Question

A girl of mass 35 kg balances herself on a
cylindrical stick of mass 3.5 kg and diameter
7 cm. Find the pressure exerted by the girl on ground
will be (Take g = 10 ms-2)​

Answer 1

Answer:

The pressure exerted by the girl on the ground is 10^5 Pa

Step-by-step explanation:

Mass of the girl = 35 kg

Mass of cylindrical stick = 3.5 kg

Diameter of the stick d = 7 cm

Therefore cross sectional area of the cylindrical stick = πr²

or, $A=\pi(\frac{d}{2})^2$

$\implies A=\frac{22}{7}\times\frac{7^2}{2^2}=38.5$ cm²

We know that

$\bosex{\text{Pressure}=\frac{\text{Force}}{\text{Area}}}$

or $P = \frac{F}{A}$

Here F will be the force acting due to the weight of the girl as well as cylindrical stick

∴ F = 35g + 3.5g = 38.5g

Therefore,

$P=\frac{38.5g}{38.5\times 10^{-4}} \text{N/m^2}$

$\implies P=10\times 10^4 \text{N/m^2}$

$\implies P=10^5 \text{ Pa}$

Thus, the pressure exerted on the ground by the girl = 10^5 Pa

Answer 2

Answer:

The pressure exerted by the girl on the ground is 10^5 Pa

Step-by-step explanation:

Mass of the girl = 35 kg

Mass of cylindrical stick = 3.5 kg

Diameter of the stick d = 7 cm

Therefore cross sectional area of the cylindrical stick = πr²

or,

cm²

We know that

or

Here F will be the force acting due to the weight of the girl as well as cylindrical stick

∴ F = 35g + 3.5g = 38.5g

Therefore,

Thus, the pressure exerted on the ground by the girl = 10^5 Pa