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Dimensions & Error Analysis

91 Questions Start Objective Physics Vol-1 Test
All Exams Objective Physics Vol-1 DPT
Q51 Objective Physics Vol-1 Significant Figures MCQ
24 Jul 2026
Concept: In multiplication and division, the final result must contain only as many significant figures as are present in the measurement with the least number of significant figures.
Multiply $107.88$ by $0.610$ and express the result with correct number of significant figures.
A.
$65.8068$
B.
$64.807$
C.
$65.81$
D.
$65.8$
Q52 Objective Physics Vol-1 Significant Figures MCQ
24 Jul 2026
Concept: Volume is calculated as $\text{Volume} = \text{length} \times \text{breadth} \times \text{thickness}$. When multiplying measured values, all quantities must be converted to the same unit, and the final result must be rounded off to the same number of significant figures as the measurement with the least number of significant figures.
The length, breadth and thickness of rectangular sheet of metal are $4.234\text{ m}$, $1.005\text{ m}$ and $2.01\text{ cm}$, respectively. The volume of the sheet upto correct significant figures is
A.
$0.0855\text{ m}^3$
B.
$0.086\text{ m}^3$
C.
$0.08556\text{ m}^3$
D.
$0.085\text{ m}^3$
Q53 Objective Physics Vol-1 Significant Figures MCQ
24 Jul 2026
Concept: The area of cross-section of a circular wire is given by $A = \pi r^2$. When multiplying or squaring measured values, the final result must be rounded off to the same number of significant figures as the measurement with the least number of significant figures.
The radius of a thin wire is $0.16\text{ mm}$. The area of cross-section of the wire (in $\text{mm}^2$) with correct number of significant figures is
A.
$0.08\text{ mm}^2$
B.
$0.080\text{ mm}^2$
C.
$0.0804\text{ mm}^2$
D.
$0.080384\text{ mm}^2$
Q54 Objective Physics Vol-1 Significant Figures MCQ
24 Jul 2026
Concept: In multiplication and division of measured values, the final result must be rounded off to have as many significant figures as are present in the measurement with the least number of significant figures.
When $97.52$ is divided by $2.54$, the correct result (considering significant figures) is
A.
$38.3937$
B.
$38.394$
C.
$65.81$
D.
$38.4$
Q55 Objective Physics Vol-1 Significant Figures MCQ
24 Jul 2026
Concept: To find the order of magnitude of a physical quantity, express it in scientific notation as $a \times 10^b$, where $1 \le a < 10$ or $0.5 \le a < 5$ depending on the convention. Standard rule for order of magnitude states that if $a \le 3.16$ ($\sqrt{10}$), then $a$ is rounded to $10^0$ and the order of magnitude is $b$. If $a > 3.16$, then $a$ is rounded to $10^1$ and the order of magnitude becomes $b + 1$.
What is the order of magnitude of $[(5.0 \times 10^{-6}) \times (5.0 \times 10^{-9})]$ with due regards to significant digits?
A.
$-14$
B.
$-15$
C.
$+15$
D.
$+1$
Q56 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
Concept: The Least Count (LC) of a vernier callipers is the difference between the length of one Main Scale Division (MSD) and one Vernier Scale Division (VSD):
$\text{LC} = 1\text{ MSD} - 1\text{ VSD}$
Alternatively, it can be calculated using the relation:
$\text{LC} = \frac{\text{Value of 1 MSD}}{\text{Total number of divisions on vernier scale}}$
In a vernier callipers, 1 main scale division is $1\text{ mm}$ and the 9th main scale division coincides with the 10th vernier scale division. Find the least count of the vernier callipers.
A.
$0.01\text{ mm}$
B.
$0.1\text{ mm}$
C.
$0.05\text{ mm}$
D.
$0.2\text{ mm}$
Q57 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
Concept: 1. The average or mean length ($x_{\text{mean}}$) is the arithmetic mean of all measured values.
