NCERT Class 9 Maths Exercise 11.4 Solutions

📘 Concept & Theory Concept Used ›

A sphere is a perfectly round three-dimensional solid in which every point on the surface is at the same distance from the centre.

The volume of a sphere gives the amount of space occupied by it.

Formula for the volume of a sphere:

\[ V = \frac{4}{3}\pi r^3 \]

where

\(V\) = Volume of the sphere
\(r\) = Radius of the sphere
\(\pi \approx \frac{22}{7}\)

🗺️ Solution Roadmap Step-by-step Plan ›

Identify the radius of the sphere.
Use the formula \(\displaystyle V=\frac{4}{3}\pi r^3\).
Substitute the given value of radius.
Calculate \(r^3\).
Simplify step by step and write the final answer with proper units.

📊 Graph / Figure Graph / Figure ›

Sphere and its radius

✏️ Solution Complete Solution ›

Step-by-step Solution · 6 steps

(i) Radius = 7 cm
We know that,\[V=\frac{4}{3}\pi r^3\]
where \(V\) is the volume and \(r\) is the radius of the sphere.

(i) Radius \(=7\) cm
Therefore, \[ \begin{aligned} V &=\frac{4}{3}\times\frac{22}{7}\times343 \\ &=\frac{4}{3}\times22\times49 \qquad \left( \because \frac{343}{7}=49 \right) \\ &=\frac{4\times22\times49}{3} \\ &=\frac{4312}{3} \\ &=1437.33 \end{aligned} \]
Hence,\[V\approx1437.33\text{ cm}^3\]
(ii) (ii) Radius \(=0.63\) m
Given,\[ r=0.63\text{ m} \]
Using the formula: \[\begin{aligned}V &=\frac{4}{3}\times\frac{22}{7}\times(0.63)^3\\&=\frac{4}{3}\times\frac{22}{7}\times0.250047 \\ &=\frac{88}{21}\times0.250047 \\ &=1.047816\end{aligned}\]
Therefore,\[V\approx1.05\text{ m}^3\]

🎯 Exam Significance Exam Significance ›

This question helps students master the direct application of the volume formula of a sphere.
Board examinations frequently ask numerical problems based on substitution into standard mensuration formulas.
Competitive entrance examinations test speed and accuracy in calculations involving cubes and decimals.
Students learn proper unit writing such as \(\text{cm}^3\) and \(\text{m}^3\), which is important for scoring full marks.
The problem strengthens conceptual understanding of three-dimensional geometry and mensuration.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Volume of a sphere is calculated using \[ V=\frac{4}{3}\pi r^3 \]
Always cube the radius before multiplication.
Use proper approximation for \(\pi\), generally \(\frac{22}{7}\).
Write all calculation steps clearly to avoid losing marks in board examinations.
Final answers for volume must always be written in cubic units.

📘 Concept & Theory Concept Used ›

When a solid object is completely immersed in water, it displaces water equal to its own volume.

Therefore, the amount of water displaced by a solid spherical ball is equal to the volume of the sphere.

Formula for the volume of a sphere:

\[ V=\frac{4}{3}\pi r^3 \]

where

\(V\) = Volume of the sphere
\(r\) = Radius of the sphere
\(\pi \approx \frac{22}{7}\)

🗺️ Solution Roadmap Step-by-step Plan ›

Find the radius using: \[ r=\frac{d}{2} \]
Use the formula: \[ V=\frac{4}{3}\pi r^3 \]
Substitute the value of radius.
Calculate \(r^3\) carefully.
Simplify step by step and write the final answer with proper cubic units.

