Q155 / 180 · KCET 2019 · Physics
A cylindrical container containing water has a small hole at height of \( \mathrm{H}=8 \mathrm{~cm} \) from the bottom
and at a depth of \( 2 \mathrm{~cm} \) from the top surface of the liquid. The maximum horizontal distance
travelled by the water before it hits the ground \( (x) \) is
and at a depth of \( 2 \mathrm{~cm} \) from the top surface of the liquid. The maximum horizontal distance
travelled by the water before it hits the ground \( (x) \) is
- A\( 4 \mathrm{~cm} \)
- Correct answerB\( 8 \mathrm{~cm} \)
- C\( 6 \mathrm{~cm} \)
- D\( 4 \sqrt{2} \mathrm{~cm} \)
Solution
(B)
\[ \begin{array}{l} v=\sqrt{2 g h}=\sqrt{2 \times 10 \times 2}=\sqrt{40} \\ t=\sqrt{\frac{2 h}{g}} \\ R=v \cdot t=\sqrt{40^{4}} \sqrt{\frac{2 \times 8}{10}}=8 \mathrm{~cm} \end{array} \]
\[ \begin{array}{l} v=\sqrt{2 g h}=\sqrt{2 \times 10 \times 2}=\sqrt{40} \\ t=\sqrt{\frac{2 h}{g}} \\ R=v \cdot t=\sqrt{40^{4}} \sqrt{\frac{2 \times 8}{10}}=8 \mathrm{~cm} \end{array} \]
Hence, the correct answer is option: 'B'.