A cylindrical vessel of 40 cm radius is completely filled with water and its capacity is 528 (dm: decimeter).

Question

A cylindrical vessel of 40 cm radius is completely filled with water and its capacity is 528 ${dm}^{3}$ (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. [2026]

Smart Achievers · Accepted Answer

(2)

$V = π r^{2} h \Rightarrow 528 \times 10^{- 3} = π {(0.4)}^{2} h$

$h = \frac{528 \times 10^{- 3} \times 7}{{(0.4)}^{2} \times 22} = 1.05 m = 105 cm$

$\Rightarrow t = \sqrt{\frac{2 H}{g}} = \sqrt{\frac{2 \times (105 + 35) \times 10^{- 2}}{10}}$

$\Rightarrow t = \sqrt{\frac{2 (140) \times 10^{- 2}}{10}} = \sqrt{28} \times 10^{- 1} sec$

$R = u t = \sqrt{2 \times 10 \times 70 \times 10^{- 2}} \times \sqrt{28} \times 10^{- 1}$

$\Rightarrow R = 140 \sqrt{2} cm$

Solution

1 $120 \sqrt{2}$

2 $140 \sqrt{2}$

3 $140 \sqrt{3}$

4 $120 \sqrt{3}$

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Solution

1 1202

2 1402

3 1403

4 1203

Ans. (2) V=πr2h⇒528×10-3=π(0.4)2h h=528×10-3×7(0.4)2×22=1.05 m=105 cm ⇒t=2Hg=2×(105+35)×10-210 ⇒t=2(140)×10-210=28×10-1 sec R=ut=2×10×70×10-2×28×10-1 ⇒R=1402 cm

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1 $120 \sqrt{2}$

2 $140 \sqrt{2}$

3 $140 \sqrt{3}$

4 $120 \sqrt{3}$