Physics 9702 Paper 41 Oct/Nov 2025: Q9-10 Explained

[00:00:00]Hey guys, uh it's IBM. I'm explaining paper 4's October November 2025 physics 9702 paper 41 stroke 43. This is going to be part five of this paper 41 stroke 43. I think in part four we stopped on question eight. So in part five want us to start from question nine.

[00:00:28]state Wayne's displacement law. Of course, we know that Wayne's displacement law states that lambda max is inversely proportional to whether where lambda max is the wavelength at peak intensity or wavelength at which maximum emission rate occurs and it is the absolute temperature or the thermodynamic temperature. So the wavelength at maximum wavelength at which maximum emission rate maximum emission rate from a occurs is inversely proportional to the thermodynamic thermodynamic temperature. A lot of students confuse lambda max to be maximum wavelength but the wavelength at which maximum intensity occurs or the wavelength at which maximum emission occurs or the wavelength at peak intensity.

[00:02:05]Figure 9.1 shows the variation of with d² d minus 2 of the radant flux intensity. Remember we know that f = l / 4<unk>i d². So it means f = l / 4<unk>i * dus 2 where d is the distance of the observer from the star. The figure 9.2 shows the variation with wavelength lambda of the rates of emission P of rad radiation by star X and the sun.

[00:02:41]The surface temperature of the sun is 5770 Kelvin.

[00:02:46]State three conclusions about star X that can be drawn from this data. The conclusions might be may be qualitative that means you can quote a number or quantitative sorry qualitative just explain or quantitative where you can quote a number use the space for any working so using displacement law we know that lambda max time temperature equals to a constant if I consider the sun it means um the wavelength of of star X from here star X peak wavelength is occurring at I think this is 0.4 I mean 4.45 so 4.45 no 4.5 not 4.45 45 4.5 So 4.5 * 10 ^ - 7 then times the temperature of X should be the same as the wavelength for the sun.

[00:03:55]The peak wavelength the peak at which maximum intensity occurs for the sun is 5.5. So this should be equal to 5.5 * 10 ^ - 7 * the temperature of the sun which is 5770.

[00:04:11]This means that so this implies that temperature of x is going to be equal to I'll just simply say 5.5 / 4.5 * 5770.

[00:04:29]So that is 7050 kel. So that is 7050 kel. So I'll say surface temperature of the star X surface temperature of star X equals to 7050 Kelvin.

[00:04:55]Then um instead of saying that you would say star X has a higher temperature than star than the sun. We would say star X has um a higher high temperature than the sun.

[00:05:23]You would even just sit from here without calculation. Wavelength is inversely proportional to temperature.

[00:05:29]You see that the wavelength of star the wavelength of star X is smaller than the wavelength of star of the sun. So it means wavelength has smaller wavelength means higher temperature than the sun.

[00:05:45]Then the other one you can talk about the luminosity. Let's check the luminosity.

[00:05:53]We know that um luminosity or radian flux intensity from the formula for radian flux intensity luminosity is going to be equal to f * 4 pi / dus 2.

[00:06:13]That's the luminosity 4 pi * f d minus 2 which is actually the gradient not the gradient 4<unk> * f dus 2 and this would be equal to f was given I think this is 6.4.

[00:06:50]So that is 6.4 4 * 10 ^ of 3 10 * 4 pi its corresponding value of dz3 * d minus 3 * 10 ^ of - 23 I'm just I just made this the sub l the subject from that formula there to check the luminosity So 6.4 exponent 3 * 4<unk>i / 3 exponent - 23.

[00:07:36]So I can say the luminosity of the star x is equal to 2.68 * 10 ^ 27.

[00:07:49]So I can sell star X is 2. 68 * 10 ^ 27 W or because luminosity is directly proportional because luminosity is directly proportional. We know that luminosity is equal to 4 pi sigma r² t ^ 4. We know that luminosity is directly proportional to temperature to the power 4. And we know that star x has a higher temperature than than the sun. So it means you can just say luminosity of x is greater than luminosity of of the sun because the temperature of the of of star x is greater. So we can instead you just say star x has a higher luminosity than the sun.

[00:09:07]and then um the last one maybe you can talk about the radius from this expression here.

[00:09:14]You can find the radius of star x star x from this expression. Since I have I know the luminosity then the radius is going to be the square root of the luminosity which is 2.68 * 10 ^ 27 / 4<unk>i stefan bosman's constant is 5.67 67 * 10 ^ -8 then temperature which is 750 raised to ^ 4.