2. The absolute error in each measurement is $\vert{}\Delta x_i\vert{} = \vert{}x_i - x_{\text{mean}}\vert{}$.
3. The mean absolute error ($\Delta x_{\text{mean}}$) is the average of the absolute errors.
4. The percentage error ($\delta x$) is calculated as $\frac{\Delta x_{\text{mean}}}{x_{\text{mean}}} \times 100\%$.
The length of a rod as measured in an experiment is found to be $2.48\text{ m}$, $2.46\text{ m}$, $2.49\text{ m}$, $2.49\text{ m}$ and $2.46\text{ m}$. Find the average length and the percentage error.
A.
$2.48\text{ m}$, $0.40\%$
B.
$2.46\text{ m}$, $0.50\%$
C.
$2.48\text{ m}$, $0.80\%$
D.
$2.50\text{ m}$, $0.40\%$
Q58 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
**Question 1**
Q59 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
Concept: 1. Mean value ($a_m$) is the arithmetic mean of all measurements.
2. Absolute error ($\Delta a_i$) in each measurement is the difference between the measured value and the mean value: $\Delta a_i = a_i - a_m$.
3. Mean absolute error ($\Delta a_{\text{mean}}$) is the arithmetic mean of the magnitudes of absolute errors: $\Delta a_{\text{mean}} = \frac{1}{n}\sum \vert{}\Delta a_i\vert{}$.
The diameter of a wire as measured by a screw gauge was found to be $2.620\text{ cm}$, $2.625\text{ cm}$, $2.630\text{ cm}$, $2.628\text{ cm}$ and $2.626\text{ cm}$. Find:
(i) mean value of diameter,
(ii) absolute error in each measurement, and
(iii) mean absolute error.
A.
$2.626\text{ cm}$; $0.006\text{ cm}, 0.001\text{ cm}, -0.004\text{ cm}, -0.002\text{ cm}, 0.000\text{ cm}$; $0.003\text{ cm}$
B.
$2.626\text{ cm}$; $-0.006\text{ cm}, -0.001\text{ cm}, +0.004\text{ cm}, +0.002\text{ cm}, 0.000\text{ cm}$; $0.003\text{ cm}$
C.
$2.625\text{ cm}$; $-0.005\text{ cm}, 0.000\text{ cm}, +0.005\text{ cm}, +0.003\text{ cm}, +0.001\text{ cm}$; $0.005\text{ cm}$
D.
$2.626\text{ cm}$; $-0.006\text{ cm}, -0.001\text{ cm}, +0.004\text{ cm}, +0.002\text{ cm}, 0.000\text{ cm}$; $0.010\text{ cm}$
Q60 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
Concept: 1. Fractional error $= \frac{\Delta a_{\text{mean}}}{a_m}$.
2. Percentage error $= \frac{\Delta a_{\text{mean}}}{a_m} \times 100\%$.
3. Expressing the result: Measured quantity $= a_m \pm \text{Percentage error}$.
For a measured diameter of a wire with mean value $2.626\text{ cm}$ and mean absolute error $0.003\text{ cm}$, calculate:
(iv) fractional error,
(v) percentage error, and
(vi) express the final result in terms of percentage error.
A.
$0.0011$; $0.11\%$; $(2.626 \pm 0.11\%)\text{ cm}$
B.
$0.003$; $0.30\%$; $(2.626 \pm 0.30\%)\text{ cm}$
C.
$0.0011$; $1.1\%$; $(2.626 \pm 1.1\%)\text{ cm}$
D.
$0.011$; $0.11\%$; $(2.626 \pm 0.003)\text{ cm}$
Q61 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
**Question 1**
Q62 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
Concept: 1. Mean value ($n_{\text{mean}}$) is the arithmetic mean of all measured values.
2. Absolute error ($\Delta n_i$) in each measurement is calculated as $\Delta n_i = n_i - n_{\text{mean}}$.
3. Mean absolute error ($\Delta n_{\text{mean}}$) is the arithmetic mean of the magnitudes of the absolute errors.
The refractive index ($n$) of glass is found to have the values $1.49$, $1.50$, $1.52$, $1.54$ and $1.48$. Calculate:
(i) the mean value of refractive index,
(ii) absolute error in each measurement, and
(iii) mean absolute error.
A.
$1.51$; $-0.02, -0.01, +0.01, +0.03, -0.03$; $0.02$
B.
$1.51$; $+0.02, +0.01, -0.01, -0.03, +0.03$; $0.02$
C.
