Answer
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Hint: Here we will proceed from calculating the volume of sand by using the formula of volume of cylinder i.e. $\pi {r^2}h$. Also we will use a formula of volume of conical heap i.e. $ = \dfrac{1}{3}\pi {r_2}^2h$ to calculate the volume of the cylindrical bucket.
Complete step-by-step answer:
Now,
Let ${r_1}$ and ${r_2}$ be the radius of cylinder and conical heap respectively.
Now we know that volume of cylinder $ = \pi {r_1}^2h$
And volume of conical heap $ = \dfrac{1}{3}\pi {r_2}^2h$
$ \Rightarrow $ Volume of bucket full of sand will be $\pi {r_1}^2h$
$ \Rightarrow $ Volume of cone $ = \dfrac{1}{3}\pi {r_2}^2h$
At first we will find volume of sand using the below mentioned formula-
$ \Rightarrow $ $\pi {r_1}^2h$
$ = \dfrac{{22}}{7} \times 18 \times 18 \times 32$
= 32585.1429------(i)
Also we know that cylindrical buckets are emptied into conical heaps so both will have the same amount of sand.
Now we will find the volume of the cylindrical bucket by using the formula of volume of conical heap.
Therefore , Volume of conical heap $ = \dfrac{1}{3}\pi {r_2}^2h$------(ii)
Comparing equation (i) and (ii) we get,
Or $\dfrac{1}{3} \times \dfrac{{22}}{7} \times {r_2}^2 \times h = 32585.1429$
Now we calculate the value of $r_2^2$
$ \Rightarrow $$r_2^2$ = 32585.1429$ \times $ 7$ \times $$\dfrac{1}{{22}} \times \dfrac{1}{8}$
Or $r_2^2 = \sqrt {1296} $
$ \Rightarrow $radius of conical heap = 36
$ \Rightarrow $(Slant height) L = $\sqrt {r_2^2 + {h^2}} $
$ \Rightarrow $$\sqrt {1296 + 576} $
$ \Rightarrow $$\sqrt {1872} $
$ \Rightarrow $L = 43.26
Since we have to find the value of slant height of conical heap to one place of decimal, the answer will be 43.3 cm.
Note: While solving this question, one can get confused with the radius of both the shapes because the radius of the conical heap will be different from the radius of the cylindrical bucket. Also we must understand that the slant height of the conical heap will be calculated through a new radius drawn from the volume of the cylindrical bucket. Hence we will get the desired result.
Complete step-by-step answer:
Now,
Let ${r_1}$ and ${r_2}$ be the radius of cylinder and conical heap respectively.
Now we know that volume of cylinder $ = \pi {r_1}^2h$
And volume of conical heap $ = \dfrac{1}{3}\pi {r_2}^2h$
$ \Rightarrow $ Volume of bucket full of sand will be $\pi {r_1}^2h$
$ \Rightarrow $ Volume of cone $ = \dfrac{1}{3}\pi {r_2}^2h$
At first we will find volume of sand using the below mentioned formula-
$ \Rightarrow $ $\pi {r_1}^2h$
$ = \dfrac{{22}}{7} \times 18 \times 18 \times 32$
= 32585.1429------(i)
Also we know that cylindrical buckets are emptied into conical heaps so both will have the same amount of sand.
Now we will find the volume of the cylindrical bucket by using the formula of volume of conical heap.
Therefore , Volume of conical heap $ = \dfrac{1}{3}\pi {r_2}^2h$------(ii)
Comparing equation (i) and (ii) we get,
Or $\dfrac{1}{3} \times \dfrac{{22}}{7} \times {r_2}^2 \times h = 32585.1429$
Now we calculate the value of $r_2^2$
$ \Rightarrow $$r_2^2$ = 32585.1429$ \times $ 7$ \times $$\dfrac{1}{{22}} \times \dfrac{1}{8}$
Or $r_2^2 = \sqrt {1296} $
$ \Rightarrow $radius of conical heap = 36
$ \Rightarrow $(Slant height) L = $\sqrt {r_2^2 + {h^2}} $
$ \Rightarrow $$\sqrt {1296 + 576} $
$ \Rightarrow $$\sqrt {1872} $
$ \Rightarrow $L = 43.26
Since we have to find the value of slant height of conical heap to one place of decimal, the answer will be 43.3 cm.
Note: While solving this question, one can get confused with the radius of both the shapes because the radius of the conical heap will be different from the radius of the cylindrical bucket. Also we must understand that the slant height of the conical heap will be calculated through a new radius drawn from the volume of the cylindrical bucket. Hence we will get the desired result.
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