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Physics NCERT Exemplar Solutions Class 11th Chapter Seven Physics Class 11th Systems of Particles and Rotational Motion

The vector sum of a system of non-collinear forces acting on a rigid body is given to be non-zero. If the vector sum of all the torques due to the system of forces about a certain point is found to be zero, does this mean that it is necessarily zero about any arbitrary point?

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Payal Gupta | Contributor-Level 10

4 months ago

This is a short answer type question as classified in NCERT Exemplar

no $\sum_{i} F_{i} \neq 0$

The sum of torques about a certain point O $\sum_{i} r_{i} \times F_{i} = 0$

The sum of torques about any other O $\sum_{i} r_{i} F_{i} = 0$

The sum of torques about any other point O’

$\sum_{i} (r_{i} - a) \times F_{i} = \sum_{i} r_{i} \times F_{i} - a \times \sum_{i} F_{i}$

Here the second term need not vanish.

Sum of all torques about any point is zero.

This is a short answer type question as classified in NCERT Exemplarno <math> <munder> <mo>∑</mo> <mrow> <mi>i</mi> </mrow> </munder> <mrow> <mrow> <msub> <mrow> <mrow> <mi>F</mi> </mrow> </mrow> <mrow> <mrow> <mi>i</mi> </mrow> </mrow> </msub> <mo>≠</mo> <mn>0</mn> </mrow> </mrow> </math> The sum of torques about a certain point O <math> <munder> <mo>∑</mo> <mrow> <mi>i</mi> </mrow> </munder> <mrow> <mrow> <msub> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mi>i</mi> </mrow> </mrow> </msub> <mo>×</mo> <msub> <mrow> <mrow> <mi>F</mi> </mrow> </mrow> <mrow> <mrow> <mi>i</mi> </mrow> </mrow> </msub> <mo>=</mo> <mn>0</mn> </mrow> </mrow> </math> The sum of torques about any other O <math> <munder> <mo>∑</mo> <mrow> <mi>i</mi> </mrow> </munder> <mrow> <mrow> <msub> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mi>i</mi> </mrow> </mrow> </msub> <msub> <mrow> <mrow> <mi>F</mi> </mrow> </mrow> <mrow> <mrow> <mi>i</mi> </mrow> </mrow> </msub> <mo>=</mo> <mn>0</mn> </mrow> </mrow> </math> The sum of torques about any other point O’ <math> <munder> <mo>∑</mo> <mrow> <mi>i</mi> </mrow> </munder> <mrow> <mrow> <mo> (</mo> <msub> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mi>i</mi> </mrow> </mrow> </msub> <mo>-</mo> <mi>a</mi> <mo>)</mo> <mo>×</mo> </mrow> </mrow> <msub> <mrow> <mrow> <mi>F</mi> </mrow> </mrow> <mrow> <mrow> <mi>i</mi> </mrow> </mrow> </msub> <mo>=</mo> <munder> <mo>∑</mo> <mrow> <mi>i</mi> </mrow> </munder> <mrow> <mrow> <msub> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mi>i</mi> </mrow> </mrow> </msub> <mo>×</mo> <msub> <mrow> <mrow> <mi>F</mi> </mrow> </mrow> <mrow> <mrow> <mi>i</mi> </mrow> </mrow> </msub> <mo>-</mo> <mi>a</mi> <mo>×</mo> <munder> <mo>∑</mo> <mrow> <mi>i</mi> </mrow> </munder> <mrow> <mrow> <msub> <mrow> <mrow> <mi>F</mi> </mrow> </mrow> <mrow> <mrow> <mi>i</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mrow> </mrow> </math> Here the second term need not vanish.Sum of all torques about any point is zero.

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A circular disc reaches from top to bottom of an inclined plane of length ‘L’ when it slips down the plane, it takes time ‘t₁’. When it rolls down the plane, it takes time t₂. The value of $\frac{t_{1}}{t_{2}} i s \sqrt[]{\frac{3}{x}}$ The value of x will be__________.

alok kumar singh

Case – I : When disk slides down

$t_{1} = \sqrt[]{\frac{2 L}{g s i n θ}} . . . . . . . . . . (i)$

Case – II : When disk rolls down

$t_{2} = \sqrt[]{\frac{2 L}{\frac{g s i n θ}{1 + β^{2}}}} = \sqrt[]{\frac{2 L}{\frac{g s i n θ}{1 + {(\frac{k}{R})}^{2}}}} = \sqrt[]{\frac{2 L}{\frac{g s i n θ}{1 + {(\frac{R / \sqrt[]{2}}{R})}^{2}}}} = \sqrt[]{\frac{2 L}{\frac{g s i n θ}{1 + \frac{1}{2}}}} = \sqrt[]{\frac{4 L}{3 g s i n θ}} . . . . . . (i i)$

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With reference to Fig. of a cube of edge a and mass m, state whether the following are true or false. (O is the centre of the cube.)

