Solve the previous problem if the lead piece is fastened on the top surface of the block and the block is to float with its upper surface just dipping into water.

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Solution

$m_{w} = 200 g, ρ_{w} = 0.8$ gm/cc,

$ρ_{p} b$ = 11.3gm/cc

Again, Mg =w

$\Rightarrow (m_{w} + m_{p} b) g = V_{w} \times ρ \times g$

[ $ρ$ = density of water]

$\Rightarrow 200 + m_{p} b = \frac{200}{0.8} \times 1$

$\Rightarrow m_{p} b = 250 - 200 = 50$ gm

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Solve the previous problem if the lead piece is fastened on the top surface of the block and the block is to float with its upper surface just dipping into water.

mw=200g,ρw=0.8 gm/cc, ρpb = 11.3gm/cc Again, Mg =w ⇒(mw+mpb)g=Vw×ρ×g [ρ = density of water] ⇒200+mpb=2000.8×1 ⇒mpb=250−200=50 gm

$m_{w} = 200 g, ρ_{w} = 0.8$ gm/cc,

$ρ_{p} b$ = 11.3gm/cc

Again, Mg =w

$\Rightarrow (m_{w} + m_{p} b) g = V_{w} \times ρ \times g$

[ $ρ$ = density of water]

$\Rightarrow 200 + m_{p} b = \frac{200}{0.8} \times 1$

$\Rightarrow m_{p} b = 250 - 200 = 50$ gm