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PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366191 Assertion :
Torque due to force is maximum when angle is $90^{\circ}$ .
Reason :
The unit of torque is 'Newton-metre'.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Torque $τ = r F \sin θ$ . Torque is maximum when the force is applied perpendicular to the lever arm (angle $θ = 90^{\circ}$ ). The unit of torque is indeed Newton-meter. Reason does not explain Assertion. So correct option is (2).

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366192 A force of $40 N$ acts on a point $B$ at the end of an L-shaped object as shown in the figure. The angle $θ$ that will produce maximum moment of the force about point $A$ is given by:
supporting img

1

\tan θ = 4

2

\tan θ = \frac{1}{4}

3

\tan θ = \frac{1}{2}

4

\tan θ = 2

Explanation:

Moment of force will be maximum when line of action of force is perpendicular to line $AB$ .
supporting img
$\tan θ = \frac{2}{4} = \frac{1}{2}$

JEE - 2018

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366193 A force $\vec{F} = 4 \hat{i} - 5 \hat{j} + 3 \hat{k}$ is acting at a point ${\vec{r}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ . The torque acting about a point ${\vec{r}}_{2} = 3 \hat{i} - 2 \hat{j} - 3 \hat{k}$ is

1

42 \hat{i} - 30 \hat{j} + 6 \hat{k}

2 Zero

3

42 \hat{i} + 30 \hat{j} - 6 \hat{k}

4

42 \hat{i} + 30 \hat{j} + 6 \hat{k}

Explanation:

Position vector of the point at which force is acting
$\begin{aligned} {\vec{r}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}, {\vec{r}}_{2} = 3 \hat{i} - 2 \hat{j} - 3 \hat{k} \\ \vec{r_{12}} = \vec{r_{1}} - \vec{r_{2}} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) - (3 \hat{i} - 2 \hat{j} - 3 \hat{k}) \\ = - 2 \hat{i} + 4 \hat{j} + 6 \hat{k} \end{aligned}$
Now $\vec{τ} = \vec{r_{12}} \times \vec{F} = (- 2 \hat{i} + 4 \hat{j} + 6 \hat{k}) \times (4 \hat{i} - 5 \hat{j} + 3 \hat{k})$
$\begin{aligned} \vec{τ} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ - 2 & 4 & 6 \\ 4 & - 5 & 3 \end{array} | \\ = \hat{i} (12 + 30) - \hat{j} (- 6 - 24) + \hat{k} (10 - 16) \\ = (42 \hat{i} + 30 \hat{j} - 6 \hat{k}) N - m \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366194 A force $\vec{F} = 5 \hat{i} + 2 \hat{j} - 5 \hat{k}$ acts on a particle whose position vector is $\vec{r} = \hat{i} - 2 \hat{j} + \hat{k}$ . What is the torque about the origin?

1

8 \hat{i} - 10 \hat{j} - 8 \hat{k}

2

8 \hat{i} + 10 \hat{j} + 12 \hat{k}

3

10 \hat{i} - 10 \hat{j} - \hat{k}

4

8 \hat{i} + 10 \hat{j} - 12 \hat{k}

Explanation:

Here, $\vec{r} = \hat{i} - 2 \hat{j} + \hat{k}$ .
$\vec{F} = 5 \hat{i} + 2 \hat{j} - 5 \hat{k}$
Torque, $\vec{τ} = \vec{r} \times \vec{F}$
$= [\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & 1 \\ 5 & 2 & - 5 \end{matrix}]$
$= \hat{i} (10 - 2) - \hat{j} (- 5 - 5) + \hat{k} (2 - (- 10)$
$= 8 \hat{i} + 10 \hat{j} + 12 \hat{k}$

KCET - 2014

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366195 A thin rod of mass $m$ and length $l$ is made to rotate about an axis passing through its centre and perpendicular to it. If its angular velocity changes from 0 to $ω$ in time $t$ , the torque acting on it is

1

\frac{m ℓ^{2} ω}{12 t}

2

\frac{m ℓ^{2} ω}{3 t}

3

\frac{m ℓ^{2} ω}{t}

4

\frac{4 m ℓ^{2} ω}{3 t}

Explanation:

Since, $τ = I α$
So, $τ = (\frac{m (2 ℓ)^{2}}{12}) (\frac{ω}{t})$ or $τ = \frac{m \times 4 ℓ^{2} \times ω}{12 \times t}$
Or $τ = \frac{4 m ℓ^{2} ω}{12 t} = (\frac{m ℓ^{2} ω}{3 t})$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366191 Assertion :
Torque due to force is maximum when angle is $90^{\circ}$ .
Reason :
The unit of torque is 'Newton-metre'.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Torque $τ = r F \sin θ$ . Torque is maximum when the force is applied perpendicular to the lever arm (angle $θ = 90^{\circ}$ ). The unit of torque is indeed Newton-meter. Reason does not explain Assertion. So correct option is (2).

