How many silver coins,1.75 ,cm in diameter an | vedclass

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(B) Coins are cylindrical in shape.Height $(h_1)$ of cylindrical coins $= 2 \,mm = 0.2 \,cm$.Radius $(r)$ of circular end of coins $= \frac{1.75}{2} =

Vedclass · Accepted Answer

$400$

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(B) Coins are cylindrical in shape.Height $(h_1)$ of cylindrical coins $= 2 \,mm = 0.2 \,cm$.Radius $(r)$ of circular end of coins $= \frac{1.75}{2} = 0.875 \,cm$.Let $n$ coins be melted to form the required cuboid.Volume of $n$ coins $=$ Volume of cuboid$n 	imes \pi 	imes r^2 	imes h_1 = l 	imes b 	imes h$$n 	imes \frac{22}{7} 	imes (0.875)^2 	imes 0.2 = 5.5 	imes 10 	imes 3.5$$n = \frac{5.5 	imes 10 	imes 3.5 	imes 7}{22 	imes (0.875)^2 	imes 0.2}$$n = \frac{192.5 	imes 7}{22 	imes 0.765625 	imes 0.2} = \frac{1347.5}{3.36875} = 400$.Therefore,the number of coins melted to form such a cuboid is $400$.

How many silver coins,$1.75 \,cm$ in diameter and of thickness $2 \,mm$,must be melted to form a cuboid of dimensions $5.5 \,cm \times 10 \,cm \times 3.5 \,cm$? [Take $\pi = \frac{22}{7}$]

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