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PHXI06:WORK ENERGY AND POWER

355547 Two forces $3 N$ and $2 N$ are at an angle $θ$ such that the resultant is $R$ . The first force is now increased to $6 N$ and the resultant becomes $2 R$ . The value of $θ$ is

1

30^{\circ}

2

60^{\circ}

3

90^{\circ}

4

120^{\circ}

Explanation:

$A = 3 N, B = 2 N$
then $R = \sqrt{A^{2} + B^{2} + 2 A B \cos θ}$
$R = \sqrt{9 + 4 + 12 \cos θ} (1)$
Now $A = 6 N, B = 2 N$ then
$2 R = \sqrt{36 + 4 + 24 \cos θ} (2)$
From (1) and (2) we get $\cos θ = - \frac{1}{2} ∴ θ = 120^{\circ}$

PHXI06:WORK ENERGY AND POWER

355548 If unit vectors $\hat{A}$ and $\hat{B}$ are inclined at an angle $θ$ , then $| \hat{A} - \hat{B} |$ is

1

2 \sin \frac{θ}{2}

2

2 \cos \frac{θ}{2}

3

2 \tan \frac{θ}{2}

4

\tan θ

Explanation:

$| \hat{A} - \hat{B} |^{2} = (\hat{A} - \hat{B}) \cdot (\hat{A} - \hat{B})$
$= \hat{A} \cdot \hat{A} - \hat{A} \cdot \hat{B} - \hat{B} \cdot \hat{A} + \hat{B} \cdot \hat{B}$
$= 1 - \hat{A} \cdot \hat{B} - \hat{A} \cdot \hat{B} + 1$
$= 2 - 2 \cos θ = 2 (1 - \cos θ)$
$= 2 (2 \sin^{2} \frac{θ}{2}) = 4 \sin^{2} \frac{θ}{2}$

PHXI06:WORK ENERGY AND POWER

355549 If $\vec{A} = 3 \hat{i} + 4 \hat{j}$ and $\vec{B} = a \hat{i} + 3 \hat{j}$ are perpendicular to each other. The value of $a$ is

1

- 4

2 4

3

- 3

4 3

Explanation:

For vectors to be perpendicular,
$\vec{A} \cdot \vec{B} = 0 \Rightarrow a = - 4$

PHXI06:WORK ENERGY AND POWER

355550 The component of vector $\vec{A} = 2 \hat{i} + 3 \hat{j}$ along the vector $\vec{B} = \hat{i} + \hat{j}$ is

1

\frac{5}{\sqrt{2}}

2

10 \sqrt{2}

3

5 \sqrt{2}

4 5

Explanation:

$\frac{\vec{A} \cdot \vec{B}}{| B |} = \frac{(2 \hat{i} + 3 \hat{j}) \cdot (\hat{i} + \hat{j})}{\sqrt{2}} = \frac{2 + 3}{\sqrt{2}} = \frac{5}{\sqrt{2}}$

PHXI06:WORK ENERGY AND POWER

355547 Two forces $3 N$ and $2 N$ are at an angle $θ$ such that the resultant is $R$ . The first force is now increased to $6 N$ and the resultant becomes $2 R$ . The value of $θ$ is

1

30^{\circ}

2

60^{\circ}

3

90^{\circ}

4

120^{\circ}

Explanation:

$A = 3 N, B = 2 N$
then $R = \sqrt{A^{2} + B^{2} + 2 A B \cos θ}$
$R = \sqrt{9 + 4 + 12 \cos θ} (1)$
Now $A = 6 N, B = 2 N$ then
$2 R = \sqrt{36 + 4 + 24 \cos θ} (2)$
From (1) and (2) we get $\cos θ = - \frac{1}{2} ∴ θ = 120^{\circ}$