📊 Graph / Figure Graph / Figure ›

A spherical ball immersed in water

✏️ Solution Complete Solution ›

Step-by-step Solution · 10 steps

The amount of water displaced by the spherical ball is equal to the volume of the sphere.
We know that,\[V=\frac{4}{3}\pi r^3\]
(i) Diameter \(=28\) cm
Given,\[d=28\text{ cm}\]
Radius: \[ \begin{aligned} r&=\frac{d}{2} \\ &=\frac{28}{2} \\ &=14\text{ cm} \end{aligned} \]
Using the formula: \[ \begin{aligned} V &=\frac{4}{3}\times\frac{22}{7}\times(14)^3\\ &=\frac{4}{3}\times\frac{22}{7}\times14\times14\times14 \\ &=\frac{4}{3}\times\frac{22}{7}\times2744\\ &=\frac{4}{3}\times22\times392 \qquad \left( \because \frac{2744}{7}=392 \right) \\ &=\frac{4\times22\times392}{3} \\ &=\frac{34496}{3} \\ &=11498.67 \end{aligned} \]
Hence, the amount of water displaced is \[11498.67\text{ cm}^3\]
(ii) Diameter \(=0.21\) m
Given,\[d=0.21\text{ m}\]
Radius:\[ \begin{aligned} r &=\frac{d}{2} \\ &=\frac{0.21}{2} \\ &=0.105\text{ m} \end{aligned} \]
Using the formula: \[ \begin{aligned} V&=\frac{4}{3}\times\frac{22}{7}\times(0.105)^3\\ &=\frac{4}{3}\times\frac{22}{7}\times0.001157625\\ &=\frac{88}{21}\times 0.001157625\\ &=0.004851 \end{aligned} \]
Therefore, the amount of water displaced is\[0.004851\text{ m}^3\]

🎯 Exam Significance Exam Significance ›

This question tests understanding of the relationship between displacement of water and volume of solids.
It strengthens the application of the sphere volume formula in real-life situations.
Board examinations frequently include direct mensuration problems involving spheres and hemispheres.
Competitive examinations check numerical accuracy involving cubes, decimals, and unit conversions.
The problem improves conceptual clarity about three-dimensional geometry and measurement.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Water displaced by a completely immersed object equals the volume of the object.
Radius is always half of the diameter.
Volume of a sphere is given by \[ V=\frac{4}{3}\pi r^3 \]
Always calculate \(r^3\) carefully to avoid arithmetic mistakes.
Final answers for volume must be written in cubic units such as \(\text{cm}^3\) or \(\text{m}^3\).

📘 Concept & Theory Concept Used ›

The mass of a solid object can be calculated if its density and volume are known.

Relation between mass, density, and volume:

\[ \text{Mass}=\text{Density}\times\text{Volume} \]

Since the object is spherical, its volume is calculated using:

\[ V=\frac{4}{3}\pi r^3 \]

where

\(V\) = Volume of the sphere
\(r\) = Radius of the sphere
\(\pi \approx \frac{22}{7}\)
\(\rho\) = Density of the material

🗺️ Solution Roadmap Step-by-step Plan ›

Find the radius by dividing the diameter by 2.
Use the sphere volume formula.
Calculate the volume step by step.
Multiply the volume by density.
Write the final answer in grams.

✏️ Solution Complete Solution ›

Step-by-step Solution · 8 steps

Given: Diameter of the metallic ball:\[d=4.2\text{ cm}\]
Density of the metal:\[\rho=8.9\text{ g/cm}^3\]
Radius is half of the diameter: \[ \begin{aligned} r &=\frac{d}{2} \\ &=\frac{4.2}{2} \\ &=2.1\text{ cm} \end{aligned} \]
Volume of the sphere: \[ \begin{aligned} V&=\frac{4}{3}\pi r^3\\ &=\frac{4}{3}\times\frac{22}{7}\times9.261 \\ &=\frac{88}{21}\times9.261 \\ &=38.808 \end{aligned} \]
Therefore,\[V\approx38.81\text{ cm}^3\]
Now,\[\text{Mass}=\text{Density}\times\text{Volume}\]
Substituting the values: \[ \begin{aligned} m &=8.9\times38.808 \\ &=345.3912 \end{aligned} \]
Hence, the mass of the metallic ball is \[345.39\text{ g}\]

🎯 Exam Significance Exam Significance ›

This problem combines concepts of mensuration and density, which are frequently tested in examinations.
Students learn how mathematical formulas are applied in physics and real-life situations.
Board examinations often ask application-based questions involving volume and mass calculations.
Competitive entrance examinations test numerical calculation skills and unit handling.
The question strengthens understanding of three-dimensional geometry and practical measurement.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Radius is always half of the diameter.
Volume of a sphere is calculated using \[ V=\frac{4}{3}\pi r^3 \]
Mass is obtained using \[ \text{Mass}=\text{Density}\times\text{Volume} \]
Proper units must be maintained throughout the calculation.
Step-by-step simplification helps avoid arithmetic mistakes in examinations.

