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This is one of our many quizzes on Volume of solids. Click on this to access more such competitive quizzes on Mensuration.

Volume of solids Quiz 2

mathematics

 A wall is needed to be made with brick having dimensions $11cm\times 1100cm\times 1100\text{cm}$. Dimensions of the wall are $121m\times 1331cm\times 11m$. The cost of each brick is Rs. 20. Find the total number of bricks required and the total cost to make the wall.

 Wall and Bricks are considered to be cuboids so the formula of volume of the cuboid is used for both. Using unitary we can find out the number of blocks and total cost.

  $\text{Volume of brick }=lbh=\left( 11\times 1100\times 1100 \right)c{{m}^{3}} \\ \text{Volume of wall }=lbh=\left( 12100\times 1331\times 1100 \right)c{{m}^{3}} \\ \text{Number of bricks used = }\dfrac{\text{Volume of wall}}{\text{Volume of brick}}\text{=}\dfrac{\text{12100 }\!\!\times\!\!\text{ 1331 }\!\!\times\!\!\text{ 1100}}{\text{11 }\!\!\times\!\!\text{ 1100 }\!\!\times\!\!\text{ 1100}}\text{=1331} \\ \text{Total cost = }1331\times 20=Rs. 22620$

 A solid cone of slant height 5cm and radius 3cm is melted. The plasma is used to make the cubic block of the side 1 cm. Let the number of the cubic block which can be formed from the plasma be x. Use $\pi =3.14$. Take exact decimal $\text{(rounding off to two places)}$ for the number of a cubic block. Find the greatest integer function for the number of cubic blocks found i.e., $\left[ x \right]$.  

 Find the volume of cone and cubic block, then find the number. Taking the number of cubic blocks in decimal then apply the concept of GIF to get the final answer.

  $\text{Height of cone = }\sqrt{{{l}^{2}}-{{r}^{2}}}=\sqrt{{{5}^{2}}-{{3}^{2}}}=4cm \\ \text{Volume of cone = }\dfrac{1}{3}\times \pi {{r}^{2}}h=\dfrac{3.14\times {{3}^{2}}\times 4}{3}=37.68c{{m}^{3}} \\ \text{The volume of small cubic = }{{l}^{3}}=1c{{m}^{3}} \\ \text{Number of cubic blocks that can be formed = }\dfrac{37.68}{1}=37.68 \\ \text{So, } x=37.68 \\ \text{GIF }\left[ x \right]=\left[ 37.68 \right]=37$

 A cylindrical bucket of radius is 21cm and height is twice the radius was filled with water up to the surface. 6 spherical balls are dropped in the bucket. Each ball has a radius that equals  $\dfrac{1}{7}$of the height of the bucket. Find the value of  $\dfrac{Volume\text{ of water displaced}}{Volume\text{ of 1 spherical ball}}\times \dfrac{\left( Volume\text{ of remaining water+Volume of 6 spherical ball} \right)}{\text{Volume of Bucket}}$ 

 Assume components of required value to be some variables and try solving the equation to get the result. The information of dimensions is given to make it harder to solve. Dimensions are not necessarily needed in this question.

  $\text{Let the volume of water displaced be } “x”,\text{ the volume of 1 spherical ball be } “y”,\text{ the volume of the bucket be } “z”. \\ \text{The volume of water displaced = }6\times \text{Volume of 1 spherical ball} \\ \Rightarrow x=6y \\ \left( Volume\text{ of remaining water+Volume of 6 spherical ball} \right) \text{= Volume of bucket }-6\times\text{ Volume of 1 spherical ball }+6\times\text{ Volume of 1 spherical ball} \\ \Rightarrow \left( Volume\text{ of remaining water+Volume of 6 spherical ball} \right) = z-6y+6y=z \\ \text{Value = }\dfrac{x}{y}\times \dfrac{z}{z}=\dfrac{6y}{y}=6$

 An atom of radius 0.21${\mu}m$ is put inside a box of a spherical jar of radius 21mm. Find the probability of finding the atom inside the jar.

 The total volume of the jar becomes the total sample of the location of the atom. The total volume of the atom becomes the favourable sample. The ratio of the above two is the answer.

  $\text{Volume of atom = }\dfrac{4}{3}\pi {{r}^{3}}=\dfrac{4}{3}\pi \times {{\left( 0.21\times {{10}^{-6}} \right)}^{3}} \\ \text{Volume of jar = }\dfrac{4}{3}\pi {{r}^{3}}=\dfrac{4}{3}\pi \times {{\left( 21\times {{10}^{-3}} \right)}^{3}} \\ \text{Probability = }\dfrac{Volume\text{ of atom}}{Volume\text{ of jar}}=\dfrac{\dfrac{4}{3}\pi \times {{\left( 0.21\times {{10}^{-6}} \right)}^{3}}}{\dfrac{4}{3}\pi \times {{\left( 21\times {{10}^{-3}} \right)}^{3}}}=\dfrac{\dfrac{4}{3}\pi \times {{\left( 21\times {{10}^{-8}} \right)}^{3}}}{\dfrac{4}{3}\pi \times {{\left( 21\times {{10}^{-3}} \right)}^{3}}}={{10}^{-15}}$

 A silver hemispherical bowl with an inner radius of 3.5 cm and thickness of 0.7cm is made up of iron. Taking the density of iron is $7.7gm/cc$, find the mass of the bowl when it is filled with water to the brim. Take the density of water is $1gm/c{{m}^{3}}$. Round off one decimal place.

 Find the volume of the total bowl and volume of the hemisphere with outer radius. Then find the weight of the total bowl and find the weight of the water that can be contained in the bowl with inner radius ‘r’. Sum both the weight to find the final answer.  

  The radius of the outer circle $= 3.5cm+0.7cm=4.2cm \\ $ The volume of the bowl is Volume of the hemisphere with outer radius - Volume of the hemisphere with inner radius $\\ \Rightarrow $ The volume of the bowl is $ \\ \Rightarrow \dfrac{22\times {{\left( 4.2 \right)}^{3}}}{2\times 7}-\dfrac{22\times {{\left( 3.5 \right)}^{3}}}{2\times 7}=49c{{m}^{3}} \\ $ The volume of the inner hemisphere is $ \\ \Rightarrow \dfrac{22\times {{\left( 3.5 \right)}^{3}}}{2\times 7}=67.4c{{m}^{3}} \\ $ Weight of bowl is $\Rightarrow 49\times 7.7=377.3gm \\ $ Weight of water is $\Rightarrow 67.4\times 1=67.4gm \\ $ Total weight is $\Rightarrow 377.3+67.4=441.7gm$