QBManager

Motion in Plane

143556 What is the torque of a force $3 \hat{i} + 7 \hat{j} + 4 \hat{k}$ about the origin, if the force acts on a particle whose position vector is $2 \hat{i} + 2 \hat{j} + 1 \hat{k}$ ?

1

\hat{i} - 5 \hat{j} + 8 \hat{k}

2

2 \hat{i} + 2 \hat{j} + 2 \hat{k}

3

\hat{i} + \hat{j} + \hat{k}

4

3 \hat{i} + 2 \hat{j} + 3 \hat{k}

Explanation:

A Given that,
$\vec{F} = 3 \hat{i} + 7 \hat{j} + 4 \hat{k}$
$\vec{r} = 2 \hat{i} + 2 \hat{j} + 1 \hat{k}$
$τ = \vec{r} \times \vec{F}$
$τ = | \begin{array}{ccc} \hat{i} \hat{j} \hat{k} \\ 221 \\ 374 \end{array} |$
$τ = \hat{i} (8 - 7) - \hat{j} (8 - 3) + \hat{k} (14 - 6)$
$τ = \hat{i} - 5 \hat{j} + 8 \hat{k}$

J and K-CET-2014

Motion in Plane

143559 A certain vector in the $x y$ plane has an $x$ component of $12 m$ and a y-component of $8 m$ . It is then rotated in the $x y$ plane so that its $x$ component is halved. Then its new $y$ component is approximately

1

14 m

2

13.11 m

3

10 m

4

2.0 m

Explanation:

B $x -$ component $= 12 cm$
$y -$ component $= 8 cm$
Length of the resultant vector (R)
$= \sqrt{x^{2} + y^{2}} = \sqrt{12^{2} + 8^{2}}$
$= \sqrt{144 + 64} = \sqrt{208}$
Now,
$x^{'} = \frac{x}{2} = \frac{12}{2} = 6 cm$
Resultant will always be constant even after the rotation So,
$\sqrt{{(x^{'})}^{2} + {(y^{'})}^{2}} = \sqrt{208}$
$(6)^{2} + {(y^{'})}^{2} = 208$
${(y^{'})}^{2} = 208 - 36 = 172$
$(y^{'}) = \sqrt{172}$
$(y^{'}) = 13.11 m$

J and K-CET-2012

Motion in Plane

143560 Figure shows three forces ${\vec{F}}_{1}, {\vec{F}}_{2}$ and ${\vec{F}}_{3}$ acting along the sides of an equilateral triangle. If the total torque acting at point ' $O$ ' (centre of the triangle) is zero then the magnitude of ${\vec{F}}_{3}$ is

1

F_{1} + F_{2}

2

F_{1} - F_{2}

3

\frac{F_{1} - F_{2}}{2}

4

\frac{F_{1}}{F_{2}}

Explanation:

A Force ${\vec{F}}_{1}$ and ${\vec{F}}_{2}$ produce anticlockwise torque while force ${\vec{F}}_{3}$ produces clockwise torque. The torque in the two directions balance each other. The perpendicular distance of the forces from the centre is the same.
$∴ F_{1} r + F_{2} r = F_{3} r$ or $F_{1} + F_{2} = F_{3}$

MHT-CET 2020

Motion in Plane

143561 Two forces each of magnitude ' $P$ ' act at right angles. Their effect is neutralized by a third force acting along their bisector in opposite direction. The magnitude of the third force is $[\cos \frac{π}{2} = 0]$

1

P

2

\frac{P}{\sqrt{2}}

3

\sqrt{2} P

4

\frac{P}{2}

Explanation:

C The third force will have magnitude equal to their resultant,

$R = \sqrt{R_{1}^{2} + R_{2}^{2} + 2 R_{1} R_{2} \cos θ}$
$R = \sqrt{P^{2} + P^{2} + 2 \cdot P \cdot P \cos 90^{\circ}}$
$R = \sqrt{2 P^{2}}$
$R = \sqrt{2} P$

MHT-CET 2020

Motion in Plane

143556 What is the torque of a force $3 \hat{i} + 7 \hat{j} + 4 \hat{k}$ about the origin, if the force acts on a particle whose position vector is $2 \hat{i} + 2 \hat{j} + 1 \hat{k}$ ?

