연속·베르누이:

$$
A_1V_1=A_2V_2,\quad P_1+\tfrac12\rho V_1^2=P_2+\tfrac12\rho V_2^2
$$

$D_1=0.20\,\text{m}, D_2=0.10\,\text{m}\Rightarrow A_1/A_2=4,\ V_2=4V_1$

$$
V_1=\sqrt{\frac{P_1-P_2}{7.5\rho}}
=\sqrt{\frac{(98-29.42)\times10^3}{7.5\times1000}}
\approx 3.02\,\text{m/s},\quad V_2\approx12.10\,\text{m/s}
$$

$\dot m=\rho A_1V_1=1000\cdot\frac{\pi(0.2)^2}{4}\cdot3.02\approx94.9\,\text{kg/s}$

x-방향 운동량 방정식(출구 속도는 좌향, $V_2=-12.10$):

$$
R_x=\dot m\,(V_2-V_1)-(P_1A_1+P_2A_2)
$$

$$
=94.9(-12.10-3.02)
$$
$$
-(98\times10^3\cdot A_1+29.42\times10^3\cdot A_2)
\approx-4.74\times10^3\,\text{N}
$$

따라서 필요한 고정력의 크기 $=4.744\,\text{kN}$

정답: ④ 4744