Explanation:
Given that,
Length of gold wire, l = 4 m
Voltage of battery, V = 1.5 V
Current, I = 4 mA
The resistivity of gold, ![\rho=2.44\times 10^{-8}\ \Omega-m](https://tex.z-dn.net/?f=%5Crho%3D2.44%5Ctimes%2010%5E%7B-8%7D%5C%20%5COmega-m)
Resistance in terms of resistivity is given by :
![R=\dfrac{\rho l}{A}](https://tex.z-dn.net/?f=R%3D%5Cdfrac%7B%5Crho%20l%7D%7BA%7D)
Also, V = IR
So,
![\dfrac{V}{I}=\dfrac{\rho l}{A}](https://tex.z-dn.net/?f=%5Cdfrac%7BV%7D%7BI%7D%3D%5Cdfrac%7B%5Crho%20l%7D%7BA%7D)
A is area of wire,
, r is radius, r = d/2 (diameter=d)
![\dfrac{V}{I}=\dfrac{\rho l}{\pi (d/2)^2}\\\\\dfrac{V}{I}=\dfrac{4\rho l}{\pi d^2}\\\\d=\sqrt{\dfrac{4\rho l I}{V\pi}} \\\\d=\sqrt{\dfrac{4\times 2.44\times 10^{-8}\times 4\times 4\times 10^{-3}}{1.5\times \pi}} \\\\d=18.2\ \mu m](https://tex.z-dn.net/?f=%5Cdfrac%7BV%7D%7BI%7D%3D%5Cdfrac%7B%5Crho%20l%7D%7B%5Cpi%20%28d%2F2%29%5E2%7D%5C%5C%5C%5C%5Cdfrac%7BV%7D%7BI%7D%3D%5Cdfrac%7B4%5Crho%20l%7D%7B%5Cpi%20d%5E2%7D%5C%5C%5C%5Cd%3D%5Csqrt%7B%5Cdfrac%7B4%5Crho%20l%20I%7D%7BV%5Cpi%7D%7D%20%5C%5C%5C%5Cd%3D%5Csqrt%7B%5Cdfrac%7B4%5Ctimes%202.44%5Ctimes%2010%5E%7B-8%7D%5Ctimes%204%5Ctimes%204%5Ctimes%2010%5E%7B-3%7D%7D%7B1.5%5Ctimes%20%5Cpi%7D%7D%20%5C%5C%5C%5Cd%3D18.2%5C%20%5Cmu%20m)
Out of four option, near option is (C) 17 μm.
Answer:
![1.62\times 10^{-8}\ \text{s}](https://tex.z-dn.net/?f=1.62%5Ctimes%2010%5E%7B-8%7D%5C%20%5Ctext%7Bs%7D)
Explanation:
= Vacuum permittivity = ![8.854\times 10^{-12}\ \text{F/m}](https://tex.z-dn.net/?f=8.854%5Ctimes%2010%5E%7B-12%7D%5C%20%5Ctext%7BF%2Fm%7D)
= Area = ![10\times 2\times 10^{-4}\ \text{m}^2](https://tex.z-dn.net/?f=10%5Ctimes%202%5Ctimes%2010%5E%7B-4%7D%5C%20%5Ctext%7Bm%7D%5E2)
= Distance between plates = 1 mm
= Changed voltage = 60 V
= Initial voltage = 100 V
= Resistance = ![1000\ \Omega](https://tex.z-dn.net/?f=1000%5C%20%5COmega)
Capacitance is given by
![C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.854\times 10^{-12}\times 10\times 2\times 10^{-4}}{1\times 10^{-3}}\\\Rightarrow C=1.7708\times 10^{-11}\ \text{F}](https://tex.z-dn.net/?f=C%3D%5Cdfrac%7B%5Cepsilon_0A%7D%7Bd%7D%5C%5C%5CRightarrow%20C%3D%5Cdfrac%7B8.854%5Ctimes%2010%5E%7B-12%7D%5Ctimes%2010%5Ctimes%202%5Ctimes%2010%5E%7B-4%7D%7D%7B1%5Ctimes%2010%5E%7B-3%7D%7D%5C%5C%5CRightarrow%20C%3D1.7708%5Ctimes%2010%5E%7B-11%7D%5C%20%5Ctext%7BF%7D)
We have the relation
![V_c=V(1-e^{-\dfrac{t}{CR}})\\\Rightarrow e^{-\dfrac{t}{CR}}=1-\dfrac{V_c}{V}\\\Rightarrow -\dfrac{t}{CR}=\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-CR\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-1.7708\times 10^{-11}\times 1000\ln(1-\dfrac{60}{100})\\\Rightarrow t=1.62\times 10^{-8}\ \text{s}](https://tex.z-dn.net/?f=V_c%3DV%281-e%5E%7B-%5Cdfrac%7Bt%7D%7BCR%7D%7D%29%5C%5C%5CRightarrow%20e%5E%7B-%5Cdfrac%7Bt%7D%7BCR%7D%7D%3D1-%5Cdfrac%7BV_c%7D%7BV%7D%5C%5C%5CRightarrow%20-%5Cdfrac%7Bt%7D%7BCR%7D%3D%5Cln%20%281-%5Cdfrac%7BV_c%7D%7BV%7D%29%5C%5C%5CRightarrow%20t%3D-CR%5Cln%20%281-%5Cdfrac%7BV_c%7D%7BV%7D%29%5C%5C%5CRightarrow%20t%3D-1.7708%5Ctimes%2010%5E%7B-11%7D%5Ctimes%201000%5Cln%281-%5Cdfrac%7B60%7D%7B100%7D%29%5C%5C%5CRightarrow%20t%3D1.62%5Ctimes%2010%5E%7B-8%7D%5C%20%5Ctext%7Bs%7D)
The time taken for the potential difference to reach the required level is
.
The amplitude of a sound<span> wave </span>determines<span> its </span>loudness<span> or volume. A larger amplitude means a louder </span>sound<span>, and a smaller amplitude means a softer </span><span>sound</span>
I dont really know but i think the first one is true super sorry
Answer:
4s
Explanation:
when a bOdy rises into the air,the time it takes to reach a particular height is the same as the time it will take the body to fall from that height to the ground.