An irregular shaped body of mass $250 g$ is put in Eureka can and water displaced is collect in measuring cylinder in which water is already at mark $20 c m^{3}$ . Now the water level rises to $60 c m^{3}$ . Calculate
(a) Volume of body
(b) Density of body in CGS and SI units.

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Solution

Volume of body $V =$ change in water levels
$⟹ V = 60 - 20 = 40 c m^{3}$

Density of body $ρ = \frac{m}{V} = \frac{250}{40} = 6.25 g m / c m^{3}$

Also $ρ = 6.25 \times 1000 = 6250 k g / m^{3}$

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An irregular shaped body of mass 250g is put in Eureka can and water displaced is collect in measuring cylinder in which water is already at mark 20cm3. Now the water level rises to 60cm3. Calculate(a) Volume of body(b) Density of body in CGS and SI units.

Volume of body V= change in water levels⟹ V=60−20=40 cm3Density of body ρ=mV=25040=6.25 gm/cm3Also ρ=6.25×1000=6250 kg/m3

An irregular shaped body of mass $250 g$ is put in Eureka can and water displaced is collect in measuring cylinder in which water is already at mark $20 c m^{3}$ . Now the water level rises to $60 c m^{3}$ . Calculate
(a) Volume of body
(b) Density of body in CGS and SI units.

Volume of body $V =$ change in water levels
$⟹ V = 60 - 20 = 40 c m^{3}$

Density of body $ρ = \frac{m}{V} = \frac{250}{40} = 6.25 g m / c m^{3}$

Also $ρ = 6.25 \times 1000 = 6250 k g / m^{3}$