Fluid Mechanics Practice Questions - DPP for NEET

 Page 1


1. (a) Force required to separate the plates
32
3
2 2 70 10 10
28
0.05 10
TA
FN
t
--
-
´ ´´
===
´
2. (c)
T T
q q
Weight of metal disc = total upward force
= upthrust force + force due to surface tension
= weight of displaced water + cos (2) qp Tr
2 cos W rT =+pq
3. (d)
22
22
21
8( )8
44
Dd
W T r rT
æö
= ´p - = ´p - ç÷
ç÷
èø
22
2() D dT =p-
4. (b) Increment in area of soap film 
21
AA =-
4 42
2 [(10 0.6) (10 0.5)] 10 2 10 m
--
= ´ ´ - ´ ´ =´
Work done 
TA = ´D
2 45
7.2 10 2 10 1.44 10 J
- --
= ´ ´´ =´
5. (a)
6. (c) Excess pressure inside soap bubble is inversely proportional
to the radius of bubble i.e. 
1
Dµ P
r
This means that bubbles A and C posses greater
pressure inside it than B. So the air will move from A
and C towards B.
7. (b)
12
12
54
20
54
rr
r
rr
´
= ==
--
cm
8. (c) The radius of resultant bubble is given by 
222
12
R r r. =+
9. (b)
h
x
b
Let the width of each plate is b and due to surface
tension liquid will rise upto height h then upward force
due to Surface tension
2 cos Tb =q
.......(i)
Weight of the liquid rises in between the plates
() Vdg bxh dg == ........(ii)
Equating (i) and (ii) we get, 2 cos T bxhdg q=
2 cos T
h
xdg
q
\=
10. (b) Mass of liquid in capillary tube
22
1 æö
=p ´r\ µ´
ç÷
èø
M RH MR
R
(As 1/) µ HR
\µ MR
. If radius becomes double then mass will
becomes twice.
11. ( d) In the satellite, the weight of the liquid column is zero.
So the liquid will rise up to the top of the tube.
12. (b)
0
(1) = -µ
c
TTt
i.e. surface tension decreases with increase in
temperature.
13. (a)
14. (d) Tension in spring 
T
= upthrust - weight of sphere
V g Vg V g Vg = s -r = hr -r (As ) s = hr
( 1) ( 1) Vg = h- r = h- mg.
15. (b)
16. (c)
17. (c) A stream lined body has less resistance due to air.
18. (a) Weight of cylinder = upthrust due to both liquids
3
2
5 4 54
æ ö æö
´ ´ = ´ ´ ´ +´´ ´
ç ÷ ç÷
è ø èø
A AL
V D g L dg dg
54
´´´ æö
Þ ´ ´ ´=
ç÷
èø
A A L dg
L Dg
5
544
Dd
Dd Þ = \=
19. (b) V elocity of efflux when the hole is at depth ,2 = h v gh
Rate of flow of water from square hole
2
1 11
2 == Q a v L gy
Rate of flow of water from circular hole
2
2 22
2 (4) = =p Q av R gy
According to problem 
12
QQ =
22
2 2 (4)
2
Þ =p Þ=
p
L
L gy R gyR
Page 2


