Question

A common hydrometer has $1$ and $0.8$ specific gravity marks $4$ cm apart. Calculate the time period of vertical oscillations when it floats in water. Neglect resistance of water.

• A. 0.6 s
• B. 0.7 s
• C. 0.65 s
• D. 0.8 s

Answer 1

The correct option is D 0.8 s

lets say $l$ is the length of the hydrometer when it is dipped in water and $A$ is the cross section area of the hydrometer, then $l+4$cm length of the hydrometer will be dipped when it is placed in liquid of specific gravity 0.8. So we have
weight of hydrometer=water displaced = liquid (0.8 s.g.) displaced
$\Rightarrow l\times A\times 1\times g=(l+4)\times A\times 0.8\times g\Rightarrow l=(l+4)0.8\Rightarrow l=16cm=.16m$
And time period of a floating cylinder (hydrometer )is given by $T=2\pi \sqrt{\frac{l}{g}}$ where $l$ is the length of the cylinder (hydrometer) dipped in water.
$\Rightarrow T=2\pi \sqrt{\frac{0.16}{9.8}}=0.8s$