Experimental Skills Previous Year Questions With Solutions

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Experimental Skills JEE Questions with Solutions | Practical Physics Prep

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.

JEE Main Previous Year Solved Questions on Experimental Skills

Download PDF

2.0Key Concepts to Remember

Vernier Calipers

It is a device, designed by a French Mathematician Pierre Vernier to measure accurately upto $(\frac{1}{10})^{t h}$ 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 $T, θ, ρ an d g$ 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	$V.C. = 1 MSD - 1 VSD$ $V . C or L . C . = M - V = M - \frac{a}{b} M = (\frac{b - a}{b}) M$
Correction of zero Error	$Correct reading = MSR + (V SR) \times (L . C .) - Z ero E rror$
Screw gauge Least Count	$L e a s t C o u n t = \frac{Pitch}{Total number of division on the circular scale}$
Undamped oscillation	$A = A_{0} = Constant E_{0} = \frac{1}{2} K A_{0}^{2}$
Damped oscillation	$A = A_{0} e^{- b t /2 m} = A_{0} e^{- λ t} E = \frac{1}{2} K A^{2} = \frac{1}{2} K [A_{0} e^{- b t /2 m}]^{2} E = (\frac{1}{2} K A_{0}^{2}) e^{- b t / m} = (E_{0}) e^{- b t / m} = E_{0} e^{- (2 λ) t}$
Metre Scale	Sum of clockwise moments = Sum of anticlockwise moments.
Young's modulus of elasticity	$Y = \frac{Stress}{Strain} = \frac{F / A}{Δ l / L} = \frac{F L}{A Δ l} = N / m^{2} or Pascals$
Pressure Balance Method	$P_{atm} = (P_{atm} - \frac{2 T}{R}) + h ρ g \Rightarrow h = \frac{2 T}{Rρ g} = \frac{2 T c o s θ _{c}}{Rρ g}$
Force Balance Method	$In equilibrium : f orce d u e t o S . T = w e i g h t o f r i se l i q u i d (2 π r) T cos θ_{c} = m g h = \frac{2 T c o s θ _{c}}{r ρ g}$
Stokes Law	F=6πηrv
Terminal Velocity	$v_{T} = \frac{2}{9} \frac{r ^{2} ( ρ - σ )}{g}$
End correction (Resonance Tube)	$x = \frac{l _{2} - 3 l _{1}}{2}$
Specific heat capacity	$S = \frac{( m _{w} + w ) ( θ _{2} - θ _{1} ) _{s}}{( m _{s} - m _{w} ) ( θ _{2} - θ _{0} ) _{w}} J/gm/°C$
Specific heat of a given liquid	Specific heat of liquid, $S_{l} = \frac{m _{s} s ( 100 - θ _{1} )}{m ( θ _{2} - θ _{1} )} - \frac{w}{m} (J g m^{- 1} ° C^{- 1})$
Metre Bridge	$U nkn o w n res i s t an ce X = R (\frac{l}{100 - l})$
Figure of Merit	$Figure of merit (k) = \frac{l}{θ} \times \frac{E}{R + G}$
Travelling Microscope	$Refractive index of the material μ = \frac{r _{3} - r _{1}}{r _{3} - r _{2}}$

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 6^th 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 4^th 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)

$L C = 0.1 mm Zero Error = 6 \times L C = 0.6 mm Reading = MSR + VSR \times LC - Zero Error = [32 mm + (0.14) mm] - 0.6 mm = 31.8 mm = 3.18 c m$

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)

$1 MSD = \frac{1}{20} cm 1 VSD = \frac{24}{25} MSD = \frac{24}{25} \times \frac{1}{20} cm Least Count = \frac{1}{20} (1 - \frac{24}{25}) cm = \frac{1}{25} \times \frac{1}{25} = \frac{1}{500} cm = 0.002 cm$

Q-3. 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) 45^th 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)

$MSR = 2.5 mm CSR = 45 \times \frac{0.5}{50} mm = 0.45 mm Diameter Reading = MSR + CSR - Zero Error = 2.5 + 0.45 - (- 0.03) mm = 2.98 mm$

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 4^th 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 6^th 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

$Zero Error = - (10 - 4) \times 0.1 mm = - 0.6 mm$

Reading=MSR+VSR -Zero Error

$Zero Error = - (10 - 4) \times 0.1 mm = - 0.6 mm$

$= 3 cm + 6 \times 0.01 - (- 0.06) = 3 + 0.06 + 0.06 = 3.12 cm$

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

$(1) 6.8 c m^{2} (2) 3.4 c m^{2} (3) 3.9 c m^{2} (4) 2.4 c m^{2}$

Solution: Ans(2)

$L C = \frac{P}{N} = \frac{0.5 mm}{50} = 0.01 mm Length of wire = 6.8 cm Diameter of wire = 1.5 mm + 7 \times L C = 1.5 + 7 \times 0.01 = 1.57 mm Curved Surface Area = π D l = 3.14 \times 6.8 \times 1.57 \times 1 0^{- 1} cm^{2} = 3.352 cm^{2} \approx 3.4 cm^{2}$

