QBManager

Motion in Plane

143582 The position vector of a particle is $r = (α \cos ω t) \hat{i} + (α \sin ω t) \hat{j}$ . The velocity vector of the particle is

1 Parallel to position vector

2 Perpendicular to position vector

3 Directed towards the origin

4 Directed away from the origin

Explanation:

B $\vec{r} = α \cos ω t \hat{i} + α \sin ω \hat{j}$
$\vec{v} = \frac{d \vec{r}}{dt} = (- α \sin ω t) ω \hat{i} + (α \cos ω t) ω \hat{j}$
$\vec{v} \cdot \vec{r} = [(- α ω \sin ω t) \hat{i} + (α ω \cos ω t) \hat{j}] \cdot [(α \cos ω t) \hat{i} + (α \sin ω t) \hat{j}$
$= - α^{2} ω \sin ω t \cdot \cos ω t + α^{2} ω \cos ω t \cdot \sin ω t$
$\vec{v} \cdot \vec{r} = 0$
$∵$ Dot product of $\vec{v}$ and $\vec{r}$ is zero
$∴$ Both are perpendicular to each other.

JCECE-2014

Motion in Plane

143583 If a vector $A$ having a magnitude of 8 is added to a vector $B$ which lies along $x$ -axis, then the resultant of two vectors lies along $y$ -axis and has magnitude twice that of $B$ . The magnitude of $B$ is

1

\frac{6}{\sqrt{5}}

2

\frac{12}{\sqrt{5}}

3

\frac{16}{\sqrt{5}}

4

\frac{8}{\sqrt{5}}

Explanation:

D Given,

$\vec{A} = 8$ units
Since, $\vec{B}$ is along $x$ - axis and resultant of two vector $\vec{C}$ lies on $y -$ axis
So, $\vec{B}$ and $\vec{C}$ are perpendicular vector.
Hence,
$| \vec{A} |^{2} = | \vec{B} |^{2} + | \vec{C} |^{2}$
$| 8 |^{2} = | B |^{2} + {2 B |}^{2}$
$64 = 5 B^{2}$
$B^{2} = \frac{64}{5}$
$∴ B = \frac{8}{\sqrt{5}}$

JCECE-2012

Motion in Plane

143584 It two forces each of $2 N$ are inclined at $60^{\circ}$ , then resultant force is:

1

2 N

2

2 \sqrt{5} N

3

2 \sqrt{3} N

4

4 \sqrt{2} N

Explanation:

C Let A \& B be two forces,
$∴ R = \sqrt{A^{2} + B^{2} + 2 AB \cos θ}$
$R = \sqrt{2^{2} + 2^{2} + 2 \times 2 \times 2 \cos 60^{\circ}}$
$(\cos 60^{\circ} = 1 / 2)$
$∴ R = \sqrt{12} = 2 \sqrt{3} N$
$∴ R = 2 \sqrt{3} N$

JCECE-2006

Motion in Plane

143586 Calculate the work done when a force $\vec{F} = 2 \hat{i} + 3 \hat{j} - 5 \hat{k}$ units acts on a body producing a displacement $\vec{s} = 2 \hat{i} + 4 \hat{j} + 3 \hat{k}$ units :

1 1 unit

2 20 unit

3 5 unit

4 zero

Explanation:

A Given that, $\vec{F} = 2 \hat{i} + 3 \hat{j} - 5 \hat{k}$
$\vec{s} = 2 \hat{i} + 4 \hat{j} + 3 \hat{k}$
$W = \vec{F} \cdot \vec{S}$
$W = (2 \hat{i} + 3 \hat{j} - 5 \hat{k}) \cdot (2 \hat{i} + 4 \hat{j} + 3 \hat{k})$
$W = 4 + 12 - 15$
$W = 1$ Unit

JCECE-2003

Motion in Plane

143587 Three forces acting on a body are shown in the figure. To have the resultant force only along the $y$ -direction, the magnitude of the minimum additional force needed along $O X$ is

