Question 10: Properties of Matter

Question

A cylindrical vessel of 40 cm radius is completely filled with water and its capacity is $528 \mathrm{dm}^3$ (dm : decimeter) The vessel is placed on a solid block of exactly same height as vessel. If a small hole is made at 70 cm below the top of water level, then horizontal range of water falling on the ground in the beginning is $\_\_\_\_$ cm .

Accepted Answer

The capacity of vessel is $528 \mathrm{dm}^3$.
Since $1 \mathrm{dm}=10 \mathrm{~cm} \Rightarrow 1 \mathrm{dm}^3=1000 \mathrm{~cm}^3$
$$ \mathrm{V}=528,000 \mathrm{~cm}^3 $$
Volume of a cylinder is $\mathrm{V}=\pi \mathrm{r}^2 \mathrm{H}$.
Given $\mathrm{r}=40 \mathrm{~cm}$
$$ 528,000=	imes H $$
$$\Rightarrow $$ $\frac{528,000}{\pi 	imes(40)^2}=\mathrm{H}$
$$\Rightarrow $$ $$ \mathrm{H}=105 \mathrm{~cm} $$
Depth of hole from top is $\mathrm{h}=70 \mathrm{~cm}$.
From Torricelli's Law velocity of efflux is given as,
$$ \mathrm{v}=\sqrt{2 \mathrm{gh}} $$
The vessel is on a block of height H and the hole is at distance $(\mathrm{H}-\mathrm{h})$ from the bottom of the vessel.
Total height from ground $\mathrm{Y}=(\mathrm{H}-\mathrm{h})+\mathrm{H}=2 \mathrm{H}-\mathrm{h}$
The time to fall is,
$$ t=\sqrt{\frac{2 Y}{g}} $$
The horizontal range of projectile is given as
$$ \mathrm{R}=\mathrm{v} 	imes \mathrm{t}=\sqrt{2 \mathrm{gh}} 	imes \sqrt{\frac{2(2 \mathrm{H}-\mathrm{h})}{\mathrm{g}}}=2 \sqrt{\mathrm{~h}(2 \mathrm{H}-\mathrm{h})} $$
The height of water level from the hole is $\mathrm{h}=70 \mathrm{~cm}$
$$ 2 \mathrm{H}-\mathrm{h}=2(105)-70=210-70=140 \mathrm{~cm} . $$
$$\Rightarrow $$ $\mathrm{R}=2 \sqrt{70 	imes 140}$
$$\Rightarrow $$ $R=2 \sqrt{70 	imes 70 	imes 2}$
$$\Rightarrow $$ $$ \mathrm{R}=2 	imes 70 	imes \sqrt{2}=140 \sqrt{2} \mathrm{~cm} $$
Therefore, the horizontal of water falling on the ground is $140 \sqrt{2} \mathrm{~cm}$.
Thus. the correct option is (B).

Answer

$$ 120 \sqrt{2} $$

Answer

$$ 140 \sqrt{2} $$

Answer

$$ 140 \sqrt{3} $$

Answer

$$ 120 \sqrt{3} $$

Properties of Matter

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