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Properties of Matter question

2011 · Shift 0 · Q71
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Properties of Matter question

2011 · Shift 0 · Q71

JEE MainPhysicsProperties of MatterMCQ+4 / −1
Water is flowing continuously from a tap having an internal diameter 8×10−3  m.8 \times {10^{ - 3}}\,\,m.8×10−3m. The water velocity as it leaves the tap is 0.4  ms−10.4\,\,m{s^{ - 1}}0.4ms−1. The diameter of the water stream at a distance 2×10−1  m2 \times {10^{ - 1}}\,\,m2×10−1m below the tap is close to :
  1. A
    7.5×10−3m7.5 \times {10^{ - 3}}m7.5×10−3m
  2. B
    9.6×10−3m9.6 \times {10^{ - 3}}m9.6×10−3m
  3. C
    3.6×10−3m3.6 \times {10^{ - 3}}m3.6×10−3m
  4. D
    5.0×10−3m5.0 \times {10^{ - 3}}m5.0×10−3m
View written solutionFree

Correct answer: C

  1. Given data
  • Diameter of tap opening: d1=8×10−3 md_1 = 8\times 10^{-3}\,\text{m}d1​=8×10−3m
  • Initial speed of water: v1=0.4 m s−1v_1 = 0.4\,\text{m s}^{-1}v1​=0.4m s−1
  • Vertical distance below tap: h=2×10−1=0.2 mh = 2\times 10^{-1} = 0.2\,\text{m}h=2×10−1=0.2m

We need the diameter of the stream at this lower point.

  1. Find speed of water after falling by height hhh

As the water falls freely under gravity, its speed increases according to

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

Substitute values:

v22=(0.4)2+2(9.8)(0.2)v_2^2 = (0.4)^2 + 2(9.8)(0.2)v22​=(0.4)2+2(9.8)(0.2) v22=0.16+3.92=4.08v_2^2 = 0.16 + 3.92 = 4.08v22​=0.16+3.92=4.08

So,

v2=4.08≈2.02 m s−1v_2 = \sqrt{4.08} \approx 2.02\,\text{m s}^{-1}v2​=4.08​≈2.02m s−1

  1. Use equation of continuity

For incompressible flow,

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

Since area A∝d2A \propto d^2A∝d2,

d12v1=d22v2d_1^2 v_1 = d_2^2 v_2d12​v1​=d22​v2​

Hence,

d2=d1v1v2d_2 = d_1\sqrt{\frac{v_1}{v_2}}d2​=d1​v2​v1​​​

Substitute values:

d2=8×10−30.42.02d_2 = 8\times 10^{-3}\sqrt{\frac{0.4}{2.02}}d2​=8×10−32.020.4​​

0.42.02≈0.198\frac{0.4}{2.02} \approx 0.1982.020.4​≈0.198

0.198≈0.445\sqrt{0.198} \approx 0.4450.198​≈0.445

Therefore,

d2≈8×10−3×0.445d_2 \approx 8\times 10^{-3} \times 0.445d2​≈8×10−3×0.445

d2≈3.56×10−3 md_2 \approx 3.56\times 10^{-3}\,\text{m}d2​≈3.56×10−3m

This is close to

3.6×10−3 m3.6\times 10^{-3}\,\text{m}3.6×10−3m

  1. Check options
  • A: 7.5×10−3 m7.5\times 10^{-3}\,\text{m}7.5×10−3m — too large
  • B: 9.6×10−3 m9.6\times 10^{-3}\,\text{m}9.6×10−3m — impossible, diameter should decrease
  • C: 3.6×10−3 m3.6\times 10^{-3}\,\text{m}3.6×10−3m — matches calculation
  • D: 5.0×10−3 m5.0\times 10^{-3}\,\text{m}5.0×10−3m — not close enough
  1. Final answer

The correct option is:

C   3.6×10−3 m\boxed{\text{C }\; 3.6\times 10^{-3}\,\text{m}}C 3.6×10−3m​

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