gas is escaping from a sperical ballom 4 cm3 per minute .how fast is the surface area shrinking when radius is 24cm

dS/dt = 0.33 cm²/min

Explanation:

Volume of sphere is given by;

V = (4/3)πr³

We are told that gas is escaping from a spherical balloon 4 cm³ per minute. Thus; dV/dt = 4 cm³/min

Now, let's find the rate at which the radius is changing.

dr/dt = (dV/dt) ÷ (dV/dr)

Now, dV/dr = 4πr²

Thus;

dr/dt = 4/(4πr²)

dr/dt = 1/(πr²)

Now, surface area of sphere is given by;

S = 4πr²

Thus, dS/dr = 8πr

Now, let's find the rate at t which area is shrinking.

dS/dt = (dS/dr) × (dr/dt)

Plugging relevant terms, we have;

dS/dt = 8πr × 1/(πr²)

dS/dt = 8/r

At r = 24 cm, we have;

dS/dt = 8/24

dS/dt = 1/3

dS/dt = 0.33 cm²/min

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .

no

sodium is solid at rt