Exercise-11.2
Maths - Exercise
Q1. Find the surface area of a sphere of radius:
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm
Let’s find the surface area for each sphere using the formula $$ SA = 4\pi r^2 $$
For \(r = 10.5~\text{cm}\): $$ \begin{aligned} SA &= 4 \pi r^2 \\ &= 4 \times \frac{22}{7} \times (10.5)^2 \\ &= 4 \times \frac{22}{7} \times 110.25 \\ &= 4 \times 22 \times 15.75 \\ &= 88 \times 15.75 \\ &= 1386~\text{cm}^2 \end{aligned} $$
For \(r = 5.6~\text{cm}\): $$ \begin{aligned} SA &= 4 \pi r^2 \\ &= 4 \times \frac{22}{7} \times (5.6)^2 \\ &= 4 \times \frac{22}{7} \times 31.36 \\ &= 4 \times 22 \times 4.48 \\ &= 88 \times 4.48 \\ &= 394.24~\text{cm}^2 \end{aligned} $$
For \(r = 14~\text{cm}\): $$ \begin{aligned} SA &= 4 \pi r^2 \\ &= 4 \times \frac{22}{7} \times 14 \times 14 \\ &= 4 \times \frac{22}{7} \times 196 \\ &= 4 \times 22 \times 28 \\ &= 88 \times 28 \\ &= 2464~\text{cm}^2 \end{aligned} $$
Therefore, the surface areas are $$1386~\text{cm}^2$$ $$394.24~\text{cm}^2$$ and $$2464~\text{cm}^2$$ respectively.
Q2. Find the surface area of a sphere of diameter:
(i) 14 cm
(ii) 21 cm
(iii) 3.5 m
Solution:
To find the surface area of a sphere, we use the formula $$SA = 4\pi r^2$$ where \(r\) is the radius of the sphere.
For a sphere with diameter \(14~\text{cm}\): The radius is $$r = \frac{14}{2} = 7~\text{cm}$$ $$ \begin{aligned} SA &= 4\pi r^2 \\ &= 4 \times \frac{22}{7} \times 7 \times 7 \\ &= 4 \times 22 \times 7 \\ &= 88 \times 7 \\ &= 616~\text{cm}^2 \end{aligned} $$
For a sphere with diameter \(21~\text{cm}\): The radius is $$r = \frac{21}{2} = 10.5~\text{cm}$$ $$ \begin{aligned} SA &= 4\pi r^2 \\ &= 4 \times \frac{22}{7} \times 10.5 \times 10.5 \\ &= 4 \times 22 \times 15.75 \\ &= 88 \times 15.75 \\ &= 1386~\text{cm}^2 \end{aligned} $$
For a sphere with diameter \(3.5~\text{m}\): The radius is $$r = \frac{3.5}{2} = 1.75~\text{m}$$ $$ \begin{aligned} SA &= 4\pi r^2 \\ &= 4 \times \frac{22}{7} \times 1.75 \times 1.75 \\ &= 4 \times \frac{22}{7} \times 3.0625 \\ &= 4 \times 22 \times 0.4375 \\ &= 88 \times 0.4375 \\ &= 38.5~\text{m}^2 \end{aligned} $$
Therefore, the surface areas of the spheres are \(616~\text{cm}^2,\;1386~\text{cm}^2\text{ and }
38.5~\text{m}^2\)
for the diameters \(14~\text{cm},\;21~\text{cm}\text{ and }3.5~\text{m}\)
respectively.
Q3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)
Solution:
The radius of the hemisphere is $$10~\text{cm}$$ The total surface area of a hemisphere includes both its curved surface and the flat circular base, so the formula is $$TSA = 3\pi r^2$$
Substituting the given values: $$ \begin{aligned} TSA &= 3\pi r^2 \\ &= 3 \times 3.14 \times 10 \times 10 \\ &= 3 \times 3.14 \times 100 \\ &= 3 \times 314 \\ &= 942~\text{cm}^2 \end{aligned} $$
Therefore, the total surface area of the hemisphere is $$942~\text{cm}^2$$
Q4. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Solution:
The radius of the balloon increases from \(7~\text{cm}\) to \(14~\text{cm}\)
Let the initial radius be $$r_1 = 7~\text{cm}$$ and the final radius be $$r_2 = 14~\text{cm}$$
The surface area of a sphere is given by $$4\pi r^2$$ Let $$TSA_1$$ be the initial surface area and $$TSA_2$$ the final surface area: $$ \begin{aligned} TSA_1 &= 4\pi r_1^2 \\ TSA_2 &= 4\pi r_2^2 \\ \end{aligned} $$
To find the ratio of the two surface areas: $$ \begin{aligned} \frac{TSA_1}{TSA_2} &= \frac{4\pi r_1^2}{4\pi r_2^2} \\ &= \left( \frac{r_1}{r_2} \right)^2 \\ &= \left( \frac{7}{14} \right)^2 \\ &= \left( \frac{1}{2} \right)^2 \\ &= \frac{1}{4} \end{aligned} $$
Therefore, the ratio of the surface areas is $$1 : 4$$
Q5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹ 16 per 100 \(cm^2\).
