A siphon in use is demonstrated in the following figure. The density of the liquid flowing in siphon is $1.5 g m / c c$ . The pressure difference between the point P and S will be

10^{5} N / m

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2 \times 10^{5} N / m

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Zero

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Infinity

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Solution

The correct option is C Zero
Answer is C.
As the both points are at the surface of liquid and these points are in the open atmosphere. So both point possess similar pressure and equal to $1 a t m$ .
Hence the pressure difference will be zero.

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A siphon in use is demonstrated in the following figure. The density of the liquid flowing in siphon is 1.5gm/cc. The pressure difference between the point P and S will be

The correct option is C ZeroAnswer is C.As the both points are at the surface of liquid and these points are in the open atmosphere. So both point possess similar pressure and equal to 1atm. Hence the pressure difference will be zero.

A siphon in use is demonstrated in the following figure. The density of the liquid flowing in siphon is $1.5 g m / c c$ . The pressure difference between the point P and S will be

The correct option is C Zero
Answer is C.
As the both points are at the surface of liquid and these points are in the open atmosphere. So both point possess similar pressure and equal to $1 a t m$ .
Hence the pressure difference will be zero.