Surface Areas and Volumes - Class 10 Mathematics - Chapter 12 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Surface Areas and Volumes | NCERT | Mathematics | Class 10
Two identical cubes of side $4 \mathrm{~cm}$ are placed one above the other, forming a cuboid. Find the total surface area of the cuboid thus formed.
(A) $96 \mathrm{~cm}^{2}$
(B) $100 \mathrm{~cm}^{2}$
(C) $160 \mathrm{~cm}^{2}$
(D) $180 \mathrm{~cm}^{2}$
The correct answer is (C) $160 \mathrm{~cm}^{2}$.
When two identical cubes, each with a side length of $4 \mathrm{~cm}$, are stacked one on top of the other, a cuboid is formed with dimensions:
Length, $l = 4 \mathrm{~cm}$
Breadth, $b = 4 \mathrm{~cm}$
Height, $h = 4 + 4 = 8 \mathrm{~cm}$
The Total Surface Area (TSA) of a cuboid is calculated with the formula: $$ TSA = 2(lb + bh + hl) $$ Plugging in the dimensions of the formed cuboid: $$ TSA = 2(4 \times 4 + 4 \times 8 + 8 \times 4) $$ $$ TSA = 2(16 + 32 + 32) $$ $$ TSA = 2 \times 80 $$ $$ TSA = 160 \mathrm{~cm}^{2} $$
Thus, the total surface area of the constructed cuboid is $160 \mathrm{~cm}^{2}$.
The height of a right circular cylinder is $10.5 \mathrm{~m}$. 3 times the sum of the areas of its 2 circular faces is twice the area of the curved surface. Find the volume of the cylinder.
A) $1661 \mathrm{~m}^{3}$ B) $1617 \mathrm{~m}^{3}$ C) $1579 \mathrm{~m}^{3}$ D) $1544 \mathrm{~m}^{3}$ E) $1602 \mathrm{~m}^{3}$ F) $1543 \mathrm{~m}^{3}$
The correct option is $\mathbf{B}$ $1617 \mathrm{~m}^{3}$.
Given the height of the cylinder $H = 10.5 , \text{m}$ and the condition that three times the sum of the areas of its two circular faces equals twice the area of the curved surface, we can set up the equation:
$$ 3 \times 2 \times \pi r^2 = 2 \times 2\pi r H $$
Simplifying this equation:
$$ 6 \pi r^2 = 4 \pi r H \ \frac{3r}{2} = H $$
Substitute $H = 10.5 , \text{m}$ into the equation:
$$ \frac{3r}{2} = 10.5 \ 3r = 21 \ r = 7 , \text{m} $$
Now, to find the volume of the cylinder:
$$ \text{Volume} = \pi r^2 H \ = \frac{22}{7} \times 7^2 \times 10.5 \ = 22 \times 49 \times 1.5 \ = 1617 , \text{m}^3 $$
Thus, the volume of the cylinder is $1617 , \text{m}^3$.
Sushant has a vessel in the form of an inverted cone, open at the top, with a height of $11$ cm and a radius of the top as $2.5$ cm, and is full of water. Metallic spherical balls, each with a diameter of $0.5$ cm, are put in the vessel due to which $\frac{2}{5}$ of the water in the vessel flows out. Find how many balls were put in the vessel. Sushant made the arrangement so that the water that flows out irrigates the flower beds. What value has been shown by Sushant?
Given Parameters:
Height ($h$) of the conical vessel: $11 , \text{cm}$
Radius ($r_1$) of the conical vessel: $2.5 , \text{cm}$
Radius ($r_2$) of the metallic spherical balls: $\frac{0.5 , \text{cm}}{2} = 0.25 , \text{cm}$
Let $n$ be the number of spherical balls that were dropped in the vessel.
Key Relationships and Formulae:
The volume of water displaced is equal to the volume of the spherical balls that were added into the vessel, given by $\frac{2}{5}$ times the volume of the cone which equates to $n$ times the volume of one spherical ball.
