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

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

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

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Physics Class 11th Physics System of Particles and Rotational Motion

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 -

$10 r a d / s^{2}$

Option 2 -

$100 r a d / s^{2}$

Option 3 -

$1000 r a d / s^{2}$

Option 4 -

$10000 r a d / s^{2}$

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 2

Detailed Solution:

Angular momentum imparted to sprinkler per

$s e c o n d = v r \frac{d m}{d t} = \frac{0.6}{60} \times 0.5 \times 1 = 0.005 N . m$

$? l \frac{d ω}{d t} = 0.005$

$? \frac{d ω}{d t} = \frac{0.005}{500 \times 10^{- 7}}$

$= 100 \frac{r a d}{s^{2}}$

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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)