Category: NEET/PMT

  • streamline, turbulent and laminar flow.

    streamline, turbulent and laminar flow.

    What is streamline flow ?

    Streamline flow—

    When liquid flows such that each particle of the liquid passing a given point moves along the such path and has the same velocity as its predecessor, the flow is called streamline flow or steady flow.

    1. What is Streamline Flow?

    Streamline flow, also called laminar flow, is a type of fluid flow in which fluid particles move along smooth, fixed paths called streamlines without mixing irregularly with one another.

    Example: The slow flow of water through a narrow pipe.

    Properties of Streamline Flow

    1. Smooth motion: Fluid particles move smoothly along definite paths.
    2. No crossing of streamlines: Two streamlines can never intersect each other.
    3. No irregular mixing: Fluid layers move smoothly without significant mixing between them.
    4. Constant velocity at a point: In steady streamline flow, the velocity of the fluid at any fixed point remains constant with time.
    5. Low velocity: It generally occurs when the fluid velocity is low.
    6. Viscosity: The viscosity of the fluid helps maintain smooth flow.

    Q.-Why two streamlines cannot cross each other?

    Ans- Two streamlines can’t cross each other , If they will do so , then the fluit particle at the point of intersection will have two different directions of motion. i.e. it cam go one way or the other destroying the steady nature of the liquid flow.   

    2. What is Turbulent Flow?

    Turbulent flow is a type of fluid flow in which fluid particles move irregularly, forming eddies and causing mixing between different layers of the fluid.

    Example: Water flowing rapidly in a river or a stream.

    Properties of Turbulent Flow

    1. Irregular motion: Fluid particles move in random and unpredictable directions.
    2. Mixing of layers: Different layers of the fluid mix with one another.
    3. Formation of eddies: Small swirling regions called eddies are formed.
    4. Fluctuating velocity: The velocity at a fixed point changes irregularly with time.
    5. High velocity: It generally occurs when the fluid velocity is high.
    6. Greater energy loss: Turbulent flow usually produces greater frictional energy losses than laminar flow.

    3. Difference Between Streamline Flow and Turbulent Flow

    Streamline FlowTurbulent Flow
    Fluid particles move smoothly along definite paths.Fluid particles move irregularly.
    There is little mixing between fluid layers.There is considerable mixing between fluid layers.
    Eddies are generally absent.Eddies are present.
    Velocity at a fixed point remains constant in steady flow.Velocity at a fixed point fluctuates with time.
    Usually occurs at low flow velocities.Usually occurs at high flow velocities.
    Energy loss is generally lower.Energy loss is generally higher.
    Example: Slow water flow through a narrow pipe.Example: Rapid water flow in a river.

    Important Note: Reynolds Number

    The type of flow in a pipe can be predicted using the Reynolds number (Re):

    Re=ρvD/η

    Where:

    • ρ = Density of the fluid
    • v= Average velocity of the fluid
    • D = Diameter of the pipe
    • η = Dynamic viscosity of the fluid

    For flow in a circular pipe:

    • Re < 2000: Generally laminar flow.
    • Re = 2000–4000: Transition region.
    • Re > 4000: Generally turbulent flow.

    In short: Streamline flow is smooth and orderly, whereas turbulent flow is irregular and involves mixing and eddies.

    Laminar flow-

    What is Laminar Flow?

    Laminar flow is a type of fluid flow in which fluid particles move smoothly in parallel layers without mixing irregularly with one another.

    Properties of Laminar Flow

    1. Fluid particles move in smooth and regular paths.
    2. Different layers of fluid slide over one another.
    3. There is no significant mixing between the layers.
    4. The velocity at any fixed point remains constant with time in steady laminar flow.
    5. It generally occurs at low fluid velocities.
    6. It usually occurs when the Reynolds number is less than 2000 in a circular pipe.

    Example: Water flowing slowly through a narrow pipe.

    Remember: Laminar flow is also known as streamline flow when the flow is smooth and orderly.

    To read another important notes of class 11th click here-

  • Objective test chapter 1 and 2 class 12th physics.

    Objective test chapter 1 and 2 class 12th physics.

