Grade 11 : Physics - Units (Set 2) | Concept of Maths

Question 1

1. A technician uses a micrometer screw gauge with a least count of $0.01 \text{ mm}$ to measure the diameter of a thin copper wire. The main scale reading is $2.5 \text{ mm}$ and the $34^{th}$ division of the circular scale coincides with the reference line. If the zero error is $-0.03 \text{ mm}$, what is the correct diameter of the wire?

A) $2.84 \text{ mm}$

B) $2.81 \text{ mm}$

C) $2.87 \text{ mm}$

D) $2.53 \text{ mm}$

Answer

Correct Answer: C

Solution: Measured Reading = $MSR + (VSR \times LC)$ Measured Reading = $2.5 \text{ mm} + (34 \times 0.01 \text{ mm}) = 2.5 + 0.34 = 2.84 \text{ mm}$. Correct Reading = Measured Reading $-$ Zero Error Correct Reading = $2.84 - (-0.03) = 2.84 + 0.03 = 2.87 \text{ mm}$.

Question 2

2. An experimental drone engine produces power $P$. In the SI base units, what are the dimensions of power?

A) $[M L^2 T^{-3}]$

B) $[M L T^{-2}]$

C) $[M L^2 T^{-2}]$

D) $[M L^{-1} T^{-2}]$

Answer

Correct Answer: A

Solution: Power $P = \frac{\text{Work}}{\text{Time}}$. Work = $\text{Force} \times \text{Distance} = [M L T^{-2}] \times [L] = [M L^2 T^{-2}]$. Power = $\frac{[M L^2 T^{-2}]}{[T]} = [M L^2 T^{-3}]$.

Question 3

3. In a binary star system, the gravitational force $F$ between two stars of masses $m_1$ and $m_2$ separated by distance $r$ is $F = \frac{G m_1 m_2}{r^2}$. What is the dimensional formula for the Universal Gravitational Constant $G$?

A) $[M^1 L^3 T^{-2}]$

B) $[M^{-1} L^3 T^{-2}]$

C) $[M^{-1} L^2 T^{-2}]$

D) $[M L^2 T^{-1}]$

Question 4

4. A chemist weighs a sample of a rare isotope. The masses recorded are $12.050 \text{ g}$, $0.0040 \text{ g}$, and $1200 \text{ g}$. What are the number of significant figures in each measurement respectively?

A) 5, 2, 4

B) 4, 1, 2

C) 5, 2, 2

D) 5, 2, Undetermined (depends on notation)

Answer

Correct Answer: D

Solution: 1. $12.050$: All non-zero digits and zeros between them or trailing zeros after decimal are significant. Total = 5. 2. $0.0040$: Leading zeros are not significant; trailing zero after decimal is. Total = 2. 3. $1200$: Trailing zeros without a decimal point are ambiguous (usually not significant unless specified).

Question 5

5. The density of a metallic cube is determined by measuring its mass and the length of its side. If the maximum error in the measurement of mass is $2\%$ and the maximum error in the measurement of side length is $1\%$, what is the maximum percentage error in the density?

A) $3\%$

B) $4\%$

C) $5\%$

D) $6\%$

Answer

Correct Answer: C

Solution: Density $\rho = \frac{m}{V} = \frac{m}{l^3}$. The relative error is $\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3 \frac{\Delta l}{l}$. Max % error = $2\% + 3(1\%) = 5\%$.

Question 6

6. The energy $E$ of a photon is given by $E = h \nu$, where $\nu$ is the frequency. In an experiment involving the photoelectric effect, what are the SI units of Planck's constant $h$?

A) $\text{J s}^{-2}$

B) $\text{J s}$

C) $\text{W s}$

D) $\text{kg m}^2 \text{s}^{-1}$

Answer

Correct Answer: B

Solution: $E = h \nu \implies h = \frac{E}{\nu}$. Unit of $E$ is Joules (J), unit of $\nu$ is $1/\text{s}$ ($\text{Hz}$). Unit of $h = \frac{\text{J}}{1/\text{s}} = \text{J s}$. Alternatively, in base units: $[M L^2 T^{-2}] [T] = [M L^2 T^{-1}]$, which is $\text{kg m}^2 \text{s}^{-1}$. Both B and D are technically correct units, but J s is the standard SI unit notation.

Question 7

7. An astronomer measures the distance to a galaxy as $10^6$ parsecs. How many light years (ly) is this approximately? (Take $1 \text{ parsec} \approx 3.26 \text{ ly}$).

A) $3.26 \times 10^6 \text{ ly}$

B) $3.08 \times 10^{22} \text{ ly}$

C) $0.306 \times 10^6 \text{ ly}$

D) $9.46 \times 10^{15} \text{ ly}$

Answer

Correct Answer: A

Solution: Distance = $10^6 \text{ pc}$. $1 \text{ pc} = 3.26 \text{ ly}$. Distance in ly = $10^6 \times 3.26 = 3.26 \times 10^6 \text{ ly}$.

