The total number of spherical bullets,each of | vedclass

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(B) Volume of the rectangular block $= 11 	imes 10 	imes 5 = 550 	ext{ m}^3$.Since $1 	ext{ m} = 10 	ext{ dm}$,$1 	ext{ m}^3 = 1000 	ext{ dm}^3

Vedclass · Accepted Answer

Less than $8800$

Vedclass · Answer

(B) Volume of the rectangular block $= 11 	imes 10 	imes 5 = 550 	ext{ m}^3$.Since $1 	ext{ m} = 10 	ext{ dm}$,$1 	ext{ m}^3 = 1000 	ext{ dm}^3$.Volume of the block $= 550 	imes 1000 = 550,000 	ext{ dm}^3$.Diameter of each spherical bullet $= 5 	ext{ dm}$,so radius $r = 2.5 	ext{ dm} = \frac{5}{2} 	ext{ dm}$.Volume of one spherical bullet $= \frac{4}{3} \pi r^3 = \frac{4}{3} 	imes \pi 	imes (\frac{5}{2})^3 = \frac{4}{3} 	imes \pi 	imes \frac{125}{8} = \frac{125\pi}{6} 	ext{ dm}^3$.Given $\pi > 3$,the volume of one bullet is $V_b = \frac{125\pi}{6} > \frac{125 	imes 3}{6} = 62.5 	ext{ dm}^3$.The number of bullets $n = \frac{	ext{Total Volume}}{	ext{Volume of one bullet}} = \frac{550,000}{V_b} Since the volume of each bullet is strictly greater than $62.5 	ext{ dm}^3$,the total number of bullets must be less than $8800$.

The total number of spherical bullets,each of diameter $5$ decimetre,that can be made by utilizing the maximum of a rectangular block of lead with $11$ $m$ length,$10$ $m$ breadth and $5$ $m$ width is (assume that $\pi > 3$)

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