A 5.0-m-diameter solid aluminum sphere is launched into space. By how much does its diameter increase? Give your answer in μm.

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Hi everyone. In this practice problem, we're being asked to calculate the change in length of a cube where we'll have a solid copper cube with a length of 4.6 m moved from the surface of Venus into space. We're being asked to calculate the change in length in micrometer of the cube. If the pressure at the surface of Venus is 93 ATM. The options given are a 310 micrometer B 1000 and 420 micrometer C 1810 micrometer and lastly d 394 micrometer. So the air pressure on P fenu will introduce stress on the volume of the copper cube which is going to be re remove removed once the cube is in space. So as the pressure decreases, the length of the cube will increase and as the pressure increases, the length of the cube will decrease. So pressure, if you remember is uh given by P equals to F divided by A where F is the force and A is the area at which the force is applied to. Therefore, there is an inverse relationship between the pressure and the length of the cube. However, at the same time, pressure can also be given as B equals to B multiplied by delta V divided by V where B is the bulk modular V is the volume and delta V is the change in volume. So using the second formula, we can get an equation by rearranging this, that delta V will equals to V multiplied by B divided by B or bach modulus. We know for a cube that V will equals to L cube or the cube of one of its uh sites length, which will mean that delta V will equals to delta L multiplied by L squared. In this case, we wanna re uh we wanna substitute this two equation into our rearranged delta V equation so that we will get delta V equals to delta L multiplied by L squared equals to L Q multiplied by B divided by B just like. So we can simplify this by taking or crossing the LS out. So that we have an equation for delta L to then just be L multiplied by B divided by B just like. So, so the bulk modules of copper B for copper will equals to 1.4 times 10 to the power of 11 pass CALS. And we know previously that the pressure at the surface of Venus or B is going to be nine P three E T M which we have to convert into past CALS by multiplying this with uh one point 01, 325 times 10 to the power of five pascals divide that by one ATM. And that will essentially give us B to be equals to 9.4, 23 times 10 to the power of six pascals. So now we have B, we have B and we know what the original length is, which is L. So we can actually start calculating delta L. So then using the equation that we have data previously, delta L will equals to L multiplied by B divided by B L is given in the problem statement to be 4.6 m. And then the pressure is 9.4 to 3 times 10 to the power of six pascals. And lastly B or the above modules is 1.4 times 10 to the power of 11 pascals for copper. They'll give us a delta L value of 3.1 times 10 to the power of negative four m are essentially uh converting that into micrometer. Delta L will then equals to 3.1 times 10 to the power of negative four m times 10 to the power of six micrometer for every one m. So divide that by one m and that will give us micrometer just like. So, so delta L or the change in length of the cube is going to be 310 micrometer which will actually correspond to option A in our answer choices. So answer or option A is going to be the answer to this particular practice problem and that'll be it for this video. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and that'll be it for this one. Thank you.

A 5.0-m-diameter solid aluminum sphere is launched into space. By how much does its diameter increase? Give your answer in μm.

Verified Solution

Key Concepts

Thermal Expansion

Coefficient of Linear Expansion

Conversion to Micrometers