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Properties Of Matter Mcq
Quiz 1
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Q.1
In Milikan's oil drop experiment a small, spherical oil drop of radius r is moving in a medium of density ρ with an instantaneous speed v. The viscous force F is [ MPPMT 1989]
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a)6πηrv
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b) ηrv
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c)6η/rv
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d)6πηrv/ρ
Explanation
Answer: (a)
Q.2
Two wires A and B are of the same material. Their lengths are in the ratio 1 :2 and the diameter are in the ratio 2:If they are pulled by the same force, then increase in length will be in the ratio [ CBSE-PMT 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
Formula for Young's modulus is Since Y and F are same there foresince D1=2D2L2=2 L1substituting the values in above we getl1:l2=1:8Answer: (c)
Q.3
When an elastic material with Young's modulus Y is subjected to stretching stress S, elastic energy stored per unit volume of the material is [1989]
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a) YS/2
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b) S2Y/2
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c) S2/2Y
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d) S/2Y
Explanation
Energy stored per unit volume=½(stress × strain)=½[stress ×(stress /Young's modulus)]=½×(stress)2 / (Young's modulus)=S2 / 2YAnswer: (c)
Q.4
The compressibility of water is 4 × 10 -5 per unit atmospheric pressure. The decrease in volume of 100 cm3 of water under a pressure of 100 atmosphere will be [ CBSE_PMT 1990]
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(a) 0.4cm3
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(b) 4x 10-5cm3
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(c) 0.025cm3
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(d) 0.004cm3
Explanation
Formula for compressibility of water Here P=100 atmK=4 × 10 -5V=1003Hence ΔV=0.4 cm3 Answer:(a)
Q.5
The terminal velocity vt of a small steel ball of radius r falling under gravity through a column of a viscous liquid of coefficient of viscosity depends on mass of the ball m, acceleration due to gravity g, coefficient of viscosity η and radius r. Which of the following relations is dimensionally correct ? [1992]
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a)
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b)
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c)
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d)
Explanation
According to Stoke's law 6πηrvt=mg hence option (c) is correct Answer: (c)
Q.6
The angle of contact between pure water and pure glass, is
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a) 0°
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b) 45°
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c) 90°
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d) 1350°
Explanation
` Angle of contact is the angle between the tangent to liquid surface at the point of contact and solid surface inside the liquid. In case of pure water and pure glass, the angle of contact is zero.Answer: (a)
Q.7
For a given material, the Young's modulus is 2.4 times that of rigidity modulus. Its Poisson ratio is [ EAMCET 1990]
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a) 2.4
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b) 1.2
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c) 0.4
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d) 0.2
Explanation
Relation between modulus of rigidity η, Young's modulus Y and Poisson Ratio σ is given by Answer: (d)
Q.8
A rectangular block of mass m and area of cross-section A floats in a liquid of density ρ. if it is given a small vertical displacement from equilibrium it undergoes oscillation with a time period T. Then
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a) T ∝ 1/√A
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b) T ∝ 1/ρ
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c)T ∝ 1/√m
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d)T ∝√ρ
Explanation
Let the body be depressed by distance x from its equilibrium position. weight of displaced liquid will produce up thrust which applies to whole body and body will oscillate.Up thrust=Volume of liquid displaced ×density × gravitational accelerationUp thrust=xA×ρ×gIf 'a' is acceleration thenma=xAρga=(xAρg)/mComparing it with standard equation for SHM we getAnswer: (a)
Q.9
A soap bubble has radius r and volume V. If the excess pressure inside the bubble is P. Then PV is proportional to [ AFMC 1997]
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a) r
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b)r2
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c)r3
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d)r4
Explanation
For soap bubble excess pressure=4T/r Answer:(b)
Q.10
A ball is falling in a lake of depth 200m creates a decrease of 0.1% in its volume at the bottom. The bulk modulus of the material of the ball will be ..[ CBSE-PMT 1997]
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a) 19.6 ×10-8 N/m2
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b) 19.6 ×1010 N/m2
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c) 19.6 ×10-10 N/m2
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d) 19.6 ×108 N/m2
Explanation
Formula for bulk modulus dv/V=0.1/100=10-3dP=hρgdP=200×103 × 9.8dP=1.960×106Bulk modulus B= P/(dv/V) Answer: (d)
Q.11
The kinetic energy of one gram molecule of a gas at normal temperature and pressure is (R=8.31 J/ mole K) [ AFMC 1998]
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a)3.4×103