$1.50$; $-0.01, 0.00, +0.02, +0.04, -0.02$; $0.03$
D.
$1.51$; $-0.02, -0.01, +0.01, +0.03, -0.03$; $0.05$
Q63 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
Concept: 1. Fractional error $= \frac{\Delta n_{\text{mean}}}{n_{\text{mean}}}$.
2. Percentage error $= \frac{\Delta n_{\text{mean}}}{n_{\text{mean}}} \times 100\%$.
3. Expressing the result: Measured quantity $= n_{\text{mean}} \pm \text{Percentage error}$.
For a measured refractive index of glass with mean value $1.51$ and mean absolute error $0.02$, calculate:
(iv) fractional error,
(v) percentage error, and
(vi) express the final result in terms of percentage error.
A.
$0.013$; $1.3\%$; $1.51 \pm 1.3\%$
B.
$0.020$; $2.0\%$; $1.51 \pm 2.0\%$
C.
$0.013$; $0.13\%$; $1.51 \pm 0.13\%$
D.
$0.130$; $1.3\%$; $1.51 \pm 0.02$
Q64 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
Concept: In the addition or subtraction of two physical quantities, the absolute error in the final result is equal to the sum of the absolute errors in the individual quantities.
If $Z = A + B$ or $Z = A - B$, then $\Delta Z = \pm (\Delta A + \Delta B)$.
The volumes of two bodies are measured to be $V_1 = (10.2 \pm 0.02)\text{ cm}^3$ and $V_2 = (6.4 \pm 0.01)\text{ cm}^3$. Calculate the sum and difference in volumes with error limits.
A.
$(16.6 \pm 0.03)\text{ cm}^3$ and $(3.8 \pm 0.03)\text{ cm}^3$
B.
$(16.6 \pm 0.01)\text{ cm}^3$ and $(3.8 \pm 0.01)\text{ cm}^3$
C.
$(16.6 \pm 0.03)\text{ cm}^3$ and $(3.8 \pm 0.01)\text{ cm}^3$
D.
$(16.6 \pm 0.02)\text{ cm}^3$ and $(3.8 \pm 0.02)\text{ cm}^3$
Q65 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
Concept: 1. The mirror formula relates focal length ($f$), object distance ($u$), and image distance ($v$):
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \implies f = \frac{uv}{u + v}$
2. To find the error in focal length ($\Delta f$), differentiate the mirror formula:
$\frac{\Delta f}{f^2} = \frac{\Delta u}{u^2} + \frac{\Delta v}{v^2} \implies \Delta f = f^2 \left(\frac{\Delta u}{u^2} + \frac{\Delta v}{v^2}\right)$
Calculate the focal length of a spherical mirror from the following observations: Object distance $u = (50.1 \pm 0.5)\text{ cm}$ and image distance $v = (20.1 \pm 0.2)\text{ cm}$. Express the focal length with error limits.
A.
$(14.3 \pm 0.15)\text{ cm}$
B.
$(14.3 \pm 0.05)\text{ cm}$
C.
$(14.3 \pm 0.25)\text{ cm}$
D.
$(14.3 \pm 0.50)\text{ cm}$
Q66 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
Concept: 1. The surface area of a sphere is given by $S = 4\pi r^2$.
2. For a quantity $S = k r^n$ where $k$ is a constant, the fractional error is given by $\frac{\Delta S}{S} = n \left(\frac{\Delta r}{r}\right)$, which leads to the absolute error $\Delta S = 2 \left(\frac{\Delta r}{r}\right) S$.
The radius of a sphere is measured to be $(2.1 \pm 0.5)\text{ cm}$. Calculate its surface area with error limits.
A.
$(55.4 \pm 26.4)\text{ cm}^2$
B.
$(55.4 \pm 13.2)\text{ cm}^2$
C.
$(55.4 \pm 5.2)\text{ cm}^2$
D.