Solution:
The inner diameter of the hemispherical brass bowl is \(10.5~\text{cm}\), so its radius is: $$ r = \frac{10.5}{2} = 5.25~\text{cm} $$
The inner surface area of the bowl is the curved surface area of a hemisphere, given by $$SA = 2\pi r^2$$ $$ \begin{aligned} SA &= 2 \pi r^2 \\ &= 2 \times \frac{22}{7} \times 5.25 \times 5.25 \\ &= 2 \times \frac{22}{7} \times 27.5625 \\ &= 2 \times 22 \times 3.9375 \\ &= 44 \times 3.9375 \\ &= 173.25~\text{cm}^2 \end{aligned} $$
The rate of tin-plating is ₹16 per $$100~\text{cm}^2$$ Therefore, the cost for tin-plating the bowl is: $$ \text{Total cost} = \frac{16}{100} \times 173.25 = 27.72 $$
Therefore, the cost of tin-plating the inside of the hemispherical bowl is ₹27.72.
Q6. Find the radius of a sphere whose surface area is 154 \(cm^2.\)
Solution:
The surface area of the sphere is given as $$154~\text{cm}^2$$ Let the radius be \(r\)
The surface area formula is $$SA = 4\pi r^2$$ $$ \begin{aligned} 154 &= 4\pi r^2 \\\\ r^2 &= \frac{154}{4\pi} \\\\ &= \frac{154}{4 \times \frac{22}{7}} \\\\ &= \frac{154 \times 7}{4 \times 22} \\\\ &= \frac{1078}{88} \\\\ &= 12.25 \\\\ r &= \sqrt{12.25} \\\\ &= 3.5~\text{cm} \end{aligned} $$
Therefore, the radius of the sphere is $$3.5~\text{cm}$$
Q7.The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Solution:
Let the radius of the moon be \(r_m\) and the radius of the earth be \(r_e\)
Given that the diameter of the moon is one fourth that of the earth,
$$
r_m = \frac{1}{4} r_e
$$
The surface area of a sphere is $$4\pi r^2$$ Let the surface area of the moon be \(SA_m\) and of the earth be \(SA_e\) $$ SA_m = 4\pi r_m^2,\\\\ SA_e = 4\pi r_e^2 $$
The ratio of their surface areas is: $$\begin{aligned} \frac{SA_m}{SA_e} &= \frac{4\pi r_m^2}{4\pi r_e^2} \\\\&= \frac{r_m^2}{r_e^2} \end{aligned}$$
Substituting $$r_m = \frac{1}{4} r_e$$ we get: $$\begin{aligned} \frac{SA_m}{SA_e} &= \left(\frac{r_m}{r_e}\right)^2 \\\\&= \left(\frac{1}{4}\right)^2 \\\\&= \frac{1}{16} \end{aligned}$$
Therefore, the ratio of the surface area of the moon to that of the earth is $$1 : 16$$
Q8.A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
Solution:
The inner radius of the hemispherical bowl is $$5~\text{cm}$$ and the steel has a thickness of $$0.25~\text{cm}$$ Therefore, the outer radius \((r_0)\) is: $$\begin{aligned} r_0 &= 5 + 0.25 \\&= 5.25~\text{cm} \end{aligned}$$
The outer curved surface area of a hemisphere is given by $$2\pi r_0^2$$ Outer curved surface area \((OCSA)\) $$ \begin{aligned} \text{OCSA}&= 2 \pi (r_0)^2 \\\\ &= 2 \times \frac{22}{7} \times 5.25 \times 5.25 \\\\ &= 2 \times \frac{22}{7} \times 27.5625 \\\\ &= 2 \times 22 \times 3.9375 \\\\ &= 44 \times 3.9375 \\\\ &= 173.25~\text{cm}^2 \end{aligned} $$
Therefore, the outer curved surface area of the bowl is $$173.25~\text{cm}^2$$
Q9.A right circular cylinder just encloses a sphere of
radius r (see Fig. 11.10). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).
Solution:
Let the radius of the sphere be \(r\).
(i) The surface area of the sphere is: $$ SA_{\text{sphere}} = 4 \pi r^2 $$
(ii) For a cylinder to just enclose a sphere, the height of the cylinder equals the diameter of the sphere, so \(h = 2r\), and the radius of the cylinder is also \(r\)
The curved surface area of the cylinder is: $$\begin{aligned} CSA_{\text{cylinder}} &= 2\pi r h \\&= 2\pi r \times 2r \\&= 4\pi r^2 \end{aligned}$$
(iii) The ratio of the surface area of the sphere to the curved surface area of the cylinder is: $$ \frac{SA_{\text{sphere}}}{CSA_{\text{cylinder}}} = \frac{4\pi r^2}{4\pi r^2} = 1 $$
Therefore, the ratio is $$1 : 1$$
