Question

a block of iron of mass 7.5 kg and dimensions 12 cm*8 cm*10 cm is kept on a table top on its base of side 12 cm*8 cm.calculate the thrust and pressure exerted on the table top.take 1 kgf=10n.

Answer 1

thrust = mxa =7.5*10=75N
pressure= thrust/area=75/0.12*0.8=75/0.096=781.25

Answer 2

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

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

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

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

$\bold{Therefore,}$

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

hope it helps you...

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