Calculation
Using the volume formulas for a cone and a sphere:
Volume of the cone: $$ V_{cone} = \frac{1}{3} \pi r_1^2 h $$
Volume of one spherical ball: $$ V_{ball} = \frac{4}{3} \pi r_2^3 $$
Setting up the equation: $$ \frac{2}{5} \times \frac{1}{3} \pi (2.5)^2 \times 11 = n \times \frac{4}{3} \pi (0.25)^3 $$ Expanding and simplifying the equation: $$ \frac{2}{5} \times \frac{1}{3} \pi \times 6.25 \times 11 = n \times \frac{4}{3} \pi \times 0.015625 $$ $$ \frac{2}{5} \times \frac{1}{3} \cdot 68.75 = n \times \frac{4}{3} \times 0.015625 $$ $$ 5.583333 = n \times 0.0208333 $$ Solving for $n$: $$ n = \frac{5.583333}{0.0208333} \approx 440 $$
Conclusion
The number of spherical balls that were dropped in the vessel is 440. Sushant made the arrangement so that the water which flows out irrigates the flower beds. This exemplifies judicious usage of water and environmental consciousness.
4 The radius of a spherical balloon increases from $7 \mathrm{~cm}$ to $14 \mathrm{~cm}$ as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Solution:
We begin by noting the initial and final radii of the spherical balloon. Let:
- Initial radius, $r_1 = 7 , \text{cm}$
- Final radius after air is pumped in, $r_2 = 14 , \text{cm}$
To find the ratio of the surface areas, we use the formula for the surface area of a sphere, which is given by: $$ S = 4\pi r^2 $$ Therefore, the surface areas of the balloon initially and finally are $4\pi r_1^2$ and $4\pi r_2^2$ respectively.
The ratio of the surface areas is then calculated as: $$ \text{Ratio} = \frac{4\pi r_1^2}{4\pi r_2^2} = \left(\frac{r_1}{r_2}\right)^2 $$ Plugging in the values, we have: $$ \left(\frac{7}{14}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} $$
Thus, the required ratio of the surface areas of the balloon in the two cases is $1:4$.
Dimensions of cylindrical overhead tank are: radius = 6 m and height = 7 m. If water is filled in at the rate of 20 liters per minute, the tank will be completely filled in how much time?
To calculate the time it will take for a cylindrical tank with a radius of 6 m and a height of 7 m to be completely filled at the rate of 20 liters per minute, follow this approach:
Volume Calculation
First, calculate the volume of the cylindrical tank using the formula: $$ V = \pi r^2 h $$ Where:
$ V $ is the volume
$ r $ is the radius of the base of the cylinder
$ h $ is the height of the cylinder
$ \pi $ approximately equals $ \frac{22}{7} $ (for calculation purpose)
Substitute the given values:
$ r = 6 $ meters
$ h = 7 $ meters
Calculating: $$ V = \frac{22}{7} \times 6^2 \times 7 $$ $$ V = \frac{22}{7} \times 36 \times 7 $$ $$ V = \frac{22}{7} \times 252 \text{ cubic meters} $$ $$ V = 792 \text{ cubic meters} $$
Conversion to Liters
Since 1 cubic meter equals 1000 liters: $$ 792 \text{ cubic meters} = 792,000 \text{ liters} $$
Calculating Time
Now, the tank is being filled at a rate of 20 liters per minute.
To find how long it takes to fill the tank completely: $$ \text{Time} = \frac{\text{Total volume in liters}}{\text{Rate of filling}} $$ $$ \text{Time} = \frac{792,000 \text{ liters}}{20 \text{ liters/minute}} $$ $$ \text{Time} = 39,600 \text{ minutes} $$
To convert this into hours: $$ \text{Time in hours} = \frac{39,600 \text{ minutes}}{60 \text{ minutes/hour}} $$ $$ \text{Time in hours} = 660 \text{ hours} $$
Conclusion
It will take 660 hours for the cylindrical tank to be completely filled at the rate of 20 liters per minute.
Dimensions of cylindrical overhead tank are: radius $=6$ m and height $=7$ m. If it is to be painted to save it from corrosion, how much area needs to be painted?
To find the area that needs to be painted on a cylindrical overhead tank with a radius of $6$ meters and a height of $7$ meters, we need to calculate the total surface area of the cylinder. This includes both the curved surface area and the area of the two circular bases (top and bottom).
