A simple pendulum measuring 40 cm in length is placed inside a deep...

The acceleration due to gravity at a depth $d$ is given by:

g' = g \left(1 -\dfrac{d}{R}\right)

Substituting $d = 1600 \text{ km}$ and $R = 6400 \text{ km}$:

g' = g\left(1 -\dfrac{1600}{6400}\right) =g\left(1 -\dfrac{1}{4}\right) = \dfrac{3g}{4}

The time period of a simple pendulum is:

$T = 2\pi$\sqrt{\dfrac{L}{g'}}

Given $L = 40 \text{ cm} = 0.4 \text{ m}$ and taking $g = 9.8 \text{ m/s}^2$:

$T = 2\pi$\sqrt{\dfrac{0.4}{\dfrac{3 \times 9.8}{4}}} =$2\pi$\sqrt{\dfrac{1.6}{29.4}}

$T = 2\pi$\sqrt{0.0544} \approx$2 \times 3.14 \times 0.233 \approx 1.466 \text{ s}$

Rounding off to two decimal places, $T \approx 1.47 \text{ s}$.