JEE Physics Errors in Measurement MCQs Set 02

Answer: (d) (αb₁ + βc₁ + γd₁ + δe₁)%

Explanation:
1. \( \frac{\Delta a}{a} = \alpha\frac{\Delta b}{b} + \beta\frac{\Delta c}{c} + \gamma\frac{\Delta d}{d} + \delta\frac{\Delta e}{e} \).
2. Quantities in the denominator contribute with a positive sign to the maximum error.

Teacher's Note:
1. Maximum error is always a sum of positive terms.
2. Each exponent multiplies the relative error of its quantity.

Answer: (d) 5%

Explanation:
1. \( g = 4\pi^2 l/T^2 \), so \( \frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T} \).
2. = 1% + 2(2%) = 5%.

Teacher's Note:
1. T appears squared, so its error counts twice.
2. 4π² is exact.

Answer: (c) C

Explanation:
1. Contribution of each quantity to the percentage error is |power| × its own percentage error.
2. Powers: A is 3, B is 1/2, C is 4 and D is 3/2. C has the largest power, so with equal individual errors it contributes the most.

Teacher's Note:
1. Here the individual percentage errors are taken to be equal.
2. The sign of the power is irrelevant for the maximum error.

Answer: (c) 1.96 m

Explanation:
1. \( h = \tfrac12 g t^2 \), so \( \Delta h = g t \Delta t \).
2. Δh = 9.8 × 2 × 0.1 = 1.96 m.

Teacher's Note:
1. Differentiating h = ½gt² gives Δh = g t Δt.
2. Here h = 19.6 m, so this is a 10% error.

Answer: (b) 5, 1, 2

Explanation:
1. 23.023: all five digits count (zeros between non-zero digits are significant).
2. 0.0003: leading zeros do not count, so 1.
3. 2.1 × 10⁻³: the digits before the power of ten, so 2.

Teacher's Note:
1. Powers of ten do not change the number of significant figures.
2. Leading zeros are only place holders.

Answer: (d) 1,29,600

Explanation:
1. 1 m/s² = \( \frac{(1/1000)\ \text{km}}{(1/3600)^2\ \text{h}^2} = \frac{3600^2}{1000} = 12960 \) km/h².
2. So 10 m/s² = 1,29,600 km/h².

Teacher's Note:
1. Convert length and time separately; time enters squared.
2. Hour is a much larger unit of time than the second, and time is squared, so the number grows by a factor of 3600² divided by 1000.

Answer: (c) (1.20 ± 0.18) cm

Explanation:
1. \( S = 2.0\cos 53^\circ = 1.20 \) cm.
2. \( \Delta S = \cos\theta\,\Delta x + x\sin\theta\,\Delta\theta \) with \( \Delta\theta = 2^\circ = 0.0349 \) rad.
3. = 0.60 × 0.2 + 2.0 × 0.80 × 0.0349 = 0.12 + 0.056 = 0.176 cm, about 0.18 cm.

Teacher's Note:
1. Angular errors must be converted to radian before using derivatives.
2. Absolute errors from independent terms add.

Answer: (c) 0.2%

Explanation:
1. \( \frac{\Delta g}{g} = \frac{\Delta l}{l} + 2\frac{\Delta T}{T} \).
2. Δl/l = 0.1 cm/100 cm = 0.1%.
3. Total time = 200 s with error 0.1 s, so ΔT/T = 0.1/200 = 0.05%.
4. Error in g = 0.1% + 2 × 0.05% = 0.2%.

Teacher's Note:
1. The error in T is the stop watch error divided over all 100 oscillations.
2. Always convert l to the same unit as its error.

Answer: (a) aα + bβ + cγ

Explanation:
1. For a product of powers, \( \frac{\Delta X}{X} = a\frac{\Delta M}{M} + b\frac{\Delta L}{L} + c\frac{\Delta T}{T} \).
2. In percentages this is aα + bβ + cγ (taking the exponents as magnitudes).

Teacher's Note:
1. The errors of different quantities never cancel in the maximum error.
2. Option (c) has the wrong form: errors are multiplied by, not divided by, the exponent.

Answer: (d) 14%

Explanation:
1. Error = 2(1%) + 3(3%) + 1(2%) + ½(2%) = 2 + 9 + 2 + 1 = 14%.

Teacher's Note:
1. √d means power 1/2.
2. Add all terms.

Answer: (a) 6%

Explanation:
1. Density = mass/volume, so percentage errors add.
2. 0.05/5.00 = 1% and 0.05/1.00 = 5%; total = 6%.

Teacher's Note:
1. Relative errors add for a quotient.
2. The volume error dominates.

