Question

calculate the pressure exerted by a brick of dimensions 0.2mX 0.3mX 0.6m that weighs 20N when kept on its sides with widest base.

Answer 1

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Pressure = 0.2m × 0.2m

= 0.04m²

Answer 2

Let the volume of the block of wood be Vcm3 and its density be dwgcm−3

So the weight of the block =Vdwg dyne, where g is the acceleration due to gravity =980cms−2

The block floats in liquid of density 0.8gcm−3 with 14th of its volume submerged.So the upward buoyant force acting on the block is the weight of displaced liquid=14V×0.8×g dyne.

Hence by cindition of floatation

V×dw×g=14×V×0.8×g

⇒dw=0.2gcm-3,

Now let the density of oil be dogcm-3

The block floats in oil with 60% of its volume submerged.So the buoyant force balancing the weight of the block is the weight of displaced oil = 60%×V×do×g dyne

Now applying the condition of floatation we get

60%×V×do×g=V×dw×g

⇒60100×V×do×g=V×0.2×g

⇒do=0.2×106=13=0.33gcm−-3