JEE Physics Units and Dimensions MCQs Set 07

Answer: (b) Pressure

Explanation:
1. Temperature, electric current and length are base quantities of the SI system.
2. Pressure is a derived quantity: force per unit area.

Teacher's Note:
1. Base quantities cannot be expressed through other quantities.
2. There are seven of them in SI.

Answer: (a) All of these

Explanation:
1. Unit symbols are written in lower case for the unit of a common name, e.g. s for second and u for unified atomic mass unit.
2. Newton is the name, N is the symbol; 'Sec' and 'a.m.u.' are not the standard symbols (s and u). So all three are incorrect.

Teacher's Note:
1. Symbols are not followed by a full stop.
2. Symbols of units named after persons start with a capital letter.

Answer: (c) Pascal

Explanation:
1. Ampere, candela and kelvin are SI base units.
2. Pascal = N m⁻² is a derived unit.

Teacher's Note:
1. Learn the seven SI base units.
2. Pascal is derived from newton and metre.

Answer: (a) \( \sqrt{AB} \)

Explanation:
1. Quantities with different dimensions can be multiplied, divided or have roots taken, but cannot be added or subtracted.
2. In \( A(1+B) \) the number 1 is added to B, which is possible only if B is dimensionless.

Teacher's Note:
1. Only like quantities can be added.
2. Products and roots are always allowed.

Answer: (d) \( E = \frac{P^2U}{2mv^2} \)

Explanation:
1. \( \frac{P^2}{m} \) and \( mv^2 \) both have dimensions of energy, so the first three are consistent.
2. \( \frac{P^2U}{mv^2} = \frac{m^2v^2U}{mv^2} = mU \), which is not energy.

Teacher's Note:
1. Check every term of an equation separately.
2. Energy is ML²T⁻².

Answer: (b) \( [ML^{11/2}T^{-2}] \)

Explanation:
1. B is added to \( x^2 \), so \( [B] = L^2 \).
2. \( [A] = [U][x^{2}]/[\sqrt x] = ML^2T^{-2}\cdot L^{3/2} = ML^{7/2}T^{-2} \).
3. \( [AB] = ML^{7/2+2}T^{-2} = ML^{11/2}T^{-2} \).

Teacher's Note:
1. Terms that are added have equal dimensions.
2. Rearrange the formula for A.

Answer: (c) All of these

Explanation:
1. Dimensional analysis cannot give dimensionless constants such as the factor 1/2 in (iii), nor sums of terms like (ii), nor trigonometric forms like (i).
2. So none of the three can be derived; they can only be checked.

Teacher's Note:
1. Dimensions only check consistency.
2. Pure numbers are invisible to dimensional analysis.

Answer: (d) \( [P^1A^{1/2}T^{-1}] \)

Explanation:
1. Let \( E = P^aA^bT^c \). Then \( ML^2T^{-2} = (MLT^{-1})^a(L^2)^bT^c \).
2. M: a = 1; L: a + 2b = 2 so b = 1/2; T: −a + c = −2 so c = −1.

Teacher's Note:
1. Equate the powers of M, L and T separately.
2. Write each new base quantity in M, L, T.

Answer: (b) 6310 m

Explanation:
1. The first, second and fourth values have 3 significant figures.
2. 6310 m has four significant figures (the trailing zero after the last non-zero digit is taken as significant here), so it is the most precisely specified.

Teacher's Note:
1. More significant figures mean a finer measurement.
2. Leading zeros are never significant.

Answer: (a) 7.3792 cm

Explanation:
1. The true value lies within ± half a least count: between 7.375 cm and 7.385 cm.
2. Only 7.3792 cm lies in this range.

Teacher's Note:
1. Reading error is of the order of the least count.
2. Check the interval 7.38 ± 0.005.

Answer: (c) 1.33%

Explanation:
1. \( R = \frac{R_1R_2}{R_1+R_2} = 2\,\Omega \) and \( \frac{\Delta R}{R^2} = \frac{\Delta R_1}{R_1^2}+\frac{\Delta R_2}{R_2^2} \).
2. \( \frac{\Delta R}{R} = \frac{R}{R_1}(1\%)+\frac{R}{R_2}(2\%) = \frac23+\frac23 = 1.33\% \).

Teacher's Note:
1. Differentiate 1/R = 1/R₁ + 1/R₂.
2. Weight each fractional error by R/Rᵢ.

Answer: (a) 4

Explanation:
1. Leading zeros are not significant.
2. The digits 1, 0, 0, 1 are significant, giving four.

Teacher's Note:
1. Zeros between non-zero digits count.
2. Zeros that only locate the decimal point do not.

Answer: (d) (31.4 ± 0.6) cm

Explanation:
1. Perimeter = 2(10.5 + 5.2) = 31.4 cm.
2. Error = 2(0.2 + 0.1) = 0.6 cm.

Teacher's Note:
1. Absolute errors add.
2. Do not forget the factor 2.

Answer: (b) 2.00 mm

Explanation:
1. All have three significant figures, so the relative precision is the same, but the absolute uncertainty (±0.005 of the unit) is smallest for millimetres.
2. Hence 2.00 mm is the most precise.

Teacher's Note:
1. Precision is about the smallest division read.
2. Compare the last digit's place value.

Answer: (c) \( \frac{1}{10N} \)

Explanation:
1. 1 VSD = \( \frac{N-1}{N} \) mm. Least count = 1 MSD − 1 VSD = \( \frac1N \) mm.
2. In cm: \( \frac{1}{10N} \) cm.

Teacher's Note:
1. Least count = one MSD minus one VSD.
2. Convert mm to cm at the end.

Answer: (d) 0.052 cm

Explanation:
1. Least count = 1 mm/100 = 0.01 mm. Reading = 0 + 52 × 0.01 = 0.52 mm.
2. 0.52 mm = 0.052 cm.

Teacher's Note:
1. Pitch divided by number of divisions is the least count.
2. Convert mm to cm.

Answer: (b) 5.124 cm

Explanation:
1. 1 MSD = 0.05 cm, 1 VSD = 2.45/50 = 0.049 cm, so least count = 0.001 cm.
2. Diameter = 5.10 + 24 × 0.001 = 5.124 cm.

Teacher's Note:
1. Main scale reading is the mark just left of the vernier zero.
2. Add n × least count.

Answer: (a) 1′

Explanation:
1. Least count = \( \frac{\text{1 MSD}}{30} = \frac{0.5^\circ}{30} = \frac{1^\circ}{60} = 1' \).

Teacher's Note:
1. Least count = MSD / number of vernier divisions.
2. 1° = 60′.

Answer: (c) B and C

Explanation:
1. Machine 1: values cluster (47.5–47.7) but are far from 50.0, so precise but less accurate.
2. Machine 2: mean about 50.2, close to 50.0, but scattered, so more accurate and less precise. Statements B and C are correct.

Teacher's Note:
1. Accuracy is closeness to the true value.
2. Precision is closeness of readings to each other.

Answer: (a) 2.6

Explanation:
1. Least count = 0.01 mm. Diameter = 1 + 47 × 0.01 = 1.47 mm = 0.147 cm.
2. Curved area = \( \pi d\,\ell = \pi\times0.147\times5.6 = 2.586\ \mathrm{cm^2} \).
3. The data have only 2 significant figures (5.6 cm), so the area is 2.6 cm².

Teacher's Note:
1. Result keeps the least number of significant figures.
2. Curved surface area = πdℓ.