Question

✯The difference between the outside and inside surface of a cylindrical metallic tube 147 cm long is 184.8 cm². If r₁ + r₂ = 6.2 cm Where r₁ and r₂ are the inner and outer radii respectively, find r₁ and r₂ .​

Answer 1

✪ Question :- ✯The difference between the outside and inside surface of a cylindrical metallic tube 147 cm long is 184.8 cm². If r₁ + r₂ = 6.2 cm Where r₁ and r₂ are the inner and outer radii respectively, find r₁ and r₂ . ?

✰ Formula used :-

➳ Difference between the surface area of the outer cylinder and surface area of the inner cylinder = 2πr₁h - 2πr₂h.

❁ Solution :-

It is given that,

➺ Height of cylinder is = 147cm.

➺ difference between the outside and inside surface = 184.8cm²

Putting values now, we get,

➪ 2πh(r₁ - r₂) = 184.8

➪ 2*(22/7)*147*(r₁ - r₂) = 184.8

➪ 44*21*(r₁ - r₂) = 184.8

➪ (r₁ - r₂) = 184.8 / 924

➪ (r₁ - r₂) = 0.2 cm. ------------- Equation (1)

Now, it is given that,

☛ r₁ + r₂ = 6.2 ----------------- Equation (2)

Adding Equation (1) and Equation (2) now, we get,

☞ (r₁ - r₂) + ( r₁ + r₂) = 0.2 + 6.2

☞ 2r₁ = 6.4

☞ r₁ = 3.2 cm.

Putting this value in Equation 1 now, we get,

☞ 3.2 - r₂ = 0.2

☞ r₂ = 3.2 - 0.2

☞ r₂ = 3 cm.

Hence, value of r₁ and r₂ are 3.2cm and 3 cm respectively..

Answer 2

Answer:

${\underline{\boxed{\sf\green{r_1 \: \& \: r_2 = 3.2 \: cm \: \& \: 3 \: cm }}}}$

Step-by-step explanation:

Given:-

• Difference between outside surface area and inside surface area of cylindrical cube = 184.8 sq.cm
• Height of cylindrical cube = 147 cm
• r1 + r2 = 6.2 cm

To find:-

• Value of r1 & r2

We know that,

$\boxed{\boxed{\sf\pink{Surface\: area \: of \: cube = 2\pi r h}}}$

A/Q,

$\Rightarrow\sf 2\pi {r}_{1}h - 2\pi {r}_{2}h = 184.8 \\\\\Rightarrow\sf 2\pi h(r_1 - r_2 ) = 184.8 \\\\\Rightarrow\sf 2 \times \dfrac{22}{7} \times 147(r_1 - r_2) = 184.8 \\\\\Rightarrow\sf 924(r_1 - r_2) = 184.8 \\\\\Rightarrow\sf r_1 - r_2 = \dfrac{184.8}{924} \\\\\Rightarrow\sf\green{r_1 - r_2 = 0.2.........(i)}$

$\sf As, r_1 + r_2 = 6.2 ...........(ii)$

Adding both equation we get:-

$\Rightarrow\sf r_1 - \cancel{ r_2 } + r_1 +\cancel{ r_2 } = 0.2 + 6.2 \\\\\Rightarrow\sf {2r}_{1} = 6.4 \\\\\Rightarrow\sf r_1 =\cancel{\dfrac{6.4}{2}} \: cm \\\\\Rightarrow{\boxed{\sf\green{ {r}_{1} = 3.2 \: cm }}}$

Putting value of r1 into (ii) we get:-

$\Rightarrow\sf 3.2 + r_2 = 6.2 \\\\\Rightarrow\sf r_2 = 6.2 - 3.2 \\\\\Rightarrow{\boxed{\sf\green{ {r}_{2} = 3\: cm }}}$

${\underline{\boxed{\therefore{\sf\blue{ {r}_{1} = 3.2 \: cm\: \& \: {r}_{2} = 3 \: cm }}}}}$