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Quiz 9
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Q.1
A thin square steel plate with each side equal to 10cm is heated by a blacksmith. The rate of radiated energy by the heated plate is 1134 watts. The temperature of the hot steel plate is... ( stefan's constant σ = 5.67×10-8 watt meter-2 K-4, emissivity of the late = 1 ) [ PMT 1995]
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a) 1000 K
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b) 1189 K
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c) 2000 K
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d) 2378 K
Explanation
From equation energy radiated per meter per second = E = σeT4 Given energy radiated form 100cm2 = 1134 watt energy radiated from 1 m2 = 1134×102 1134×102 = 5.67×10-8×1 (T)4 T4 = 200×1010 T4 = 2×1012 T = 1.189 ×103 Answer: (b)
Q.2
If the colour of star is changing from red to blue it means [ RajPMT 1996]
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a) The star is going away from the earth
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b) The star is coming near the earth
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c) The star is stationary
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d) Non of the above
Explanation
Answer: (b)
Q.3
If a liquid is heated in space under no gravity , the transfer of heat will take place by process of [ Raj.PMT 1996]
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a) conduction
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b) convection
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c) Radiation
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d) Cannot be heated in absence of gravity
Explanation
Answer: (c)
Q.4
A body radiates energy 5W at temperature of 127°. If the temperature is increased to 927 °C then it radiates energy at the rate of [ BHU 1995]
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a) 410 W
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b) 81 W
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c) 405 W
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d) 200W
Explanation
E ∝ T4 E ∝ (127+273)4 E ∝ (400)4 eq(1) E' ∝ (927+273)4 E' ∝ (1200)4 eq(2) from equation 1 and equation 2 we get E' = E( 1200 /400 ) 4 E' = 5×81 = 405 w Answer:(c)
Q.5
If the temperature of sun is doubled [ raj. PMT 1996]
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a) Emission will become four times
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b) Emission will become double
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c) Mainly it will emit ultra violet radiations
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d) Mainly it will emit infra red radiations
Explanation
Answer: (c)
Q.6
Two rods of same length and material transfers given amount of heat in 12 sec, when they are joined end to end. But when they are joined lengthwise, then they will transfer same heat in same conditions in [ BHU 1998]
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a) 24 sec
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b) 3 sec
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c) 1.5 sec
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d) 48 sec
Explanation
Amount of heat transferred Q ∝ At/d Q is same in both the cases, lenth becomes half and area becomes doubled of initial Answer: (b)
Q.7
An iron tyre is to be fitted on a wooden wheel 1.0 m in diameter, the diameter of the tyre is 6 mm smaller than that of the wheel. The tyre should be heated so that its temperature increases by a minimum of :.. ( coefficient of expansion of iron is 3.6×10-5 per °C) [ CPMT 1989]
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a) 167°C
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b) 334°C
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c) 500 °C
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d) 1000 °C
Explanation
circumference l = πD = 3.14×100 = 314 cm Δl = πΔD = 3.14×0.6 Coefficient of linear expansion α = γ / 3 = 1.2×105 Answer: (c)
Q.8
A metal ball immersed in alcohol weighs W1 at 0°C and W2 at 95°C. the coefficient of cubical expansion of the metal is less than that of alcohol. Assuming that the density of the metal is large compared to that of alcohol, it can be shown that [ CPMT 1998]
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a) W1 > W2
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b) W1 = W2
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c) W1 < W2
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d) W2 = W1 /2
Explanation
At 59°C, the density of alcohol decreases, therefore upthrust decreases and apparent weight increases W2 > W1. Answer:(c)
Q.9
Two rods of lengths l1 and l2 are made of materials whose coefficients of linear expansions are α1 and αIf the difference between the two lengths is independent of temperature [ EAMCET 1995]
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a)
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b)
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c)
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d)
Explanation
Given l2 - l1 = constant ∴ Δl2 = Δl1 α1l1Δθ = α2l2Δθ l1 / l2 = α2 / α1 Answer: (b)
Q.10
The temperature of an isotropic cubical solid of length L, density d and coefficient of expansion α per degree Kelvin, is raised by 10°C. Then, at this temperature, to a good approximation ..[ IIT 1995]
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a) Lengt is L(1 +10α)
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b) Total surface area is L2 ( 1 +20α)
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c) density is d (1 + 30α)
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d) density is d/(1+30α)
Explanation
From formula Lt = L0( 1 +αt) Lt = L0( 1 + 10t) Answer: (a)
Q.11
A ring shaped piece of metal is heated . If the material expands, the hole will [ EAMCET 1994]
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a) expand
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b) contract
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c) expand or contracts depending on the width of ring
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d) expand or contract depending on the value of the coefficient of expansion
Explanation
Answer: (a)