1. (a) Force required to separate the plates
32
3
2 2 70 10 10
28
0.05 10
TA
FN
t
--
-
´ ´´
===
´
2. (c)
T T
q q
Weight of metal disc = total upward force
= upthrust force + force due to surface tension
= weight of displaced water + cos (2) qp Tr
2 cos W rT =+pq
3. (d)
22
22
21
8( )8
44
Dd
W T r rT
æö
= ´p - = ´p - ç÷
ç÷
èø
22
2() D dT =p-
4. (b) Increment in area of soap film 
21
AA =-
4 42
2 [(10 0.6) (10 0.5)] 10 2 10 m
--
= ´ ´ - ´ ´ =´
Work done 
TA = ´D
2 45
7.2 10 2 10 1.44 10 J
- --
= ´ ´´ =´
5. (a)
6. (c) Excess pressure inside soap bubble is inversely proportional
to the radius of bubble i.e. 
1
Dµ P
r
This means that bubbles A and C posses greater
pressure inside it than B. So the air will move from A
and C towards B.
7. (b)
12
12
54
20
54
rr
r
rr
´
= ==
--
cm
8. (c) The radius of resultant bubble is given by 
222
12
R r r. =+
9. (b)
h
x
b
Let the width of each plate is b and due to surface
tension liquid will rise upto height h then upward force
due to Surface tension
2 cos Tb =q
.......(i)
Weight of the liquid rises in between the plates
() Vdg bxh dg == ........(ii)
Equating (i) and (ii) we get, 2 cos T bxhdg q=
2 cos T
h
xdg
q
\=
10. (b) Mass of liquid in capillary tube
22
1 æö
=p ´r\ µ´
ç÷
èø
M RH MR
R
(As 1/) µ HR
\µ MR
. If radius becomes double then mass will
becomes twice.
11. ( d) In the satellite, the weight of the liquid column is zero.
So the liquid will rise up to the top of the tube.
12. (b)
0
(1) = -µ
c
TTt
i.e. surface tension decreases with increase in
temperature.
13. (a)
14. (d) Tension in spring 
T
= upthrust - weight of sphere
V g Vg V g Vg = s -r = hr -r (As ) s = hr
( 1) ( 1) Vg = h- r = h- mg.
15. (b)
16. (c)
17. (c) A stream lined body has less resistance due to air.
18. (a) Weight of cylinder = upthrust due to both liquids
3
2
5 4 54
æ ö æö
´ ´ = ´ ´ ´ +´´ ´
ç ÷ ç÷
è ø èø
A AL
V D g L dg dg
54
´´´ æö
Þ ´ ´ ´=
ç÷
èø
A A L dg
L Dg
5
544
Dd
Dd Þ = \=
19. (b) V elocity of efflux when the hole is at depth ,2 = h v gh
Rate of flow of water from square hole
2
1 11
2 == Q a v L gy
Rate of flow of water from circular hole
2
2 22
2 (4) = =p Q av R gy
According to problem 
12
QQ =
22
2 2 (4)
2
Þ =p Þ=
p
L
L gy R gyR
62
DPP/ P 21
20. (a) Let A = cross-section of tank
52.5 cm
3 m
A
B
v
a = cross-section hole
V = velocity with which level decreases
v = velocity of efflux
From equation of continuity = Þ=
av
av AVV
A
By using Bernoulli’s theorem for energy per unit volume
Energy per unit volume at point A
= Energy per unit volume at point B
22
11
0
22
+r +r = + +r Pgh VPv
22
22
2 2 10 (3 0.525)
50( / sec)
1 (0.1)
1
´ ´-
Þ= ==
-
æö
-
ç÷
èø
gh
vm
a
A
21. (c) If the liquid is incompressible then mass of liquid
entering through left end, should be equal to mass of liquid
coming out from the right end.
1 2 12
M m m Av Av 1.5A.v \ = + Þ =+
A 3 A 1.5 1.5A.v v 1 m / s Þ ´ = ´ + Þ=
22. ( b)
4
2
3
F 6.28 10
T 5 10 N/m
2r
2 3.14 2 10
-
-
-
´
= = =´
p
´ ´´
23. (d) At critical temperature ( ) c
T 370 C 643 K , ==
o
the
surface tension of water is zero.
24. (d)
25. (b)
Bottom Surface
P P. > So bubble rises upward.
At constant temperature 
( )
1
V Boyle 's law
P
µ
Since as the bubble rises upward, pressure decreases, then
from above law volume of bubble will increase i.e. its size
increases.
26. (a)
27. (d).
F = P
atm
 × Area = 10
5
 × 1 × 10
–6
 = 0.1 N
atm
2T
F P A 0.10023 N
r
æö
= + ´=
ç÷
èø
28. (a) Since the excess pressure due to surface tension is
inversely proportional to its radius, it follows that
smaller the bubble, greater is the excess pressure. Thus
when the larger and the smaller bubbles are put in
communication, air starts passing from the smaller into
the large bubble because excess pressure inside the
former is greater than inside the latter. As a result, the
smaller bubble shrinks and the larger one swells.
29. ( b) Statement-1 is True, Statement-2 is True; Statement-2
is NOT a correct explanation for Statement-1.
30. (a) In a stream line flow of a liquid, according to equation
of continuity,
av = constant
Where a is the area of cross-section and v is the velocity
of liquid flow. When water flowing in a broader pipe
enters a narrow pipe, the area of cross-section of water
decreases therefore the velocity of water increases.