
If 1 cubic cm of metal weighs 21 gms, then the weight of a metal pipe of length 1 metre with a bore of 3cm and in which thickness of the metal is 1cm is
(A) 8800 kg
(B) 21000 kg
(C) 26.4 kg
(D) 25.8 kg
Answer
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Hint:We are given various measurements of the dimensions of a metal pipe along with its weight. We shall first analyze these measurements and formulate the net volume of metal used to make the pipe. Then, we will plug in the values and calculate the volume of metal used. Further, using this information, we will find the weight of the metal pipe.
Complete step by step solution:
Given that the measurement of the bore of the pipe is 3cm, we understand that the inner diameter of the pipe is 3cm. Since, the diameter of any circular object is twice of its radius, therefore, we get
$\Rightarrow radius=\dfrac{diameter}{2}$
$\Rightarrow {{r}_{1}}=\dfrac{3}{2}cm$
$\therefore {{r}_{1}}=1.5cm$
Also, the thickness of the metal of the pipe is 1cm. This implies that the outer radius of the pipe is the sum of the inner diameter and the thickness of the pipe.
Outer radius, ${{r}_{2}}={{r}_{1}}+thickness$
$\Rightarrow {{r}_{2}}=1.5+1$
$\Rightarrow {{r}_{2}}=2.5cm$
Thus, the inner radius $\left( {{r}_{1}} \right)$ of the pipe is 1.5cm and the outer radius $\left( {{r}_{2}} \right)$ is 2.5cm.
The volume, $V$ of this pipe which is cylindrical in shape is given as:
\[V=\pi \left( r_{2}^{2}-r_{1}^{2} \right)h\]
Here, $h=1m=100cm$ as the length of pipe is 1m.
Substituting all the values, we get
\[\begin{align}
& \Rightarrow V=\pi \left( {{\left( 2.5 \right)}^{2}}-{{\left( 1.5 \right)}^{2}} \right)100 \\
& \Rightarrow V=100\pi \left( 6.25-2.25 \right) \\
& \Rightarrow V=100\pi \left( 4 \right) \\
\end{align}\]
Taking $\pi =\dfrac{22}{7}$, we have
$\Rightarrow V=400\dfrac{22}{7}c{{m}^{3}}$
We have been given that $1c{{m}^{3}}=21grams$ in the problem.
Multiplying $400\dfrac{22}{7}$ on both sides, we get
$\Rightarrow 400\dfrac{22}{7}\times 1c{{m}^{3}}=400\dfrac{22}{7}\times 21grams$
$\Rightarrow 400\dfrac{22}{7}c{{m}^{3}}=26400grams$
Using a unitary method, we shall now find the weight of the pipe in kilograms.
Since, $1000grams=1kg$,
$\begin{align}
& \Rightarrow 1gram=\dfrac{1}{1000}kg \\
& \Rightarrow 26400grams=26400\times \dfrac{1}{1000}kg \\
\end{align}$
$\Rightarrow 26400grams=26.4kg$
Therefore, the weight of the metal pipe is 26.4 kg.
Hence, the correct option is (C) 26.4 kg.
Note:
We use the unitary methods when we know the relation between two different quantities. Unitary method helps us to use that relation and find the value of as much amount of that quantity as we require. We first find the value of a single unit of any quantity with respect to the other one. Then we multiply the number of units which we require with the relative value calculated before to find the required amount of quantity.
Complete step by step solution:
Given that the measurement of the bore of the pipe is 3cm, we understand that the inner diameter of the pipe is 3cm. Since, the diameter of any circular object is twice of its radius, therefore, we get
$\Rightarrow radius=\dfrac{diameter}{2}$
$\Rightarrow {{r}_{1}}=\dfrac{3}{2}cm$
$\therefore {{r}_{1}}=1.5cm$
Also, the thickness of the metal of the pipe is 1cm. This implies that the outer radius of the pipe is the sum of the inner diameter and the thickness of the pipe.
Outer radius, ${{r}_{2}}={{r}_{1}}+thickness$
$\Rightarrow {{r}_{2}}=1.5+1$
$\Rightarrow {{r}_{2}}=2.5cm$
Thus, the inner radius $\left( {{r}_{1}} \right)$ of the pipe is 1.5cm and the outer radius $\left( {{r}_{2}} \right)$ is 2.5cm.
The volume, $V$ of this pipe which is cylindrical in shape is given as:
\[V=\pi \left( r_{2}^{2}-r_{1}^{2} \right)h\]
Here, $h=1m=100cm$ as the length of pipe is 1m.
Substituting all the values, we get
\[\begin{align}
& \Rightarrow V=\pi \left( {{\left( 2.5 \right)}^{2}}-{{\left( 1.5 \right)}^{2}} \right)100 \\
& \Rightarrow V=100\pi \left( 6.25-2.25 \right) \\
& \Rightarrow V=100\pi \left( 4 \right) \\
\end{align}\]
Taking $\pi =\dfrac{22}{7}$, we have
$\Rightarrow V=400\dfrac{22}{7}c{{m}^{3}}$
We have been given that $1c{{m}^{3}}=21grams$ in the problem.
Multiplying $400\dfrac{22}{7}$ on both sides, we get
$\Rightarrow 400\dfrac{22}{7}\times 1c{{m}^{3}}=400\dfrac{22}{7}\times 21grams$
$\Rightarrow 400\dfrac{22}{7}c{{m}^{3}}=26400grams$
Using a unitary method, we shall now find the weight of the pipe in kilograms.
Since, $1000grams=1kg$,
$\begin{align}
& \Rightarrow 1gram=\dfrac{1}{1000}kg \\
& \Rightarrow 26400grams=26400\times \dfrac{1}{1000}kg \\
\end{align}$
$\Rightarrow 26400grams=26.4kg$
Therefore, the weight of the metal pipe is 26.4 kg.
Hence, the correct option is (C) 26.4 kg.
Note:
We use the unitary methods when we know the relation between two different quantities. Unitary method helps us to use that relation and find the value of as much amount of that quantity as we require. We first find the value of a single unit of any quantity with respect to the other one. Then we multiply the number of units which we require with the relative value calculated before to find the required amount of quantity.
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