1

\hat{i} - 5 \hat{j} + 8 \hat{k}

2

2 \hat{i} + 2 \hat{j} + 2 \hat{k}

3

\hat{i} + \hat{j} + \hat{k}

4

3 \hat{i} + 2 \hat{j} + 3 \hat{k}

Explanation:

A Given that,
$\vec{F} = 3 \hat{i} + 7 \hat{j} + 4 \hat{k}$
$\vec{r} = 2 \hat{i} + 2 \hat{j} + 1 \hat{k}$
$τ = \vec{r} \times \vec{F}$
$τ = | \begin{array}{ccc} \hat{i} \hat{j} \hat{k} \\ 221 \\ 374 \end{array} |$
$τ = \hat{i} (8 - 7) - \hat{j} (8 - 3) + \hat{k} (14 - 6)$
$τ = \hat{i} - 5 \hat{j} + 8 \hat{k}$

J and K-CET-2014

Motion in Plane

143559 A certain vector in the $x y$ plane has an $x$ component of $12 m$ and a y-component of $8 m$ . It is then rotated in the $x y$ plane so that its $x$ component is halved. Then its new $y$ component is approximately

1

14 m

2

13.11 m

3

10 m

4

2.0 m

Explanation:

B $x -$ component $= 12 cm$
$y -$ component $= 8 cm$
Length of the resultant vector (R)
$= \sqrt{x^{2} + y^{2}} = \sqrt{12^{2} + 8^{2}}$
$= \sqrt{144 + 64} = \sqrt{208}$
Now,
$x^{'} = \frac{x}{2} = \frac{12}{2} = 6 cm$
Resultant will always be constant even after the rotation So,
$\sqrt{{(x^{'})}^{2} + {(y^{'})}^{2}} = \sqrt{208}$
$(6)^{2} + {(y^{'})}^{2} = 208$
${(y^{'})}^{2} = 208 - 36 = 172$
$(y^{'}) = \sqrt{172}$
$(y^{'}) = 13.11 m$

J and K-CET-2012

Motion in Plane

143560 Figure shows three forces ${\vec{F}}_{1}, {\vec{F}}_{2}$ and ${\vec{F}}_{3}$ acting along the sides of an equilateral triangle. If the total torque acting at point ' $O$ ' (centre of the triangle) is zero then the magnitude of ${\vec{F}}_{3}$ is

1

F_{1} + F_{2}

2

F_{1} - F_{2}

3

\frac{F_{1} - F_{2}}{2}

4

\frac{F_{1}}{F_{2}}

Explanation:

A Force ${\vec{F}}_{1}$ and ${\vec{F}}_{2}$ produce anticlockwise torque while force ${\vec{F}}_{3}$ produces clockwise torque. The torque in the two directions balance each other. The perpendicular distance of the forces from the centre is the same.
$∴ F_{1} r + F_{2} r = F_{3} r$ or $F_{1} + F_{2} = F_{3}$

MHT-CET 2020

Motion in Plane

143561 Two forces each of magnitude ' $P$ ' act at right angles. Their effect is neutralized by a third force acting along their bisector in opposite direction. The magnitude of the third force is $[\cos \frac{π}{2} = 0]$

1

P

2

\frac{P}{\sqrt{2}}

3

\sqrt{2} P

4

\frac{P}{2}

Explanation:

C The third force will have magnitude equal to their resultant,

$R = \sqrt{R_{1}^{2} + R_{2}^{2} + 2 R_{1} R_{2} \cos θ}$
$R = \sqrt{P^{2} + P^{2} + 2 \cdot P \cdot P \cos 90^{\circ}}$
$R = \sqrt{2 P^{2}}$
$R = \sqrt{2} P$

MHT-CET 2020

Motion in Plane

143556 What is the torque of a force $3 \hat{i} + 7 \hat{j} + 4 \hat{k}$ about the origin, if the force acts on a particle whose position vector is $2 \hat{i} + 2 \hat{j} + 1 \hat{k}$ ?

