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Q.1
Assuming the sun to have a spherical outer surface of radius r, radiating like a black body at temperature t°C, the power received by a unit surface, (normal to the incident rays) at a distance R from the center of the sun is…. [ CBSE-PMT 2007]where σ is the Stefan's constant.
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a)
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b)
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c)
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d)
Explanation
Power radiated by the sun at t°C=σT44πr2 Power received by a unit surface Since temperature is in t°C we get Answer: (c)
Q.2
A black body at 227°C radiates heat at the rate of 7 cals/cm2s. At a temperature of 727°C, the rate of heat radiated in the same units will be: [ CBSE-PMT 2009]
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a)50
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b) 112
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c)80
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d)60
Explanation
According to Stefan's law E=σT4Bay taking the ratio E=112 cal / cm2sAnswer: (b)
Q.3
Radiation from which of the following sources, approximates black body radiation best? [ CBSE-PMT 2002]
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a) A tungsten lamp
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b) Sodium flame
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c)Hot lamp black
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d)A hole in a cavity, maintained at constant temperature
Explanation
Answer: (d)
Q.4
A black body at 1227°C emits radiations with maximum intensity at a wavelength of 5000Å. If the temperature of the body is increased by 1000°C, the maximum intensity will be observed at [2006]
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a) 5000 Å (Angstrom)
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b) 6000 Å (Angstrom)
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c) 3000 Å (Angstrom)
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d)4000 Å (Angstrom)
Explanation
Applying Wein's displacement law, λmT=constant5000 × (1227+273)=(2227 + 273) × λm∴ λ=3000 Å (Angstrom) Answer:(c)
Q.5
Which of the following circular rods (given radius r and length l), each made of the same material and whose ends are maintained at the same temperature will conduct most heat? [ CBSE-PMT 2005]
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a) r=r0 ; l=l0
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b) r=2r0 ; l=l0
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c) r=r0 ; l=2l0
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d) r=2r0 ; l=2l0
Explanation
W know that thermal resistance R=l / KA Heat flow will be maximum when thermal resistance is minimum from the given option option (b) have minimum resistance Answer: (b)
Q.6
Mercury thermometer can be used to measure temperature up to [ CBSE-PMT 1992]
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a)260°C
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b) 100°C
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c)357°C
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d)500°C
Explanation
Mercury thermometer can measure temperature from -38.9° C to 356.7° CAnswer: (c)
Q.7
Two rods of thermal conductivities K1 and K2 cross-sections A1 and A2 and specific heats S1 and S2 are of equal lengths. The temperatures of two ends of each rod are T1, and TThe rate of flow of heat at the steady state will be equal if[ CBSE-PMT 2002]
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a)
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b) K1A1=K2A2
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c) K1S1=K2S2
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d) A1S1=A2S2
Explanation
Let L be the length of Rod Rate of flow of heat for one rod=Rate of flow from other rodIn steady state Rate of flow of heat from one rod=Rate of flow of heat from other rod∴ K1A1=K2A2Answer: (b)
Q.8
Consider a compound slab consisting of two different materials having equal thick nesses and thermal conductivities K and 2K, respectively. The equivalent thermal conductivity of the slab is [ CBSE-PMT 2003]
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a) (4/3) K
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b) (2/3) K
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c)√3 K
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d)3 K
Explanation
In series, thermal conductivity Answer:(a)
Q.9
A cylindrical rod having temperature T1 , and T2 at its end. The rate of flow of heat is Q1 cal/sec. If all the linear dimensions are doubled keeping temperature constant, then the rate of flow of heat Q2 will be [ CBSE-PMT 2001]
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a) 4Q1
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b) 2Q1
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c) Q1 / 4
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d) Q1 / 2
Explanation
Rate of flow of heat is given by Dimensions of area A=[L2], dimension of distance L=[L] ∴ H ∝ L ⇒ H2=2H1Answer: (b)
Q.10
A vessel containing water is given a constant acceleration 'a' towards the right along a straight horizontal path. Which of the following diagram in fiure represents the surface [ IIT 1981]
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a)
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b)
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c)
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d)none of these
Explanation
Answer: (c)
Q.11
The following four wires are made of the same meterial. Which of these will have the largest extension when the same tension is applied [IIT 1981]
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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
Young's modulusSince T/Y constant ∴ δl ∝ l/Al/A is largeest for first caseAnswer: (a)
Q.12
A U tube of uniform cross section is partially filled with liquid I and another liquid II which does not mix with liquid I is poured into one side as shown in figure. It is found that the liquid levels of the two sides of the tube are same, while the level of liquid I has risen by 2 cm. If the specific gravity of liquid is 1.1, the specific gravity of liquid II must be [ IIT 1983]
