If f varies jointly as q^2 and h, and f=96 when q=4 and h=3 then the value of q when f = 48 and h = 6 is 2
Solution:
Given that, f varies jointly as q^2 and h
[tex]\begin{array}{l}{\text { Then, } f \alpha q^{2} \times h} \\ {\rightarrow f=k \times q^{2} \times h \text { where } k \text { is proportionality constant. }}\end{array}[/tex]
And f=96 when q=4 and h=3
Now substitute f, q, h values in above formula
[tex]\begin{array}{l}{\rightarrow 96=\mathrm{k} \times 4^{2} \times 3} \\\\ {\rightarrow 32=\mathrm{k} \times 16} \\\\ {\rightarrow \mathrm{k}=2}\end{array}[/tex]
[tex]\text { Then, formula is } f=2 \times q^{2} \times h[/tex]
We have to find q when f=48 and h=6
[tex]\begin{array}{l}{\text { So, } 48=2 \times q^{2} \times 6} \\\\ {\rightarrow 48=q^{2} \times 12} \\\\ {\rightarrow q^{2}=4} \\\\ {\rightarrow q=2}\end{array}[/tex]
Hence, the value of q is 2
