Physics · Fluid Mechanics

JEE Advanced 2020 — Paper 2 — Question 1

A train with cross-sectional area StS_{t} is moving with speed vtv_{t} inside a long tunnel of cross-sectional area S0( S0=4 St)\mathrm{S}_{0}\left(\mathrm{~S}_{0}=4 \mathrm{~S}_{\mathrm{t}}\right). Assume that almost all the air (density ρ\rho ) in front of the train flows back between its sides and the walls of the tunnel. Also, the air flow with respect to the train is steady and laminar. Take the ambient pressure and that inside the train to be p0\mathrm{p}_{0}. If the pressure in the region between the sides of the train and the tunnel walls is pp, then p0−p=72Nρvt2p_{0}-p=\frac{7}{2 N} \rho v_{t}^{2}. The value of NN is ____\_\_\_\_

Answer: 9

Numerical answer — enter this value.

Step-by-step solution

The velocity of air flow relative to the train between its sides and the walls of the tunnel is v

v×3 St=vt4 St\mathrm{v} \times 3 \mathrm{~S}_{\mathrm{t}}=\mathrm{v}_{\mathrm{t}} 4 \mathrm{~S}_{\mathrm{t}}

v=43vtv=\frac{4}{3} v_{t}

Now,

P+12ρv2=P0+12ρvt2P+\frac{1}{2} \rho v^{2}=P_{0}+\frac{1}{2} \rho v_{t}^{2}

P0−P=12ρ(v2−vt2)=12ρvt2(169−1)P_{0}-P=\frac{1}{2} \rho\left(v^{2}-v_{t}^{2}\right)=\frac{1}{2} \rho v_{t}^{2}\left(\frac{16}{9}-1\right)

P0−P=718ρvt2P_{0}-P=\frac{7}{18} \rho v_{t}^{2}

Hence N=9\mathrm{N}=9

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2020
Paper
Paper 2
Subject
Physics
Chapter
Fluid Mechanics
Topic
Bernoulli's Equation and its Applications