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366192 A force of $40 N$ acts on a point $B$ at the end of an L-shaped object as shown in the figure. The angle $θ$ that will produce maximum moment of the force about point $A$ is given by:
supporting img

1

\tan θ = 4

2

\tan θ = \frac{1}{4}

3

\tan θ = \frac{1}{2}

4

\tan θ = 2

Explanation:

Moment of force will be maximum when line of action of force is perpendicular to line $AB$ .
supporting img
$\tan θ = \frac{2}{4} = \frac{1}{2}$

JEE - 2018

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366193 A force $\vec{F} = 4 \hat{i} - 5 \hat{j} + 3 \hat{k}$ is acting at a point ${\vec{r}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ . The torque acting about a point ${\vec{r}}_{2} = 3 \hat{i} - 2 \hat{j} - 3 \hat{k}$ is

1

42 \hat{i} - 30 \hat{j} + 6 \hat{k}

2 Zero

3

42 \hat{i} + 30 \hat{j} - 6 \hat{k}

4

42 \hat{i} + 30 \hat{j} + 6 \hat{k}

Explanation:

Position vector of the point at which force is acting
$\begin{aligned} {\vec{r}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}, {\vec{r}}_{2} = 3 \hat{i} - 2 \hat{j} - 3 \hat{k} \\ \vec{r_{12}} = \vec{r_{1}} - \vec{r_{2}} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) - (3 \hat{i} - 2 \hat{j} - 3 \hat{k}) \\ = - 2 \hat{i} + 4 \hat{j} + 6 \hat{k} \end{aligned}$
Now $\vec{τ} = \vec{r_{12}} \times \vec{F} = (- 2 \hat{i} + 4 \hat{j} + 6 \hat{k}) \times (4 \hat{i} - 5 \hat{j} + 3 \hat{k})$
$\begin{aligned} \vec{τ} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ - 2 & 4 & 6 \\ 4 & - 5 & 3 \end{array} | \\ = \hat{i} (12 + 30) - \hat{j} (- 6 - 24) + \hat{k} (10 - 16) \\ = (42 \hat{i} + 30 \hat{j} - 6 \hat{k}) N - m \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366194 A force $\vec{F} = 5 \hat{i} + 2 \hat{j} - 5 \hat{k}$ acts on a particle whose position vector is $\vec{r} = \hat{i} - 2 \hat{j} + \hat{k}$ . What is the torque about the origin?

1

8 \hat{i} - 10 \hat{j} - 8 \hat{k}

2

8 \hat{i} + 10 \hat{j} + 12 \hat{k}

3

10 \hat{i} - 10 \hat{j} - \hat{k}

4

8 \hat{i} + 10 \hat{j} - 12 \hat{k}

Explanation:

Here, $\vec{r} = \hat{i} - 2 \hat{j} + \hat{k}$ .
$\vec{F} = 5 \hat{i} + 2 \hat{j} - 5 \hat{k}$
Torque, $\vec{τ} = \vec{r} \times \vec{F}$
$= [\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & 1 \\ 5 & 2 & - 5 \end{matrix}]$
$= \hat{i} (10 - 2) - \hat{j} (- 5 - 5) + \hat{k} (2 - (- 10)$
$= 8 \hat{i} + 10 \hat{j} + 12 \hat{k}$

KCET - 2014

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366195 A thin rod of mass $m$ and length $l$ is made to rotate about an axis passing through its centre and perpendicular to it. If its angular velocity changes from 0 to $ω$ in time $t$ , the torque acting on it is

1

\frac{m ℓ^{2} ω}{12 t}

2

\frac{m ℓ^{2} ω}{3 t}

3

\frac{m ℓ^{2} ω}{t}

4

\frac{4 m ℓ^{2} ω}{3 t}

Explanation:

Since, $τ = I α$
So, $τ = (\frac{m (2 ℓ)^{2}}{12}) (\frac{ω}{t})$ or $τ = \frac{m \times 4 ℓ^{2} \times ω}{12 \times t}$
Or $τ = \frac{4 m ℓ^{2} ω}{12 t} = (\frac{m ℓ^{2} ω}{3 t})$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366191 Assertion :
Torque due to force is maximum when angle is $90^{\circ}$ .
Reason :
The unit of torque is 'Newton-metre'.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Torque $τ = r F \sin θ$ . Torque is maximum when the force is applied perpendicular to the lever arm (angle $θ = 90^{\circ}$ ). The unit of torque is indeed Newton-meter. Reason does not explain Assertion. So correct option is (2).

