11.SOLUTIONS TO CONCEPTS

Page 1

SOLUTIONS TO CONCEPTS CHAPTER 11 1.

Gravitational force of attraction, GMm F= r2 =

2.

6.67  10 11  10  10 –7 = 6.67 × 10 N (0.1)2

To calculate the gravitational force on ‘m’ at unline due to other mouse.

FOD =

G  m  4m 8Gm2 = (a / r 2 )2 a2

FOI =

G  m  2m 6Gm2 = (a / r 2 )2 a2 G  m  2m

FOB =

=

2 2

(a / r )

Gmm

FOA =

=

2 2

(a / r )

m

A

E

B

2m

m

4Gm2 a2

4m

F

D

C

3m

2Gm2 a2 2

Resultant FOF =

 Gm2   Gm2  64 2   36 2   a   a     

Resultant FOE =

 Gm2   Gm 2  64 2   4 2   a   a 

2

2

= 10

Gm2 a2

2

= 2 5

Gm 2 a2

The net resultant force will be, 2

2

F=

 Gm2   Gm 2   Gm2  100 2   20 2   2 2   20 5  a   a   a        2

=

 Gm2   120  40 5 =   a2   

2

 Gm2   (120  89.6)   a2   

Gm2 Gm 2 40.4 = 4 2 2 2 a a a) if ‘m’ is placed at mid point of a side =

3.

then FOA =

4Gm2 in OA direction a2

A m

4Gm 2 in OB direction a2 Since equal & opposite cancel each other

m

FOB =

Foc =

Gm

2

B

2

r / 2a 3

O

2

=

4Gm in OC direction 3a 2

Net gravitational force on m =

A

4Gm 2

m

a2

b) If placed at O (centroid) the FOA =

C m

m

3Gm2 Gm 2 = (a / r3 ) a2

B

11.1

m O m

C m


Chapter 11

3Gm2 a2

FOB =

 Resultant F =

2

2

2  2  3Gm 2     2 3Gm   1 = 3Gm 2 2 2 2    a  a   a  2

3Gm2 , equal & opposite to F, cancel a2 Net gravitational force = 0 Since FOC =

4.

2 Gm2 ˆi  Gm sin 60ˆj cos 60 4a 2 4a 2

FCB =

2

M

A

B

2

Gm Gm cos 60 ˆi  sin 60ˆj  4a 2 4a 2

FCA =

 F = FCB + FCA

C

2 2  2Gm 2 ˆj =  2Gm r3 = r3Gm sin 60 4a 2 2 4a 2 4a 2 Force on M at C due to gravitational attraction.

=

5.

FCB =

Gm2 ˆ j 2R 2

B

A

2

 GM ˆ i 4R 2

FCD =

R D

2  GM2 ˆj  GM sin 45ˆj cos 45 4R 2 4R 2 So, resultant force on C,

FCA =

 FC = 

C

= FCA + FCB + FCD

GM2  1 ˆ GM2   i  2  4R 2  4R 2 2

 1 ˆ  2   j 2 

GM2 2 2 1 4R 2

FC =

 mv 2 For moving along the circle, F = R or 6.

MV 2 GM2 = or V = 2 2  1 R 4R 2 GM

R  h

2

=

GM  2 2  1  R  4 

49.358  1011 6.67  10 11  7.4  10 22 = 2 6 (1740  1000 )  10 2740  2740  10 6

49.358  1011 –2 2 = 65.8 × 10 = 0.65 m/s 0.75  1013 The linear momentum of 2 bodies is 0 initially. Since gravitational force is internal, final momentum is also zero. So (10 kg)v1 = (20 kg) v2 Or v1 = v2 …(1) Since P.E. is conserved =

7.

 6.67  10 11  10  20 –9 = –13.34×10 J 1 When separation is 0.5 m, Initial P.E. =

11.2


Chapter 11 –13.34 × 10

–9

+0=

 13.34  10 9 2 2 + (1/2) × 10 v1 + (1/2) × 20 v2 (1/ 2)

–9

–9

2

…(2)

2

 – 13.34 × 10 = -26.68 ×10 + 5 v1 + 10 v2 2 –9 –9  – 13.34 × 10 = -26.68 ×10 + 30 v2

13.34  10 9 –10 = 4.44 × 10 30 –5  v2 = 2.1 × 10 m/s. –5 So, v1 = 4.2 × 10 m/s. In the semicircle, we can consider, a small element of d then R d = (M/L) R d = dM. GMRdm M F= LR 2  2GMm dF3 = 2 dF since = sin  d LR d 2

 v2 =

8.

