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LuckyWell [14K]
4 years ago
5

Two drums of the same size and same height are taken.

Physics
1 answer:
Gelneren [198K]4 years ago
3 0

Answer:

i) The pressure acting on the base of <em>B</em> will be half the pressure acting on the base of <em>A</em>

ii) The pressure acting on the base of <em>B</em> will be the same as the pressure acting on the base of <em>A</em>

iii) The pressure on the base of drum <em>A</em> will be slightly less than the pressure on the base of drum <em>B</em>

Explanation:

The pressure acting on the base of the drum, P = h·ρ·g

Where;

h = The level of the liquid in the drum

h_{max} = The height of the drums

ρ = The density of the liquid in the drum

g = The acceleration due to gravity ≈ 9.81 m/s²

i) If <em>A</em> is completely filled, we have h_A = h_{max}

Therefore, P_A = h_{max}×\rho_{liquid}×g

If <em>B</em> is half filled, we have, h_B =  (1/2)·h_{max}

P_B = (1/2) × h_{max}×\rho_{liquid}×g

Therefore, P_B = (1/2) × P_A

The pressure acting on the base of <em>B</em> will be half the pressure acting on the base of <em>A</em>

ii) If both <em>A</em> and <em>B</em> are each filled with water (the same liquid), then the pressure on their bases will be P_A = h_{max}×\rho_{water}×g = P_B, the same, given that the acceleration due to gravity, <em>g</em>, is constant and the same in Nepal and India

iii) If <em>A</em> is filled with water, and <em>B</em> is filled with salty water, we have that, the density of salty water is slightly higher than water, therefore, we get;

P_A = h_{max}×\rho_{water}×g <  P_B =

The pressure on the base of drum <em>A</em> will be less than the pressure on the base of drum <em>B.</em>

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Explanation:

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An aluminum wire is held between two clamps under zero tension at room temperature. Reducing the temperature, which results in a
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Answer:

\frac{\Delta L}{L} =5.37\times 10^{-4}

Explanation:

Given:

  • cross sectional area of the wire, A=5.75\times 10^{-6}\ m^2
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  • young's modulus of the material, E=7\times 10^{10}\ N.m^{-2}
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<u>We have mathematical expression for strain as:</u>

\frac{\Delta L}{L} =\frac{\sigma}{E} ...............................(1)

and since, \sigma =\frac{T}{A}

where, T = tension force in the wire

equation (1) becomes:

\frac{\Delta L}{L} =\frac{T}{A.E} ............................(2)

<u>Also velocity ofwave in tensed wire:</u>

v=\sqrt{\frac{T}{\mu} } ...................................(3)

where: \mu= linear mass density of the wire

\therefore \mu=\rho \times A

Now, equation (3) becomes

v=\sqrt{\frac{T}{\rho \times A} }

T=v^2.\rho \times A ............................(4)

Using eq. (2) & (4) for tension T

v^2.\rho \times A=A.E\times \frac{\Delta L}{L}

\frac{\Delta L}{L} =\frac{v^2.\rho}{E}

putting the respective values

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The volume of the sample is 48.39 cm³.

<h3 /><h3>What is volume?</h3>

The term volume refers to the amount of three-dimensional space taken up by an item or a closed surface

To calculate the volume of the sample, we use the formula below

Formula:

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Where:

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From the question,

Given:

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Substitute these values into equation 1

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Hence, the volume of the sample is 48.39 cm³.

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