The outer circumference of a circular race track is 528 metre. The track is every where 14 metre wide. Cost of levelling the track at the rate of ₹10 per sq. metre is:

Let outer radius = $$R$$ m and inner radius = $$r=(R-14)$$ m

Outer circumference = $$2\pi R=528$$

=> $$2\times\frac{22}{7}\times R=528$$

=> $$R=528\times\frac{7}{44}=84$$ m

Thus, inner radius = $$84-14=70$$ m

=> Area of track = $$\pi(R^2-r^2)$$

= $$\frac{22}{7}(R+r)(R-r)$$

= $$\frac{22}{7}(84+70)(84-70)$$

= $$\frac{22}{7}\times154\times14=6776$$ $$m^2$$

$$\therefore$$ Total cost of levelling = $$6776\times10=Rs.$$ $$67,760$$

=> Ans - (D)