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JEE Physics
Experimental Skills Previous Year Questions With Solutions

Experimental Skills Previous Year Questions With Solutions

1.0Introduction

The Experimental Skills  assesses a proficiency in  applying theoretical principles to practical scenarios, with an emphasis on precision, accuracy, and data interpretation. It involves familiarity with standard laboratory apparatus, techniques of measurement (such as vernier calipers, screw gauge, and stopwatches), and analysis of experimental data including plotting graphs, calculating slopes, and identifying sources of error. Candidates are expected to understand significant figures, error propagation, and methods to minimize uncertainties, reflecting real-world laboratory conditions. Mastery in this area enhances a student's problem-solving ability in  Physics  practical-based questions.

2.0Key Concepts to Remember

Vernier Calipers

  • It is a device, designed by a French Mathematician Pierre Vernier to measure accurately upto (101​)th of a millimetre.Vernier calipers are precision measuring instruments used to measure the internal and external dimensions, as well as depths, of an object with high accuracy. They consist of a main scale and a sliding vernier scale that allows measurements typically accurate to 0.1 mm or 0.02 mm, depending on the design.

Uses:

  • Measuring the external diameter or width of objects (using the main jaws)
  • Measuring the internal diameter of hollow objects like tubes (using the smaller jaws)
  • Measuring the depth of containers or holes (using the depth probe/rod)
  • Widely used in laboratories, mechanical engineering, and metalworking for precise measurements

Screw Gauge

  • A screw gauge is a precision instrument used to measure small lengths or thicknesses with high accuracy, typically up to 0.01 mm.

Metre Scale

  • If a beam or rod is balanced under the action of different forces, then about point of balance or equilibrium position :

Sum of clockwise moments = Sum of anticlockwise moments.

Searle’s Apparatus: It is used to determine Young's modulus of elasticity of the material of a given wire.

Capillary Tube and Capillarity: Capillarity is the tendency of a liquid to rise or fall in a narrow tube due to surface tension.This rise or fall is called capillary action.

Zurin's Law : The height of rise of liquid in a capillary tube is inversely proportional to the radius of the capillary tube, if T,θ,ρ and g are constant or rh=constant

Resonance Tube

  • A resonance tube is a device used to determine the speed of sound in air and the frequency of a tuning fork. It consists of a vertical tube partially filled with water and a tuning fork vibrating above it. Resonance occurs when the air column’s natural frequency matches the tuning fork’s frequency, producing a loud sound. The length of the resonating air column is then used to calculate the speed of sound.

Metre Bridge 

  • A metre bridge is a device based on the principle of a Wheatstone bridge, used to measure unknown resistance.

Parallax

  • Parallax is the apparent shift in position of two objects at different distances when the eye is moved sideways.

Travelling Microscope 

  • A travelling microscope is a precision instrument used for measuring small distances with high accuracy, typically in laboratory experiments. It consists of a microscope mounted on a sliding carriage that moves horizontally or vertically along a graduated scale, allowing accurate measurement of the position of objects or changes in displacement.

3.0Important Formulas

Vernier Constant or Least count 

V.C. = 1 MSD - 1 VSD

V.C or L.C.=M−V=M−ba​M=(bb−a​)M

Correction of zero Error

Correct reading=MSR+(VSR)×(L.C.)−ZeroError 

Screw gauge Least Count

Least Count=Total number of division on the circular scalePitch​

Undamped oscillation

A=A0​=ConstantE0​=21​KA02​

Damped oscillation

A=A0​e−bt/2m=A0​e−λtE=21​KA2=21​K[A0​e−bt/2m]2E=(21​KA02​)e−bt/m=(E0​)e−bt/m=E0​e−(2λ)t

Metre Scale

Sum of clockwise moments = Sum of anticlockwise moments.