1

\frac{\sqrt{3}}{4} N

2

\sqrt{13} N

3

0.5 N

4

1.5 N

Explanation:

C $∵ R = \sqrt{{(\sum F_{x})}^{2} + {(\sum F_{y})}^{2}}$
Let the additional force $F$ be directed along the positive $x$ -direction.
Taking $X$ -component, the total force should be zero.
Let $F$ be the magnitude of minimum force which must be along $x$ -direction, by resolving the vector we get-
$1 \times \cos 60^{\circ} + 2 \sin 30^{\circ} + F - 4 \sin 30^{\circ} = 0$
$\frac{1}{2} + 1 + F - 2 = 0$
$F = 1 / 2 = 0.5 N$

COMEDK 2019

Motion in Plane

143582 The position vector of a particle is $r = (α \cos ω t) \hat{i} + (α \sin ω t) \hat{j}$ . The velocity vector of the particle is

1 Parallel to position vector

2 Perpendicular to position vector

3 Directed towards the origin

4 Directed away from the origin

Explanation:

B $\vec{r} = α \cos ω t \hat{i} + α \sin ω \hat{j}$
$\vec{v} = \frac{d \vec{r}}{dt} = (- α \sin ω t) ω \hat{i} + (α \cos ω t) ω \hat{j}$
$\vec{v} \cdot \vec{r} = [(- α ω \sin ω t) \hat{i} + (α ω \cos ω t) \hat{j}] \cdot [(α \cos ω t) \hat{i} + (α \sin ω t) \hat{j}$
$= - α^{2} ω \sin ω t \cdot \cos ω t + α^{2} ω \cos ω t \cdot \sin ω t$
$\vec{v} \cdot \vec{r} = 0$
$∵$ Dot product of $\vec{v}$ and $\vec{r}$ is zero
$∴$ Both are perpendicular to each other.

JCECE-2014

Motion in Plane

143583 If a vector $A$ having a magnitude of 8 is added to a vector $B$ which lies along $x$ -axis, then the resultant of two vectors lies along $y$ -axis and has magnitude twice that of $B$ . The magnitude of $B$ is

1

\frac{6}{\sqrt{5}}

2

\frac{12}{\sqrt{5}}

3

\frac{16}{\sqrt{5}}

4

\frac{8}{\sqrt{5}}

Explanation:

D Given,

$\vec{A} = 8$ units
Since, $\vec{B}$ is along $x$ - axis and resultant of two vector $\vec{C}$ lies on $y -$ axis
So, $\vec{B}$ and $\vec{C}$ are perpendicular vector.
Hence,
$| \vec{A} |^{2} = | \vec{B} |^{2} + | \vec{C} |^{2}$
$| 8 |^{2} = | B |^{2} + {2 B |}^{2}$
$64 = 5 B^{2}$
$B^{2} = \frac{64}{5}$
$∴ B = \frac{8}{\sqrt{5}}$

JCECE-2012

Motion in Plane

143584 It two forces each of $2 N$ are inclined at $60^{\circ}$ , then resultant force is:

1

2 N

2

2 \sqrt{5} N

3

2 \sqrt{3} N

4

4 \sqrt{2} N

Explanation:

C Let A \& B be two forces,
$∴ R = \sqrt{A^{2} + B^{2} + 2 AB \cos θ}$
$R = \sqrt{2^{2} + 2^{2} + 2 \times 2 \times 2 \cos 60^{\circ}}$
$(\cos 60^{\circ} = 1 / 2)$
$∴ R = \sqrt{12} = 2 \sqrt{3} N$
$∴ R = 2 \sqrt{3} N$

JCECE-2006

Motion in Plane

143586 Calculate the work done when a force $\vec{F} = 2 \hat{i} + 3 \hat{j} - 5 \hat{k}$ units acts on a body producing a displacement $\vec{s} = 2 \hat{i} + 4 \hat{j} + 3 \hat{k}$ units :

