Question

Questions -_-
1]. A shopkeeper have one theracal Laddoo of radius 5 cm. with the same amount of material, how many laddoo of radius 2.5 cm can be made.

2].find the amount of water displaced by a solid spherical ball of a diameter 4.2 cm. when it is completely immersed in water.

3]. A sphere and a right circular cylinder of the same radius have equal volume. By what percentage does the diameter of the cylinder exceed it's height.​

Answer 1

Solution : 1

$\sf \: Radius \: of \: the \: spherical \: laddoo = 5 \: cm$

$\sf \: \therefore Volume \: of \: big \: spherical \: laddoo \: \implies \: \sf \: \dfrac{4}{3} \pi \: \times 5 \times 5 \times 5 cm ^{3}$

$\sf \: Radius \: of \: small \: spherical \: laddoo = 2.5 \: cm$

$\sf \: \therefore Volume \: of \: small \: spherical \: laddoo \: \implies \: \sf \: \dfrac{4}{3} \pi \: \times 2.5 \times 2.5 \times 2.5 cm ^{3}$

$\sf \: \therefore Number \: of \: spherical \: laddoo \: \sf \: \implies \: \dfrac{\dfrac{4}{3} \pi \: \times 5 \times 5 \times 5}{\dfrac{4}{3} \pi \: \times 2.5 \times 2.5 \times 2.5 } \: = \: 8$

$\underline{ \boxed{ \pink{\sf \: Hence, \: with \: same \: amount \: of \: material, 8 \: laddoos \: can \: be \: made. }}}$

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Solution : 2

$\sf \: Here, diameter \: of \: solid \: spherical \: ball = 4.2 \: cm$

$\sf \therefore Radius \: of \: solid \: spherical \: ball = \dfrac{4.2}{2} = 2.1 \: cm$

$\sf \: Volume \: of \: solid \: spherical \: ball = \dfrac{4}{3} \pi r^{3}$

$\implies \: \sf \: \dfrac{4}{3} \times \dfrac{22}{7} \times 2.1 \times 2.1 \times 2.1$

$\implies \: \sf \: 38.808 \: cm ^{3}$

$\underline{ \boxed{ \pink{\sf Answer = 38.808 \: cm ^{3} }}}$

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Solution : 3

A sphere and a right circular cylinder have same radius and equal volume.

$\sf \: Volume \: of \: sphere = Volume \: of \: right \: circular \: cylinder$

$\implies \: \sf \: \dfrac{4}{3} \pi r^{3} = \pi r^{2} h$

$\implies \: \sf \: \dfrac{4}{3} r = h$

$\implies \: \sf r = \dfrac{3}{4} h$

$\sf \: \therefore Diameter \: of \: cylinder = 2 \times r = \dfrac{3}{2} h$

$\sf \: Total \: increase = \bigg( \dfrac{3}{2} - 1 \bigg)h = \dfrac{1}{2} h$

$\sf \: Increase\: \% = \dfrac{\dfrac{1}{2} h}{h} \times 100$

$\sf \: Increase\: \% = 50 \: \%$

$\boxed{ \pink{ \sf \: Hence, \: diameter \: of \: cylinder \: exceed \: it's \: height \: by \: 50 \%}}$