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Properties of Matter question

2019 · 10 Apr · Shift 2 · Q57
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Properties of Matter question

2019 · 10 Apr · Shift 2 · Q57

JEE MainPhysicsProperties of MatterMCQ+4 / −1
Water from a tap emerges vertically downwards with an initial speed of 1.0 ms–1 . The cross-sectional area of the tap is 10–4 m2. Assume that the pressure is constant throughout the stream of water and that the flow is streamlined. The cross-sectional area of the stream, 0.15 m below the tap would be : (Take g = 10 ms–2)
  1. A
    5 × 10–4 m2
  2. B
    2 × 10–5 m2
  3. C
    5 × 10–5 m2
  4. D
    1 × 10–5 m2
View written solutionFree

Correct answer: C

  1. Given data
  • Initial speed at the tap: v1=1.0 m s−1v_1 = 1.0\,\text{m s}^{-1}v1​=1.0m s−1
  • Cross-sectional area at the tap: A1=10−4 m2A_1 = 10^{-4}\,\text{m}^2A1​=10−4m2
  • Vertical distance below the tap: h=0.15 mh = 0.15\,\text{m}h=0.15m
  • Acceleration due to gravity: g=10 m s−2g = 10\,\text{m s}^{-2}g=10m s−2

We need the cross-sectional area A2A_2A2​ of the stream at a point 0.15 0.15\,0.15m below the tap.

  1. Find the speed of water 0.15 m below the tap

Since pressure is constant throughout the stream and flow is streamlined, we can use Bernoulli / kinematics along the stream.

Using

v22=v12+2ghv_2^2 = v_1^2 + 2ghv22​=v12​+2gh

Substitute values:

v22=(1)2+2(10)(0.15)v_2^2 = (1)^2 + 2(10)(0.15)v22​=(1)2+2(10)(0.15) v22=1+3=4v_2^2 = 1 + 3 = 4v22​=1+3=4 v2=2 m s−1v_2 = 2\,\text{m s}^{-1}v2​=2m s−1
  1. Apply equation of continuity

For incompressible flow,

A1v1=A2v2A_1 v_1 = A_2 v_2A1​v1​=A2​v2​

So,

A2=A1v1v2A_2 = \frac{A_1 v_1}{v_2}A2​=v2​A1​v1​​

Substitute values:

A2=10−4×12A_2 = \frac{10^{-4} \times 1}{2}A2​=210−4×1​ A2=5×10−5 m2A_2 = 5 \times 10^{-5}\,\text{m}^2A2​=5×10−5m2
  1. Match with options
A2=5×10−5 m2A_2 = 5 \times 10^{-5}\,\text{m}^2A2​=5×10−5m2

This corresponds to Option C.

  1. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They match.

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