Question

A block of iron of mass 8 kg and of dimensions 10 cm x 8 cm x 5cm is kept on a table top on its base of side 8 cm x 5 cm.

Calculate : (i) thrust (ii) pressure exerted on the table top.

(1 kgf = 10 N)​

Answer 1

We are Given :-

A block of iron whose mass = 8 Kg and having a dimension = 10 cm × 8 cm × 5 cm

We are required to calculate

1. Thurst.

2. Pressure on the top of table.

For Thurst, When a body is at rest then there is a Normal force acting to it and it is perpendicular to it. This normal force is equal to self at 'mg'. Here 'm' is the mass of body and 'g' is the acceleration due to gravity acting on the body towards earth.

N = F = mg cosØ ( N = Normal Force)

F = 8 × 10 × cos90°

F = 80 × 1

F = 80 N

Dimension = 10 cm × 8 cm × 5 cm

Area of table = (8 × 5) cm²

A = 40 cm²

A = 40/10000 m²

A = 0.004 m²

Now, we can find pressure on the body.

Pressure = Force/Area

P = 80/0.004

P = 20000 Pa

Hence,

Thurst = Force = F = 80 N

Pressure = P = 20000 Pa

Answer 2

$\bold{(a) \: thrust}$

$\bold{Force \: = \: Mass \: × \: acceleration \: due \: to \: gravity} \\ \bold{ = \: 7.5 \times 10} \\ \bold{ = 75N \: .... \: \: ans}$

$\bold{(b) \: pressure \: exerted \: on \: tabletop}$

$\bold{Area \: of \: the \: tabletop \: = \: 12 × 8} \\ \bold{ \: \: \: = 96 {cm}^{2} \: to \: {m}^{2} } \\ \bold{ \: \: \: = \: \frac{96}{1000} {m}^{2} } \\ \bold{ \: \: \: = \: 0.0096 {m}^{2} }$

$\bold{Therefore,}$

$\bold{Pressure \: = \: \frac{force}{area} } \\ \bold{ \: \: \: \: \: \: \: \: = \: \frac{75}{0.0096} } \\ \bold{ \: \: \: \: \: = \: \frac{75 \times 10000}{96} } \\ \bold{ \: \: \: \: \: = \:\frac{ \cancel{75} \: \: \: ²⁵ \times \cancel{10000} \: \: ⁶²⁵}{ \cancel{96} \: \: ²} } \\ \bold{ \: \: \: = \: \frac{25 \times 625}{2} } \\ \bold \red{ \: \: \: = \: 7182.5 \: Pa \: \: ....ans}$

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