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b) 2.97×103
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c)1.2×103
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d)0.66×103
Explanation
Kinetic energy of one mole of gas molecule=(3/2)RT=(3/2)(8.31)(273)=3.4×103JAnswer: (a)
Q.12
A big drop of radius R is formed by 1000 small droplets of water, the radius of small drop is [ AFMC 1998]
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a) R/10
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b) R/100
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c)R/500
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d)R/1000
Explanation
If radius of small drop is 'r' and big drop is 'R' when small drop combines then volume of big drop=1000( volume of small drop)(4/3)π×R3=(1000)(4/3)π×r3R=10r r=R/10Answer: (a)
Q.13
An iron rod of length 2m and cross -sectional area of 50 mm2is stretched by 0.5mm, when a mass of 250kg is hung from its lower end. Young's modulus of iron rod is [ AFMC 1999]
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a) 19.6×1020N/m2
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b) 19.6×1018N/m2
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c)19.6×1010N/m2
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d)19.6×1015N/m2
Explanation
Formula for Young's modulus is F=mg=250(10)=2500NA=50×10-6l=0.5×10-3l=2 msubstituting values in above equation we get Y=20×1010 N/m2Approx 20×1010 N/m2 Answer:(c)
Q.14
The rain drops are spherical shape due to [AFMC 1999]
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a) residual pressure
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b) thrust on the drop
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c) surface tension
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d) viscosity
Explanation
Answer: (c)
Q.15
10 gm of ice cubes at 0°C are released in a tumbler (water equivalent 55 g) at 40°C. Assuming that negligible heat is taken from the surroundings, the temperature of water in the tumbler becomes nearly (L=80 cal/g) [1988]
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a)31°C
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b) 22°C
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c)19°C
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d)15°C
Explanation
Let the final temperature be T Heat gained by ice=mL + m × S × (T-0)Heat gained by ice=10 × 80 + 10×1× ( T )Heat lost by water=55 ×1×(40 - T) By using law of clorimetery800 + 10T=55 × (40-T) ∴ T=21.54°C=22°CAnswer: (b)
Q.16
The two ends of a rod of length L and a uniform cross-sectional area A are kept at two temperatures T1 and T2 (T1 > T2). The rate of heat transfer, dQ/dt through the rod in a steady state is given by:
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a)
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b)
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c)
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d)
Explanation
T1 - T2 is the temperature difference Answer: (c)
Q.17
A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat Q in time t. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod, when placed in thermal contact with the two reservoirs in time t? [ CBSE-PMT 2010]
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a) Q/4
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b) Q/16
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c)2Q
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d)Q/2
Explanation
The rate of flow of heat is given by Q/t=K.A.( ΔT/l)Area of Original rod A=πR2Area of new rod A'=πR2 / 4 ∴ A'/A=1/4 Volume of orginal rod=Volume of new rodπR2 l=π (R/2)2l'∴ l/l'=1/4 ∴ Q'/Q=A'l / A l'=(1/4)(1/4)=1/16 ∴ Q'=Q/16 Answer:(b)
Q.18
Thermal capacity of 40 g of aluminium (s=0.2 cal /gK) is [ CBSE-PMT 1990]
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a) 168joule/ °C
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b) 672joule/°C
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c) 840joule/°C
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d) 33.3 joule/°C
Explanation
Thermal capacity=ms=40×0.2=8 cal/°C Thermal capacity=4.2× 8 cal=33.6joules/°C Answer: (d)
Q.19
If the temperature of the sun is doubled, the rate of energy received on earth will be increased by a factor of [ CBSE-PMT 1993]
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a)2
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b) 4
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c)8
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d)16
Explanation
Amount of energy radiated ∝ T4Thus Amount energy radiated ∝ (2T)4=16TAnswer: (d)
Q.20
A body cools from 50.0°C to 48°C in 5s. How long will it take to cool from 40.0°C to 39°C? Assume the temperature of surroundings to be 30.0°C and Newton's law of cooling to be valid, [ CBSE-PMT 1994]
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a) 2.5s
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b) 10s
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c)20s
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d) 5s
Explanation
Rate of cooling ∝ temperature difference between system and surroundingAs the temperature difference is halved, the rate of cooling will also be halved. Hence two time increase in time takenAnswer: (b)
Q.21
A beaker full of hot water is kept in a room. If it cools from 80°C to 75°C in t1 minutes, from 75° C to 70°C in t2 minutes and from 70°C to 65°C in t3 minutes, then [ CBSE-PMT 1995]
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a)t1=t2=t3
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b)t1 < t2=t3
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c)t1
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d)t1 >t2 > t3
Explanation
Let θ0 be the temperature of the surrounding thensimilarly,from above it is cleart1 < t2 < t3 Answer:(c)
