Question

The number of solid spheres, each of diameter 6 cm that could be moulded to form a solid metal cylinder of height 45 cm and diameter 4 cm, is
(a)3
(b)4
(c)5
(d)6

Answer 1

Answer:

The number of solid spheres is 5  .

Among the given options option (c) 5 is  the correct answer.

Step-by-step explanation:

Given :  

Height of a solid cylinder, h = 45 cm

Diameter of solid sphere = 6 cm

Diameter of a solid cylinder = 4 cm  

Radius of solid sphere ,R  = 6/2 = 3 cm

Radius of a solid cylinder , r = 4/2 = 2 cm  

Let 'n’ be the number of solid spheres  

Volume of Cylinder = n × Volume of sphere

n = Volume of Cylinder / Volume of sphere

n = (πr²h) /(4/3πR³)

n = (R²h) /(4/3r³)

n = (2² × 45)/ 4/3 × 3³

n = (4 × 45)/(4 × 27/3)

n = 45 /( 9)

n = 5  

Number of solid spheres = 5  

Hence, the number of solid spheres is 5  

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

$\mathfrak\pink{Question}$

The number of solid spheres, each of diameter 6 cm that could be moulded to form a solid metal cylinder of height 45 cm and diameter 4 cm, is

(a)3

(b)4

(c)5

(d)6

$\mathfrak\purple{Solution}$

Let the number of spheres be x

$\bold{Diameter\: of \:sphere= 6\:m}$

$\bold{Radius\: of \:sphere= \frac{6}{2} \:m}$

Now,

$\bold{Volume\: of \:sphere= Number\: of \:sphere× Volume \:of \:one sphere}$

$\bold{\pi \: r {}^{2} h} = \bold{x \times \frac{4}{3} \pi \: r {}^{3} }$

$\bold{\pi \times 2 \times 2 \times 45 = x \times \frac{4}{3} \times \pi \times r {}^{3} }$

$\bold{180 = x \times \frac{4}{3} \times 27 }$

$\bold{x = \frac{180}{36} } \:$

$\bold{x = 5}$

Therefore,

$\bold\blue{Option\:C\:is\:answer}$