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366192 A force of $40 N$ acts on a point $B$ at the end of an L-shaped object as shown in the figure. The angle $θ$ that will produce maximum moment of the force about point $A$ is given by:
supporting img

1

\tan θ = 4

2

\tan θ = \frac{1}{4}

3

\tan θ = \frac{1}{2}

4

\tan θ = 2

Explanation:

Moment of force will be maximum when line of action of force is perpendicular to line $AB$ .
supporting img
$\tan θ = \frac{2}{4} = \frac{1}{2}$

JEE - 2018

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366193 A force $\vec{F} = 4 \hat{i} - 5 \hat{j} + 3 \hat{k}$ is acting at a point ${\vec{r}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ . The torque acting about a point ${\vec{r}}_{2} = 3 \hat{i} - 2 \hat{j} - 3 \hat{k}$ is

1

42 \hat{i} - 30 \hat{j} + 6 \hat{k}

2 Zero

3

42 \hat{i} + 30 \hat{j} - 6 \hat{k}

4

42 \hat{i} + 30 \hat{j} + 6 \hat{k}

Explanation:

Position vector of the point at which force is acting
$\begin{aligned} {\vec{r}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}, {\vec{r}}_{2} = 3 \hat{i} - 2 \hat{j} - 3 \hat{k} \\ \vec{r_{12}} = \vec{r_{1}} - \vec{r_{2}} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) - (3 \hat{i} - 2 \hat{j} - 3 \hat{k}) \\ = - 2 \hat{i} + 4 \hat{j} + 6 \hat{k} \end{aligned}$
Now $\vec{τ} = \vec{r_{12}} \times \vec{F} = (- 2 \hat{i} + 4 \hat{j} + 6 \hat{k}) \times (4 \hat{i} - 5 \hat{j} + 3 \hat{k})$
$\begin{aligned} \vec{τ} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ - 2 & 4 & 6 \\ 4 & - 5 & 3 \end{array} | \\ = \hat{i} (12 + 30) - \hat{j} (- 6 - 24) + \hat{k} (10 - 16) \\ = (42 \hat{i} + 30 \hat{j} - 6 \hat{k}) N - m \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366194 A force $\vec{F} = 5 \hat{i} + 2 \hat{j} - 5 \hat{k}$ acts on a particle whose position vector is $\vec{r} = \hat{i} - 2 \hat{j} + \hat{k}$ . What is the torque about the origin?

1

8 \hat{i} - 10 \hat{j} - 8 \hat{k}

2

8 \hat{i} + 10 \hat{j} + 12 \hat{k}

3

10 \hat{i} - 10 \hat{j} - \hat{k}

4

8 \hat{i} + 10 \hat{j} - 12 \hat{k}

Explanation:

Here, $\vec{r} = \hat{i} - 2 \hat{j} + \hat{k}$ .
$\vec{F} = 5 \hat{i} + 2 \hat{j} - 5 \hat{k}$
Torque, $\vec{τ} = \vec{r} \times \vec{F}$
$= [\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & 1 \\ 5 & 2 & - 5 \end{matrix}]$
$= \hat{i} (10 - 2) - \hat{j} (- 5 - 5) + \hat{k} (2 - (- 10)$
$= 8 \hat{i} + 10 \hat{j} + 12 \hat{k}$

KCET - 2014

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366195 A thin rod of mass $m$ and length $l$ is made to rotate about an axis passing through its centre and perpendicular to it. If its angular velocity changes from 0 to $ω$ in time $t$ , the torque acting on it is

1

\frac{m ℓ^{2} ω}{12 t}

2

\frac{m ℓ^{2} ω}{3 t}

3

\frac{m ℓ^{2} ω}{t}

4

\frac{4 m ℓ^{2} ω}{3 t}

Explanation:

Since, $τ = I α$
So, $τ = (\frac{m (2 ℓ)^{2}}{12}) (\frac{ω}{t})$ or $τ = \frac{m \times 4 ℓ^{2} \times ω}{12 \times t}$
Or $τ = \frac{4 m ℓ^{2} ω}{12 t} = (\frac{m ℓ^{2} ω}{3 t})$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366191 Assertion :
Torque due to force is maximum when angle is $90^{\circ}$ .
Reason :
The unit of torque is 'Newton-metre'.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Torque $τ = r F \sin θ$ . Torque is maximum when the force is applied perpendicular to the lever arm (angle $θ = 90^{\circ}$ ). The unit of torque is indeed Newton-meter. Reason does not explain Assertion. So correct option is (2).

