Samacheer Kalvi Class 10 Maths Chapter 7 Mensuration Exercise 7

Question 1. The barrel of a fountain-pen cylindrical in shape is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used for writing 330 words on an average. How many words can be written using a bottle of ink containing one-fifth of a litre?
Answer:
Length of the pen \( (h) = 7 \) cm
Radius of the pen \( (r) = \frac{5}{2} \) mm
\( \implies = \frac{5}{2 \times 10} \) cm \( = \frac{1}{4} \) cm
Volume of the barrel of a pen \( = \) Volume of the cylinder
\( = \pi r^2 h \) cu. units
\( = \frac{22}{7} \times \frac{1}{4} \times \frac{1}{4} \times 7 \) cm\(^3 \)
\( \implies = \frac{11}{8} \) cm\(^3 \)
To convert to liters, we use the conversion \( 1 \) liter \( = 1000 \) cm\(^3 \), so \( 1 \) cm\(^3 = \frac{1}{1000} \) liter:
\( = \frac{11}{8} \times \frac{1}{1000} \) lit
\( \implies = \frac{11}{8000} \) lit
For \( \frac{11}{8000} \) lit, words written \( = 330 \)
Now, we want to find the number of words for \( \frac{1}{5} \) litre. We set up a proportion:
No. of words \( = 330 \times \frac{\frac{1}{5}}{\frac{11}{8000}} \)
\( \implies = 330 \times \frac{1}{5} \times \frac{8000}{11} \)
\( \implies = \frac{330}{11} \times \frac{8000}{5} \)
\( \implies = 30 \times 1600 \)
\( \implies = 48000 \)
Therefore, the number of words that can be written is 48000.
In simple words: First, find the volume of the pen's ink barrel. Then, convert this volume to liters. You know how many words the pen's ink writes. Now, calculate how many words can be written with one-fifth of a liter of ink from the bottle.

🎯 Exam Tip: Always ensure unit consistency; convert all measurements to the same unit (e.g., cm) before performing calculations to avoid errors.

Question 2. A hemispherical tank of radius 1.75 m is full of water. It is connected with a pipe which empties the tank at the rate of 7 litres per second. How much time will it take to empty the tank completely?
Answer:
Radius of the hemispherical tank \( = 1.75 \) m
Volume of the tank \( = \frac{2}{3} \pi r^3 \) cu. units
\( = \frac{2}{3} \times \frac{22}{7} \times 1.75 \times 1.75 \times 1.75 \)
To simplify calculations, convert \( 1.75 \) to a fraction: \( 1.75 = \frac{175}{100} = \frac{7}{4} \).
\( = \frac{2}{3} \times \frac{22}{7} \times \frac{7}{4} \times \frac{7}{4} \times \frac{7}{4} \)
\( = \frac{2 \times 22 \times 7 \times 7 \times 7}{3 \times 7 \times 4 \times 4 \times 4} \)
\( \implies = \frac{11 \times 49}{3 \times 4 \times 4} \) m\(^3 \)
\( \implies = \frac{539}{48} \) m\(^3 \)
We know that \( 1 \) m\(^3 = 1000 \) liters. So, convert the volume to liters:
\( = \frac{539}{48} \times 1000 \) lit
\( \implies = 11.229166 \times 1000 = 11229.17 \) lit (approximately)
The pipe empties the tank at the rate of 7 liters per second.
Time taken \( = \frac{\text{Total volume in liters}}{\text{Rate of emptying}} \)
\( = \frac{11229.17}{7} = 1604.17 \) seconds
To convert seconds to minutes, divide by 60:
\( = \frac{1604.17}{60} \approx 26.736 \) minutes
Rounding to the nearest whole minute, the time taken is approximately 27 minutes.
In simple words: First, find the total water volume in the tank using the hemisphere formula. Change this volume from cubic meters to liters. Divide the total liters by the rate at which water leaves (liters per second). This gives you the time in seconds. Finally, convert seconds to minutes.

🎯 Exam Tip: Remember the conversion factor \( 1 \) m\(^3 = 1000 \) liters, which is essential for solving problems involving liquid capacities.