1

\hat{i} - 5 \hat{j} + 8 \hat{k}

2

2 \hat{i} + 2 \hat{j} + 2 \hat{k}

3

\hat{i} + \hat{j} + \hat{k}

4

3 \hat{i} + 2 \hat{j} + 3 \hat{k}

Explanation:

A Given that,
$\vec{F} = 3 \hat{i} + 7 \hat{j} + 4 \hat{k}$
$\vec{r} = 2 \hat{i} + 2 \hat{j} + 1 \hat{k}$
$τ = \vec{r} \times \vec{F}$
$τ = | \begin{array}{ccc} \hat{i} \hat{j} \hat{k} \\ 221 \\ 374 \end{array} |$
$τ = \hat{i} (8 - 7) - \hat{j} (8 - 3) + \hat{k} (14 - 6)$
$τ = \hat{i} - 5 \hat{j} + 8 \hat{k}$

J and K-CET-2014

Motion in Plane

143559 A certain vector in the $x y$ plane has an $x$ component of $12 m$ and a y-component of $8 m$ . It is then rotated in the $x y$ plane so that its $x$ component is halved. Then its new $y$ component is approximately

1

14 m

2

13.11 m

3

10 m

4

2.0 m

Explanation:

B $x -$ component $= 12 cm$
$y -$ component $= 8 cm$
Length of the resultant vector (R)
$= \sqrt{x^{2} + y^{2}} = \sqrt{12^{2} + 8^{2}}$
$= \sqrt{144 + 64} = \sqrt{208}$
Now,
$x^{'} = \frac{x}{2} = \frac{12}{2} = 6 cm$
Resultant will always be constant even after the rotation So,
$\sqrt{{(x^{'})}^{2} + {(y^{'})}^{2}} = \sqrt{208}$
$(6)^{2} + {(y^{'})}^{2} = 208$
${(y^{'})}^{2} = 208 - 36 = 172$
$(y^{'}) = \sqrt{172}$
$(y^{'}) = 13.11 m$

J and K-CET-2012

Motion in Plane

143560 Figure shows three forces ${\vec{F}}_{1}, {\vec{F}}_{2}$ and ${\vec{F}}_{3}$ acting along the sides of an equilateral triangle. If the total torque acting at point ' $O$ ' (centre of the triangle) is zero then the magnitude of ${\vec{F}}_{3}$ is

1

F_{1} + F_{2}

2

F_{1} - F_{2}

3

\frac{F_{1} - F_{2}}{2}

4

\frac{F_{1}}{F_{2}}

Explanation:

A Force ${\vec{F}}_{1}$ and ${\vec{F}}_{2}$ produce anticlockwise torque while force ${\vec{F}}_{3}$ produces clockwise torque. The torque in the two directions balance each other. The perpendicular distance of the forces from the centre is the same.
$∴ F_{1} r + F_{2} r = F_{3} r$ or $F_{1} + F_{2} = F_{3}$

MHT-CET 2020

Motion in Plane

143561 Two forces each of magnitude ' $P$ ' act at right angles. Their effect is neutralized by a third force acting along their bisector in opposite direction. The magnitude of the third force is $[\cos \frac{π}{2} = 0]$

1

P

2

\frac{P}{\sqrt{2}}

3

\sqrt{2} P

4

\frac{P}{2}

Explanation:

C The third force will have magnitude equal to their resultant,

$R = \sqrt{R_{1}^{2} + R_{2}^{2} + 2 R_{1} R_{2} \cos θ}$
$R = \sqrt{P^{2} + P^{2} + 2 \cdot P \cdot P \cos 90^{\circ}}$
$R = \sqrt{2 P^{2}}$
$R = \sqrt{2} P$

MHT-CET 2020

Motion in Plane

143556 What is the torque of a force $3 \hat{i} + 7 \hat{j} + 4 \hat{k}$ about the origin, if the force acts on a particle whose position vector is $2 \hat{i} + 2 \hat{j} + 1 \hat{k}$ ?