Q.12
A pendulum of brass has period of 1 sec. at 20°C. Coefficient of linear expansion of brass is 1.93×10-5 / °C. The clock gets delayed in week at 30° by ..[ Raj.PET 1997]
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a) 8 sec
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b) 58s
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c) 224 s
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d) 504 s
Explanation
Periodic c time of pendulum Thus it gets delayed by 9.62 × 10-5 sec in none second Delayed in week = 9.62× 10-5×7×24×3600 sec Delayed in week = 58.18 sec Answer: (b)
Q.13
A beaker is filled with water at 4°C. At one time the temperature is increased by few degrees above 4°C and at another time it is decreased by few degree below 4°C. One shall observe that
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a) the level remains constant in each case
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b) water overflows in both cases
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c) water overflows in the latter case, while its level comes down in the previous case
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d) in previous case water overflows while in latter case its level comes down
Explanation
Answer: (b)
Q.14
A container of volume 1 m3 is divided into two equal compartments by a partition. One of these compartments contains an ideal gas at 300K. The other compartment is vacuum. The whole system is thermally isolated from its surroundings. The partition is removed and gas expands to occupy the whole volume of the container. It temperature now would be ..[ UGET 1995]
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a) 300 K
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b) 250K
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c) 200 K
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d) 100 K
Explanation
This is free expansion therefore temperature remains unchanged. Answer: (a)
Q.15
A long cylindrical vessel of volume V and linear coefficient of expansion α contains a liquid. The level of liquid has not changed then coefficient of cubical expansion of liquid is .. [ EAMCET 1986]
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a) (V -α) / V
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b) (V + α) / V
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c) V / (V -α)
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d) 3α
Explanation
If γ is coefficient of cubical expansion of liquid then ΔV = γVt Comparing equations we get γ = 3α Answer:(d)
Q.16
A rod of length 30 cm made of material A expands by 0.075 when its temperature is raised from 0°C to 100 °C. Another rod of a different metal B having same length expands by 0.045cm, for the same change in temperature. A third rod of the same length is composed of two parts, one of metal A and other of metal B. This rod expands by 0.065 cm, for the same change in temperature. The portion made of metal A has the length [ CPMT 1991]
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a) 20 cm
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b) 10 cm
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c) 15 cm
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d) 18 cm
Explanation
Δl = l(αt) For Rod A 0.075 = 30×100× α1 α1 = 2.5×10-5 For Rod B 0.045 = 30×100× α2 α2 = 1.5×10-5 Third rod C both rod are connected Δl1 + Δl2 = 0.065 0.065 = 2.5×10-5×l1×100+1.5×10-5×l2×100 2.5l1+1.5×l2 = 65 --eq(1) given l1+l2 = 30 --eq(2) on solving equation 1 and equation 2 we get l2 = 20 cm Answer: (a)
Q.17
A bimetallic strip having two strips each of thickness 'd' are heated by Δt°C. If their coefficient of linear expansions are α1 and α2 respectively, then on heating they will bend in the form of arc whose radius is
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a)
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b)
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c)
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d)
Explanation
Initial length of strips is lLet bimetallic strip bend by angle θ Let l1 the length of first strip after heating let R1 be the radius of arc for metal 1thus θ=l1 / R1similarly for second metal θ=l2 / R2l2 - l1=(R2 - R1 )θl(1+α1Δt) - l(1 +α2Δt)=(R2 -R1 )θR1 - R2=dl(α1 - α2)Δt=dθ(α1 - α2)Δt / d=θ / l (α2 - α1)Δt / d=R Answer: (a)
Q.18
The moment of inertia of an object is l and the coefficient of linear expansion is α. If the temperature of the object is changed by Δθ , then the change in its moment of inertia is
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a) αlΔθ
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b) 2αlΔθ
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c)4αlΔθ
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d)αlΔθ/2
Explanation
Moment of inertia I=ML2ΔI=2MLΔL Δ I=2MLαLΔθ Δ I=2Iα(Δθ)Answer: (b)
Q.19
A heat flux of 4000 J/s is to be passed through a copper rod of length of 10cm and area of cross section 100 sq. cm. The thermal conductivity of copper is 400 W/m. °C. The two ends of this rod must be kept at a temperature difference of [ MPPMT 1999]
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a) 1°C
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b) 10°C
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c)100 °C
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d)1000 °C
Explanation
Area of cross section A=100×10-4=10-2Length L=0.1 mK=400 W/m °C.From the equation for heat flow Answer:(c)
Q.20
A sphere of density ρ, specific heat capacity c and radius r, is hung by a thermally insulated thread in an enclosure which is kept at lower temperature than the sphere. the temperature of the sphere starts to drop at the rate which depends up on the temperature difference between the sphere and the enclosure and the nature of the surface of the sphere and is proportional to ..
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a) c/ r3ρ
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b) 1 / r3ρc
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c) 3r3ρc
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d) 1 / rρc
Explanation
Energy lost is due to radiation thus Answer: (d)
Q.21
The following four wires are made of the samematerial. Which of these will have the largestextension when the same tension is applied?
