Question

the length of stem of a hydrometer is 10 cm and volume is 10cm3. While floating in pure water it sinks
to 5 cm mark. The mass of hydrometer is 200 gm.What will be density of liquid in which the hydrometer:
1. sinks to top of stem
2. sinks to bottom of stem

Answer 1

the length of stem of a hydrometer is $10\mathrm{cm}$ and volume is $10\mathrm{cm}^3$ . While floating in pure water it sinks

to $5\mathrm{cm}$ mark. The mass of hydrometer is $200\mathrm{gm}$ . What will be density of liquid in which the hydrometer:

1. sinks to top of stem

2. sinks to bottom of stem

Solution:

Let the volume of bulb is V. When hydrometr sinks to the mark $h = 5$ cm of stem.

Using the law of Floatation (figure 1):

$$F _ {1} = m g$$ $$F _ {1} = \rho_ {w} g (V + V _ {5})$$

(2)in(1): $\rho_w g(V + V_5) = mg$

$$V = \frac {m - \rho_ {w} V _ {5}}{\rho_ {w}} = \frac {2 0 0 g - 1 \frac {g}{c m ^ {3}} \cdot 5 c m ^ {3}}{1 \frac {g}{c m ^ {3}}} = 1 9 5 c m ^ {3}$$

1. Sinks to the top of the stem:

Law of Floatation (figure 2):

$$F _ {2} = m g$$ $$F _ {2} = \rho_ {L} g (V + V _ {1 0})$$

(2) $in(1): \rho_L g(V + V_{10}) = mg$

$$\rho_ {L} = \frac {m}{V + V _ {1 0}} = \frac {2 0 0 g}{1 9 5 c m ^ {3} + 1 0 c m ^ {3}} = 0. 9 7 6 \frac {g}{c m ^ {3}}$$

2. Sinks to the bottom of the stem:

Law of Floatation (figure 3):

$$F _ {3} = m g$$ $$F _ {3} = \rho_ {L}, g (V + V _ {1 0})$$

(2) $in(1): \rho_L, g(V + V_{10}) = mg$

$$\rho_ {L ^ {\prime}} = \frac {m}{V} = \frac {2 0 0 g}{1 9 5 c m ^ {3}} = 1. 0 2 6 \frac {g}{c m ^ {3}}$$

Answer: 1. $\rho_{L} = 0.976\frac{g}{cm^{3}}$

2. $\rho_{L^{\prime}} = 1.026\frac{g}{cm^{3}}$