(a) The moment of inertia of cube about z-axis is I_z = I_x + I_y

(b) The moment of inertia of cube about z′ is I_z =I_z +ma²/2

(c) The moment of inertia of cube about z″ is I_z+ma²/2 (d) I_x = I_y

Payal Gupta

This is a multiple choice type question as classified in NCERT Exemplar

a, b, d

a) according to the perpendicular axes theorem statement 1 is wrong

b) As z’|z so distance between them = a $\frac{\sqrt[]{2}}{2} = \frac{a}{\sqrt[]{2}}$

So according to parallel axes theorem I_z’=I_z+m (a/ $\sqrt[]{2}$ )²= I_z+ma²/2

Hence b is true

c) z’ is not parallel to z hence Parallel axes does not applied so statement is false

d) as x and y axes are symmetrical . hence I_x=I_yso d is true

Figure 7.5 shows a lamina in x-y plane. Two axes z and z′ pass perpendicular to its plane. A force F acts in the plane of lamina at point P as shown. Which of the following are true? (The point P is closer to z′-axis than the z-axis.)

(a) Torque τ caused by F about z axis is along ˆ -k.

(b) Torque τ′ caused by F about z′ axis is along ˆ -k.

(c) Torque τ caused by F about z axis is greater in magnitude than that about z axis.

(d) Total torque is given be τ = τ + τ′.

Payal Gupta

This is a multiple choice type question as classified in NCERT Exemplar

b, c

a) When r>r’

Torque about z-axis t=r $\times$ F

b) t’=r’ $\times F$ which is along negative z axis

c) t_z=Fr = magnitude of torque about z axis where r is perpendicular between F and z axis so torque along positive z axis is greater than negative z axis.

d) We are always calculating resultant torque about common axis. Hence total torque not equal to combination of torque along both axis of z, because they are not on common axis.

The net external torque on a system of particles about an axis is zero. Which of the following are compatible with it ?

(a) The forces may be acting radially from a point on the axis.

(b) The forces may be acting on the axis of rotation.

(c) The forces may be acting parallel to the axis of rotation.

(d) The torque caused by some forces may be equal and opposite to that caused by other forces

Payal Gupta

This is a multiple choice type question as classified in NCERT Exemplar

a, b, c, d

As we know torque = r $\times$ F = rFsin $θ$

a) when forces act radially angle =0 hence torque =0

b) when forces are acting on the axis of rotation r=0 torque=0

c) when forces acting parallel to the axis of rotation angle =0 so torque =0

d) when torque by forces are equal and opposite torque net = t₁-t₂=0

Figure shows two identical particles 1 and 2, each of mass m, moving in opposite directions with same speed v along parallel lines. At a particular instant, r₁ and r₂ are their respective position vectors drawn from point A which is in the plane of the parallel lines.

Choose the correct options:

(a) Angular momentum l₁ of particle 1 about A is I = mvd₁

(b) Angular momentum l2 of particle 2 about A is l₂= mvr₂

(c) Total angular momentum of the system about A is l= mv(r₁+r₂)

(d) Total angular momentum of the system about A is l = mv(d₂-d₁)⊗ represents a unit vector coming out of the page. ⊗ represents a unit vector going into the page.

Payal Gupta

This is a multiple choice type question as classified in NCERT Exemplar

(a), (b)As we know L= r $\times$ p where r is position vector and p is the linear momentum . the direction of L is perpendicular to both r and p by right hand rule.

For particle 1

I₁=r_{1
$\times$}mv is out of the plane of the paper and perpendicular to r₁ and p . similarly I₂=r_{2
$\times$}m (-v) is into the plane of the paper and perpendicular to r₂ and -p.

Hence total angular momentum

L= L₁+L₂= I₁=r_{1
$\times$}mv+ (r_{2
$\times$}m (-v)

L= mvd₁-mvd₂ as d₂>d₁

So total angular momentum will be inwards so I = l = mv (d₂-d₁)⊗

L= mvd₁-mvd₂ as d₂>d₁

So total angular momentum will be inwards so I = l = mv (d₂-

...more

This is a multiple choice type question as classified in NCERT Exemplar

(a), (b)As we know L= r $\times$ p where r is position vector and p is the linear momentum . the direction of L is perpendicular to both r and p by right hand rule.

For particle 1

I₁=r_{1
$\times$}mv is out of the plane of the paper and perpendicular to r₁ and p . similarly I₂=r_{2
$\times$}m (-v) is into the plane of the paper and perpendicular to r₂ and -p.

Hence total angular momentum

L= L₁+L₂= I₁=r_{1
$\times$}mv+ (r_{2
$\times$}m (-v)

L= mvd₁-mvd₂ as d₂>d₁

So total angular momentum will be inwards so I = l = mv (d₂-d₁)⊗

L= mvd₁-mvd₂ as d₂>d₁

So total angular momentum will be inwards so I = l = mv (d₂-d₁)?

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The vector sum of a system of non-collinear forces acting on a rigid body is given to be non-zero. If the vector sum of all the torques due to the system of forces about a certain point is found to be zero, does this mean that it is necessarily zero about any arbitrary point?

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