/2

F =

0

2GMm 2GMm  cos 0 / 2  sin d  LR LR

L

 d R

m

GMm 2GMm 2GMm 2GMm ( 1) = = = LR LR L L / A L2 A small section of rod is considered at ‘x’ distance mass of the element = (M/L). dx = dm G(dm)  1 = dE2 dE1 = d2  x 2  = –2

9.

Resultant dE = 2 dE1 sin  G(dm) d 2  GM  d dx =2× 2 =  2 2 2 d  x2 d x L d  x 2  d2  x 2    Total gravitational field

 

L/2

E=

 

2

 x2

dE1 d 

M

2Gmd dx

 Ld 0

 dE2

O

3/2

dx

x

x

a

Integrating the above equation it can be found that, 2GM  E= d L2  4d2 10. The gravitational force on ‘m’ due to the shell of M2 is 0. R  R2 M is at a distance 1 2 Then the gravitational force due to M is given by GM1m 4GM1m = = (R1  R 2 / 2 (R1  R 2 )2

R2

M1 R1 m

3

11. Man of earth M = (4/3) R  Man of the imaginary sphere, having 3 Radius = x, M = (4/3)x  or

M x3 = 3 M R

 Gravitational force on F = or F =

m

GMm m2

x

GMmx GMx 3m =  3 2 R x R3 11.3

m2


Chapter 11 12. Let d be the distance from centre of earth to man ‘m’ then D=

 R2   = (1/2) x 2     4 

4x 2  R2

m

x R/2

M be the mass of the earth, M the mass of the sphere of radius d/2. 3 Then M = (4/3) R  3 M = (4/3)d 

d

O

M d3 = 3 M R  Gravitational force is m, or

n

Gmm GMmd Gd3Mm = = 2 d R 3 d2 R3 So, Normal force exerted by the wall = F cos. GMm GMmd R = (therefore I think normal force does not depend on x) =  2d 2R 2 R3 13. a) m is placed at a distance x from ‘O’. If r < x , 2r, Let’s consider a thin shell of man

 d F

F=

dm =

m 4 3 mx 3   x = ( 4 / 3)r 2 3 r3

Thus

dm =

 R/2

x

R M

mx 3 r3

m

r O

G md m Gmx Gmx 3 / r 3 = =  2 2 x x r3 b) 2r < x < 2R, then F is due to only the sphere. Gmm F= x  r 2 Then gravitational force F =

c) if x > 2R, then Gravitational force is due to both sphere & shell, then due to shell, GMm F= x  R 2 due to the sphere =

Gmm

x  r 2

So, Resultant force =

Gm m

x  r 2

+

GMm

x  R 2

14. At P1, Gravitational field due to sphere M =

GM

3a  a 

2

=

GM 16a 2

At P2, Gravitational field is due to sphere & shell,

a 49 P1

a GM GM GM  1 1   61  GM  + = =     2 2 2 a a  36 25  (a  4a  a) ( 4a  a ) 2  900  a P2 15. We know in the thin spherical shell of uniform density has gravitational field at its internal point is zero.

=

At A and B point, field is equal and opposite and cancel each other so Net field is zero.

A A B

Hence, EA = EB 16. Let 0.1 kg man is x m from 2kg mass and (2 – x) m from 4 kg mass. 4  0 .1 2  0 .1  =– 2 ( 2  x )2 x 11.4

B


Chapter 11 or

0 .4 0.2 =– 2 ( 2  x )2 x

or

2 1 2 2 = or (2 – x) = 2 x ( 2  x )2 x2

or 2 – x = 2 x or x(r2 + 1) = 2 2 or x = = 0.83 m from 2kg mass. 2.414 17. Initially, the ride of  is a To increase it to 2a, 2

2

m a

a

2

Gm Gm 3Gm  = m m a 2a a 2a 18. Work done against gravitational force to take away the particle from sphere, work done =

G  10  0.1 6.67  10 11  1 –10 = = 6.67 × 10 J 0.1 0.1 1 10 1  19. E = (5 N/kg) ˆi + (12 N/kg) ˆj   a) F = E m = 2kg [(5 N/kg) ˆi + (12 N/kg) ˆj ] = (10 N) ˆi + (12 N) ˆj  F = 100  576 = 26 N   b) V = E r   At (12 m, 0), V = – (60 J/kg) ˆi V = 60 J

