14.SOLUTIONS TO CONCEPTS by Aditya Bhardwaj

SOLUTIONS TO CONCEPTS CHAPTER 14 1.

F = mg

F A L Strain = L FL L F Y=   AL L YA  = stress = mg/A e = strain = /Y Compression L = eL F L FL y=  L  A L AY Lsteel = Lcu and Asteel = Acu F Stress of cu Fcu A g a) = cu  1  Stress of st A cu Fg Fst Stress =

3. 4.

b) Strain = 5.

Lst FstL st A cu Ycu   ( Lcu = Ist ; Acu = Ast) lcu A st Yst FcuIcu

F  L      L st AYst F  L      L cu AYcu AYcu Y strain steel wire F ( A cu  A st )  cu   Strain om copper wire AYst F Yst

Stress in lower rod =

T1 m g  g  1  w = 14 kg A1 A1

Stress in upper rod =

T2 m g  m1g  wg  w = .18 kg  2 Au Au

For same stress, the max load that can be put is 14 kg. If the load is increased the lower wire will break first. T1 m1g  g 8  = 8  10  w = 14 kg A1 A1

T2 m g  m1g  g 8  2 = 8  10  0 = 2 kg Au Au

7. 8. 9.

The maximum load that can be put is 2 kg. Upper wire will break first if load is increased. F L Y A L F L YA L Y  F L A L m2g – T = m2a …(1) and T – F = m1a …(2) m gF a= 2 m1  m2 14.1

Chapter-14 From equation (1) and (2), we get

m2 g 2(m1  m2 )

Again, T = F + m1a  T

m2 g m2 g m2 g  2m1m2g  m1  2 2 2(m1  m2 ) 2(m1  m2 )

Now Y = 

FL L F   A L L AY

T a

L (m22  2m1m2 )g m2 g(m2  2m1 )   L 2(m1  m2 )AY 2AY(m1  m2 )

m2 m2g

10. At equilibrium  T = mg When it moves to an angle , and released, the tension the T at lowest point is  T = mg +

mv 2 r

The change in tension is due to centrifugal force T =

mv 2 …(1) r

 Again, by work energy principle, 1  mv 2 – 0 = mgr(1 – cos) 2 2  v = 2gr (1 – cos)

…(2) m[2gr(1  cos )]  2mg(1  cos ) So, T  r  F = T YA L YA L F= = 2mg – 2mg cos   2mg cos  = 2mg – L L YA L = cos  = 1 –  L(2mg) 11. From figure cos  =

x  x2  = 1  2  l  l 

1/ 2

x 2  l2 =x/l … (1) Increase in length L = (AC + CB) – AB 2 2 1/2 Here, AC = (l + x ) 2 2 1/2 So, L = 2(l + x ) – 100 …(2) F l …(3) Y= A l From equation (1), (2) and (3) and the freebody diagram, 2l cos = mg. FL L F  12. Y =  AL L Ay

D / D D L   D L L / L A 2r Again,  A r 2r  A   r =

14.2

A T

L

Tx 

L C mg

B T

Chapter-14

Pv  v   P = B  v  v 

13. B =

14. 0 

m m  V0 Vd

d V0  0 Vd

so,

vol.strain =

V0  Vd V0

0 gh V  gh 1– d = 0 (V0  Vd ) / V0 V0 B

B= 

…(1)

vD  0 gh   1   v0  B 

…(2)

Putting value of (2) in equation (1), we get d 1 1  d   0  (1  0 gh / B) 0 1  0 gh / B

F A Lateral displacement = l. F=Tl 2THg 4Tg 2Tg a) P  b) P  c) P  r r r a) F = P0A b) Pressure = P0 + (2T/r) F = PA = (P0 + (2T/r)A c) P = 2T/r 2T F = PA = A r 2T cos  2T cos  a) hA  b) hB  rA  g rB g

15.  

16. 17. 18.

19.

20. hHg 

h 

c) hC 

2THg cos Hg rHgg

2T cos  where, the symbols have their usual meanings. r g

Hg cos  h T     hHg THg  cos Hg 21. h 

2T cos  rC g

2T cos  rg

2T r P = F/r 2 23. A = r 4 3 4 3 24. R  r  8 3 3  r = R/2 = 2 Increase in surface energy = TA – TA 22. P =

14.3

Chapter-14 25. h =

2T cos  2T cos  , h = rg rg

hrg 2T –1 So,  = cos (1/2) = 60°. 2T cos  26. a) h = rg  cos  =

b) T  2r cos  = r h    g

hrg  2T –3 27. T(2l) = [1  (10 )  h]g 2 28. Surface area = 4r 29. The length of small element = r d  dF = T  r d  considering symmetric elements, dFy = 2T rd . sin [dFx = 0] 

 cos  =

/2

so, F = 2Tr

 sin d = 2Tr[cos ]

/2 0

=T2r

Tension  2T1 = T  2r  T1 = Tr 30. a) Viscous force = 6rv

4 b) Hydrostatic force = B =   r 3 g 3 4 c) 6 rv +   r 3 g = mg 3

m    g  2 r 2 (  )g 2 2  (4 / 3)r 3   r v= 9  9 n 31. To find the terminal velocity of rain drops, the forces acting on the drop are, 3 i) The weight (4/3) r g downward. 3 ii) Force of buoyancy (4/3) r g upward. iii) Force of viscosity 6  r v upward. Because,  of air is very small, the force of buoyancy may be neglected. Thus,

4 6  r v =   r 2 g 3 32. v =

2r 2 g  9

R vD R= D 

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14.4