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Q.

A cylindrical pot is slowly filled with water. The centre of mass of the empty pot is at a height of $10 c m,$ the mass of the pot is $1 k g,$ and its inner area is $0.4 m^{2} .$ What is the height (in cm) of the water in it, if the centre of mass of the system is at the lowest position?
(Take density of water $1000 k g / m^{3}$ )

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Detailed Solution

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The position of centre of mass of the system is $y_{c m} .$
$y_{c m} = \frac{m_{1} \times H + m_{2} \times \frac{h}{2}}{m_{1} + m_{2}}$
Where $m_{1} = 1 k g, m_{2} = (0.4 \times h \times 10^{3}) k g$
$= (400 h) k g$
Question Image
$y_{c m} = \frac{1 \times H + (400 h) \times \frac{h}{2}}{(1 + 400 h)} = \frac{H + 200 h^{2}}{(1 + 400 h)}$
for $y_{c m}$ to be lowest (minimum)
$\frac{d y_{c m}}{d h} = 0$
200 h² + h –H = 0
h = 2 cm

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A cylindrical pot is slowly filled with water. The centre of mass of the empty pot is at a height of 10 cm, the mass of the pot is 1 kg, and its inner area is 0.4 m2. What is the height (in cm) of the water in it, if the centre of mass of the system is at the lowest position? (Take density of water 1000 kg/m3 )