In a Calcutta laboratory, the height of the mercury column in a barometer was recorded as 75 cm . Determine this pressure in SI and CGS units using the following data: specific gravity of mercury = 13.6 , density of

Correct answer: A. 10 × 10^4 N/m^2 , 10 × 10^5 dyne/cm^2 Given h = 75 cm = 0.75 m Density of mercury, = 13.6 × 10^3 kg/m^3 Acceleration due to gravity, g = 9.8 m/s^2 In SI units, the pressure is given by: P = h g = 0.75 × 13.6 × 10^3 × 9.8 P = 99.96 × 10^3 N/m^2 10 × 10^4 N/m^2 In CGS units: h = 75 cm = 13.6 g/cm^3 g = 9.8 m/s^2 = 980 cm/s^2 P = h g = 75 × 13.6 × 980 P = 999600 dyne/cm^2 10 × 10^5 dyne/cm^2

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Introduction to Physics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Introduction to Physics

In a Calcutta laboratory, the height of the mercury column in a...

In a Calcutta laboratory, the height of the mercury column in a barometer was recorded as

75 \text{ cm}

. Determine this pressure in SI and CGS units using the following data: specific gravity of mercury

= 13.6

, density of water

= 10^3 \text{ kg/m}^3

, and

g = 9.8 \text{ m/s}^2

at Calcutta. Use the formula Pressure

= h\rho g

in usual symbols.

Options

A $10 \times 10^4 \text{ N/m}^2$ , $10 \times 10^5 \text{ dyne/cm}^2$
B $10 \times 10^4 \text{ N/m}^2$ , $10 \times 10^4 \text{ dyne/cm}^2$
C $1.0 \times 10^4 \text{ N/m}^2$ , $1.0 \times 10^5 \text{ dyne/cm}^2$
D $10 \times 10^5 \text{ N/m}^2$ , $10 \times 10^6 \text{ dyne/cm}^2$

Correct answer

A. $10 \times 10^4 \text{ N/m}^2$ , $10 \times 10^5 \text{ dyne/cm}^2$

Step-by-step solution

Given

h = 75 \text{ cm} = 0.75 \text{ m}

Density of mercury,

\rho = 13.6 \times 10^3 \text{ kg/m}^3

Acceleration due to gravity,

g = 9.8 \text{ m/s}^2

In SI units, the pressure is given by:

P = h\rho g = 0.75 \times 13.6 \times 10^3 \times 9.8

P = 99.96 \times 10^3 \text{ N/m}^2 \approx 10 \times 10^4 \text{ N/m}^2

In CGS units:

h = 75 \text{ cm}

\rho = 13.6 \text{ g/cm}^3

g = 9.8 \text{ m/s}^2 = 980 \text{ cm/s}^2

P = h\rho g = 75 \times 13.6 \times 980

P = 999600 \text{ dyne/cm}^2 \approx 10 \times 10^5 \text{ dyne/cm}^2

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