📘 Concept & Theory Concept Used ›

The Earth and the Moon are considered spherical in shape.

The volume of a sphere is given by:

\[ V=\frac{4}{3}\pi r^3 \]

Since volume depends on the cube of the radius, if the radius changes in a certain ratio, the volume changes in the cube of that ratio.

If

\[ r_1:r_2=a:b \]

then

\[ V_1:V_2=a^3:b^3 \]

🗺️ Solution Roadmap Step-by-step Plan ›

Use the relation between the diameters of the Moon and Earth.
Convert the diameter ratio into the radius ratio.
Use the sphere volume formula.
Find the ratio of volumes using the cube of the radius ratio.
Write the required fraction clearly.

📊 Graph / Figure Graph / Figure ›

Metallic spherical ball

✏️ Solution Complete Solution ›

Step-by-step Solution · 14 steps

Let
- \(r_e\) = Radius of the Earth
- \(r_m\) = Radius of the Moon
According to the question:
Diameter of the Moon\[=\frac{1}{4}\times \text{Diameter of the Earth}\]
Since radius is half of diameter, the same ratio applies to the radii.
Therefore,\[r_m=\frac{1}{4}r_e\]
Equivalently,\[r_e=4r_m\]
Volume of a sphere:\[V=\frac{4}{3}\pi r^3\]
Volume of the Earth:\[V_e=\frac{4}{3}\pi r_e^3\]
Volume of the Moon:\[V_m=\frac{4}{3}\pi r_m^3\]
Taking the ratio: \[ \begin{aligned} \frac{V_e}{V_m} &=\frac{\dfrac{4}{3}\pi r_e^3}{\dfrac{4}{3}\pi r_m^3} \\ &=\frac{r_e^3}{r_m^3} \\ &=\left(\frac{r_e}{r_m}\right)^3 \end{aligned} \]
Since \[r_e=4r_m\]
we get \[ \begin{aligned} \frac{V_e}{V_m} &=\left(\frac{4r_m}{r_m}\right)^3 \\ &=4^3 \\ &=64 \end{aligned} \]
Therefore,\[\frac{V_m}{V_e}=\frac{1}{64}\]
Hence, the volume of the Moon is\[\frac{1}{64}\] of the volume of the Earth.

🎯 Exam Significance Exam Significance ›

This question develops understanding of proportional reasoning in geometry.
Students learn that volume changes according to the cube of linear dimensions.
Board examinations frequently ask ratio-based mensuration problems.
Competitive examinations test conceptual clarity about scaling of areas and volumes.
The problem improves algebraic manipulation and formula application skills.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Volume of a sphere depends on the cube of its radius.
If radii are in the ratio \(a:b\), then volumes are in the ratio \(a^3:b^3\).
Diameter ratio and radius ratio remain the same.
Common terms cancel while taking ratios of volumes.
Careful step-by-step simplification helps avoid mistakes in ratio problems.

📘 Concept & Theory Concept Used ›

A hemispherical bowl is half of a sphere.

The capacity of the bowl is equal to the volume of the hemisphere.

Formula for the volume of a hemisphere:

\[ V=\frac{2}{3}\pi r^3 \]

where

\(V\) = Volume of the hemisphere
\(r\) = Radius of the hemisphere
\(\pi \approx \frac{22}{7}\)

Conversion relation:

\[ 1000\text{ cm}^3=1\text{ litre} \]

🗺️ Solution Roadmap Step-by-step Plan ›

Find the radius from the given diameter.
Use the hemisphere volume formula.
Calculate \(r^3\) carefully.
Find the volume in cubic centimetres.
Convert cubic centimetres into litres.

