Question

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➟ Answer with explaination​

Answer 1

$\large\underline{\sf{Solution-}}$

Let us suppose that

• m be the mass of the particle.

• v be the speed of the particle.

• r be the radius of the circular path.

We know,

☆ Centrifugal force is given by

$\rm :\longmapsto\:F_c = \dfrac{m {v}^{2} }{r} - - - (1)$

☆ Let N be the normal force acting on the particle and angle θ made by the conical surface with the horizontal.

So,

☆ Horizontal component of normal force is given by

$\rm :\longmapsto\:N_x = N \: sin \theta - - - (2)$

and

☆ Vertical component of normal force is given by

$\rm :\longmapsto\:N_y = N \: cos \theta - - - (3)$

Now,

According to statement,

$\rm :\longmapsto\:F_c = N_x$

$\rm :\longmapsto\:\dfrac{m {v}^{2} }{r} = N \: sin\theta - - (4)$

Also,

According to statement,

$\rm :\longmapsto\:N_y \: = \: W$

$\rm :\longmapsto\:N \: cos\theta \: = mg - - - (5)$

where,

• m = mass of the particle

• g = acceleration due to gravity

☆ On dividing equation (4) by equation (5), we get

$\rm :\longmapsto\:\dfrac{N \: sin\theta}{N \: cos\theta} = \dfrac{\dfrac{m {v}^{2} }{r} }{ \: \: \: mg \: \: \: }$

$\bf :\longmapsto\:tan\theta \: = \: \dfrac{ {v}^{2} }{rg} - - - (6)$

Now,

☆ From the figure,

$\rm :\longmapsto\:tan(90 \degree \: - \theta) = \dfrac{r}{h}$

$\rm :\longmapsto\:cot\theta = \dfrac{r}{h}$

$\bf :\longmapsto\:tan\theta = \dfrac{h}{r} - - - (7)$

So,

☆ On comparing equation (6) and equation (7), we get

$\rm :\longmapsto\:\dfrac{ {v}^{2} }{rg} = \dfrac{h}{r}$

$\rm :\longmapsto\: {v}^{2} = gh$

$\bf\implies \:v = \sqrt{gh}$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bf \: Option \: (3) \: is \: correct}$

Answer 2

Answer:

$\large\underline{\sf{Solution-}}$

Let us suppose that

m be the mass of the particle.

v be the speed of the particle.

r be the radius of the circular path.

We know,

☆ Centrifugal force is given by

$\rm :\longmapsto\:F_c = \dfrac{m {v}^{2} }{r} - - - (1)$

☆ Let N be the normal force acting on the particle and angle θ made by the conical surface with the horizontal.

So,

☆ Horizontal component of normal force is given by

$\rm :\longmapsto\:N_x = N \: sin \theta - - - (2)$

and

☆ Vertical component of normal force is given by

$\rm :\longmapsto\:N_y = N \: cos \theta - - - (3)$

Now,

According to statement,

$\rm :\longmapsto\:F_c = N_x$

$\rm :\longmapsto\:\dfrac{m {v}^{2} }{r} = N \: sin\theta - - (4)$

Also,

According to statement,

$\rm :\longmapsto\:N_y \: = \: W$

$\rm :\longmapsto\:N \: cos\theta \: = mg - - - (5)$

where,

m = mass of the particle

g = acceleration due to gravity

☆ On dividing equation (4) by equation (5), we get

$\rm :\longmapsto\:\dfrac{N \: sin\theta}{N \: cos\theta} = \dfrac{\dfrac{m {v}^{2} }{r} }{ \: \: \: mg \: \: \: }$

$\bf :\longmapsto\:tan\theta \: = \: \dfrac{ {v}^{2} }{rg} - - - (6)$

Now,

☆ From the figure,

$\rm :\longmapsto\:tan(90 \degree \: - \theta) = \dfrac{r}{h}$

$\rm :\longmapsto\:cot\theta = \dfrac{r}{h}$

$\bf :\longmapsto\:tan\theta = \dfrac{h}{r} - - - (7)$

So,

☆ On comparing equation (6) and equation (7), we get

$\rm :\longmapsto\:\dfrac{ {v}^{2} }{rg} = \dfrac{h}{r}$

$\rm :\longmapsto\: {v}^{2} = gh$

$\bf\implies \:v = \sqrt{gh}$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bf \: Option \: (3) \: is \: correct}$