Volume and Surface Area Class 10 Question PYQ

Given:

• Total length = 5 cm
• Diameter = 2.8 cm ⇒ Radius = 1.4 cm
• 45 gulabjamuns
• Syrup = 30% of volume

Step 1: Height of cylindrical part

$$h=5-2(1.4)=2.2\text{ cm}$$

Step 2: Volume of one gulabjamun

It consists of a cylinder and two hemispheres (= one sphere).

$$V=\pi r^2h+\frac{4}{3}\pi r^3$$

Using $r=1.4$r=1.4 and $h=2.2$h=2.2:

$$V=\frac{22}{7}(1.4)^2(2.2)+\frac{4}{3}\times\frac{22}{7}(1.4)^3$$

$$V=13.55+11.50=25.05\text{ cm}^3$$

Step 3: Syrup in 45 gulabjamuns

$$30\%\times45\times25.05=338.2\text{ cm}^3$$

Answer: 338 cm³ (approximately)

Given:

• Cylinder: Radius = 6 cm, Height = 8 cm
• Conical cavity: Radius = 6 cm, Height = 8 cm
• π = 3.14

Volume of Remaining Solid

Cylinder volume:

$$\pi r^2h=3.14\times6^2\times8=904.32$$

Cone volume:

$$\frac13\pi r^2h=\frac13\times3.14\times6^2\times8=301.44$$

Remaining volume:

$$904.32-301.44=602.88\text{ cm}^3$$

Total Surface Area

Slant height:

$$l=\sqrt{6^2+8^2}=10\text{ cm}$$

Surface Area:

Answer:

• Volume = 602.88 cm³
• Total Surface Area = 602.88 cm²

A solid metallic sphere of diameter 21 cm is melted and recast into smaller cones, each having diameter 3.5 cm and height 3 cm.

Given:

• Sphere radius = 10.5 cm
• Cone radius = 1.75 cm
• Cone height = 3 cm

Step 1: Volume of the sphere

$$V=\frac{4}{3}\pi r^3$$

$$=\frac{4}{3}\times\frac{22}{7}\times(10.5)^3$$

$$=4851\text{ cm}^3$$

Step 2: Volume of one cone

$$V=\frac{1}{3}\pi r^2h$$

$$=\frac{1}{3}\times\frac{22}{7}\times(1.75)^2\times3$$

$$=\frac{22}{7}\times\left(\frac{7}{4}\right)^2=\frac{77}{8}=9.625\text{ cm}^3$$

Step 3: Number of cones

$$\frac{4851}{9.625}=504$$

Final Answer

504 cones can be formed.

Formula

$$\text{Volume of sphere}=\frac{4}{3}\pi r^3$$

Since the metal is melted, the total volume remains the same.

$$\frac{4}{3}\pi(3)^3=\frac{4}{3}\pi(1.5)^3+\frac{4}{3}\pi(2)^3+\frac{4}{3}\pi r^3$$

Cancel the common factor $\frac{4}{3}\pi$ which is cancel out in both LHS and RHS. Then,

$$3^3=1.5^3+2^3+r^3$$

$$27=3.375+8+r^3$$

$$r^3=27-11.375=15.625$$

$$r=\sqrt[3]{15.625}=2.5\text{ cm}$$

Final Answer

The radius of the third spherical ball is 2.5 cm.

A spherical ball of diameter 21 cm is melted and recast into cubes of side 1 cm.

Step 1: Radius of the sphere

$r=\frac{21}{2}=10.5\text{ cm}$

Step 2: Volume of the sphere

$$V=\frac{4}{3}\pi r^3$$

$$V=\frac{4}{3}\times\frac{22}{7}\times(10.5)^3$$

$$V=4851\text{ cm}^3$$

Step 3: Volume of one cube

$$V=(1)^3=1\text{ cm}^3$$

Step 4: Number of cubes

$$\frac{4851}{1}=4851$$

Final Answer: 4851 cubes

A sphere of radius 8 cm is melted to make lead balls, each of radius 1 cm.

Formula

Volume of a sphere:

$$V=\frac{4}{3}\pi r^3$$

Since the metal is melted, the total volume remains the same.

$$\text{Number of balls}=\frac{\frac{4}{3}\pi(8)^3}{\frac{4}{3}\pi(1)^3}$$

Cancel the common terms:

$$=\frac{8^3}{1^3}=512$$

Final Answer

512 lead balls can be made.

Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm. The water level rises by 5.6 cm.

Given:

• Radius of marble = 0.7 cm
• Radius of beaker = 3.5 cm
• Rise in water level = 5.6 cm

Step 1: Volume of water displaced

The rise in water level equals the volume of the marbles.

$$V=\pi r^2h$$

$$=\frac{22}{7}\times(3.5)^2\times5.6$$

$$=215.6\text{ cm}^3$$

Step 2: Volume of one marble

$$V=\frac{4}{3}\pi r^3$$

$$=\frac{4}{3}\times\frac{22}{7}\times(0.7)^3$$

$$=\frac{2156}{1500}\approx1.437\text{ cm}^3$$

Step 3: Number of marbles

$$\frac{215.6}{1.437}=150$$

Final Answer

150 marbles should be dropped into the beaker.