✏️ Solution Complete Solution ›

Step-by-step Solution · 6 steps

Given:
Diameter of the hemispherical bowl:

\[ d=10.5\text{ cm} \]

Radius is half of the diameter:

\[ \begin{aligned} r &=\frac{d}{2} \ &=\frac{10.5}{2} \ &=5.25\text{ cm} \end{aligned} \]
Step-by-step Solution
Volume of a hemisphere: \[ \begin{aligned} V&=\frac{2}{3}\pi r^3\\ &=\frac{2}{3}\times\frac{22}{7}\times 5.25\times 5.25 \times 5.25\\ &=\frac{2}{3}\times\frac{22}{7}\times144.703125\\ &=\frac{44}{21}\times144.703125 \\ &=303.1875 \end{aligned} \]
Therefore,\[V\approx303.19\text{ cm}^3\]
Since \[1000\text{ cm}^3=1\text{ litre}\]
Capacity of the bowl: \[ \begin{aligned} \text{Capacity} &=\frac{303.1875}{1000} \\ &=0.3031875 \end{aligned} \]
Hence, the hemispherical bowl can hold approximately \(0.303\text{ litre}\) of milk

🎯 Exam Significance Exam Significance ›

This problem strengthens understanding of the hemisphere volume formula.
Students learn practical applications of mensuration in capacity measurement.
Board examinations often include conversion-based questions involving litres and cubic centimetres.
Competitive examinations test numerical accuracy and unit conversion skills.
The question improves conceptual understanding of three-dimensional solids and their capacities.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

A hemisphere is half of a sphere.
Volume of a hemisphere is given by \[ V=\frac{2}{3}\pi r^3 \]
Radius is half of the diameter.
Always convert cubic centimetres into litres when capacity is asked.
Careful step-by-step simplification helps avoid calculation mistakes.

📘 Concept & Theory Concept Used ›

The hemispherical tank is hollow because it is made from an iron sheet of certain thickness.

Therefore, the volume of iron used is equal to:

\[ \text{Volume of outer hemisphere}- \text{Volume of inner hemisphere} \]

Volume of a hemisphere:

\[ V=\frac{2}{3}\pi r^3 \]

where \(r\) is the radius of the hemisphere.

🗺️ Solution Roadmap Step-by-step Plan ›

Find the outer radius of the hemispherical tank.
Calculate the volume of the outer hemisphere.
Calculate the volume of the inner hemisphere.
Subtract both volumes to obtain the volume of iron used.

📊 Graph / Figure Graph / Figure ›

Hemispherical iron tank

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

Given
Inner radius of the hemispherical tank:\[r_i=1\text{ m}\]
Thickness of the iron sheet:\[1\text{ cm}=0.01\text{ m}\]
Step-by-step Solution
Therefore, outer radius: \[ \begin{aligned} r_o &=r_i+0.01 \\ &=1+0.01 \\ &=1.01\text{ m} \end{aligned} \]
Volume of a hemisphere:\[ V=\frac{2}{3}\pi r^3 \]
Volume of iron used: \[ \begin{aligned} V &=\frac{2}{3}\pi r_o^3-\frac{2}{3}\pi r_i^3 \\ &=\frac{2}{3}\pi\left(r_o^3-r_i^3\right) \end{aligned} \]
Substituting the values: \[ \begin{aligned} V &=\frac{2}{3}\times\frac{22}{7}\left[(1.01)^3-(1)^3\right]\\ &=\frac{2}{3}\times\frac{22}{7}\times (1.030301-1)\\ &=\frac{2}{3}\times\frac{22}{7}\times0.030301\\ &=\frac{44}{21}\times0.030301\\ &=0.063488 \end{aligned} \]
Hence, the volume of iron used is approximately \(0.0635\text{ m}^3\)

🎯 Exam Significance Exam Significance ›

This problem develops understanding of hollow solids and thickness concepts.
Students learn how to calculate material used in manufacturing containers and tanks.
Board examinations frequently ask application-based mensuration questions involving hollow hemispheres.
Competitive entrance examinations test conceptual clarity about inner and outer dimensions.
The question strengthens calculation accuracy involving decimals and cubic measurements.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Volume of material used in a hollow object is found by subtraction of volumes.
Volume of a hemisphere is given by \[ V=\frac{2}{3}\pi r^3 \]
Always convert all measurements into the same unit before calculation.
Outer radius equals inner radius plus thickness.
Step-by-step simplification helps avoid arithmetic mistakes in examination problems.

📘 Concept & Theory Concept Used ›

The surface area and volume of a sphere are related through the radius.

Surface area of a sphere:

\[ S=4\pi r^2 \]

Volume of a sphere:

\[ V=\frac{4}{3}\pi r^3 \]

First, we find the radius using the surface area formula. Then we substitute the radius into the volume formula.

