# NIELIT 2019 Feb Scientist D - Section D: 1

149 views

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 ?

1. $30$ cm
2. $33.33$ cm
3. $20.33$ cm
4. $25$ cm

recategorized

Ans : Option B
172 points 1 4
Let 'h' m be the rise in the water level of the swimming pool when 150 men take a dip.
Then, volume of water displaced = Displacement of water by 150 men
⇒90 m ×40 m ×h m =150×8
⇒h=150×8/90×40​=33.33 cm
18 points 1
According to the Archimedes Principle,

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

Let increase in height be $h$ m,

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

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

$3600h = 150\times 8 \\ \implies h = \frac{150\times8}{3600}~m = 33.33~cm$
518 points 1 2 12

## Related questions

1
104 views
A conical tent is to accommodate $10$ persons. Each person must have $6$ $m$^{2}$space to sit and$30m$^{2}$ of air to breath. What will be height of cone ? $37.5$ $m$ $150$ $m$ $75$ $m$ $15$ $m$
2
107 views
If $A$ be the area of a right angled triangle and $b$ be one of the sides containing the right angle, then the length of altitude on the hypotenuse is : $\frac{2Ab}{\sqrt{4b^{4}+A^{2}}}$ $\frac{Ab}{\sqrt{b^{4}+4A^{2}}}$ $\frac{2Ab}{\sqrt{b^{4}+4A^{2}}}$ $\frac{Ab}{\sqrt{4b^{4}+A^{2}}}$
In an acute angled triangle $ABC$, if $\tan \left(A+B-C \right)=1$ and $\sec \left(B+C-A \right)=2$, Find angle $A$. $60^\circ$ $45^\circ$ $30^\circ$ $90^\circ$
What will be area of the rhombus with equations of sides $ax \pm$ $by \pm c$ = $1$ ? $\frac{3c^{2}}{ab}$sq. units $\frac{4c^{2}}{ab}$sq. units $\frac{2c^{2}}{ab}$sq. units $\frac{c^{2}}{ab}$sq. units
If $\left (-4, 0 \right), \left(1, -1 \right)$ are two vertices of a triangle whose area is $4$ Sq units then its third vertex lies on : $y=x$ $5x+y+12=0$ $x+5y-4=0$ $x-5y+4=0$