Physics NCERT Exemplar Solutions Class 11th Chapter Seven

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4 months ago

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Payal Gupta

Contributor-Level 10

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 2 2 = a 2

So according to parallel axes theorem Iz'=Iz+m (a/ 2 )2= Iz+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 Ix=Iy so d is true

d) as x and y axes are symmetrical . hence Ix=Iy so d is true

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Payal Gupta

Contributor-Level 10

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

b, c

a) When r>r'

Torque about z-axis t=r * F

b) t'=r' * F which is along negative z axis

c) tz=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.

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4 months ago

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Payal Gupta

Contributor-Level 10

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

a, b, c, d

As we know torque = r * 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

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Payal Gupta

Contributor-Level 10

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

(a), (b)As we know L= r * 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=r1 * mv is out of the plane of the paper and perpendicular to r1 and p . similarly I2=r2 * m (-v) is into the plane of the paper  and perpendicular to r2 and -p.

Hence total angular momentum

L= L1+L2= I1=r1 * mv+ (r2 * 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)?

...more

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Payal Gupta

Contributor-Level 10

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

(a), (c) For general rotational motion where axis of rotation is not symmetric . angular momentum L and angular velocity w need not be parallel. For general translational motion momentum p=mv hence p and v are always parallel.

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Payal Gupta

Contributor-Level 10

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

(a) As no external torque acts on the system, angular momentum should be conserved hence Iw=constant

Where I is moment of inertia of the system and w is the angular velocity of the system.

So I1w1= I2w2

Where w1 and w2 are angular velocities before and after jumping.

Iw= I 2 w 2

w2=2w

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Payal Gupta

Contributor-Level 10

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

(a) Density is given as ρ (x) = a (1+bx2)

Where a and b are constant and 0 x 1

Let b=0 ρ (x) = a= constant

Hence COM will be at x =0.5m

Putting b=0 ain above option only a gives this answer.

We should not check options by putting a=0 because ρ = 0  for a=0

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Payal Gupta

Contributor-Level 10

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

(c) First the centre of mass of the system was on the dotted line and was shifted towards Q from the centre. When mass is removed it will be on the same line but shifed away from the centre and below 3rd quadrant.

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Payal Gupta

Contributor-Level 10

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

(b) When the small piece Q is removed and glued to the centre of the plate . the mass closer to the z axis . hence moment of inertia decreases.

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Payal Gupta

Contributor-Level 10

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

(d) we know that angular acceleration α = d w d t , given w=constant

where w is the angular velocity of the disc

α = d w d t = 0 d t = 0

Hence angular acceleration is zero.

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