The rate of flow of glycerine of density 1.25 × 103 kgm -3 through the conical section of a pipe if the radii of its ends are 0.1 m and 0.04 m and the pressure drop across its length 10 Nm -2 is

Question

The rate of flow of glycerine of density 1.25 × 103 kgm -3 through the conical section of a pipe if the radii of its ends are 0.1 m and 0.04 m and the pressure drop across its length 10 Nm -2 is (A) 6.93 × 10-4 m 3 s -1 (B) 7.8 × 10-4 m 3 s -1 (C) 10.4 × 10-5 m 3 s -1 (D) 14.5 × 10-5 m 3 s -1

Tardigrade Admin · Accepted Answer

From Bernoullis theorem p1+(1/2) ρ v12=p2+(1/α) ρ v22 ∴ p1-p2=(1/2) ρ(v22-v12) ∴ 10=(1/2) × 1.25 × 103(v22-v12) ∴ v22-v12=(10 × 2/1.25 × 103) =16 × 10-3 ...(i) Also from equation of continuity =A1 v1=A2 v2 π r12 v1=π r22 v2 ∴ (v1/v2)=[(r2/r1)]2=(0.04/0.1)=0.4 v1=0.4 v2 ... (ii) Substituting this value in Eq. (i) v22-(0.4 v2)2=16 × 10-3 v2=1.38 × 10-1=0.138 ms -1 Rate of flow of glycrine v=A2 v2 =π r22 v2 =6.93 × 10-4 m3 s-1

Q. The rate of flow of glycerine of density $1.25 \times 1 0^{3} k g m^{- 3}$ through the conical section of a pipe if the radii of its ends are $0.1 m$ and $0.04 m$ and the pressure drop across its length $10 N m^{- 2}$ is

$6.93 \times 1 0^{- 4} m^{3} s^{- 1}$

$7.8 \times 1 0^{- 4} m^{3} s^{- 1}$

$10.4 \times 1 0^{- 5} m^{3} s^{- 1}$

$14.5 \times 1 0^{- 5} m^{3} s^{- 1}$

Q. The rate of flow of glycerine of density 1.25×103kgm−3 through the conical section of a pipe if the radii of its ends are 0.1m and 0.04m and the pressure drop across its length 10Nm−2 is

6.93×10−4m3s−1

7.8×10−4m3s−1

10.4×10−5m3s−1

14.5×10−5m3s−1

Q. The rate of flow of glycerine of density $1.25 \times 1 0^{3} k g m^{- 3}$ through the conical section of a pipe if the radii of its ends are $0.1 m$ and $0.04 m$ and the pressure drop across its length $10 N m^{- 2}$ is

$6.93 \times 1 0^{- 4} m^{3} s^{- 1}$

$7.8 \times 1 0^{- 4} m^{3} s^{- 1}$

$10.4 \times 1 0^{- 5} m^{3} s^{- 1}$

$14.5 \times 1 0^{- 5} m^{3} s^{- 1}$