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 $\sin^{2} 6 P$ proportional to $v^{2} R$ proportional to $g$

Answer 1

Answer: (A)

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