Q.34. A cylindrical vessel of 40 cm radius is completely filled with water and its capacity is 528 d

Q: What is the correct answer for this exam Physics Q34?

The correct answer is C. Given radius $r = 40 cm = 0.4 m$. Capacity $V = 528 dm^3 = 528 \times 10^{-3} m^3$. The volume of a cylinder is $V = \pi r^2 h$. So, $h = \frac{V}{\pi r^2} = \frac{528 \times 10^{-3}}{\frac{22}{7} \times (0.4)^2} = \frac{528 \times 10^{-3} \times 7}{22 \times 0.16} = \frac{3696 \times 10^{-3}}{3.52} = 1.05 m = 105 cm$. The hole is made at $h_1 = 70 cm = 0.7 m$ below the top water level. The height

Q: Q.34. A cylindrical vessel of 40 cm radius is completely filled with water and its capacity is 528 dm (dm: decimeter). The vessel is placed on a solid block of exactly same height as vessel. If a…

The correct answer is C. Given radius $r = 40 cm = 0.4 m$. Capacity $V = 528 dm^3 = 528 \times 10^{-3} m^3$. The volume of a cylinder is $V = \pi r^2 h$. So, $h = \frac{V}{\pi r^2} = \frac{528 \times 10^{-3}}{\frac{22}{7} \times (0.4)^2} = \frac{528 \times 10^{-3} \times 7}{22 \times 0.16} = \frac{3696 \times 10^{-3}}{3.52} = 1.05 m = 105 cm$. The hole is made at $h_1 = 70 cm = 0.7 m$ below the top water level. The height

Q: What are the options for this this exam Physics question?

A. $120\sqrt{2}$ · B. $140\sqrt{2}$ · C. $140\sqrt{3}$ · D. $120\sqrt{3}$

Question

Q.34. A cylindrical vessel of 40 cm radius is completely filled with water and its capacity is 528 dm (dm: decimeter). The vessel is placed on a solid block of exactly same height as vessel. If a small hole is made at 70 cm below the top of water level, then horizontal range of water falling on the ground in the beginning is _______ cm.

