Question

Water is flowing continuously from a tap having an internal diameter $8\times {10}^{-3}\mathrm{m}$. The water velocity as it leaves the tap is $0.4{\mathrm{ms}}^{-1}$. The diameter of the water stream at a distance $2\times {10}^{-1}\mathrm{m}$ below the tap is close to.

• A. $5.0\times {10}^{-3}\mathrm{m}$
• B. $7.5\times {10}^{-3}\mathrm{m}$
• C. $9.6\times {10}^{-3}\mathrm{m}$
• D. $3.6\times {10}^{-3}\mathrm{m}$

Answer 1

The correct option is D

$3.6\times {10}^{-3}\mathrm{m}$

Step 1: Given parameters

Internal diameter, $d_1=8\times {10}^{-3}\mathrm{m}$

Initial water velocity, $v_1=0.4{\mathrm{ms}}^{-1}$

Height, $h=2\times {10}^{-1}\mathrm{m}$

Step 2: Calculate the final water velocity

Apply the third equation of motion,

$\begin{array}{rcl}{v}_2^2& =& v_1^2+2gh\\ & \Rightarrow & v_2^2={\left(0.4\right)}^2+2\times 10\times 2\times {10}^{-1}\\ & \Rightarrow & v_2^2=0.16+4\\ & \Rightarrow & v_2^2=4.16\\ & \Rightarrow & v_2=\sqrt{4.16}\\ & \Rightarrow & v_2=2.04{\mathrm{ms}}^{-1}\end{array}$

Here,

$v_2$ is the final velocity.

$g$ is the acceleration due to gravity.

Step 3: Find the diameter of the water stream at a given distance

Apply equation of continuity,

$\begin{array}{rcl}{a}_1{v}_1& =& a_2{v}_2\end{array}$

Where, $a_1$ and $a_2$ is the flow area.

$v_1$ and $v_2$ is the flow velocity.

$\begin{array}{rcl}{a}_1{v}_1& =& a_2{v}_2\\ & \Rightarrow & \mathrm{\pi }\times {\left(8\times {10}^{-3}\right)}^2\times 0.4=\mathrm{\pi }\times {\mathrm{d}}_2^2\times 2\\ & \Rightarrow & {\mathrm{d}}_2^2={\left(8\times {10}^{-3}\right)}^2\times 0.4\times \frac{1}{2}\\ & \Rightarrow & {\mathrm{d}}_2^2=12.8\times {10}^{-6}\\ & \Rightarrow & {\mathrm{d}}_2=\sqrt{12.8\times {10}^{-6}}\\ & \Rightarrow & {\mathrm{d}}_2=3.6\times {10}^{-3}\mathrm{m}\end{array}$

Therefore, the required diameter is $3.6\times {10}^{-3}\mathrm{m}$.

Hence, option (A) is correct.