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PHXI05:LAWS OF MOTION

363303 The block is in equilibrium:
$(i) T_{1} = \frac{m g \cos β}{\sin (α + β)} (i i) T_{2} = \frac{m g \cos α}{\sin (α + β)}$
$(i i i) \frac{T_{1}}{T_{2}} = \frac{\cos β}{\cos α} (i v) \frac{T_{1}}{T_{2}} = \frac{\cos α}{\cos β}$
Find the correct option
supporting img

1

(i), (i i)

2

(i), (i i), (i i i)

3

(i i), (i i i)

4

(i i i), (i v)

Explanation:

$\frac{T_{1}}{\sin (90^{0} + β)} = \frac{T_{2}}{\sin (90^{0} + α)} = \frac{m g}{\sin [180^{0} - (α + β)]}$
$\frac{T_{1}}{\cos β} = \frac{T_{2}}{\cos α} = \frac{m g}{\sin (α + β)}$
$T_{1} = \frac{m g \cos β}{\sin (α + β)}$
$T_{2} = \frac{m g \cos α}{\sin (α + β)}$
supporting img

PHXI05:LAWS OF MOTION

363304 A body of mass 60 $k g$ suspended by means of three strings. $P$ , $Q$ and $R$ as shown in the figure is in equilibrium. The tension in the string $P$ is:
supporting img

1

130.9 g N

2

60 g N

3

50 g N

4

103.9 g N

Explanation:

Method - 1
The free body diagram of mass $M$ is shown in figure. Let $T$ be the tension in the string $P$ .
Taking component of forces
$R \cos θ = M g \Rightarrow R \cos 60^{0} = M g (1)$
$and R \sin 60^{0} = T (2)$
By eqs. (1) and (2), we get
$\Rightarrow \tan 60^{\circ} = \frac{T}{M g} \Rightarrow T = M g \tan 60^{\circ}$
$\Rightarrow T = 60 \times g \times \sqrt{3} = 103.9 g N$
Method - 2
Using lami’s theorem
$\frac{R}{\sin 90^{\circ}} = \frac{P}{\sin (90^{\circ} + 30^{\circ})} = \frac{60 g}{\sin (90^{\circ} + 60^{\circ})}$
$\Rightarrow P = 103.9 g N$
supporting img

PHXI05:LAWS OF MOTION

363305 A block of mass $m$ is connected with two strings, as shown in the figure. The ratio of $\frac{T_{1}}{T_{2}}$ is
supporting img

1

\frac{1}{\sqrt{2}}

2 2

3

\sqrt{2} - 1

4

\sqrt{2}

Explanation:

The components of tension are shown in the figure.
supporting img
$\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$
As the block is in horizontal equilibrium
$T_{1} \cos 60^{\circ} = T_{2} \cos 45^{\circ}$
$\frac{T_{1}}{T_{2}} = \frac{\cos 45^{\circ}}{\cos 60^{\circ}} = \frac{(1 / \sqrt{2})}{(1 / 2)} \sqrt{2}$

PHXI05:LAWS OF MOTION

363306 A mass $M$ is hung with a light inextensible string as shown in the figure. Find the tension of the horizontal string.
supporting img

1

\sqrt{2} M g

2

\sqrt{3} M g

3

2 M g

4

3 M g

Explanation:

As there is a load at $P$ , so tension in $A P$ and
$P B$ will be different. Let these be $T_{1}$ and $T_{2}$
respectively. For vertical equilibrium of $P$ ,
supporting img

$T_{2} \cos 60^{\circ} = M g i . e ., T_{2} = 2 M g (1)$
and for horizontal equilibrium of $P$ ,
$T_{1} = T_{2} \sin 60^{\circ} = T_{2} (\sqrt{3} / 2)$
Substituting the value of $T_{2}$ from Eq. (1),
$T_{1} = (2 M g) \times (\sqrt{3} / 2) = \sqrt{3} M g$

