Question

plzzzz help............ ​

Answer 1

Answer:

The value of x is 2

Explanation:

$\rule{300}{1.5}$

Given, we have a given equation of potential energy, and we need to evaluate it when the force is zero. So, from the relation of Potential energy and force we know that,

$\longrightarrow\sf F=-\Bigg[\dfrac{dU}{dx}\Bigg]$

Here,

F Denotes Force.

U Denotes Potential energy.

Solving,

$\longrightarrow\sf F=-\Bigg[\dfrac{d}{dx}\Bigg(\dfrac{a}{r^{2}}-\dfrac{b}{r}\Bigg)\Bigg]\\\\\\\\\longrightarrow\sf F=-\left[\dfrac{d\;\Big(ar^{-2}\Big)}{dx}-\dfrac{d\;\Big(br^{-1}\Big)}{dx}\right]$

Differentiating it,

$\longrightarrow\sf F=-\Bigg[-2\times ar^{-3}-\bigg(-1\times br^{-2}\bigg)\Bigg]\\\\\\\\\longrightarrow\sf F=-\Bigg[\dfrac{-2a}{r^{3}}+\dfrac{b}{r^{2}}\Bigg]\\\\\\\\\longrightarrow\sf F=\Bigg[\dfrac{2a}{r^{3}}-\dfrac{b}{r^{2}}\Bigg]$

Now, the Force on the particle is Zero.

$\longrightarrow\sf 0=\Bigg[\dfrac{2a}{r^{3}}-\dfrac{b}{r^{2}}\Bigg]\\\\\\\\\longrightarrow\sf \dfrac{2a}{r^{3}}=\dfrac{b}{r^{2}}\\\\\\\\\longrightarrow\sf \dfrac{2a}{r}=b\\\\\\\\\longrightarrow\sf r=\dfrac{2\;a}{b}$

Comparing with r = xa/b

$\longrightarrow\sf \dfrac{x\;a}{b}=\dfrac{2\;a}{b}\\\\\\\\\longrightarrow\sf x=2\\\\\\\\\longrightarrow\large{\underline{\boxed{\red{\sf x=2}}}}$

Therefore, Value of x is 2.

$\rule{300}{1.5}$

Answer 2

Given,

The potential of the particle in the force field is:

u= A/r² − b/r

For stable in equilibrium,

f=du/dr =0

f=du/dr =02a/r³ +b/r³

B = 2A/b

Now,

xa/b = 2a/b

x = 2

--thanks for ur valuable ques--

Hope it will help you.