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a) 1.12
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b) 1.1
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c)1.05
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d)1.0
Explanation
Level of liquids is same in both the arms thus Pressure at A=Pressure at B hρ1g=hρ2g ⇒ ρ1=ρ2 Answer:(b)
Q.13
A homogeneus solid cylinder of length L ( L < H/2), cross-sectional area A/5 is immersed such that it flots with its axis verical at the liquid-liquid interface with length L/4 in the denser liquid as shown in the figure. The lower density liquid is open to atmosphere having pressure PThen density D of solid is given by [ IIT 1995]
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a) 5d/4
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b) 4d/5
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c) 4d
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d) d/5
Explanation
Weight of cylinder=Upthrust due to upper liquid + Upthrust due to lower liquid D=5d/4 Answer: (a)
Q.14
A large open tank has two holes in the wall. One is square hole of side L at a depth y from the top and the other circular hole of radius R at depth 4y from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are same. Then R is equal to [ IIT 2005]
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a)L/√2π
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b) 2πL
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c)L
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d)L/2π
Explanation
Formula for rate of flow=(area) ×(velocity) Velocity=√(2gx), Here x is the height from topEquating the rate of flow, we have Answer: (a)
Q.15
A metal ball of surface area 200 square cm, temperature 527 °C is surrounded by a vessel at 27°C. If the emissivity of the metal is 0.4, then the rate of loss of heat from the ball is σ=5.67×10-8 S.I. unit
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a) 108 joule approximately
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b) 168 joule approximately
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c)182 joule approximately
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d)192 joule approximately
Explanation
From fomula E=σeA(T4 - T04)E=5.67×10-8×0.4×200× [8004 - 3004 E=182 joule approximatelyAnswer: (c)
Q.16
A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R. The volume of the remaining cylinder is V and its mass M. It is suspended by a string in a liquid of density ρ where it stays verical. The upper surface of the cylinder is at a depth h below the liquid surface. The force on the bottom of the cylinder by the liquid is [ IIT 2001]
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a) Mg
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b) Mg- Vρg
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c)Mg+πR2hρg
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d)ρg(V+πR2h)
Explanation
According to Archimedes principleUpthrust=Wt. of fluid displacedFbottom - Ftop=Vρg Fbottom=Ftop + VρgFbottom=P1×A + Vρg Fbottom=(hρg) × ( πR2) + Vρg Fbottom=ρg[ πR2h + V]Answer: (d)
Q.17
A wooden block, with coin placed on its top, floats in water as shown in figure. The distance l and h are shown here. After some time the coin falls into the water. then [ IIT 2002]
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a)l decreases and h increases
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b) l increases and h decreases
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c)both l and h increases
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d)both l and h decreases
Explanation
l decreases as the block moves up. h also decreases because when the coin is in water it will displace a volume of water, equal to its own volume. Whereas when it is on the block it displaces more than to own volume ( because density of coin is greater than the density of water ) Answer:(d)
Q.18
The adjacent graph shows the extension (Δl) of a wire of length 1 m suspended from the top of a roof at one end and with load W connected to the other end. If the cross sectional area of the wire is 10-6 m2, calculate the Young modulus of the material of the wire [ IIT 2003]
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a) 2 ×1011 N/m
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b) 2 ×10-11 N/m
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c) 3 ×10-12 N/m
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d) 2 ×10-13 N/m
Explanation
From the formula of Young's modulus Answer: (a)
Q.19
Water is filled in a container upto height 3m. A small hole of area 'a' is punched in the wall of the container at a height 52.5cm from bottom. the cross sectional area of the container is A. If a/A=0.1, then v2 is ( where v is the velocity of water coming out of hole) [ IIT 2005]
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a)50
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b) 51
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c)48
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d)51.5
Explanation
h=3-0.525=2.475The square of the velocity of flux Answer: (a)
Q.20
When temperature of gas is 20° C and pressure is changed from P1=1.01×105 Pa to P2=1.165×105 Pa then the volume changed by 10%. the bulk modulus is [ IIT 2005]
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a) 1.55×105Pa
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b) 0.115×105Pa
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c)1.4×105Pa
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d)1.01×105Pa
Explanation
Formula for bulk modulus ΔP=1.165×105 - 1.01×105ΔP=0.155×105ΔV/V=0.1 givenOn substituting values in equation B=1.55×105PaAnswer: (a)
Q.21
A glass tube of uniform internal radius (r) has a valve separating the two identical ends. Initially, the valves is in a tightly closed position. End 1 has a hemispherical soap bubble of radius r. End 2 has sub-hemispherical soap bubble as as shown in figure. Just after opening the valve [ IIT 2008]
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a)air from end 1 flows towards end 2. No change in the volume of the soap bubbles
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b)air from end 1 flows towards end 2. volume of the soap bubble at end 1 decreases
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c)no changes occurs