PHXI05:LAWS OF MOTION

363303 The block is in equilibrium:
$(i) T_{1} = \frac{m g \cos β}{\sin (α + β)} (i i) T_{2} = \frac{m g \cos α}{\sin (α + β)}$
$(i i i) \frac{T_{1}}{T_{2}} = \frac{\cos β}{\cos α} (i v) \frac{T_{1}}{T_{2}} = \frac{\cos α}{\cos β}$
Find the correct option
supporting img

1

(i), (i i)

2

(i), (i i), (i i i)

3

(i i), (i i i)

4

(i i i), (i v)

Explanation:

$\frac{T_{1}}{\sin (90^{0} + β)} = \frac{T_{2}}{\sin (90^{0} + α)} = \frac{m g}{\sin [180^{0} - (α + β)]}$
$\frac{T_{1}}{\cos β} = \frac{T_{2}}{\cos α} = \frac{m g}{\sin (α + β)}$
$T_{1} = \frac{m g \cos β}{\sin (α + β)}$
$T_{2} = \frac{m g \cos α}{\sin (α + β)}$
supporting img

PHXI05:LAWS OF MOTION

363304 A body of mass 60 $k g$ suspended by means of three strings. $P$ , $Q$ and $R$ as shown in the figure is in equilibrium. The tension in the string $P$ is:
supporting img

1

130.9 g N

2

60 g N

3

50 g N

4

103.9 g N

Explanation:

Method - 1
The free body diagram of mass $M$ is shown in figure. Let $T$ be the tension in the string $P$ .
Taking component of forces
$R \cos θ = M g \Rightarrow R \cos 60^{0} = M g (1)$
$and R \sin 60^{0} = T (2)$
By eqs. (1) and (2), we get
$\Rightarrow \tan 60^{\circ} = \frac{T}{M g} \Rightarrow T = M g \tan 60^{\circ}$
$\Rightarrow T = 60 \times g \times \sqrt{3} = 103.9 g N$
Method - 2
Using lami’s theorem
$\frac{R}{\sin 90^{\circ}} = \frac{P}{\sin (90^{\circ} + 30^{\circ})} = \frac{60 g}{\sin (90^{\circ} + 60^{\circ})}$
$\Rightarrow P = 103.9 g N$
supporting img

PHXI05:LAWS OF MOTION

363305 A block of mass $m$ is connected with two strings, as shown in the figure. The ratio of $\frac{T_{1}}{T_{2}}$ is
supporting img

1

\frac{1}{\sqrt{2}}

2 2

3

\sqrt{2} - 1

4

\sqrt{2}

Explanation:

The components of tension are shown in the figure.
supporting img
$\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$
As the block is in horizontal equilibrium
$T_{1} \cos 60^{\circ} = T_{2} \cos 45^{\circ}$
$\frac{T_{1}}{T_{2}} = \frac{\cos 45^{\circ}}{\cos 60^{\circ}} = \frac{(1 / \sqrt{2})}{(1 / 2)} \sqrt{2}$

PHXI05:LAWS OF MOTION

363306 A mass $M$ is hung with a light inextensible string as shown in the figure. Find the tension of the horizontal string.
supporting img

1

\sqrt{2} M g

2

\sqrt{3} M g

3

2 M g

4

3 M g

Explanation:

As there is a load at $P$ , so tension in $A P$ and
$P B$ will be different. Let these be $T_{1}$ and $T_{2}$
respectively. For vertical equilibrium of $P$ ,
supporting img

$T_{2} \cos 60^{\circ} = M g i . e ., T_{2} = 2 M g (1)$
and for horizontal equilibrium of $P$ ,
$T_{1} = T_{2} \sin 60^{\circ} = T_{2} (\sqrt{3} / 2)$
Substituting the value of $T_{2}$ from Eq. (1),
$T_{1} = (2 M g) \times (\sqrt{3} / 2) = \sqrt{3} M g$