1

\hat{i} - 5 \hat{j} + 8 \hat{k}

2

2 \hat{i} + 2 \hat{j} + 2 \hat{k}

3

\hat{i} + \hat{j} + \hat{k}

4

3 \hat{i} + 2 \hat{j} + 3 \hat{k}

Explanation:

A Given that,
$\vec{F} = 3 \hat{i} + 7 \hat{j} + 4 \hat{k}$
$\vec{r} = 2 \hat{i} + 2 \hat{j} + 1 \hat{k}$
$τ = \vec{r} \times \vec{F}$
$τ = | \begin{array}{ccc} \hat{i} \hat{j} \hat{k} \\ 221 \\ 374 \end{array} |$
$τ = \hat{i} (8 - 7) - \hat{j} (8 - 3) + \hat{k} (14 - 6)$
$τ = \hat{i} - 5 \hat{j} + 8 \hat{k}$

J and K-CET-2014

Motion in Plane

143559 A certain vector in the $x y$ plane has an $x$ component of $12 m$ and a y-component of $8 m$ . It is then rotated in the $x y$ plane so that its $x$ component is halved. Then its new $y$ component is approximately

1

14 m

2

13.11 m

3

10 m

4

2.0 m

Explanation:

B $x -$ component $= 12 cm$
$y -$ component $= 8 cm$
Length of the resultant vector (R)
$= \sqrt{x^{2} + y^{2}} = \sqrt{12^{2} + 8^{2}}$
$= \sqrt{144 + 64} = \sqrt{208}$
Now,
$x^{'} = \frac{x}{2} = \frac{12}{2} = 6 cm$
Resultant will always be constant even after the rotation So,
$\sqrt{{(x^{'})}^{2} + {(y^{'})}^{2}} = \sqrt{208}$
$(6)^{2} + {(y^{'})}^{2} = 208$
${(y^{'})}^{2} = 208 - 36 = 172$
$(y^{'}) = \sqrt{172}$
$(y^{'}) = 13.11 m$

J and K-CET-2012

Motion in Plane

143560 Figure shows three forces ${\vec{F}}_{1}, {\vec{F}}_{2}$ and ${\vec{F}}_{3}$ acting along the sides of an equilateral triangle. If the total torque acting at point ' $O$ ' (centre of the triangle) is zero then the magnitude of ${\vec{F}}_{3}$ is

1

F_{1} + F_{2}

2

F_{1} - F_{2}

3

\frac{F_{1} - F_{2}}{2}

4

\frac{F_{1}}{F_{2}}

Explanation:

A Force ${\vec{F}}_{1}$ and ${\vec{F}}_{2}$ produce anticlockwise torque while force ${\vec{F}}_{3}$ produces clockwise torque. The torque in the two directions balance each other. The perpendicular distance of the forces from the centre is the same.
$∴ F_{1} r + F_{2} r = F_{3} r$ or $F_{1} + F_{2} = F_{3}$

MHT-CET 2020

Motion in Plane

143561 Two forces each of magnitude ' $P$ ' act at right angles. Their effect is neutralized by a third force acting along their bisector in opposite direction. The magnitude of the third force is $[\cos \frac{π}{2} = 0]$

1

P

2

\frac{P}{\sqrt{2}}

3

\sqrt{2} P

4

\frac{P}{2}

Explanation:

C The third force will have magnitude equal to their resultant,

$R = \sqrt{R_{1}^{2} + R_{2}^{2} + 2 R_{1} R_{2} \cos θ}$
$R = \sqrt{P^{2} + P^{2} + 2 \cdot P \cdot P \cos 90^{\circ}}$
$R = \sqrt{2 P^{2}}$
$R = \sqrt{2} P$

MHT-CET 2020

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