🗺️ Solution Roadmap Step-by-step Plan ›

Use the surface area formula to find the radius.
Solve for \(r^2\).
Find the value of \(r\).
Substitute the radius into the sphere volume formula.
Simplify step by step and write the final answer with cubic units

✏️ Solution Complete Solution ›

Step-by-step Solution · 10 steps

Given
Surface area of the sphere:\[ S=154\text{ cm}^2\]
Step-by-step Solution
Surface area formula: \[ S=4\pi r^2 \]
Substituting the given value:
\[ \begin{aligned} 154 &=4\times\frac{22}{7}\times r^2 \end{aligned} \]
Simplifying: \[ \begin{aligned} 154 &=\frac{88}{7}r^2 \end{aligned} \]
Multiplying both sides by \(7\): \[ \begin{aligned} 154\times7 &=88r^2 \\ 1078 &=88r^2 \end{aligned} \]
Dividing both sides by \(88\): \[ \begin{aligned} r^2 &=\frac{1078}{88} \\ &=12.25 \end{aligned} \]
Taking square root: \[ \begin{aligned} r &=\sqrt{12.25} \\ &=3.5\text{ cm} \end{aligned} \]
Calculation of Volume
\[ \begin{aligned} V&=\frac{4}{3}\pi r^3\\ &=\frac{4}{3}\times\frac{22}{7}\times(3.5)^3\\ &=\frac{4}{3}\times\frac{22}{7}\times42.875\\ &=\frac{88}{21}\times42.875\\ &=179.67 \end{aligned} \]
Hence, the volume of the sphere is\[179.67\text{ cm}^3\]

🎯 Exam Significance Exam Significance ›

This question develops understanding of the relationship between surface area and volume.
Students learn how to derive unknown dimensions using formulas.
Board examinations frequently ask multi-step mensuration problems.
Competitive entrance examinations test algebraic manipulation and numerical calculation skills.
The problem strengthens conceptual understanding of spherical geometry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Surface area of a sphere is given by \[ S=4\pi r^2 \]
Volume of a sphere is given by \[ V=\frac{4}{3}\pi r^3 \]
Always find the radius first before calculating the volume.
Proper simplification helps reduce arithmetic errors.
Final answers for volume must always be written in cubic units.

📘 Concept & Theory Concept Used ›

The inside of the dome is hemispherical in shape.

The white-washed area is the curved surface area of the hemisphere.

Curved surface area of a hemisphere:

\[ \text{CSA}=2\pi r^2 \]

Volume of a hemisphere:

\[ V=\frac{2}{3}\pi r^3 \]

We first find the inside surface area using the cost and rate, then calculate the radius and finally the volume.

🗺️ Solution Roadmap Step-by-step Plan ›

Find the inside surface area using: \[ \text{Area}=\frac{\text{Cost}}{\text{Rate}} \]
Use the curved surface area formula of a hemisphere to find the radius.
Substitute the radius into the hemisphere volume formula.
Calculate the volume step by step.
Write the final answers with correct units.

📊 Graph / Figure Graph / Figure ›

Hemispherical dome

✏️ Solution Complete Solution ›

Step-by-step Solution · 15 steps

Given
Cost of white-washing:\[\text{₹ }\,4989.60\]
Rate of white-washing:\[\text{₹ }\,20\text{ per m}^2\]
(i) Inside Surface Area of the Dome
Inside surface area: \[ \begin{aligned} \text{Area} &=\frac{\text{Cost}}{\text{Rate}} \\ &=\frac{4989.60}{20} \\ &=249.48\text{ m}^2 \end{aligned} \]
Hence, the inside surface area of the dome is \[249.48\text{ m}^2\]
(ii) Volume of the Air Inside the Dome
Let the radius of the hemispherical dome be \(r\) metres.
Curved surface area of a hemisphere:\[\text{CSA}=2\pi r^2\]
Substituting the area: \[ \begin{aligned} 249.48 &=2\times\frac{22}{7}\times r^2 \end{aligned} \]
Simplifying: \[ \begin{aligned} 249.48 &=\frac{44}{7}r^2 \end{aligned} \]
Multiplying both sides by \(7\):
\[ \begin{aligned} 249.48\times7 &=44r^2 \\ 1746.36 &=44r^2 \end{aligned} \]
Dividing both sides by \(44\): \[ \begin{aligned} r^2 &=\frac{1746.36}{44} \\ &=39.69 \end{aligned} \]
Taking square root: \[ \begin{aligned} r &=\sqrt{39.69} \\ &=6.3\text{ m} \end{aligned} \]
Volume of a hemisphere: \[ V=\frac{2}{3}\pi r^3 \]
Substituting \(r=6.3\): \[ \begin{aligned} V &&=\frac{2}{3}\times\frac{22}{7}\times(6.3)^3\\ &=\frac{2}{3}\times\frac{22}{7}\times250.047\\ &=\frac{44}{21}\times250.047\\ &=523.908 \end{aligned} \]
Hence, the volume of air inside the dome is approximately \(523.91\text{ m}^3\)

