$\therefore\text{k}_1=\frac{\text{Mg}}{\text{l}_1}\cdots\text{(iv)}$
$\text{k}_2=\frac{\text{Mg}}{\text{l}_2}\cdots\text{(v)}$
Dividing equation (iv) by equation (iii) we find$\frac{\text{k}_1}{\text{k}}=\frac{\text{l}}{\text{l}_1}=\frac{\text{l}_1+\text{l}_2}{\text{l}_1}=1+\frac{\text{l}_2}{\text{l}_1}$
From equation (ii) we find$\frac{\text{l}_1}{\text{l}_2}=\text{n}$
$\therefore\frac{\text{k}_1}{\text{k}}=1+\frac{1}{\text{n}}$
$\text{k}_1=\Big(\frac{\text{n}+1}{\text{n}}\Big)\text{k}$ From equation (ii) and (iii), we find:$\frac{\text{k}_2}{\text{k}}=\frac{\text{l}}{\text{l}_2}=\frac{\text{l}_1+\text{l}_2}{\text{l}_2}=\frac{\text{l}_1}{\text{l}_2}+1$
From equation (ii) we have $\frac{\text{l}_1}{\text{l}_2}=\text{n}$.$\therefore\frac{\text{k}_2}{\text{k}}=(\text{n}+1)$
$\therefore\text{k}_2=\text{k}(\text{n}+1)$
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