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In a swimming-pool $90$ m by $40$ m, $150$ men take a dip. If the average displacement of water by a man is $8$ cubic metres, what will be rise in water level ?

- $30$ cm
- $33.33$ cm
- $20.33$ cm
- $25$ cm

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0
votes

In a swimming-pool $90$ m by $40$ m, $150$ men take a dip. If the average displacement of water by a man is $8$ cubic metres, what will be rise in water level ?

- $30$ cm
- $33.33$ cm
- $20.33$ cm
- $25$ cm

See all

0
votes

According to the Archimedes Principle,

$\text{total rise in volume}=\text{total volume displaced by submerged objects}$

Let increase in height be $h$ m,

$\text{total rise in volume}= 90\times40\times h~m^3 = 3600h~m^3$

$\text{total volume displaced} = 150\times8~m^3$

$3600h = 150\times 8 \\ \implies h = \frac{150\times8}{3600}~m = 33.33~cm $

$\text{total rise in volume}=\text{total volume displaced by submerged objects}$

Let increase in height be $h$ m,

$\text{total rise in volume}= 90\times40\times h~m^3 = 3600h~m^3$

$\text{total volume displaced} = 150\times8~m^3$

$3600h = 150\times 8 \\ \implies h = \frac{150\times8}{3600}~m = 33.33~cm $

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0
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Let the height of the pool be “h”, and the rise of the water level be “x” , then according to the question , we can write as follows : –

Final volume- Initial Volume= Displacement of water by 150 men

$90 \times 40 \times (h+x) - 90\times40\times x= 150 \times 8 \hspace{1mm} m^3$

Solving the equation above for “x” will give 33.33 cm

So,option B is right

Final volume- Initial Volume= Displacement of water by 150 men

$90 \times 40 \times (h+x) - 90\times40\times x= 150 \times 8 \hspace{1mm} m^3$

Solving the equation above for “x” will give 33.33 cm

So,option B is right

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