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Quiz 10
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Q.1
An Two non-mixing liquids of densities ρ and nρ (n > 1) are put in a container. The height ofeach liquid is h. A solid cylinder of length L and density d is put in this container. The cylinder floatswith its axis vertical and length pL(p < 1) in thedenser liquid. The density d is equal to :-
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a) {1 + (n + 1)p}ρ
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b) {2 + (n + 1)p}ρ
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c) {2 + (n – 1)p}ρ
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d) {1 + (n – 1)p}ρ
Explanation
Let A be the cross-sectional area Down ward gravitational force = upward buoyant force due to both the liquid LAdg = (pL)A(nρ)g + (1-p)LA ρg ⇒ d= (1-p) ρ + pn ρ d= [1 + (n-1)p]ρ Answer:(c)
Q.2
A rectangular film of liquid is extended from(4 cm × 2 cm) to (5 cm × 4 cm). If the work doneis 3 × 10–4 J, the value of the surface tension of theliquid is [NEET II – 2016]
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a) 0.2 Nm–1
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b) 8.0 Nm–1
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c) 0.250 Nm–1
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d) 0.125 Nm–1
Explanation
Since it is film , it have two free surface ΔA = 20 cm2 – 8 cm2 = 12 cm2 = 12×10-4m W = T(2ΔA) Answer:(d)
Q.3
Three liquids of densities ρ1, ρ2 and ρ3(with ρ1 > ρ2 > ρ3), having the same value of surfacetension T, rise to the same height in three identicalcapillaries. The angles of contact θ1, θ2 and θ3 obey… [ NEET II – 2016]
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a)
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b)
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c)
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d)
Explanation
From the formula for capillary h, r, T are constant As ρ1 > ρ2 > ρ3 ⇒ cosθ1 > cosθ2 > cosθ3 ⇒ θ1 < θ2 < θ3 Since water rises implies that θ is less than π/2 0 < θ1 < θ2 < θ3 < π/2 Answer:(d)
Q.4
A U tube with both ends open to the atmosphere, ispartially filled with water. Oil, which is immiscible withwater, is poured into one side until it stands at adistance of 10 mm above the water level on the otherside. Meanwhile the water rises by 65 mm from itsoriginal level (see diagram). The density of the oil is ..[ NEET 2017]
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a) 650 kg m-3
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b) 425 kg m-3
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c) 800 kg m-3
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d) 928 kg m-3
Explanation
At point B and C pressure is equal P due to oil = pressure due to water hOρog = hwρg height of oil = 65+65+10 = 140 mm = 0.14 m height of water hw = 65+65 =130 mm = 0.13 m density of water = 1000 kg/m3 Answer:(d)
Q.5
The bulk modulus of a spherical object is ‘B’. If it issubjected to uniform pressure ‘p’, the fractionaldecrease in radius is…[NEET 2017]
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a)
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b)
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c)
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d)
Explanation
Answer:(d)
Q.6
A uniform cylinder of length L and mass M having cross - sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half submerged in a liquid of density σ at equilibrium position. The extension x0 of the spring when it is in equilibrium is … [ IIT Mains -2013]Here k is spring constant
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a)
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b)
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c)
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d)
Explanation
Down ward force = Mg Upward force due to restoring force = kx0 Upward force due to buoyancy = Volume submerged × density of liquid = ( AL/2) ×σg At equilibrium downward force = Upward force Answer:(c)
Q.7
Assume that a drop of liquid evaporates by decrease in its surface energy. So that its temperatureremains unchanged. What should be the minimum radius of the drop for this to be possible? Thesurface tension is T, density of liquid is ρ and L is its latent heat of vaporization.…[ IIT Mains 2013]
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a)
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b)
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c)
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d)
Explanation
W = T ΔS Surface of sphere = 4πr2 ; thus ΔS = 8πr Let dm be the mass evaporated dm =4πr2ρ Energy utilised for evaporation = 4πr2ρL T8πr= 4πr2ρL Answer:(d)
Q.8
One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is
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a) 0.25
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b) 0.50
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c) 2.00
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d) 4.00
Explanation
Material of the road is same so Young modulus is same and force acting on rods is same Answer:(c)
Q.9
than one correct answer Q279) A solid sphere of radius R and density ρ is attached to one end of a mass-less spring of force constant k. The other end of the spring is connected to another solid sphere of radius R and density 3ρ. The complete arrangement is placed in a liquid of density 2ρ and isallowed to reach equilibrium. The correct statement(s) is (are)
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a) the net elongation of the spring is
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b) the net elongation of the spring is
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c) the light sphere is partially submerged.