PHXI05:LAWS OF MOTION

363303 The block is in equilibrium:
$(i) T_{1} = \frac{m g \cos β}{\sin (α + β)} (i i) T_{2} = \frac{m g \cos α}{\sin (α + β)}$
$(i i i) \frac{T_{1}}{T_{2}} = \frac{\cos β}{\cos α} (i v) \frac{T_{1}}{T_{2}} = \frac{\cos α}{\cos β}$
Find the correct option
supporting img

1

(i), (i i)

2

(i), (i i), (i i i)

3

(i i), (i i i)

4

(i i i), (i v)

Explanation:

$\frac{T_{1}}{\sin (90^{0} + β)} = \frac{T_{2}}{\sin (90^{0} + α)} = \frac{m g}{\sin [180^{0} - (α + β)]}$
$\frac{T_{1}}{\cos β} = \frac{T_{2}}{\cos α} = \frac{m g}{\sin (α + β)}$
$T_{1} = \frac{m g \cos β}{\sin (α + β)}$
$T_{2} = \frac{m g \cos α}{\sin (α + β)}$
supporting img

PHXI05:LAWS OF MOTION

363304 A body of mass 60 $k g$ suspended by means of three strings. $P$ , $Q$ and $R$ as shown in the figure is in equilibrium. The tension in the string $P$ is:
supporting img

1

130.9 g N

2

60 g N

3

50 g N

4

103.9 g N

Explanation:

Method - 1
The free body diagram of mass $M$ is shown in figure. Let $T$ be the tension in the string $P$ .
Taking component of forces
$R \cos θ = M g \Rightarrow R \cos 60^{0} = M g (1)$
$and R \sin 60^{0} = T (2)$
By eqs. (1) and (2), we get
$\Rightarrow \tan 60^{\circ} = \frac{T}{M g} \Rightarrow T = M g \tan 60^{\circ}$
$\Rightarrow T = 60 \times g \times \sqrt{3} = 103.9 g N$
Method - 2
Using lami’s theorem
$\frac{R}{\sin 90^{\circ}} = \frac{P}{\sin (90^{\circ} + 30^{\circ})} = \frac{60 g}{\sin (90^{\circ} + 60^{\circ})}$
$\Rightarrow P = 103.9 g N$
supporting img

PHXI05:LAWS OF MOTION

363305 A block of mass $m$ is connected with two strings, as shown in the figure. The ratio of $\frac{T_{1}}{T_{2}}$ is
supporting img

1

\frac{1}{\sqrt{2}}

2 2

3

\sqrt{2} - 1

4

\sqrt{2}

Explanation:

The components of tension are shown in the figure.
supporting img
$\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$
As the block is in horizontal equilibrium
$T_{1} \cos 60^{\circ} = T_{2} \cos 45^{\circ}$
$\frac{T_{1}}{T_{2}} = \frac{\cos 45^{\circ}}{\cos 60^{\circ}} = \frac{(1 / \sqrt{2})}{(1 / 2)} \sqrt{2}$

PHXI05:LAWS OF MOTION

363306 A mass $M$ is hung with a light inextensible string as shown in the figure. Find the tension of the horizontal string.
supporting img

1

\sqrt{2} M g

2

\sqrt{3} M g

3

2 M g

4

3 M g

Explanation:

As there is a load at $P$ , so tension in $A P$ and
$P B$ will be different. Let these be $T_{1}$ and $T_{2}$
respectively. For vertical equilibrium of $P$ ,
supporting img

$T_{2} \cos 60^{\circ} = M g i . e ., T_{2} = 2 M g (1)$
and for horizontal equilibrium of $P$ ,
$T_{1} = T_{2} \sin 60^{\circ} = T_{2} (\sqrt{3} / 2)$
Substituting the value of $T_{2}$ from Eq. (1),
$T_{1} = (2 M g) \times (\sqrt{3} / 2) = \sqrt{3} M g$