Question 3. Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius r units.
Answer:
To obtain the maximum volume, the cone carved out of a solid hemisphere must have its base radius equal to the hemisphere's radius and its height also equal to the hemisphere's radius.
Radius of a cone \( = \) Radius of a hemisphere \( = r \) unit
Height of a cone \( = r \) units (since the maximum height will be the radius of the hemisphere)
Maximum volume of the cone \( = \frac{1}{3} \pi r^2 h \) cu. units
Substitute \( h = r \) into the formula:
\( = \frac{1}{3} \pi r^2 \times r \) cu. units
\( \implies = \frac{1}{3} \pi r^3 \) cu. units
In simple words: If you carve the biggest possible cone from a hemisphere, the cone's base radius and its height will both be the same as the hemisphere's radius. Use the cone volume formula, replacing height with radius, to find this maximum volume.

🎯 Exam Tip: For optimization problems like this, visualize the geometric relationship between the two shapes to correctly identify parameters such as height and radius.

Question 4. An oil funnel of the tin sheet consists of a cylindrical portion 10 cm long attached to a frustum of a cone. If the total height is 22 cm, the diameter of the cylindrical portion by 8 cm and the diameter of the top of the funnel be 18 cm, then find the area of the tin sheet required to make the funnel.
Answer:
Total height of oil funnel \( = 22 \) cm
Height of the cylindrical portion \( (H_{\text{cyl}}) = 10 \) cm
Height of the frustum \( (h_{\text{frustum}}) = \text{Total height} - H_{\text{cyl}} = 22 - 10 = 12 \) cm
Diameter of the cylindrical portion \( = 8 \) cm
Radius of the cylindrical portion \( (r_{\text{cyl}}) = \frac{8}{2} = 4 \) cm
This radius is also the bottom radius of the frustum, so \( r_{\text{frustum,bottom}} = 4 \) cm.
Diameter of the top of the funnel \( = 18 \) cm
Top radius of the funnel (frustum) \( (R_{\text{frustum,top}}) = \frac{18}{2} = 9 \) cm

First, calculate the slant height \( (l) \) of the frustum:
\( l = \sqrt{h_{\text{frustum}}^2 + (R_{\text{frustum,top}} - r_{\text{frustum,bottom}})^2} \)
\( l = \sqrt{12^2 + (9 - 4)^2} \)
\( l = \sqrt{144 + 5^2} \)
\( l = \sqrt{144 + 25} \)
\( l = \sqrt{169} = 13 \) cm

The area of the tin sheet required is the sum of the Curved Surface Area (C.S.A) of the frustum and the C.S.A of the cylinder.
Area \( = \text{C.S.A of frustum} + \text{C.S.A of cylinder} \)
\( \text{C.S.A of frustum} = \pi (R_{\text{frustum,top}} + r_{\text{frustum,bottom}}) l \)
\( \text{C.S.A of cylinder} = 2 \pi r_{\text{cyl}} H_{\text{cyl}} \)
Total Area \( = \pi (9 + 4) \times 13 + 2 \pi \times 4 \times 10 \)
\( = \pi (13 \times 13) + \pi (8 \times 10) \)
\( = \pi (169) + \pi (80) \)
\( = \pi (169 + 80) \)
\( = \pi (249) \)
\( = \frac{22}{7} \times 249 \)
\( = 782.57 \) cm\(^2 \) (approximately)
The total area of tin sheet required to make the funnel is 782.57 cm\(^2 \).
In simple words: Break the funnel into a cylindrical part and a cone frustum. Find the height and radii for each section. Calculate the frustum's slant height. Then, find the curved surface area of both the cylinder and the frustum. Add these two areas together to get the total tin sheet needed.

🎯 Exam Tip: Always draw a simple diagram to visualize the components (cylinder and frustum) and their respective dimensions, which helps in applying the correct formulas.

Question 5. Find the number of coins, 1.5 cm in diameter and 2 mm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
Answer:
For the large cylinder:
Diameter \( = 4.5 \) cm, so Radius \( (R) = \frac{4.5}{2} \) cm
Height \( (H) = 10 \) cm
Volume of the cylinder \( V_{\text{cyl}} = \pi R^2 H \)
\( = \pi \times \left(\frac{4.5}{2}\right)^2 \times 10 \)
\( = \pi \times \frac{4.5 \times 4.5}{4} \times 10 \) cm\(^3 \)
To make calculations easier, convert decimals to fractions: \( 4.5 = \frac{9}{2} \).
\( V_{\text{cyl}} = \pi \times \left(\frac{9}{4}\right)^2 \times 10 = \pi \times \frac{81}{16} \times 10 \) cm\(^3 \)

