The pressure in a water tap at the base of a building is 3×106dyne cm−2 and on its top it is 1.6×106dyne cm−2 and on its top building is approximately

Pbase−Ptop=hρwg ⇒ 3-1.6×106=h(1)(1000)⇒ h=1.4×1061000cm⇒ h=1400 cm⇒ h=14 m

The pressure in a water tap at the base of a building is 3×106dyne cm−2 and on its top it is 1.6×106dyne cm−2 and on its top building is approximately

Pbase−Ptop=hρwg ⇒ 3-1.6×106=h(1)(1000)⇒ h=1.4×1061000cm⇒ h=1400 cm⇒ h=14 m

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The pressure in a water tap at the base of a building is $3 \times 10^{6} d y n e c m^{- 2}$ and on its top it is $1.6 \times 10^{6} d y n e c m^{- 2}$ and on its top building is approximately

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7 m

14 m

70 m

140 m

answer is B.

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Detailed Solution

$P_{base} - P_{top} = h ρ_{w} g$

$\Rightarrow$ $(3 - 1.6) \times 10^{6} = h (1) (1000)$

$\Rightarrow$ $h = \frac{1.4 \times 10^{6}}{1000} c m$

$\Rightarrow$ $h = 1400 c m$

$\Rightarrow$ $h = 14 m$

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The pressure in a water tap at the base of a building is 3×106dyne cm−2 and on its top it is 1.6×106dyne cm−2 and on its top building is approximately

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The pressure in a water tap at the base of a building is $3 \times 10^{6} d y n e c m^{- 2}$ and on its top it is $1.6 \times 10^{6} d y n e c m^{- 2}$ and on its top building is approximately