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Sati [7]
3 years ago
12

A pressure of 7x10^5N/m is applied to all surfaces of a copper cube (of sides 25 cm) what is the fractional change in volume of

a cube? ( for copper B= 14x10^10N/m)
Physics
1 answer:
AnnyKZ [126]3 years ago
5 0

Answer:

The correct solution is "5\times 10^{-4} %".

Explanation:

The given values are:

Pressure,

\Delta P=7\times 10^5 \ N/m

for copper,

B=14\times 10^{10} \ N/m

As we know,

The Bulk Modulus (B) = \frac{\Delta P}{-\frac{\Delta V}{V} }

or,

The decrease in volume will be:

= (\frac{\Delta V}{V})\times 100 \ percent

then,

= \frac{\Delta P}{B}\times 100 \ percent

On putting the values, we get

= \frac{7\times 10^5}{14\times 10^{10}}\times 100 \ percent

= 5\times 10^{-4} \ percent

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Answer:

f = 12 cm

Explanation:

<u>Center of Curvature</u>:

The center of that hollow sphere, whose part is the spherical mirror, is known as the ‘Center of Curvature’ of  mirror.

<u>The Radius of Curvature</u>:

The radius of that hollow sphere, whose part is the spherical mirror, is known as the ‘Radius of Curvature’ of  mirror. It is the distance from pole to the center of curvature.

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The focal length of the spherical (concave) mirror is approximately equal to half of the radius of curvature:

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Therefore,

f = \frac{24\ cm}{2}

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