Question

A graph is plotted between the temperature of a copper cube in ${ }^{\circ} C$ versus ${ }^{\circ} F$. The sine of the angle made by the graph with ${ }^{\circ} F$ axis is :

• A. $\frac{2}{\sqrt{106}}$
• B. $\frac{3}{\sqrt{106}}$
• C. $\frac{4}{\sqrt{106}}$
• D. $\frac{5}{\sqrt{106}}$

Answer 1

Answer: (D)

$T_{C}=\frac{5}{9}\left(T_{F}-32\right)$
$T_{C}=\frac{5}{9} T_{F}-\frac{160}{9}$
Thus graph between ${ }^{\circ} C$ and ${ }^{\circ} F$ can be drawn as

From graph we have $\sin \theta=\frac{A C}{A B}=\frac{160 / 9}{\sqrt{\left(\frac{160}{9}\right)^{2}+(32)^{2}}}$
$\Rightarrow =\frac{160 \times 9}{9 \times 32 \sqrt{106}}=\frac{5}{\sqrt{106}}$