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(B) The total cost of repairing the road is $Rs. 22176$ at a rate of $Rs. 1$ per $sq. m$.Therefore,the area of the circular road is $\frac{22176}{1} =

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$28$

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(B) The total cost of repairing the road is $Rs. 22176$ at a rate of $Rs. 1$ per $sq. m$.Therefore,the area of the circular road is $\frac{22176}{1} = 22176 \ m^2$.Let the inner radius be $r = 112 \ m$ and the outer radius be $R$.The area of the circular road is given by $\pi(R^2 - r^2) = 22176$.Substituting the values: $\frac{22}{7} 	imes (R^2 - 112^2) = 22176$.$R^2 - 12544 = 22176 	imes \frac{7}{22}$.$R^2 - 12544 = 1008 	imes 7 = 7056$.$R^2 = 7056 + 12544 = 19600$.$R = \sqrt{19600} = 140 \ m$.The width of the road is $R - r = 140 - 112 = 28 \ m$.

Around a circular garden,a circular road is to be repaired which costs $Rs. 22176$ at the rate of $Rs. 1$ per $sq. m$. If the inner radius is $112$ $m$,find the width of the circular road in $m$.

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