If $x$ varies inversely as square of $y$. Given that $y = 2$ for $x = 1$. The value of $x$ for $y = 6$ will be equal to:
A.3
B.9
C.$\dfrac{1}{3}$
D.$\dfrac{1}{9}$
Answer
599.1k+ views
Hint: We will first use the proportionality relation between $x$ and $y$. Then, we will remove the proportionality by introducing a constant. Next, we will substitute the given values to find the value of the constant. Finally, we will find the value of $x$ by substituting the value of constant and the value of $y$ in the relation.
Formula used:
If two quantities $a$ and $b$ are inversely related then $a \propto \dfrac{1}{b}$.
Complete step-by-step answer:
We are given that $x$ varies inversely as square of $y$. This means that $x$ is proportional to the inverse of ${y^2}$. Mathematically, we can write it as
$x \propto \dfrac{1}{{{y^2}}}$ ………………………….$\left( 1 \right)$
We now have to remove the sign of proportionality from the relation $\left( 1 \right)$ by introducing a constant. Let us take the constant of proportionality as $n$. Thus,
$ \Rightarrow x = \dfrac{n}{{{y^2}}}$ ………………………….$\left( 2 \right)$
Now, we have to find the value of the constant of proportionality, i.e., $n$.
We can do this by substituting the given values i.e., $y = 2$ for $x = 1$ in equation $\left( 2 \right)$. This gives us
$ \Rightarrow 1 = \dfrac{n}{{{2^2}}} \\
\Rightarrow 1 = \dfrac{n}{4} \\ $
Multiplying both side by 4, we get
$ \Rightarrow n = 4$.
Now substituting $n = 4$ in equation $\left( 2 \right)$, we get
$x = \dfrac{4}{{{y^2}}}$ ……………………………………..$\left( 3 \right)$
We are supposed to find the value of $x$ when $y = 6$.
Let us substitute $y = 6$ in equation $\left( 3 \right)$. Therefore, w get
$ x = \dfrac{4}{{{6^2}}} \\
\Rightarrow x = \dfrac{4}{{36}} \\ $
We can reduce $\dfrac{4}{{36}}$ to the simplest form. Here, 4 is the common factor and $36 = 4 \times 9$. Hence, we have
$x = \dfrac{1}{9}$
Therefore, the correct option is option D.
Note: Proportionality is of two types i.e. direct proportional and inversely proportional. Two variables are said to be directly proportional if both variables change in the same proportion, i.e., if one increases, then the other variable also increases. If two variables are inversely proportional, then both variables change in the opposite sense i.e., if one increases, then the other decreases and vice versa.
Formula used:
If two quantities $a$ and $b$ are inversely related then $a \propto \dfrac{1}{b}$.
Complete step-by-step answer:
We are given that $x$ varies inversely as square of $y$. This means that $x$ is proportional to the inverse of ${y^2}$. Mathematically, we can write it as
$x \propto \dfrac{1}{{{y^2}}}$ ………………………….$\left( 1 \right)$
We now have to remove the sign of proportionality from the relation $\left( 1 \right)$ by introducing a constant. Let us take the constant of proportionality as $n$. Thus,
$ \Rightarrow x = \dfrac{n}{{{y^2}}}$ ………………………….$\left( 2 \right)$
Now, we have to find the value of the constant of proportionality, i.e., $n$.
We can do this by substituting the given values i.e., $y = 2$ for $x = 1$ in equation $\left( 2 \right)$. This gives us
$ \Rightarrow 1 = \dfrac{n}{{{2^2}}} \\
\Rightarrow 1 = \dfrac{n}{4} \\ $
Multiplying both side by 4, we get
$ \Rightarrow n = 4$.
Now substituting $n = 4$ in equation $\left( 2 \right)$, we get
$x = \dfrac{4}{{{y^2}}}$ ……………………………………..$\left( 3 \right)$
We are supposed to find the value of $x$ when $y = 6$.
Let us substitute $y = 6$ in equation $\left( 3 \right)$. Therefore, w get
$ x = \dfrac{4}{{{6^2}}} \\
\Rightarrow x = \dfrac{4}{{36}} \\ $
We can reduce $\dfrac{4}{{36}}$ to the simplest form. Here, 4 is the common factor and $36 = 4 \times 9$. Hence, we have
$x = \dfrac{1}{9}$
Therefore, the correct option is option D.
Note: Proportionality is of two types i.e. direct proportional and inversely proportional. Two variables are said to be directly proportional if both variables change in the same proportion, i.e., if one increases, then the other variable also increases. If two variables are inversely proportional, then both variables change in the opposite sense i.e., if one increases, then the other decreases and vice versa.
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