Young's modulus of elasticity

Y=StrainStress​=Δl/LF/A​=AΔlFL​=N/m2 or Pascals

Pressure Balance Method

Patm​=(Patm​−R2T​)+hρg⇒h=Rρg2T​=Rρg2Tcosθc​​

Force Balance Method 

In equilibrium:forceduetoS.T=weightofriseliquid(2πr)Tcosθc​=mgh=rρg2Tcosθc​​

Stokes Law

F=6πηrv

Terminal Velocity

vT​=92​gr2(ρ−σ)​

End correction (Resonance Tube)

x=2l2​−3l1​​

Specific heat capacity

S=(ms​−mw​)(θ2​−θ0​)w​(mw​+w)(θ2​−θ1​)s​​ J/gm/°C

Specific heat of a given liquid

Specific heat of liquid,

Sl​=m(θ2​−θ1​)ms​s(100−θ1​)​−mw​(Jgm−1°C−1)

Metre Bridge

Unknown resistanceX=R(100−ll​)

Figure of Merit 

Figure of merit (k)=θl​×R+GE​

Travelling Microscope

Refractive index of the material μ=r3​−r2​r3​−r1​​

4.0Past Year Questions with Solutions on Experimental Skills:JEE (Mains)

Q-1.In an experiment with Vernier callipers of least count 0.1 mm, when two jaws are joined together the zero of Vernier scale lies right to the zero of the main scale and 6th division of Vernier scale coincides with the main scale division. While measuring the diameter of a spherical bob, the zero of vernier scale lies in between 3.2 cm and 3.3 cm marks, and 4th division of vernier scale coincides with the main scale division. The diameter of bob is measured as :

1) 3.18 cm             2) 3.25 cm                 3) 3.26 cm                4) 3.22 cm

Solution: Ans(1)

LC=0.1mmZero Error=6×LC=0.6mmReading=MSR+VSR×LC−Zero Error=[32mm+(0.14)mm]−0.6mm=31.8mm=3.18cm


Q-2.A travelling microscope has 20 divisions per cm on the main scale while its Vernier scale has total 50 divisions and 25 Vernier scale divisions are equal to 24 main scale divisions, what is the least count of the travelling microscope ?

a) 0.001 cm              b) 0.002 mm                 c) 0.002 cm                 d) 0.005 cm       

Solution: Ans(c)

1MSD=201​cm1VSD=2524​MSD=2524​×201​cmLeast Count=201​(1−2524​)cm=251​×251​=5001​cm=0.002cm 


Q-3. In an experiment to find out the diameter of wire using screw gauge, the following observations were noted :

.In an experiment to find out the diameter of wire using screw gauge, the following observations were noted :

(A) Screw moves 0.5 mm on main scale in one complete rotation

(B) Total divisions on circular scale = 50

(C) Main scale reading is 2.5 mm

(D) 45th division of circular scale is in the pitch line

(E) Instrument has 0.03 mm negative error

Then the diameter of wire is :

12.92 mm                22.54 mm                 3 2.98 mm                   43.45 mm       

Solution: Ans(3) 

MSR=2.5mmCSR=45×500.5​mm=0.45mmDiameter Reading=MSR+CSR−Zero Error=2.5+0.45−(−0.03)mm=2.98mm


Q-4.In a Vernier Calipers. 10 divisions of Vernier scale is equal to the 9 divisions of main scale. When both jaws of Vernier calipers touch each other, the zero of the Vernier scale is shifted to the left of zero of the main scale and 4th Vernier scale division exactly coincides with the main scale reading. One main scale division is equal to 1 mm. While measuring the diameter of a spherical body, the body is held between two jaws. It is now observed that zero of the Vernier scale lies between 30 and 31 divisions of main scale reading and 6th Vernier scale division exactly. coincides with the main scale reading. The diameter of the spherical body will be 

1) 3.02 cm                2) 3.06 cm          3) 3.10 cm              4) 3.20 cm     

Solution: Ans(3)

