Question

एक धातु की पाइप 28 सेंटीमीटर लंबी है आंतरिक और brahe chetrafal ka antar 176 सेंटीमीटर square है आंतरिक और बाह्य त्रिज्या ज्ञात कीजिए यदि पाइप 352 सेंटीमीटर क्यूब धातु से बनी है
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Answer 1

Given:

The metal pipe is 28 cm long

The difference between the outer lateral surface area and the inner lateral surface area of a cylindrical metal pipe is 176 cm²

If the pipe is made of 352 cm³ of metal

To find:

The inner and outer radius of the pipe

Solution:

Let's assume,

"r₁" → the inner radius of the pipe

"r₂" → the outer radius of the pipe

The height of the hollow cylindrical pipe, h = 28 cm

The inner lateral surface area of the cylindrical pipe = $2 \pi r_1h$

and

The outer lateral surface area of the cylindrical pipe = $2 \pi r_2h$

We are given that the difference between the inner and the outer lateral surface areas of the cylindrical pipe = 176 cm²

So, we can write as:

$2\pi r_2h - 2\pi r_1h = 176$

⇒ $2\pi h (r_2 - r_1) = 176$

⇒ $(r_2 - r_1) = \frac{176}{2\pi h}$

⇒ $(r_2 - r_1) = \frac{176}{2\times \frac{22}{7} \times 28}$

⇒ $(r_2 - r_1) = \frac{176}{176}$

⇒ $(r_2 - r_1) = 1$ ..... (i)

Also, we are given that the pipe consists of 352 cm³ of metal

i.e., The volume of the hollow cylindrical pipe = 352 cm³

⇒ $\pi (r_2^2 - r_1^2)h = 352$

⇒ $\pi (r_2 - r_1) (r_2 + r_1)h = 352$

substituting from (i)

⇒ $\frac{22}{7} \times 1 \times (r_2 + r_1)\times 28 = 352$

⇒ $(r_2 + r_1) = \frac{352}{4\times 22}$

⇒ $(r_2 + r_1) = 4$ ..... (ii)

Now, adding equation (i) & (ii), we get

r₂ - r₁ = 4

r₂ - r₁ = 1

---------------

2r₂ = 5

---------------

∴ r₂ = $\frac{5}{2} = 2.5\:cm$

Substituting r₂ in eq. (i), we get

r₂ - r₁ = 1

⇒ 2.5 - r₁ = 1

⇒ r₁ = 2.5 - 1

⇒ r₁ = 1.5 cm

Thus,

$\boxed{\bold{The \:inner\:radius\:of\:the\:pipe\:is\:\underline{1.5\:cm}}}.\\\boxed{\bold{The \:outer\:radius\:of\:the\:pipe\:is\:\underline{2.5\:cm}}}.$

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