91Ó°ÊÓ

The given data is listed below as-

• The time taken by raindrop to fall is, t = 3.0 s
• Constant is$\mathrm{k}=2.00\frac{\mathrm{g}}{\mathrm{m}}$

The relation between raindrop mass and the distance that it has fallen is given by

m = kx

Here, m is the mass of raindrop, and x is the distance covered by raindrop that has fallen down.

(a)

The following differential equation is given:

$\mathrm{xg}=\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dt}}+{\mathrm{v}}^2$…â¶Ä¦â¶Ä¦â¶Ä¦(1)

The solution of above equation is:

v = at ……(2)

Now, acceleration is first derivative of velocity with respect to time.

$\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}$…..(3)

And velocity is the first derivative of position with respect to time.

$\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}$

dx = vdt ……

Substitute v from equation (2) ,in the above equation

dx = at dt

Integrate the above equation-

$$\begin{gathered} {∫}_{{\mathrm{x}}_0}^{\mathrm{x}}\mathrm{dx}=\mathrm{a}{∫}_0^{\mathrm{t}}\mathrm{t}\mathrm{dt} \\ \mathrm{x}-{\mathrm{x}}_0=\mathrm{a}\frac{{\mathrm{t}}^2}{2} \end{gathered}$$

Here, ${\mathrm{x}}_0$is the initial position of rain drop-

Take starting point as the origin, then substitute ${\mathrm{x}}_0=0$in above equation.

Therefore, $\mathrm{x}=\frac{1}{2}{\mathrm{at}}^2$…â¶Ä¦â¶Ä¦..(4)

Substitute v from equation (2), $\frac{\mathrm{dv}}{\mathrm{dt}}$from equation (3) and x from equation (4) in equation (1)

$$\begin{gathered} \frac{1}{2}a{t}^2.g=\frac{1}{2}a{t}^2.a+a^2{t}^2 \\ =\frac{3}{2}{a}^2{t}^2 \end{gathered}$$

Therefore, $\mathrm{a}=\frac{\mathrm{g}}{3}$

Thus, using the proposed solution for v , the acceleration is $\mathrm{a}=\frac{\mathrm{g}}{3}$.

(b)

Substitute t = 3.0 s and value of a in equation (4) to obtain the value of value of x

$$\begin{gathered} \mathrm{x}{|}_{\mathrm{t}=3\mathrm{s}}=\frac{1}{2}\left(\frac{9.8\mathrm{m}/{\mathrm{s}}^2}{3}\right){\left(3.00\mathrm{s}\right)}^2 \\ =14.7\mathrm{m} \end{gathered}$$

Thus, the distance the rain drop has fallen in t = 3.0 s is 14.7 m.

(c)

The relation between raindrop mass and the distance that it has fallen is given by

m = kx ……….(5)

Where, $\mathrm{k}=2.00\frac{\mathrm{g}}{\mathrm{m}}$

Now, to get the mass of rain drop at t = 3.0 s , put value of x and k in equation (5)

Therefore,

$$\begin{gathered} \mathrm{m}{|}_{\mathrm{t}=3\mathrm{s}}=\left(2.00\frac{\mathrm{g}}{\mathrm{m}}\right)×\left(14.7\mathrm{m}\right) \\ =29.4\mathrm{g} \end{gathered}$$

Thus, the mass of raindrop at t = 3.0 s is 29.4 g.