The tank has a volume of
, where
is its height and
is its radius.
At any point, the water filling the tank and the tank itself form a pair of similar triangles (see the attached picture) from which we obtain the following relationship:
![\dfrac26=\dfrac rh\implies r=\dfrac h3](https://tex.z-dn.net/?f=%5Cdfrac26%3D%5Cdfrac%20rh%5Cimplies%20r%3D%5Cdfrac%20h3)
The volume of water in the tank at any given time is
![V=\dfrac\pi3r^2h](https://tex.z-dn.net/?f=V%3D%5Cdfrac%5Cpi3r%5E2h)
and can be expressed as a function of the water level alone:
![V=\dfrac\pi3\left(\frac h3\right)^2h=\dfrac\pi{27}h^3](https://tex.z-dn.net/?f=V%3D%5Cdfrac%5Cpi3%5Cleft%28%5Cfrac%20h3%5Cright%29%5E2h%3D%5Cdfrac%5Cpi%7B27%7Dh%5E3)
Implicity differentiating both sides with respect to time
gives
![\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9h^2\,\dfrac{\mathrm dh}{\mathrm dt}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20dV%7D%7B%5Cmathrm%20dt%7D%3D%5Cdfrac%5Cpi9h%5E2%5C%2C%5Cdfrac%7B%5Cmathrm%20dh%7D%7B%5Cmathrm%20dt%7D)
We're told the water level rises at a rate of
at the time when the water level is
, so the net change in the volume of water
can be computed:
![\dfrac{\mathrm dV}{\mathrm dt}=\dfrac\pi9(200\,\mathrm{cm})^2\left(20\,\dfrac{\rm cm}{\rm min}\right)=\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20dV%7D%7B%5Cmathrm%20dt%7D%3D%5Cdfrac%5Cpi9%28200%5C%2C%5Cmathrm%7Bcm%7D%29%5E2%5Cleft%2820%5C%2C%5Cdfrac%7B%5Crm%20cm%7D%7B%5Crm%20min%7D%5Cright%29%3D%5Cdfrac%7B800%2C000%5Cpi%7D9%5C%2C%5Cdfrac%7B%5Cmathrm%7Bcm%7D%5E3%7D%7B%5Crm%20min%7D)
The net rate of change in volume is the difference between the rate at which water is pumped into the tank and the rate at which it is leaking out:
![\dfrac{\mathrm dV}{\mathrm dt}=(\text{rate in})-(\text{rate out})](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20dV%7D%7B%5Cmathrm%20dt%7D%3D%28%5Ctext%7Brate%20in%7D%29-%28%5Ctext%7Brate%20out%7D%29)
We're told the water is leaking out at a rate of
, so we find the rate at which it's being pumped in to be
![\dfrac{800,000\pi}9\,\dfrac{\mathrm{cm}^3}{\rm min}=(\text{rate in})-10,500\,\dfrac{\mathrm{cm}^3}{\rm min}](https://tex.z-dn.net/?f=%5Cdfrac%7B800%2C000%5Cpi%7D9%5C%2C%5Cdfrac%7B%5Cmathrm%7Bcm%7D%5E3%7D%7B%5Crm%20min%7D%3D%28%5Ctext%7Brate%20in%7D%29-10%2C500%5C%2C%5Cdfrac%7B%5Cmathrm%7Bcm%7D%5E3%7D%7B%5Crm%20min%7D)
![\implies\text{rate in}\approx289,753\,\dfrac{\mathrm{cm}^3}{\rm min}](https://tex.z-dn.net/?f=%5Cimplies%5Ctext%7Brate%20in%7D%5Capprox289%2C753%5C%2C%5Cdfrac%7B%5Cmathrm%7Bcm%7D%5E3%7D%7B%5Crm%20min%7D)
B. Universal Gravitational Law
Answer:
Power dissipated in resistor 532 ohm is 0.503 watt
Explanation:
We have given in first case resistance ![R_1=206ohm](https://tex.z-dn.net/?f=R_1%3D206ohm)
Power dissipated in this resistance is ![P_1=1.30watt](https://tex.z-dn.net/?f=P_1%3D1.30watt)
Power dissipated in the resistor is equal to ![P=\frac{v_{rms}}^2{R}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7Bv_%7Brms%7D%7D%5E2%7BR%7D)
We have to find the power dissipated in the resistor is 1.30 watt
From the relation we can say that ![\frac{P_1}{P_2}=\frac{R_2}{R_1}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1%7D%7BP_2%7D%3D%5Cfrac%7BR_2%7D%7BR_1%7D)
![\frac{1.3}{P_2}=\frac{532}{206}](https://tex.z-dn.net/?f=%5Cfrac%7B1.3%7D%7BP_2%7D%3D%5Cfrac%7B532%7D%7B206%7D)
![P_2=0.503watt](https://tex.z-dn.net/?f=P_2%3D0.503watt)
So power dissipated in resistor 532 ohm is 0.503 watt