From a sprinkler of length 1 m , water is throw out with 0.5 m / s tangentially at the rate of 0.6 k g / m i n . If diameter of sprinkler is 8 c m and moment of inertia is 500 g c m 2 , the rate at which angular speed of sprinkler will increases is (assume there in no resistance to motion)

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <mn>10</mn> <mi> </mi> <mi>r</mi> <mi>a</mi> <mi>d</mi> <mo>/</mo> <msup> <mrow> <mrow> <mi>s</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> </span></p>
Option 2 - <p><span class="mathml" contenteditable="false"> <math> <mn>100</mn> <mi> </mi> <mi>r</mi> <mi>a</mi> <mi>d</mi> <mo>/</mo> <msup> <mrow> <mrow> <mi>s</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> </span></p>
Option 3 - <p><span class="mathml" contenteditable="false"> <math> <mn>1000</mn> <mi> </mi> <mi>r</mi> <mi>a</mi> <mi>d</mi> <mo>/</mo> <msup> <mrow> <mrow> <mi>s</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> </span></p>
Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mn>10000</mn> <mi> </mi> <mi>r</mi> <mi>a</mi> <mi>d</mi> <mo>/</mo> <msup> <mrow> <mrow> <mi>s</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> </span></p>
4 Views|Posted 4 months ago
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1 Answer
A
4 months ago
Correct Option - 2
Detailed Solution:

Angular momentum imparted to sprinkler per

s e c o n d = v r d m d t = 0.6 60 * 0.5 * 1 = 0.005 N . m

? l d ω d t = 0.005

? d ω d t = 0.005 500 * 10 - 7

= 100 r a d s 2

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Physics System of Particles and Rotational Motion 2025

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