1 1 unit

2 20 unit

3 5 unit

4 zero

Explanation:

A Given that, $\vec{F} = 2 \hat{i} + 3 \hat{j} - 5 \hat{k}$
$\vec{s} = 2 \hat{i} + 4 \hat{j} + 3 \hat{k}$
$W = \vec{F} \cdot \vec{S}$
$W = (2 \hat{i} + 3 \hat{j} - 5 \hat{k}) \cdot (2 \hat{i} + 4 \hat{j} + 3 \hat{k})$
$W = 4 + 12 - 15$
$W = 1$ Unit

JCECE-2003

Motion in Plane

143587 Three forces acting on a body are shown in the figure. To have the resultant force only along the $y$ -direction, the magnitude of the minimum additional force needed along $O X$ is

1

\frac{\sqrt{3}}{4} N

2

\sqrt{13} N

3

0.5 N

4

1.5 N

Explanation:

C $∵ R = \sqrt{{(\sum F_{x})}^{2} + {(\sum F_{y})}^{2}}$
Let the additional force $F$ be directed along the positive $x$ -direction.
Taking $X$ -component, the total force should be zero.
Let $F$ be the magnitude of minimum force which must be along $x$ -direction, by resolving the vector we get-
$1 \times \cos 60^{\circ} + 2 \sin 30^{\circ} + F - 4 \sin 30^{\circ} = 0$
$\frac{1}{2} + 1 + F - 2 = 0$
$F = 1 / 2 = 0.5 N$

COMEDK 2019

Motion in Plane

143582 The position vector of a particle is $r = (α \cos ω t) \hat{i} + (α \sin ω t) \hat{j}$ . The velocity vector of the particle is

1 Parallel to position vector

2 Perpendicular to position vector

3 Directed towards the origin

4 Directed away from the origin

Explanation:

B $\vec{r} = α \cos ω t \hat{i} + α \sin ω \hat{j}$
$\vec{v} = \frac{d \vec{r}}{dt} = (- α \sin ω t) ω \hat{i} + (α \cos ω t) ω \hat{j}$
$\vec{v} \cdot \vec{r} = [(- α ω \sin ω t) \hat{i} + (α ω \cos ω t) \hat{j}] \cdot [(α \cos ω t) \hat{i} + (α \sin ω t) \hat{j}$
$= - α^{2} ω \sin ω t \cdot \cos ω t + α^{2} ω \cos ω t \cdot \sin ω t$
$\vec{v} \cdot \vec{r} = 0$
$∵$ Dot product of $\vec{v}$ and $\vec{r}$ is zero
$∴$ Both are perpendicular to each other.

JCECE-2014

Motion in Plane

143583 If a vector $A$ having a magnitude of 8 is added to a vector $B$ which lies along $x$ -axis, then the resultant of two vectors lies along $y$ -axis and has magnitude twice that of $B$ . The magnitude of $B$ is

1

\frac{6}{\sqrt{5}}

2

\frac{12}{\sqrt{5}}

3

\frac{16}{\sqrt{5}}

4

\frac{8}{\sqrt{5}}

Explanation:

D Given,

$\vec{A} = 8$ units
Since, $\vec{B}$ is along $x$ - axis and resultant of two vector $\vec{C}$ lies on $y -$ axis
So, $\vec{B}$ and $\vec{C}$ are perpendicular vector.
Hence,
$| \vec{A} |^{2} = | \vec{B} |^{2} + | \vec{C} |^{2}$
$| 8 |^{2} = | B |^{2} + {2 B |}^{2}$
$64 = 5 B^{2}$
$B^{2} = \frac{64}{5}$
$∴ B = \frac{8}{\sqrt{5}}$

JCECE-2012

Motion in Plane

143584 It two forces each of $2 N$ are inclined at $60^{\circ}$ , then resultant force is:

1

2 N

2

2 \sqrt{5} N

3

2 \sqrt{3} N

4

4 \sqrt{2} N

Explanation:

