Voltage must decrease by a factor of 3 to reduce the current in the resistor by a factor of 3
Answer: Option A
<u>Explanation:</u>
According to ohms law, the voltage passed in a circuit will be directly proportional to the current flow in the circuit. So the resistance of the resistor will act as proportionality constant in ohms law. Let us consider V as original voltage producing a current I in resistor of resistance R. Then according to ohms law:
![V=I \times R](https://tex.z-dn.net/?f=V%3DI%20%5Ctimes%20R)
Now the revised current flow in the circuit is reduced by a factor of 3, So the new current is
in the same resistance R, then the new voltage will be
![V^{\prime}=\frac{I \times R}{3}=\frac{V}{3}](https://tex.z-dn.net/?f=V%5E%7B%5Cprime%7D%3D%5Cfrac%7BI%20%5Ctimes%20R%7D%7B3%7D%3D%5Cfrac%7BV%7D%7B3%7D)
So, the voltage must be reduced by a factor of 3 to reduce the current in the resistor by a factor of 3.
Answer:
The DNA in bacteria exists as unpacked blobs.
The DNA in animals and plants are neatly packed.
<h3>EXPLAMATION :</h3>
I don't really have an explanation.
The DNA in animals and plants are neatly packed into chromosomes, whereas the DNA in bacteria exists as an unpacked blob in the cell
Chefs change the way they cook and their spices
Answer:
The rate of transfer of heat is 0.119 W
Solution:
As per the question:
Diameter of the fin, D = 0.5 cm = 0.005 m
Length of the fin, l =30 cm = 0.3 m
Base temperature, ![T_{b} = 75^{\circ}C](https://tex.z-dn.net/?f=T_%7Bb%7D%20%3D%2075%5E%7B%5Ccirc%7DC)
Air temperature, ![T_{infty} = 20^{\circ}](https://tex.z-dn.net/?f=T_%7Binfty%7D%20%3D%2020%5E%7B%5Ccirc%7D)
k = 388 W/mK
h = ![20\ W/m^{2}K](https://tex.z-dn.net/?f=20%5C%20W%2Fm%5E%7B2%7DK)
Now,
Perimeter of the fin, p = ![\pi D = 0.005\pi \ m](https://tex.z-dn.net/?f=%5Cpi%20D%20%3D%200.005%5Cpi%20%5C%20m)
Cross-sectional area of the fin, A = ![\frac{\pi}{4}D^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%7D%7B4%7DD%5E%7B2%7D)
A = ![\frac{\pi}{4}(0.5\times 10^{-2})^{2} = 6.25\times 10^{- 6}\pi \ m^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%7D%7B4%7D%280.5%5Ctimes%2010%5E%7B-2%7D%29%5E%7B2%7D%20%3D%206.25%5Ctimes%2010%5E%7B-%206%7D%5Cpi%20%5C%20m%5E%7B2%7D)
To calculate the heat transfer rate:
![Q_{f} = \sqrt{hkpA}tanh(ml)(T_{b} - T_{infty})](https://tex.z-dn.net/?f=Q_%7Bf%7D%20%3D%20%5Csqrt%7BhkpA%7Dtanh%28ml%29%28T_%7Bb%7D%20-%20T_%7Binfty%7D%29)
where
![m = \sqrt{\frac{hp}{kA}} = \sqrt{\frac{20\times 0.005\pi}{388\times 6.25\times 10^{- 6}\pi}} = 41.237](https://tex.z-dn.net/?f=m%20%3D%20%5Csqrt%7B%5Cfrac%7Bhp%7D%7BkA%7D%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B20%5Ctimes%200.005%5Cpi%7D%7B388%5Ctimes%206.25%5Ctimes%2010%5E%7B-%206%7D%5Cpi%7D%7D%20%3D%2041.237)
Now,
![Q_{f} = \sqrt{20\times 388\times 0.005\pi\times 6.25\times 10^{- 6}\pi}tanh(41.237\times 0.3)(75 - 20) = 0.119\ W](https://tex.z-dn.net/?f=Q_%7Bf%7D%20%3D%20%5Csqrt%7B20%5Ctimes%20388%5Ctimes%200.005%5Cpi%5Ctimes%206.25%5Ctimes%2010%5E%7B-%206%7D%5Cpi%7Dtanh%2841.237%5Ctimes%200.3%29%2875%20-%2020%29%20%3D%200.119%5C%20W)