Question 8

8. A mechanic applies a torque $\tau$ to a bolt using a wrench. The torque is defined as the cross product of position vector and force. What is the dimensional formula of torque?

A) $[M L T^{-2}]$

B) $[M L^2 T^{-1}]$

C) $[M L^2 T^{-2}]$

D) $[M L^3 T^{-2}]$

Answer

Correct Answer: C

Solution: Torque $\tau = \vec{r} \times \vec{F}$. Dimensions: $[\tau] = [L] \times [M L T^{-2}] = [M L^2 T^{-2}]$. Note: Torque has the same dimensions as Work and Energy.

Question 9

9. In a lab experiment to find the acceleration due to gravity $g$ using a simple pendulum, the formula used is $T = 2\pi \sqrt{l/g}$. If the error in length $l$ is $1\%$ and the error in time period $T$ is $2\%$, the maximum error in $g$ is:

A) $3\%$

B) $4\%$

C) $1\%$

D) $5\%$

Answer

Correct Answer: D

Solution: $T = 2\pi \sqrt{l/g} \implies T^2 = 4\pi^2 \frac{l}{g} \implies g = 4\pi^2 \frac{l}{T^2}$. Relative error $\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2 \frac{\Delta T}{T}$. Max % error = $1\% + 2(2\%) = 1\% + 4\% = 5\%$.

Question 10

10. The viscous force $F$ acting on a sphere of radius $r$ moving with velocity $v$ through a fluid is $F = 6\pi \eta r v$. What are the SI units of the coefficient of viscosity $\eta$?

A) $\text{kg m s}^{-1}$

B) $\text{kg m}^{-1} \text{s}^{-1}$

C) $\text{kg m}^2 \text{s}^{-2}$

D) $\text{N m s}^{-2}$

Answer

Correct Answer: B

Solution: $\eta = \frac{F}{6\pi r v}$. Units: $\frac{[M L T^{-2}]}{[L][L T^{-1}]} = [M L^{-1} T^{-1}]$. In SI: $\text{kg m}^{-1} \text{s}^{-1}$.

Question 11

11. A physicist proposes a new formula for the speed of a wave in a string: $v = \sqrt{\frac{T^a}{\mu^b}}$, where $T$ is tension (force) and $\mu$ is linear mass density (mass per unit length). For the formula to be dimensionally correct, what must be the values of $a$ and $b$?

A) $a=1, b=-1$

B) $a=2, b=2$

C) $a=1, b=1$

D) $a=1/2, b=1/2$

Answer

Correct Answer: C

Solution: $[v] = [L T^{-1}]$. $[T] = [M L T^{-2}]$. $[\mu] = [M L^{-1}]$. Equating: $[L T^{-1}] = \left( \frac{[M L T^{-2}]^a}{[M L^{-1}]^b} \right)^{1/2}$. $[L^2 T^{-2}] = [M^{a-b} L^{a+b} T^{-2a}]$. From $T$: $-2 = -2a \implies a=1$. From $M$: $0 = a-b \implies b=a=1$.

Question 12

12. In nuclear physics, the radius of a nucleus is measured in Fermis (femtometers). If a nucleus has a radius of $5 \text{ fm}$, what is this value in meters?

A) $5 \times 10^{-15} \text{ m}$

B) $5 \times 10^{-12} \text{ m}$

C) $5 \times 10^{-9} \text{ m}$

D) $5 \times 10^{-18} \text{ m}$

Answer

Correct Answer: A

Solution: 1 Fermi (fm) = $10^{-15}$ meters. So, $5 \text{ fm} = 5 \times 10^{-15} \text{ m}$.

Question 13

13. The Solar Constant $S$ is defined as the solar energy incident per unit area per unit time on the Earth's upper atmosphere. What are the dimensions of $S$?

A) $[M L^2 T^{-3}]$

B) $[M L T^{-2}]$

C) $[M L^0 T^{-2}]$

D) $[M T^{-3}]$

Answer

Correct Answer: D

Solution: $S = \frac{\text{Energy}}{\text{Area} \times \text{Time}}$. Dimensions: $S = \frac{[M L^2 T^{-2}]}{[L^2][T]} = [M T^{-3}]$.

Question 14

14. During a collision, the impulse $J$ delivered to an object is the change in its momentum. Which of the following has the same dimensions as impulse?

A) Force

B) Momentum

C) Energy

D) Pressure

Answer

Correct Answer: B

Solution: Impulse $J = \int F dt = \Delta p$. Dimensions of Impulse = $[F][T] = [M L T^{-2}][T] = [M L T^{-1}]$. Dimensions of Momentum $p = mv = [M][L T^{-1}] = [M L T^{-1}]$.

