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Physics NEET MCQ
Quiz 3
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Q.1
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
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a) Rises
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b) Falls
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c)Remains unchanged
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d)none
Explanation
Volume of liquid displaced more than the volume of water formed. Therefore level of water falls Answer:(b)
Q.2
1 kg of cotton and iron in air are transferred to vacuum and weighed again then.
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a) Cotton and iron will weigh same
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b) Iron will weigh more than cotton
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c) Cotton will weigh more than iron
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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.3
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
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a)Remain unchanged
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b) Rise
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c)Fall
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d)Rise or fall depending on the number of stones unloaded.
Explanation
Answer: (c)
Q.4
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
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c)0.173 kg
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d)0.388 kg
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a)12×10⁻³kg
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b) 12×10⁻³kg
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.5
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
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a)
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b)
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c)
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d)
Explanation
Pressure at height can be obtain by integrating given equation Answer:(b)
Q.6
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]
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a) The balance A will read more than 2kg
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b) The balance B will read less than 5kg
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c) The balance A will read less than 2kg and B will read more than 5 kg
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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.7
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]
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a)3.3
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b) 6.4
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c)7.2
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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.8
The principle of the operation of hydraulic press is based on
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a) Boyle's law
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b) Pascal's law
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c)Dalton's law of partial pressure
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d)Newtons law of gravitation
Explanation
Answer: (b)
Q.9
Pressure applied to enclosed fluid is
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a) Increased and applied to every part of the fluid
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b) Diminished and transmitted to wall of container
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c)Increased in proportion to the mass of the fluid and then transmitted
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d)Transmitted unchanged to every portion of the fluid and walls of containing vessel
Explanation
Answer:(d)
Q.10
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
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a) The reading of the balance will be grater when the parrot flies in the cage.
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b) the reading of the balance will be lesser when the parrot flies in the cage
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c) the reading will remain unchanged
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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.11
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
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a)4 : 1
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b) 2 : 1
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c)1 : 1
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d)16 : 1
Explanation
Upthrust depends on the volume of liqid dispalce , since volume of both container is same upthrust is sameAnswer: (c)
Q.12
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]
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a) 90 dcm
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b) 60 dcm
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c)40 dcm
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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.13
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]
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a) 1/5
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b) 1/3
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c)3/5
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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.14
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]
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a)0.1 m
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b) 9.8 m
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c)0.98 m
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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.15
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]
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a)9 kg
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b) 10 kg
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c)11 kg
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d)12 kg
Explanation
Answer: (c)
Q.16
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]
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a) A has more loss of weight
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b) B has more loss of weight
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c)Both have same loss of weight
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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.17
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]
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a) 5 m
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b) 10 m
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c)15 m
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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.18
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
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a) m(y-x)
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b) (y-x)/m
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c) my(x-y)
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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.19
Hook's law essentially defines [ MPPMT 1998]
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a)Stress
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b) Strain
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c)Yield point
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d)Elastic limit
Explanation
Answer: (d)
Q.20
Bulk modulus was first defined by [ CPMT 1987]
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a) Young
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b) Bulk
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c)Maxwell
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d)none
Explanation
Answer: (c)
Q.21
The beam of metal supported at the two ends is loaded at centre. The depression at the centre is proportional to
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a) Y2
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b) Y
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c)1/Y
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d)1/Y2
Explanation
Answer:(c)
Q.22
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]
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a) 0.005 m
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b) 0.01 m
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c) 0.02 m
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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)
Q.23
Two rods of different materials having coefficient of linear expansion α1 , α2 and Young's modulii Y1 , Y2 respectively are fixed between two rigid massive walls. the rods are heated such that they undergoes the same increase in temperature. there is no bending of rods. If α1 : α2=2 : 3, the thermal stress developed in two rods are equal provided Y1:Y2 is equal to [ IIT 1989]
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a)3 : 2
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b) 1 : 1
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c)2 : 3
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d)4 : 9
Explanation
From the formula you modulus Y=Stress / StrainStress=Y ×Strain Stress=Y (Δl / l ) Thus Y(δ l)=Stress × l Since stress and length of rod is same Y(Δl )=constant Now increase in length of rod=αΔθ here Δθ is increase in temperature Y1(Δl )=Y1(Δl )Y1(α1Δθ)=Y2 (α2Δθ) Y1 /Y2=α2 /α1=3/2 Answer: (a)
Q.24
A spherical ball contracts in volume by 0.01%, when subjected to a normal uniform pressure of 100 atmospheres. The bulk modulus of its material in dynes/cm2 is
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a) 10 ×1012
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b) 100 ×1012
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c)1 ×1012
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d)2 ×1012
Explanation
We know that Bulk modulus Answer: (c)
Q.25
Two wires A and B are of the same material. their lengths are in ratio 1:2 and the diameters are in ratio 2:If they are pulled by the same force, the increase in length will be in ratio [ MPPMT 1988]
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a)2 : 1
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b) 1 : 4
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c)1 : 8
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d)8 : 1
Explanation
Elongation e Force is same thus e ∝ l/r2 Answer:(c)
Q.26
A thick copper rope of density 1.5×103 kg/m3 and Young's modulus 5×106 N/m2, 8 m in length is hung from the ceiling of a room. The increase in its length due to its own weight is [ CPMT 1998]
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d) 9.6 m
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a) 9.6×10⁻² m
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b) 19.2×10⁻⁷ m
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c) 9.6×10⁻³ m
Explanation
Average force on the rod=weight of rod /2 F=πr2L×ρg / 2 Thus increase in length from Young’s modulus formula Answer: (a)
Q.27
The upper end of a wire of radius 4mm and length 100 cm is clamped and its other end is twisted through an angle of 30°. Then angle of shear is [ NCERT 1990]
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a)12°
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b) 0.12°
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c)1.2°
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d)0.012°
Explanation
We know that θ=Φr / l here θ=angle of shear Φ=angle of twist 30°θ=(30×0.4 ) / 100=0.12°Answer: (b)
Q.28
A cube at temperature 0°C is compressed equally from all sides by an external pressure P. By what amount should its temperature be raised to bring it back to the size it had before the external pressure was applied. the bulk modulus of the material of the cube is K and the coefficient of linear expansion is α [ CPMT 1998]
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a) P / Kα
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b) P/ 3Kα
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c)3Pα / K
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d) 3K/ P
Explanation
We know that thermal stress=P=KγΔθ θ=P / (Kγ)=P / ( 3Kα) Here γ=3 α Answer: (b)
Q.29
A steel ring of radius r and cross-section area 'A' is fitted onto a wooden disc of radius R(R>r). If young's modulus be E, then the force with which the steel ring is expanded is [ AP 1986]
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a)
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b)
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c)
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d)
Explanation
Extension=(R-r), original=r Strain=(R-r) / r Stress=Foce /Area=F/A Young Modulus E=Stress/ Strain Answer: (b)
Q.30
The extension of wire by the application of load is 3mm. The extension in a wire of the same material and length but half the radius by the same load [ CPMT 1990]
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a)12 mm
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b) 0.75 mm
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c)6 mm
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d)15 mm
Explanation
if e is elongation e ∝ (F l) / r2 Fand l are same thus e ∝ (1/r2)given r2=½ r1 Answer: (a)
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