\(Z\) is partly constant and partly varies inversely as the square of \(d\). when \(d=1, z=11\) and when \(d=2 . z=5\). Find the value of \(=\) when \(d=4 .\)
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Correct Answer: Option B
Explanation:
$$
=a d i=\alpha \frac{b}{d^{2}} \Rightarrow Z=a+\frac{b}{d^{2}}
$$
when \(d=1 . z=11\)
\(11=a+b\) .............. (1)
when \(d=2 .=5 \Rightarrow\)
\begin{aligned}
&\text { Sothat } 5=a+\frac{b}{4} \\
&\left\{\begin{array}{l}
20=4 a+b \\
9=a+b
\end{array}\right. \\
&b=11-3=8 ; \text { when } d=4 \\
&==a+\frac{b}{d^{2}}=3+\frac{8}{16}=3.5
\end{aligned}
$$
=a d i=\alpha \frac{b}{d^{2}} \Rightarrow Z=a+\frac{b}{d^{2}}
$$
when \(d=1 . z=11\)
\(11=a+b\) .............. (1)
when \(d=2 .=5 \Rightarrow\)
\begin{aligned}
&\text { Sothat } 5=a+\frac{b}{4} \\
&\left\{\begin{array}{l}
20=4 a+b \\
9=a+b
\end{array}\right. \\
&b=11-3=8 ; \text { when } d=4 \\
&==a+\frac{b}{d^{2}}=3+\frac{8}{16}=3.5
\end{aligned}