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Quiz 4
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Q.1
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]
0%
a)3 : 2
0%
b) 1 : 1
0%
c)2 : 3
0%
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.2
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
0%
a) 10 ×1012
0%
b) 100 ×1012
0%
c)1 ×1012
0%
d)2 ×1012
Explanation
We know that Bulk modulus Answer: (c)
Q.3
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]
0%
a)2 : 1
0%
b) 1 : 4
0%
c)1 : 8
0%
d)8 : 1
Explanation
Elongation e Force is same thus e ∝ l/r2 Answer:(c)
Q.4
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]
0%
a) 9.6×10-2 m
0%
b) 19.2×10-7 m
0%
c) 9.6×10-3 m
0%
d) 9.6 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.5
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]
0%
a)12°
0%
b) 0.12°
0%
c)1.2°
0%
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.6
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]
0%
a) P / Kα
0%
b) P/ 3Kα
0%
c)3Pα / K
0%
d) 3K/ P
Explanation
We know that thermal stress=P=KγΔθ θ=P / (Kγ)=P / ( 3Kα) Here γ=3 α Answer: (b)
Q.7
When a force is applied on a wire of uniform cross-sectional area 3×10-6 m2 and length 4m, the increase in length is 1mm. Energy stored in it will be [ Y=2×1011 N/m2] [ MP PET 1995]
0%
a) 6250 Joule
0%
b) 0.177 Joule
0%
c)0.075 Joule
0%
d)0.150 Joule
Explanation
∴ Energy Stored=½ F × lEnergy Stored=½ × 1.5×102 ×10-3 Energy Stored=0.075 J Answer:(c)
Q.8
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]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Extension=(R-r), original=r Strain=(R-r) / r Stress=Foce /Area=F/A Young Modulus E=Stress/ Strain Answer: (b)
Q.9
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]
0%
a)12 mm
0%
b) 0.75 mm
0%
c)6 mm
0%
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)
Q.10
A material has Poisson's ratio 0.if a uniform rod of it suffers a longitudinal strain of 2×10-3, then the percentage change in volume is [ AP1987]
0%
a) 0.6
0%
b) 0.4
0%
c)0.2
0%
d)zero
Explanation
Poisson ratio σ=lateral Strain / Longitudinal Strain lateral strain (ΔR/R)=σ × longitudinal StrainΔR /R=0.5×2×10-3 Volume of rod V=πR2lΔV=π[2RΔR)l + R2Δl] Thus percentage change in volume is zeroAnswer: (d)
Q.11
The poisson ratio can not have the value [ EAMCET 1989]
0%
a) 0.7
0%
b) 0.2
0%
c)0.1
0%
d)0.5
Explanation
Answer:(a)
Q.12
Streamline flow is more likely for liquids with
0%
a) High density an low viscosity
0%
b) Low density and high viscosity
0%
c)High density and high viscosity
0%
d)Low density and low viscosity
Explanation
Answer: (b)
Q.13
The compressibility of water is 4×10-5 per unit atmospheric pressure. The decrease in volume of 100cm3 of water under a pressure of 100 atmospheres will be [ MPPMT 1990]
0%
a) 0.4 cm3
0%
b) 4×10-5 cm3
0%
c)0.025 cm3
0%
d)0.04 cm3
Explanation
Given compressibility=4×10-5 ( atmospheric pressure)-1One atmospheric pressure=106 dynes/cm2Compressibility=4×10-5 ×10-6 cm2=4×10-11 cm2 / dynes Bulk modulus B=1/ Compressibility Bulk modulus B=1 /4×10-11=(1/4)1011 dyne/cem2 Formula for bulk modulus Answer: (a)
Q.14
Two cylinder A and B are made of the same material. the length and radii of the two cylinder are in the ratio of 1 :Both are twisted by the same external torque. the ratio of the angle of twist of A and B is
0%
a) 1: 2
0%
b) 1 : 4
0%
c)1 : 8
0%
d)8 : 1
Explanation
