Question

The formula for horizontal range of a projectile is $S=\frac{v^{2} \sin ^{2} \theta}{2 g}$ 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.