Calculate the Area of the Circular Bases:
The area of one circle (either top or bottom) is given by the formula: $$ \text{Area of one circle} = \pi r^2, $$ where ( r ) is the radius. For our cylinder: $$ \text{Area of one circle} = \pi \times 6^2 = 36\pi , \text{square meters}. $$
Since there are two circles (top and bottom), the total area for both circles is: $$ \text{Total area for two circles} = 2 \times 36\pi = 72\pi , \text{square meters}. $$
Calculate the Curved Surface Area:
The curved surface of a cylinder can be thought of as a rectangle if it was "unrolled" or laid out flat. The length of this rectangle would be the circumference of the circle (which is the distance around the circular base), and the width would be equal to the height of the cylinder. Thus, the formula for the curved surface area is: $$ \text{Curved Surface Area} = 2\pi r h, $$ where ( h ) is the height. For our cylinder: $$ \text{Curved Surface Area} = 2\pi \times 6 \times 7 = 84\pi , \text{square meters}. $$
Sum the Areas:
The total surface area that needs to be painted is the sum of the area of two circles and the curved surface area: $$ \text{Total Surface Area} = 72\pi + 84\pi = 156\pi , \text{square meters}. $$
Converting the total surface area from terms of ( \pi ) into numerical values, using ( \pi \approx 3.14159 ): $$ \text{Total Surface Area} \approx 156 \times 3.14159 = 490.068 , \text{square meters}. $$
So, the total area that needs to be painted on the cylindrical tank is approximately 490 square meters.
Dimensions of cylindrical overhead tank are: radius $= 6 \mathrm{~m}$ and height $= 7 \mathrm{~m}$. The capacity (in litres) of the overhead tank is:
To calculate the capacity of a cylindrical overhead tank using its given dimensions (radius $= 6 , \text{m}$ and height $= 7 , \text{m}$), we follow these steps:
Calculate the Volume: The formula for the volume $V$ of a cylinder is given by: $$ V = \pi r^2 h $$ where $r$ is the radius and $h$ is the height of the cylinder. Substituting the given values: $$ V = \pi \times (6 , \text{m})^2 \times 7 , \text{m} $$ Here, $\pi$ can be approximated as $\frac{22}{7}$ for calculations: $$ V = \frac{22}{7} \times 6^2 \times 7 $$ Simplifying further: $$ V = \frac{22}{7} \times 36 \times 7 = 22 \times 36 = 792 , \text{m}^3 $$
Convert to Liters: Knowing that $1 , \text{m}^3 = 1000 , \text{liters}$, we convert the volume from cubic meters to liters: $$ \text{Capacity} = 792 , \text{m}^3 \times 1000 , \text{liters/m}^3 = 792,000 , \text{liters} $$
Therefore, the capacity of the overhead tank is 792,000 liters.
Selvis house has an overhead tank in the shape of a cylinder. This is filled by pumping water from a sump (an underground tank) which is in the shape of a cuboid. The sump has dimensions $1.57 \mathrm{~m} \times 1.44 \mathrm{~m} \times 95 \mathrm{~cm}$. The overhead tank has its radius $60 \mathrm{~cm}$ and height $95 \mathrm{~cm}$. Find the height of the water left in the underground tank after the overhead tank has been completely filled with water from the underground tank which had been full. Compare the capacity of both tanks. (Take $\pi = 22/7$)
A. $1:2$
B. $2:1$
C. $1:4$
D. $4:1$
Selvi's house features an overhead tank in the shape of a cylinder and a sump, an underground tank in the shape of a cuboid. We must determine the height of the water left in the sump after filling the cylindrical overhead tank from the full sump, and compare the capacity (volume) of both tanks.
Dimensions Given
Sump (Cuboid): Length = $1.57 , \text{m}$, Breadth = $1.44 , \text{m}$, Height = $95 , \text{cm}$ (which is $0.95 , \text{m}$).
Overhead Tank (Cylinder): Radius = $60 , \text{cm}$ (which is $0.6 , \text{m}$), Height = $95 , \text{cm}$ (which is $0.95 , \text{m}$).
Calculating Volumes
Volume of the Sump:$$ \text{Volume} = \text{Length} \times \text{Breadth} \times \text{Height} = 1.57 \times 1.44 \times 0.95 , \text{m}^3 $$
Volume of the Overhead Tank:Using the formula for the volume of a cylinder, $V = \pi r^2 h$, we substitute $\pi = \frac{22}{7}$ for better approximation: $$ \text{Volume} = \pi \times (0.6)^2 \times 0.95 = \frac{22}{7} \times 0.36 \times 0.95 , \text{m}^3 $$
After substituting and simplifying, the ratio of the volumes (Overhead Tank : Sump) turns out to be $\frac{1}{2}$ implying that the overhead tank's volume is half of the sump's volume.
Height of Water Left in Sump
Knowing that the overhead tank, which occupied half of the sump’s volume, is now completely filled, the remaining volume in the sump is also half of its initial total volume. Given the base area of the sump hasn't changed, the height of the water will proportionally be half.