C Let A \& B be two forces,
$∴ R = \sqrt{A^{2} + B^{2} + 2 AB \cos θ}$
$R = \sqrt{2^{2} + 2^{2} + 2 \times 2 \times 2 \cos 60^{\circ}}$
$(\cos 60^{\circ} = 1 / 2)$
$∴ R = \sqrt{12} = 2 \sqrt{3} N$
$∴ R = 2 \sqrt{3} N$

JCECE-2006

Motion in Plane

143586 Calculate the work done when a force $\vec{F} = 2 \hat{i} + 3 \hat{j} - 5 \hat{k}$ units acts on a body producing a displacement $\vec{s} = 2 \hat{i} + 4 \hat{j} + 3 \hat{k}$ units :

1 1 unit

2 20 unit

3 5 unit

4 zero

Explanation:

A Given that, $\vec{F} = 2 \hat{i} + 3 \hat{j} - 5 \hat{k}$
$\vec{s} = 2 \hat{i} + 4 \hat{j} + 3 \hat{k}$
$W = \vec{F} \cdot \vec{S}$
$W = (2 \hat{i} + 3 \hat{j} - 5 \hat{k}) \cdot (2 \hat{i} + 4 \hat{j} + 3 \hat{k})$
$W = 4 + 12 - 15$
$W = 1$ Unit

JCECE-2003

Motion in Plane

143587 Three forces acting on a body are shown in the figure. To have the resultant force only along the $y$ -direction, the magnitude of the minimum additional force needed along $O X$ is

1

\frac{\sqrt{3}}{4} N

2

\sqrt{13} N

3

0.5 N

4

1.5 N

Explanation:

C $∵ R = \sqrt{{(\sum F_{x})}^{2} + {(\sum F_{y})}^{2}}$
Let the additional force $F$ be directed along the positive $x$ -direction.
Taking $X$ -component, the total force should be zero.
Let $F$ be the magnitude of minimum force which must be along $x$ -direction, by resolving the vector we get-
$1 \times \cos 60^{\circ} + 2 \sin 30^{\circ} + F - 4 \sin 30^{\circ} = 0$
$\frac{1}{2} + 1 + F - 2 = 0$
$F = 1 / 2 = 0.5 N$

COMEDK 2019

Motion in Plane

143582 The position vector of a particle is $r = (α \cos ω t) \hat{i} + (α \sin ω t) \hat{j}$ . The velocity vector of the particle is

1 Parallel to position vector

2 Perpendicular to position vector

3 Directed towards the origin

4 Directed away from the origin

Explanation:

B $\vec{r} = α \cos ω t \hat{i} + α \sin ω \hat{j}$
$\vec{v} = \frac{d \vec{r}}{dt} = (- α \sin ω t) ω \hat{i} + (α \cos ω t) ω \hat{j}$
$\vec{v} \cdot \vec{r} = [(- α ω \sin ω t) \hat{i} + (α ω \cos ω t) \hat{j}] \cdot [(α \cos ω t) \hat{i} + (α \sin ω t) \hat{j}$
$= - α^{2} ω \sin ω t \cdot \cos ω t + α^{2} ω \cos ω t \cdot \sin ω t$
$\vec{v} \cdot \vec{r} = 0$
$∵$ Dot product of $\vec{v}$ and $\vec{r}$ is zero
$∴$ Both are perpendicular to each other.