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d) the light sphere is completely submerged
Explanation
Gravitational force on spheres = Vρ +V(3ρ) =4Vρg Buoyant force = 2V(2ρ) = 4Vρg As Gravitational force = Buoyant force thus spheres will remain submerged option “d” correct and option “c” fails Let elongation be x then restoring force = kx i)Now force on sphere of density ρ Gravitational = Vρg(down) Buoyant force =V2ρg(up) Restoring force = kx (down) Net force on sphere of density ρ is F =Vρg - V2ρg +kx = -Vρg + kx ii)Now force on sphere of density 2ρ Gravitational = V3ρg (down) Buoyant force =V2ρg(up) Restoring force =- kx (upward) Net force on sphere of density ρis F’ =V3ρg - V2ρg +kx = Vρg - kx Since spheres are in equilibrium F = F’ -Vρg + kx = Vρg – kx 2kx = 2Vρg Option “a” correct hence option “b” wrong Answer:(a, d)
Q.10
A glass capillary tube is of the shape of a truncated cone with an apex angle α so that itstwo ends have cross sections of different radii. When dipped in water vertically, waterrises in it to a height h, where the radius of its cross section is b. If the surface tension of water is S, its density is ρ , and its contact angle with glass is θ, the value of h will be (g is the acceleration due to gravity)[ IIT Advance 2014]
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a)
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b)
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c)
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d)
Explanation
Excess pressure on concave side of meniscus = 2S/R Balanced by height of water column = hρg Answer:(d)
Q.11
RAPH A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube ofuniform cross section is connected to the nozzle. The other end of the tube is in a small liquidcontainer. As the piston pushes air through the nozzle, the liquid from the container rises intothe nozzle and is sprayed out. For the spray gun shown, the radii of the piston and the nozzleare 20 mm and 1 mm respectively. The upper end of the container is open to the atmosphere.[ IIT Advance 2014] Q281) If the piston is pushed at a speed of 5 mms−1, the air comes out of the nozzle with a speed of
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a) 0.1 ms−1
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b) 1 ms−1
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c) 2 ms−1
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d) 8 ms−1
Explanation
Mass of air = ρa Av Av = Volume Mass of liquid = ρl Av From law of conservation of energy Answer:(a)
Q.12
than one correct option Q282) In plotting stress versus strain curves for two materials P and Q, a student by mistake puts strain on the y-axis and stress on the x-axis as shown in the figure. Then the correct statement(s) is (are) …[ IIT Advance 2015]
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a) P has more tensile strength than Q
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b) P is more ductile than Q
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c) P is more brittle than Q
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d) The Young’s modulus of P is more than that of Q
Explanation
For same stress, strain in P is more. ∴ P is more ductile Answer:(a, b)
Q.13
A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its own gravity. If P(r) is the pressure at r(r < R), then the correct option(s) is(are) …[ IIT Advance 2015]
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a) P(r = 0) = 0
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b)
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c)
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d)
Explanation
We know that dp = -g(r) ρ dr Integrating above equation Take P at =0 option b is correct and option c is correct Answer:(b, c)
Q.14
than one correct answer Q284) Two spheres P and Q of equal radii have densities ρ1 and ρ2 , respectively. The spheres are connected by a massless string and placed in liquids L1 and L2 of densities σ1 and σ2 and viscosities η1 and η2, respectively. They float in equilibrium with the sphere P in L1 and sphere Q in L2 and the string being taut (see figure). If sphere P alone in L2 has terminal velocity P V and Q alone in L1 has terminal velocity V Q, then
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a)
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b)
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c)
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d)
Explanation
Sphere P alone in L2 have terminal speed vp Sphere P Down forces Weight :Vρ1g Tension T Upward Buoyant force : Vσ1 Since Sphere P is in equilibrium velocity is zero, no viscous force Vρ1g +T = Vg σ1 … (i) Sphere Q Down forces Weight :Vρ2g Upward force: Tension T Upward Buoyant force : Vσ1 Vρ2g = Vg σ2 + T Vρ2g – T = Vg σ2 …(ii) From (i) and (ii) Vρ1g + Vρ2g= Vg σ1 + Vg σ2 ρ1 + ρ2 = σ1 + σ2 ρ1- σ2 = -( ρ2 - σ2) …(iii) From (iii) and (A) we get Option a is correct From (A) and (iii) since signs are opposite Option d is correct Answer:(a, d)
Q.15
Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is such that the instantaneous density ρ remains uniform throughout the volume. The rate of fractional change in density is constant. The velocity v of any point on the surface of the expanding sphere is proportional to … [ IIT Advance 2017]
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a) R3
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b) 1/R
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c) R
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d) R2/3
Explanation
Answer:(c)
0 h : 0 m : 1 s
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