Original height = $0.95 , \text{m}$ (fully filled), thus after transferring half: $$ \text{Remaining Height} = \frac{0.95}{2} , \text{m} = 0.475 , \text{m} , (\text{or } 47.5 , \text{cm}) $$
Answer: The height of the water left in the sump is 47.5 cm.
The capacity comparison between the overhead tank and the sump stands at a ratio of 1:2. Given answer options, the correct choice reflecting this comparison is:
A. $1:2$
Thus, upon completely filling the overhead tank, half of the water in the full sump remains, effectively leaving a height of $47.5$ cm of water in the sump.
If a metallic cube of edge $1$ cm is drawn into a wire of diameter $3.5$ mm, then find the length of the wire.
To find the length of the wire when a metallic cube with edge 1 cm is drawn into a wire of diameter 3.5 mm, we use the principle of volume conservation. The volume of the cube equates to the volume of the cylindrical wire formed.
Calculate the volume of the cube: Since the cube has an edge of $1$ cm, its volume ($V_1$) will be: $$ V_1 = \text{side}^3 = 1^3 = 1 \text{ cm}^3 $$
Calculate the volume of the wire: First, convert the diameter of the wire into centimeters: $$ \text{Diameter} = 3.5 \text{ mm} = 0.35 \text{ cm} $$ The radius ($r$) of the wire is half of this: $$ r = \frac{0.35}{2} = 0.175 \text{ cm} $$ The formula for the volume of a cylinder ($V_2$) is: $$ V_2 = \pi r^2 h $$ where $h$ is the length of the wire we need to find.
Equating volumes: Since $V_1 = V_2$: $$ 1 = \pi (0.175)^2 h $$ Simplify the radius squared: $$ (0.175)^2 = 0.030625 $$ Thus, the equation for $h$ is: $$ 1 = \pi \times 0.030625 \times h $$ $$ h = \frac{1}{\pi \times 0.030625} $$ Using $\pi \approx 3.14159$, we find: $$ h \approx \frac{1}{3.14159 \times 0.030625} \approx 10.36 \text{ meters} $$
Thus, the length of the wire is approximately 10.36 meters.
The perimeter of a rectangle is $82$ m and its length is $30$ m. Find the breadth of the rectangle.
To find the breadth of a rectangle when the perimeter is given as $82$ meters and the length as $30$ meters, we first need to recall the formula for the perimeter of a rectangle. The formula is: $$ P = 2(L + B) $$ where $P$ represents the perimeter, $L$ is the length, and $B$ is the breadth of the rectangle.
Given:
Perimeter, $P = 82$ m
Length, $L = 30$ m
We can substitute these values into the perimeter formula: $$ 82 = 2(30 + B) $$
To solve for $B$, we first divide both sides by $2$: $$ 41 = 30 + B $$
Next, isolate $B$ by subtracting $30$ from both sides: $$ B = 41 - 30 $$ $$ B = 11 \text{ meters} $$
Thus, the breadth of the rectangle is $11$ meters.
A heap of rice is in the form of a cone of base radius $4 \mathrm{~m}$ and height $3 \mathrm{~m}$. How much canvas cloth is required to cover the heap?
To determine how much canvas cloth is needed to cover the heap of rice modeled as a cone, we focus on computing the surface area of the cone. The key dimensions provided are:
Base radius ($r$) = $4 , \mathrm{m}$
Height ($h$) = $3 , \mathrm{m}$.
First, let's calculate the slant height ($l$) of the cone, since the cloth will cover the lateral surface area, not the base. The slant height forms the hypotenuse of a right triangle where the height of the cone and the radius of the base are the other two sides.
Using the Pythagorean theorem: $$ l^2 = r^2 + h^2 $$
Substitute the known values: $$ l^2 = 4^2 + 3^2 = 16 + 9 = 25 $$ $$ l = \sqrt{25} = 5 , \mathrm{m} $$
The slant height ($l$) is $5 , \mathrm{m}.
Next, calculate the lateral surface area of the cone (which is what the canvas will cover). The formula for the lateral surface area of a cone is: $$ A = \pi r l $$ Substitute the known values: $$ A = \pi \times 4 \times 5 = 20\pi , \mathrm{m}^2 $$
Thus, the area of the canvas cloth required to cover the heap of rice is $20\pi , \mathrm{m}^2.
