Answer
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Hint: Manometer is a device used for measuring the pressure in liquids. One of the most common types of manometers are called U-tube manometers which consist of a U-shaped tube. The pressure is calculated by measuring the height difference between the two limbs of the manometer.
Complete step by step answer:
Let us consider a U-tube manometer partially filled with water, as shown:
When oil is added to one side, the water on the other limb rises by 25 cm as shown.
Let h be the height to which the oil is filled.
If the level of water in one end rises by 25 cm at point B, the water level in the other end decreases by 25 cm at point A when the oil is added on top of it.
At the points A and B, the pressure will be equal.
Pressure at A = Pressure at B
${P_A} = {P_B}$
Pressure at point A,${P_A} = {\rho _{oil}} \times g \times (h + 25)$
Pressure at point B,${P_B} = {\rho _{water}} \times g \times 25$
Equating, we get –
${\rho _{oil}} \times g \times (h + 25)$= ${\rho _{water}} \times g \times 25$
Given, specific gravity of oil = 0.8
Specific gravity, $\dfrac{{{\rho _{oil}}}}{{{\rho _{water}}}} = 0.8$
\[
{\rho _{oil}} \times{g} \times (h + 25) = {\rho _{water}} \times {g} \times 25 \\
\dfrac{{{\rho _{oil}}}}{{{\rho _{water}}}} \times (h + 25) = 25 \\
Substituting, \\
0.8 \times (h + 25) = 25 \\
\Rightarrow (h + 25) = \dfrac{{25}}{{0.8}} \\
\Rightarrow h + 25 = 31.25 \\
\Rightarrow h = 31.25 - 25 \\
\Rightarrow h = 6.25cm \\
\]
Thus, the correct option is Option A.
Note: When the manometer is used to measure the pressure, the pressure, so recorded, is called absolute pressure. This pressure includes the atmospheric pressure. Now, if we subtract the atmospheric pressure from the absolute pressure, we get something called as gauge pressure.
Complete step by step answer:
Let us consider a U-tube manometer partially filled with water, as shown:
When oil is added to one side, the water on the other limb rises by 25 cm as shown.
Let h be the height to which the oil is filled.
If the level of water in one end rises by 25 cm at point B, the water level in the other end decreases by 25 cm at point A when the oil is added on top of it.
At the points A and B, the pressure will be equal.
Pressure at A = Pressure at B
${P_A} = {P_B}$
Pressure at point A,${P_A} = {\rho _{oil}} \times g \times (h + 25)$
Pressure at point B,${P_B} = {\rho _{water}} \times g \times 25$
Equating, we get –
${\rho _{oil}} \times g \times (h + 25)$= ${\rho _{water}} \times g \times 25$
Given, specific gravity of oil = 0.8
Specific gravity, $\dfrac{{{\rho _{oil}}}}{{{\rho _{water}}}} = 0.8$
\[
{\rho _{oil}} \times{g} \times (h + 25) = {\rho _{water}} \times {g} \times 25 \\
\dfrac{{{\rho _{oil}}}}{{{\rho _{water}}}} \times (h + 25) = 25 \\
Substituting, \\
0.8 \times (h + 25) = 25 \\
\Rightarrow (h + 25) = \dfrac{{25}}{{0.8}} \\
\Rightarrow h + 25 = 31.25 \\
\Rightarrow h = 31.25 - 25 \\
\Rightarrow h = 6.25cm \\
\]
Thus, the correct option is Option A.
Note: When the manometer is used to measure the pressure, the pressure, so recorded, is called absolute pressure. This pressure includes the atmospheric pressure. Now, if we subtract the atmospheric pressure from the absolute pressure, we get something called as gauge pressure.
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