Answer: (d) \( 2\pi\sqrt{\frac{M}{\eta L}} \)

Explanation:
1. Shear: restoring force \( F = \eta L^2 \cdot \frac{x}{L} = \eta L x \), so the force constant is \( k = \eta L \).
2. Hence \( T = 2\pi\sqrt{M/k} = 2\pi\sqrt{M/(\eta L)} \).
3. Dimension check: ηL has the unit N/m, so M/(ηL) has the unit s², as required.

Teacher's Note:
1. Modulus of rigidity = shear stress/shear strain, with shear strain x/L.
2. Dimensional analysis alone also selects option (d).

Answer: (d) 3.07 cm

Explanation:
1. Least count = 1 mm − 0.9 mm = 0.1 mm.
2. Zero error = + 7 × 0.1 = + 0.7 mm.
3. Main scale reading just before the vernier zero = 31 mm; vernier reading = 4 × 0.1 = 0.4 mm; observed = 31.4 mm.
4. Corrected length = 31.4 − 0.7 = 30.7 mm = 3.07 cm.

Teacher's Note:
1. Positive zero error is subtracted from the reading.
2. If the zero is slightly left of 3.2 cm, the main scale reading is 3.1 cm, not 3.2 cm.

Answer: (a) 3.64 mm

Explanation:
1. Least count = 1 mm/50 = 0.02 mm.
2. Zero error: with 44 on the line (before the zero is reached), the reading is 6 divisions short of 50, i.e. − 6 × 0.02 = − 0.12 mm.
3. Observed = 3 mm + 26 × 0.02 mm = 3.52 mm.
4. Corrected thickness = 3.52 − (− 0.12) = 3.64 mm.

Teacher's Note:
1. If the circular scale reads more than half the divisions when closed, the zero error is negative: subtract (50 − reading) divisions.
2. A negative zero error is added to the observed reading.

Answer: (b) \( 5\pi/18\ \% \)

Explanation:
1. \( \mu = \frac{\sin\frac{A+\delta}{2}}{\sin\frac{A}{2}} \). With A fixed, \( \frac{\Delta\mu}{\mu} = \frac12\cot\frac{A+\delta}{2}\,\Delta\delta \).
2. \( \frac{A+\delta}{2} = 45^\circ \), so \( \cot 45^\circ = 1 \) and \( \Delta\delta = 1^\circ = \frac{\pi}{180} \) rad.
3. \( \frac{\Delta\mu}{\mu} = \frac{\pi}{360} \), i.e. \( \frac{\pi}{360}\times 100 = \frac{5\pi}{18}\ \% \).

Teacher's Note:
1. Take logs and differentiate with respect to δ only because A is exact.
2. Convert the angular error to radian.

Answer: (a) A and C only

Explanation:
1. T = 40/20 = 2 s. The error in the total time is 1 s over 20 oscillations, so ΔT = 1/20 = 0.05 s: (A) is true, (B) false.
2. \( \frac{\Delta g}{g} = 2\frac{\Delta T}{T} = 2 \times \frac{0.05}{2} = 5\% \) (the length is exact): (C) is true, (D) false.

Teacher's Note:
1. The least count applies to the total time, so divide by the number of oscillations.
2. Only the T term contributes since l is exact.

Answer: (b) A and D only

Explanation:
1. Least count = 0.5 mm/50 = 0.01 mm: (A) true, (C) false.
2. Main scale reading = 5 × 0.5 mm = 2.5 mm (each division is one pitch).
3. Reading = 2.5 + 34 × 0.01 = 2.84 mm: (D) true, (B) false.

Teacher's Note:
1. On a screw gauge each main scale division equals one pitch.
2. Least count = pitch divided by the number of circular divisions.

Answer: (a) \( Q\left(n\frac{\Delta X}{X} + m\frac{\Delta Y}{Y}\right) \)

Explanation:
1. Take logs: \( \ln Q = n\ln X - m\ln Y \). Differentiating, \( \frac{dQ}{Q} = n\frac{dX}{X} - m\frac{dY}{Y} \).
2. For the maximum error all terms are taken with a positive sign, so ΔQ = Q(nΔX/X + mΔY/Y).

Teacher's Note:
1. Logarithmic differentiation gives the rule for any product of powers.
2. The sign is made positive for the worst case.

Answer: (b) 2 × 10⁻⁶ s

Explanation:
1. Error of each clock = 10⁻¹⁶ × 10¹⁰ s = 10⁻⁶ s.
2. In the worst case one runs fast and the other slow, so the difference is 2 × 10⁻⁶ s.

Teacher's Note:
1. Maximum difference of two readings equals the sum of their errors.
2. Do not forget that there are two clocks.

Answer: (b) 1 mm

Explanation:
1. Pitch = least count × number of divisions = 0.005 × 200 = 1 mm.

Teacher's Note:
1. Least count = pitch / number of circular divisions.
2. The same relation holds for the screw gauge.