NEET Physics: Experimental Skills · Chapter 2

Experiments in Electricity, Optics and Electronics

What to remember

Metre bridge: S = R(100 − l)/l, with R in the left gap and S in the right gap and l measured from the left end. Keep the balance point near 50 cm, interchange R and S to remove end errors, and use ρ = SA/L = Sπd²/4L.
Half-deflection method: G = RS/(R − S) and figure of merit k = E/[(R + G)θ] (current per division).
Prism: the i–δ graph has a single minimum; at minimum deviation μ = sin[(A + δm)/2]/sin(A/2). Glass slab: μ = real depth/apparent depth.
Diodes: a p-n junction conducts in forward bias after a knee voltage (about 0.7 V for Si, about 0.3 V for Ge); a Zener diode is used in reverse bias and keeps a nearly constant voltage after breakdown.

Resistivity with a metre bridge

Aim. To find the resistivity of the material of a wire.

Principle. A metre bridge is a Wheatstone bridge with a uniform wire 100 cm long. A known resistance R is in the left gap and the unknown resistance S (the wire) is in the right gap. A jockey touches the wire at a point that gives zero deflection in the galvanometer. If l is the balance length from the left end:

R/S = l/(100 − l), so S = R(100 − l)/l.

Resistivity: ρ = S × A/L = S π d²/(4L), where d = diameter of the wire (screw gauge) and L = length of the wire (metre scale).

Readings. Use several values of R so that l lies between 40 cm and 60 cm. For each R note l, and then interchange R and S and note the new balance length. Calculate S each time and take the mean.

Why the null point is near the middle. The percentage error in S is proportional to 1/[l(100 − l)]. This is smallest at l = 50 cm, so the balance point should be close to the middle.

End errors. The copper strips at the ends have a small resistance, so the true balance lengths differ slightly from the observed ones. Interchanging R and S and averaging cancels most of this error.

Precautions. Insert the key only while taking a reading, because a continuous current heats the wire and changes its resistance. Tap the jockey lightly; do not slide or press it hard, as this makes the wire non-uniform. Clean and tighten all connections. Measure the diameter at several places along the wire.

Worked example 1. R = 40 Ω is in the left gap and the balance point is at 40.0 cm. S = 40 × 60/40 = 60 Ω.

Worked example 2. A wire of diameter 0.70 mm and length 0.55 m has S = 5.0 Ω. With π = 22/7: A = (22/7)(0.35 × 10⁻³)² = 3.85 × 10⁻⁷ m². ρ = 5.0 × 3.85 × 10⁻⁷/0.55 = 3.5 × 10⁻⁶ Ω m.

Ohm's law and the half-deflection method

Resistance of a wire by Ohm's law.

Principle: V = IR. The V–I graph of an ohmic conductor is a straight line through the origin. Resistance R = slope of the V–I graph = ΔV/ΔI (not the slope of the I–V graph, which is 1/R).
Circuit: a cell, a rheostat and a key in series with the wire and an ammeter (series). The voltmeter is connected in parallel across the wire only.
Readings: change the rheostat to get five or six pairs of V and I. Plot V on the y-axis and I on the x-axis.
Precautions: use small currents so that the wire does not heat (heating raises the resistance and bends the graph), check the zero errors of the meters, and close the key only while taking readings.

Galvanometer: resistance by half-deflection.

Aim: to find the resistance G of a galvanometer and its figure of merit k.
Circuit: a cell E, a high resistance box R in series with the galvanometer and a key K₁. A shunt resistance box S is connected across the galvanometer by a key K₂.
Procedure: with K₂ open, adjust R until the deflection θ is large (about full scale or a convenient number of divisions). Close K₂ and adjust S until the deflection becomes θ/2.
Result: G = RS/(R − S). If R is much larger than S, then G ≈ S.
Figure of merit: k = I/θ = E/[(R + G)θ], in A per division, where θ is the deflection with only R in series.
Precautions: R should be large (several thousand ohms) so that the current is small and nearly independent of G, E should be steady, and R should be kept the same while S is adjusted.

Worked example 3. R = 5100 Ω and S = 100 Ω for half deflection: G = 5100 × 100/5000 = 102 Ω. For E = 3 V, R = 2900 Ω, G = 100 Ω and θ = 25 divisions, k = 3/(3000 × 25) = 4.0 × 10⁻⁵ A per division.

Quantity	Formula	Remark
Resistance (Ohm's law)	R = ΔV/ΔI	Slope of the V–I graph
Galvanometer resistance	G = RS/(R − S)	S gives half deflection
Figure of merit	k = E/[(R + G)θ]	A per division

Focal length by the parallax method

General idea. Parallax is the relative shift between two objects at different distances when the eye is moved sideways. If the tip of the image and the tip of the object needle do not shift relative to each other, they are at the same position.

Concave mirror. Place a bright object pin on an optical bench in front of the mirror, beyond the focus. Place an image pin so that the real inverted image coincides with it (no parallax). Note u and v (distances from the pole, taken as positive magnitudes). f = uv/(u + v). For the plot of 1/u against 1/v the intercepts are each 1/f. The simplest value comes from u = v = 2f = R, when the pin and its image are side by side at the centre of curvature.

Convex lens. The same set-up with a lens in a holder. 1/f = 1/v − 1/u with signs; in magnitudes f = uv/(u + v). If u = v = 30 cm, f = 15 cm. The power of the lens in dioptres is P = 1/f (f in metres).

Convex mirror using a convex lens.

1. A convex lens forms a real image I of a pin on the bench.
2. A convex mirror is placed between the lens and I and moved until the pin and its image coincide (no parallax). Then the rays fall on the mirror normally, so I is at the centre of curvature of the mirror.
3. The distance of the mirror from I is R, so f = R/2.

For example, I at 40 cm from the lens and the mirror at 28 cm gives R = 12 cm and f = 6 cm.

Precautions. Keep the tips of the pins at the same height as the centre of the mirror or lens. Keep the eye about 25 cm or more from the pins. Remove the parallax tip to tip, not base to base. The optical bench should be horizontal and rigid. Measure from the pole of the mirror or the optical centre of the lens, and take a mean of several readings.

Worked example 4. Concave mirror: u = 30 cm and v = 60 cm. f = 30 × 60/90 = 20 cm. Check: 1/30 + 1/60 = 3/60 = 1/20.

Prism and glass slab

Prism: i–δ curve.

Aim: to plot the graph of the angle of deviation δ against the angle of incidence i for a triangular prism and to find the refractive index.
Description of the graph: as i increases from a small value, δ first falls, reaches a single minimum value δm, and then rises. For every δ larger than δm there are two angles of incidence (i and the angle of emergence e, interchanged), because the ray path is reversible.
At minimum deviation, i = e and r₁ = r₂ = A/2, and the ray inside an equilateral prism is parallel to the base.
Refractive index: μ = sin[(A + δm)/2]/sin(A/2).
Precautions: draw the incident ray with at least two pins at least 10 cm apart, use the same pin tips for the emergent ray, and keep the pins vertical.

Worked example 5. A = 60° and δm = 30°. μ = sin 45°/sin 30° = 0.707/0.5 = 1.41. For A = 60° and δm = 60°, μ = sin 60°/sin 30° = 1.73.

Glass slab with a travelling microscope.

Principle: a glass slab of real thickness t seems thinner when viewed from above. μ = real depth/apparent depth.
Procedure: put a mark on a sheet of paper and focus the microscope on it; read R₁. Place the slab over the mark and refocus on the image of the mark; read R₂. Sprinkle a little lycopodium powder on the top surface and focus on it; read R₃.
Real depth = R₃ − R₁. Apparent depth = R₃ − R₂. μ = (R₃ − R₁)/(R₃ − R₂).
The least count of the vernier scale of the microscope is typically 0.001 cm. Precautions: turn the screw in one direction only to avoid backlash error, and remove parallax between the cross-wire and the object.

Reading	Meaning	Example
R₁	Mark without the slab	2.000 cm
R₂	Image of the mark through the slab	2.800 cm
R₃	Top surface (powder)	4.400 cm
μ	(R₃ − R₁)/(R₃ − R₂)	2.400/1.600 = 1.50

Diodes: p-n junction and Zener

p-n junction diode characteristics.

Circuit for forward bias: a cell, a rheostat (potential divider), the diode with its p side towards the positive terminal, a milliammeter in series and a voltmeter across the diode. Keep a protective resistor in series.
Forward bias: the current is very small up to the knee (threshold) voltage and then rises rapidly. The knee is about 0.7 V for silicon and about 0.3 V for germanium.
Reverse bias: the p side is connected to the negative terminal. Only a very small leakage current flows (microamperes), so a microammeter is used. The depletion region becomes wider.
Graph: plot the I–V curve with the forward part on the right and the reverse part on the left. The two parts use different scales for current (mA and μA).
Precautions: do not exceed the rated current; keep the connections tight; increase the voltage in small steps.

Zener diode characteristic.

Circuit: the Zener diode is connected in reverse bias with a series resistor so that the current is limited.
Description of the curve: the reverse current is almost zero until the reverse breakdown voltage (Zener voltage Vz) is reached. Then the current rises sharply while the voltage stays almost constant.
Dynamic resistance in breakdown = ΔV/ΔI, which is small.
Use: voltage regulation, because the voltage across it stays near Vz.

Worked example 6. In breakdown the voltage rises from 5.60 V to 5.80 V while the current rises from 10 mA to 30 mA. Dynamic resistance = 0.20/0.020 = 10 Ω.

Diode	Bias used	Typical feature
p-n junction	Forward and reverse	Knee at about 0.7 V (Si), about 0.3 V (Ge)
Zener	Reverse	Sharp rise at breakdown voltage Vz

Identifying components with a multimeter

Set the multimeter to a resistance range and touch the probes to the two terminals, then reverse the probes.

Component	Behaviour on the resistance range
Resistor	The same reading in both directions; the value matches its marked or colour-code value
Diode	Low resistance in one polarity and very high resistance in the other
LED	Like a diode: conducts in one polarity only, and gives light when it conducts
Capacitor	The needle first swings towards low resistance and then moves back to a very high value; the effect repeats after reversing the probes

Precautions: do not test a component while it is connected in a circuit, and discharge a capacitor before testing.

Formula summary.

Experiment	Formula	Note
Metre bridge	S = R(100 − l)/l	l near 50 cm
Resistivity	ρ = Sπd²/4L	d by screw gauge
Ohm's law	R = ΔV/ΔI	Slope of V–I graph
Half deflection	G = RS/(R − S)	k = E/[(R + G)θ]
Mirror and lens	f = uv/(u + v)	Magnitudes of u and v
Convex mirror	f = R/2	R from the lens method
Prism	μ = sin[(A + δm)/2]/sin(A/2)	At minimum deviation
Slab	μ = real depth/apparent depth	R₃ − R₁ over R₃ − R₂
Zener	r = ΔV/ΔI	Small in breakdown

NEET traps

In the metre bridge formula, l is measured from the end of the wire next to the known resistance R. Swapping the sides inverts the ratio.
The null point is kept near 50 cm to reduce percentage error, not because the wire is thinnest there.
The slope of a V–I graph is R; the slope of an I–V graph is 1/R.
G = RS/(R − S), not R + S and not RS/(R + S).
The figure of merit is the current per division, not the divisions per ampere.
A convex mirror cannot form a real image of a real object; a convex lens is used to supply converging rays.
Parallax is tested by moving the eye sideways, not by keeping it fixed.
In the prism graph, δ does not rise continuously; it has one minimum. At the minimum i = e.
Apparent depth is smaller than real depth: μ = real/apparent, not apparent/real.
Silicon has the larger knee voltage (about 0.7 V); germanium has about 0.3 V.

One-liners

A metre bridge is a Wheatstone bridge with a 100 cm uniform wire.
The key of a metre bridge is plugged in only while taking readings.
Interchanging R and S reduces end errors.
Resistivity ρ = SA/L with A = πd²/4.
A voltmeter is always connected in parallel with the wire.
Large series resistance is used at the start of the half-deflection method.
Half deflection is obtained by adjusting the shunt S.
A concave mirror with u = v has u = v = R = 2f.
A convex mirror is tested with a convex lens.
At minimum deviation the ray in a prism is symmetric about the prism.
Lycopodium powder gives a fine mark on the top face of the slab.
A travelling microscope is moved with the screw in one direction to avoid backlash.
The reverse current of a p-n diode is in microamperes.
A Zener diode keeps nearly constant voltage after breakdown.
A capacitor reads low resistance first and then a very high resistance on a multimeter.

Practice questions

In a metre bridge a 40 Ω resistance is in the left gap and the unknown resistance S is in the right gap. The balance point is at 40.0 cm from the left end. S is:
1. 60 Ω
2. 26.7 Ω
3. 100 Ω
4. 40 Ω
Answer
A. 60 Ω
S = R(100 − l)/l = 40 × 60/40 = 60 Ω.
In the metre bridge experiment the balance point is kept near the middle of the wire because:
1. end errors vanish exactly at the middle
2. the current in the galvanometer is maximum there
3. the percentage error in the unknown resistance is least there
4. the wire is thinnest at the middle
Answer
C. the percentage error in the unknown resistance is least there
Percentage error in S is smallest when l(100 − l) is largest, which is at l = 50 cm.
A wire of diameter 0.70 mm and length 0.55 m has a resistance of 5.0 Ω found with a metre bridge. Take π = 22/7. The resistivity of its material is:
1. 3.5 × 10⁻⁹ Ω m
2. 3.5 × 10⁻⁶ Ω m
3. 1.4 × 10⁻⁵ Ω m
4. 7.0 × 10⁻⁶ Ω m
Answer
B. 3.5 × 10⁻⁶ Ω m
ρ = SA/L, A = π(0.35 × 10⁻³)² = 3.85 × 10⁻⁷ m². ρ = 5 × 3.85 × 10⁻⁷/0.55 = 3.5 × 10⁻⁶ Ω m.
In an Ohm's law experiment the V–I graph is a straight line through the origin. It passes through the points (I = 0.30 A, V = 1.5 V) and (I = 0.60 A, V = 3.0 V). The resistance is:
1. 4.5 Ω
2. 0.2 Ω
3. 2.5 Ω
4. 5 Ω
Answer
D. 5 Ω
R = slope of the V–I graph = ΔV/ΔI = 1.5/0.30 = 5 Ω.
To find the resistance of a wire using Ohm's law, the ammeter and voltmeter are connected respectively:
1. in series with the wire, and in parallel with the wire
2. in parallel with the wire, and in series with the wire
3. both in series with the wire
4. both in parallel with the wire
Answer
A. in series with the wire, and in parallel with the wire
The ammeter must carry the same current as the wire. The voltmeter measures the potential difference across it.
In the half-deflection method a resistance R = 5100 Ω is in series with the galvanometer, and a shunt S = 100 Ω across it reduces the deflection to half. The galvanometer resistance G is:
1. 98 Ω
2. 102 Ω
3. 5000 Ω
4. 51 Ω
Answer
B. 102 Ω
G = RS/(R − S) = 5100 × 100/5000 = 102 Ω.
The figure of merit of a galvanometer is:
1. the resistance of the galvanometer per division
2. the voltage needed for full-scale deflection
3. the number of divisions produced by one ampere
4. the current required to produce a deflection of one scale division
Answer
D. the current required to produce a deflection of one scale division
Figure of merit k = I/θ, the current per division.
In the u–v method with a concave mirror a real image is obtained at 60 cm from the mirror for an object at 30 cm. The focal length of the mirror is:
1. 45 cm
2. 40 cm
3. 20 cm
4. 10 cm
Answer
C. 20 cm
With magnitudes, f = uv/(u + v) = 30 × 60/90 = 20 cm. Check: 1/30 + 1/60 = 1/20.
With a convex lens, the needle and its real image are found at the same distance of 30 cm from the lens by the parallax method. The focal length of the lens is:
1. 60 cm
2. 30 cm
3. 10 cm
4. 15 cm
Answer
D. 15 cm
u = v = 2f = 30 cm, so f = 15 cm.
A convex lens forms a real image of a pin at 40 cm from the lens. A convex mirror is placed between the lens and the image, 28 cm from the lens, and the pin and its image are made to coincide. The focal length of the convex mirror is:
1. 3 cm
2. 6 cm
3. 12 cm
4. 34 cm
Answer
B. 6 cm
The image point is then the centre of curvature of the mirror. R = 40 − 28 = 12 cm, so f = R/2 = 6 cm.
A prism of angle 60° has a minimum deviation of 30°. Its refractive index is:
1. 1.41
2. 0.71
3. 0.52
4. 1.73
Answer
A. 1.41
μ = sin[(A + δm)/2]/sin(A/2) = sin 45°/sin 30° = 0.707/0.5 = 1.41.
The graph of angle of deviation against angle of incidence for a triangular prism:
1. is a straight line through the origin
2. falls continuously
3. falls to a single minimum value and then rises
4. rises continuously
Answer
C. falls to a single minimum value and then rises
Deviation decreases to a minimum value δm and then increases.
With a travelling microscope the readings are: focused on a mark on the table, 2.000 cm; focused on the image of the mark seen through a glass slab, 2.800 cm; focused on powder on the top face of the slab, 4.400 cm. The refractive index of the slab is:
1. 0.67
2. 1.50
3. 3.00
4. 1.25
Answer
B. 1.50
Real thickness = 4.400 − 2.000 = 2.400 cm. Apparent = 4.400 − 2.800 = 1.600 cm. μ = 2.400/1.600 = 1.50.
In the forward-bias characteristic of a silicon p-n junction diode, the current begins to rise sharply at a voltage of about:
1. 0 V exactly
2. 0.3 V
3. 5 V
4. 0.7 V
Answer
D. 0.7 V
The knee (threshold) voltage of silicon is about 0.7 V.
In the reverse breakdown region of a Zener diode, as the current increases the voltage across the diode:
1. stays almost constant
2. reverses its sign
3. falls to zero
4. rises in proportion to the current
Answer
A. stays almost constant
The breakdown curve is almost vertical: current changes a lot while the voltage stays near the Zener voltage.
A component is tested with a multimeter on the resistance range. The needle first shows a low resistance and then moves slowly to a very high resistance, and the effect repeats on reversing the probes. The component is:
1. a diode
2. a LED
3. a capacitor
4. a resistor
Answer
C. a capacitor
A capacitor charges from the meter battery, so the apparent resistance rises to a very high value.
Given below are two statements: Statement I: In a metre bridge, R and S are interchanged and the mean of the two results is taken to reduce end errors. Statement II: The jockey should be pressed hard and dragged along the wire.
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
Interchanging reduces end errors. The jockey must be tapped lightly, not dragged.
Given below are two statements: Statement I: In an Ohm's law experiment a large current is passed so that the wire heats up for accurate readings. Statement II: The V–I graph of a heated wire remains a straight line.
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
Heating changes the resistance, so the current should be small and the key closed only while taking readings. The graph then bends. Both statements are incorrect.
Given below are two statements: Statement I: The slope of an I–V graph of a wire gives its resistance. Statement II: The slope of a V–I graph of a wire gives its resistance.
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 slope of the I–V graph is 1/R. The slope of the V–I graph is R.
Given below are two statements: Statement I: In the half-deflection method the shunt S is adjusted until the deflection falls to half. Statement II: The galvanometer resistance is then G = R + S.
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
G = RS/(R − S), not R + S.
Which of the following statements about the parallax method are correct? 1. When there is no parallax, the tip of the image needle and the tip of the object needle do not move relative to each other when the eye is moved sideways. 2. The image needle is placed at the position of the real image. 3. The eye is kept fixed at one position while testing for parallax.
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
The eye is moved from side to side to test parallax, so 3 is wrong.
Given below are two statements: Statement I: For a prism, two different angles of incidence (other than the minimum deviation position) give the same deviation. Statement II: At minimum deviation, the ray inside an equilateral prism is parallel to its base.
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
By reversibility the angles of incidence and emergence can be interchanged. At minimum deviation the path is symmetric.
Given below are two statements: Statement I: The angle of deviation of a prism increases continuously as the angle of incidence increases. Statement II: At minimum deviation the angle of incidence equals the angle of emergence.
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
Deviation first decreases to a minimum and then increases. At the minimum, i = e.
Which of the following statements about the refractive index of a glass slab by a travelling microscope are correct? 1. Lycopodium powder is sprinkled on the top surface so that the microscope can be focused on it. 2. Apparent depth equals real depth multiplied by μ. 3. Real depth is found from the readings on the mark and on the top surface.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
C. 1 and 3 only
Apparent depth = real depth/μ, so 2 is wrong.
Given below are two statements: Statement I: A p-n junction diode in forward bias needs a series resistance to limit the current. Statement II: The knee voltage of a germanium diode is about 0.7 V and that of a silicon diode about 0.3 V.
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
Statement II has the values swapped: silicon about 0.7 V, germanium about 0.3 V.
Which of the following statements about identifying components with a multimeter are correct? 1. A resistor gives the same reading when the probes are reversed. 2. A diode gives a low resistance in one polarity and a very high resistance in the other. 3. An LED also conducts in one polarity only.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
D. 1, 2 and 3
All three statements are correct.
Assertion (A): In a metre bridge the balance point is kept near the middle of the wire. Reason (R): The percentage error in S = R(100 − l)/l is least when l is close to 50 cm.
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
Percentage error varies as 1/[l(100 − l)], which is smallest at l = 50 cm. R explains A.
Assertion (A): A Zener diode is used in reverse bias for voltage regulation. Reason (R): A silicon diode has a higher forward knee voltage than a germanium diode.
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 knee voltage has no bearing on the reverse breakdown use.
Assertion (A): A convex lens is used to find the focal length of a convex mirror. Reason (R): A convex mirror forms a real image of a real object.
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 convex mirror always forms a virtual image of a real object, so R is false. The lens supplies converging rays.
Assertion (A): A p-n junction diode carries a large current in reverse bias below breakdown. Reason (R): Reverse bias widens the depletion region.
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
Only a very small leakage current flows in reverse bias, so A is false. R is true.
Assertion (A): At minimum deviation, μ = sin[(A + δm)/2]/sin(A/2). Reason (R): At minimum deviation the path inside the prism is symmetric, with r₁ = r₂ = A/2.
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
With i = (A + δm)/2 and r = A/2, μ = sin i/sin r. R explains A.
Assertion (A): The dynamic resistance of a Zener diode in breakdown is very high. Reason (R): In breakdown the current changes sharply for a small change in voltage.
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
ΔV/ΔI is small in breakdown, so A is false. R is true.
Match List I with List II. List I: a. Resistor b. Diode c. Capacitor d. LED. List II: i. Same reading with either polarity of the probes ii. Conducts in one direction only iii. Needle swings towards low resistance, then returns to a very high value iv. Conducts in one direction only and gives light when it conducts
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-ii, b-i, c-iii, d-iv
Answer
B. a-i, b-ii, c-iii, d-iv
The resistor is non-polar, the diode and LED are one-way, and the capacitor charges.
Match List I with List II. List I: a. Metre bridge b. Half-deflection method c. Prism d. Glass slab with a travelling microscope. List II: i. S = R(100 − l)/l ii. G = RS/(R − S) iii. μ = sin[(A + δm)/2]/sin(A/2) iv. μ = real depth/apparent depth
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-i, b-iii, c-ii, d-iv
Answer
B. a-i, b-ii, c-iii, d-iv
Each formula belongs to the named experiment.
Match List I with List II. List I: a. Ohm's law, V against I b. Prism, deviation against incidence c. Forward characteristic of a p-n diode d. Reverse characteristic of a Zener diode. List II: i. Straight line through the origin ii. Curve with a single minimum iii. Almost no current at first, then a sharp rise beyond the knee voltage iv. Almost no current until a sudden sharp rise at the breakdown voltage
1. a-i, b-ii, c-iii, d-iv
2. a-iii, b-ii, c-i, d-iv
3. a-i, b-ii, c-iv, d-iii
4. a-ii, b-i, c-iii, d-iv
Answer
A. a-i, b-ii, c-iii, d-iv
Each shape follows from the physics of the device or experiment.
Match List I with List II. List I: a. Metre bridge, R = 10 Ω in the left gap, balance at 25 cm b. Half-deflection, R = 30 Ω, S = 10 Ω c. Prism, A = 60°, δm = 60°, μ d. Slab, real depth 3.0 cm, apparent depth 2.0 cm, μ. List II: i. 15 Ω ii. 30 Ω iii. 1.5 iv. 1.73
1. a-ii, b-i, c-iii, d-iv
2. a-ii, b-i, c-iv, d-iii
3. a-ii, b-iv, c-i, d-iii
4. a-i, b-ii, c-iv, d-iii
Answer
B. a-ii, b-i, c-iv, d-iii
a: S = 10 × 75/25 = 30 Ω. b: G = 30 × 10/20 = 15 Ω. c: sin 60°/sin 30° = 1.73. d: 3.0/2.0 = 1.5.
Match List I with List II. List I: a. Metre bridge b. Zener diode c. Travelling microscope d. Half-deflection method. List II: i. Keep the balance point near the middle ii. Keep the reverse current within the rated limit using a series resistor iii. Turn the screw in one direction only to avoid backlash error iv. Keep a large resistance in series at the start
1. a-i, b-ii, c-iii, d-iv
2. a-i, b-ii, c-iv, d-iii
3. a-ii, b-i, c-iii, d-iv
4. a-i, b-iii, c-ii, d-iv
Answer
A. a-i, b-ii, c-iii, d-iv
Each precaution belongs to the named experiment.
Match List I with List II. List I: a. Silicon diode, forward knee b. Germanium diode, forward knee c. Typical glass slab, refractive index d. Concave mirror of radius 30 cm, focal length. List II: i. about 0.3 V ii. about 0.7 V iii. 15 cm iv. about 1.5
1. a-ii, b-i, c-iii, d-iv
2. a-i, b-ii, c-iv, d-iii
3. a-iv, b-i, c-ii, d-iii
4. a-ii, b-i, c-iv, d-iii
Answer
D. a-ii, b-i, c-iv, d-iii
Silicon about 0.7 V, germanium about 0.3 V, glass about 1.5, f = R/2 = 15 cm.
In the breakdown region of a Zener diode the voltage rises from 5.60 V to 5.80 V while the current rises from 10 mA to 30 mA. The dynamic resistance is:
1. 0.1 Ω
2. 1.0 Ω
3. 10 Ω
4. 100 Ω
Answer
C. 10 Ω
R = ΔV/ΔI = 0.20 V/0.020 A = 10 Ω.
In a metre bridge the unknown resistance S is in the left gap and a 15 Ω resistance is in the right gap. The balance point is 70.0 cm from the left end. S is:
1. 6.4 Ω
2. 35 Ω
3. 50 Ω
4. 15 Ω
Answer
B. 35 Ω
S/15 = l/(100 − l) = 70/30, so S = 15 × 7/3 = 35 Ω.
A voltmeter reads 2.4 V and an ammeter reads 0.30 A for a wire of length 1.0 m and diameter 0.40 mm. Take π = 3.14. The resistivity of the wire material is about:
1. 4.0 × 10⁻⁶ Ω m
2. 2.5 × 10⁻⁷ Ω m
3. 1.0 × 10⁻⁶ Ω m
4. 1.0 × 10⁻⁹ Ω m
Answer
C. 1.0 × 10⁻⁶ Ω m
R = 2.4/0.30 = 8 Ω. A = π(0.2 × 10⁻³)² = 1.256 × 10⁻⁷ m². ρ = RA/L = 1.0 × 10⁻⁶ Ω m.
In a parallax experiment a needle at 15 cm from a convex lens has its real image at 30 cm from the lens. The power of the lens is:
1. 10 D
2. 0.1 D
3. 20 D
4. 5 D
Answer
A. 10 D
f = uv/(u + v) = 15 × 30/45 = 10 cm = 0.10 m. P = 1/f = 10 D.
A prism of angle 60° shows a minimum deviation of 37°. Given sin 48.5° = 0.75 and sin 30° = 0.50, the refractive index of the prism material is:
1. 1.33
2. 1.50
3. 0.67
4. 0.75
Answer
B. 1.50
i = (A + δm)/2 = 48.5°. μ = sin 48.5°/sin 30° = 0.75/0.50 = 1.50.
A pin below a glass slab of real thickness 3.0 cm appears to be raised by 1.0 cm when viewed from above through the slab. The refractive index of the glass is:
1. 3.0
2. 1.0
3. 2.0
4. 1.5
Answer
D. 1.5
Apparent depth = 3.0 − 1.0 = 2.0 cm. μ = 3.0/2.0 = 1.5.
A cell of emf 3 V is connected through a resistance of 2900 Ω to a galvanometer of resistance 100 Ω, giving a deflection of 25 divisions. The figure of merit is:
1. 4.0 × 10⁻⁵ A per division
2. 1.2 × 10⁻¹ A per division
3. 4.0 × 10⁻² A per division
4. 4.1 × 10⁻⁵ A per division
Answer
A. 4.0 × 10⁻⁵ A per division
I = 3/(2900 + 100) = 1.0 × 10⁻³ A. k = I/θ = 1.0 × 10⁻³/25 = 4.0 × 10⁻⁵ A per division.

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