torque required to prode twist of Φ in a wire of length 'l', radius 'r' and modulus of rigidity η is given by equationGiven torque is same , material is same r2 / r1=2 l2 / l1=2 from above formula and taking ratios we get Answer:(d)
Q.15
Two cylinder A and B of the same material have same length, the radii of A and B are in the ratio of 1:The two are joined end to end as shown in figure. The upper end of A is rigidly fixed. The lower end of B is twisted through an angle θ, the angle of twist of cylinder A is
0%
a) θ
0%
b) (15/16) θ
0%
c) (16/17)θ
0%
d) (17/16) θ
Explanation
The twist at upper end of A=0. the twist at the point is Φ, the twist at the lowest end=θ. Then,given τA=τB Answer: (c)
Q.16
In steel the Young's modulus and strain at the breaking point are 2×1011Nm-2 and 0.15 respectively. The stress at the break point for steel is therefore [ MPPET 1990]
0%
a)1.33 × 1011Nm-2
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b) 1.33×1012Nm-2
0%
c)7.5×10-3 Nm-2
0%
d)3×1010Nm-2
Explanation
We know Y=stress / strain Stress=Y ( strain) Stress=(0.15) (2×1011)=3×1010Nm-2 Answer: (d)
Q.17
Which of the following statement is correct? [ MPPET 1992]
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a) Hook's law is applicable only within elastic limit
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b) The adiabatic and isothermal elastic constant of gas are equal
0%
c)Young's modulus is dimensionless
0%
d)Stress multiplied by strain is equal to the stored energy
Explanation
Answer: (a)
Q.18
The force required to stretch a steel wire of 1cm2 cross-section to 1.1 times its length would be ( Y=2×1011 Nm-2) MPPER 1992]
0%
a) 2×106 N
0%
b) 2×103 N
0%
c)2×10-6
0%
d)2×10-7 N
Explanation
Increase in length=1.1(l) - l=0.1lFrom the formula of Young's modulus Answer:(a)
Q.19
Which of the following substance possesses the highest elasticity [ MPPMT 1992]
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a) Rubber
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b) Glass
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c) Steel
0%
d) Copper
Explanation
Answer: (c)
Q.20
Which of the following quantities does not have the unit of force per unit area [ MPPMT]
0%
a)Stress
0%
b) Strain
0%
c)Young's modulus of elasticity
0%
d)Pressure
Explanation
Answer: (b)
Q.21
A rod of length l and radius is joined to a rod of length l/2 and radius r/2 of same material. The free end of small rod is fixed to rigid base and the free end of larger rod is given a twist of θ°, the twist angle at the joint will be
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a) θ/4
0%
b) θ/2
0%
c)(5/6) θ
0%
d)(8/9) θ
Explanation
Twist at upper end of small rod A=0Twist at the joint is=ΦTwist at lower end of big rod B=θ since torque acting on the rods is same Answer: (d)
Q.22
As steel wire has length 2.0 m, radius 1.0 mm and Y=20×1010 N/mA sphere of mass 10 kg is attached to one end of the wire ,which is then whirled in a vertical circle with an angular velocity of 2 revolutions per sec. The elongation of wire when the mass is at lowest point of the path is nearly [ AMU 1995]
0%
a)1.0 mm
0%
b) 2.0 mm
0%
c)0.1 mm
0%
d)0.01 mm
Explanation
Tension at lower end=Centrifugal force + weight of sphereT=mlω2 + mg here m=10 kgω=2 revolution/sec.l=2.0 m T=10×2×(2π×2) + 10×10 T=420 N Answer:(a)
Q.23