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 6^th Vernier division coincides with a main scale division. The diameter of the bob will be_____10^–2 cm

$Solution: Ans (412 \times 1 0^{- 2} cm) 10 VSD = 9 MSD \Rightarrow 1 VSD = 0.9 MSD \Rightarrow L C = 1 mm = 0.1 mm = 0.01 cm Positive Zero Error = 4 \times 0.01 = 0.04 cm$

$Negative Zero Error = 4.1 cm + 6 \times 0.01 = 4.12 cm = 412 \times 1 0^{- 2} cm$

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)

$Measured Diameter = MSR + V SR L C = 1.7 + 0.01 \times 5 = 1.75 Corrected = Measured - Error = 1.75 - (- 0.05) = 1.80 cm = 180 \times 1 0^{- 2} cm$

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 )

$50 VSD = 49 MSD 1 VSD = \frac{49}{50} MSD Least Count = 1 MSD - 1 VSD = (1 - \frac{49}{50}) MSD = \frac{1}{50} MSD 1 MSD = \frac{1}{40} cm Least Count = \frac{1}{50 \times 40} cm = \frac{1}{200} cm = \frac{1}{2} \times 1 0^{- 5} m = 0.5 \times 1 0^{- 5} m = 5 \times 1 0^{- 6} m$

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)

$20 MSD = 1 cm 1 MSD = \frac{1}{20} cm 10 VSD = 9 MSD 1 VSD = \frac{9}{10} MSD = \frac{9}{10} \times \frac{1}{20} = \frac{9}{200} cm V C = 1 MSD - 1 VSD = \frac{1}{20} - \frac{9}{200} = \frac{1}{200} cm = \frac{1}{200} \times 10 mm = 5 \times 1 0^{- 2} mm$

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:

$300 MSD = 15 cm 1 MSD = \frac{15}{300} cm = 0.05 cm 25 VSD = 24 MSD 1 VSD = \frac{24}{25} MSD L C = 1 MSD - 1 VSD L C = 1 MSD - \frac{24}{25} MSD = \frac{1}{25} MSD L C = \frac{1}{25} \times 0.05 = 0.002 cm$

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.

JEE Main Previous Year Solved Questions on Experimental Skills

Download PDF

2.0Key Concepts to Remember

Vernier Calipers

It is a device, designed by a French Mathematician Pierre Vernier to measure accurately upto $(\frac{1}{10})^{t h}$ 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 $T, θ, ρ an d g$ 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	$V.C. = 1 MSD - 1 VSD$ $V . C or L . C . = M - V = M - \frac{a}{b} M = (\frac{b - a}{b}) M$
Correction of zero Error	$Correct reading = MSR + (V SR) \times (L . C .) - Z ero E rror$
Screw gauge Least Count	$L e a s t C o u n t = \frac{Pitch}{Total number of division on the circular scale}$
Undamped oscillation	$A = A_{0} = Constant E_{0} = \frac{1}{2} K A_{0}^{2}$
Damped oscillation	$A = A_{0} e^{- b t /2 m} = A_{0} e^{- λ t} E = \frac{1}{2} K A^{2} = \frac{1}{2} K [A_{0} e^{- b t /2 m}]^{2} E = (\frac{1}{2} K A_{0}^{2}) e^{- b t / m} = (E_{0}) e^{- b t / m} = E_{0} e^{- (2 λ) t}$
Metre Scale	Sum of clockwise moments = Sum of anticlockwise moments.
Young's modulus of elasticity	$Y = \frac{Stress}{Strain} = \frac{F / A}{Δ l / L} = \frac{F L}{A Δ l} = N / m^{2} or Pascals$
Pressure Balance Method	$P_{atm} = (P_{atm} - \frac{2 T}{R}) + h ρ g \Rightarrow h = \frac{2 T}{Rρ g} = \frac{2 T c o s θ _{c}}{Rρ g}$
Force Balance Method	$In equilibrium : f orce d u e t o S . T = w e i g h t o f r i se l i q u i d (2 π r) T cos θ_{c} = m g h = \frac{2 T c o s θ _{c}}{r ρ g}$
Stokes Law	F=6πηrv
Terminal Velocity	$v_{T} = \frac{2}{9} \frac{r ^{2} ( ρ - σ )}{g}$
End correction (Resonance Tube)	$x = \frac{l _{2} - 3 l _{1}}{2}$
Specific heat capacity	$S = \frac{( m _{w} + w ) ( θ _{2} - θ _{1} ) _{s}}{( m _{s} - m _{w} ) ( θ _{2} - θ _{0} ) _{w}} J/gm/°C$
Specific heat of a given liquid	Specific heat of liquid, $S_{l} = \frac{m _{s} s ( 100 - θ _{1} )}{m ( θ _{2} - θ _{1} )} - \frac{w}{m} (J g m^{- 1} ° C^{- 1})$
Metre Bridge	$U nkn o w n res i s t an ce X = R (\frac{l}{100 - l})$
Figure of Merit	$Figure of merit (k) = \frac{l}{θ} \times \frac{E}{R + G}$
Travelling Microscope	$Refractive index of the material μ = \frac{r _{3} - r _{1}}{r _{3} - r _{2}}$

