Question

$n=\frac{1}{2\pi }\sqrt{\frac{8T}{lA}}$ ; make 'A' the subject of formula.

• A. $A=\frac{2T}{l{\pi }^{2}n}$
• B. $A=\frac{T}{l{\pi }^2{n}^2}$
• C. $A=\frac{2T}{l{\pi }^2{n}^2}$
• D. $A=\frac{T}{l{\pi }^{2}n}$

Answer 1

The correct option is C $A=\frac{2T}{l{\pi }^2{n}^2}$

Given, $n=\frac{1}{2\pi }\sqrt{\frac{8T}{lA}}$

$\Rightarrow 2\pi n=\sqrt{\frac{8T}{lA}}$

Squaring both sides,

$\Rightarrow 4{\pi }^2{n}^2=\frac{8T}{lA}$

$\Rightarrow A=\frac{8T}{4l{\pi }^2{n}^2}$

$\Rightarrow A=\frac{2T}{l{\pi }^2{n}^2}$