Question

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Answer 1

Step-by-step explanation:

Given :

Dimensions of the cuboidal solid block = 30 cm × 40 cm × 50cm.

and

Force applied by block = 50 N

We have to find :

The maximum pressure that will be exerted by the block on ground.

Solution :

Dimensions of the cuboidal solid block

= 30 cm × 40 cm × 50cm.

= 0.3 m × 0.4 m × 0.5 m. [ in meter ]

Now,

Area of the first face =

A_1A

1

= 0.3 × 0.4 = 0.12 m²

֎ Area of the second face =A_2A

2

= 0.4 × 0.5 = 0.20 m²

֎ Area of the third face =

A_3A

3

= 0.5 × 0.3 = 0.15 m²

Here we can see that, out of the above three faces , A_1A

1

is minimum.

Here, we consider the minimum value of area to calculate the maximum pressure because

we know that,

\begin{gathered}P ∝ \frac{1}{A} < /p > < p > \\ \end{gathered}

P∝

A

1

</p><p>

The relation between Pressure (P) , Force(F) and Area (A) is given by ,

\begin{gathered}\: {\large{{ \bf{Pressure= \dfrac{Force}{Area}}}}}\\ \implies \sf \: P = \dfrac{F}{A_1}\\ \implies \sf \: P = \dfrac{50 \: N}{0.12 \: {m}^{2} } \\ \\ \implies \red{\boxed{ \green{ \bf{P = 416.66 \: N{m}^{ - 2} }} }}\: \end{gathered}

Pressure=

Area

Force

⟹P=

A

1

F

⟹P=

0.12m

2

50N

⟹

P=416.66Nm

−2

∴ The maximum pressure applied by the given cuboidal solid block is 416.6 N/m².

Answer 2

Answer:

Given :

Dimensions of the cuboidal solid block = 30 cm × 40 cm × 50cm.

and

Force applied by block = 50 N

We have to find :

The maximum pressure that will be exerted by the block on ground.

Solution :

Dimensions of the cuboidal solid block

= 30 cm × 40 cm × 50cm.

= 0.3 m × 0.4 m × 0.5 m. [ in meter ]

Now,

֎ Area of the first face =

$A_1$ = 0.3 × 0.4 = 0.12 m²

֎ Area of the second face =

$A_2$ = 0.4 × 0.5 = 0.20 m²

֎ Area of the third face =

$A_3$ = 0.5 × 0.3 = 0.15 m²

Here we can see that, out of the above three faces , $A_1$ is minimum.

Here, we consider the minimum value of area to calculate the maximum pressure because

we know that,

$P ∝ \frac{1}{A} </p><p>$

The relation between Pressure (P) , Force(F) and Area (A) is given by ,

$\: {\large{{ \bf{Pressure= \dfrac{Force}{Area}}}}}\\ \implies \sf \: P = \dfrac{F}{A_1}\\ \implies \sf \: P = \dfrac{50 \: N}{0.12 \: {m}^{2} } \\ \\ \implies \red{\boxed{ \green{ \bf{P = 416.66 \: N{m}^{ - 2} }} }}\:$

∴ The maximum pressure applied by the given cuboidal solid block is 416.6 N/m².