Pls give me a relevant answer with proper steps
Answers
Answer:
As the figure has not specified the mass,
Let us assume that the given body has a mass "m".
\huge{\bold{\underline{Explanation:-}}}
Explanation:−
Let's assume that the system is in equilibrium.
\large{\bold{Resolving \: The \: Tension(T_2) \: into \: components.}}ResolvingTheTension(T
2
)intocomponents.
\large{\bold{After \: resolving, \: it's \: one \: component \: will \: be \: (T_2 \sin 30 \degree) \: vertically \: upwards.}}Afterresolving,it
′
sonecomponentwillbe(T
2
sin30°)verticallyupwards.
\large{\bold{and, \: the \: other \: component \: will \: be \: (T_2 \cos 30 \degree) horizontally.}}and,theothercomponentwillbe(T
2
cos30°)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}∴F=W
\large{\boxed{ F = mg \: -----(1)}}
F=mg−−−−−(1)
Now,
CASE-1
As the system is in equilibrium,
therefore,
\large{T_2 \sin 30 \degree = F}T
2
sin30°=F
From equation (1)
\large{ \implies T_2 \sin 30 \degree = mg}⟹T
2
sin30°=mg
As sin 30° = 1/2
\large{ \implies T_2 \times \dfrac{1}{2} = mg}⟹T
2
×
2
1
=mg
\large{ \implies T_2 = 2 \times mg \: -----(2)}⟹T
2
=2×mg−−−−−(2)
\huge{\boxed{\boxed{T_2 = 2mg}}}
T
2
=2mg
CASE-2
As the system is in equilibrium,
therefore,
\large{T_1 = T_2 \cos 30 \degree}T
1
=T
2
cos30°
From equation (2) Substituting the value of \large{T_2}T
2
in the above equation,
\large{ \implies T_1 = 2mg . \cos 30 \degree}⟹T
1
=2mg.cos30°
as cos 30° = √3/2
\large{ \implies T_1 = 2mg \times \dfrac{ \sqrt{3}}{2}}⟹T
1
=2mg×
2
3
\large{ \implies T_1 = \cancel{2}mg \times \dfrac{ \sqrt{3}}{ \cancel{2}}}⟹T
1
=
2
mg×
2
3
\huge{\boxed{\boxed{T_1 = \sqrt{3} mg}}}
T
1
=
3
mg
So, the \large{\bold{T_1}}T
1
is √3 mg ,
and the \large{\bold{T_2}}T
2
is 2mg.
#refer the attachment for figure
