JEE Mains · Physics · STD 11 - Motion in straight line
- JEE Mains 2024
- Hard
- Numerical
A body moves on a frictionless plane starting from rest. If \(\mathrm{S}_{\mathrm{n}}\) is distance moved between \(\mathrm{t}=\mathrm{n}-1\) and \(\mathrm{t}\) \(=\mathrm{n}\) and \(\mathrm{S}_{\mathrm{n}-1}\) is distance moved between \(\mathrm{t}=\mathrm{n}-2\) and \(t=n-1\), then the ratio \(\frac{S_{n-1}}{S_n}\) is \(\left(1-\frac{2}{x}\right)\) for \(n\) \(=10\). The value of \(x\) is _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(19\)
Step-by-step Solution
Detailed explanation
\(\mathrm{S}_{\mathrm{n}}=\frac{1}{2} \mathrm{a}(2 \mathrm{n}-1)=\frac{19 \mathrm{a}}{2}\)
\(\mathrm{~S}_{\mathrm{n}-1}=\frac{1}{2} \mathrm{a}(2 \mathrm{n}-3)=\frac{17 \mathrm{a}}{2}\)
\(\frac{\mathrm{S}_{\mathrm{n}-1}}{\mathrm{~S}_{\mathrm{n}}}=\frac{17}{19}=1-\frac{2}{\mathrm{x}} \Rightarrow \mathrm{x}=19\)
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