91影视

Radius of a drop of water,$r=10渭\mathrm{m}$

Height of the cloud,$h=3.0\mathrm{km}$

Radius of the cloud, $R=1.0\mathrm{km}$

Water drops in one cubic centimeter of cumulus cloud are $50\mathrm{to}500$.

Density of water, $p=1000\mathrm{kg}/{\mathrm{m}}^3$.

The density of a material (in this case water) is defined as the mass per unit volume.

The expression for density is given as:

$p=\frac{m}{V}$ 鈥� (i)

Here, $p$is the density, $m$is the mass and $V$is the volume.

The volume of the cloud is calculated as:

$V={\mathrm{蟺搁}}^{2}h$

$$\begin{gathered} =3.14脳{\left(1.0\mathrm{km}脳\frac{1000\mathrm{m}}{1\mathrm{km}}\right)}^2脳\left(3.0\mathrm{km}脳\frac{1000\mathrm{m}}{1\mathrm{km}}\right) \\ =9.42脳{10}^9脳{\mathrm{m}}^3 \end{gathered}$$

Since $1{\mathrm{cm}}^3={10}^{鈥�6}{\mathrm{m}}^3$and the number of water droplets in one cubic cm are 50 to 500 therefore, the number of water droplets in one cubit meter are $50脳{10}^6$to $500脳{10}^6$.

So, the number of total water droplets in entire cloud are,

$\left(9.42脳{10}^9{\mathrm{m}}^3\right)脳\left(50脳{10}^6\right)=4.71脳{10}^{17}$to $\left(9.42脳{10}^9{\mathrm{m}}^3\right)脳\left(500脳{10}^6\right)=4.71脳{10}^{18}$

The volume of each drop is calculated as:

role="math" localid="1654517265789" $$\begin{gathered} V_{\mathrm{drop}}=\frac{4}{3}{\mathrm{蟺谤}}^3 \\ =\frac{4}{\mathrm{尾尘}}脳\mathrm{尾尘}14脳\left(10脳\frac{{10}^{鈥�6}\mathrm{m}}{1}\right) \\ =4.20脳{10}^{鈥�15}{\mathrm{m}}^3 \end{gathered}$$

Now, multiply the volume of the each drop with the number of drops to find the total volume of the water in the cloud.

role="math" localid="1654517499825" $$\begin{gathered} V_{\mathrm{water}}=4.20脳{10}^{鈥�15}{\mathrm{m}}^3脳=4.71脳{10}^{17} \\ 鈮�2.0脳{10}^3{\mathrm{m}}^3 \end{gathered}$$

Or,

$$\begin{gathered} V_{\mathrm{water}}=4.20脳{10}^{鈥�15}{\mathrm{m}}^3脳=4.71脳{10}^{18} \\ 鈮�2.0脳{10}^4{\mathrm{m}}^3 \end{gathered}$$

Thus, the total volume of water in the cloud is from$2.0脳{10}^3{\mathrm{m}}^3$to$2.0脳{10}^4{\mathrm{m}}^3$.

Since, $1\mathrm{L}={10}^3{\mathrm{cm}}^3\mathrm{or}{10}^{鈥�3}{\mathrm{m}}^3$therefore, the number of bottles filled will be,

$$\begin{gathered} N=\frac{2.0脳{10}^3{\mathrm{m}}^3}{{10}^{鈥�3}{\mathrm{m}}^3} \\ =2.0脳{10}^6 \end{gathered}$$

Or,

$$\begin{gathered} N=\frac{2.0脳{10}^4{\mathrm{m}}^3}{{10}^{鈥�3}{\mathrm{m}}^3} \\ =2.0脳{10}^7 \end{gathered}$$

Thus, the number of bottles filled are $2.0脳{10}^6$ to$2.0脳{10}^7$.

Using equation (i), the mass of water is given as:

$m=V脳p$

Substitute the values in the above expression.

role="math" localid="1654518531625" $$\begin{gathered} m=2.0脳{10}^3{\mathrm{m}}^3脳{10}^3\mathrm{kg}/{\mathrm{m}}^3 \\ =2.0脳{10}^6\mathrm{kg} \end{gathered}$$

Or,

$$\begin{gathered} m=2.0脳{10}^4{\mathrm{m}}^3脳{10}^3\mathrm{kg}/{\mathrm{m}}^3 \\ =2.0脳{10}^7\mathrm{kg} \end{gathered}$$

Thus, the mass of the water in cloud is from $2.0脳{10}^6\mathrm{kg}$to $2.0脳{10}^7\mathrm{kg}$.