Question

A metal block of mass 20 g is kept on a floor. The contact area of metal block with the floor is 0.005 m². The pressure exerted by the metal
block on the floor is (Take, g = 10 m/s2)
0 40 Pa
O 4 kPa
O 2 kPa
O 200 Pa​

Answer 1

• Mass of metal block, m = 20 g.

$1 \: g \: = {10}^{ - 3} \: kg \\ \\ 20\: g \: =20 \times {10}^{ - 3} \: kg \\ \\ m =20 \times {10}^{ - 3} \: kg$

• Area of contact, A = 0.005 m²

According to the result of Newton's second law of motion, f = ma

Here a = g , acceleration due to gravity.

• g = 10 m/s².

f = mg

$f = 20 \times { 10 }^{ - 3} \times 10 \\ \\ = 200 \times {10}^{ - 3} \\ \\ = 2 \times 100 \times {10}^{ - 3} \\ \\ = 2 \times {10}^{2} \times {10}^{ - 3} \\ \\ = 2 \times {10}^{(2 - 3)} \\ \\ = 2 \times {10}^{ - 1} \\ \\ = 2 \times \frac{1}{10} \\ \\f = 0.2 \: newton$

Force exerted is 0.2 N.

Pressure = Force / Area

P = f/A

$P = \dfrac{0.2}{0.005} \\ \\ = \dfrac{0.2}{0.005} \times \dfrac{1000}{1000} \\ \\ = \dfrac{0.2 \times 1000}{0.005 \times 1000} \\ \\ = \frac{200}{5} \\ \\ P= 40 \: pa$

The pressure exerted is 40 Pascal.

Option 1 is right.

Answer 2

Mass (m) = 20 g = 0.02 kg

Acceleration (g) = 10 m/s²

Area of contact (x) = 0.005 m²

________________________

$\rm Pressure(P) = \dfrac{m \times g}{x}$

$\rm P= \dfrac{0.02 \times 10}{0.005}$

$\rm P = \dfrac{0.2}{0.005}$

$\rm{\underline{\underline{ P = \red{40 \: Pa}}}}$