The time period of a bob performing simple harmonic motion in water...

Given, density of bob, $ρ = \dfrac{4}{3} \times {10}^{3}kgm^{- 3}$
Density of water, $\sigma = {10}^{3}kgm^{- 3}$

If $g^{'}$ be gravitational acceleration in water then,

$g^{'} = g\left(1 - \dfrac{σ}{ρ}\right) = g\left(1 - \dfrac{103}{\dfrac{4}{3} \times {10}^{3}}\right) = \dfrac{g}{4}$

As, ${T}_{air} = 2\pi\sqrt{\dfrac{I}{g}}$,

Similarly, ${T}_{water} = 2\pi\sqrt{\dfrac{I}{g '}} \left[{T}_{water} =2s\right]\left(g^{'} = \dfrac{g}{4}\right)</p><p>2 = 2\pi\sqrt{\dfrac{I}{g / 4}} =2\pi\sqrt{\dfrac{I}{g}} \cdot2$

$2 = 2. {T}_{air}$

$⇒ {T}_{air} = 1s$