Question

17. A particle of mass 200 g is attached to an ideal
string of length 1.30 m whose upper end is fixed to
ceiling. The particle is made to revolve in a horizontal
circle of radius 50 cm. The tension in the string is
(g = 10 m/s)



​

Answer 1

Answer:

By Circular Motion

T×(5/13)=(1/5)V^2/1.3

now

by Vertical Forces are Equal

So T×(12/13)=2N

Now T=13/6 N

But this second Equation is enough.

But if we want to find the Velocity

then 5/6=(2/13) ×V^2

Because Of that V^2=65/12

Now V=sq.root (65/12)

Answer 2

The tension in the string is 5.2 N.

Explanation:

Given that,

Mass = 200 g

Length = 1.30 m

Radius = 50 cm

We need to calculate the height

Using Pythagorean theorem

$h=\sqrt{(l)^2-(r)^2}$

Put the value into the formula

$h=\sqrt{(130)^2-(50)^2}$

$h=120\ cm$

We need to calculate the angle

Using formula of angle

$\tan\theta=\dfrac{h}{r}$

Put the value into the formula

$\tan\theta=\dfrac{120}{50}$

$\theta=\tan^{-1}(\dfrac{120}{50})$

$\theta=67.38^{\circ}$

We need to calculate the tension in the string

Using force component

$T\cos\theta=mg$

$T=\dfrac{mg}{\cos\theta}$

$T=\dfrac{0.2\times10}{\cos67.38}$

$T=5.2\ N$

Hence, The tension in the string is 5.2 N.

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