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JEE MAIN 2024
05-04-24 S1
Question
A body moves on a frictionless plane starting from rest. If $S_n$ is distance moved between $t=n-1$ and $t =\mathrm{n}$ and $\mathrm{S}_{\mathrm{n}-1}$ is distance moved between $\mathrm{t}=\mathrm{n}-2$ and $\mathrm{t}=\mathrm{n}-1$, then the ratio $\frac{\mathrm{S}_{\mathrm{n}-1}}{\mathrm{~S}_{\mathrm{n}}}$ is $\left(1-\frac{2}{\mathrm{x}}\right)$ for $\mathrm{n}=$ 10. The value of $x$ is $\_\_\_\_$
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Solution
$\begin{aligned} & S_n=\frac{1}{2} a(2 n-1)=\frac{19 a}{2} \\ & S_{n-1}=\frac{1}{2} a(2 n-3)=\frac{17 a}{2} \\ & \frac{S_{n-1}}{S_n}=\frac{17}{19}=1-\frac{2}{x} \Rightarrow x=19\end{aligned}$
Question Tags
JEE Main
Physics
Medium
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