12th Grade > Physics
MECHANICAL PROPERTIES OF FLUIDS MCQs
Total Questions : 29
| Page 2 of 3 pages
Answer: Option B. -> The water will not come out
:
B
Water rises to the top of the capillary tube and adjusts its meniscus shape and forms spherical surface but it can't come outside.
:
B
Water rises to the top of the capillary tube and adjusts its meniscus shape and forms spherical surface but it can't come outside.
Answer: Option A. -> 80 Kg
:
A
Volume of log of wood V=massdensity=120600=0.2m3
Let x weight that can be put on the log of wood.
So weight of the body = (120+x)×10N
Weight of displaced liquid = Vρg=0.2×103×10N
The body will just sink in liquid if the weight of the body will be equal to the weight of displaced liquid.
(120+x)×10N=0.2×103×10N
⇒x=80kg
:
A
Volume of log of wood V=massdensity=120600=0.2m3
Let x weight that can be put on the log of wood.
So weight of the body = (120+x)×10N
Weight of displaced liquid = Vρg=0.2×103×10N
The body will just sink in liquid if the weight of the body will be equal to the weight of displaced liquid.
(120+x)×10N=0.2×103×10N
⇒x=80kg
Answer: Option D. -> Water flowing out through a hole in the side of a tank
:
D
When water flows out of a hole in a tank, it is in contact with the edges of the hole over a very small area, and therefore viscosity effects are small
:
D
When water flows out of a hole in a tank, it is in contact with the edges of the hole over a very small area, and therefore viscosity effects are small
Answer: Option B. -> 20
:
B
v=√2gh=√2×10×20=20m/s
:
B
v=√2gh=√2×10×20=20m/s
Answer: Option D. -> Because of capillarity
:
D
Due to capillarity it absorbs the ink.
:
D
Due to capillarity it absorbs the ink.
Answer: Option C. -> 1.0 m/s
:
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.
∴M=m1+m2⇒Av1=Av2+1.5A.v
⇒A×3=A×1.5+1.5A.v⇒v=1m/s
:
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.
∴M=m1+m2⇒Av1=Av2+1.5A.v
⇒A×3=A×1.5+1.5A.v⇒v=1m/s
Answer: Option D. -> 4×5×5×5 gf
:
D
Apparent weight
=V(ρ−σ)g=l×b×h×(5−1)×g
=(5×5×5×4×g) dyne = (4×5×5×5)gf
:
D
Apparent weight
=V(ρ−σ)g=l×b×h×(5−1)×g
=(5×5×5×4×g) dyne = (4×5×5×5)gf
Answer: Option B. -> 1.94
:
B
Apparent weight = V(ρ−σ)g=Mρ(ρ−σ)g
=M⟮1−σρ⟯g=2.1⟮1−0.810.5⟯g=1.94gN
=1.94Kg−wt
:
B
Apparent weight = V(ρ−σ)g=Mρ(ρ−σ)g
=M⟮1−σρ⟯g=2.1⟮1−0.810.5⟯g=1.94gN
=1.94Kg−wt
Answer: Option B. -> Liquid of high viscosity and low density flowing through a pipe of small radius
:
B
For streamline flow, Reynold's number NR∝rρη should be less. For less value of NR, radius and density should be small and viscosity should be high.
:
B
For streamline flow, Reynold's number NR∝rρη should be less. For less value of NR, radius and density should be small and viscosity should be high.
Answer: Option B. -> 4Tr
:
B
The pressure inside the bubble is more than outside. Inside pressure -outside pressure =excess pressure. In the case of bubble excess pressure = 4S/R. where s is the surface tension and r is the radius.
△P=4Tr
:
B
The pressure inside the bubble is more than outside. Inside pressure -outside pressure =excess pressure. In the case of bubble excess pressure = 4S/R. where s is the surface tension and r is the radius.
△P=4Tr