$(27.7 \pm 26.4)\text{ cm}^2$
Q67 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
Concept: Volume is calculated using the formula $V = \frac{m}{\rho}$, where $m$ is mass and $\rho$ is density. For quantities related by division, the maximum relative error in the calculated quantity is the sum of the relative errors in the individual measurements:
$\frac{\Delta V}{V} = \frac{\Delta m}{m} + \frac{\Delta \rho}{\rho}$
The mass and density of a solid sphere are measured to be $(12.4 \pm 0.1)\text{ kg}$ and $(4.6 \pm 0.2)\text{ kg m}^{-3}$ respectively. Calculate the volume of the sphere with error limits.
A.
$(2.7 \pm 0.05)\text{ m}^3$
B.
$(2.7 \pm 0.14)\text{ m}^3$
C.
$(2.7 \pm 0.28)\text{ m}^3$
D.
$(2.7 \pm 0.50)\text{ m}^3$
Q68 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
Concept: For small fractional changes, the percentage change in a physical quantity $X = k L^n$ (where $k$ and $n$ are constants) is related to the percentage change in length $L$ by:
$\frac{\Delta X}{X} \times 100\% = n \left(\frac{\Delta L}{L} \times 100\%\right)$
A thin copper wire of length $L$ increases in length by $2\%$ when heated from $T_1$ to $T_2$. If a copper cube having side $10L$ is heated from $T_1$ to $T_2$, what will be the percentage change in:
(i) area of one face of the cube, and
(ii) volume of the cube?
A.
$2\%$ and $4\%$
B.
$4\%$ and $6\%$
C.
$4\%$ and $8\%$
D.
$6\%$ and $6\%$
Q69 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
Concept: For a physical quantity $T = k l^a g^b$, where $k$ is a constant, the maximum percentage error in $T$ is given by the sum of the absolute power-weighted percentage errors of the individual variables:
$\frac{\Delta T}{T} \times 100\% = \vert{}a\vert{} \left(\frac{\Delta l}{l} \times 100\%\right) + \vert{}b\vert{} \left(\frac{\Delta g}{g} \times 100\%\right)$
Calculate the percentage error in the determination of the time period of a simple pendulum given by $T = 2\pi \sqrt{\frac{l}{g}}$, where $l$ and $g$ are measured with $\pm 1\%$ and $\pm 2\%$ errors, respectively.
A.
$3\%$
B.
$1.5\%$
C.
$2\%$
D.
$0.5\%$
Q70 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
Concept: For a physical quantity $Z = \frac{A^a B^b}{C^c D^d}$, the maximum relative percentage error is given by:
$\frac{\Delta Z}{Z} \times 100\% = a \left(\frac{\Delta A}{A} \times 100\%\right) + b \left(\frac{\Delta B}{B} \times 100\%\right) + c \left(\frac{\Delta C}{C} \times 100\%\right) + d \left(\frac{\Delta D}{D} \times 100\%\right)$
The relative error is the fractional value obtained by dividing the percentage error by $100$: $\frac{\Delta Z}{Z} = \frac{\text{Percentage Error}}{100}$.
Find the relative error in $Z$, if $Z = \frac{A^4 B^{1/3}}{C D^{3/2}}$ and the percentage error in the measurements of $A$, $B$, $C$ and $D$ are $4\%$, $2\%$, $3\%$ and $1\%$, respectively.
A.
$0.2116$
B.
$0.1621$
C.
$21.16$
D.
$0.02116$
Q71 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
Concept: The Least Count (LC) of a spherometer is calculated using the formula:
$\text{Least Count} = \frac{\text{Pitch}}{\text{Total number of divisions on the circular scale}}$
where Pitch is the distance moved by the spindle on the main scale for one complete rotation of the disc.
A spherometer has 100 equal divisions marked along the periphery of its disc and one full rotation of the disc advances on the main scale by $0.01\text{ cm}$. The least count of this system is
A.
$10^{-2}\text{ cm}$
B.
$10^{-4}\text{ cm}$
C.
$10^{-5}\text{ cm}$
D.
$10^{-1}\text{ cm}$
Q72 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
Concept: When calculating the arithmetic mean of measured values, the final result should be rounded off to the same number of decimal places as the measurement with the least number of decimal places (or reported according to standard rules of significant figures in addition/division).
Three measurements are made as $18.425\text{ cm}$, $7.21\text{ cm}$ and $5.0\text{ cm}$. The mean of measurements should be written as
A.
$10.212\text{ cm}$
B.
$10.21\text{ cm}$
C.
$10.22\text{ cm}$
D.
$10.2\text{ cm}$
Q73 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
Concept: The radius $r$ of a circle is related to its diameter $D$ by the linear relation:
$r = \frac{D}{2}$
Since $2$ is an exact constant without any measurement uncertainty, the fractional or percentage error in the radius is identical to the percentage error in the diameter:
$\frac{\Delta r}{r} \times 100\% = \frac{\Delta D}{D} \times 100\%$
If error in measuring diameter of a circle is $4\%$, the error in measuring radius of the circle would be
A.
$2\%$
B.
$8\%$
C.
$4\%$
D.
$1\%$
Q74 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
Concept: When two identical or different measured quantities are added together, the total magnitude is the sum of their individual values, and the absolute error in the result is the sum of the absolute errors of each individual measurement:
$L_{\text{net}} = L_1 + L_2$
$\Delta L_{\text{net}} = \pm (\Delta L_1 + \Delta L_2)$
The length of a rod is $(11.05 \pm 0.2)\text{ cm}$. What is the net length of the system of rods, when these two rods are joined side by side?
A.
$(22.1 \pm 0.05)\text{ cm}$
B.
$(22.1 \pm 0.4)\text{ cm}$
C.
$(22.10 \pm 0.05)\text{ cm}$
D.
$(22.1 \pm 0.1)\text{ cm}$
Q75 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
Concept: 1. Velocity is calculated as $v = \frac{s}{t}$, where $s$ is distance and $t$ is time.
2. For division, the relative error in velocity is the sum of the relative errors in distance and time:
$\frac{\Delta v}{v} = \frac{\Delta s}{s} + \frac{\Delta t}{t}$
3. The absolute error is given by $\Delta v = v \left(\frac{\Delta s}{s} + \frac{\Delta t}{t}\right)$.
A body travels uniformly a distance of $(13.8 \pm 0.2)\text{ m}$ in a time $(4.0 \pm 0.3)\text{ s}$. The velocity of the body within error limit is
A.
$(3.45 \pm 0.2)\text{ ms}^{-1}$
B.
$(3.45 \pm 0.5)\text{ ms}^{-1}$
C.
$(3.45 \pm 0.4)\text{ ms}^{-1}$
D.
$(3.45 \pm 0.3)\text{ ms}^{-1}$
Q76 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
Concept: For a physical quantity $V = k l^n$ (where $k$ is a constant and $n$ is a power), the relative percentage error in $V$ is related to the relative percentage error in $l$ by:
$\frac{\Delta V}{V} \times 100\% = n \left(\frac{\Delta L}{l} \times 100\%\right)$
A cuboid has volume $V = l \times 2l \times 3l$, where $l$ is the length of one side. If the relative percentage error in the measurement of $l$ is $1\%$, then the relative percentage error in measurement of $V$ is
A.
$18\%$
B.
$6\%$
C.
$3\%$
D.
$1\%$
Q77 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
Concept: 1. Pressure is defined as force per unit area: $P = \frac{F}{A} = \frac{F}{L^2}$.
2. The maximum fractional or percentage error in a calculated quantity $P = F L^{-2}$ is given by the sum of the percentage errors:
$\frac{\Delta P}{P} \times 100\% = \frac{\Delta F}{F} \times 100\% + 2 \left(\frac{\Delta L}{L} \times 100\%\right)$
A force $F$ is applied on a square plate of side $L$. If the percentage error in the determination of $L$ is $2\%$ and that in $F$ is $4\%$, what is the permissible error in pressure?
A.
$8\%$
B.
$6\%$
C.
$4\%$
D.
$2\%$
Q78 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
Concept: Joule's law of heating states that heat generated is $H = I^2 R t$. For a quantity given by $H = I^a R^b t^c$, the maximum relative percentage error is calculated as the sum of the absolute power-weighted percentage errors:
$\frac{\Delta H}{H} \times 100\% = 2 \left(\frac{\Delta I}{I} \times 100\%\right) + 1 \left(\frac{\Delta R}{R} \times 100\%\right) + 1 \left(\frac{\Delta t}{t} \times 100\%\right)$
The heat generated in a wire depends directly on the resistance, current and time. If the error in measuring the above are $1\%$, $2\%$ and $1\%$, respectively. The maximum error in measuring the heat is
A.