100g 10cm

=

10kg

  At (0, 5 m), V = – (60 J/kg) ˆj V = – 60 J  c)  V =

(1,2,5 )

 E mdr =  (10N)ˆi  (24N)ˆj r  

(12,5 ) ( 0,0 )

( 0,0 )

= – (120 J ˆi + 120 J ˆi ) = 240 J 0,5m  d)  v = – r(10N ˆi  24Nˆj ) 12m,0 

= –120 ˆj + 120 ˆi = 0 20. a) V = (20 N/kg) (x + y)

MLT 2 M1L3 T 2M1 ML2 T 2 GM = L or = M R L M 0 2 –2 0 2 –2 Or M L T = M L T  L.H.S = R.H.S  b) E( x, y ) = – 20(N/kg) ˆi – 20(N/kg) ˆj   c) F = E m = 0.5kg [– (20 N/kg) ˆi – (20 N/kg) ˆj = – 10N ˆi - 10 N ˆj   | F | = 100  100 = 10 2 N  21. E = 2 ˆi + 3 ˆj The field is represented as tan 1 = 3/2 3j 5/3 2 Again the line 3y + 2x = 5 can be represented as  tan 2 = – 2/3 2j 5/2 m1 m2 = –1 Since, the direction of field and the displacement are perpendicular, is done by the particle on the line. 11.5


Chapter 11 22. Let the height be h GM GM (1/2) 2 = (R  h)2 R 2

Or 2R = (R + h)

2

Or 2 R = R + h Or h = (r2 – 1)R 23. Let g be the acceleration due to gravity on mount everest.

 2h  g = g1   R  17696   2 =9.8 1   = 9.8 (1 – 0.00276) = 9.773 m/s  6400000  24. Let g be the acceleration due to gravity in mine. d  Then g= g 1    R 640   2 = 9.8 1   = 9.8 × 0.9999 = 9.799 m/s 3  6400  10  25. Let g be the acceleration due to gravity at equation & that of pole = g 2 g= g –  R –5 2 3 = 9.81 – (7.3 × 10 ) × 6400 × 10 = 9.81 – 0.034 2 = 9.776 m/s 2 mg = 1 kg × 9.776 m/s = 9.776 N or 0.997 kg The body will weigh 0.997 kg at equator. 2 …(1) 26. At equator, g = g –  R Let at ‘h’ height above the south pole, the acceleration due to gravity is same.

 2h  Then, here g = g 1   R 

…(2)

 2h  2  g -  R = g 1   R  or 1 

2h 2R = 1 R g

  2

2

7.3  10  5  6400  10 3 2R 2 = = 11125 N = 10Km (approximately)  2  9.81 2g 27. The apparent ‘g’ at equator becomes zero. 2 i.e. g = g –  R = 0 2 or g =  R or h =

or  =

g = R

9 .8 = 6400  10 3

1.5  10 6 = 1.2 × 10

2 2  3.14 –6 = = 1.5 × 10 sec. = 1.41 hour  1.2  10  3 28. a) Speed of the ship due to rotation of earth v = R 2 b) T0 = mgr = mg – m R 2  T0 – mg = m R c) If the ship shifts at speed ‘v’ 2 T = mg – m R

–3

rad/s.

T=

11.6

To A

A


Chapter 11

 v  R 2  R = T0 -    R2    v 2  2R 2  2Rv   m = T0 –    R    T = T0 + 2v m 29. According to Kepler’s laws of planetary motion, 2 3 T R Tm

2

Te 2



3

R es 3

3

 Rms   R es 

R ms

2

  1.88        1  

R ms 2/3 = (1.88) = 1.52 R es

30. T = 2

r3 GM

3.84  10 

5 3

27.3 = 2 × 3.14 or 2.73 × 2.73 = or M =

6.67  10 11  M

2  3.14  3.84  10 5 6.67  10 11  M

3

2  (3.14 )2  (3.84)3  1015 24 = 6.02 × 10 kg 11 2 3.335  10 (27.3 )

 mass of earth is found to be 6.02 × 10

24

kg.