For one coin (which is also a small cylinder):
Diameter \( = 1.5 \) cm, so Radius \( (r) = \frac{1.5}{2} \) cm
Thickness \( (h) = 2 \) mm \( = \frac{2}{10} \) cm \( = \frac{1}{5} \) cm
Volume of one coin \( V_{\text{coin}} = \pi r^2 h \)
\( = \pi \times \left(\frac{1.5}{2}\right)^2 \times \frac{1}{5} \)
\( = \pi \times \frac{1.5 \times 1.5}{4} \times \frac{1}{5} \) cm\(^3 \)
Again, convert decimals to fractions: \( 1.5 = \frac{3}{2} \).
\( V_{\text{coin}} = \pi \times \left(\frac{3}{4}\right)^2 \times \frac{1}{5} = \pi \times \frac{9}{16} \times \frac{1}{5} \) cm\(^3 \)

Number of coins \( = \frac{\text{Volume of the cylinder}}{\text{Volume of one coin}} \)
\( = \frac{\pi \times \frac{81}{16} \times 10}{\pi \times \frac{9}{16} \times \frac{1}{5}} \)
Cancel out \( \pi \) and \( \frac{1}{16} \):
\( = \frac{81 \times 10}{9 \times \frac{1}{5}} \)
\( = \frac{81 \times 10 \times 5}{9} \)
\( = 9 \times 10 \times 5 \)
\( = 450 \)
Therefore, 450 coins are needed.
In simple words: First, calculate the volume of the large cylinder that is formed. Then, calculate the volume of a single coin. Make sure all units are consistent (e.g., centimeters) before starting. The number of coins is found by dividing the total volume of the large cylinder by the volume of one coin.

🎯 Exam Tip: Remember that when materials are melted and recast, the total volume remains constant, which is the key principle to solve such problems.

Question 6. A hollow metallic cylinder whose external radius is 4.3 cm and internal radius is 1.1 cm and the whole length is 4 cm is melted and recast into a solid cylinder of 12 cm long. Find the diameter of a solid cylinder.
Answer:
For the hollow metallic cylinder:
External radius \( (R) = 4.3 \) cm
Internal radius \( (r) = 1.1 \) cm
Length \( (h) = 4 \) cm
Volume of hollow cylinder \( = \pi h (R^2 - r^2) \)
\( = \pi \times 4 \times (4.3^2 - 1.1^2) \)
\( = \pi \times 4 \times (4.3 - 1.1)(4.3 + 1.1) \) (using the identity \( a^2 - b^2 = (a-b)(a+b) \))
\( = \pi \times 4 \times (3.2)(5.4) \) cm\(^3 \)

For the solid cylinder:
Length \( (H) = 12 \) cm
Let the radius of the solid cylinder be \( x \).
Volume of solid cylinder \( = \pi x^2 H \)
\( = \pi x^2 \times 12 \) cm\(^3 \)

Since the hollow cylinder is melted and recast into a solid cylinder, their volumes are equal:
Volume of solid cylinder \( = \) Volume of hollow cylinder
\( \pi x^2 \times 12 = \pi \times 4 \times 3.2 \times 5.4 \)
Divide both sides by \( \pi \):
\( 12 x^2 = 4 \times 3.2 \times 5.4 \)
\( x^2 = \frac{4 \times 3.2 \times 5.4}{12} \)
\( x^2 = \frac{4 \times 32 \times 54}{12 \times 10 \times 10} \)
\( x^2 = \frac{1 \times 32 \times 54}{3 \times 100} \)
\( x^2 = \frac{1728}{300} = \frac{576}{100} \)
\( x = \sqrt{\frac{576}{100}} \)
\( x = \frac{24}{10} = 2.4 \) cm
The radius of the solid cylinder is \( 2.4 \) cm.
Diameter of the solid cylinder \( = 2 \times x = 2 \times 2.4 = 4.8 \) cm.
In simple words: When a hollow cylinder is melted and reshaped into a solid cylinder, the total amount of material (volume) stays the same. First, calculate the volume of the hollow cylinder using its external and internal radii and height. Then, set this volume equal to the volume of the new solid cylinder and solve for its radius. Finally, double the radius to find the diameter.

