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

Answer 1

Answer: (A)

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