JEE Physics Units and Measurements MCQs Set 11

Answer: (b) 1

Explanation:
1. Leading zeros only show the position of the decimal point and are not significant.
2. The only significant digit is 7, so there is 1 significant figure.

Teacher's Note:
1. Zeros before the first non-zero digit never count.
2. Writing 0.007 g as 7 mg shows clearly that the digit count is 1.

Answer: (a) 3

Explanation:
1. In scientific notation only the digits of the coefficient count.
2. Coefficient 2.64 has digits 2, 6 and 4, so 3 significant figures.

Teacher's Note:
1. The power of ten never affects the number of significant figures.
2. Converting units changes the power of ten but not the significant figures.

Answer: (d) 4

Explanation:
1. Leading zero before 2 is not significant.
2. Digits 2, 3, 7 and the trailing zero after the decimal point are significant, giving 4.

Teacher's Note:
1. Trailing zeros after a decimal point show the precision of the instrument.
2. So 0.2370 is more precise than 0.237.

Answer: (c) 4

Explanation:
1. The zeros before 6 are leading zeros and are not significant.
2. The zero between 6 and 3 is enclosed between non-zero digits so it counts.
3. Significant digits: 6, 0, 3, 2, which is 4.

Teacher's Note:
1. Zeros between non-zero digits are always significant.
2. Leading zeros are removed when the number is written in scientific notation: \( 6.032\times10^{-4} \).

Answer: (b) 5, 1, 2

Explanation:
1. 23.023: all five digits count because the zero lies between non-zero digits.
2. 0.0003: only the digit 3 counts, so 1.
3. \( 2.1\times10^{-3} \): digits 2 and 1, so 2.

Teacher's Note:
1. Apply the three rules in order: non-zero digits, zeros in between, and leading zeros.
2. Check each number separately.

Answer: (a) 44.0 kg

Explanation:
1. Multiply: 25 × 1.76 = 44.0 kg.
2. The number 25 here is an exact count, so the result keeps the 3 significant figures of 1.76.
3. Hence 44.0 kg.

Teacher's Note:
1. Exact counts have unlimited significant figures and do not restrict the answer.
2. 44 kg would show only two significant figures.

Answer: (d) \( 1.7\times10^{-6} \) m³

Explanation:
1. \( V=a^3=(1.2)^3\times10^{-6}=1.728\times10^{-6} \) m³.
2. The edge has 2 significant figures, so the volume is rounded to 2: \( 1.7\times10^{-6} \) m³.

Teacher's Note:
1. In multiplication the answer carries as many significant figures as the least accurate factor.
2. Powers count as repeated multiplication.

Answer: (c) 0.0393 kg

Explanation:
1. The first three significant digits are 3, 9, 2 and the next digit is 7.
2. Since 7 is greater than 5, the last kept digit goes up: 0.0393 kg.

Teacher's Note:
1. Count significant digits from the first non-zero digit.
2. The leading zeros are not counted.

Answer: (b) \( 4.08\times10^{8} \) s

Explanation:
1. The discarded digit is exactly 5 and the digit before it, 8, is even.
2. By the rule it is dropped without changing 8, giving \( 4.08\times10^{8} \) s.

Teacher's Note:
1. The even-odd rule avoids a systematic upward bias in repeated rounding.
2. If the preceding digit were odd, it would be raised by 1.

Answer: (a) \( 4.74\times10^{-6} \) kg

Explanation:
1. The discarded digit is exactly 5 and the preceding digit 3 is odd.
2. So the 3 is raised by 1 to make it even: 4.74.

Teacher's Note:
1. The even-odd rule applies only when the removed part is exactly 5 followed by nothing.
2. Always keep the power of ten unchanged.

Answer: (d) (0.17 ± 0.01) cm

Explanation:
1. Difference of diameters = 4.23 − 3.89 = 0.34 cm with error 0.01 + 0.01 = 0.02 cm.
2. Wall thickness is half of this difference: 0.17 cm with error 0.02/2 = 0.01 cm.

Teacher's Note:
1. Wall thickness is the difference of the radii, not of the diameters.
2. Errors in a difference add, and halving a quantity halves its absolute error.

Answer: (c) ± 0.2 cm²

Explanation:
1. Area = 10.0 × 1.00 = 10.0 cm².
2. \( \Delta A=l\,\Delta b+b\,\Delta l=10.0(0.01)+1.00(0.1)=0.2 \) cm².

Teacher's Note:
1. For a product, add the fractional errors: \( \frac{\Delta A}{A}=\frac{0.1}{10.0}+\frac{0.01}{1.00}=2\% \).
2. 2% of 10.0 cm² is 0.2 cm².

