Class 11th Overview Physics Physics Laws of Motion

A 20 kg block B is suspended from a cord attached to a 40 kg cart A. Find the ratio of the acceleration of block in cases (i) and (ii) shown in the figure immediately after the system is released from rest. (Neglect friction)

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math> </span></p>

Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mn>3</mn> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> </mrow> </math> </span></p>

Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math> </span></p>

Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>2</mn> <mroot> <mrow> <mn>2</mn> </mrow> <mrow></mrow> </mroot> </mrow> </mfrac> </mrow> </math> </span></p>

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Answered by

Raj Pandey | Contributor-Level 9

a year ago

Correct Option - 4

Detailed Solution:

Case I:

$T - N = 40 a$

and $20 g - T = 20 a$

Also $N = 20 a$

After simplifying, we get

$a = \frac{g}{4}$

Acceleration of block $B, = \sqrt[]{2} a = \frac{g}{2 \sqrt[]{2}} .$

Case II:

$T = 40 a$

and $20 g - T = 20 a$

After simplifying above equation, we get

$a = g / 3$

Ratio $= \frac{g / 2 \sqrt[]{2}}{g / 3} = \frac{3}{2 \sqrt[]{2}} .$

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