Answer
Verified
399k+ views
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.
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Which type of bond is stronger ionic or covalent class 12 chemistry CBSE
What organs are located on the left side of your body class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
When people say No pun intended what does that mea class 8 english CBSE