Q.22
The radiant energy from the sun, incident normally at the surface of earth is 20 k cal/ m2 min. What would have been the radiant energy, incident normally on the earth, if the sun had a temperature, twice of the present one? [ CBSE-PMT 1998]
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a) 160kcal/m2min
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b) 40kcal/m2min
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c) 320k cal/m2 min
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d) 80kcal/m2min
Explanation
According to Stefan's law E ∝ σ εAT4 Thus E∝T4 ∴ E2=320 kcal/m2min.Answer: (c)
Q.23
If 1 g of steam is mixed with 1 g of ice, then the resultant temperature of the mixture is [ CBSE-PMT 1999]
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a)270°C
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b) 230°C
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c)100°C
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d)50°C
Explanation
Heat available with 1g steam to condense into 1g of water at 100°C=536 calHeat enquired by ice to reach temperature of 100°C=mL + mcΔθ =1× + 1×1×(100-0)=180 calFrom above it is clear that all the steam will not condensed and ice will attain temperature of 100°; so the temperature of mixture will be 100°CAnswer: (c)
Q.24
The presence of gravitational field is required for the heat transfer by.... [ CBSE-PMT 2000]
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a) conduction
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b) stirring of liquids
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c)natural convection
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d)radiation
Explanation
In convection, temperature gradient exists in the vertical direction and not in the horizontal direction. So, up and down movement of particles takes place which depends on the weight and gravityAnswer: (c)
Q.25
A black body has maximum wavelength λm at temperature 2000 K. Its corresponding wavelength at temperature 3000 K will be [2001]
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a) (3/2) λm
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b) (2/3) λm
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c)(4/9) λm
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d)(9/4) λm
Explanation
According to Wein's displacement law λmT=2.88 × 10-3When T=2000Kλm2000=2.88 × 10-3 --eq(1)When T=3000K λ'm3000=2.88 × 10-3 --eq(2)Dividing eq(1) by eq(2) we getλ'm=(2/3) λm Answer:(b)
Q.26
Wien's law is concerned with [ CBSE-PMT 2002]
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a) relation between emissivity and absorptivity of a radiating surface
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b) total radiation, emitted by a hot surface
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c) an expression for spectral distribution of energy of a radiation from any source
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d) a relation between the temperature of a black body and the wavelength at which there is maximum radiant energy per unit wavelength
Explanation
According to Wein's displacement law λmT=2.88 × 10-3 or constant Answer: (d)
Q.27
If λm denotes the wavelength at which the radiative emission from a black body at a temperature T K is maximum, then [2004]
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a)λm ∝ T-1
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b) λm ∝ T4
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c)λm independent of T
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d) λm ∝ T1
Explanation
According to Wein's displacement law λm T=constant∴ λm ∝ T-1Answer: (a)
Q.28
The upper end of wire 1 meter long and 2 mm radius is clamped. the lower end is twisted through an angle of 45°. The angle of shear is [ MPPMT 1990]
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a)0.09°
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b) 0.9°
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c)9°
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d)90°
Explanation
Angle of shear=θ=Φr / l θ=( 45×2×10-3) / 1 θ=90×10-3=0.09°Answer: (a)
Q.29
On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are 39° W and 239° W respectively. What will be the temperature on the new scale, corresponding to a temperature of 39°C on the Celsius scale? [ CBSE-PMT 2008]
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a) 78° W
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b) 117°W
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c)200° W
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d) 139°W
Explanation
let L.F.P is Lower Fixed Point U.H.P be Upper fixed point x be the measurement at that scale Then we have If C and W be the measurement on Celsius and W scale then , Now C=39° given or From the data given we can get equation W = 2 × °C + 39 Answer: (b)
Q.30
The coefficient of linear expansions of brass and steel are α1 and α2 respectively. When we take a brass rod of length l1 and a steel rod of length l2 at 0°C, then the difference in their lengths (l2 –l1) will remain the same at all temperatures if [ CBSE-PMT 1999]
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a)α1 l1=α2 l2
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b) α1 l2=α2 l1
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c)α12 l2=α22 l1
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d)α1 l22=α2 l12
Explanation
Let ΔT be increase in the temperature of brass wire. Then length of brass wirel'=l1( 1 + α1ΔT)similarly, length of steel wire when temperature is increased by ΔTl"=l2 ( 1+ α2ΔTAccording to equation l'-l"=l1 - l2 l1+l1α1ΔT - l2 +l2α2ΔT=l1 - l2as ΔT ≠ 0l1α1=l2α2 Answer:(a)
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