    TEST CHAPTER-1 , 2

    SET-1 Electric charges and field

    1. Two small identical sphere having charges +10μC attract each other with a force of F newton. If they are kept in contact and then separated by the same distance, the new force between them is: (a)F/6                         (b) 16F                   (c) 16F/9                    (d) 9F                   
    2. A charge q is placed at the mid-point of the line joining two equal charges Q. The system of three charges will be in equilibrium when q has the value: (a)Q/4                         (b) Q/2                 (c) -Q/4                        (d) -Q/2           
    3. An electric dipole is placed in a non-uniform electric field. Then net:(a)force experienced is zero while torque is not zero(b)force experienced is zero and torque is also zero(c)both force and torque are not zero(d)force experienced is not zero while torque is zero                                    
    4. The total electric flux leaving spherical surface of radius 1cm and surrounding an electric dipole is: (a)q/ε0                       (b) zero                     (c) 2q/ε0                            (d) 8πr2q/ε0               
    5. Two particles of masses m and 2m with charge 2q and 2q are placed in a uniform electric field E and allowed to move for the same time. The ratio of their kinetic energies will be:(a)2 : 1                       (b) 8 : 1                     (c) 4 : 1                      (d) 1 : 4                
    6. Electric charges q , q, -2q are placed at the corners of an equilateral triangle ABC of side l. The magnitude of electric dipole moment of the system is: (a)ql                              (b) 2ql                      (c) √3ql                      (d) 4ql                
    7. A particle of mass m and charge q is placed at rest in a uniform electric field E and then released. The KE attained by the particle after moving a distance y is: (a)qEy2                          (b) qE2y                    (c) qEy                        (d) q2Ey           

    To download complete question paper click here-

  • Diffraction of light

    Diffraction of light

    What is diffraction of light?

    In this topic we will discuss what is diffraction of light , using single slit experiment formation of central max , position of secondary maxima and minima.

    Diffraction of Light – Single Slit Experiment

    diffraction single slit

    1. What is diffraction of light?

    Diffraction of light is the phenomenon of bending and spreading of light waves when they pass through a narrow slit or around the edge of an obstacle.

    Diffraction is clearly observed when the size of the slit or obstacle is comparable to the wavelength of light.

    To read the complete topic click here-

  • What is photo electric effect?

    What is photo electric effect?

    In this topic we will explain the photoelectric effect and also will explain hertz’s observation and Hallwach’s and Lenard’s observation.

    Photoelectric Effect: Hertz, Hallwachs and Lenard’s Observations

    1. What is the Photoelectric Effect?

    The photoelectric effect is the phenomenon in which electrons are emitted from the surface of a metal when electromagnetic radiation of a suitable frequency falls on it.

    The emitted electrons are called photoelectrons, and the current produced by their motion is called photoelectric current.

    Turning Points in Physics Notes — AQA A-Level

    Example: When ultraviolet light falls on a clean zinc plate, electrons may be emitted from its surface.

    Important terms

    • Photoelectrons: Electrons emitted from a metal surface due to incident light.
    • Work function (ϕ): The minimum energy required to remove an electron from a metal surface.
    • Threshold frequency (ν0): The minimum frequency of light required to produce photoelectric emission.
    • Stopping potential (Vs): The minimum reverse potential required to stop even the fastest photoelectrons.

    Einstein’s photoelectric equation is

    hν=ϕ+Kmax⁡

    where:

    • h= Planck’s constant
    • ν = frequency of incident light
    • ϕ = work function of the metal
    • Kmax⁡ = maximum kinetic energy of emitted electrons

    Since ϕ=hν0

    Kmax⁡=h(ν−ν0)

    This equation explains the main features of the photoelectric effect.

    2. Hertz’s Observation (1887)

    Radio — Whiting-Robertsdale Historical Society

    Heinrich Hertz

    Heinrich Hertz was conducting experiments to demonstrate electromagnetic waves. During his experiments, he observed that ultraviolet radiation helped an electric spark pass between two metal electrodes.

    Experimental observation

    1. Hertz produced an electric spark between two metal electrodes.
    2. He observed that ultraviolet light falling on the electrodes made the spark easier to produce.
    3. When the ultraviolet radiation was blocked, the spark became more difficult to produce.

    Conclusion

    Hertz’s observation indicated that ultraviolet radiation could help release electrons from a metal surface, facilitating the discharge.

    Importance: His observation was an early discovery of the photoelectric effect.

    3. Hallwachs’s Observation (1888)

    Wilhelm Hallwachs

    Wilhelm Hallwachs studied the effect of ultraviolet light on electrically charged metal plates, especially a clean zinc plate.