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 6^th 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 4^th 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)

$L C = 0.1 mm Zero Error = 6 \times L C = 0.6 mm Reading = MSR + VSR \times LC - Zero Error = [32 mm + (0.14) mm] - 0.6 mm = 31.8 mm = 3.18 c m$

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)

$1 MSD = \frac{1}{20} cm 1 VSD = \frac{24}{25} MSD = \frac{24}{25} \times \frac{1}{20} cm Least Count = \frac{1}{20} (1 - \frac{24}{25}) cm = \frac{1}{25} \times \frac{1}{25} = \frac{1}{500} cm = 0.002 cm$

Q-3. 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) 45^th 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)

$MSR = 2.5 mm CSR = 45 \times \frac{0.5}{50} mm = 0.45 mm Diameter Reading = MSR + CSR - Zero Error = 2.5 + 0.45 - (- 0.03) mm = 2.98 mm$

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 4^th 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 6^th 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

$Zero Error = - (10 - 4) \times 0.1 mm = - 0.6 mm$

Reading=MSR+VSR -Zero Error

$Zero Error = - (10 - 4) \times 0.1 mm = - 0.6 mm$

$= 3 cm + 6 \times 0.01 - (- 0.06) = 3 + 0.06 + 0.06 = 3.12 cm$

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

$(1) 6.8 c m^{2} (2) 3.4 c m^{2} (3) 3.9 c m^{2} (4) 2.4 c m^{2}$

Solution: Ans(2)

$L C = \frac{P}{N} = \frac{0.5 mm}{50} = 0.01 mm Length of wire = 6.8 cm Diameter of wire = 1.5 mm + 7 \times L C = 1.5 + 7 \times 0.01 = 1.57 mm Curved Surface Area = π D l = 3.14 \times 6.8 \times 1.57 \times 1 0^{- 1} cm^{2} = 3.352 cm^{2} \approx 3.4 cm^{2}$

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 6^th Vernier division coincides with a main scale division. The diameter of the bob will be_____10^–2 cm

$Solution: Ans (412 \times 1 0^{- 2} cm) 10 VSD = 9 MSD \Rightarrow 1 VSD = 0.9 MSD \Rightarrow L C = 1 mm = 0.1 mm = 0.01 cm Positive Zero Error = 4 \times 0.01 = 0.04 cm$

$Negative Zero Error = 4.1 cm + 6 \times 0.01 = 4.12 cm = 412 \times 1 0^{- 2} cm$

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)

$Measured Diameter = MSR + V SR L C = 1.7 + 0.01 \times 5 = 1.75 Corrected = Measured - Error = 1.75 - (- 0.05) = 1.80 cm = 180 \times 1 0^{- 2} cm$

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 )

$50 VSD = 49 MSD 1 VSD = \frac{49}{50} MSD Least Count = 1 MSD - 1 VSD = (1 - \frac{49}{50}) MSD = \frac{1}{50} MSD 1 MSD = \frac{1}{40} cm Least Count = \frac{1}{50 \times 40} cm = \frac{1}{200} cm = \frac{1}{2} \times 1 0^{- 5} m = 0.5 \times 1 0^{- 5} m = 5 \times 1 0^{- 6} m$

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)

$20 MSD = 1 cm 1 MSD = \frac{1}{20} cm 10 VSD = 9 MSD 1 VSD = \frac{9}{10} MSD = \frac{9}{10} \times \frac{1}{20} = \frac{9}{200} cm V C = 1 MSD - 1 VSD = \frac{1}{20} - \frac{9}{200} = \frac{1}{200} cm = \frac{1}{200} \times 10 mm = 5 \times 1 0^{- 2} mm$

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:

$300 MSD = 15 cm 1 MSD = \frac{15}{300} cm = 0.05 cm 25 VSD = 24 MSD 1 VSD = \frac{24}{25} MSD L C = 1 MSD - 1 VSD L C = 1 MSD - \frac{24}{25} MSD = \frac{1}{25} MSD L C = \frac{1}{25} \times 0.05 = 0.002 cm$

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Experimental Skills Previous Year Questions With Solutions

1.0Introduction

2.0Key Concepts to Remember

Vernier Calipers

Screw Gauge

Metre Scale

3.0Important Formulas

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

Table of Contents

Experimental Skills Previous Year Questions With Solutions

1.0Introduction

2.0Key Concepts to Remember

Vernier Calipers

Screw Gauge

Metre Scale

3.0Important Formulas

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

Table of Contents