Answer

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**Hint:**Here, we need to find the number of lead shots that can be made. We will find out the volume of the cuboid and the volume of 1 lead shot. Using the volume of 1 lead shot and the volume of the cuboid, we will form an equation. We will then solve the equation to find the number of lead shots that can be made.

**Formula used:**We will use the following formulas for solving the question:

1.The volume of a cuboid is given by \[l \times b \times h\], where \[l\] is the length, \[b\] is the breadth, and \[h\] is the height.

2.The volume of a sphere is given by the formula \[\dfrac{4}{3}\pi {r^3}\], where \[r\] is the radius of the sphere.

**Complete step-by-step answer:**Let the number of lead shots that can be made be \[x\].

Let us convert all the given dimensions to centimetres.

We know that 1 millimetre is equal to \[\dfrac{1}{{10}}\] centimetre.

Therefore, we get

Diameter of each lead shot \[ = 3{\rm{ mm}} = 3 \times \dfrac{1}{{10}}{\rm{ cm}} = \dfrac{3}{{10}}{\rm{ cm}}\]

Now, the radius is half of its diameter, that is \[r = \dfrac{d}{2}\].

Substituting \[\dfrac{3}{{10}}\] in the expression, we get

\[r = \dfrac{{\dfrac{3}{{10}}}}{2} = \dfrac{3}{{20}}{\rm{ cm}}\]

Now, we need to find out the volume of the cuboid.

The volume of a cuboid is given by the formula \[l \times b \times h\], where \[l\] is the length, \[b\] is the breadth, and \[h\] is the height.

Substituting \[l = 9{\rm{ cm}}\], \[b = 11{\rm{ cm}}\], and \[h = 12{\rm{ cm}}\] in the formula, we get

Volume of the wall \[ = 9 \times 11 \times 12{\rm{ c}}{{\rm{m}}^3}\]

Next, we need to find out the volume of each lead shot.

We know that each lead shot is in the shape of a sphere.

The volume of a sphere is given by the formula \[\dfrac{4}{3}\pi {r^3}\], where \[r\] is the radius of the sphere.

Substituting \[r = \dfrac{3}{{20}}{\rm{ cm}}\]in the formula, we get

Volume of 1 spherical lead shot \[ = \dfrac{4}{3}\pi {\left( {\dfrac{3}{{20}}} \right)^3}{\rm{ c}}{{\rm{m}}^3} = \dfrac{4}{3} \times \pi \times \dfrac{3}{{20}} \times \dfrac{3}{{20}} \times \dfrac{3}{{20}}{\rm{ c}}{{\rm{m}}^3}\]

Now, we will find the number of lead shots that can be made using the cuboid.

The volume of lead in the cuboid is used to make the lead shots. This means that the volume of lead in the cuboid is equal to the total volume of lead in \[x\] spherical lead shots.

Therefore, the volume of the cuboid is equal to the product of the volume of each lead shot multiplied by the number of lead shots.

Thus, we get the equation

Volume of cuboid \[ = \] Volume of 1 lead shot \[ \times \] Number of lead shots

Substituting volume of cuboid as \[9 \times 11 \times 12{\rm{ c}}{{\rm{m}}^3}\], volume of 1 lead shot as \[\dfrac{4}{3} \times \pi \times \dfrac{3}{{20}} \times \dfrac{3}{{20}} \times \dfrac{3}{{20}}{\rm{ c}}{{\rm{m}}^3}\], and number of lead shots as \[x\], we get

\[ \Rightarrow 9 \times 11 \times 12 = \dfrac{4}{3} \times \pi \times \dfrac{3}{{20}} \times \dfrac{3}{{20}} \times \dfrac{3}{{20}} \times x\]

We will simplify this equation to get the value of \[x\].

Rewriting the equation, we get

\[ \Rightarrow \dfrac{{9 \times 11 \times 12}}{{\dfrac{4}{3} \times \pi \times \dfrac{3}{{20}} \times \dfrac{3}{{20}} \times \dfrac{3}{{20}}}} = x\]

Substituting \[\pi = \dfrac{{22}}{7}\], we get

\[ \Rightarrow \dfrac{{9 \times 11 \times 12}}{{\dfrac{4}{3} \times \dfrac{{22}}{7} \times \dfrac{3}{{20}} \times \dfrac{3}{{20}} \times \dfrac{3}{{20}}}} = x\]

Simplifying the equation, we get

\[\begin{array}{l} \Rightarrow \dfrac{3}{{\dfrac{1}{7} \times \dfrac{1}{{10}} \times \dfrac{1}{{20}} \times \dfrac{1}{{20}}}} = x\\ \Rightarrow 3 \times 7 \times 10 \times 20 \times 20 = x\end{array}\]

Multiplying the terms, we get

\[ \Rightarrow x = 84000\]

**\[\therefore\] We get the number of lead shots that can be made from the cuboid as 84000.**

**Note:**We have left the volumes as the products instead of simplifying them by multiplication to get a large number. This is because leaving them as product helps to simplify the final equation \[9 \times 11 \times 12 = \dfrac{4}{3} \times \pi \times \dfrac{3}{{20}} \times \dfrac{3}{{20}} \times \dfrac{3}{{20}} \times x\] much easier.

We need to ensure that all units are in either millimetres or centimetres. We can make a mistake by calculating the volume of lead shot in millimetres, and volume of cuboid in centimetres, and then dividing them, thus getting the wrong answer.

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