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The length of a semi-circular tunnel is 2 km and diameter is 7 m. Find the expenditure for digging the tunnel at the rate of Rs. 600 per${m^3}$. Find the expenditure for plastering inner side of the tunnel at the rate of Rs. 50 per ${m^2}$. $\left( {\pi = \dfrac{{22}}{7}} \right)$

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Last updated date: 13th Jun 2024
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Answer
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Hint:
Start the solution by calculating the volume of the semi-circular tunnel and then multiply it with the given rate of digging a tunnel . Also, we will find the area of the tunnel and then multiply it with the cost of plastering 1 metre square of the tunnel. We will be getting two answers for this question.

Complete step by step solution:
We have been given that the length of the semi-circular tunnel, say $l$ is equal to 2km, which is equal to 2000m.
And the diameter of the tunnel is 7m.
We know that the diameter is double the radius.
Then, the radius of the semi-circular tunnel, say $r$ is $\dfrac{7}{2}m$.
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Now, the volume of the semi-circle is half the volume of the circular tunnel.
Therefore, the volume of the circular tunnel is
$ = \dfrac{1}{2} \times \pi {r^2}l$
On substituting the values of the $\pi = \dfrac{{22}}{7},r = \dfrac{7}{2},l = 2000$, we will get,
$
   = \dfrac{1}{2} \times \left( {\dfrac{{22}}{7}} \right)\left( {\dfrac{7}{2}} \right)\left( {\dfrac{7}{2}} \right)\left( {2000} \right) \\
   \Rightarrow 11 \times 0.5 \times 3.5 \times 2000 \\
   \Rightarrow 38500{m^2} \\
$
Then, the expenditure of digging the tunnel at the rate of Rs. 600 per${m^3}$.
$
   = 38500 \times 600 \\
   = {\text{Rs}}{\text{. }}2,31,00,000 \\
$
Now, we will calculate the cost of plastering the inner side.
The inner area of the tunnel is $\pi rl$
$
   = \dfrac{{22}}{7} \times \dfrac{7}{2} \times 2000 \\
   = 22000{m^2} \\
$

Hence, the total expenditure on plastering is $ = 22000 \times 50$
$ = {\text{Rs}}{\text{. }}11,00,000$


Note:
We must draw the corresponding diagram of the figure as it helps in solving the problem. And the shape of the tunnel is that of a cylinder. We will apply the formula of mensuration depending on the shape of the object.