PHXI05:LAWS OF MOTION

363303 The block is in equilibrium:
$(i) T_{1} = \frac{m g \cos β}{\sin (α + β)} (i i) T_{2} = \frac{m g \cos α}{\sin (α + β)}$
$(i i i) \frac{T_{1}}{T_{2}} = \frac{\cos β}{\cos α} (i v) \frac{T_{1}}{T_{2}} = \frac{\cos α}{\cos β}$
Find the correct option
supporting img

1

(i), (i i)

2

(i), (i i), (i i i)

3

(i i), (i i i)

4

(i i i), (i v)

Explanation:

$\frac{T_{1}}{\sin (90^{0} + β)} = \frac{T_{2}}{\sin (90^{0} + α)} = \frac{m g}{\sin [180^{0} - (α + β)]}$
$\frac{T_{1}}{\cos β} = \frac{T_{2}}{\cos α} = \frac{m g}{\sin (α + β)}$
$T_{1} = \frac{m g \cos β}{\sin (α + β)}$
$T_{2} = \frac{m g \cos α}{\sin (α + β)}$
supporting img

PHXI05:LAWS OF MOTION

363304 A body of mass 60 $k g$ suspended by means of three strings. $P$ , $Q$ and $R$ as shown in the figure is in equilibrium. The tension in the string $P$ is:
supporting img

1

130.9 g N

2

60 g N

3

50 g N

4

103.9 g N

Explanation:

Method - 1
The free body diagram of mass $M$ is shown in figure. Let $T$ be the tension in the string $P$ .
Taking component of forces
$R \cos θ = M g \Rightarrow R \cos 60^{0} = M g (1)$
$and R \sin 60^{0} = T (2)$
By eqs. (1) and (2), we get
$\Rightarrow \tan 60^{\circ} = \frac{T}{M g} \Rightarrow T = M g \tan 60^{\circ}$
$\Rightarrow T = 60 \times g \times \sqrt{3} = 103.9 g N$
Method - 2
Using lami’s theorem
$\frac{R}{\sin 90^{\circ}} = \frac{P}{\sin (90^{\circ} + 30^{\circ})} = \frac{60 g}{\sin (90^{\circ} + 60^{\circ})}$
$\Rightarrow P = 103.9 g N$
supporting img

PHXI05:LAWS OF MOTION

363305 A block of mass $m$ is connected with two strings, as shown in the figure. The ratio of $\frac{T_{1}}{T_{2}}$ is
supporting img

1

\frac{1}{\sqrt{2}}

2 2

3

\sqrt{2} - 1

4

\sqrt{2}

Explanation:

The components of tension are shown in the figure.
supporting img
$\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}$
As the block is in horizontal equilibrium
$T_{1} \cos 60^{\circ} = T_{2} \cos 45^{\circ}$
$\frac{T_{1}}{T_{2}} = \frac{\cos 45^{\circ}}{\cos 60^{\circ}} = \frac{(1 / \sqrt{2})}{(1 / 2)} \sqrt{2}$

PHXI05:LAWS OF MOTION

363306 A mass $M$ is hung with a light inextensible string as shown in the figure. Find the tension of the horizontal string.
supporting img

1

\sqrt{2} M g

2

\sqrt{3} M g

3

2 M g

4

3 M g

Explanation:

As there is a load at $P$ , so tension in $A P$ and
$P B$ will be different. Let these be $T_{1}$ and $T_{2}$
respectively. For vertical equilibrium of $P$ ,
supporting img

$T_{2} \cos 60^{\circ} = M g i . e ., T_{2} = 2 M g (1)$
and for horizontal equilibrium of $P$ ,
$T_{1} = T_{2} \sin 60^{\circ} = T_{2} (\sqrt{3} / 2)$
Substituting the value of $T_{2}$ from Eq. (1),
$T_{1} = (2 M g) \times (\sqrt{3} / 2) = \sqrt{3} M g$

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