Physics, asked by Anonymous, 1 year ago

find the value of T1 and T2 from the figure

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Answered by knigam941
6

Answer:

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Answered by ShivamKashyap08
10

\huge{\bold{\underline{\underline{....Answer....}}}}

\huge{\bold{\underline{Given:-}}}

As the figure has not specified the mass,

Let us assume that the given body has a mass "m".

\huge{\bold{\underline{Explanation:-}}}

Let's assume that the system is in equilibrium.

\large{\bold{Resolving \: The \:  Tension(T_2) \: into \: components.}}

\large{\bold{After \: resolving, \:  it's \: one \: component \:  will \:  be \: (T_2  \sin 30 \degree) \:  vertically \: upwards.}}

\large{\bold{and, \: the  \: other \: component  \: will  \: be \: (T_2 \cos 30 \degree) horizontally.}}

As we have assumed the system in equilibrium.

The Force (F) will be equal to the weight of the given block.

\large{ \therefore F = W}

\large{\boxed{ F = mg \: -----(1)}}

Now,

CASE-1

As the system is in equilibrium,

therefore,

\large{T_2 \sin 30 \degree = F}

From equation (1)

\large{ \implies T_2 \sin 30 \degree = mg}

As sin 30° = 1/2

\large{ \implies T_2  \times \dfrac{1}{2} = mg}

\large{ \implies T_2 = 2 \times mg \: -----(2)}

\huge{\boxed{\boxed{T_2 = 2mg}}}

CASE-2

As the system is in equilibrium,

therefore,

\large{T_1 = T_2 \cos 30 \degree}

From equation (2) Substituting the value of \large{T_2} in the above equation,

\large{ \implies T_1 = 2mg . \cos 30 \degree}

as cos 30° = √3/2

\large{ \implies T_1 = 2mg \times \dfrac{ \sqrt{3}}{2}}

\large{ \implies T_1 =  \cancel{2}mg \times \dfrac{ \sqrt{3}}{ \cancel{2}}}

\huge{\boxed{\boxed{T_1 = \sqrt{3} mg}}}

So, the \large{\bold{T_1}} is 3 mg ,

and the \large{\bold{T_2}} is 2mg.

#refer the attachment for figure.

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