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Properties of Matter question

2023 · 12 Apr · Shift 1 · Q64
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  5. /2023 · 12 Apr · Shift 1 · Q64

Properties of Matter question

2023 · 12 Apr · Shift 1 · Q64

JEE MainPhysicsProperties of MatterNumerical+4 / −1
Glycerin of density 1.25×103 kg m−31.25 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}1.25×103 kg m−3 is flowing through the conical section of pipe The area of cross-section of the pipe at its ends are 10 cm210 \mathrm{~cm}^{2}10 cm2 and 5 cm25 \mathrm{~cm}^{2}5 cm2 and pressure drop across its length is 3 Nm−23 ~\mathrm{Nm}^{-2}3 Nm−2. The rate of flow of glycerin through the pipe is x×10−5 m3 s−1x \times 10^{-5} \mathrm{~m}^{3} \mathrm{~s}^{-1}x×10−5 m3 s−1. The value of xxx is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 4

  1. Given data
  • Density of glycerin: ρ=1.25×103 kg m−3\rho = 1.25 \times 10^3\,\text{kg m}^{-3}ρ=1.25×103kg m−3
  • Area at one end: A1=10 cm2=10×10−4=10−3 m2A_1 = 10\,\text{cm}^2 = 10 \times 10^{-4} = 10^{-3}\,\text{m}^2A1​=10cm2=10×10−4=10−3m2
  • Area at the other end: A2=5 cm2=5×10−4 m2A_2 = 5\,\text{cm}^2 = 5 \times 10^{-4}\,\text{m}^2A2​=5cm2=5×10−4m2
  • Pressure drop: P1−P2=3 N m−2P_1 - P_2 = 3\,\text{N m}^{-2}P1​−P2​=3N m−2
  • Flow rate: Q=x×10−5 m3s−1Q = x \times 10^{-5}\,\text{m}^3\text{s}^{-1}Q=x×10−5m3s−1

Since the pipe is conical and no height difference is mentioned, we apply Bernoulli's equation between the two ends.

  1. Use continuity equation

For incompressible flow,

A1v1=A2v2=QA_1 v_1 = A_2 v_2 = QA1​v1​=A2​v2​=Q

So,

v1=QA1,v2=QA2v_1 = \frac{Q}{A_1}, \qquad v_2 = \frac{Q}{A_2}v1​=A1​Q​,v2​=A2​Q​
  1. Apply Bernoulli's equation

At same height,

P1+12ρv12=P2+12ρv22P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2P1​+21​ρv12​=P2​+21​ρv22​

Therefore,

P1−P2=12ρ(v22−v12)P_1 - P_2 = \frac{1}{2}\rho \left(v_2^2 - v_1^2\right)P1​−P2​=21​ρ(v22​−v12​)

Substitute v1=Q/A1v_1 = Q/A_1v1​=Q/A1​ and v2=Q/A2v_2 = Q/A_2v2​=Q/A2​:

3=12(1.25×103)[(QA2)2−(QA1)2]3 = \frac{1}{2}(1.25\times10^3)\left[\left(\frac{Q}{A_2}\right)^2 - \left(\frac{Q}{A_1}\right)^2\right]3=21​(1.25×103)[(A2​Q​)2−(A1​Q​)2]
  1. Substitute areas
3=12(1.25×103)Q2[1(5×10−4)2−1(10−3)2]3 = \frac{1}{2}(1.25\times10^3)Q^2\left[\frac{1}{(5\times10^{-4})^2} - \frac{1}{(10^{-3})^2}\right]3=21​(1.25×103)Q2[(5×10−4)21​−(10−3)21​]

Now,

(5×10−4)2=25×10−8=2.5×10−7(5\times10^{-4})^2 = 25\times10^{-8} = 2.5\times10^{-7}(5×10−4)2=25×10−8=2.5×10−7

so

1(5×10−4)2=12.5×10−7=4×106\frac{1}{(5\times10^{-4})^2} = \frac{1}{2.5\times10^{-7}} = 4\times10^6(5×10−4)21​=2.5×10−71​=4×106

Also,

(10−3)2=10−6,1(10−3)2=106(10^{-3})^2 = 10^{-6}, \qquad \frac{1}{(10^{-3})^2} = 10^6(10−3)2=10−6,(10−3)21​=106

Hence,

3=12(1.25×103)Q2(4×106−106)3 = \frac{1}{2}(1.25\times10^3)Q^2(4\times10^6 - 10^6)3=21​(1.25×103)Q2(4×106−106) 3=12(1.25×103)Q2(3×106)3 = \frac{1}{2}(1.25\times10^3)Q^2(3\times10^6)3=21​(1.25×103)Q2(3×106)
  1. Simplify
12(1.25×103)(3×106)=0.625×103×3×106\frac{1}{2}(1.25\times10^3)(3\times10^6) = 0.625\times10^3 \times 3\times10^621​(1.25×103)(3×106)=0.625×103×3×106 =1.875×109= 1.875\times10^9=1.875×109

So,

3=1.875×109Q23 = 1.875\times10^9 Q^23=1.875×109Q2 Q2=31.875×109=1.6×10−9Q^2 = \frac{3}{1.875\times10^9} = 1.6\times10^{-9}Q2=1.875×1093​=1.6×10−9 Q=1.6×10−9=4×10−5 m3s−1Q = \sqrt{1.6\times10^{-9}} = 4\times10^{-5}\,\text{m}^3\text{s}^{-1}Q=1.6×10−9​=4×10−5m3s−1

Thus,

Q=x×10−5 m3s−1Q = x \times 10^{-5}\,\text{m}^3\text{s}^{-1}Q=x×10−5m3s−1

so

x=4x = 4x=4
  1. Final answer

The required value is:

4\boxed{4}4​
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