[00:09:49]So we have 4<unk> 5.67 6 7 exponent - 8 * uh 70 50 power 4.

[00:10:07]Then we have 2.68 exponent 27ide by the answer.

[00:10:15]Then root of the answer that's 1.23 23 * 10 ^ of 369 1.23 * 10 ^ of 9 m. So I can set the radius of start x = to 1.23 * 10 ^ of 9 m.

[00:10:46]Okay. So that is question 90. I think it's still continuing.

[00:10:50]Star X is in a galaxy that is moving away. So it's wavelength must be really shifted because it's moving away. It should be red shifted.

[00:11:00]So just with the reason how line of star X in the figure 9.2 would appear differently if it had been obtained from data measured on the Earth. So because it's moving away from the Earth, it's more already shifted. to light light from start X is already shifted is already shifted. So wavelength at which uh peak emission occurs or maximum emission wavelength at which peak emission rate occurs like that. Which peak emission rate occurs would be greater would be greater using observed data.

[00:12:24]So using observed data from the earth because the light is already shifted means the observed wavelength is going to be greater. Wave length at which peak emission rate occurs is going to be greater using the data observed from the earth. Okay. Question 10.

[00:12:42]Define specific acostic impedance that is zed is equal to density time c. So which is the product of um the density of a medium and the speed and the speed of ultrasound.

[00:13:22]in the medium product of the density of the medium and the speed of ultrasound in the medium that is acostic impedance. Explain how ultrasound waves are detected by P a piso electric crystal. Remember the same transducer which produces the ultrasound is the same transducer which picks them. So the ultrasound which is incident of the transducer maybe from the body um makes the crystal of the transducer to vibrate. The vibrations of the crystal are the ones which create an alternate aid across the crystal which is detected.

[00:14:04]So ultrasound WS instant on the crystal causes the crystal to vibrate.

[00:14:38]causes the crystal to vibrate. The vibrations of the crystal cause an induced EMF.

[00:15:05]across.

[00:15:11]So the vibration will cause an an EMF to be induced across the crystal or an EMF to be set up across the crystal. which EMF is what is detected by electronic circuits as the ultrasound which has been which which is incident on the transducer again after coming from the body. The table shows the specific acoustic impedance Z for body tissue water and steel.

[00:15:38]Capture the intensity reflection coefficient that is called alpha for ultrasound incident on a water steel boundary. So we know that alpha = to z 2 - z1 / z 2 + z1 and this should be squared that is the intensity reflection coefficient. So this is going to be let's say this is z 2 this is z this is z1 so this is going to be 4.04 * 10 ^ 7 - 1.48 * 10 ^ 6 / 4.04 04 * 10 ^ 7 + 1.48 * 10 ^ of 6 that this should be squared.

[00:16:28]So uh this is of course the same as if I want to ignore the power of six I'll just say 4.04 04 + 0.148 divide by 4.04 sorry I was supposed to say 4.04 minus 0.148 if I change to the power of 7 divide by 4.04 04 plus uh 0.148.

[00:17:10]So this is squared 0.864 that is to three significant figures. So the intensity reflection because they quoted everything here to three significant figures. So it will be wise for me to go to 0.864.

[00:17:35]Explain without calculation what is likely to happen when ultrasound is instant on a body tissue water boundary. Tissue water boundary. Now when you look at the um zed for body tissue and zed for water they approximately the same. So Z for water is approximately the same as Z for body tissue. Which means when I substitute in alpha equals to of course intensity reflected over the incident intensity this is going to be um approximately equal to zero because when you substitute if Z water is the same as Z uh Z water is the same as Z tissue.

[00:18:21]Z what is the same as Z tissue? It means that when you substitute in the equation Z 2 - Z 1 / Z2 + Z1 you get approximately zero. So which means intensity reflected is approximately zero. And because intensity transmitted over the incident is equal to 1us alpha. It means intensity transmitted approximately equal to one.

[00:18:45]that would mean more transmission um more transmission less reflection. So we can say Z values Z values or acostic impedance values are very similar are very similar.

[00:19:11]Almost none of the ultrasound will be reflected.

[00:19:32]So almost all the ultra sound will be transmitted through the boundary.

[00:20:06]Okay, I think that marks the end of this paper. See you in paper 42. Bye-bye.