Question

​a particle inside a hollow sphere of radius r, having coefficient of friction 1/√3 can rest upto a height of

r/4

r/8

0.134 r

3r/8

Answer 1

The answer will be 0.134r.. to find the shortcut trick and detailed explanation see my YouTube channel named Physics Learner ( Shankar Mitra)

Answer 2

Answer:The height is 0.134 r.

Explanation:

Given that,

Coefficient of friction $\mu= \dfrac{1}{\sqrt{3}}$

Radius = r

According to figure,

$Cos\ \theta= \dfrac{r-h}{r}$.....(I)

Equivalent force is

$F_{f}= \mu mg cos\theta$.....(II)

$F= mg sin\theta$......(III)

Now, $F_{f}=F$

$\mu\ mg cos\theta = mg sin\theta$

$\dfrac{1}{\sqrt{3}}= tan\theta$

$\theta = 30^{0}$

Now, from equation (I)

$cos\theta = \dfrac{r-h}{r}$

$cos 30^{0}=\dfrac{r-h}{r}$

$\dfrac{\sqrt{3}}{2}= \dfrac{r-h}{r}$

$h = \dfrac{(2-\sqrt{3})r}{2}$

$h = 0.134\ r$

Hence, The height is 0.134 r.