3

31. T = 2

r GM

9.4  10

2

or (27540) = (6.28) or M = 32. a) V = =

3

 103 11 6.67  10  M

 27540 = 2 × 3.14

3

9.4  10 

6 2

2

6.67  10 11  M

(6.28)2  (9.4 )3  1018 23 = 6.5 × 10 kg. 11 2 6.67  10  (27540 ) gr 2 r h

GM = r h

9.8  ( 6400  10 3 )2 6

10  (6.4  2)

3

= 6.9 × 10 m/s = 6.9 km/s

2

b) K.E. = (1/2) mv 6 10 = (1/2) 1000 × (47.6 × 10 ) = 2.38 × 10 J GMm c) P.E. =  (R  h) =–

40  1013 6.67  10 11  6  10 24  10 3 10 = – = – 4.76 × 10 J 8400 (6400  2000 )  10 3

d) T =

2(r  h) 2  3.14  8400  10 3 2 = = 76.6 × 10 sec = 2.1 hour 3 V 6.9  10 11.7


Chapter 11 33. Angular speed f earth & the satellite will be same 2 2 = Te Ts or

or

1 = 24  3600

1 2

or 12 I 3600 = 3.14

(R  h)3 gR 2

(R  h)2 (12  3600 )2 = gR 2 (3.14 )2

or

(R  h)3 gR 2

(6400  h)3  109 (12  3600 )2 = 9.8  (6400 )2  10 6 (3.14 )2

(6400  h)3  10 9 4 = 432 × 10 9 6272  10 3 4 or (6400 + h) = 6272 × 432 × 10 4 1/3 or 6400 + h = (6272 × 432 × 10 ) 4 1/3 or h = (6272 × 432 × 10 ) – 6400 = 42300 cm. b) Time taken from north pole to equator = (1/2) t or

( 43200  6400)3

= (1/2) × 6.28

2

10  (6400 )  10

6

= 3.14

( 497 )3  10 6 (64)2  1011

497  497  497 = 6 hour. 64  64  10 5 34. For geo stationary satellite, 4 r = 4.2 × 10 km h = 3.6 × 104 km Given mg = 10 N = 3.14

 R2   mgh = mg   R  h2   

2  6400  103 = 10   6400  10 3  3600  10 3 

 4096  = = 0.23 N 2 17980  

R 23

35. T = 2

gR12

Or T = 4 2

R2

3 2

gR1 3

Or g =

42 R 2 T 2 R12

 Acceleration due to gravity of the planet is =

3

42 R 2 T 2 R12

A

36. The colattitude is given by . OAB = 90° – ABO Again OBC =  = OAB 6400 8  sin  = = 42000 53  = sin

–1

Colatitude



O 

 8  –1   = sin 0.15.  53 

11.8

B

C


Chapter 11 37. The particle attain maximum height = 6400 km. On earth’s surface, its P.E. & K.E.

  GMm  2 Ee = (1/2) mv +    R  In space, its P.E. & K.E.

…(1)

 GMm  Es =   +0  Rh   GMm  …(2) Es =    2R   Equating (1) & (2) GMm GMm 1   mv 2 =  2R R 2

( h = R)

1  1 2   Or (1/2) mv = GMm   2 R R   2

Or v = =

GM R

6.67  10 11  6  10 24 6400  10 3

40.02  1013 6.4  10 6 7 8 = 6.2 × 10 = 0.62 × 10 =

4

Or v = 0.62  10 8 = 0.79 × 10 m/s = 7.9 km/s. 38. Initial velocity of the particle = 15km/s Let its speed be ‘v’ at interstellar space.  GMm 3 2 2 (1/2) m[(15 × 10 ) – v ] = dx R x2

 1 3 2 2  (1/2) m[(15 × 10 ) – v ] = GMm    x R 6

2

 (1/2) m[(225 × 10 ) – v ] =

GMm R

2  6.67  10 11  6  10 24 6400  10 3 40.02 2 6 8  v = 225 × 10 – × 10 32 2 6 8 8  v = 225 × 10 – 1.2 × 10 = 10 (1.05) 4 Or v = 1.01 × 10 m/s or = 10 km/s 24 39. The man of the sphere = 6 × 10 kg. 8 Escape velocity = 3 × 10 m/s 6

2

 225 × 10 – v =

Vc = Or R = =

2GM R 2GM Vc 2

2  6.67  10 11  6  10 24

3  10 

8 2

=

80.02 –3 –3 × 10 = 8.89× 10 m  9 mm. 9  11.9


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