    Experiment

    Photoelectric Effect
    1. A clean zinc plate was connected to a negatively charged gold-leaf electroscope.
    2. The electroscope’s gold leaf diverged because the plate and electroscope carried negative charge.
    3. Ultraviolet light was allowed to fall on the zinc plate.
    4. The gold leaf gradually collapsed, indicating that the electroscope was losing its negative charge.

    Observation and conclusion

    • Negatively charged zinc plate: It loses negative charge when exposed to suitable ultraviolet light.
    • Positively charged zinc plate: It does not readily discharge through photoelectric emission because the emitted electrons are attracted back to the positively charged plate.
    • Uncharged zinc plate: It can become positively charged after photoelectrons escape.

    Conclusion: Ultraviolet light causes electrons to leave the zinc surface.

    This provided clear evidence that light can cause the emission of electrons from a metal.

    4. Lenard’s Observation (1902)

    Nobel Fisica 1905 > Philipp Eduard Anton von Lenard | Russian history, Historical illustration, Giclee print

    Philipp Lenard

    Philipp Lenard investigated photoelectric emission using a metal surface and an apparatus in which emitted electrons could be collected and measured.

    Experimental setup

    Doodlingo - Learn AI, Computer Science, Physics & Startups Through Doodles

    In a typical arrangement:

    • Light falls on a metal cathode.
    • Photoelectrons are emitted from the cathode.
    • A collector electrode (anode) collects the electrons.
    • The resulting photoelectric current is measured using an external circuit.

    Main observations of Lenard

    1. Photoelectric current depends on light intensity.

    For light of frequency above the threshold frequency, increasing the intensity generally increases the number of emitted electrons per second. Therefore, the photoelectric current increases.

    2. Maximum kinetic energy depends on frequency.

    Increasing the frequency of incident light increases the maximum kinetic energy of photoelectrons, provided the frequency is above the threshold.

    3. Existence of stopping potential.

    A reverse potential can be applied to oppose the motion of photoelectrons. At a particular reverse potential, even the fastest electrons cannot reach the collector, and the current becomes zero.

    Kmax⁡=eVsKmax​=eVs​​

    Here, ee is the magnitude of the electron’s charge and VsVs​ is the stopping potential.

    4. Threshold frequency exists.

    Below a certain minimum frequency, no photoelectrons are emitted, even if the intensity of light is increased.

    Conclusion: Lenard’s experiments established important properties of photoelectric emission and showed that the energy of photoelectrons is related to the frequency of incident light.

    5. Comparison of Hertz, Hallwachs and Lenard

    ScientistMain observationSignificance
    Hertz (1887)UV light made an electric spark easier to produce.Early discovery of photoelectric emission.
    Hallwachs (1888)UV light discharged a negatively charged zinc plate.Demonstrated that electrons leave the metal surface.
    Lenard (1902)Studied emitted electrons, their current, kinetic energy and stopping potential.Established important experimental laws of the photoelectric effect.

    6. Key results to remember for exams

    Intensity of light

    Increasing intensity increases photoelectric current, provided the frequency is above the threshold.

    Frequency of light

    Increasing frequency increases the maximum kinetic energy of photoelectrons.

    Threshold frequency

    Below the threshold frequency, photoelectric emission does not occur, regardless of intensity in the ordinary one-photon photoelectric effect.

    Stopping potential

    The stopping potential depends on the maximum kinetic energy, and therefore on the frequency of light for a given metal.

  • Objective class 12th Book part-1

    Objective class 12th Book part-1

    Introduction to Objective Physics for Class 12
    “Objective Physics” by Nayan Jha meets the demand for objective questions in Class 12 Physics exams.
    Key Features of This Book:
    Comprehensive Question Types:
    Multiple Choice Questions (MCQs) with answers,
    Assertion–Reason Questions with answers,
    Case Study Based Questions with answers,

    Focus on Conceptual Understanding: Each question enhances students’ conceptual understanding of Physics.
    Problem-Solving Skill Development: The book strengthens students’ ability to solve various Physics problems.
    Detailed Solutions:
    Every question includes a detailed solution.
    Solutions explain the correct answer and underlying Physics principles.
    Target Audience and Benefits:
    This book is for:
    Class 12th Board Examination Preparation: It offers effective preparation for board exams.
    NEET and JEE Aspirants: Concept-based questions benefit students preparing for NEET and JEE.
    How This Book Helps Students:
    Chapter-wise Practice: Chapter-organized questions allow students to assess preparation and identify improvement areas.
    Beyond Memorization: The aim is to foster deeper understanding, helping students understand concepts, learn applications, and build confidence in solving objective questions.
    Author’s Philosophy:
    “Concepts + Practice = Success — Your Success is My Mission.”
    — Nayan Jha
    Physics by Nayan Jha

    To download the complete book (pdf) click here-

  • Alternating current objective questions with answers

    In this blog physics classes by Nayan jha providing objective questions and answers , which contain multiple choice , assertion reason and case study based questions with explanation( complete solution.

