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Answer :
The correct answer is option: C.43.2m.
To solve this problem, we can use the formula for the maximum safe speed V on a banked road:
[tex]V=\sqrt{r g \tan \theta}[/tex]
Where:
V = maximum safe speed
r = radius of curvature
g = acceleration due to gravity
θ = angle of banking
Given that the angle of banking θ remains constant, but the maximum safe speed is to be increased by 20%, we can write:
[tex]V_{\text {new }}=1.2 V_{\text {old }}[/tex]
Substituting the expression for V into this equation:
[tex]\sqrt{r g \tan \theta_{\text {new }}}=1.2 \sqrt{ r g \tan \theta_{\text {old }}$$[/tex]
Since g and θ are constants, they cancel out. Squaring both sides gives:
[tex]r g \tan \theta_{\text {new }}=1.44 \times r g \tan \theta_{\text {old }}[/tex]
Given that the radius of curvature [tex]r_{old[/tex]=30 m, we need to find [tex]r_{new[/tex]. Rearranging the equation:
[tex]r_{\text {new }}=\frac{1.44 \times r_{\text {old }}}{\tan \theta_{\text {new }}}[/tex]
Since [tex]\tan \theta_{\text {new }}=\tan \theta_{\text {old }}[/tex] (angle of banking remains unchanged), we can substitute [tex]r_{\text {old }}=30m[/tex] and [tex]\tan \theta_{\text {old }}=\frac{1}{\sqrt{3}}[/tex] into the equation to find [tex]r_{\text {new }}:[/tex]
[tex]\begin{aligned}& r_{\text {new }}=\frac{1.44 \times 30}{\frac{1}{\sqrt{3}}} \\& r_{\text {new }}=43.2 \mathrm{~m}\end{aligned}[/tex]
So, the correct answer is option C.
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