1MSD=1 mm

9MSD=10 VSD

1VSD=0.9MSD=0.9 mm

LC of Vernier Caliper=1-0.9=0.1 mm=0.01 cm

Zero Error=−(10−4)×0.1mm=−0.6mm

Reading=MSR+VSR -Zero Error

Zero Error=−(10−4)×0.1mm=−0.6mm

=3cm+6×0.01−(−0.06)=3+0.06+0.06=3.12cm

Nearest given answer is 3.10

Q-5.A screw gauge of pitch 0.5mm is used to measure the diameter of uniform wire of length 6.8cm, the main scale reading is 1.5 mm and circular scale reading is 7. The calculated curved surface area of wire to appropriate significant figures is :[Screw gauge has 50 divisions on the circular scale

(1)6.8 cm2     (2)3.4 cm2      (3)3.9 cm2   (4)2.4 cm2

Solution: Ans(2)

LC=NP​=500.5mm​=0.01mmLength of wire=6.8cmDiameter of wire=1.5mm+7×LC =1.5+7×0.01=1.57mmCurved Surface Area=πDl=3.14×6.8×1.57×10−1cm2=3.352cm2≈3.4 cm2


Q-6.The one division of main scale of vernier callipers reads 1 mm and 10 divisions of Vernier scale is equal to the 9 divisions on main scale. When the two jaws of the instrument touch each other the zero of the Vernier lies to the right of zero of the main scale and its fourth division coincides with a main scale division. When a spherical bob is tightly placed between the two jaws, the zero of the Vernier scale lies in between 4.1 cm and 4.2 cm and 6th Vernier division coincides with a main scale division. The diameter of the bob will be_____10–2 cm

Solution: Ans(412×10−2cm)10VSD=9MSD⇒1VSD=0.9MSD⇒LC=1mm=0.1mm=0.01cmPositive Zero Error=4×0.01=0.04cm

Negative Zero Error=4.1cm+6×0.01=4.12cm=412×10−2cm


Q-7.The vernier constant of Vernier callipers is 0.1 mm and it has zero error of (–0.05) cm. While measuring diameter of a sphere, the main scale reading is 1.7 cm and coinciding vernier division is 5. The corrected diameter will be_____ × 10–2 cm.

Solution: Ans(180)

Measured Diameter=MSR+VSR LC=1.7+0.01×5=1.75Corrected=Measured−Error=1.75−(−0.05)=1.80cm=180×10−2cm


Q-8.A travelling microscope is used to determine the refractive index of a glass slab. If 40 divisions are there in 1 cm on the main scale and 50 Vernier scale divisions are equal to 49 main scale divisions, then the least count of the travelling microscope is ______ × 10–6 m.

Solution: Ans(5 ✕  10-6 m )

50VSD=49MSD 1VSD=5049​MSDLeast Count=1MSD−1VSD=(1−5049​)MSD=501​MSD1MSD=401​cmLeast Count=50×401​cm=2001​cm=21​×10−5m=0.5×10−5m=5×10−6m


Q-9.In a vernier caliper, each cm on the main scale is divided into 20 equal parts. If tenth vernier scale division coincides with ninth main scale division. Then the value of the vernier constant will be .............. × 10–2 mm.

Solution:Ans(5)

20MSD=1cm1MSD=201​cm10VSD=9MSD 1VSD=109​MSD=109​×201​=2009​cmVC=1MSD−1VSD=201​−2009​=2001​cm=2001​×10mm=5×10−2mm


Q-10.For the determination of refractive index of glass slab, a travelling microscope is used whose main scale contains 300 equal divisions equal to 15 cm. The vernier scale attached to the microscope has 25 divisions equal to 24 divisions of the main scale. The least count (LC) of the travelling microscope is (in cm) :

1) 0.001                2) 0.002          3) 0.0005                 4) 0.0025    

Solution:

300MSD=15cm1MSD=30015​cm=0.05cm25VSD=24MSD1VSD=2524​MSDLC=1MSD−1VSDLC=1MSD−2524​MSD=251​MSDLC=251​×0.05=0.002cm


Table of Contents


  • 1.0Introduction
  • 2.0Key Concepts to Remember
  • 2.1Vernier Calipers
  • 2.2Screw Gauge
  • 2.3Metre Scale
  • 3.0Important Formulas
  • 4.0Past Year Questions with Solutions on Experimental Skills:JEE (Mains)

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