### Discussion :: GATE Exam Questions - Section 1 (Q.No.5)

Mohan said: (Dec 13, 2013) | |

From Fluid kinematics. For radial flow the normal acceleration is An = dvn/ds+Vr^{2}/r,Acceleration is proportionate to V ^{2}/r, therefore 1/r^{5}. |

Ram said: (Sep 24, 2014) | |

When the acceleration was take place on frist line. |

Dileep Barnwal said: (Dec 19, 2014) | |

Acceleration a = v*(dv/ds). => ao < 1/r5. |

Chandra said: (Jul 6, 2015) | |

I cannot understand this concept. Can anybody elaborate me please? |

Aparna said: (Nov 13, 2015) | |

v2/r but; v = 1/r2 so v = 1/r4+1; v = 1/r5. |

Yoga said: (Dec 18, 2015) | |

Acceleration a = V2/r = (1/r2)2/r = 1/r5. |

Deepak said: (Jun 14, 2016) | |

I agree with you @Aparna. |

Teja said: (Sep 8, 2016) | |

Nice @Aparna. |

Pran said: (Sep 17, 2016) | |

Please give me conceptual & answer. |

Md Shams Zeya said: (Nov 21, 2016) | |

I can not understand this, Please help me to get it. |

Madhu Kumar said: (Feb 7, 2017) | |

I agree with your answer @Aparna. |

Nathan said: (Feb 8, 2017) | |

What condition to apply r/2 in V in the question? Please anyone tell me. |

Tulasi Sivaprasad said: (Mar 9, 2017) | |

Please explain clearly. |

Kumari Juhi said: (Mar 13, 2017) | |

a = V2/r = (1/r2)2/r = 1/r5. |

Vishal Babu Jaiswal said: (Jun 5, 2017) | |

For radial flow : a =v2/r. V=k/r2, hence put the value in above equation, then a = k2/r5. |

Deepanchakaravarthy said: (Jun 7, 2017) | |

Thanks @Aparna. |

Maha said: (Aug 23, 2017) | |

For radial flow, a = V^2/r. Here, V = 1/r^2 hence, a = [(1/r^2)^2]/r, = (1/r^4)/r, a = 1/r^5. |

Swapan said: (Oct 11, 2017) | |

Acceleration a= v2/r--- eq-01 Given velocity v=1/r2, putting the value v in eq-. no 01, therefore,a = v.v/r= ((1/r2).(1/r2))/r =(1/r4)/r = 1/r4 x 1/r = 1/r5. |

M.Naresh said: (Dec 17, 2017) | |

We know the radial acc is=V2/r .(1/(r2)2*r=1/r5. |

Niki said: (Feb 8, 2018) | |

I agree @Maha. |

Nanu said: (Mar 11, 2018) | |

dr/dt = - constant / r*r. To find the acceleration, we want to differentiate the velocity with respect to time. That is: a = d/dt(dr/dt). = - constant * d/dt(1/r*r) Now, if f is a function of r, then d/dt(f) = df/dr * dr/dt and so; a = - constant * d/dt(1/r*r) = - constant * ( -2 / r*r*r) * (-constant) / r*r. a = -2 * velocity^2 / r = - 2 * constant^2 / r*r*r*r*r. So the acceleration is proportional to 1 / (r*r*r*r*r). |

Maha said: (May 5, 2018) | |

Thanks for explaining the answer. |

Phani said: (May 6, 2018) | |

Simple V = d/t, Given V = 1/r^2 and d = r, Substitute t = r^3, Now a = V/t = 1/r^5. |

Desalegn Degu said: (Feb 26, 2019) | |

For this a = V^2/r. Here, V = 1/r^2 hence, a = [(1/r^2)^2]/r, = (1/r^4)/r. a = 1/r^5. |

Aki said: (Sep 21, 2019) | |

Well explained, Thanks @Desalegn Degu. |

Blackie said: (Jan 5, 2021) | |

Yes, 1/r5 is correct. |

Gourab said: (Aug 10, 2021) | |

a= v(dv/ds) + dv/dt As steady flow is taking place, dv/dt=0. a= v (dv/ds). now, v= k/r2. dv/ds= dv/dr= -2k/ r3. therefore, a varies as 1/r5. |

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