Case study (4 Marks)

Question 14 Marks

In the derivation of $P_1-P_2=P g z$, it was assumed that the liquid is incompressible. Why will this equation not be strictly valid for a compressible liquid?

Answer

For gases which are compressible we have already derived that $\text{P}=\text{p}_0\text{e}-\frac{\text{MgA}}{\text{RT}}$

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Question 24 Marks

Suppose the density of air at Madras is $\mathrm{P}_0$ and atomospheric pressure is $\mathrm{P}_0$. If we go up, the density and the pressure both decrease. Suppose we wish to calculate the pressure at a height 10 km above Madras. If we use the equation $\mathrm{P}_0$ - $\mathrm{P}=$ pogz, will we get a pressure more than the actual or less than the actual? Neglect the variation in g. Does your answer change if you also consider the variation in g?

Answer

Less than actual. it will again be less than actual.

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Question 34 Marks

While watering a distant plant, a gardener partially water than in fresh closes the exit hole of the pipe by putting his finger on it. Explain why this results in the water stream goirig to a larger distance.

Answer

Area reduces, therefore speed increase, therefore range increases.

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Question 44 Marks

Refer to the previous problem. Suppose, 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. Calculate the volume of the cavities left that will allow the weights given in that problem.

Answer

Let the volume of the cavities be v,
The total volume of the ornament V = volume of gold + volume of cavities
$=\frac{36}{19.3}+\text{V}\ \text{cm}^3$
The volume of the water displaced is also $Vcm^3$
The mass of the water displaced $=\text{V}\times1\text{gram}=\text{V}\text{gram}$
The weight of the water displaced = Vg
Hence Vg = (36 - 34)g
$\Rightarrow\text{V}=2$
$\Rightarrow\frac{36}{19.3} +\text{v}=2$
$\Rightarrow\text{V}=2-\frac{36}{19.3}=2-1.865$
$\Rightarrow\text{V}=0.135\text{cm}^3$

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Physics Case study (4 Marks)

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