Question 15

15. A surveyor measures three lengths: $15.42 \text{ m}$, $1.2 \text{ m}$, and $10.035 \text{ m}$. What is the total length expressed to the correct number of significant figures?

A) $26.7 \text{ m}$

B) $26.655 \text{ m}$

C) $26.66 \text{ m}$

D) $27 \text{ m}$

Answer

Correct Answer: A

Solution: Sum: $15.42 + 1.2 + 10.035 = 26.655$. Rule for addition: The result must be rounded to the same number of decimal places as the measurement with the fewest decimal places. $1.2$ has only 1 decimal place. So, $26.655$ rounds to $26.7$.

Question 16

16. The force between two long parallel wires carrying currents $I_1$ and $I_2$ is $F/L = \frac{\mu_0 I_1 I_2}{2\pi d}$. What are the dimensions of the permeability of free space $\mu_0$?

A) $[M L T^{-2} A^{-1}]$

B) $[M L^2 T^{-2} A^{-2}]$

C) $[M L T^{-2} A^{-2}]$

D) $[M L^0 T^{-2} A^{-1}]$

Answer

Correct Answer: C

Solution: $\mu_0 = \frac{2\pi d (F/L)}{I_1 I_2}$. Dimensions: $[\mu_0] = \frac{[L] ([M L T^{-2}]/[L])}{[A][A]} = \frac{[M L T^{-2}]}{[A^2]} = [M L T^{-2} A^{-2}]$.

Question 17

17. In fluid dynamics, the pressure gradient is defined as the change in pressure per unit length. Its dimensional formula is:

A) $[M L^{-1} T^{-2}]$

B) $[M L^{-2} T^{-2}]$

C) $[M L^2 T^{-2}]$

D) $[M L^{-2} T^{-1}]$

Answer

Correct Answer: B

Solution: Pressure Gradient = $\frac{dP}{dx}$. Pressure $P = [M L^{-1} T^{-2}]$. Distance $x = [L]$. Dimension = $\frac{[M L^{-1} T^{-2}]}{[L]} = [M L^{-2} T^{-2}]$.

Question 18

18. A physical quantity $Z$ is related to four observables $a, b, c$ and $d$ as $Z = \frac{a^2 b^3}{c \sqrt{d}}$. The percentage errors of measurement in $a, b, c$ and $d$ are $1\%, 3\%, 2\%$ and $4\%$ respectively. What is the percentage error in $Z$?

A) $15\%$

B) $10\%$

C) $13\%$

D) $12\%$

Answer

Correct Answer: A

Solution: $\frac{\Delta Z}{Z} = 2\frac{\Delta a}{a} + 3\frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{1}{2}\frac{\Delta d}{d}$. % error = $2(1\%) + 3(3\%) + 2\% + \frac{1}{2}(4\%)$ % error = $2 + 9 + 2 + 2 = 15\%$.

Question 19

19. In an electric circuit, the potential difference $V$ is measured. What are the dimensions of $V$ in terms of base SI units ($M, L, T, A$)?

A) $[M L^2 T^{-3} A^{-2}]$

B) $[M L T^{-3} A^{-1}]$

C) $[M L^2 T^{-3} A^{-1}]$

D) $[M L^2 T^{-2} A^{-1}]$

Answer

Correct Answer: C

Solution: $V = \frac{\text{Work}}{\text{Charge}} = \frac{W}{Q}$. $W = [M L^2 T^{-2}]$. $Q = I \times t = [A][T]$. $V = \frac{[M L^2 T^{-2}]}{[A T]} = [M L^2 T^{-3} A^{-1}]$.

Question 20

20. The surface tension of a soap solution is measured by the force acting per unit length of an imaginary line drawn on the surface. What are its dimensions?

A) $[M L T^{-2}]$

B) $[M L^0 T^{-1}]$

C) $[M L^{-1} T^{-2}]$

D) $[M T^{-2}]$

Answer

Correct Answer: D

Solution: Surface Tension $S = \frac{\text{Force}}{\text{Length}}$. Dimensions: $S = \frac{[M L T^{-2}]}{[L]} = [M T^{-2}]$.

Accepted Answer

Correct Answer: B

Solution: From $F = \frac{G m_1 m_2}{r^2}$, we have $G = \frac{F r^2}{m_1 m_2}$. Dimensions: $G = \frac{[M L T^{-2}][L^2]}{[M][M]} = \frac{[M L^3 T^{-2}]}{[M^2]} = [M^{-1} L^3 T^{-2}]$.

Grade 11 : Physics - Units (Set 2) | Concept of Maths