MCQGeeks
0 : 0 : 1
CBSE
JEE
NTSE
NEET
English
UK Quiz
Quiz
Driving Test
Practice
Games
NEET
Physics NEET MCQ
Properties Of Matter Mcq
Quiz 3
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Q.1
16 cm3 of water flows per second through a capillary tube of radius a cm and length l cm when connected to a pressure head of h cm of water. If a tube of the same length and radius a/2 cm is connected to the same pressure head, the quantity of water following through the tube per second will be [ CPMT 1998]
0%
a) 16 cm3
0%
b) 4 cm3
0%
c) 1 cm3
0%
d) 8 cm3
Explanation
Same as question Q163; Here Q=16cm3 Answer: (c)
Q.2
A jar is filled with two non-mixing liquids 1 and 2 having densities ρ1 and ρ2 respectively. A solid ball, made of a material of density ρ3, is dropped in the jar. It comes to equilibrium in the position shown in figure. Which of the following is true for ρ1 , ρ2 ,ρ3? [ AIEEE 2008]
0%
a)ρ3 < ρ1 < ρ2
0%
b) ρ1 > ρ3 >ρ2
0%
c)ρ1 < ρ2 < ρ3
0%
d)ρ1 < ρ3 < ρ2
Explanation
From the figure it is clear that liquid 1 floats on liquid 2. ThE lighter liquid floates over heavier liquid. Therefore we can conclude that ρ1 < ρ2 Also ρ3 < ρ2 otherwise the ball would have sink to the bottom of the jar. also ρ3 > ρ1 otherwise the ball would have floated in liquid 1. From above discussion we conclude option 'd' is correctAnswer: (d)
Q.3
Two wires are made of same material and have the same volume. However wire 1 has cross-section area A and wire 2 has crosssectional area 3A. If the length of wire 1 increases by Δx on appling force F, how much force is needded to stretch wire 3 by same amount
0%
a) 4F
0%
b) 6F
0%
c)9F
0%
d)F
Explanation
For wire 1 For wire 2 Since vloume is same wire 2 will have length=l/3 Answer: (c)
Q.4
In order that a floating object be in stable equilibrium, its centre of buoyancy should be
0%
a) vertically above its centre of gravity
0%
b) vertically below its centre of gravity
0%
c)Horizontally line with its centre of gravity
0%
d)May be anywhere
Explanation
Answer:(b)
Q.5
A large air bubble of radius r rises from the bottom of a lake to the surface. The depth of the lake is 7H. If the atmospheric pressure is equal to that of water height H, the radius of the bubble is
0%
a) r
0%
b) 2r
0%
c) 3r
0%
d) 8r
Explanation
Let ρ be the density of water Total pressure on the surface of lake P2=Hρg Let R be the radius of bubble at surface Total pressure at depth=P1=7H×ρg + Hρg=8Hρg Applying Boyle's law P1V1=P2V2(7H×ρg + Hρg=8Hρg) × (4/3) πr3=Hρg× (4/3) πR3 8H r3=HR3 R=2rAnswer: (b)
Q.6
A block of ice is floating on water contained in a beaker. When all ice melts, the level of water
0%
a) Rises
0%
b) Falls
0%
c)Remains unchanged
0%
d)none
Explanation
Answer: (c)
Q.7
A body is just floating in a liquid 99 their densities are equal). If the body is slightly pressed down and released it will
0%
a) start oscillating
0%
b) sink to the bottom
0%
c)comes back to the same position immediately
0%
d)come back to the same position immediately.
Explanation
Answer: (b)
Q.8
A block of ice in which a piece of stone is embedded is floating on water contained in a beaker. When all the ice melts the level of water
0%
a) Rises
0%
b) Falls
0%
c)Remains unchanged
0%
d)none
Explanation
Volume of liquid displaced more than the volume of water formed. Therefore level of water falls Answer:(b)
Q.9
1 kg of cotton and iron in air are transferred to vacuum and weighed again then.
0%
a) Cotton and iron will weigh same
0%
b) Iron will weigh more than cotton
0%
c) Cotton will weigh more than iron
0%
d) None of the above
Explanation
Cotton will weigh more as volume of 1kg cotton is much more than the volume of 1kg steel. the up thrust on cotton is more. On transferring it to vacuum, up thrust vanishes. Answer: (c)
Q.10
A boat carrying a number of stones is floating in a water tank. If the stone are unloaded into water level in the tank will
0%
a)Remain unchanged
0%
b) Rise
0%
c)Fall
0%
d)Rise or fall depending on the number of stones unloaded.