PHXI06:WORK ENERGY AND POWER

355548 If unit vectors $\hat{A}$ and $\hat{B}$ are inclined at an angle $θ$ , then $| \hat{A} - \hat{B} |$ is

1

2 \sin \frac{θ}{2}

2

2 \cos \frac{θ}{2}

3

2 \tan \frac{θ}{2}

4

\tan θ

Explanation:

$| \hat{A} - \hat{B} |^{2} = (\hat{A} - \hat{B}) \cdot (\hat{A} - \hat{B})$
$= \hat{A} \cdot \hat{A} - \hat{A} \cdot \hat{B} - \hat{B} \cdot \hat{A} + \hat{B} \cdot \hat{B}$
$= 1 - \hat{A} \cdot \hat{B} - \hat{A} \cdot \hat{B} + 1$
$= 2 - 2 \cos θ = 2 (1 - \cos θ)$
$= 2 (2 \sin^{2} \frac{θ}{2}) = 4 \sin^{2} \frac{θ}{2}$

PHXI06:WORK ENERGY AND POWER

355549 If $\vec{A} = 3 \hat{i} + 4 \hat{j}$ and $\vec{B} = a \hat{i} + 3 \hat{j}$ are perpendicular to each other. The value of $a$ is

1

- 4

2 4

3

- 3

4 3

Explanation:

For vectors to be perpendicular,
$\vec{A} \cdot \vec{B} = 0 \Rightarrow a = - 4$

PHXI06:WORK ENERGY AND POWER

355550 The component of vector $\vec{A} = 2 \hat{i} + 3 \hat{j}$ along the vector $\vec{B} = \hat{i} + \hat{j}$ is

1

\frac{5}{\sqrt{2}}

2

10 \sqrt{2}

3

5 \sqrt{2}

4 5

Explanation:

$\frac{\vec{A} \cdot \vec{B}}{| B |} = \frac{(2 \hat{i} + 3 \hat{j}) \cdot (\hat{i} + \hat{j})}{\sqrt{2}} = \frac{2 + 3}{\sqrt{2}} = \frac{5}{\sqrt{2}}$

PHXI06:WORK ENERGY AND POWER

355547 Two forces $3 N$ and $2 N$ are at an angle $θ$ such that the resultant is $R$ . The first force is now increased to $6 N$ and the resultant becomes $2 R$ . The value of $θ$ is

1

30^{\circ}

2

60^{\circ}

3

90^{\circ}

4

120^{\circ}

Explanation:

$A = 3 N, B = 2 N$
then $R = \sqrt{A^{2} + B^{2} + 2 A B \cos θ}$
$R = \sqrt{9 + 4 + 12 \cos θ} (1)$
Now $A = 6 N, B = 2 N$ then
$2 R = \sqrt{36 + 4 + 24 \cos θ} (2)$
From (1) and (2) we get $\cos θ = - \frac{1}{2} ∴ θ = 120^{\circ}$

PHXI06:WORK ENERGY AND POWER

355548 If unit vectors $\hat{A}$ and $\hat{B}$ are inclined at an angle $θ$ , then $| \hat{A} - \hat{B} |$ is

1

2 \sin \frac{θ}{2}

2

2 \cos \frac{θ}{2}

3

2 \tan \frac{θ}{2}

4

\tan θ

Explanation:

$| \hat{A} - \hat{B} |^{2} = (\hat{A} - \hat{B}) \cdot (\hat{A} - \hat{B})$
$= \hat{A} \cdot \hat{A} - \hat{A} \cdot \hat{B} - \hat{B} \cdot \hat{A} + \hat{B} \cdot \hat{B}$
$= 1 - \hat{A} \cdot \hat{B} - \hat{A} \cdot \hat{B} + 1$
$= 2 - 2 \cos θ = 2 (1 - \cos θ)$
$= 2 (2 \sin^{2} \frac{θ}{2}) = 4 \sin^{2} \frac{θ}{2}$