$8\%$
B.
$6\%$
C.
$18\%$
D.
$12\%$
Q79 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
Concept: 1. Kinetic energy $K$ is related to momentum $p$ and mass $m$ by the formula:
$K = \frac{p^2}{2m}$
2. When the change in a variable is large (such as $100\%$), relative error formulas using differentiation do not apply. Instead, calculate the new value directly:
$p' = p + 100\% \text{ of } p = 2p$
3. Find the new kinetic energy $K'$ and compute the percentage increase:
$\text{Percentage Error} = \frac{K' - K}{K} \times 100\%$
If the error in the measurement of momentum of a particle is $(+100\%)$, then the error in the measurement of kinetic energy is
A.
$100\%$
B.
$200\%$
C.
$300\%$
D.
$400\%$
Q80 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
Concept: 1. The volume of a spherical ball is given by $V = \frac{4}{3}\pi r^3$.
2. For a quantity $V = k r^n$ where $k$ is a constant, the relative percentage error is given by $\frac{\Delta V}{V} \times 100\% = n \left(\frac{\Delta r}{r} \times 100\%\right)$.
The radius of a ball is $(5.2 \pm 0.2)\text{ cm}$. The percentage error in the volume of the ball is (approximately)
A.
$11\%$
B.
$4\%$
C.
$7\%$
D.
$9\%$
Q81 Objective Physics Vol-1 Error Analysis MCQ
24 Jul 2026
Concept: For two resistors $R_1$ and $R_2$ connected in parallel, the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$, or $R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$.
The absolute error in parallel combination is calculated using:
$\Delta R_{eq} = R_{eq}^2 \left(\frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}\right)$
The percentage error in equivalent resistance is $\frac{\Delta R_{eq}}{R_{eq}} \times 100\%$.
The values of two resistors are $(5.0 \pm 0.2)\text{ k}\Omega$ and $(10.0 \pm 0.1)\text{ k}\Omega$. What is the percentage error in the equivalent resistance when they are connected in parallel?
A.
$2\%$
B.
$5\%$
C.
$7\%$
D.
$3\%$
Q82 Objective Physics Vol-1 Chapter Exercise MCQ
24 Jul 2026
Concept: According to the principle of homogeneity of dimensions, physical quantities can be added or subtracted only if they have the same dimensions. Additionally, exponents in exponential functions must be dimensionless. Quantities with different dimensions can be multiplied or divided to form a new physical quantity.
If dimensions of $A$ and $B$ are different, then which of the following operation is valid?
A.
$A / B$
B.
$e^{-A/B}$
C.
$A - B$
D.
$A + B$
Q83 Objective Physics Vol-1 Chapter Exercise MCQ
24 Jul 2026
Concept: Significant figures reflect the precision of a measurement. Rules for determining significant figures:
1. All non-zero digits are significant.
2. Zeros preceding the first non-zero digit are leading zeros and are not significant.
3. Trailing zeros to the right of the decimal point in a measured quantity are significant.
4. Exponential factors ($10^n$) in scientific notation do not affect the number of significant figures.
The diameter of a wire is measured to be $0.0250 \times 10^{-4}\text{ m}$. The number of significant figures in the measurement is
A.
five
B.
four
C.
three
D.
nine
Q84 Objective Physics Vol-1 Chapter Exercise MCQ
24 Jul 2026
Concept: Electromotive force (emf) is defined as the work done per unit charge in moving a charge around a circuit, which is given by $V = W / q$. Electric potential is also defined as the work done per unit charge ($V = W / q$). Since both represent energy per unit charge, their dimensions are identical.
Dimensional formula for electromotive force is same as that for
A.
potential
B.
current
C.
force
D.
energy
Q85 Objective Physics Vol-1 Chapter Exercise MCQ
24 Jul 2026
Concept: Rules for determining significant figures:
1. All non-zero digits are significant.
2. Zeros preceding the first non-zero digit are leading zeros and are not significant.
3. Trailing zeros to the right of the decimal point are significant.
The number of significant figures in $0.06900$ is
A.