Explanation
Answer: (c)
Q.11
An alloy of gold and copper weighs 0.2kg in air and 0.188 kg in water. The densities of gold and copper are 19.3×103 kg m-3 and 8.93×103kg m-3 respectively. The amount of gold in block is nearly
0%
a)12×10-3kg
0%
b) 12×10-3kg
0%
c)0.173 kg
0%
d)0.388 kg
Explanation
Loss of weight=wt of water displaced 0.2 -0.188=volume × density 0.012=V× 1000 volume of alloy V=12ccLet m=mass of goldVolume of gold=m/density=m/19.3∴ 200-m=mass of copper Volume of copper=(200-m) / 8.93 Volume of copper + Volume of Gold=Volume of alloy (200-m) / 8.93 + m/19.3=12 m=0.173 kgAnswer: (c)
Q.12
The density of atmospheric air varies with height above the ground according to the relation ρ=ρ0 e-λh , where λ is constant. The pressure at height h is given by
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Pressure at height can be obtain by integrating given equation Answer:(b)
Q.13
the spring balance A reads 2kg with block m suspended from it. A balance B reads 5kg when a beaker with liquid is put on the pan of the balance. the two balances are now so arranged that hanging mass is inside the liquid in the beaker as shown. In this situation [ IIT 1985]
0%
a) The balance A will read more than 2kg
0%
b) The balance B will read less than 5kg
0%
c) The balance A will read less than 2kg and B will read more than 5 kg
0%
d) the balance A and B will read 2kg and 5 kgs respectively
Explanation
Due to up thrust balance A will read less while balance B will read more than 5 kg Answer: (c)
Q.14
A vessel contains oil ( density=0.8 gm/cm3) over mercury ( density=13.6 gm/cm3). A homogeneous sphere floats with half of its volume immersed in mercury and the other half in oil. Then density of material of the sphere in gm/cm3 is [ IIT 1988]
0%
a)3.3
0%
b) 6.4
0%
c)7.2
0%
d)2..8
Explanation
half of sphere immersed in mercury and half in oil thus there will be up thrust due to both liquidsLet d be the density of sphereUp thrust=Mass of mercury displaced + Mass of oil displaced Up thrust=(V/2)×13.6 ×g + (V/2)× 0.8×g Down ward gravitations force on sphere=(V)×d×g Downward gravitational force=Up thrust (V)×d×g=(V/2)×13.3 ×g + (V/2)× 0.8×g d=7.2 gm/cm3Answer: (c)
Q.15
The principle of the operation of hydraulic press is based on
0%
a) Boyle's law
0%
b) Pascal's law
0%
c)Dalton's law of partial pressure
0%
d)Newtons law of gravitation
Explanation
Answer: (b)
Q.16
Pressure applied to enclosed fluid is
0%
a) Increased and applied to every part of the fluid
0%
b) Diminished and transmitted to wall of container
0%
c)Increased in proportion to the mass of the fluid and then transmitted
0%
d)Transmitted unchanged to every portion of the fluid and walls of containing vessel
Explanation
Answer:(d)
Q.17
A parrot is in a wire cage which is hanging from a spring balance. Initially the parrot sits in the cage and in the second instant the parrot flies inside the cage
0%
a) The reading of the balance will be grater when the parrot flies in the cage.
0%
b) the reading of the balance will be lesser when the parrot flies in the cage
0%
c) the reading will remain unchanged
0%
d) None of the above
Explanation
As the up thrust does not act on the base but is transmitted to air ( Pascal law) so reading in balance will be lesser Answer: (b)
Q.18
Two vessels A nad B have same base area and contain water to the same height, but the mass of water in A is four times that in B. The ratio of the liquid thrust at the base of A to that of B is
0%
a)4 : 1
0%
b) 2 : 1
0%
c)1 : 1
0%
d)16 : 1
Explanation
Upthrust depends on the volume of liqid dispalce , since volume of both container is same upthrust is sameAnswer: (c)
Q.19
A wooden rod of uniform cross-section and length 120cm, having density d is hanged at the bottom of a tank which is filled with water to a height of 40cm. Under equilibrium conditions, it makes an angle of 60° with the vertical. The centre of buoyancy of the rod is located at a distance ( measured from the hinge, along the length ) of [ SCRA 1994]
0%
a) 90 dcm
0%
b) 60 dcm
0%
c)40 dcm
0%
d)20 dcm
Explanation
Length of rod immersed in water L=40 / cos60=80 cm ∴ Upthrust=W'=length immersed × Area×density of liquid × gUpthrust W'=80×A×1×gWeight of rod W=120A×d×g here d is density of woodRod is in rotational equilibrium Taking moment of force about O We know Moment of force=force × perpendicular distance W'×x sin60=W × 60sin60 80×A×1×g×x sin60=120A×d×g× 60sin60 x=120×60d / 80 x=90d cm Answer: (a)
Q.20
A cubical block of wood of specific gravity 0.5 and chunk of concrete of specific gravity 2.5 are fastened together. The ratio of the mass of wood to the mass of concrete, which makes the combination to floate with its entire volume submerged under water is [ SCRA 1994]