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a) Length = 50 cm, diameter = 0.5 mm
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b) Length = 100 cm, diameter = 1 mm
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c) Length = 200 cm, diameter = 2 mm
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d) Length = 300 cm, diameter = 3 mm
Explanation
Since wires are made of same Young modulus is same Given same tension applied or F is same Ratio of l/d2 is largest for option (a) Answer:(a)
Q.22
The wettability of a surface by a liquid dependsprimarily on
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a) Viscosity
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b) Surface tension
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c) Density
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d) Angle of contact between the surface and theliquid
Explanation
If angle of contact > 90, liquid rise in capillary , if less than 90O liquid fall down in capillary. Answer:(d)
Q.23
Copper of fixed volume 'V' is drawn intowireof length 'L '.When thiswire is subjected to aconstant force 'F', the extension produced in the wire in 'ΔL '.Which of the following graphsis a straight line?[ AIMPT 2014]
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a) ΔL versus 1/L2
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b) ΔL versus L
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c) ΔL versus 1/L
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d) ΔL versus L2
Explanation
As Volume of rod is fixed A depends on length of rod V = A×L A = V/L, substituting value of A in above Young’s modulus equation Thus ΔL versus L2 gives straight line Answer:(d)
Q.24
Acertain number of spherical drops of a liquidof radius 'r' coalesce to forma single drop ofradius 'R' and volume 'V'. If 'T' is the surfacetension of the liquid, then … [ AIPMT 2014 ]
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a)
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b) Energy is neither released nor absorbed
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c)
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d)
Explanation
Energy of big drop, E= T× S E1 =T×(4πR2 ) Energy of n small drops E2 =T×(4πr2 ) Since surface are is reduced due to coalesce energy is released Energy released = E2 – E1 E=T×(n4πr2-4πR2 ) Now R3=nr3 or n= R3/r3 Answer:(a)
Q.25
The approximate depth of an ocean is 2700 m. The compressibility of water is 45.4 × 10-11 Pa-1 anddensity of water is 103 kg/mWhat fractional compression of water will be obtained at the bottom ofthe ocean?
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a) 1.4 × 10-2
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b) 0.8 × 10-2
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c) 1.0 × 10-2
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d) 1.2 × 10-2
Explanation
But P = hρg Answer:(d)
Q.26
A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 mAssuming that the pressure inside the house is atmospheric pressure, the force exerted by thewind on the roof and the direction of the force will be : … [ AIPMT 2015] (ρair = 1.2 kg/m3)
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a) 2.4 × 105 N, downwards
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b) 4.8 × 105 N, downwards
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c) 4.8 × 105 N, upwards
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d) 2.4 × 105 N, upwards
Explanation
From Bernoulli's equation, height is same h1=h2 , v1 velocity of air in room is zero Force = Pressure × area (upward) P=960×250=240000 N P= 2.4×105 N Answer:(d)
Q.27
The cylindrical tube of spray pump has radius R, one end of which has n fine holes, each of radius r. If the speed of the liquid in the tube is V, the speed of the ejection of the liquid through the holes is…[ Re. AIPMT 2015]
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a)
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b)
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c)
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d)
Explanation
Av= constant Answer:(c)
Q.28
The Young's modulus of steel is twice that of brass. Two wires of same length and of same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights added to the steel and brass wires must be in the ratio of : …[Re AIPMT 2015]
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a) 1 : 1
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b) 1:2
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c) 2:1
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d) 4:1
Explanation
YS = 2YB Ratio of weights = 2:1 Answer:(c)
Q.29
Water rises to height 'h' in capillary tube. If the lengthof capillary tube above the surface of water is madeless than 'h', then – [ Re AIPMT 2015] Solution:
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a) water does not rise at all.
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b) water rises upto the tip of capillary tube and thenstarts overflowing like afountain.
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c) water rises upto the top of capillary tube and stays there without overflowing.
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d) water rises upto a point a little below the topand stays there.
Explanation
Water won't overflow. That should be obvious since doing so would create a constant flow, constantly using energy, but without any energy input. Put another way, that would be a perpetual motion machine, one you could actually extract free power from. The same force that pulls the water along the inside of the capillary tube also holds it there when it reaches the end. This force doesn't just pull upward, it pulls the water along the glass. At a certain height, the weight of the water column ballances this pull. In that case the pull is upward since there is water below but not above. In given case, there is nothing to pull the water column higher when it reaches the end of the tube, so it just stays there. The pull only exists at the boundary between water and not-water. The bulk of the water in the tube isn't being pulled any particular way. Answer: (c) Answer:(c)
Q.30
The value of coefficient of volume expansion ofglycerin is 5 × 10-4 K-The fractional change inthe density of glycerin for a rise of 40°C in itstemperature, is :-
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a) 0.010
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b) 0.015
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c) 0.020
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d) 0.025
Explanation
Let ρ be initial density and ρ’ be final density Fractional change in density= Density(ρ) = mass(m)/volume(V) Answer:(c)
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