Two wires of the same material have lengths in ratio 1:2 and their radii are in the ratio 1 :√If they are stretched by applying equal force, the increase in their lengths will be in the ratio.. [ MPPMT 1994]
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a) 2 : √2
0%
b) √2 : 2
0%
c) 1 : 1
0%
d) 1 : 2
Explanation
We know that increase in length is given by formula Given that material same so Y is same, Force is same for both wire Answer: (c)
Q.24
A uniform cube is subjected to volume compression. If each side is increased by 1% then bulk strain is .. [ EAMCET 1995]
0%
a)0.01
0%
b) 0.06
0%
c)0.02
0%
d)0.03
Explanation
Strain=ΔV/V=(3Δl)/ lGiven increase in length h=1%=0.01 Bulk Strain=3 (0.01)=0.03Answer: (d)
Q.25
A cylinder tree has a breaking stress of 106 N/mThe density of the tree is 2×104 kg/mThe maximum possible height of the tree is then [ CET 1994]
0%
a) 25 m
0%
b) 150 m
0%
c)5.0 m
0%
d)100 m
Explanation
Breaking stress=hdgd=density , h is height 106=h ×2×104×10 h=5 m Answer: (c)
Q.26
A force of 103 N stretches the length of a hanging wire by 1 millimeter. The force required to stretch a wire of same material and length but having four times the diameter by 1 millimetre is [ PMT 1995]
0%
a) 4×103 N
0%
b) 16×103 N
0%
c)(1/4)×103 N
0%
d)(1/16)×103 N
Explanation
Increase in length is given by Same material hence Y is constant, Length same, but diameter is four times Answer:(b)
Q.27
When a weight of 10 kg is suspended from a copper wire of length 3 metres and diameter 0.4 mm, it length increases by 2.4 cm. If the diameter of the wire is doubled, then the extension in its length will be [ MPPMT 1994]
0%
a) 9.6 cm
0%
b) 4.8 cm
0%
c) 1.2 cm
0%
d) 0.6cm
Explanation
Increase in length is given by force is same in both the cases, length is same , material is same only diameter is doubled Answer: (d)
Q.28
When a 4 kg mass is hung vertically on a light spring that obeys Hook's law, the spring stretches by 2cm. The work required to be done by an external agent in stretching this spring by 5cm will be ( g=9.8 m/s2[ PMT 1995]
0%
a)4.900 Joule
0%
b) 2.450 Joule
0%
c)0.495 Joule
0%
d)0.245 Joule
Explanation
Force on spring=mg F=4×9.8 By Hook's Law F=k x F=k×24×9.8=k×2k=2×9.8 Now potential energy of spring=½ k x2 Potential energy=½ × (2×9.8)×5×10-2 Potential energy=-2.45 Joule Now according to work energy theoram Change in potential energy=Work done Answer: (c)
Q.29
A 2 m long rod of radius 1cm, which is fixed from one end is given twist of 0.8 radians. The shear strain developed will be [ Raj.PET 1997]
0%
a) 0.002
0%
b) 0.004
0%
c)0.008
0%
d)0.016
Explanation
we know that angle of shear θ=Φr / l Here Φ is angle of twist θ=(0.8×0.01) / 2=0.004 radianAnswer: (b)
Q.30
A body of mass 10 kg is attached to a wire 30 cm long. It breaking stress is 4.8 ×107 Nm-The area of cross-section of the wire is 10-6 mWhat is the maximum angular velocity with which it can be rotated in horizontal circle [ CPMT 1998]
0%
a) 1 rad/s
0%
b) 2 rad/s
0%
c)4 rad/s
0%
d)8/rad/s
Explanation
When rotated in horizontal plane centrifugal force will be on the wire if it is more than breaking Force wire will break Force for braking=Breaking stress × Area of wireThus centrifugal force=Force for braking mlω2=Breaking stress × A ω2=[Breaking stress × A]/ ml Given Breaking stress=4.8 ×107Nm-2 Area A=10-6 m2length l=0.3 mω2=4.8 ×107× 10-6]/ 10×0.3 ω2=16 ω=4 rad/s Answer:(c)
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