of copper.

\text{ Mass of copper}=\text{ Moles of copper}\times \text{ Molar mass of copper}

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