0%
a) 1/5
0%
b) 1/3
0%
c)3/5
0%
d)2/3
Explanation
let mc=mass of concrete mw=mass of wood Volume of concrete=mc / 2.5 Volume of wood=mw / 0.5 Upthrust=Volume of concrete × density of liquid displace×g + Volume of wood × density of liquid displaced ×g Upthrust=( mc / 2.5) ×1×g + (mw / 0.5) ×1×g Weight of object=( mc + mw)g In equilibrium Upthrust=Weight of object ( mc / 2.5) ×1×g + (mw / 0.5) ×1×g=( mc + mw)g on solving we get mc / m w=3/5 Answer:(c)
Q.21
In a surface tension experiment with a capillary tube water rises upto 0.1m. If the same experiment is repeated on a n artificial satellite which revolving around the earth, water will rise in capillary tube upto height of [ CPMT 1998]
0%
a)0.1 m
0%
b) 9.8 m
0%
c)0.98 m
0%
d)full length of capillary tube
Explanation
We know that surface tension exerts upward force hence there is rise in water column, in satellite there is no gravitational force to updown the waterAnswer: (d)
Q.22
A beaker contain water is placed on a scale balance which indicates a mass of 0.5kg. A glass sphere of mass 0.2kg is suspended from a string and immersed in water such that it does not touch the beaker. If the relative density of glass is 2.5, then the reading on the balance will change to [ SCRA 1994]
0%
a) 0.62 kg
0%
b) 0.58 kg
0%
c) 0.50kg
0%
d) 0.42 kg
Explanation
density of sphere=relative density of sphere / density of water density of sphere=2.5 / 13=2.5×10-3 Volume of sphere=mass / density of sphere Volume of sphere=0.2 / 2.5×10-3 Upthrust=Volume × density of liquid ×g Upthrust=0.2 / 2.5×10-3×103g Upthrust=2/25 kg=0.08kg ∴ reaction of upthrust on the bottom of beaker=0.08 reading on scale=0.5+.08=0.58kg Answer: (b)
Q.23
A man is carrying a block of a certain substance ( of density 1000 kg/m3) weighing 1 kg in left hand and a bucket filled with water and weighing 10 kg in his right hand. He drops the block into bucket. How much load does he carry in his right hand now [ UGET 1995]
0%
a)9 kg
0%
b) 10 kg
0%
c)11 kg
0%
d)12 kg
Explanation
Answer: (c)
Q.24
Two rods A and B of same material and length, have their electric resistance in ratio 1:2, When both rods are dipped in water, the correct statement will be [ Raj. PMT 1997]
0%
a) A has more loss of weight
0%
b) B has more loss of weight
0%
c)Both have same loss of weight
0%
d)Loss ratio will be in the ratio 1:2
Explanation
Electric resistance is givn by formula R=ρ(l/A) for other wire R'=ρ ( l/A')R/R'=A'/A∴ 1 : 2=A': AImplies that Area of first(A) is two times the area of second(B) thus volume of first resistance is more than second∴ Upthrust on A is more than B. Loss of weight of A is more Answer: (a)
Q.25
The volume of air bubble becomes three times as it rises from the bottom of lake to its surface. Assuming atmospheric pressure to be 75 cm of Hg and the density of water is to be 0.1 times of the density of mercury, the depth of the lake is [ AMU 1995]
0%
a) 5 m
0%
b) 10 m
0%
c)15 m
0%
d)20 m
Explanation
Pressure at depth P1=Pressure due to depth + Atmospheric pressure P1=H (0.1d)g + 75dg V1=V pressure at surface P2=75dg Volume V2=3 V (given) According to Boyle's law P1V1=P2 V2 [H (0.1d)g + 75dg] V=(75dg) (3V)H(0.1) +75=(75)(3) H=15m Answer:(c)
Q.26
the density of ice is x gm/cm3 and that of water is y gm/cm3 when m gram of ice melts, then the change in volume is
0%
a) m(y-x)
0%
b) (y-x)/m
0%
c) my(x-y)
0%
d) (m/y) - (m/x)
Explanation
Volume of ice=mass/ density=m/ x density of water=m/y Thus change in volume ( final - initial volume )=m/y - m/ x Answer: (d)
Q.27
Hook's law essentially defines [ MPPMT 1998]
0%
a)Stress
0%
b) Strain
0%
c)Yield point
0%
d)Elastic limit
Explanation
Answer: (d)
Q.28
Bulk modulus was first defined by [ CPMT 1987]
0%
a) Young
0%
b) Bulk
0%
c)Maxwell
0%
d)none
Explanation
Answer: (c)
Q.29
The beam of metal supported at the two ends is loaded at centre. The depression at the centre is proportional to
0%
a) Y2
0%
b) Y
0%
c)1/Y
0%
d)1/Y2
Explanation
Answer:(c)
Q.30
A wire is stretched by 0.1m by a certain force F. Another wire of same material whose diameter and lengths are doubled to the original wire is stretched by the same force. Then its elongation will be .. [ EAMCET 1995]
0%
a) 0.005 m
0%
b) 0.01 m
0%
c) 0.02 m
0%
d) 0.02 m
Explanation
Elongation e F is constant Given 0.01=kl/ r2 For second wire e=kl' / r'2 given l'=2l and r'=2r e=k(2l) / (2r)2 e=kl/ r2 ×(1/2) e=0.01× (1/2)=0.005 mAnswer: (a)
0 h : 0 m : 1 s
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Support mcqgeeks.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page