🎯 Exam Tip: Remember to use the difference of squares formula \( (a^2 - b^2) = (a-b)(a+b) \) to simplify calculations when finding the volume of hollow objects.

Question 7. The slant height of a frustum of a cone is 4 m and the perimeter of circular ends are 18 m and 16 m. Find the cost of painting its curved surface area at Rs. 100 per sq. m.
Answer:
Slant height of a frustum \( (l) = 4 \) m

For the top circular end:
Perimeter \( = 18 \) m
\( 2 \pi R = 18 \)
\( R = \frac{18}{2 \pi} = \frac{9}{\pi} \) m
If we use \( \pi = \frac{22}{7} \), then \( R = \frac{9}{22/7} = \frac{9 \times 7}{22} = \frac{63}{22} \) m

For the bottom circular end:
Perimeter \( = 16 \) m
\( 2 \pi r = 16 \)
\( r = \frac{16}{2 \pi} = \frac{8}{\pi} \) m
If we use \( \pi = \frac{22}{7} \), then \( r = \frac{8}{22/7} = \frac{8 \times 7}{22} = \frac{56}{22} \) m

Curved Surface Area (C.S.A) of the frustum \( = \pi l (R + r) \)
\( = \pi \times 4 \times \left(\frac{9}{\pi} + \frac{8}{\pi}\right) \)
\( = \pi \times 4 \times \frac{17}{\pi} \)
\( = 4 \times 17 \)
\( = 68 \) sq. m.
Cost of painting \( = \text{C.S.A} \times \text{Rate per sq. m} \)
\( = 68 \times \text{Rs. } 100 \)
\( = \text{Rs. } 6800 \)
The total cost of painting the curved surface area is Rs. 6800.
In simple words: First, use the given perimeters to find the radii of the top and bottom circles of the frustum. Then, use these radii along with the slant height to calculate the curved surface area of the frustum. Finally, multiply this area by the painting cost per square meter to get the total cost.

🎯 Exam Tip: Notice that \( \pi \) often cancels out when you calculate radii from perimeters and then use them in area formulas for frustums, simplifying the calculation.

Question 8. A hemispherical hollow bowl has material of volume cubic \( \frac{436 \pi}{3} \) cubic cm. Its external diameter is 14 cm. Find its thickness.
Answer:
External diameter of the hollow hemispherical bowl \( = 14 \) cm
External radius \( (R) = \frac{14}{2} = 7 \) cm
Volume of material in a hollow hemisphere \( = \frac{2}{3} \pi (R^3 - r^3) \), where \( r \) is the internal radius.
Given volume of material \( = \frac{436 \pi}{3} \) cm\(^3 \)

Set the volume formula equal to the given volume:
\( \frac{2}{3} \pi (R^3 - r^3) = \frac{436 \pi}{3} \)
Multiply both sides by \( \frac{3}{\pi} \):
\( 2 (R^3 - r^3) = 436 \)
Divide by 2:
\( R^3 - r^3 = \frac{436}{2} \)
\( R^3 - r^3 = 218 \)
Substitute the value of \( R = 7 \) cm:
\( 7^3 - r^3 = 218 \)
\( 343 - r^3 = 218 \)
\( r^3 = 343 - 218 \)
\( r^3 = 125 \)
\( r = \sqrt[3]{125} \)
\( r = 5 \) cm
The internal radius of the bowl is 5 cm.
Thickness of the hemisphere \( = \text{External radius} - \text{Internal radius} \)
\( = R - r = 7 - 5 = 2 \) cm.
In simple words: First, find the external radius of the bowl. Then, use the given volume of material and the formula for a hollow hemisphere to find the internal radius. Subtract the internal radius from the external radius to get the thickness of the bowl.

🎯 Exam Tip: Clearly distinguish between external and internal radii, and remember the correct formula for the volume of a hollow sphere or hemisphere.