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366192 A force of $40 N$ acts on a point $B$ at the end of an L-shaped object as shown in the figure. The angle $θ$ that will produce maximum moment of the force about point $A$ is given by:
supporting img

1

\tan θ = 4

2

\tan θ = \frac{1}{4}

3

\tan θ = \frac{1}{2}

4

\tan θ = 2

Explanation:

Moment of force will be maximum when line of action of force is perpendicular to line $AB$ .
supporting img
$\tan θ = \frac{2}{4} = \frac{1}{2}$

JEE - 2018

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366193 A force $\vec{F} = 4 \hat{i} - 5 \hat{j} + 3 \hat{k}$ is acting at a point ${\vec{r}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ . The torque acting about a point ${\vec{r}}_{2} = 3 \hat{i} - 2 \hat{j} - 3 \hat{k}$ is

1

42 \hat{i} - 30 \hat{j} + 6 \hat{k}

2 Zero

3

42 \hat{i} + 30 \hat{j} - 6 \hat{k}

4

42 \hat{i} + 30 \hat{j} + 6 \hat{k}

Explanation:

Position vector of the point at which force is acting
$\begin{aligned} {\vec{r}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}, {\vec{r}}_{2} = 3 \hat{i} - 2 \hat{j} - 3 \hat{k} \\ \vec{r_{12}} = \vec{r_{1}} - \vec{r_{2}} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) - (3 \hat{i} - 2 \hat{j} - 3 \hat{k}) \\ = - 2 \hat{i} + 4 \hat{j} + 6 \hat{k} \end{aligned}$
Now $\vec{τ} = \vec{r_{12}} \times \vec{F} = (- 2 \hat{i} + 4 \hat{j} + 6 \hat{k}) \times (4 \hat{i} - 5 \hat{j} + 3 \hat{k})$
$\begin{aligned} \vec{τ} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ - 2 & 4 & 6 \\ 4 & - 5 & 3 \end{array} | \\ = \hat{i} (12 + 30) - \hat{j} (- 6 - 24) + \hat{k} (10 - 16) \\ = (42 \hat{i} + 30 \hat{j} - 6 \hat{k}) N - m \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366194 A force $\vec{F} = 5 \hat{i} + 2 \hat{j} - 5 \hat{k}$ acts on a particle whose position vector is $\vec{r} = \hat{i} - 2 \hat{j} + \hat{k}$ . What is the torque about the origin?

1

8 \hat{i} - 10 \hat{j} - 8 \hat{k}

2

8 \hat{i} + 10 \hat{j} + 12 \hat{k}

3

10 \hat{i} - 10 \hat{j} - \hat{k}

4

8 \hat{i} + 10 \hat{j} - 12 \hat{k}

Explanation:

Here, $\vec{r} = \hat{i} - 2 \hat{j} + \hat{k}$ .
$\vec{F} = 5 \hat{i} + 2 \hat{j} - 5 \hat{k}$
Torque, $\vec{τ} = \vec{r} \times \vec{F}$
$= [\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & 1 \\ 5 & 2 & - 5 \end{matrix}]$
$= \hat{i} (10 - 2) - \hat{j} (- 5 - 5) + \hat{k} (2 - (- 10)$
$= 8 \hat{i} + 10 \hat{j} + 12 \hat{k}$

KCET - 2014

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366195 A thin rod of mass $m$ and length $l$ is made to rotate about an axis passing through its centre and perpendicular to it. If its angular velocity changes from 0 to $ω$ in time $t$ , the torque acting on it is

1

\frac{m ℓ^{2} ω}{12 t}

2

\frac{m ℓ^{2} ω}{3 t}

3

\frac{m ℓ^{2} ω}{t}

4

\frac{4 m ℓ^{2} ω}{3 t}

Explanation:

Since, $τ = I α$
So, $τ = (\frac{m (2 ℓ)^{2}}{12}) (\frac{ω}{t})$ or $τ = \frac{m \times 4 ℓ^{2} \times ω}{12 \times t}$
Or $τ = \frac{4 m ℓ^{2} ω}{12 t} = (\frac{m ℓ^{2} ω}{3 t})$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366191 Assertion :
Torque due to force is maximum when angle is $90^{\circ}$ .
Reason :
The unit of torque is 'Newton-metre'.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Torque $τ = r F \sin θ$ . Torque is maximum when the force is applied perpendicular to the lever arm (angle $θ = 90^{\circ}$ ). The unit of torque is indeed Newton-meter. Reason does not explain Assertion. So correct option is (2).

