Question

Calculate the maximum pressure developed by 20 N of a brick of dimensions 25cm into 15 cm into 10 cm
Please Give the right answer with explaination because tomorrow I have test....
I will Mark You As *BRAINLEST*​​

Answer 1

ANSWER:

$\sf Maximum \ pressure = 1.33 \times {10}^{ 3} \ Pa$

GIVEN:

• Dimensions of brick is 25cm × 15 cm × 10 cm .

• 20 N acts on the brick.

TO FIND:

• The maximum pressure developed on the brick.

EXPLANATION:

$\boxed{\bold{\large{\blue{Pressure = \dfrac{Force}{Area}}}}}$

According to the question, three areas are possible and so three pressure are possible.

$\sf A_1 = 25 \times 15$

$\sf A_1 = 325 \ {cm}^{2}$

$\boxed{ \bold{\large{ \red{1 \ m = 10^{-2}\ cm}}}}$

$\boxed{ \bold{\large{ \green{1 \ m^2 = 10^{ - 4}\ cm^2}}}}$

$\sf A_1 = 325 \times {10}^{ - 4} \ {m}^{2}$

$\sf A_2 = 15 \times 10$

$\sf A_2 = 150 \ {cm}^{2}$

$\boxed{ \bold{\large{ \orange{1 \ m^2 = 10^{ - 4}\ cm^2}}}}$

$\sf A_2 = 150 \times {10}^{ - 4} \ {m}^{2}$

$\sf A_3 = 25 \times 10$

$\sf A_3 = 250 \ {cm}^{2}$

$\boxed{ \bold{\large{ \purple{1 \ m^2 = 10^{ - 4}\ cm^2}}}}$

$\sf A_3 = 250 \times {10}^{ - 4} \ {m}^{2}$

As Pressure varies inversely with area, lowest area will offer the largest pressure.

$\sf \small As\ A_2\ is\ smaller,\ it \ will\ experience\ larger\ pressure.$

$\sf P_2 = \dfrac{F}{A_2}$

$\sf F = 20 \ N$

$\sf A_2 = 150 \times 10^{-4} \ m^2$

$\sf P_2 = \dfrac{20}{150 \times {10}^{ - 4} }$

$\sf P_2 = \dfrac{4}{30 \times {10}^{ - 4} }$

$\sf P_2 = \dfrac{4}{3 \times {10}^{ - 3} }$

$\sf P_2 = \dfrac{4}{3} \times {10}^{ 3}$

$\sf P_2 = 1.33 \times {10}^{ 3} \ Pa$

HENCE THE LARGEST PRESSURE = 1.33 × 10³ Pa.