🎯 Exam Significance Exam Significance ›

This problem connects mensuration with practical applications such as painting and construction.
Students learn how to derive dimensions using cost and area relationships.
Board examinations frequently ask application-based questions involving hemispheres.
Competitive entrance examinations test formula application and multi-step calculations.
The question improves understanding of surface area and volume relationships in three-dimensional geometry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Curved surface area of a hemisphere is \[ \text{CSA}=2\pi r^2 \]
Volume of a hemisphere is \[ V=\frac{2}{3}\pi r^3 \]
Area can be calculated using: \[ \text{Area}=\frac{\text{Cost}}{\text{Rate}} \]
Always calculate the radius before finding volume.
Step-by-step simplification improves accuracy in board examinations.

📘 Concept & Theory Concept used ›

When solid spheres are melted and recast into another sphere, the total volume remains conserved.

Volume of a sphere:

\[ V=\frac{4}{3}\pi r^3 \]

Surface area of a sphere:

\[ S=4\pi r^2 \]

Since the new sphere is formed by melting 27 identical spheres, the total volume of all small spheres equals the volume of the new sphere.

🗺️ Solution Roadmap Step-by-step Plan ›

Find the volume of one small sphere.
Multiply by 27 to get the total volume.
Equate the total volume with the volume of the new sphere.
Find the radius of the new sphere.
Use the surface area formulas to determine the ratio \(S:S'\).

✏️ Solution Complete Solution ›

Step-by-step Solution · 17 steps

Given
Each small sphere has:
- Radius \(=r\)
- Surface area \(=S\)
(i) radius r′ of the new sphere,
Surface area of one sphere:\[S=4\pi r^2\]
Volume of one small sphere:\[V=\frac{4}{3}\pi r^3\]
Since there are 27 spheres, total volume: \[ \begin{aligned} V_{\text{total}} &=27\times\frac{4}{3}\pi r^3 \\ &=36\pi r^3 \end{aligned} \]
Let the radius of the new sphere be \(r^\prime\).
Volume of the new sphere:\[\frac{4}{3}\pi (r^\prime)^3\]
Since volume remains conserved during melting: \[ \begin{aligned} 36\pi r^3 &=\frac{4}{3}\pi (r^\prime)^3 \end{aligned} \]
Cancelling \(\pi\) from both sides: \[ \begin{aligned} 36r^3 &=\frac{4}{3}(r^\prime)^3 \end{aligned} \]
Multiplying both sides by \(3\): \[ \begin{aligned} 108r^3 &=4(r^\prime)^3 \end{aligned} \]
Dividing both sides by \(4\):
\[ \begin{aligned} (r^\prime)^3 &=27r^3 \end{aligned} \]
Taking cube root on both sides: \[ \begin{aligned} r^\prime &=\sqrt[3]{27r^3} \\ &=3r \end{aligned} \]
Hence,\[r^\prime=3r\]
(ii) Ratio of Surface Areas \(S:S^\prime\)
Surface area of the new sphere: \[ \begin{aligned} S^\prime &=4\pi (r^\prime)^2 \\ &=4\pi (3r)^2 \\ &=4\pi\times9r^2 \\ &=36\pi r^2 \end{aligned} \]
Original surface area:\[S=4\pi r^2\]
Therefore, \[ \begin{aligned} S:S^\prime &=4\pi r^2:36\pi r^2 \\ &=1:9 \end{aligned} \]
Hence,\[S:S^\prime=1:9\]

