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Physics NEET MCQ
Quiz 10
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Q.1
A beaker contain water is placed on a scale balance which indicates a mass of 0.5kg. A glass sphere of mass 0.2kg is suspended from a string and immersed in water such that it does not touch the beaker. If the relative density of glass is 2.5, then the reading on the balance will change to [ SCRA 1994]
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a) 0.62 kg
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b) 0.58 kg
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c) 0.50kg
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d) 0.42 kg
Explanation
density of sphere=relative density of sphere / density of water density of sphere=2.5 / 13=2.5×10⁻³ Volume of sphere=mass / density of sphere Volume of sphere=0.2 / 2.5×10⁻³ Upthrust=Volume × density of liquid ×g Upthrust=0.2 / 2.5×10⁻³×103g Upthrust=2/25 kg=0.08kg ∴ reaction of upthrust on the bottom of beaker=0.08 reading on scale=0.5+.08=0.58kg Answer: (b)
Q.2
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]
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a) 6250 Joule
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b) 0.177 Joule
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c)0.075 Joule
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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.3
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]
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a) 0.6
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b) 0.4
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c)0.2
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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.4
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]
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a) 0.4 cm3
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c)0.025 cm3
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d)0.04 cm3
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b) 4×10⁻⁵ 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.5
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]
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a)4.900 Joule
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b) 2.450 Joule
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c)0.495 Joule
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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.6
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]
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a) 1 rad/s
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b) 2 rad/s
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c)4 rad/s
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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)
Q.7
The work in splitting a drop of water of 1mm radius into 106 droplets is ( surface tension of water=72×10⁻³ J /m2 ) [ MPPMT 1994]
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a)9.98×10⁻⁵ Joule
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b) 8.95×10⁻⁵ Joule
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c)5.89×10⁻⁵ Joule
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d)5.98×10⁻⁵ Joule
Explanation
Let R be the radius of big sphere and r be the radius of small sphere , volmue of big sphere=n× volume of small sphere Thus R=n1/3r ot r=R/ n1/3 Change in Surface area=Area of small spheres - Area of big sphere ΔS=n×4πr2 - 4πR2 Substituting value of r in terms of R in above equation we get ΔS=4πR2 [ n1/3 -1]Work done in making small drops W=Increase in surface are × surface tensionW=4πR2T[ n1/3 - 1]W=4π×10-6×72×10⁻³×99W=8.95×10⁻⁵ JouleAnswer: (b)
Q.8
Water rises in a capillary tube to a certain height such that the upward force due to surface tension is balanced by 75×10⁻⁴N force due to the weight of the liquid. If the surface tension of water is 6×10⁻²Nm-1, the inner circumference of the capillary must be [ CPMT 1988]
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d)12.5×10⁻²
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a) 1.25×10⁻² m
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b) 0.50 ×10⁻² m
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c)6.5×10⁻² m
Explanation
Upward force due to surface tension=2πrTcosθDown ward force=W2πrTcosθ=W2×3.14×6×10⁻²×r=75×10⁻⁴ r=1.99 ×10⁻²Circumference=2πr Circumference=2×3.14×1.99×10⁻²Circumference=12.5×10⁻² Answer:(d)
Q.9
The clouds float in the atmosphere because of
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a) their low temperature
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b) their low viscosity
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c) their low density
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d) creation of low pressure
Explanation
Clouds can hold an enormous amount of water. When this water falls as rain it clearly has a significant mass so why don't clouds fall? In fact, the small water droplets that make up clouds do fall slowly. However, the drag force of the air dominates over the gravitational force for small particles. The drag force increases as the size of an object decreases. The force needed to move a sphere through a viscous medium is given by Stokes's law, F = 6πηRv. Here, R is the radius of the sphere, v is the velocity, and η is the viscosity. The viscosity of air is about 0.018×10⁻³ Pa·s and the viscosity of water is about 1.8×10⁻³³ Pa·s. Answer: (b)
Q.10
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.11
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.12
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.13
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.14
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.15
A jar is filled with two non-mixing liquids 1 and 2 having densities ρ1 and ρ2 respectively. A solid ball, made of a material of density ρ3, is dropped in the jar. It comes to equilibrium in the position shown in figure. Which of the following is true for ρ1 , ρ2 ,ρ3? [ AIEEE 2008]
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a)ρ3 < ρ1 < ρ2
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b) ρ1 > ρ3 >ρ2
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c)ρ1 < ρ2 < ρ3
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d)ρ1 < ρ3 < ρ2
Explanation
From the figure it is clear that liquid 1 floats on liquid 2. ThE lighter liquid floates over heavier liquid. Therefore we can conclude that ρ1 < ρ2 Also ρ3 < ρ2 otherwise the ball would have sink to the bottom of the jar. also ρ3 > ρ1 otherwise the ball would have floated in liquid 1. From above discussion we conclude option 'd' is correctAnswer: (d)
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