    To download the pdf all questions and its answer click here-

    For question numbers 86-100, two statements are given-one labelled Assertion (A) and the other labelled Reason (R).
    Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
    (a) Both A and R are true and R is the correct explanation of A
    (b) Both A and R are true but R is NOT the correct explanation of A
    (c) A is true but R is false
    (d) A is false and R is also false

    1. Assertion (A) : Long distance transmission
      of A.C. is carried out at extremely high voltage.
      Reason (R): For large distance, voltage has
      to be large….
    2. Assertion (A) : An electric lamp connected
      in series with a variable capacitor and A.C.
      source, its brightness increases with increase
      in capacitance.
      Reason (R): Capacitive reactance decreases
      with increase in capacitance of capacitor.
    3. Assertion (A) : A transformer cannot work
      on D.C. supply.
      Reason (R): D.C. changes neither in magnitude
      nor in direction.
    4. Assertion (A) : Step-down transformer
      increases the current.
      Reason (R): Transformer obeys the law of
      conservation of energy.

    To download complete assignment of these questions and answer (PDF)click here-

  • Test magnetism and electromagnetic induction

    There are 20 questions in the question paper , in which 14 are objective, (1 marker) and other 6 are subjective which carry 2 marks each .

    This question paper covers the chapter magnetism and electromagnetic induction . It is just for the practice the questions of the chapters magnetism and electromagnetic induction.

    1. Define the term magnetic moment of a current loop . Write the expression for the magnetic moment when an electron revolve at speed v around the   orbit of radius r in hydrogen atom .

     2 .  Define magnetic susceptibility of the materials . Name two elements , one having positive and other having negative susceptibility . what does negative susceptibility signify ?

      3. A bar magnet of magnetic moment M is aligned parallel to the direction of a uniform magnetic field B. What is the work done to turn the magnet, so as the align its magnetic moment  (i) Opposite to the field direction  (ii) Normal to the field direction?

    4. Two circular coils one of small radius r and other of very large radius R are placed coaxially with center coinciding. Obtain the mutual inductance of the arrangement.

    To download the complete question paper click here –

  • Test NEET /JEE Chapter 1 -4 (class 11th)

    ( chapter 1 -4 class 11th)

    There are 20 questions in this question paper. Students have to solve it in 25 minutes only .

    Test NEET/JEE (ch-1,2,3,4)

    1. A  new system of units is proposed in which unit of mass is A kg . unit of length is B kg and unit of time is C s , What will be the value of 5 j in new system?

    (a) 5AB2C-2 (b) 5A-1B-2C2 (c) 5 A-2 B-1 C-2 (d) 5 A-1 B-1 C2

    2. If linear momentum P , area A and  time t is taken as fundamental units then the dimensional formula of  energy is (a) P1 A-1 t-1  (b) P2 A1 t1   (c) P1 A-1/2 t1   (d) P1 A1/2 t-1.

    3.  If E , m,L  and G denotes energy , mass , length and gravitational constant then the ( EL2 /m5G2) has the dimension of   (a) mass (b) length (c) time (d) angle.

    4. The period of oscillation of time period T is 2π(l/g)1/2 measured value of l is 10 cm is known as 1mm accuracy ant time for 100 0scillations of the pendulum is found to be 50 s using a wrist watch of 1s resolution . What is the accuracy in the measurement of g ?

    (a)2% (b) 3% (c) 4% (d) 5% .

    5. Two resistors of resistance R1=100+-3 Ohm R2=200+-4 Ohm then the equivalent resistance of parallel combination is (a) 66.7 +-1.8  (b) 66.7+- 4  (c) 66.7+-3  (d)66.7+-7 Ohm.

    6. A cube has a side of length 1.2 x 10-2 its volume upto correct significant figure is (a) 1.7×10 -6 (b) 1.73×10 -6 (c) 1.78×10 -6(d) 1.732×10 -6.

