A circular disc of area 0.49 pi , m^2 rolls d | vedclass

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(B) The area of the circular disc is given by $A = \pi r^2 = 0.49 \pi \, m^2$.By equating the two,we get $r^2 = 0.49$,which implies $r = 0.7 \, m$.The

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(B) The area of the circular disc is given by $A = \pi r^2 = 0.49 \pi \, m^2$.By equating the two,we get $r^2 = 0.49$,which implies $r = 0.7 \, m$.The circumference of the disc is $C = 2 \pi r = 2 	imes \frac{22}{7} 	imes 0.7 = 4.4 \, m$.The total distance covered is $1.76 \, km = 1760 \, m$.The number of revolutions is given by the ratio of the total distance to the circumference: $	ext{Number of revolutions} = \frac{1760}{4.4} = 400$.

$A$ circular disc of area $0.49 \pi \, m^2$ rolls down a length of $1.76 \, km$. The number of revolutions it makes is:

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