Question

solid spheres of d 6cm are dropped into a cylindrical beaker containing some water and are fully submerged. if the diameter of the beaker is 18 cm qnd the water rises by 40 cm find the number of solid spheres dropped in the water

Answer 1

Aиѕωєr —

90 Solid spheres were dropped in water .

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Cσrrєcт Qυєѕтiσи —

Solid spheres of diameter 6cm are dropped into a cylindrical beaker containing some water and are fully submerged. If the diameter of the beaker is 18 cm and the water rises by 40 cm ,find the number of solid spheres dropped in the water.

Hєrє —

Diameter of spheres ( d ) = 6 cm

Diameter of cylindrical breaker ( D ) = 18 cm

Height of water rises in the beaker ( h ) = 40 cm

Hєиcє —

❍ Radius of lead sphere ( r ) = $\sf{\dfrac{d}{2}}$

$\qquad\qquad\qquad\qquad\sf{ \cancel{=\dfrac{6}{2}}}$

$\qquad\qquad\qquad\qquad\sf{= 3 \:cm }$

❍ Radius of cylindrical beaker ( R ) = $\sf{\dfrac{D}{2}}$

$\qquad\qquad\qquad\qquad\sf{\cancel{=\dfrac{18}{2}} }$

$\qquad\qquad\qquad\qquad\sf{ =9\:cm}$

❍ The volume rises in water in the beaker :

$\qquad\qquad\qquad\qquad{\boxed{\sf{V =\pi r^2 h }}}$

❍ The volume of one sphere :

$\qquad\qquad\qquad\qquad\boxed{\sf{ v=\dfrac{4}{3}\pi r^3}}$

★ Hence, volume of n spheres

$\qquad\qquad\qquad\qquad\sf{V =nv=\dfrac{4}{3}n\pi r^3 }$

❍ Let the volume of water in the beaker rises by 40 cm by dropping n lead spheres of diameter 6 cm.

$\qquad\rm{\implies Volume\:of\:n\: spheres=increased\: volume\:of\:water\:in\:the\: water}$

$\qquad\rm{\implies \dfrac{4}{3}n\pi r^3 =\pi R^2h}$

$\qquad\rm{\implies n = \dfrac{3\pi R^2h}{4\pi r^3}}$

$\qquad\rm{\implies n =\dfrac{3R^2 h}{4r^4}}$

$\qquad\rm{\implies n = \dfrac{3\times 9^2\times 40}{4\times 3^3}}$

$\qquad\rm{\implies n = \dfrac{3\times 81\times 40}{4\times 27}}$

$\qquad\rm{\implies n = \cancel{\dfrac{9720}{108}}}$

$\qquad\bf{\implies n = 90 }$

$\huge{\dag}$Hence , for rising the water by 40cm , 90 solid spheres were dropped in the water .

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