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Properties of Matter question

2022 · 25 Jun · Shift 1 · Q64
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Properties of Matter question

2022 · 25 Jun · Shift 1 · Q64

JEE MainPhysicsProperties of MatterNumerical+4 / −1
The velocity of upper layer of water in a river is 36 kmh −-− 1. Shearing stress between horizontal layers of water is 10 −-− 3 Nm −-− 2. Depth of the river is ‾\underline{\hspace{2cm}}​ m. (Co-efficient of viscosity of water is 10 −-− 2 Pa.s)
Numerical answer
View written solutionFree

Correct answer: 100

  1. Given data
  • Velocity of upper layer: 36 km h−136\ \text{km h}^{-1}36 km h−1
  • Shearing stress: 10−3 N m−210^{-3}\ \text{N m}^{-2}10−3 N m−2
  • Coefficient of viscosity: η=10−2 Pa⋅s\eta = 10^{-2}\ \text{Pa·s}η=10−2 Pa⋅s

We need to find the depth of the river, say hhh.

  1. Convert velocity into SI unit

36 km h−1=36×10003600=10 m s−136\ \text{km h}^{-1} = 36\times \frac{1000}{3600} = 10\ \text{m s}^{-1}36 km h−1=36×36001000​=10 m s−1

So, the velocity difference across the depth is

Δv=10 m s−1\Delta v = 10\ \text{m s}^{-1}Δv=10 m s−1

  1. Use the formula for viscous shear stress

For horizontal layers of a fluid,

τ=ηdvdy\tau = \eta \frac{dv}{dy}τ=ηdydv​

Assuming velocity changes uniformly from 000 at the bottom to 10 m s−110\ \text{m s}^{-1}10 m s−1 at the top over depth hhh,

dvdy=10h\frac{dv}{dy} = \frac{10}{h}dydv​=h10​

Hence,

10−3=10−2⋅10h10^{-3} = 10^{-2}\cdot \frac{10}{h}10−3=10−2⋅h10​

  1. Solve for hhh

10−3=10−1h10^{-3} = \frac{10^{-1}}{h}10−3=h10−1​

h=10−110−3=102=100 mh = \frac{10^{-1}}{10^{-3}} = 10^2 = 100\ \text{m}h=10−310−1​=102=100 m

  1. Final answer

100\boxed{100}100​

  1. Comparison with stored answer

Stored correct answer = 100100100

My derived answer also is 100100100, so they agree.

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