JCECE-2014

Motion in Plane

143583 If a vector $A$ having a magnitude of 8 is added to a vector $B$ which lies along $x$ -axis, then the resultant of two vectors lies along $y$ -axis and has magnitude twice that of $B$ . The magnitude of $B$ is

1

\frac{6}{\sqrt{5}}

2

\frac{12}{\sqrt{5}}

3

\frac{16}{\sqrt{5}}

4

\frac{8}{\sqrt{5}}

Explanation:

D Given,

$\vec{A} = 8$ units
Since, $\vec{B}$ is along $x$ - axis and resultant of two vector $\vec{C}$ lies on $y -$ axis
So, $\vec{B}$ and $\vec{C}$ are perpendicular vector.
Hence,
$| \vec{A} |^{2} = | \vec{B} |^{2} + | \vec{C} |^{2}$
$| 8 |^{2} = | B |^{2} + {2 B |}^{2}$
$64 = 5 B^{2}$
$B^{2} = \frac{64}{5}$
$∴ B = \frac{8}{\sqrt{5}}$

JCECE-2012

Motion in Plane

143584 It two forces each of $2 N$ are inclined at $60^{\circ}$ , then resultant force is:

1

2 N

2

2 \sqrt{5} N

3

2 \sqrt{3} N

4

4 \sqrt{2} N

Explanation:

C Let A \& B be two forces,
$∴ R = \sqrt{A^{2} + B^{2} + 2 AB \cos θ}$
$R = \sqrt{2^{2} + 2^{2} + 2 \times 2 \times 2 \cos 60^{\circ}}$
$(\cos 60^{\circ} = 1 / 2)$
$∴ R = \sqrt{12} = 2 \sqrt{3} N$
$∴ R = 2 \sqrt{3} N$

JCECE-2006

Motion in Plane

143586 Calculate the work done when a force $\vec{F} = 2 \hat{i} + 3 \hat{j} - 5 \hat{k}$ units acts on a body producing a displacement $\vec{s} = 2 \hat{i} + 4 \hat{j} + 3 \hat{k}$ units :

1 1 unit

2 20 unit

3 5 unit

4 zero

Explanation:

A Given that, $\vec{F} = 2 \hat{i} + 3 \hat{j} - 5 \hat{k}$
$\vec{s} = 2 \hat{i} + 4 \hat{j} + 3 \hat{k}$
$W = \vec{F} \cdot \vec{S}$
$W = (2 \hat{i} + 3 \hat{j} - 5 \hat{k}) \cdot (2 \hat{i} + 4 \hat{j} + 3 \hat{k})$
$W = 4 + 12 - 15$
$W = 1$ Unit

JCECE-2003

Motion in Plane

143587 Three forces acting on a body are shown in the figure. To have the resultant force only along the $y$ -direction, the magnitude of the minimum additional force needed along $O X$ is

1

\frac{\sqrt{3}}{4} N

2

\sqrt{13} N

3

0.5 N

4

1.5 N

Explanation:

C $∵ R = \sqrt{{(\sum F_{x})}^{2} + {(\sum F_{y})}^{2}}$
Let the additional force $F$ be directed along the positive $x$ -direction.
Taking $X$ -component, the total force should be zero.
Let $F$ be the magnitude of minimum force which must be along $x$ -direction, by resolving the vector we get-
$1 \times \cos 60^{\circ} + 2 \sin 30^{\circ} + F - 4 \sin 30^{\circ} = 0$
$\frac{1}{2} + 1 + F - 2 = 0$
$F = 1 / 2 = 0.5 N$

COMEDK 2019

Motion in Plane

143582 The position vector of a particle is $r = (α \cos ω t) \hat{i} + (α \sin ω t) \hat{j}$ . The velocity vector of the particle is

1 Parallel to position vector

2 Perpendicular to position vector

3 Directed towards the origin

4 Directed away from the origin

Explanation:

B $\vec{r} = α \cos ω t \hat{i} + α \sin ω \hat{j}$
$\vec{v} = \frac{d \vec{r}}{dt} = (- α \sin ω t) ω \hat{i} + (α \cos ω t) ω \hat{j}$
$\vec{v} \cdot \vec{r} = [(- α ω \sin ω t) \hat{i} + (α ω \cos ω t) \hat{j}] \cdot [(α \cos ω t) \hat{i} + (α \sin ω t) \hat{j}$
$= - α^{2} ω \sin ω t \cdot \cos ω t + α^{2} ω \cos ω t \cdot \sin ω t$
$\vec{v} \cdot \vec{r} = 0$
$∵$ Dot product of $\vec{v}$ and $\vec{r}$ is zero
$∴$ Both are perpendicular to each other.

