Question

Cross-section of a railway tunnel is a rectangle 6 m broad and 8 m high surmounted by a semicircle as shown in the figure. The tunnel is 35 m long. Find the cost of plastering the internal surface of the tunnel (excluding the floor) at the rate of Rs. 2.25 per ${m^2}$.

Answer 1

Hint: In this question remember to use the formula of the circumference of semicircle i.e. C =$\pi r$ Here r is the radius of the semicircle now multiply the length of the tunnel by the sum of two times the height of the rectangle with the circumference of the semicircle, using these instructions will help you to approach towards the solution of the question.

Complete step by step answer:
According to the given information we have a railway tunnel shaped as a rectangle 6 m broad and 8 m high and the rectangle is surmounted by a semicircle and the tunnel is 35 m long.
So Height of tunnel = 8 m, breadth of the tunnel = 6 m, length of tunnel = 35 m and the radius of the semi-circle = 3 m
To find the internal surface of the tunnel first we have to find the circumference of semicircle
We know that the formula of circumference of formula i.e. $\pi r$
Substituting the given values in the formula of circumference of semicircle we get
Circumference of semicircle = $\dfrac{{22}}{7} \times 3$
$ \Rightarrow $ Circumference of semi-circle = $\dfrac{{66}}{7}$ m
So the internal surface area of tunnel = length of the tunnel (circumference semicircle + 2 times the height of tunnel)
Substituting the value in the above equation we get
Internal surface area of tunnel = $35\left( {8 + 8 + \dfrac{{66}}{7}} \right)$
$ \Rightarrow $ Internal surface area of tunnel = \[35\left( {\dfrac{{178}}{7}} \right)\]
$ \Rightarrow $ Internal surface area of tunnel = \[890{m^2}\]
So the cost of plastering the internal surface = 2.25 per ${m^2}$ $ \times $ internal surface area of tunnel
Substituting the values in the above equation we get
Cost of plastering the internal surface = 2.25 $ \times $ 890
$ \Rightarrow $Cost of plastering the internal surface = $\dfrac{{225}}{{100}}$ $ \times $ 890
$ \Rightarrow $Cost of plastering the internal surface = Rs. 20002.50
So the cost of plastering the total internal surface of the tunnel is equal to Rs. 20002.50

Note: In the above question to find the cost plastering the internal surface of the tunnel we have length of tunnel, height of tunnel and radius of semicircle surmounted on the rectangle so we know that we only require the tunnel surface area to find the cost of plastering tunnel, when we add the circumference of the semicircle and the twice the height of rectangle it results the surface area of tunnel for length 1 m but in the above problem we have to find the internal surface area of tunnel of length 35 m so to find the internal surface area of tunnel we multiplied the sum of circumference of semicircle and two times the height of rectangle and the reason behind not including the base of rectangle is because we don’t have to plaster the floor of tunnel and the reason we multiplied the length of tunnel by circumference so we got the 35 times of circumference which is the total internal surface of tunnel.