    7. T he position of the object moving along x-axis is given by x= a+bt2, where a=8.5 b=2.5 and t is measured in seconds. The average velocity of the object between t=2s to t=4 s  in m/s  is (a) 5 (b) 10 (c) 15 (d) 20 .

    8. A player throw a ball vertically upwards with velocity u at highest pint , (a) both velocity and acceleration are zero (b) The velocity of the ball is u and acceleration is zero (c)The velocity of the ball is 0 but acceleration is g (d) the velocity of the ball is u and acceleration is g .

    9. If a lift is moving upward with velocity 5m/s a girl throw a ball upward with speed 50m/s the ball will returns on her hand in time (a)5s (b)10s (c) 15s (d) 20s.

    10. A boy sliding on the smooth inclined plane requires 4 s to reach on the ground starting from rest from the top . How much time will it take to cover one fourth the distance starting from rest at the top?

    (a) 1s (b) 4s (c)2s (d)16s .

    11. A particle is released from rest from a tower of height 3h . the ratio of the intervals of time to cover three equal heights h is (a) t1:t2: t3 = 3:2:1 (b) t1:t2: t3 =1: (√2-1)(:√3-2) (c) t1:t2: t3 =√3:√2:1 (d) t1:t2: t3 = 1:√2-1): (√3-√2).

    12. A ball A is thrown upward  with velocity u  at the same time another ball B is released from rest at height H,  at time t the speed of A relative to B is (a) u (b) u-2gt (c) (u2-2gh)1/2 (d)u-gt.

    13. A vehicle travel half the distance L with speed V1 and other half with speed V2 then its average speed is , (a) (V1 + V2)/2 (B) (2V1+V2)/V1+ V2 (C) 2V1V2/V1+V2 (D) L(V1+ V2)/V1V2 .

    14. Two vectors    and A+B = C then A and B are (a) parallel (b) perpendicular (c) anti-parallel (d) null vector.

    15. If resultant of two equal vectors is equal to the either forces Then the angle between these two vectors is ,(a) 900 (b) 1200 (c) 600 (d) 300

    16. if dot product  of two vectors   is null vector then the angle between these two vectors is   (a) 0 (b) 1200 (c) 900 (d) 450 .

    17. If  then  then which of the following statement is  correct ( a) vector A is perpendicular to  vector B (b) vector A is parallel to vector C (c) vector b is parallel to vector C (D) vector A is perpendicular to vector C .

    18.Two vectors  magnitude of A,B and C are 3,5,4 respectively then  (a)the angle between A and B is 900 . angle between A and C is 900. (c) angle between B and C is 900(d) angle between A and B is 00.

    19. If unit vector of a vector is represented as 0.5  – 0. 8  +c  then the value of c is (a) √0.01  (b) √0.11  (c) √0.39 (d)  1.

    20. The component of a vector A is along X-axis will have maximum value if (a)vector A is along positive Y axis   (b)vector A along positive x-axis  (c) vector A makes angle 450 with the x-axis  (d) A is along negative X aixis.

  • How to select coaching center for JEE and NEET

    Physics classes by Nayan jha provoding information ,How to select coaching centre for NEET and JEE

    Students/parents as you know these days most of the parent or students are looking to settle their child as quick as possible. And IIT (Indian institute of technology) and medical colleges like AIIMS play the great role to do so.

    As we know every student who appears for the 10th class exams has one big question about his life that which stream to select for further education. Once the decision for the stream is made, which is Science. Then next question is   medicine or engineering? The answer to this question marks the beginning of a dream career. Students can decide this based on what subject interests them. If students scores good  in NEET leads to admission to a good medical college, whereas a good score in IIT JEE will help students get admission to a reputed engineering college.

    Both the exams are national-level entrance exams and hold a lot of importance concerning students’ higher education. One more doubt that arises when deciding between the two is which exam is easier to crack. Now, and for this selection of good coaching center is required to score good marks to take the admission in reputed colleges, every student who appears for the 10th class exams has one question about which stream to select for further education. Once the decision for the stream is made, the stream is Science. The question that comes next is medicine or engineering? The answer to this question marks the beginning of a dream career. Students can decide this based on what subject interests them. A good score in NEET leads to admission to a good medical college, whereas a good score in IIT JEE will help students get admission to a reputed engineering college.