JCECE-2014

Motion in Plane

143583 If a vector $A$ having a magnitude of 8 is added to a vector $B$ which lies along $x$ -axis, then the resultant of two vectors lies along $y$ -axis and has magnitude twice that of $B$ . The magnitude of $B$ is

1

\frac{6}{\sqrt{5}}

2

\frac{12}{\sqrt{5}}

3

\frac{16}{\sqrt{5}}

4

\frac{8}{\sqrt{5}}

Explanation:

D Given,

$\vec{A} = 8$ units
Since, $\vec{B}$ is along $x$ - axis and resultant of two vector $\vec{C}$ lies on $y -$ axis
So, $\vec{B}$ and $\vec{C}$ are perpendicular vector.
Hence,
$| \vec{A} |^{2} = | \vec{B} |^{2} + | \vec{C} |^{2}$
$| 8 |^{2} = | B |^{2} + {2 B |}^{2}$
$64 = 5 B^{2}$
$B^{2} = \frac{64}{5}$
$∴ B = \frac{8}{\sqrt{5}}$

JCECE-2012

Motion in Plane

143584 It two forces each of $2 N$ are inclined at $60^{\circ}$ , then resultant force is:

1

2 N

2

2 \sqrt{5} N

3

2 \sqrt{3} N

4

4 \sqrt{2} N

Explanation:

C Let A \& B be two forces,
$∴ R = \sqrt{A^{2} + B^{2} + 2 AB \cos θ}$
$R = \sqrt{2^{2} + 2^{2} + 2 \times 2 \times 2 \cos 60^{\circ}}$
$(\cos 60^{\circ} = 1 / 2)$
$∴ R = \sqrt{12} = 2 \sqrt{3} N$
$∴ R = 2 \sqrt{3} N$

JCECE-2006

Motion in Plane

143586 Calculate the work done when a force $\vec{F} = 2 \hat{i} + 3 \hat{j} - 5 \hat{k}$ units acts on a body producing a displacement $\vec{s} = 2 \hat{i} + 4 \hat{j} + 3 \hat{k}$ units :

1 1 unit

2 20 unit

3 5 unit

4 zero

Explanation:

A Given that, $\vec{F} = 2 \hat{i} + 3 \hat{j} - 5 \hat{k}$
$\vec{s} = 2 \hat{i} + 4 \hat{j} + 3 \hat{k}$
$W = \vec{F} \cdot \vec{S}$
$W = (2 \hat{i} + 3 \hat{j} - 5 \hat{k}) \cdot (2 \hat{i} + 4 \hat{j} + 3 \hat{k})$
$W = 4 + 12 - 15$
$W = 1$ Unit

JCECE-2003

Motion in Plane

143587 Three forces acting on a body are shown in the figure. To have the resultant force only along the $y$ -direction, the magnitude of the minimum additional force needed along $O X$ is

1

\frac{\sqrt{3}}{4} N

2

\sqrt{13} N

3

0.5 N

4

1.5 N

Explanation:

C $∵ R = \sqrt{{(\sum F_{x})}^{2} + {(\sum F_{y})}^{2}}$
Let the additional force $F$ be directed along the positive $x$ -direction.
Taking $X$ -component, the total force should be zero.
Let $F$ be the magnitude of minimum force which must be along $x$ -direction, by resolving the vector we get-
$1 \times \cos 60^{\circ} + 2 \sin 30^{\circ} + F - 4 \sin 30^{\circ} = 0$
$\frac{1}{2} + 1 + F - 2 = 0$
$F = 1 / 2 = 0.5 N$

COMEDK 2019