A bucket is in the form of a frustum of a cone with a capacity of $12308.8 \mathrm{~cm}^{3}$ of water. The radii of the top and bottom circular ends are $20 \mathrm{~cm}$ and $12 \mathrm{~cm}$ respectively. Find the height of the bucket and the area of the metal sheet used in its making. [Use $\pi = 3.14$.]
To calculate the height and the area of the metal sheet used in making a bucket shaped as a frustum of a cone, we need to follow these steps:
Step 1: Calculate the Height of the Bucket
Given:
Volume of the bucket, $V = 12308.8 , \text{cm}^3$
Radius of the top, $R = 20 , \text{cm}$
Radius of the bottom, $r = 12 , \text{cm}$
$\pi = 3.14$
The formula for the volume of a frustum of a cone is: $$ V = \frac{1}{3} \pi h (R^2 + Rr + r^2) $$
Solving for the height $h$, we rearrange the formula: $$ h = \frac{3V}{\pi (R^2 + Rr + r^2)} $$
Inserting the value: $$ h = \frac{3 \times 12308.8}{3.14 \times (20^2 + 20 \times 12 + 12^2)} $$ $$ h = \frac{36926.4}{3.14 \times (400 + 240 + 144)} $$ $$ h = \frac{36926.4}{3.14 \times 784} $$ $$ h = \frac{36926.4}{2462.96} \approx 15 , \text{cm} $$
Step 2: Calculate the Area of the Metal Sheet
To find the area of the metal sheet used in forming the bucket, we need to determine the lateral surface area of the frustum plus the area of the circular base.
Lateral Surface Area of a frustum: $$ A = \pi \ell (r + R), $$ where $\ell$ is the slant height calculated by: $$ \ell = \sqrt{h^2 + (R - r)^2} $$
Inserting values: $$ \ell = \sqrt{15^2 + (20 - 12)^2} = \sqrt{225 + 64} = \sqrt{289} = 17 , \text{cm} $$
Now, calculate the lateral surface area: $$ A = 3.14 \times 17 \times (20 + 12) = 3.14 \times 17 \times 32 \approx 1709.28 , \text{cm}^2 $$
Circular Base Area: $$ A_{\text{base}} = \pi r^2 = 3.14 \times 12^2 \approx 452.16 , \text{cm}^2 $$
Adding both areas to find the total metal sheet area: $$ A_{\text{total}} = 1709.28 , \text{cm}^2 + 452.16 , \text{cm}^2 \approx 2161.44 , \text{cm}^2 $$
Conclusion
Height of the bucket: $15 , \text{cm}$
Area of the metal sheet: $2161.44 , \text{cm}^2$
This breakdown details how the dimensions and material requirements for the bucket can be derived using mathematical principles related to the frustum of a cone.
If the lengths and two diagonals of a rectangle are each increased by 6%, find % increase in its breadth.
(A) 5%
(B) 6%
(C) 7%
(D) 8%
The problem involves a rectangle whose lengths and diagonals are increased by 6%, and we need to determine the percentage increase in its breadth.
Start by considering a rectangle with length $L$ and breadth $B$. The formula for the diagonal $D$ of a rectangle is derived from the Pythagorean theorem:
$$ D = \sqrt{L^2 + B^2} $$
If each side length and diagonal are increased by 6%, the new length $L'$ and breadth $B'$ would be:
$$ L' = 1.06L, \quad B' = tB $$
We need to find $t$, which represents the factor increase in the breadth $B$. The new diagonal $D'$ will be:
$$ D' = 1.06D $$
Using the Pythagorean theorem for the modified dimensions, we get:
$$ D' = \sqrt{(L')^2 + (B')^2} = \sqrt{(1.06L)^2 + (tB)^2} $$
Since $D' = 1.06D$ and $D = \sqrt{L^2 + B^2}$, substituting these values gives:
$$ 1.06 \sqrt{L^2 + B^2} = \sqrt{1.06^2 L^2 + t^2 B^2} $$
Squaring both sides and simplifying:
$$ 1.1236 (L^2 + B^2) = 1.1236 L^2 + t^2 B^2 $$
We can then manipulate the equation to solve for $t$:
$$ t^2 B^2 = 1.1236 B^2 \quad \text{(after grouping similar terms)} $$
$$ t^2 = 1.1236 $$
$$ t = \sqrt{1.1236} $$
A calculation shows that $t = 1.06$, meaning the breadth increases by the same factor as the length and diagonals. Therefore, the percentage increase in the breadth is also 6%.
Thus, the correct answer is:
(B) 6%
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