Question

A spherical metalic shell with 10 cm diameter external diameter, weighs 1789* 1/3 g. find the thickness of the shell if the density is 7 g/cm3

Answer 1

d=10cm
So, external r= d/2= 5 cm (Let external radius= ro)
Let internal radius= ri
Weight= $1789 \frac{1}{3}$ g= 5368g
Density (D)= 7g/cm³
Volume of the sphere= Outer vol- Inner vol 
= $( \frac{4}{3} \pi r_o^3) - ( \frac{4}{3} \pi r_i^3 )$
$\frac{4}{3} \pi (r_o^3 - r_i^3)$
$\frac{4}{3} * \frac{22}{7}(5^3-r_i^3)$
$\frac{88}{21} (125-r_i^3)$ Let this be V
Therefore, for 1cm³ of Vol., weight= 7g
So for V cm³, wt =7*V g
=$7* \frac{88}{21} (125-r_i^3)$
$\frac{88}{3} (125-r_i^3)$ g
But, given wt = 5368 g
So, $5368=\frac{88}{3} (125-r_i^3)$
or, $\frac{5368*3}{88} =125-r_i^3$
$183-125=r_i^3=58$
$r_i= \sqrt[3]{58} = 3.87 cm$
Now, thickness= $r_o-r_i= 5-3.87= 1.13cm$

Answer 2

Answer:

Mass of the spherical metallic shell = 1789 1/3 g

= 5368/3 g

Density = 7 g/cm³  

Volume of the spherical metallic shell = Mass/density

⇒ 5368/7 ÷ 7

= 255.61 cm³

Now,

External diameter = 10 cm

So, external radius = 5 cm

Let the internal radius be 'r'

Now,

According to the question.

4/3π(R³ - r³) = 255.61 cm³

⇒ 5³ - r³ = (255.61*3)/(4*3.14)

⇒ 125 - r³ = 766.83/12.56

⇒ 125 - r³ = 61.05

⇒ r³ = 125 - 61.05

⇒ r³ = 63.95

⇒ r = ∛63.95

⇒ r = 3.9989 or 4 (Approx)

So, the thickness = 5 - 4  

= 1 cm

Answer.

Step-by-step explanation:

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