
There are 10 million meter in a quadrant of the earth’s circumference. Find approximately the distance at which the top of the Eiffel tower should be visible, its height being 300 meters.
Answer
531.9k+ views
Hint: To obtain the approx distance at which the Eiffel tower will be visible we will use the radius of earth formula. Firstly we know that the distance of the Eiffel tower which is visible will be obtain by the formula ${{d}_{\max }}=\sqrt{2hr}$ where $h,r$ is the height and the radius of the earth respectively. So we will get the radius of earth by the formula of circumference of earth and by substituting it in the ${{d}_{max}}$ formula we will get our desired answer.
Complete step-by-step solution:
The circumference of the earth is given as:
Circumference of quadrant $={{10}^{7}}$……$\left( 1 \right)$
Height $=300m$…..$\left( 2 \right)$
We know,
Circumference of quadrant $=\dfrac{1}{4}\times 2\pi r$
Substitute the value of circumference from equation (1) in above we get,
$\begin{align}
& \Rightarrow {{10}^{7}}=\dfrac{1}{4}\times 2\pi \times r \\
& \Rightarrow \dfrac{{{10}^{7}}\times 4}{2\pi }=r \\
\end{align}$
$\therefore r=\dfrac{{{10}^{7}}\times 2}{\pi }$…..$\left( 3 \right)$
The distance at which the top of the Eiffel tower is visible is given by:
${{d}_{\max }}=\sqrt{2hr}$
Substituting the value from equation (2) and (3) we get,
$\begin{align}
& {{d}_{\max }}=\sqrt{2\times 300\times \dfrac{{{10}^{7}}\times 2}{\pi }} \\
& {{d}_{\max }}=\sqrt{2\times 300\times \dfrac{2\times {{10}^{7}}}{\dfrac{22}{7}}} \\
& {{d}_{\max }}=\sqrt{\dfrac{2\times 300\times 2\times {{10}^{7}}\times 7}{22}} \\
& {{d}_{\max }}=2\times {{10}^{4}}\times 3.09m \\
\end{align}$
If we convert the above value in kilometer we get,
$\begin{align}
& \Rightarrow {{d}_{max}}=\dfrac{2\times {{10}^{4}}\times 3.09}{1000} \\
& \therefore {{d}_{\max }}\approx 61.8km \\
\end{align}$
Hence the distance at which the Eiffel tower is visible is $61.8km$
Note: The circumference of a circle shape object always has the formula $2\pi r$ but in this question it is given that quadrant circumference of earth is 10 millions this is one common mistake that can happen while solving the question. The answer can be in any unit but always try to simplify it to a bigger unit.
Complete step-by-step solution:
The circumference of the earth is given as:
Circumference of quadrant $={{10}^{7}}$……$\left( 1 \right)$
Height $=300m$…..$\left( 2 \right)$
We know,
Circumference of quadrant $=\dfrac{1}{4}\times 2\pi r$
Substitute the value of circumference from equation (1) in above we get,
$\begin{align}
& \Rightarrow {{10}^{7}}=\dfrac{1}{4}\times 2\pi \times r \\
& \Rightarrow \dfrac{{{10}^{7}}\times 4}{2\pi }=r \\
\end{align}$
$\therefore r=\dfrac{{{10}^{7}}\times 2}{\pi }$…..$\left( 3 \right)$
The distance at which the top of the Eiffel tower is visible is given by:
${{d}_{\max }}=\sqrt{2hr}$
Substituting the value from equation (2) and (3) we get,
$\begin{align}
& {{d}_{\max }}=\sqrt{2\times 300\times \dfrac{{{10}^{7}}\times 2}{\pi }} \\
& {{d}_{\max }}=\sqrt{2\times 300\times \dfrac{2\times {{10}^{7}}}{\dfrac{22}{7}}} \\
& {{d}_{\max }}=\sqrt{\dfrac{2\times 300\times 2\times {{10}^{7}}\times 7}{22}} \\
& {{d}_{\max }}=2\times {{10}^{4}}\times 3.09m \\
\end{align}$
If we convert the above value in kilometer we get,
$\begin{align}
& \Rightarrow {{d}_{max}}=\dfrac{2\times {{10}^{4}}\times 3.09}{1000} \\
& \therefore {{d}_{\max }}\approx 61.8km \\
\end{align}$
Hence the distance at which the Eiffel tower is visible is $61.8km$
Note: The circumference of a circle shape object always has the formula $2\pi r$ but in this question it is given that quadrant circumference of earth is 10 millions this is one common mistake that can happen while solving the question. The answer can be in any unit but always try to simplify it to a bigger unit.
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