5 Marks Questions

Question 15 Marks

An ornament weighing $36g$ in air, weighs only $34g$ in water. Assuming that some copper is mixed with gold to prepare the ornament, find the amount of copper in it. Specific gravity of gold is $19.3$ and that of copper is $8.9$.

Answer

Given:
Weight of ornament in air, $W_{air} = 36g $ Weight of ornament in water $, W_{water} = 34g$
Specific gravity of gold, $\rho_\text{g}=19.3\text{g}/\text{cc}$ Specific gravity of copperm, $\rho_\text{c}=8.9\text{g}/\text{cc}$
Let $v_c$ be the volume of copper and $v_g$ be the valume $o$ gold Loss of weight $-$ wight if displacement water $($buayanve$)36-34=2\text{g}$
$\text{W}=\text{v}_\text{g}\times \rho_\text{g}\times\text{g}+\text{v}\text{c}\times \rho \times \text{g}$
$36\text{g}=19.3\text{v}_\text{g}\times \text{g}+8.9\text{v}_\text{c}\times \text{g}$
$\Rightarrow \text{m}_\text{g}+\text{m}_\text{c}=36\dots(1)$
$\Rightarrow \Big(\frac{\text{m}_\text{g}}{\rho_\text{g}}+\frac{\text{m}_\text{c}}{\rho\text{c}}\Big)\rho _\text{w}\times \text{g}=2\times \text{g}$
$\Rightarrow \Big(\frac{\text{m}_\text{g}}{19.3}+\frac{\text{m}_\text{c}}{8.9}\Big)=2\dots(2)$
Solving $(1)$ and $(2)$ Mass of gold in ornament, $m_g = 33.75$ Mass of copper in ornament,$ m_c = 2.225$
Since, the goldsmith argues that he has not mixed copper or any other material with gold, rather some cavities might have been left inside the ornament.
Now, $\Rightarrow \Big(\frac{33.75}{19.3}+\text{V}_\text{cavity}\Big)=2$
$\Rightarrow \Big(\frac{\text{m}_\text{g}}{\rho_\text{g}}+\text{V}_{\text{cavity}}\Big)\rho_\text{w}\times \text{g}=2\times \text{g}$
$\Rightarrow \text{V}_\text{cavity}=0.251$

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Question 25 Marks

A cubical metal block of edge 12cm floats in mercury with one fifth of the height inside the mercury. Water is poured till the surface of the block is just immersed in it. Find the height of the water column to be poured. Specific gravity of mercury = 13.6.

Answer

Length of the edge of the block $\rho_{\text{Hg}}=13.6\text{gm/cc}$ Given that, intialy $\frac{1}{5}$ of block is inside mercury. Let $\rho\text{b}\rightarrow$ density of block in gm/cc.$\therefore(\text{x})^2\times\rho_\text{b}\times\text{g}=(\text{x})^2\times\Big(\frac{\text{x}}{5}\Big)\times\rho_{\text{Hg}}\times\text{g}$
$\Rightarrow12^3\times\rho\text{b}=12^2\times\frac{12}{5}\times13.6$
$\Rightarrow\rho_\text{b}=\frac{13.6}{5}\text{gm}/\text{cc}$
After water poured, Let x = Height of water column.$\text{v}_\text{b}=\text{v}_\text{Hg}+\text{v}_\text{w}=12^3$
Where $\text{v}_\text{Hg}$ and $\text{v}_\text{w}$ are volume of block inside mercury and water respectively$\therefore(\text{v}_\text{b}\times\rho_\text{b}\times\text{g})=(\text{v}_\text{Hg} \times\rho_{\text{Hg}}\times\text{g})+(\text{v}_\text{w}\times\rho_\text{w}\times\text{g)}$
$\Rightarrow(\text{v}_\text{Hg}+\text{v}_\text{w})\rho_\text{b}=\text{v}_\text{Hg}\times\rho_{\text{Hg}}\times+\text{v}_\text{w}\times\rho_\text{w}.$
$\Rightarrow(\text{v}_\text{Hg}+\text{v}_\text{w})\times\frac{13.6}{5}=\text{v}_\text{Hg}\times13.6+\text{v}_\text{w}\times1$
$\Rightarrow(12)^2\times\frac{13.6}{5}=(12-\text{x})\times(12)^2\times13.6+(12)^2\times1$
$\Rightarrow\text{x}=10.4\text{cm}$

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Question 35 Marks

Water leaks out from an open tank through a hole of area $2\ mm^2$ in the bottom. Suppose water is filled up to a height of $80\ cm$ and the area of cross $-$ section of the tank is $0.4m^2$. The pressure at the open surface and at the hole are equal to the atmospheric pressure. Neglect the small velocity of the water near the open surface in the tank.

Find the initial speed of water coming out of the hole.
Find the speed of water coming out when half of water has leaked out.
Find the volume of water leaked out during a time interval dt after the height remained is $h$. Thus find the decrease in height dh in terms of $h$ and dt.
From the result of part $c$ find the time required for half of the water to leak out.

Answer

From the Bernoulli's theorem,

$\text{p}+\rho\text{gh}+\frac{1}{2}\rho\text{v}^2=\text{p}\ '+\rho\text{gh}\ '+\frac{1}{2}\rho\text{v}'^\text{2}$
Here $\text{p}=\text{p}\ ',\text{h}=80\text{cm}=0.80\text{m},\text{h}' =0,\text{v}=0,$
Hence,
$\Rightarrow\rho\text{gh}=\frac{1}{2}\rho\text{v}\ '$
$\Rightarrow2\text{v}'^2=2\text{gh}$
$\Rightarrow\text{v}'=\sqrt{(2\text{gh})}$
$=\sqrt{(2\times10\times0.8)} \ {\text{Taking}\ g = 10m/s^2}$
$=\sqrt{16}=4\text{m}/\text{s}$

When half the water has leaked out, $h = 0.40m$

Now $\text{v}'\sqrt{(2\text{gh})}$
$=\sqrt{(2\times10\times0.40)}$
$=\sqrt{8}\text{m}/\text{s}.$

If at any instant t, the height remaining in the tank is $h,$ then the volume of the water leaked in a small interval dt is $dQ.$

$dQ =$ Discharge rate $\times dt =$ Area $\times$ speed of water $\times dt$
$\text{dQ}=\text{A}\times\text{v}\ '\times\text{dt}=\big(2\text{mm}^2\big)\sqrt{(2\text{gh})}\text{dt}$

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