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lesantik [10]
2 years ago
15

A massless, hollow sphere of radius R is entirely filled with a fluid such that its density is p. This same hollow sphere is now

compressed so that its radius is R/2, and then it is entirely filled with the same fluid as before. As such, what is the density of the compressed sphere?
a. 8p
b. p/8
c. p/4
d. 4p
Physics
1 answer:
Vesnalui [34]2 years ago
7 0

Answer:

a. 8p

Explanation:

We are given that

Radius of hollow sphere , R1=R

Density of hollow sphere=\rho

After compress

Radius of hollow sphere, R2=R/2

We have to find density of the compressed sphere.

We know that

Density=\frac{mass}{volume}

Mass=Density\times volume=Constant

Therefore,\rho_1 V_1=\rho_2V_2

Volume of sphere=\frac{4}{3}\pi r^3

Using the formula

\rho\times \frac{4}{3}\pi R^3=\rho_2\times \frac{4}{3}\pi (R/2)^3

\rho R^3=\rho_2\times \frac{R^3}{8}

\rho_2=8\rho

Hence, the density of  the compressed sphere=8\rho

Option a is correct.

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As the electron is revolving in a circular path, it experiences a centripetal force which is balanced by the electrostatic force between the electron and the nucleus.

centripetal force = \frac{mv^{2}}{r}

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where, k be the Coulombic constant, k = 9 x 10^9 Nm^2 / C^2

So, balancing both the forces we get

\frac{kq^{2}}{r^{2}}=\frac{mv^{2}}{r}

v=\sqrt{\frac{kq^{2}}{mr}}

v=\sqrt{\frac{9\times 10^{9}\times1.6\times 10^{-19}\times 1.6\times 10^{-19}}{9.1\times 10^{-31}\times 5.92\times10^{-11}}}

v = 2.068 x 10^6 m / s

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