If the angular speed of the earth is increased so much that the objects start flying from the equator, then the length of the day will be nearly.
A. $1\dfrac{1}{2}\text{hour}$
B. $8\text{ hour}$
C. $18\text{ hour}$
D. $24\text{ hour}$

Hint: This problem can be solved by using the formula for the variation of the acceleration due to gravity with angular speed of rotation of the earth and equating the acceleration due to gravity as zero at the equator (the condition for things to fly off) and find the angular speed. The angular speed will be inversely proportional to the time period and can be compared to the original time period and angular rotation of the earth.

Formula used:
The acceleration due to gravity $g'$ at a point on the earth’s surface is given by,
$g'=g\left( 1-\dfrac{{{\omega }^{2}}R}{g}{{\cos }^{2}}\theta \right)$
where $\omega $ is the angular speed of rotation of the earth, $R$ is the radius of the earth, $\theta $ is the angle made by the line joining the point on the earth’s surface to its center with the horizontal equator and $g$ is the acceleration due to gravity at the poles.

Complete step by step answer:
We will solve this problem by applying the formula of variation of the acceleration due to gravity at different points on the earth’s surface depending on their latitude.
The acceleration due to gravity $g'$ at a point on the earth’s surface is given by,
$g'=g\left( 1-\dfrac{{{\omega }^{2}}R}{g}{{\cos }^{2}}\theta \right)$ --(1)
where $\omega $ is the angular speed of rotation of the earth, $R$ is the radius of the earth, $\theta $ is the angle made by the line joining the point on the earth’s surface to its center with the horizontal equator and $g$ is the acceleration due to gravity at the poles.
Now, since, according to the question, objects fly off at the equator,
$g'=0$, $\theta ={{0}^{0}}$ (at equator) and
let the angular speed be $\omega '$.
Therefore, putting these information in (1), we get,
$0=g\left( 1-\dfrac{\omega {{'}^{2}}R}{g}{{\cos }^{2}}{{0}^{0}} \right)=g\left( 1-\dfrac{\omega {{'}^{2}}R}{g}{{1}^{2}} \right)=g\left( 1-\dfrac{\omega {{'}^{2}}R}{g} \right)$ $\left( \because \cos {{0}^{0}}=1 \right)$
$\therefore 1-\dfrac{\omega {{'}^{2}}R}{g}=0$ $\left( \because g\ne 0 \right)$
$\therefore 1=\dfrac{\omega {{'}^{2}}R}{g}$
$\therefore \omega {{'}^{2}}=\dfrac{g}{R}$
Square rooting both sides, we get,
$\sqrt{\omega {{'}^{2}}}=\sqrt{\dfrac{g}{R}}$
$\therefore \omega =\sqrt{\dfrac{g}{R}}$ --(2)
Now, the time period of rotation $T$ is
$T=\dfrac{2\pi }{\omega }$
Using (2), we get,
$T=\dfrac{2\pi }{\sqrt{\dfrac{g}{R}}}=2\pi \sqrt{\dfrac{R}{g}}=2\pi \times \sqrt{\dfrac{6.4\times {{10}^{6}}}{9.8}}=5077.58s$ $\left( \because g=9.8m/{{s}^{2}},R=6.4\times {{10}^{6}}m \right)$
$\therefore T=\dfrac{5077.58}{3600}=1.41\text{ hour }\approx \text{1}\dfrac{1}{2}\text{ hour}$ $\left( \because 1\text{ hour = }3600s \right)$
Therefore, the length of the day will be nearly $1\dfrac{1}{2}\text{ hour}$.
Therefore, the correct option is A) $1\dfrac{1}{2}\text{ hour}$.

Note: A good way to check whether the students have written the correct formula of the acceleration due to gravity at different points on the surface of the earth is to think in this way: the earth can be compared to a bar magnet. The bar magnet has maximum magnetic force at its poles and similarly, the acceleration due to gravity will be maximum at the poles of the earth and least at the equator.
Students can use this concept and check whether they are getting the correct results when they are considering the poles and the equator respectively. This can help avoid unnecessary silly mistakes.