A small block slides down on a smooth inclined plane, starting from rest at time <math> <mi>t</mi> <mo>=</mo> <mn>0</mn> </math> . Let <math> <msub> <mrow> <mrow> <mi>S</mi> </mrow> </mrow> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> </msub> </math> be the distance travelled by the block in the interval <math> <mi>t</mi> <mo>=</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </math> to <math> <mi>t</mi> <mo>=</mo> <mi>n</mi> </math> . Then, the ratio <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>S</mi> </mrow> </mrow> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>S</mi> </mrow> </mrow> <mrow> <mrow> <mi>n</mi> <mo>+</mo> <mn>1</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </math> is:

S n = Distance in n th sec. i.e. t = n - 1 to t = n  S n + 1 = Distance in ( n + 1 ) th sec.i.e. t = n to t = n + 1 So as we know S n = a 2 2 n - 1 a = acceleration S n S n + 1 = a 2 2 n - 1 a 2 2 n + 1 - 1 = 2 n - 1 2 n + 1  S n S n + 1 = 2 n - 1 2 n + 1

A small block slides down on a smooth inclined plane, starting from rest at time <math> <mi>t</mi> <mo>=</mo> <mn>0</mn> </math> . Let <math> <msub> <mrow> <mrow> <mi>S</mi> </mrow> </mrow> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> </msub> </math> be the distance travelled by the block in the interval <math> <mi>t</mi> <mo>=</mo> <mi>n</mi> <mo>-</mo> <mn>1</mn> </math> to <math> <mi>t</mi> <mo>=</mo> <mi>n</mi> </math> . Then, the ratio <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>S</mi> </mrow> </mrow> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>S</mi> </mrow> </mrow> <mrow> <mrow> <mi>n</mi> <mo>+</mo> <mn>1</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </math> &nbsp;is: &lt;!-- [if !supportLineBreakNewLine]--&gt; &lt;!--[endif]--&gt;

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Physics Class 11th Physics Motion in Plane

A small block slides down on a smooth inclined plane, starting from rest at time $t = 0$ . Let $S_{n}$ be the distance travelled by the block in the interval $t = n - 1$ to $t = n$ . Then, the ratio $\frac{S_{n}}{S_{n + 1}}$ is:

Option 1 -

$\frac{2 n - 1}{2 n}$

Option 2 -

$\frac{2 n - 1}{2 n + 1}$

Option 3 -

$\frac{2 n + 1}{2 n - 1}$

Option 4 -

$\frac{2 n}{2 n - 1}$

3 Views | Posted 3 weeks ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

3 weeks ago

Correct Option - 4

Detailed Solution:

$S_{n} =$ Distance in $n^{th}$ sec. i.e. $t = n - 1$ to $t = n$

$S_{n} + 1 =$ Distance in $(n + 1)^{th}$ sec.

i.e. $t = n$ to $t = n + 1$

So as we know

$S n = \frac{a}{2} 2 n - 1 a =$ acceleration

$\frac{S_{n}}{S_{n + 1}} = \frac{\frac{a}{2} 2 n - 1}{\frac{a}{2} 2 n + 1 - 1} = \frac{2 n - 1}{2 n + 1}$

$\frac{S_{n}}{S_{n + 1}} = \frac{2 n - 1}{2 n + 1}$

<math> <msub> <mrow> <mrow> <mi mathvariant="normal">S</mi> </mrow> </mrow> <mrow> <mrow> <mi mathvariant="normal">n</mi> </mrow> </mrow> </msub> <mo>=</mo> </math> Distance in <math> <msup> <mrow> <mrow> <mi mathvariant="normal">n</mi> </mrow> </mrow> <mrow> <mrow> <mtext>th</mtext> <mtext> </mtext> </mrow> </mrow> </msup> </math> sec. i.e. <math> <mi mathvariant="normal">t</mi> <mo>=</mo> <mi mathvariant="normal">n</mi> <mo>-</mo> <mn>1</mn> </math> to <math> <mi mathvariant="normal">t</mi> <mo>=</mo> <mi mathvariant="normal">n</mi> </math>  <math> <msub> <mrow> <mrow> <mi mathvariant="normal">S</mi> </mrow> </mrow> <mrow> <mrow> <mi mathvariant="normal">n</mi> </mrow> </mrow> </msub> <mo>+</mo> <mn>1</mn> <mo>=</mo> </math> Distance in <math> <mo>(</mo> <mi mathvariant="normal">n</mi> <mo>+</mo> <mn>1</mn> <msup> <mrow> <mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mtext>th</mtext> <mtext> </mtext> </mrow> </mrow> </msup> </math>  sec.i.e. <math> <mrow> <mi mathvariant="normal"> </mi> </mrow> <mi>t</mi> <mo>=</mo> <mi>n</mi> </math> to <math> <mi>t</mi> <mo>=</mo> <mi>n</mi> <mo>+</mo> <mn>1</mn> </math> So as we know <math> <mi mathvariant="normal">S</mi> <mi mathvariant="normal">n</mi> <mo>=</mo> <mfrac> <mrow> <mrow> <mi mathvariant="normal">a</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mn>2</mn> <mi mathvariant="normal">n</mi> <mo>-</mo> <mn>1</mn> <mrow> <mi mathvariant="normal"> </mi> </mrow> <mi mathvariant="normal">a</mi> <mo>=</mo> </math> acceleration <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>S</mi> </mrow> </mrow> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>S</mi> </mrow> </mrow> <mrow> <mrow> <mi>n</mi> <mo>+</mo> <mn>1</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>a</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mn>2</mn> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>a</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mn>2</mn> <mi>n</mi> <mo>+</mo> <mn>1</mn> <mo>-</mo> <mn>1</mn> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>2</mn> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi>n</mi> <mo>+</mo> <mn>1</mn> </mrow> </mrow> </mfrac> </math>  <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>S</mi> </mrow> </mrow> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>S</mi> </mrow> </mrow> <mrow> <mrow> <mi>n</mi> <mo>+</mo> <mn>1</mn> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>2</mn> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <mi>n</mi> <mo>+</mo> <mn>1</mn> </mrow> </mrow> </mfrac> </math>

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A small block slides down on a smooth inclined plane, starting from rest at time $t = 0$ . Let $S_{n}$ be the distance travelled by the block in the interval $t = n - 1$ to $t = n$ . Then, the ratio $\frac{S_{n}}{S_{n + 1}}$ is:

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A small block slides down on a smooth inclined plane, starting from rest at time t = 0 . Let S n be the distance travelled by the block in the interval t = n - 1 to t = n . Then, the ratio S n S n + 1 is:

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A small block slides down on a smooth inclined plane, starting from rest at time $t = 0$ . Let $S_{n}$ be the distance travelled by the block in the interval $t = n - 1$ to $t = n$ . Then, the ratio $\frac{S_{n}}{S_{n + 1}}$ is: