Experiments in Measurement, Mechanics and Heat

What to remember

• Least count of a vernier = 1 MSD − 1 VSD = (value of 1 MSD)/N; least count of a screw gauge = pitch/number of circular scale divisions. The corrected reading is always observed reading − zero error (zero error carries its sign).
• Simple pendulum: T = 2π√(l/g), so a graph of l against T² is a straight line through the origin with slope g/4π². The length is measured to the centre of the bob and the angle is kept small.
• Formulae to know by heart: Y = MgL/(πr²l); T = rhρg/(2cosθ) (T = rhρg/2 for water in glass); η = 2r²(ρ − σ)g/(9v); v = 2f(l₂ − l₁); heat lost = heat gained.
• In the resonance tube, l₁ + e = λ/4 and l₂ + e = 3λ/4, so the end correction is e = (l₂ − 3l₁)/2 and it cancels in l₂ − l₁.

Vernier callipers

Aim. To measure length, the external and internal diameter of a hollow cylinder, and the depth of a vessel.

Principle. N vernier scale divisions (VSD) are equal to (N − 1) main scale divisions (MSD). One VSD is therefore slightly shorter than one MSD.

• Least count (LC) = 1 MSD − 1 VSD = (1 MSD)/N.
• For 1 MSD = 1 mm and N = 10: LC = 0.1 mm = 0.01 cm.
• For 1 MSD = 1 mm and N = 20 (20 VSD = 19 MSD): LC = 0.05 mm.

Reading. Total reading = main scale reading (MSR, just left of the vernier zero) + n × LC, where n is the number of the vernier division that coincides with a main scale division.

Zero error. Close the jaws and note the reading.

Case	Position of vernier zero	Zero error	Correction
Positive	To the right of the main scale zero	+ n × LC (n = coinciding division)	Subtract it
Negative	To the left of the main scale zero	− (N − m) × LC (m = coinciding division)	Add its magnitude

Corrected reading = observed reading − zero error (with sign).

Uses. The lower jaws give the external diameter, the upper jaws give the internal diameter, and the thin strip at the end gives the depth of a vessel.

Readings and errors. Take at least three readings at different positions and average them. Sources of error: zero error, parallax while reading the coinciding division, and pressing the jaws too hard. Precautions: clean the jaws, move the slider gently, tighten the screw before reading, and record the zero error before and after.

Worked example 1. LC = 0.01 cm. The jaws closed show the 3rd vernier division coinciding, so the zero error is +0.03 cm. A rod gives MSR = 2.4 cm and the 6th vernier division coincides. Observed = 2.4 + 6 × 0.01 = 2.46 cm. Corrected = 2.46 − 0.03 = 2.43 cm.

Screw gauge

Aim. To find the diameter of a wire or the thickness of a thin sheet.

Principle. A screw of known pitch moves along its axis when turned. The circular scale divides the pitch into equal parts.

• Pitch = distance moved by the screw in one complete rotation.
• LC = pitch/number of circular scale divisions.
• For pitch 1 mm and 100 divisions: LC = 0.01 mm. For pitch 0.5 mm and 50 divisions: LC = 0.01 mm.

Reading. Total reading = pitch scale reading (PSR) + circular scale reading (CSR) × LC. Corrected reading = observed − zero error.

Case (studs in contact)	Circular scale position	Zero error
Positive	Zero of circular scale is below the reference line; the n-th division is at the line	+ n × LC
Negative	Zero of circular scale is above the reference line; the m-th division is at the line	− (N − m) × LC

Precautions. Use the ratchet (safety clutch) to close the studs gently so that the wire is not squeezed. Remove backlash by turning the screw in one direction. Measure the wire at several places and in two perpendicular directions, because a wire may not be perfectly circular.

Worked example 2. Pitch 0.5 mm, 50 divisions, so LC = 0.01 mm. With studs touching, the 46th division is at the reference line, so zero error = −(50 − 46) × 0.01 = −0.04 mm. A wire gives PSR = 2.5 mm and CSR = 28. Observed = 2.5 + 0.28 = 2.78 mm. Corrected = 2.78 − (−0.04) = 2.82 mm.

Simple pendulum and the metre scale

Aim (pendulum). To find the acceleration due to gravity g from the relation between length l and time period T.

Principle. For small amplitude, T = 2π√(l/g), so g = 4π²l/T². A graph of l (y-axis) against T² (x-axis) is a straight line through the origin. Its slope is l/T² = g/4π², so g = 4π² × slope. Taking the slope from the graph reduces random error.

Procedure and readings. Hang a small, dense bob on a thin, inextensible thread. The length l = length of thread + radius of bob (measured to the centre of the bob). Displace the bob by a small angle (a few degrees), release it and time 20 to 50 oscillations starting from the mean position. T = total time/number of oscillations. Repeat for five or six lengths.

Amplitude and energy dissipation. If the amplitude is plotted against time, the curve falls smoothly (an exponential-like decay). The energy of the oscillation is slowly lost to air resistance and friction at the support, so the amplitude decreases while the time period stays almost the same for small angles.

Errors and precautions. Avoid air draughts, use a rigid support, make sure the bob moves in a vertical plane (no spinning), and count the oscillations carefully. Percentage error in g = Δl/l + 2ΔT/T (maximum).

Worked example 3. A student measures 96.0 cm from the suspension point to the top of a bob of diameter 4.0 cm. T = 2.0 s. Effective length = 96.0 + 2.0 = 98.0 cm. With π² ≈ 10, g = 4π²l/T² = 4 × 10 × 0.98/4 = 9.8 m s⁻².

Metre scale and mass by moments. Principle of moments: at balance, the sum of clockwise moments equals the sum of anticlockwise moments about the knife edge. Balance a uniform metre scale on a knife edge at the 50 cm mark. Its own weight then acts at the pivot and gives no torque. Hang the unknown mass M at distance a from the pivot and a known mass m at distance b on the other side, and adjust until the scale is horizontal. Then M × a = m × b, so M = m b/a. If the pivot is not at 50 cm, include the weight of the scale acting at its 50 cm mark. Precaution: measure the distances from the pivot to the centre of each hanging mass.

Young's modulus of a wire

Aim. To find Young's modulus Y of the material of a metallic wire.

Principle. Y = stress/strain = (F/A)/(l/L). For a wire of radius r, length L, stretched by load M with extension l: Y = MgL/(πr²l).

Set-up (Searle's method, outline). Two identical wires hang from a rigid support. The reference wire carries a constant load and compensates for temperature change and sagging of the support. The experimental wire carries a hanger to which slotted masses are added. A spirit level and micrometer screw measure the small extension. The radius of the wire is measured with a screw gauge, and the length L with a metre scale.

Graph. Load (M) against extension (l) is a straight line within the elastic limit. Slope = M/l, so Y = (slope × gL)/(πr²).

Errors and precautions. Do not exceed the elastic limit. Use a long, thin wire to get a measurable extension. Keep the wire free of kinks. Average the loading and unloading readings. The radius enters as r², so it must be measured most carefully; a small error in r gives a large error in Y.

Worked example 4. Wire radius 0.7 mm, length 1.54 m, load 3.0 kg, extension 1.0 mm. Take g = 10 m s⁻² and π = 22/7. πr² = 1.54 × 10⁻⁶ m². Y = 30 × 1.54/(1.54 × 10⁻⁶ × 10⁻³) = 3.0 × 10¹⁰ N m⁻².

Surface tension and viscosity

Surface tension by capillary rise.

• Principle: T = rhρg/(2cosθ). For water and clean glass θ is about 0, so T = rhρg/2.
• r = radius of the capillary (measured with a travelling microscope or vernier), h = height of the water column above the level in the vessel, ρ = density of water, g = 9.8 m s⁻².
• The surface tension of water at room temperature is about 7 × 10⁻² N m⁻¹.
• Effect of detergent: it lowers the surface tension of water, so the capillary rise decreases (h ∝ T/r).
• Precautions: clean the tube thoroughly, keep it vertical, and measure h from the level in the vessel to the lowest point of the meniscus. Impurities and temperature change T.

Viscosity by terminal velocity of a ball.

• Principle: a ball falling in a liquid reaches a terminal velocity v when weight − buoyancy = viscous drag (Stokes' law, F = 6πηrv).
• Result: η = 2r²(ρ − σ)g/(9v), where ρ = density of the ball and σ = density of the liquid.
• Reading: v = distance between two marks on a tall jar/time taken. The marks are placed well below the surface so that the ball has reached uniform velocity.
• Precautions: use a wide jar and drop the ball along the axis (to avoid wall effects), wet the ball first so that no air bubble sticks, and keep the temperature constant because η changes strongly with temperature.
• Effect of changes: v ∝ r² for the same ball material and liquid, and v ∝ 1/η.

Worked example 5. A ball of radius 1.0 mm and density 2000 kg m⁻³ falls with v = 0.020 m s⁻¹ in a liquid of density 1100 kg m⁻³. With g = 10 m s⁻²: η = 2 × 10⁻⁶ × 900 × 10/(9 × 0.02) = 0.1 Pa s.

Speed of sound by the resonance tube

Aim. To find the speed of sound in air at room temperature using a resonance tube and tuning forks of known frequency.

Principle. An air column closed at one end (the water surface) is set in resonance by a vibrating fork. A node forms at the water surface and an antinode forms slightly above the open end.

• First resonance: l₁ + e = λ/4.
• Second resonance: l₂ + e = 3λ/4.
• Subtracting: l₂ − l₁ = λ/2, so λ = 2(l₂ − l₁) and v = fλ = 2f(l₂ − l₁).
• End correction: e = (l₂ − 3l₁)/2. It is also about 0.6 times the radius of the tube.

Procedure. Hold a vibrating fork over the open end. Lower the water level (by moving the reservoir) until the loudest sound is heard, and note l₁. Raise the length further to get l₂. Repeat with another fork.

Errors and precautions. Strike the fork on a rubber pad (not a hard surface). Keep the tube vertical and hold the fork above the tube axis without touching it. Find the position of maximum loudness carefully (approach it from above and below and average). Using l₂ − l₁ removes the end correction.

Example: a 512 Hz fork gives l₁ = 16.0 cm and l₂ = 50.0 cm, so λ = 68 cm and v = 348 m s⁻¹.

Specific heat capacity by the method of mixtures

Aim. To find the specific heat capacity of a solid and of a liquid.

Principle. If no heat is lost to the surroundings, heat lost by the hot body equals heat gained by the cooler bodies.

Solid. A solid of mass m_s at temperature T₁ (heated in steam) is dropped quickly into water of mass m_w at T₂ in a calorimeter of mass m_c and specific heat c_c. The final temperature is T.

m_s c_s (T₁ − T) = (m_w c_w + m_c c_c)(T − T₂), so c_s = (m_w c_w + m_c c_c)(T − T₂)/[m_s (T₁ − T)].

Liquid. A solid of known specific heat (for example copper, about 400 J kg⁻¹ K⁻¹) is heated and dropped into the liquid in the calorimeter. The same equation is used with the liquid in place of water:

c_l = m_s c_s (T₁ − T)/[m_l (T − T₂)] when the calorimeter's heat capacity is negligible.

Errors and precautions. Transfer the hot solid quickly, stir gently, use a polished and lagged (insulated) calorimeter, take the initial temperature of the water just before dropping the solid, and read the highest steady temperature.

Worked example 6. 150 g of metal at 100 °C is dropped into 100 g of water at 20 °C, and the final temperature is 30 °C. Ignore the calorimeter. 0.150 × c × 70 = 0.100 × 4200 × 10, so c = 400 J kg⁻¹ K⁻¹.

Formula summary.

Experiment	Formula	Least count / key relation
Vernier	Reading = MSR + n × LC	LC = 1 MSD/N
Screw gauge	Reading = PSR + CSR × LC	LC = pitch/N
Pendulum	g = 4π²l/T²	Slope of l–T² = g/4π²
Moments	M a = m b	Pivot at 50 cm
Young's modulus	Y = MgL/πr²l	r² most sensitive
Surface tension	T = rhρg/2	θ ≈ 0 for water
Viscosity	η = 2r²(ρ − σ)g/9v	v ∝ r²
Resonance tube	v = 2f(l₂ − l₁)	e = (l₂ − 3l₁)/2
Mixtures	m_s c_s ΔT_s = (m_w c_w + m_c c_c)ΔT_w	c_w = 4200 J kg⁻¹ K⁻¹

NEET traps

• A positive zero error is subtracted; a negative zero error is added. The sign is easily reversed under time pressure.
• In a vernier, 1 VSD is smaller than 1 MSD (for N VSD = (N − 1) MSD). Do not confuse LC with the size of one VSD.
• Pitch is the distance for one full turn; least count is the distance for one circular scale division.
• The pendulum slope of the l–T² graph is g/4π², not 4π²/g. The line passes through the origin.
• Length of the pendulum is measured to the centre of the bob, not to its top or its bottom.
• In Y = MgL/(πr²l), r is the radius, not the diameter. Using the diameter makes the answer 4 times too large.
• Capillary rise increases with a narrower tube and falls when detergent is added.
• Terminal velocity decreases when viscosity increases and increases with the square of the radius.
• The difference l₂ − l₁ equals λ/2, not λ/4. The first resonance length alone gives λ/4 − e.
• In the method of mixtures the calorimeter's heat capacity is included unless it is stated to be negligible.

One-liners

• LC of a vernier with 10 VSD = 9 MSD and 1 MSD = 1 mm is 0.1 mm.
• The depth of a vessel is measured with the thin strip of the vernier callipers.
• Screw gauge with pitch 1 mm and 100 divisions has LC 0.01 mm.
• Always use the ratchet of a screw gauge so the wire is not squeezed.
• A pendulum amplitude falls with time because energy is lost to air and the support.
• Small amplitude means only a few degrees.
• The l–T² graph of a simple pendulum passes through the origin.
• Young's modulus is a property of the material, not of the wire's size.
• A reference wire in Searle's method compensates for temperature change and sagging.
• Water has a contact angle of about zero on clean glass.
• Detergent reduces the surface tension of water.
• The ball in the viscosity experiment is dropped along the axis of a wide jar.
• Sound travels at about 340 m s⁻¹ in air at room temperature.
• The end correction cancels when l₂ − l₁ is used.
• The specific heat of water is 4200 J kg⁻¹ K⁻¹.

Practice questions

1. In a vernier callipers, 10 vernier scale divisions coincide with 9 main scale divisions, and one main scale division is 1 mm. What is the least count?

   1. 0.01 mm
   2. 1.0 mm
   3. 0.9 mm
   4. 0.1 mm
   Answer

   D. 0.1 mm

   Least count = 1 MSD − 1 VSD = 1 mm − 0.9 mm = 0.1 mm.

2. A vernier callipers has a negative zero error of 0.03 cm. To get the true reading, the correction applied to the observed reading is:

   1. −0.03 cm (subtracted)
   2. 0.03 cm multiplied into the reading
   3. +0.03 cm (added)
   4. none, because zero error matters only for internal diameter
   Answer

   C. +0.03 cm (added)

   Correction = −(zero error) = −(−0.03 cm) = +0.03 cm, so it is added.

3. A vernier callipers has least count 0.01 cm. With the jaws closed it shows a zero error of −0.02 cm. While measuring a rod the main scale reads 3.2 cm and the 7th vernier division coincides with a main scale division. The corrected length is:

   1. 3.25 cm
   2. 3.29 cm
   3. 3.27 cm
   4. 3.31 cm
   Answer

   B. 3.29 cm

   Observed = 3.2 + 7 × 0.01 = 3.27 cm. Corrected = observed − (−0.02) = 3.29 cm.

4. The pitch of a screw gauge is the distance:

   1. moved by the screw along its axis in one complete rotation of the circular scale
   2. moved by the screw for a rotation of one circular scale division
   3. between two adjacent divisions of the circular scale
   4. between the two studs when the jaws are fully open
   Answer

   A. moved by the screw along its axis in one complete rotation of the circular scale

   Pitch is the linear advance of the screw per full turn. The advance for one circular division is the least count.

5. A screw gauge has least count 0.01 mm and a positive zero error of 0.03 mm. A wire gives a pitch scale reading of 4 mm and a circular scale reading of 36 divisions. The diameter of the wire is:

   1. 4.33 mm
   2. 4.39 mm
   3. 4.30 mm
   4. 4.36 mm
   Answer

   A. 4.33 mm

   Observed = 4 + 36 × 0.01 = 4.36 mm. Corrected = 4.36 − 0.03 = 4.33 mm.

6. In the simple pendulum experiment a graph of length l (in m) against T² (in s²) is a straight line through the origin with slope 0.25 m s⁻². Taking π² ≈ 10, the value of g is:

   1. 1.0 m s⁻²
   2. 160 m s⁻²
   3. 10 m s⁻²
   4. 2.5 m s⁻²
   Answer

   C. 10 m s⁻²

   l = (g/4π²) T², so slope = g/4π². g = 4π² × 0.25 = 4 × 10 × 0.25 = 10 m s⁻².

7. In the simple pendulum experiment, the effective length of the pendulum is measured from the point of suspension to:

   1. the top of the bob
   2. the centre of the bob
   3. the bottom of the bob
   4. the point where the thread is tied to the bob's hook only
   Answer

   B. the centre of the bob

   The effective length is the thread length plus the radius of the bob, which reaches the centre of the bob.

8. A wire of radius 0.7 mm and length 1.54 m is stretched by a load of 3.0 kg and its extension is 1.0 mm. Take g = 10 m s⁻² and π = 22/7. Young's modulus of the wire is:

   1. 1.5 × 10¹⁰ N m⁻²
   2. 7.5 × 10⁹ N m⁻²
   3. 3.0 × 10⁷ N m⁻²
   4. 3.0 × 10¹⁰ N m⁻²
   Answer

   D. 3.0 × 10¹⁰ N m⁻²

   Y = MgL/(πr²l). πr² = (22/7)(0.7 × 10⁻³)² = 1.54 × 10⁻⁶ m². Y = 30 × 1.54 /(1.54 × 10⁻⁶ × 10⁻³) = 3.0 × 10¹⁰ N m⁻².

9. Water rises 3.0 cm in a clean glass capillary tube of radius 0.5 mm. Take the angle of contact as zero, density of water 1000 kg m⁻³ and g = 10 m s⁻². The surface tension of water is:

   1. 0.0375 N m⁻¹
   2. 0.15 N m⁻¹
   3. 0.075 N m⁻¹
   4. 0.0075 N m⁻¹
   Answer

   C. 0.075 N m⁻¹

   T = rhρg/2 = (0.5 × 10⁻³)(3 × 10⁻²)(1000)(10)/2 = 0.075 N m⁻¹.

10. When a little detergent is added to water, the height of capillary rise in the same clean tube:

    1. remains the same, because the radius is unchanged
    2. decreases, because the surface tension of water decreases
    3. increases, because the density increases
    4. increases, because the surface tension increases
    Answer

    B. decreases, because the surface tension of water decreases

    h = 2T cosθ/(rρg). Detergent lowers T, so h decreases.

11. A ball of radius 1.0 mm and density 2000 kg m⁻³ falls with a uniform speed of 0.020 m s⁻¹ in a liquid of density 1100 kg m⁻³. Take g = 10 m s⁻². The coefficient of viscosity of the liquid is:

    1. 0.1 Pa s
    2. 0.4 Pa s
    3. 0.9 Pa s
    4. 0.05 Pa s
    Answer

    A. 0.1 Pa s

    η = 2r²(ρ − σ)g/(9v) = 2 × 10⁻⁶ × 900 × 10/(9 × 0.02) = 0.018/0.18 = 0.1 Pa s.

12. In a resonance tube experiment a tuning fork of frequency 480 Hz gives the first resonance at an air column length of 17 cm and the second at 53 cm. The speed of sound is:

    1. 691.2 m s⁻¹
    2. 326.4 m s⁻¹
    3. 172.8 m s⁻¹
    4. 345.6 m s⁻¹
    Answer

    D. 345.6 m s⁻¹

    λ = 2(l₂ − l₁) = 2 × 36 cm = 0.72 m. v = fλ = 480 × 0.72 = 345.6 m s⁻¹.

13. In a resonance tube experiment the first and second resonances occur at air column lengths of 20.0 cm and 62.0 cm. The end correction is:

    1. 2.0 cm
    2. 1.0 cm
    3. 0.5 cm
    4. 3.0 cm
    Answer

    B. 1.0 cm

    e = (l₂ − 3l₁)/2 = (62 − 60)/2 = 1.0 cm. Check: λ/4 = 21 cm = l₁ + e.

14. In the method of mixtures, the principle used is that:

    1. specific heat of the hot body equals that of the water
    2. the final temperature is the average of the two initial temperatures
    3. heat lost by the hot body is always twice the heat gained by the cold body
    4. heat lost by the hot body equals heat gained by the cold bodies, if no heat is lost to the surroundings
    Answer

    D. heat lost by the hot body equals heat gained by the cold bodies, if no heat is lost to the surroundings

    Calorimetry: heat lost = heat gained, assuming an isolated system.

15. A uniform metre scale is balanced on a knife edge at its 50 cm mark. An object is hung at the 30 cm mark and a 50 g mass is hung at the 80 cm mark, and the scale balances. The mass of the object is:

    1. 120 g
    2. 50 g
    3. 75 g
    4. 33.3 g
    Answer

    C. 75 g

    Distances from the knife edge: object 20 cm, known mass 30 cm. M × 20 = 50 × 30, so M = 75 g.

16. 150 g of a metal at 100 °C is dropped into 100 g of water at 20 °C in a calorimeter of negligible heat capacity. The final steady temperature is 30 °C. Specific heat of water is 4200 J kg⁻¹ K⁻¹. The specific heat of the metal is:

    1. 400 J kg⁻¹ K⁻¹
    2. 200 J kg⁻¹ K⁻¹
    3. 350 J kg⁻¹ K⁻¹
    4. 900 J kg⁻¹ K⁻¹
    Answer

    A. 400 J kg⁻¹ K⁻¹

    Heat lost = heat gained: 0.150 × c × 70 = 0.100 × 4200 × 10, so c = 4200/10.5 = 400 J kg⁻¹ K⁻¹.

17. Given below are two statements: Statement I: A positive zero error of a vernier callipers is subtracted from the observed reading. Statement II: When 10 vernier divisions equal 9 main scale divisions, one vernier division is smaller than one main scale division.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Positive error means vernier zero is right of main zero, so the correction is subtracted. 1 VSD = 0.9 MSD, which is smaller.

18. Given below are two statements: Statement I: The zero error of a screw gauge is found with the two studs kept fully apart. Statement II: The zero correction has the opposite sign to the zero error.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    The zero error is found with the studs in contact, so Statement I is wrong. Statement II is correct.

19. Given below are two statements: Statement I: The graph of length l against T² for a simple pendulum is a straight line through the origin. Statement II: Its slope is 4π²/g.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    l = (g/4π²)T², so the line passes through the origin but the slope is g/4π², not 4π²/g.

20. Given below are two statements: Statement I: Young's modulus is a property of the material, so it does not depend on the radius or length of the wire. Statement II: The load on the wire must not exceed the elastic limit during the experiment.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Y = stress/strain is a material property. Beyond the elastic limit the stress–strain relation is no longer linear.

21. Given below are two statements: Statement I: Water rises less in a capillary tube of smaller radius. Statement II: Adding detergent increases the surface tension of water.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    h ∝ 1/r, so a narrower tube gives a larger rise. Detergent lowers surface tension. Both statements are wrong.

22. Which of the following statements about the terminal velocity method for viscosity are correct? 1. The speed is measured only after the ball has reached uniform velocity. 2. The ball is released near the axis of a wide cylinder to reduce wall effects. 3. The terminal velocity of the same ball increases when the viscosity of the liquid increases.

    1. 1 and 2 only
    2. 1 and 3 only
    3. 2 and 3 only
    4. 1, 2 and 3
    Answer

    A. 1 and 2 only

    v = 2r²(ρ − σ)g/9η, so v decreases when η increases. Statement 3 is wrong.

23. Which of the following statements about the resonance tube experiment are correct? 1. The first resonance occurs when l₁ + e = λ/4. 2. The second resonance occurs when l₂ + e = 3λ/4. 3. The difference l₂ − l₁ equals λ/4.

    1. 1 and 2 only
    2. 2 and 3 only
    3. 1 and 3 only
    4. 1, 2 and 3
    Answer

    A. 1 and 2 only

    l₂ − l₁ = 3λ/4 − λ/4 = λ/2, so statement 3 is wrong.

24. Given below are two statements: Statement I: In the method of mixtures the heat capacity of the calorimeter is always ignored. Statement II: The calorimeter is lagged to reduce heat loss to the surroundings.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    The calorimeter's heat capacity is included unless it is negligible. Lagging reduces heat loss, so only Statement II is correct.

25. Given below are two statements: Statement I: When a uniform metre scale is balanced at its 50 cm mark, its own weight produces no torque about the knife edge. Statement II: At balance the sum of clockwise moments equals the sum of anticlockwise moments about the knife edge.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    The weight of a uniform scale acts at the 50 cm mark, at the pivot. Both statements are correct.

26. Which of the following statements about the simple pendulum experiment are correct? 1. The amplitude may be large, because the time period does not depend on amplitude. 2. The timing is started when the bob crosses the mean position. 3. A thin, inextensible thread and a small heavy bob are used.

    1. 1 and 2 only
    2. 2 and 3 only
    3. 1 and 3 only
    4. 1, 2 and 3
    Answer

    B. 2 and 3 only

    T is independent of amplitude only for small angles (a few degrees), so 1 is wrong.

27. Assertion (A): For a given main scale, the least count of a vernier callipers decreases when the number of vernier divisions increases. Reason (R): Least count = value of one main scale division divided by the number of vernier divisions.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    LC = 1 MSD/N, so a larger N gives a smaller LC. R explains A.

28. Assertion (A): Young's modulus of a metal is the same for a thick and a thin wire of the same material. Reason (R): In Searle's method a reference wire carries a constant load to compensate for temperature change and sagging of the support.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    Both are true, but the reference wire does not explain why Y is the same for different thicknesses.

29. Assertion (A): For small angles, the time period of a simple pendulum does not depend on amplitude. Reason (R): The time period of a simple pendulum depends on the mass of the bob.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    A is true. T = 2π√(l/g) is independent of the mass, so R is false.

30. Assertion (A): The heat capacity of the calorimeter is always ignored in the method of mixtures. Reason (R): The calorimeter also absorbs heat from the hot body.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    D. A is false but R is true

    A is false: its heat capacity is included unless negligible. R is true.

31. Assertion (A): A screw gauge of pitch 1 mm with 100 circular divisions has a least count of 0.01 mm. Reason (R): Least count equals pitch multiplied by the number of circular divisions.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    LC = 1 mm/100 = 0.01 mm, so A is true. LC = pitch divided by the number of divisions, so R is false.

32. Assertion (A): Detergent added to water increases the capillary rise in a clean tube. Reason (R): Detergent lowers the surface tension of water.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    D. A is false but R is true

    Lower surface tension reduces the rise, so A is false. R is true.

33. Match List I with List II. List I: a. Vernier callipers b. Screw gauge c. Resonance tube d. Simple pendulum. List II: i. Depth of a vessel ii. Diameter of a thin wire iii. Speed of sound in air iv. Acceleration due to gravity

    1. a-i, b-ii, c-iv, d-iii
    2. a-i, b-ii, c-iii, d-iv
    3. a-ii, b-i, c-iii, d-iv
    4. a-iii, b-ii, c-i, d-iv
    Answer

    B. a-i, b-ii, c-iii, d-iv

    The vernier depth strip measures depth, the screw gauge measures thin wires, the resonance tube gives v, the pendulum gives g.

34. Match List I with List II. List I: a. Young's modulus of a wire b. Terminal velocity method c. Capillary rise d. Simple pendulum. List II: i. T = rhρg/2 ii. η = 2r²(ρ − σ)g/9v iii. Y = MgL/πr²l iv. g = 4π²l/T²

    1. a-ii, b-iii, c-i, d-iv
    2. a-iii, b-ii, c-iv, d-i
    3. a-iii, b-ii, c-i, d-iv
    4. a-iii, b-i, c-ii, d-iv
    Answer

    C. a-iii, b-ii, c-i, d-iv

    Each formula belongs to the experiment named in the list.

35. Match List I with List II. A screw gauge has 100 circular scale divisions and least count 0.01 mm. With the studs in contact, the reference line coincides with the following circular scale division. List I: a. 3rd division b. 96th division c. 97th division d. 5th division. List II (zero error): i. +0.03 mm ii. −0.04 mm iii. −0.03 mm iv. +0.05 mm

    1. a-i, b-iii, c-ii, d-iv
    2. a-i, b-ii, c-iii, d-iv
    3. a-i, b-ii, c-iv, d-iii
    4. a-iii, b-ii, c-i, d-iv
    Answer

    B. a-i, b-ii, c-iii, d-iv

    3rd gives +3 × 0.01 = +0.03. 96th gives −(100 − 96) × 0.01 = −0.04. 97th gives −0.03. 5th gives +0.05 mm.

36. Match List I with List II. List I: a. Method of mixtures b. Resonance tube c. Stokes' law d. Moments with a metre scale. List II: i. Heat lost = heat gained ii. e = (l₂ − 3l₁)/2 iii. F = 6πηrv iv. Clockwise moment = anticlockwise moment

    1. a-ii, b-i, c-iii, d-iv
    2. a-i, b-iii, c-ii, d-iv
    3. a-i, b-ii, c-iv, d-iii
    4. a-i, b-ii, c-iii, d-iv
    Answer

    D. a-i, b-ii, c-iii, d-iv

    Each relation is the principle or result of the named experiment.

37. Match List I with List II. List I: a. Viscosity of a liquid b. Young's modulus c. Surface tension of water d. Specific heat of a liquid. List II: i. Ball released along the axis of a wide jar ii. Wire loaded within the elastic limit iii. Clean capillary tube held vertical iv. Calorimeter lagged and the liquid stirred

    1. a-iii, b-ii, c-i, d-iv
    2. a-i, b-ii, c-iv, d-iii
    3. a-i, b-ii, c-iii, d-iv
    4. a-ii, b-i, c-iii, d-iv
    Answer

    C. a-i, b-ii, c-iii, d-iv

    Each precaution is the standard one for the named experiment.

38. Match List I with List II. Take π² ≈ 10. List I: a. Vernier with 20 divisions equal to 19 main divisions of 1 mm: least count b. Screw gauge, pitch 1 mm, 50 circular divisions: least count c. Simple pendulum of length 1.00 m and period 2.00 s: g d. Resonance tube, fork 340 Hz, l₂ − l₁ = 0.50 m: speed of sound. List II: i. 0.05 mm ii. 0.02 mm iii. 10 m s⁻² iv. 340 m s⁻¹

    1. a-i, b-ii, c-iv, d-iii
    2. a-i, b-ii, c-iii, d-iv
    3. a-i, b-iii, c-ii, d-iv
    4. a-ii, b-i, c-iii, d-iv
    Answer

    B. a-i, b-ii, c-iii, d-iv

    LC = 1/20 = 0.05 mm; 1/50 = 0.02 mm; g = 4π²l/T² = 40/4 = 10; v = 2 × 340 × 0.50 = 340 m s⁻¹.

39. A screw gauge has pitch 0.5 mm and 50 circular divisions. With the studs touching, the 46th circular division is in line with the reference line. A wire reads 1.5 mm on the pitch scale and 24 on the circular scale. The diameter of the wire is:

    1. 1.96 mm
    2. 1.54 mm
    3. 1.78 mm
    4. 1.74 mm
    Answer

    C. 1.78 mm

    LC = 0.5/50 = 0.01 mm. Zero error = −(50 − 46) × 0.01 = −0.04 mm. Corrected = 1.5 + 0.24 + 0.04 = 1.78 mm.

40. In a pendulum experiment the length is measured as 1.000 m with an uncertainty of 1 mm, and 20 oscillations take 40.0 s with an uncertainty of 0.2 s. The maximum percentage error in g is about:

    1. 0.6%
    2. 2.2%
    3. 0.5%
    4. 1.1%
    Answer

    D. 1.1%

    g = 4π²l/T². Δl/l = 0.1%. ΔT/T = 0.2/40 = 0.5%. Error = 0.1 + 2 × 0.5 = 1.1%.

41. A student measures the length of a pendulum from the suspension point to the top of a spherical bob and gets 96.0 cm. The bob has diameter 4.0 cm. The period is 2.0 s. Taking π² ≈ 10, the correct value of g is:

    1. 9.6 m s⁻²
    2. 9.4 m s⁻²
    3. 10.2 m s⁻²
    4. 9.8 m s⁻²
    Answer

    D. 9.8 m s⁻²

    Effective length = 96.0 + 2.0 = 98.0 cm = 0.98 m. g = 4π²l/T² = 4 × 10 × 0.98/4 = 9.8 m s⁻².

42. In a clean glass capillary tube water rises 3.0 cm. The experiment is repeated with a tube of half the radius and with detergent added that halves the surface tension of water. The angle of contact stays zero. The new rise is:

    1. 1.5 cm
    2. 6.0 cm
    3. 3.0 cm
    4. 12 cm
    Answer

    C. 3.0 cm

    h = 2T/(rρg) ∝ T/r. T halves and r halves, so T/r is unchanged and h stays 3.0 cm.

43. A steel ball of radius 1.0 mm reaches a terminal velocity of 1.0 cm s⁻¹ in a liquid. A steel ball of radius 2.0 mm is dropped in the same liquid. Its terminal velocity is:

    1. 4.0 cm s⁻¹
    2. 8.0 cm s⁻¹
    3. 2.0 cm s⁻¹
    4. 16 cm s⁻¹
    Answer

    A. 4.0 cm s⁻¹

    v ∝ r² for the same material and liquid. Radius doubled, so v becomes 4 × 1.0 = 4.0 cm s⁻¹.

44. In a resonance tube experiment with a 400 Hz fork the first resonance is at 20 cm. The end correction is 1.5 cm. The speed of sound is:

    1. 172 m s⁻¹
    2. 344 m s⁻¹
    3. 320 m s⁻¹
    4. 296 m s⁻¹
    Answer

    B. 344 m s⁻¹

    l₁ + e = λ/4 = 21.5 cm, so λ = 0.86 m. v = fλ = 400 × 0.86 = 344 m s⁻¹.

45. A copper ball of mass 100 g and specific heat 400 J kg⁻¹ K⁻¹ is heated to 100 °C and dropped into 200 g of a liquid at 20 °C in a calorimeter of negligible heat capacity. The final temperature is 30 °C. The specific heat of the liquid is:

    1. 2800 J kg⁻¹ K⁻¹
    2. 700 J kg⁻¹ K⁻¹
    3. 1400 J kg⁻¹ K⁻¹
    4. 5600 J kg⁻¹ K⁻¹
    Answer

    C. 1400 J kg⁻¹ K⁻¹

    0.1 × 400 × 70 = 0.2 × c × 10, so c = 2800/2 = 1400 J kg⁻¹ K⁻¹.