PHXI06:WORK ENERGY AND POWER

355549 If $\vec{A} = 3 \hat{i} + 4 \hat{j}$ and $\vec{B} = a \hat{i} + 3 \hat{j}$ are perpendicular to each other. The value of $a$ is

1

- 4

2 4

3

- 3

4 3

Explanation:

For vectors to be perpendicular,
$\vec{A} \cdot \vec{B} = 0 \Rightarrow a = - 4$

PHXI06:WORK ENERGY AND POWER

355550 The component of vector $\vec{A} = 2 \hat{i} + 3 \hat{j}$ along the vector $\vec{B} = \hat{i} + \hat{j}$ is

1

\frac{5}{\sqrt{2}}

2

10 \sqrt{2}

3

5 \sqrt{2}

4 5

Explanation:

$\frac{\vec{A} \cdot \vec{B}}{| B |} = \frac{(2 \hat{i} + 3 \hat{j}) \cdot (\hat{i} + \hat{j})}{\sqrt{2}} = \frac{2 + 3}{\sqrt{2}} = \frac{5}{\sqrt{2}}$

PHXI06:WORK ENERGY AND POWER

355547 Two forces $3 N$ and $2 N$ are at an angle $θ$ such that the resultant is $R$ . The first force is now increased to $6 N$ and the resultant becomes $2 R$ . The value of $θ$ is

1

30^{\circ}

2

60^{\circ}

3

90^{\circ}

4

120^{\circ}

Explanation:

$A = 3 N, B = 2 N$
then $R = \sqrt{A^{2} + B^{2} + 2 A B \cos θ}$
$R = \sqrt{9 + 4 + 12 \cos θ} (1)$
Now $A = 6 N, B = 2 N$ then
$2 R = \sqrt{36 + 4 + 24 \cos θ} (2)$
From (1) and (2) we get $\cos θ = - \frac{1}{2} ∴ θ = 120^{\circ}$

PHXI06:WORK ENERGY AND POWER

355548 If unit vectors $\hat{A}$ and $\hat{B}$ are inclined at an angle $θ$ , then $| \hat{A} - \hat{B} |$ is

1

2 \sin \frac{θ}{2}

2

2 \cos \frac{θ}{2}

3

2 \tan \frac{θ}{2}

4

\tan θ

Explanation:

$| \hat{A} - \hat{B} |^{2} = (\hat{A} - \hat{B}) \cdot (\hat{A} - \hat{B})$
$= \hat{A} \cdot \hat{A} - \hat{A} \cdot \hat{B} - \hat{B} \cdot \hat{A} + \hat{B} \cdot \hat{B}$
$= 1 - \hat{A} \cdot \hat{B} - \hat{A} \cdot \hat{B} + 1$
$= 2 - 2 \cos θ = 2 (1 - \cos θ)$
$= 2 (2 \sin^{2} \frac{θ}{2}) = 4 \sin^{2} \frac{θ}{2}$

PHXI06:WORK ENERGY AND POWER

355549 If $\vec{A} = 3 \hat{i} + 4 \hat{j}$ and $\vec{B} = a \hat{i} + 3 \hat{j}$ are perpendicular to each other. The value of $a$ is

1

- 4

2 4

3

- 3

4 3

Explanation:

For vectors to be perpendicular,
$\vec{A} \cdot \vec{B} = 0 \Rightarrow a = - 4$

PHXI06:WORK ENERGY AND POWER

355550 The component of vector $\vec{A} = 2 \hat{i} + 3 \hat{j}$ along the vector $\vec{B} = \hat{i} + \hat{j}$ is

1

\frac{5}{\sqrt{2}}

2

10 \sqrt{2}

3

5 \sqrt{2}

4 5

Explanation:

$\frac{\vec{A} \cdot \vec{B}}{| B |} = \frac{(2 \hat{i} + 3 \hat{j}) \cdot (\hat{i} + \hat{j})}{\sqrt{2}} = \frac{2 + 3}{\sqrt{2}} = \frac{5}{\sqrt{2}}$

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