SAFE HANDS

Assignment for Revision by NKO on :

1. (b)

Sol. M.I of system about

where I1 = moment of inertia of ring about diameter, I2

= I3 = M.I. of inertia of ring about a tangent in a plane

2. (a)

Sol. Let is the moment of inertia of square plate

about the axis which is passing through the centre and

perpendicular to the plane.

[By the theorem of perpendicular axis]

[As AB, A' B' and CD,

C' D' are symmetric axis]

Hence

3. (c)

Sol. The moment of inertia of system about AB side of

triangle

4. (c)

Sol. Moment of inertia of rod AB about its centre and

perpendicular to the length = = I

Now moment of inertia of the rod about the axis which

is passing through O and perpendicular to the plane of

hexagon Irod= [From the theorem of

parallel axes]

Now the moment of inertia of system Isystem

=

Isystem = 5 (12 I) = 60 I [As ]

5. (c)

Sol. Moment of inertia of sphere about it diameter

[As = ]

I =

6. (c)

Sol. Moment of inertia of circular disc about an axis

passing through centre and normal to the circular face

[As

]

or If mass and thickness are

constant.

'YY 321 IIII

222

2

3

2

3

2

1mRmRmRI 2

2

7mR

ZI

'''' DCCDBAABZ IIIII

'''' 2222 DCCDBAABZ IIIII

lII ABCD

CBA IIII

200 mx

2

2

3

am 2

4

3ma

12

2ml

Iml 122

22

12mx

ml

6

5

2

3

12

22

2 mllm

ml

6

566

2

rod

mlI 25ml

Iml 122

2

5

2MRI

23

3

4

5

2RR

VM 3

3

4R

5

15

8R 55

105

176

715

228RR

2

2

1MRI

t

MM

2

1VM tR 2

t

MR 2

t

MI

2

2

1I

A

B

A B

C

D

C

D

B

C m

A

a a x

a

m m

A B

O

x

l

l l

SAFE HANDS

Assignment for Revision by NKO on :

So, in the problem [As ]

7. (a)

Sol. Initial angular momentum of the system about

point O

= Linear momentum × Perpendicular distance of linear

momentum from the axis of rotation

....(i)

Final angular momentum of the system about point O

....(ii)

Applying the law of conservation of angular

momentum

8. (c)

Sol. Kinetic energy = 2304

J.

9. (b)

Sol. Velocity at the bottom (v)

.

10. (b)

Sol. According to law of conservation of energy

.

11. (c)

Sol. N – mg = ………….. (i)

N = mg +

Or ………… (ii)

12. (a)

Sol. In the process of attaching the objects with the

ring no external torque is applied on the system.

Therefore the angular momentum of the system

remains constant.

Li = Lf

Ii i = If f

Iring . = {(Iring) + (IA) + (IB)}

Putting Iring = MR2, IA = IB = mR2 we obtain

=

13. (c)

Sol. The torque applied in both cases are equal in

magnitude, that is equal to rF

Therefore

A

B

B

A

d

d

I

I BA II BA dd

2

LMv

21 II )( 21 II

22

22

LM

LM

2

22

2

LM

LMv

L

v

2

2

1IKR

2)120()32.0(2

1

ghgh

R

K

gh

3

4

2

11

2

1

2

2

2

22)(2

1mrImgh

2

2

mrI

mgh

r

mv2

mg

2

5mg

2

3mg

2/1

2/1m2

2

1mgI

2

1 2

2

1mg

3

ml

2

1 22

mg3l

mv2

m2M

M

rF 21

21 BA II

SAFE HANDS

Assignment for Revision by NKO on :

Since

14. (b)

Sol. Total energy

K = KR + KT = I2 + mv2

= 2 + mr22

= mr22 + mr2 2 = mr22

Now, rotational kinetic energy

KR = I2 = mr22

= =

15. (a)

Sol. Given:

Angular acceleration,

When angular acceleration ( ) is zero, then the torque on

the wheel becomes zero ( )

or t = 1 s

A

B

I

I

2

1

12

1

,II BA

2

1

2

1

2

1

2mr5

2

2

1

5

1

2

1

10

7

2

1

5

1

K

KR

22

22

mr10

7

mr5

1

7

2

23 t6t2)t(

t12t6dt

d 2

12t12dt

d2

2

12t12dt

d2

2

012t12

1 B 6 c 11 C

2 A 7 a 12 A

3 C 8 c 13 C

4 C 9 b 14 B

5 c 10 b 15 a

Q No Assignment 6 (Chetan Sir)

01 2

02 4

03 1

04 1

05 1

06 1

07 3

08 3

09 2

10 3

11 3

12 3

13 3

14 3

15 4

SAFE HANDS ASSIGNMENT -06 ROTATIONAL MOTION

1. 2 2. 2 3. 1 4. 2 5. 4 6. 3 7. 4 8. 1 9. 1 10. 2 11. 4 12. 2 13. 3 14. 3 15. 4

Ans key:straight line and pos(assignment6)

1 c

2 d

3 d

4 a

5 c

6 d

7 d

8 c

9 b

10 d

11 5 12 8

13 2

14 6

15 5

Assignment-6_AnswerKey Date:- March 26, 2020 By:- Ashok Kumar

Q1. D Q6. B Q11. C

Q2. C Q7. D Q12. B

Q3. D Q8. B Q13. C

Q4. B Q9. B Q14. A

Q5. D Q10. A Q15. A

XI th Assignment SIX TRANSPORT IN PLANTS + MINERAL NUTRITION

ANSWERS 01. (3)

02. (2)

03. (4)

04. (2)

05. (3)

06. (3)

07. (2)

08. (3)

09. (4)

10. (1)

11. (3)

12. (3)

13. (2)

14. (3)

15. (1)

16. (2)

17. (3)

18. (2)

19. (3)

20. (4)

21. (1)

22. (4)

23. (2)

24. (1)

25. (2)

26. (3)

27. (4)

28. (4)

29. (1)

30. (2)

Answer Key:

Assignment -6

Transport in Plants & Mineral nutrition

1-3 2-2 3-3 4-4 5-3 6-3

7-2 8-1 9-2 10-1 11-3 12-2

13-4 14-2 15-1 16-2 17-1 18-1

19-2 20-4 21-1 22-2 23-3 24-1

25-2 26-3 27-1 18-3 29-1 30-1

Answer key on 06 test for C18 (Transport and Mineral nutrition )by Monika Deshmukh

Q.No. Ans. Q.No Ans. Q.No Ans. Q.No Ans. Q.No Ans.

01 c 07 b 13 c 19 b 25 d

02 b 08 c 14 c 20 c 26 a

03 c 09 d 15 a 21 b 27 a

04 b 10 a 16 a 22 d 28 A

05 b 11 c 17 c 23 b 29 B

06 a 12 a 18 a 24 d 30 a

CHEMISTRY ASSIGNMENT 6 ANS KEY WILL BE UPLOADED ON

TOMORROW 28th MARCH 2020.