Question 9. The volume of a cone is \( 1005\frac{5}{7} \) cu. cm. The area of its base is \( 201\frac{1}{7} \) sq. cm. Find the slant height of the cone.
Answer:
Given:
Volume of a cone \( V = 1005\frac{5}{7} = \frac{1005 \times 7 + 5}{7} = \frac{7035 + 5}{7} = \frac{7040}{7} \) cu. cm
Area of the base of a cone \( A = 201\frac{1}{7} = \frac{201 \times 7 + 1}{7} = \frac{1407 + 1}{7} = \frac{1408}{7} \) sq. cm

We know the formula for the volume of a cone: \( V = \frac{1}{3} \times \text{Base Area} \times h \)
Substitute the given values:
\( \frac{7040}{7} = \frac{1}{3} \times \frac{1408}{7} \times h \)
To find \( h \), multiply both sides by \( \frac{3 \times 7}{1408} \):
\( h = \frac{7040}{7} \times \frac{3 \times 7}{1408} \)
\( h = \frac{7040 \times 3}{1408} \)
\( h = 5 \times 3 \)
\( h = 15 \) cm
So, the height of the cone is 15 cm.

Now, we use the base area to find the radius \( (r) \):
Area of base \( A = \pi r^2 \)
\( \frac{1408}{7} = \frac{22}{7} r^2 \)
\( r^2 = \frac{1408}{7} \times \frac{7}{22} \)
\( r^2 = \frac{1408}{22} \)
\( r^2 = 64 \)
\( r = \sqrt{64} \)
\( r = 8 \) cm
So, the radius of the cone is 8 cm.

Finally, we find the slant height \( (l) \) using the Pythagorean theorem:
\( l = \sqrt{h^2 + r^2} \)
\( l = \sqrt{15^2 + 8^2} \)
\( l = \sqrt{225 + 64} \)
\( l = \sqrt{289} \)
\( l = 17 \) cm
The slant height of the cone is 17 cm.
In simple words: First, use the cone's volume and base area to calculate its height. Then, use the base area to find the radius of the cone. Finally, use the calculated height and radius to find the slant height by applying the Pythagorean theorem.

🎯 Exam Tip: Remember that the volume formula for a cone relates height, radius, and base area directly, making it easy to find missing dimensions if two are known.

Question 10. A metallic sheet in the form of a sector of a circle of radius 21 cm has a central angle of 216°. The sector is made into a cone by bringing the bounding radii together. Find the volume of the cone formed.
Answer:
When the sector of a circle is formed into a cone:
The radius of the sector \( = 21 \) cm. This becomes the slant height \( (l) \) of the cone, so \( l = 21 \) cm.
The arc length of the sector becomes the circumference of the base of the cone.
Let \( R \) be the radius of the cone.
Circumference of cone base \( = 2 \pi R \)
Arc length of sector \( = \frac{\theta}{360^\circ} \times 2 \pi l \)
Equating them:
\( 2 \pi R = \frac{216^\circ}{360^\circ} \times 2 \pi \times 21 \)
Divide both sides by \( 2 \pi \):
\( R = \frac{216}{360} \times 21 \)
Simplify the fraction \( \frac{216}{360} \) by dividing both by 72: \( \frac{3}{5} \).
\( R = \frac{3}{5} \times 21 \)
\( R = \frac{63}{5} = 12.6 \) cm
The radius of the cone is 12.6 cm.

Next, calculate the height \( (h) \) of the cone using the slant height \( (l) \) and radius \( (R) \):
\( h = \sqrt{l^2 - R^2} \) (from Pythagorean theorem)
\( h = \sqrt{21^2 - (12.6)^2} \)
\( h = \sqrt{441 - 158.76} \)
\( h = \sqrt{282.24} \)
\( h = 16.8 \) cm
The height of the cone is 16.8 cm.

Finally, calculate the volume of the cone:
Volume of cone \( V = \frac{1}{3} \pi R^2 h \)
\( V = \frac{1}{3} \times \frac{22}{7} \times (12.6)^2 \times 16.8 \)
\( V = \frac{1}{3} \times \frac{22}{7} \times 12.6 \times 12.6 \times 16.8 \)
\( V = \frac{22}{21} \times 12.6 \times 12.6 \times 16.8 \)
\( V = 2794.176 \) cm\(^3 \)
Rounding to two decimal places, the volume of the cone formed is 2794.18 cm\(^3 \).
In simple words: When a circular sector is folded into a cone, the sector's radius becomes the cone's slant height, and the sector's arc length forms the cone's base circumference. Use these facts to find the cone's base radius. Then, find the cone's height using the slant height and base radius. Finally, calculate the cone's volume.

🎯 Exam Tip: Remember the critical relationships: sector radius becomes cone slant height, and sector arc length becomes cone base circumference. These are key for transforming 2D sector problems into 3D cone problems.