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Question Number 87179 by peter frank last updated on 03/Apr/20

Commented by peter frank last updated on 03/Apr/20

qn 2

$${qn}\:\mathrm{2} \\ $$

Commented by Rio Michael last updated on 03/Apr/20

2. volume of iron = (m/ρ) = ((360 g)/(6 gcm^(−3) )) = 60 cm^3 ⇒ volume of submerge = 30 cm^3 but T + F_(upthrust ) = W ⇒ T = mg − F_(upthrust ) = 0.36 kg × 9.8ms^(−2) − 30 × 10^(−6) m^3 × 6000 kgm^(−3) × 9.8 T = 3.528 − 1.764 = 1.764 N

$$\mathrm{2}.\:\:\mathrm{volume}\:\mathrm{of}\:\mathrm{iron}\:=\:\frac{{m}}{\rho}\:=\:\frac{\mathrm{360}\:\mathrm{g}}{\mathrm{6}\:\mathrm{gcm}^{−\mathrm{3}} }\:=\:\mathrm{60}\:\mathrm{cm}^{\mathrm{3}} \\ $$$$\:\Rightarrow\:\mathrm{volume}\:\mathrm{of}\:\mathrm{submerge}\:=\:\mathrm{30}\:\mathrm{cm}^{\mathrm{3}} \: \\ $$$$\:\:\:\:\:\mathrm{but}\:\:\mathrm{T}\:+\:\mathrm{F}_{{upthrust}\:} \:=\:\mathrm{W} \\ $$$$\:\Rightarrow\:\:\:{T}\:=\:\mathrm{mg}\:−\:\mathrm{F}_{{upthrust}\:} \:=\:\mathrm{0}.\mathrm{36}\:\mathrm{kg}\:×\:\mathrm{9}.\mathrm{8ms}^{−\mathrm{2}} \:−\:\mathrm{30}\:×\:\mathrm{10}^{−\mathrm{6}} \mathrm{m}^{\mathrm{3}} \:×\:\mathrm{6000}\:\mathrm{kgm}^{−\mathrm{3}} \:×\:\mathrm{9}.\mathrm{8} \\ $$$$\:\:\mathrm{T}\:=\:\mathrm{3}.\mathrm{528}\:−\:\mathrm{1}.\mathrm{764}\:=\:\mathrm{1}.\mathrm{764}\:\mathrm{N} \\ $$

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