FirstInTest · Accepted Answer

Given radius $r = 40 cm = 0.4 m$. Capacity $V = 528 dm^3 = 528 	imes 10^{-3} m^3$. The volume of a cylinder is $V = \pi r^2 h$. So, $h = \frac{V}{\pi r^2} = \frac{528 	imes 10^{-3}}{\frac{22}{7} 	imes (0.4)^2} = \frac{528 	imes 10^{-3} 	imes 7}{22 	imes 0.16} = \frac{3696 	imes 10^{-3}}{3.52} = 1.05 m = 105 cm$. The hole is made at $h_1 = 70 cm = 0.7 m$ below the top water level. The height of water above the hole is $h_1 = 0.7 m$. The height of the hole from the ground is $h_2 = h - h_1 = 1.05 m - 0.7 m = 0.35 m$. The velocity of efflux is $v = \sqrt{2gh_1} = \sqrt{2 	imes 10 	imes 0.7} = \sqrt{14} m/s$. The time taken for the water to fall to the ground is $t = \sqrt{\frac{2h_2}{g}} = \sqrt{\frac{2 	imes 0.35}{10}} = \sqrt{0.07} s$. The horizontal range $R = v t = \sqrt{14} 	imes \sqrt{0.07} = \sqrt{14 	imes 0.07} = \sqrt{0.98} m$. This is approximately $\sqrt{1} m = 1 m = 100 cm$. Let's recheck the calculation from the provided solution. The solution states $h = 1.05m = 105cm$. The hole is at $70cm$ below the top. So $h_1 = 70cm$. The height of the vessel is $105cm$. The vessel is placed on a solid block of exactly same height as vessel. This means the total height from the ground to the top of water is $2h = 2 	imes 105 = 210 cm$. The hole is $70 cm$ below the top of water. So the height of water above the hole is $h_1 = 70 cm$. The height of the hole from the ground is $H_{total} - h_1 = 210 - 70 = 140 cm$. This interpretation is different from the solution. Let's follow the solution's interpretation: $h = 105cm$. Hole at $70cm$ below top. So $h_1 = 70cm$. The height of the hole from the ground is $h_2 = h - h_1 = 105 - 70 = 35cm$. Velocity of efflux $v = \sqrt{2gh_1} = \sqrt{2 	imes 10 	imes 0.7} = \sqrt{14} m/s$. Time of flight $t = \sqrt{\frac{2h_2}{g}} = \sqrt{\frac{2 	imes 0.35}{10}} = \sqrt{0.07} s$. Range $R = v t = \sqrt{14} 	imes \sqrt{0.07} = \sqrt{0.98} m = 98 cm$. This is not among the options. Let's re-evaluate the solution's calculation: $t = \sqrt{\frac{2(105+35) 	imes 10^{-2}}{10}} = \sqrt{\frac{2 	imes 140 	imes 10^{-2}}{10}} = \sqrt{\frac{2.8}{10}} = \sqrt{0.28} s$. This implies the total height from which water falls is $105+35 = 140 cm$. This means the vessel is on a block of height $35cm$. The total height of the water level from the ground is $105+35 = 140cm$. The hole is $70cm$ below the top. So $h_1 = 70cm$. The height of the hole from the ground is $h_2 = 140 - 70 = 70cm$. So $v = \sqrt{2gh_1} = \sqrt{2 	imes 10 	imes 0.7} = \sqrt{14} m/s$. $t = \sqrt{\frac{2h_2}{g}} = \sqrt{\frac{2 	imes 0.7}{10}} = \sqrt{0.14} s$. $R = vt = \sqrt{14} 	imes \sqrt{0.14} = \sqrt{1.96} = 1.4 m = 140 cm$. This matches option (3) if it was $140\sqrt{2}$ or something. The solution states $R = ut = \sqrt{2 	imes 10 	imes 70 	imes 10^{-2}} 	imes \sqrt{28 	imes 10^{-1}} = (\sqrt{14}) (\sqrt{28}) 	imes 10^{-1} = 14\sqrt{2} 	imes 10^{-1} m = 1.4\sqrt{2} m = 140\sqrt{2} cm$. This matches option (2). The interpretation of the problem statement is key here. "The vessel is placed on a solid block of exactly same height as vessel." means the total height of the top of the water from the ground is $h_{vessel} + h_{block} = h + h = 2h$. So total height is $2 	imes 105 = 210 cm$. Hole is $70 cm$ below top. So $h_1 = 70 cm$. Height of hole from ground $h_2 = 210 - 70 = 140 cm$. $v = \sqrt{2gh_1} = \sqrt{2 	imes 10 	imes 0.7} = \sqrt{14} m/s$. $t = \sqrt{\frac{2h_2}{g}} = \sqrt{\frac{2 	imes 1.4}{10}} = \sqrt{0.28} s$. $R = vt = \sqrt{14} 	imes \sqrt{0.28} = \sqrt{14 	imes 0.28} = \sqrt{3.92} m = \sqrt{392} cm$. This is not matching. Let's stick to the solution's calculation logic. $h = 105 cm$. Hole at $70 cm$ below top. So $h_1 = 70 cm$. The height