PG Brainstormer - 8C (MECHANICS) - Solutions635525474080922557.pdf

7/26/2019 PG Brainstormer - 8C (MECHANICS) - Solutions635525474080922557.pdf

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PG Brainstormer - 8C Solutions

Section - A

Single Choice Correct Type Questions (Q. No. 1 to 20)In this section 20 questions are there and each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is

correct. Marking Scheme - +3 for RIGHT answer and -1 for WRONG answer.

1. (D)

Sol. Min Linear velocity when pure rolling will start.

v(t) = (t)R

v0 gt= (

0+t)R mgR=

2

5mR2 =

5

2

g

R

v0 gt=

0R+

5

2

g

R

t R t= 2(v

0

0R )/7g.

2. (A)

Sol. x=3

l[Distance of centre of mass from the vertex]

N1=Mg [Equilibrium in vertical direction]

N1=N

2[Equilibrium in horizotal direction]

N1

Mg

30

N2

x

N 1

x cos (30 + )

N2= Mg

Condition for rotational equilibrium (about pointA)

Mg(x) cos(30 + ) N2(lsin ) = 0

= cot11

23

3. (D)

Sol. It pure rolling case

20 fr= 10 a (1)

20 N

10 kg

r = 0.1 m

fr 0.1 =

2

5 10 0.1 0.1

0.1

a (2)

fr 0.1 =

2

5 10 0.1 0.1

1

0.1

(20 )

10

rf

fr

=40

3N

4. (C)

Sol. I=

2 2 2 2

4 3 3 2

ml ml ml ml =

217

12

ml

By conservation of energy we use 4mgl/2 =1

2I2

=48

17

g

L.

PG Brainstormer - 8 CRIGID BODY DYNAMICS & CONSERVATION LAWS

An Ultimate Tool to understand advanced High School Physics

byASHISH ARORASir

Time Allow ed 60 M in

M aximum M arks 60

Solutions


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5. (C)

Sol. By conservation of angular momentum, when radius shrinks to half angular velocity of sphere increases to 4 times.

Now the kinetic energy of sphere is lost in work done by friction of the liquid medium.

6. (B)

Sol. Axis of pure rotation will be on the normals to both the straight lines and the intersection of these two normals will

be the co-ordinate of axis of pure rotation. The equations of normals of the two given lines are -

2y+ 3x= 16

3y 4x= 1

7. (B)

Sol. Let the component of vector in each direction isx,then direction cosines are

cos=30

x, cos=

30

x, cos =

30

x

we use cos2+ cos2 + cos2= 1

givesx= 10 3 .

8. (D)

Sol. If we write the equation of rotational motion of sphere about point P

then we have to apply pseudo force m2Ron centre of mass because

this axis has an upward acceleration = 2R towards O.

O

P

C

mg+m2R

So we use

IP= (mg+ m2R)d.

2

2

5 ( )

7

d g R

R

9. (B)

Sol. As shown in figure we can use

v= Rcos cos = v/R

h=RRcos v

R v

h = R - v/

10. (B)

Sol. Using conservation of angular momentum about new centre of mass after ball sticks to stick -

mv0 4

l=

221

12 16

lml m

+

2

16

ml this gives = 0

6

5

v

l

NOTE : In the problem it is not given that ball sticks to stick so if we solve the problem by taking coefficient of restitution zero

without considering ball sticks to stick then also we will get the same result because both the cases involve same

analysis.

11. (B)

Sol. Fcos 30 fs= ma

cm (1)

Fsin 30 + mg=N (2)

fsR=

2

3mR2 (3)

For pure rolling, acm

=R

15sf

m

=

3

2

sf

mRR


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or fs= 6N

fs

sN

i.e., s

6

20 5 3= 0.2

= min

= 0.2.

12. (D)

Sol. vcm

= acm

t

=f

tm

We know that after some time pure rolling will start so vcm

tgraph is linear upto the time pure rolling starts and

afterwards vcm

becomes constant as there no slipping and friction reduces to zero.

13. (B)

Sol. According to law of conservation of angular momentum

mv0RI = 0

mv0R= m

2

2

R

14. (B)

Sol.

15. (D)

Sol. About point P, angular momentum is given ascm ( )L I r p

Ltotal

=2

2

MR (mv)

2

R= 0

16. (C)

Sol. Let a is acceleration & is angular acceleration of sphere thena=R

4MgRcos 60 fR=2

5MR2

2Mgf=2

5Ma (i)

F=4Mg

P

O60

0

R

f& f=Ma

a=10

7g f=

10

7Mg

f (Mg+ 4Mg) 10

7Mg 5mg

min

=2

7

17. (B)

Sol. As there is no sliding, point of contact is at rest. It means point of contact does not move. Hence the work done by

static friction force is zero (static friction acts at the point of contact)


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18. (B)

Sol. For toppling of block net torque on it about the front edge/COM cm

> 0

Here we use

mg2

h>N

2

a

a

h

h/2 a/2

N

f = ms

N= mg

(mg) (h/2) > mg(a/2)

Solving we get > a/h

19. (A)

Sol. Applying work energy theorm between initial and final position of rod, we use

Work done by gravity on rod in its motion = (KE)horizontal position

mglsin =21 (2 )

2 3

m l

2

=3 sin

2

g

l

20. (C)

Sol. 3

4

lP

0cos =

2 2

12 16

ml ml

R

P0sin

P0cos

vcm

P2

P1

= 036 cos

7

P

ml

vcm

= (l/4) = 09 cos

7

P

m

If P1and P

2are the impulses by the hinge along x and y directions on rod as shown in figure we use Impulse

momentum equation and we can write -

In X direction P1= P

0sin 9 kg - m/s

In Y direction P0cos - P

2= mv

cm P

2=2 kg - m/s

Thus total impulse by hinge is PH= 2 2

1 2P P = 85 kg - m/s

* * * * *

PG Brainstormer - 8C (MECHANICS) - Solutions635525474080922557.pdf

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