I have a container that is 20cm tall and filled fully with water. The lift accelerates up at 10 $m / s^{2}$ . What is the pressure difference between the top surface of the water and the bottom of the container(in pascal)? Take density of water as 1kg/lit.

2 \times 10^{4}

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8 \times 10^{4}

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4 \times 10^{4}

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16 \times 10^{4}

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Solution

The correct option is C $4 \times 10^{4}$
Since the container is accelerating up, effective $g$ will be
$g^{'} = g + a = 10 m / s^{2} + 10 m / s^{2} = 20 m / s^{2}$

The pressure difference from top to bottom will be
$h ρ g^{'} = 0.2 \times 1000 \times 20$
$= 4 \times 10^{4} P a$

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I have a container that is 20cm tall and filled fully with water. The lift accelerates up at 10 m/s2. What is the pressure difference between the top surface of the water and the bottom of the container(in pascal)? Take density of water as 1kg/lit.

The correct option is C 4×104 Since the container is accelerating up, effective g will be g′=g+a=10 m/s2+10 m/s2=20 m/s2 The pressure difference from top to bottom will be hρg′ = 0.2×1000×20 =4×104 Pa

I have a container that is 20cm tall and filled fully with water. The lift accelerates up at 10 $m / s^{2}$ . What is the pressure difference between the top surface of the water and the bottom of the container(in pascal)? Take density of water as 1kg/lit.

The correct option is C $4 \times 10^{4}$
Since the container is accelerating up, effective $g$ will be
$g^{'} = g + a = 10 m / s^{2} + 10 m / s^{2} = 20 m / s^{2}$

The pressure difference from top to bottom will be
$h ρ g^{'} = 0.2 \times 1000 \times 20$
$= 4 \times 10^{4} P a$