of the hole from the ground is $h_2 = 105 + 35 = 140 cm$ (this implies the block height is $35cm$, not $105cm$). This is confusing. Let's assume the solution's values for $h_1$ and $h_2$ are correct based on some interpretation. $h_1 = 70 cm = 0.7 m$. $h_2 = 105 + 35 = 140 cm = 1.4 m$. (This implies the block height is $35cm$, not the same as vessel height). $v = \sqrt{2gh_1} = \sqrt{2 	imes 10 	imes 0.7} = \sqrt{14} m/s$. $t = \sqrt{\frac{2h_2}{g}} = \sqrt{\frac{2 	imes 1.4}{10}} = \sqrt{0.28} s$. $R = vt = \sqrt{14} 	imes \sqrt{0.28} = \sqrt{3.92} = 1.979 m = 197.9 cm$. This is not matching. Let's use the solution's values directly: $h_1 = 70 cm$, $h_2 = 105+35 = 140 cm$. $v = \sqrt{2gh_1} = \sqrt{2 	imes 10 	imes 0.7} = \sqrt{14} m/s$. $t = \sqrt{\frac{2h_2}{g}} = \sqrt{\frac{2 	imes 1.4}{10}} = \sqrt{0.28} s$. $R = vt = \sqrt{14} 	imes \sqrt{0.28} = \sqrt{3.92} = 1.979 m = 197.9 cm$. The solution's calculation $R = (\sqrt{14})(\sqrt{28}) 	imes 10^{-1} = 14\sqrt{2} 	imes 10^{-1} m = 1.4\sqrt{2} m = 140\sqrt{2} cm$. This implies $h_1 = 140 cm$ and $h_2 = 280 cm$ (or some other combination that leads to $\sqrt{14}$ and $\sqrt{28}$ terms). Given the options, $140\sqrt{2}$ is option (2). The calculation in the solution seems to use $h_1 = 70 cm$ and $h_2 = 140 cm$. $v = \sqrt{2gh_1} = \sqrt{2 	imes 10 	imes 0.7} = \sqrt{14} m/s$. $t = \sqrt{\frac{2h_2}{g}} = \sqrt{\frac{2 	imes 1.4}{10}} = \sqrt{0.28} s$. $R = \sqrt{14} 	imes \sqrt{0.28} = \sqrt{3.92} = 1.979 m$. The solution's calculation is $R = \sqrt{2 	imes 10 	imes 70 	imes 10^{-2}} 	imes \sqrt{2 	imes (105+35) 	imes 10^{-2} / 10}$. This is $R = \sqrt{14} 	imes \sqrt{0.28} = \sqrt{3.92}$. The solution's $R = (\sqrt{14})(\sqrt{28}) 	imes 10^{-1}$ is incorrect based on the previous steps. Let's assume the answer is $140\sqrt{2}$ and the calculation is $R = \sqrt{2gh_1} \sqrt{2h_2/g} = \sqrt{4h_1h_2}$. If $h_1 = 70cm$ and $h_2 = 140cm$, then $R = \sqrt{4 	imes 0.7 	imes 1.4} = \sqrt{3.92} m$. This is not $140\sqrt{2}$. The solution's final step $R = (\sqrt{14})(\sqrt{28}) 	imes 10^{-1}$ is likely a typo and should be $R = \sqrt{14} 	imes \sqrt{0.28}$. Let's re-evaluate the problem statement. "The vessel is placed on a solid block of exactly same height as vessel." So total height of water from ground is $H_{total} = h_{vessel} + h_{block} = h + h = 2h = 2 	imes 105 = 210 cm$. Hole is $70 cm$ below top. So $h_1 = 70 cm$. Height of hole from ground is $h_2 = H_{total} - h_1 = 210 - 70 = 140 cm$. $v = \sqrt{2gh_1} = \sqrt{2 	imes 10 	imes 0.7} = \sqrt{14} m/s$. $t = \sqrt{\frac{2h_2}{g}} = \sqrt{\frac{2 	imes 1.4}{10}} = \sqrt{0.28} s$. $R = vt = \sqrt{14} 	imes \sqrt{0.28} = \sqrt{3.92} m = 197.9 cm$. This is not matching any option. Given the solution's final answer is $140\sqrt{2}$ cm, let's work backward. If $R = 140\sqrt{2} cm = 1.4\sqrt{2} m$. Then $R^2 = (1.4\sqrt{2})^2 = 1.96 	imes 2 = 3.92 m^2$. We have $R = \sqrt{4h_1h_2}$. So $4h_1h_2 = 3.92$. $h_1h_2 = 0.98$. If $h_1 = 0.7 m$, then $h_2 = 0.98/0.7 = 1.4 m$. This means $h_1 = 70 cm$ and $h_2 = 140 cm$. This is consistent with the interpretation that the total height of the water level from the ground is $210 cm$, and the hole is $70 cm$ below the top, making its height from the ground $140 cm$. So the calculation $R = \sqrt{4h_1h_2} = \sqrt{4 	imes 0.7 	imes 1.4} = \sqrt{3.92} m = 1.4\sqrt{2} m = 140\sqrt{2} cm$. This matches option (2). The solution's intermediate steps are a bit confusing, but the final result $140\sqrt{2}$ is consistent with the problem statement if interpreted as $h_1 = 70 cm$ and $h_2 = 140 cm$.

Answer

$120\sqrt{2}$

Answer

$140\sqrt{2}$

Answer

$140\sqrt{3}$

Answer

$120\sqrt{3}$

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