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366192 A force of $40 N$ acts on a point $B$ at the end of an L-shaped object as shown in the figure. The angle $θ$ that will produce maximum moment of the force about point $A$ is given by:
supporting img

1

\tan θ = 4

2

\tan θ = \frac{1}{4}

3

\tan θ = \frac{1}{2}

4

\tan θ = 2

Explanation:

Moment of force will be maximum when line of action of force is perpendicular to line $AB$ .
supporting img
$\tan θ = \frac{2}{4} = \frac{1}{2}$

JEE - 2018

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366193 A force $\vec{F} = 4 \hat{i} - 5 \hat{j} + 3 \hat{k}$ is acting at a point ${\vec{r}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ . The torque acting about a point ${\vec{r}}_{2} = 3 \hat{i} - 2 \hat{j} - 3 \hat{k}$ is

1

42 \hat{i} - 30 \hat{j} + 6 \hat{k}

2 Zero

3

42 \hat{i} + 30 \hat{j} - 6 \hat{k}

4

42 \hat{i} + 30 \hat{j} + 6 \hat{k}

Explanation:

Position vector of the point at which force is acting
$\begin{aligned} {\vec{r}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}, {\vec{r}}_{2} = 3 \hat{i} - 2 \hat{j} - 3 \hat{k} \\ \vec{r_{12}} = \vec{r_{1}} - \vec{r_{2}} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) - (3 \hat{i} - 2 \hat{j} - 3 \hat{k}) \\ = - 2 \hat{i} + 4 \hat{j} + 6 \hat{k} \end{aligned}$
Now $\vec{τ} = \vec{r_{12}} \times \vec{F} = (- 2 \hat{i} + 4 \hat{j} + 6 \hat{k}) \times (4 \hat{i} - 5 \hat{j} + 3 \hat{k})$
$\begin{aligned} \vec{τ} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ - 2 & 4 & 6 \\ 4 & - 5 & 3 \end{array} | \\ = \hat{i} (12 + 30) - \hat{j} (- 6 - 24) + \hat{k} (10 - 16) \\ = (42 \hat{i} + 30 \hat{j} - 6 \hat{k}) N - m \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366194 A force $\vec{F} = 5 \hat{i} + 2 \hat{j} - 5 \hat{k}$ acts on a particle whose position vector is $\vec{r} = \hat{i} - 2 \hat{j} + \hat{k}$ . What is the torque about the origin?

1

8 \hat{i} - 10 \hat{j} - 8 \hat{k}

2

8 \hat{i} + 10 \hat{j} + 12 \hat{k}

3

10 \hat{i} - 10 \hat{j} - \hat{k}

4

8 \hat{i} + 10 \hat{j} - 12 \hat{k}

Explanation:

Here, $\vec{r} = \hat{i} - 2 \hat{j} + \hat{k}$ .
$\vec{F} = 5 \hat{i} + 2 \hat{j} - 5 \hat{k}$
Torque, $\vec{τ} = \vec{r} \times \vec{F}$
$= [\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & 1 \\ 5 & 2 & - 5 \end{matrix}]$
$= \hat{i} (10 - 2) - \hat{j} (- 5 - 5) + \hat{k} (2 - (- 10)$
$= 8 \hat{i} + 10 \hat{j} + 12 \hat{k}$

KCET - 2014

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

366195 A thin rod of mass $m$ and length $l$ is made to rotate about an axis passing through its centre and perpendicular to it. If its angular velocity changes from 0 to $ω$ in time $t$ , the torque acting on it is

1

\frac{m ℓ^{2} ω}{12 t}

2

\frac{m ℓ^{2} ω}{3 t}

3

\frac{m ℓ^{2} ω}{t}

4

\frac{4 m ℓ^{2} ω}{3 t}

Explanation:

Since, $τ = I α$
So, $τ = (\frac{m (2 ℓ)^{2}}{12}) (\frac{ω}{t})$ or $τ = \frac{m \times 4 ℓ^{2} \times ω}{12 \times t}$
Or $τ = \frac{4 m ℓ^{2} ω}{12 t} = (\frac{m ℓ^{2} ω}{3 t})$

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