Question

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Answer 1

1. ${T}_{2}Sin{60}^{\circ} = {T}_{1}Sin{30}^{\circ}$

$\dfrac{\sqrt{3}{T}_{2}}{2} = \dfrac{{T}_{1}}{2}$

Therefore,

${T}_{2} = \dfrac{{T}_{1}}{\sqrt{3}}$ ----(1)

Again,

${T}_{1}Cos{30}^{\circ} + {T}_{2}Cos{60}^{\circ} = 200N$

$\dfrac{\sqrt{3}{T}_{1}}{2} + \dfrac{{T}_{1}}{2}\times\dfrac{1}{2} = 200$

$\dfrac{6{T}_{1}+2{T}_{1}}{4\sqrt{3}}=200$

$\bf{T}_{1} = 100\sqrt{3}\:N$

Putting the value in equation (1) We will get,

$\bf{T}_{2} = 100\:N$

2. As the same tension is acting in both the strings.

2T CosØ = mg

T = mg/2CosØ

3. Here first we will take the horizontal component of the body then we will take vertical component of the body.

${T}_{1}Cos{45}^{\circ} = {T}_{2}Cos{30}^{\circ}$

$\dfrac{{T}_{1}}{\sqrt{2}} = \dfrac{\sqrt{3}{T}_{2}}{2}$

${T}_{1} = \dfrac{\sqrt{3}{T}_{2}}{\sqrt{2}}$ ----(1)

Again,

${T}_{1}Sin{45}^{\circ}+{T}_{2}Sin{30}^{\circ}=300\:N$ ----(2)

Putting the value of ${T}_{1}$ in eq(2).

$\dfrac{\sqrt{3}{T}_{2}}{\sqrt{2}} \times \dfrac{1}{\sqrt{2}} + \dfrac{{T}_{2}}{2} \: = 300$

Solving this we will get,

$\bf{T}_{2} = 300(\sqrt{3}-1)$

$\bf{T}_{2} = 215.2 N$

Putting the value of ${T}_{2}$ in equation we get,

${T}_{1} = \dfrac{\sqrt{3}}{\sqrt{2}} \times 300(\sqrt{3}-1)$

$\bf{T}_{1} = 263.5\:N$

Answer 2

Answer:

1. {T}_{2}Sin{60}^{\circ} = {T}_{1}Sin{30}^{\circ}T

2

Sin60

∘

=T

1

Sin30

∘

\dfrac{\sqrt{3}{T}_{2}}{2} = \dfrac{{T}_{1}}{2}

2

3

T

2

=

2

T

1

Therefore,

{T}_{2} = \dfrac{{T}_{1}}{\sqrt{3}}T

2

=

3

T

1

----(1)

Again,

{T}_{1}Cos{30}^{\circ} + {T}_{2}Cos{60}^{\circ} = 200NT

1

Cos30

∘

+T

2

Cos60

∘

=200N

\dfrac{\sqrt{3}{T}_{1}}{2} + \dfrac{{T}_{1}}{2}\times\dfrac{1}{2} = 200

2

3

T

1

+

2

T

1

×

2

1

=200

\dfrac{6{T}_{1}+2{T}_{1}}{4\sqrt{3}}=200

4

3

6T

1

+2T

1

=200

\bf{T}_{1} = 100\sqrt{3}\:NT

1

=100

3

N

Putting the value in equation (1) We will get,

\bf{T}_{2} = 100\:NT

2

=100N

2. As the same tension is acting in both the strings.

2T CosØ = mg

T = mg/2CosØ

3. Here first we will take the horizontal component of the body then we will take vertical component of the body.

{T}_{1}Cos{45}^{\circ} = {T}_{2}Cos{30}^{\circ}T

1

Cos45

∘

=T

2

Cos30

∘

\dfrac{{T}_{1}}{\sqrt{2}} = \dfrac{\sqrt{3}{T}_{2}}{2}

2

T

1

=

2

3

T

2

{T}_{1} = \dfrac{\sqrt{3}{T}_{2}}{\sqrt{2}}T

1

=

2

3

T

2

----(1)

Again,

{T}_{1}Sin{45}^{\circ}+{T}_{2}Sin{30}^{\circ}=300\:NT

1

Sin45

∘

+T

2

Sin30

∘

=300N ----(2)

Putting the value of {T}_{1}T

1

in eq(2).

\dfrac{\sqrt{3}{T}_{2}}{\sqrt{2}} \times \dfrac{1}{\sqrt{2}} + \dfrac{{T}_{2}}{2} \: = 300

2

3

T

2

×

2

1

+

2

T

2

=300

Solving this we will get,

\bf{T}_{2} = 300(\sqrt{3}-1)T

2

=300(

3

−1)

\bf{T}_{2} = 215.2 NT

2

=215.2N

Putting the value of {T}_{2}T

2

in equation we get,

{T}_{1} = \dfrac{\sqrt{3}}{\sqrt{2}} \times 300(\sqrt{3}-1)T

1

=

2

3

×300(

3

−1)

\bf{T}_{1} = 263.5\:NT

1

=263.5N

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