5
B.
4
C.
2
D.
3
Q86 Objective Physics Vol-1 Chapter Exercise MCQ
24 Jul 2026
Concept: When adding or subtracting decimal numbers, the result must be rounded off to the same number of decimal places as the quantity having the least number of decimal places (the least precise measurement).
The sum of the numbers $436.32$, $227.2$ and $0.301$ in appropriate significant figures is
A.
$663.821$
B.
$664$
C.
$663.8$
D.
$663.82$
Q87 Objective Physics Vol-1 Chapter Exercise MCQ
24 Jul 2026
Concept: Magnetic flux $\Phi$ is defined as the product of magnetic field $B$ and area $A$, given by $\Phi = B A \cos\theta$. Magnetic field $B$ can be derived from the Lorentz force equation $F = q v B$, which gives $B = F / (q v)$. Substituting $B$ into the flux formula and using base units gives the dimensional formula for magnetic flux as $[M L^2 T^{-2} A^{-1}]$.
The dimensional formula for magnetic flux is
A.
$[M L^2 T^{-2} A^{-1}]$
B.
$[M L^3 T^{-2} A^{-2}]$
C.
$[M^0 L^{-2} T^{-2} A^{-2}]$
D.
$[M L^2 T^{-1} A^2]$
Q88 Objective Physics Vol-1 Chapter Exercise MCQ
24 Jul 2026
Concept: The dimensional formula of a physical quantity shows how fundamental quantities like mass ($M$), length ($L$), and time ($T$) are combined to represent that quantity. By substituting the values of exponents $a$, $b$, and $c$, we can determine the dimensions of the quantity and compare them with standard physical quantities:
1. Force: $[M^1 L^1 T^{-2}]$
2. Pressure: $[M^1 L^{-1} T^{-2}]$
3. Velocity: $[M^0 L^1 T^{-1}]$
4. Acceleration: $[M^0 L^1 T^{-2}]$
If the dimensions of a physical quantity are given by $[M^a L^b T^c]$, then the physical quantity will be
A.
force, if $a = 0, b = -1, c = -2$
B.
pressure, if $a = 1, b = -1, c = -2$
C.
velocity, if $a = 1, b = 0, c = -1$
D.
acceleration, if $a = 1, b = 1, c = -2$
Q89 Objective Physics Vol-1 Chapter Exercise MCQ
24 Jul 2026
Concept: According to Coulomb's Law, the electrostatic force between two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by $F = \frac{k \cdot q_1 \cdot q_2}{r^2}$. Rearranging this equation to solve for the constant $k$ gives $k = \frac{F \cdot r^2}{q_1 \cdot q_2}$. Substituting the SI units for force ($\text{N}$), distance ($\text{m}$), and charge ($\text{C}$) yields the SI unit for $k$.
What is the units of $k = \frac{1}{4 \pi \varepsilon_0}$?
A.
$\text{C}^2\text{N}^{-1}\text{m}^{-2}$
B.
$\text{N-m}^2\text{C}^{-2}$
C.
$\text{N-m}^2\text{C}^2$
D.
Unitless
Q90 Objective Physics Vol-1 Chapter Exercise MCQ
24 Jul 2026
Concept: When multiplying or dividing physical quantities, the final result should retain as many significant figures as there are in the original measurement with the least number of significant figures.
The radius of a circle is $2.12\text{ m}$. Its area according to the rule of significant figures is
A.
$14.1124\text{ m}^2$
B.
$14.112\text{ m}^2$
C.
$14.11\text{ m}^2$
D.
$14.1\text{ m}^2$
Q91 Objective Physics Vol-1 Chapter Exercise MCQ
24 Jul 2026
Concept: According to Ohm's Law, potential difference is calculated as $V = I R$. When multiplying measured numbers, the final result must be rounded off to the same number of significant figures as the measurement with the fewest significant figures.
If the value of resistance is $10.845\ \Omega$ and the value of current is $3.23\text{ A}$, the value of potential with significant numbers would be
A.
$35.0\text{ V}$
B.
$3.50\text{ V}$
C.
$35.029\text{ V}$
D.
$35.03\text{ V}$