Question

The formula for horizontal range of a projectile is $S=\frac{v^2{\sin }^2\theta }{2g}$ where $v$ is initial speed, $\theta$ angle of inclination and $g$ is acceleration due to gravity. The Wheatstone bridge shown in figure can be used to determine the range if the following arrangement is made

• A. $Q$ proportional to $v^{2}P$ proportional to $gR$ proportional to ${\sin }^2\theta$
• B. $Q$ proportional to $gP$ proportional to $v^{2}R$ proportional to ${\sin }^2\theta$
• C. $Q$ proportional to $gP$ proportional to ${\sin }^2\theta R$ proportional to $v^2$
• D. $Q$ proportional to ${\sin }^{2}6P$ proportional to $v^{2}R$ proportional to $g$

Answer 1

Answer: (A)

For the given Wheatstone bridge
$\frac{P}{R}=\frac{Q}{S}S=\frac{Q\times R}{P}$
$S=\frac{v^2{\sin }^2\theta }{2g}$.