Answer: (b) 5%

Explanation:
1. \( \rho=m/a^3 \).
2. \( \frac{\Delta\rho}{\rho}=\frac{\Delta m}{m}+3\frac{\Delta a}{a}=2\%+3(1\%)=5\% \).

Teacher's Note:
1. A power in a formula multiplies the percentage error of that quantity.
2. Errors always add in the worst case, even for a quotient.

Answer: (a) ± 8%

Explanation:
1. \( \frac{\Delta g}{g}=\frac{\Delta L}{L}+2\frac{\Delta T}{T} \).
2. = 2% + 2(3%) = 8%.

Teacher's Note:
1. T appears squared, so its percentage error is doubled.
2. \( 4\pi^2 \) is an exact constant and adds no error.

Answer: (d) 0.8%

Explanation:
1. Error in the time period is the error of the total time divided by 20, so the fractional error is the same as for 25 s.
2. \( \frac{0.2}{25}\times100=0.8\% \).

Teacher's Note:
1. Dividing a measured time by the exact count 20 does not change the percentage error.
2. Measuring many oscillations lowers the error in one period in absolute terms.

Answer: (c) 5%

Explanation:
1. \( \frac{\Delta V}{V}=\frac{0.1}{5}+\frac{0.1}{10}+\frac{0.1}{5} \).
2. = 0.02 + 0.01 + 0.02 = 0.05, which is 5%.

Teacher's Note:
1. Each dimension contributes its own fractional error.
2. Small dimensions contribute larger fractional errors.

Answer: (b) 13%

Explanation:
1. \( \frac{\Delta t}{t}=\frac{\Delta x}{x}+2\frac{\Delta y}{y}+3\frac{\Delta z}{z} \).
2. = 1 + 2(3) + 3(2) = 13%.

Teacher's Note:
1. Powers in the numerator and denominator both add to the maximum error.
2. Never subtract errors for the worst case.

Answer: (a) 4%

Explanation:
1. \( \rho=\dfrac{m}{\pi r^2 l} \).
2. Fractional errors: m is 1%, r is 1% (counted twice) and l is 1%.
3. Total = 1 + 2(1) + 1 = 4%.

Teacher's Note:
1. Compute each percentage error first, then weight it by its power.
2. Here all three measurements have 1% error.

Answer: (d) 0.005 cm

Explanation:
1. Mean = (1.324 + 1.326 + 1.334 + 1.336)/4 = 1.330 cm.
2. Absolute deviations: 0.006, 0.004, 0.004, 0.006.
3. Mean absolute error = 0.020/4 = 0.005 cm.

Teacher's Note:
1. Absolute deviations are taken as positive numbers.
2. The mean error is the average of these deviations.

Answer: (c) 0.4%

Explanation:
1. Mean 1.330 cm and mean absolute error 0.005 cm.
2. Percentage error = (0.005/1.330)×100 ≈ 0.38%, which is about 0.4%.

Teacher's Note:
1. Relative error is the mean absolute error divided by the mean value.
2. Multiply by 100 to express it as a percentage.

Answer: (b) 2.5%

Explanation:
1. \( \frac{\Delta R}{R}=\frac{\Delta V}{V}+\frac{\Delta I}{I}=\frac{0.1}{5.0}+\frac{0.01}{2.0} \).
2. = 0.02 + 0.005 = 0.025, i.e. 2.5%.

Teacher's Note:
1. For a quotient, the fractional errors add.
2. The resistance value itself is 5.0/2.0 = 2.5 ohm.

Answer: (a) (10.0 ± 0.2) m/s²

Explanation:
1. \( g=4\pi^2 l/T^2=40(1.00)/4.00=10.0 \) m/s².
2. \( \frac{\Delta g}{g}=\frac{0.01}{1.00}+2\frac{0.01}{2.00}=2\% \).
3. \( \Delta g=0.02\times10.0=0.2 \) m/s².

Teacher's Note:
1. Compute the value and the percentage error separately.
2. Squared time doubles its fractional error.

Answer: (d) 1.3 m/s

Explanation:
1. \( \Delta v=2f_0(\Delta l_2+\Delta l_1)=2(325)(0.1+0.1)\text{ cm}\cdot\text{Hz} \).
2. = 130 cm/s = 1.3 m/s.
3. (Also v = 2(325)(0.500) = 325 m/s.)

Teacher's Note:
1. Errors in a difference add.
2. Convert 130 cm/s to 1.3 m/s before choosing.