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d)air from end 2 flows towards end 1. Volume of the soap bubble at end 1 increases
Explanation
We know that excess pressure in a soap bubble is inversely proportional to its radius. The soap bubble at end 1 has small radius compared to the soap bubble at end 2 ( given). Therefore excess pressure at 1 is more. As the value is opened, air flows from end 1 to end 2 and the volume of soap bubble at end 1 decreases Answer:(b)
Q.22
A spring of force constant 800N/m has an extension of 5 cm. The work done in extending it from 5cm to 15 cm is [ AIEEE 2002]
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a) 16J
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b) 8J
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c) 32J
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d) 24 J
Explanation
Small amount of work in extending the spring by dx is dW=kxdx On integrating above from x=0.05m to 0.15m we will get work done Answer: (b)
Q.23
A wire fixed at the upper end stretched by length l by applying a force F. The work done in stretching is [ AIEEE 2004]
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a)2Fl
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b) Fl
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c)F/2l
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d)Fl/2
Explanation
Work done by constant force in displacing the object by distance l=Change in potential energy Change in potential energy=½ × stress × strain × volume Answer: (d)
Q.24
Spherical ball of radius 'R' are falling in a viscus fluid of viscosity 'η' with a velocity 'v'. The retarding viscous force acting on the spherical ball is [ AIEEE 2004]
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a) inversely proportional to both radius 'R' and velocity 'v'
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b) directly proportional to both radius 'R' and velocity 'v'
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c)directly proportional to 'R' but inversely proportional to 'v'
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d)inversely proportional to 'R' but directly proportional to velocity 'v'
Explanation
From Stoke's law Viscus force F=6πηrv hence F is directly proportional to radius and velocityAnswer: (b)
Q.25
Two soap bubbles of different radii are connected by tube [ AIEEE 2004]
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a) air flows from the smaller bubble to the bigger
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b) air flows from bigger bubble to the smaller bubble till the size are interchanged
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c)air flows from the bigger bubble to the smaller bubble till the size becomes equal
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d)there is no flow of air
Explanation
We know that pressure inside the air bubble is inversely proportional to radius thus bubble with smaller radius will have higher internal pressure Air will move from smaller to bigger bubble Answer:(a)
Q.26
If 'S' is stress and 'Y' is young's modulus of material of a wire, the energy stored in the wire per unit volume is [ AIEEE 2005]
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a) S2 / 2Y
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b) 2S2 Y
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c) S/ 2Y
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d) 2Y/S2
Explanation
Energy stored per unit volume E=½ × Stress × Strain We know that Y=Stress/ Strain OR Strain=Stress/ YE=½ × Stress × ( Stress/ Y) E=½ ( S2 / Y) Answer: (a)
Q.27
A 20cm long capillary tube is dipped in water. The water rises up to 8 cm. If the entire arrangement is up in a freely falling elevator the length of water column in the capillary tube will be [ AIEEE 2005]
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a)10 cm
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b) 8 cm
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c)20 cm
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d)4 cm
Explanation
During freefall there is no gravitational force on water thus entire tube of 20cm will be filledAnswer: (c)
Q.28
A wire elongated by l mm when load W is hanged from it. If the wire goes over a pulley and two weights W each are hung at the two ends, the elongation of the wire will be ( in mm) [AIEEE 2006]
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a) l
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b) 2l
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c)zero
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d)l/2
Explanation
Case 1 At equilibrium T=W Young's modulusCase 2At equilibrium T=W∴ From above equations elongation is same Answer: (a)
Q.29
If the terminal speed of a sphere of gold ( density=19.5 kg/m3) is 0.2 m/s in a viscous liquid ( density=1.5 kg/m3), find the terminal speed of a sphere of silver ( density=10.5kg/m3) of the same size in the same liquid [ AIEEE 2006]
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a) 0.4 m/s
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b) 0.133 m/s
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c)0.1 m/s
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d)0.2 m/s
Explanation
Terminal velocity here ρ1 is density of sphere and ρ2 is the density of liquid Let vT2 be the terminal velocity of sphere of silver vT1 be the terminal velocity of sphere of gold=0.2 m/sBy taking the ratio of terminal velocity we get Answer:(c)
Q.30
A sphere solid ball of volume V is made of a material of density ρIt is falling through a liquid of density ρ2 . Assume that the liquid applies a viscous force on the ball that is proportional to square of its speed v,Fviscous=-kv2 (k > 0). The terminal velocity of the ball is [ AIEEE 2008]
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a)
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b)
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c)
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d)
Explanation
The condition for terminal velocity is Weight=Buoyant force + Viscous force Weight=V×ρ1g Buoyant force=V×ρ×g Viscous force=kv2( only magnitude is considered as negative sign show the direction) V×ρ1g=V×ρ×g + kv2 on simplification we get option 'a' Answer: (a)
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