🎯 Exam Significance Exam Significance ›

This question develops understanding of conservation of volume during melting and recasting.
Students learn the relationship between radius, volume, and surface area.
Board examinations frequently ask recasting problems involving spheres and hemispheres.
Competitive examinations test algebraic manipulation and ratio concepts.
The problem strengthens conceptual understanding of scaling in three-dimensional geometry.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Volume remains conserved during melting and recasting.
Volume of a sphere is \[ V=\frac{4}{3}\pi r^3 \]
Surface area of a sphere is \[ S=4\pi r^2 \]
If radius changes by a factor \(k\), volume changes by \(k^3\) and surface area changes by \(k^2\).
Step-by-step simplification helps avoid errors in ratio and recasting problems.

📘 Concept & Theory Concept Used ›

The capsule is spherical in shape.

The amount of medicine needed to fill the capsule is equal to the volume of the sphere.

Volume of a sphere:

\[ V=\frac{4}{3}\pi r^3 \]

where

\(V\) = Volume of the sphere
\(r\) = Radius of the sphere
\(\pi \approx \frac{22}{7}\)

🗺️ Solution Roadmap Step-by-step Plan ›

Find the radius from the given diameter.
Use the sphere volume formula.
Calculate \(r^3\) carefully.
Substitute the values into the formula.
Write the final answer in cubic millimetres.

📊 Graph / Figure Graph / Figure ›

Spherical Medicine Capsule

✏️ Solution Complete Solution ›

Step-by-step Solution · 6 steps

Given
Diameter of the capsule:\[d=3.5\text{ mm}\]
Step-by-step Solution
Radius is half of the diameter: \[ \begin{aligned} r &=\frac{d}{2} \\ &=\frac{3.5}{2} \\ &=1.75\text{ mm} \end{aligned} \]
Volume of a sphere:\[V=\frac{4}{3}\pi r^3\]
Substituting \(r=1.75\text{ mm}\):
\[ \begin{aligned} V &=\frac{4}{3}\times\frac{22}{7}\times(1.75)^3\\ &=\frac{4}{3}\times\frac{22}{7}\times5.359375\\ &=\frac{88}{21}\times5.359375\\ &=22.4583 \end{aligned} \]
Hence, the amount of medicine required to fill the capsule is approximately\[22.46\text{ mm}^3\]

🎯 Exam Significance Exam Significance ›

This question demonstrates the practical application of volume in medicine and pharmacy.
Students learn how geometric formulas are used in real-life situations.
Board examinations frequently include direct application problems based on spheres.
Competitive entrance examinations test numerical calculation and unit handling skills.
The problem strengthens understanding of three-dimensional geometry and mensuration.

🔑 Key Takeaways Key Takeaways ›

Key Takeaways · 5 points

Radius is always half of the diameter.
Volume of a sphere is calculated using \[ V=\frac{4}{3}\pi r^3 \]
Always calculate \(r^3\) carefully to avoid arithmetic mistakes.
Final answers for volume must be written in cubic units.
Step-by-step simplification improves accuracy in examinations.

SURFACE AREAS AND VOLUMES

(i) Radius \(=7\) cm

Chapter Complete!

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SURFACE AREAS AND VOLUMES

(i) Radius \(=7\) cm

Chapter Complete!

Recent posts

SURFACE AREAS AND VOLUMES — Learning Resources

Frequently Asked Questions

Q 01 What is Academia Aeternum?

Q 02 Which subjects are available on Academia Aeternum?

Q 03 Are the MCQs useful for competitive exams like JEE and NEET?

Q 04 Is the content based on NCERT textbooks?

Q 05 Is the website free to use?

Q 06 How do MCQs help students prepare for exams?

Q 07 Do the questions include explanations?

Q 08 Can teachers use Academia Aeternum resources in classrooms?

Q 09 Is the platform suitable for self study?

Q 10 Does the website help with quick revision before exams?

Q 11 Are explanations written in simple language?

Q 12 Are there different difficulty levels of questions?

Q 13 How often is new content added?

Q 14 Can regular MCQ practice improve exam performance?

Q 15 Why should students practice objective questions regularly?

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