    Both the exams are national-level entrance exams and hold a lot of importance concerning students’ higher education. One more doubt that arises when and where should we join the coaching center or tuitions . Here is the answer – You just watch the video given below-

  • Questions based on surface tension

    This assignment contains some important questions based on the surface tension, in which there are some objective and some numericals based questions.

    these questions are for the practice of the topic surface tension and surface energy, excess pressure inside the liquid drop, soap bubble , ascent formula etc.

    Question based on surface tension and surface energy

    Multiple choice questions. (Only one option is correct)

    1. The angle of contact doesn’t depend upon,     (a)  Temperature         (b)Surface impurity      (c) Cohesive force      (d) The inclination of surface in contact.   Ans -d

    2. When a capillary tube is immersed in a liquid , then the liquid of mass m rises in the capillary tube .If capillary tube of double radius is taken then the mass f same liquid rising in the tube is , ,     (a)   M/2        (b)   2M   (c) M/4     (d) 4M .                                                      Ans- b

    3. What will be the work done to increasing the radius of the soap bubble from r/2 to 2r . If surface tension of soap solution is T ? ,     (a) 3πr2T   (b) 30πr2T    (c)    9πr2T   (d) 12πr2T   . Ans-b

    4. pressure inside soap bubble are 1.02 atm and 1.05 atm respectively . then the ratio of their surface area is (a) 125/8 (b) 25/4  (c)5/2   (d)2/5.   Ans-b

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    5.The property utilized in the manufacture of lead shots is (a)specific gravity of lead   (b)specific weight of the liquid lead  (c) compressibility of the liquid lead  (d) surface tension of the liquid lead. Ans-d

    6. Two soap bubbles are connected by a tube as shown in figure , what happens when stopper s is removed ?

    (a)size of A increase   (b) size of B is decreased   (c) both takes the same size (d) size of bubble A decreases and size of bubble B increases.                     Ans-d

    7. Two soap bubble of radii 2cm and 4 cm are join to form a double bubble in air , then the radius of curvature of the interface is , (a)2√5 cm  (b)2cm  (c)4  cm (d)  2√3cm  .  Ans-c

    8. A square wire frame of  side l is floating on the surface of the liquid of surface tension T . The force required to pull out the frame from the liquid is  (a) Tl  (b) 2Tl  (c) 4Tl  (d) 8Tl . Ans-d

    9. Two drops of liquid coalesce to form a single drop . In this process energy is (a) Absorbed  (b) Evolved  (c) Either absorbed or evolved    (d) Neither absorbed nor evolved.  Ans-b

    10. Surface tension of a liquid , at critical temperature is (a)Maximum   (b)Unchanged  (c)  Zero (d) none of these.                                                                                                           Ans-c

    Important numericals based on this topic –

    TO WATCH THE VERY IMPORTANT TOPIC OF SURFACE TENSION WHICH IS EXCESS PRESUURE INSIDE THE SOAP BUBBLE AND LIQUD DROP CLICK ON THE VIDEO LINK GIVEN BELOW-

    1.The surface tension of the soap solution is 30 x 10-3N/m .How much work is done to increase the radius of the soap bubble from 1.5 cm to 3 cm .     Ans- 5.091 x 10-4 j.

    2. A square glass plate of length 10cm and thickness 0.4cm, weighs 40 g in air . It is held vertically such that its lower edge rests on water surface. What is the apparent weight of glass plate now? (S.T. of water is 0.073 N/m).                                                     Ans-0.407N.

    3. The radii of two air bubbles are in the ratio 4:5 .Find the ratio of excess pressure inside them . Also compare the works done in Blowing these bubbles.             Ans-5/4  ; 16/25 . .

    4. A circular plate of radius 4cm and weight W is made to rest on the surface of water if a minimum pull of   W+F is required to clear the plate off the water surface then find F , given surface tension of water is 0.072N/M.                       Ans-0.018N .

    5. Water rises up in a glass capillary upto a height of 10cm , while mercury falls down by 3.5 cm in the same tube. If the angle of contact for water – glass and mercury-glass are is taken as 00 and 1350 respectively . compare the surface tension of water and mercury. [ density of water and mercury are 103 and 13.6×103kg/m3.].

    6. A canister has a small hole at its bottom .What penetrates into the canister when its base is at a depth of 40cm from the surface of water. If surface tension of water is 73.5 dyne/cm, find the radius of the hole .     Ans- 0.0375mm.

    To get the solution of all numerical questions given above click here–

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