第9类RD Sharma解决方案–第19章直圆柱体的表面积和体积–练习19.1

问题1：直圆柱体的弯曲表面积为4.4 m ² 。如果圆柱体的半径为0.7 m。找到它的高度。

解决方案：

Given, radius (r) = 0.7 m and Curved Surface Area(C.S.A) = 4.4 m² 

By formula, C.S.A = 2πrh where, h = height of the cylinder

Putting values in the formula,

4.4 m²  = 2 * (22/7) * 0.7 * h (using π = 22/7)

h = 1 m 

问题2：在热水加热系统中，有一个长28 m，直径5 cm的圆柱管。找到系统中的总辐射面。

解决方案：

Given, length (h) = 28 m and diameter = 5cm

So, radius(r) = 2.5cm = 0.025 m

Total radiating surface of the system is nothing but the curved surface area of the cylinder.

C.S.A = 2πrh

        = 2 * (22/7) * 28 * (0.025)   (using π = 22/7)

        = 4.4 m²  

Hence, the total radiating surface of the system is 4.4 m²

问题3：圆柱支柱直径为50厘米，高度为3.5 m。^{找出以每m 2} 12.50卢比的比率喷涂支柱曲面的成本。

解决方案：

Given, diameter = 50 cm (so, radius(r) = 25 cm = 0.25 m) and height (h) = 3.5 m

C.S.A = 2πrh

        = 2 * (22/7) * (0.25) * 3.5   (using π = 22/7)

        = 5.5 m²

Cost of painting the pillar = cost of painting per m² * painting cost of per m²

Cost = 5.5 m² * 12.50 ₨/m²

Cost = ₨ 68.75

问题4：要求从金属薄板制造一个高度为1 m，底直径为140 cm的密闭圆柱罐。同一张纸需要多少平方米?

解决方案：

Given, height (h) = 1 m and diameter = 140 cm (so, radius = 70 cm = 0.7 m)

Area of the sheet required(A) = total surface area (T.S.A) of the cylinder 

We know that T.S.A = 2πr(h + r)  

A = 2 * (22/7) * (0.7) * (0.7 + 1)   (using π = 22/7)

A = 2 * (22/7) * (0.7) * (1.7)

A = 7.48 m²

问题5：实心圆柱体的总表面积为462 cm ² 。它的弯曲表面积是其总表面积的三分之一。找到圆柱体的半径和高度。

解决方案：

Given, T.S.A = 462 cm² and C.S.A = (T.S.A)/3

Let us assume radius = r, height = h of the given cylinder.

Given, C.S.A = (T.S.A)/3 = (462 cm² / 3) = 154 cm²

Remaining area (R) = area of the top and bottom of the cylinder 

R = T.S.A – C.S.A 

    = 462 cm² – 154 cm²

    = 308 cm²

We know that, R = 2πr²

308 cm² = 2 * (22/7) * r²        (using π = 22/7)

r² = 49 cm²

r = 7 cm

Now, we know that C.S.A = 2πrh

154 = 2 * (22/7) * 7 * h

h = 3.5 cm

问题6：两侧开口的空心圆柱体的总表面积为4620 cm ² ，基环的面积为115.5 cm ² ，高度为7 cm。找到圆柱体的厚度。

解决方案：

Given, total surface area(T) = 4620 cm² , area of the base ring(R) = 115.5 cm²  and height (h) = 7 cm

Total surface area here means the curved surface area of the cylinder in the outside and the inside.

Let the inner radius be r and the thickness be t.

Then the outer radius (r₂) = r + t

R = π * {r₂ ² – r²}  —- (i)

T = 2πrh + 2πr₂h + 2 * R

4620 = 2πh(r + r₂) + 231 

2π * 7 * (r + r₂) = 4389 cm²

π(r + r₂) = 313.5 cm  — (ii)

From eqn (i) 

R = π * (r + r₂) * (r₂ – r)

Putting the value from eqn (ii)

115.5 cm² = 313.5 * t

t = 0.3684 cm

问题7：假定罐的高度和半径分别为7.5 m和3.5 m，求出圆柱体的总表面积与其弯曲表面积之间的比率。

解决方案：

Given, height = 7.5 m and radius = 3.5 m

Ratio (R) = T.S.A / C.S.A

R = (2πr(h + r)) / (2πrh)

R = (r + h)/h

R = (7.5 + 3.5)/7.5

R = 11/7.5 = 22/15