Units & Measurements
If force (F), velocity (V) and time (T) are considered as fundamental physical quantity, then dimensional formula of density will be :
Given below are two statements :
Statements I : Astronomical unit (Au), Parsec (Pc) and Light year (ly) are units for measuring astronomical distances.
Statements II : $\mathrm{Au} < \mathrm{Parsec} (\mathrm{Pc}) < \mathrm{ly}$
In the light of the above statements, choose the most appropriate answer from the options given below:
In an experiment with vernier callipers of least count $0.1 \mathrm{~mm}$, when two jaws are joined together the zero of vernier scale lies right to the zero of the main scale and 6th division of vernier scale coincides with the main scale division. While measuring the diameter of a spherical bob, the zero of vernier scale lies in between $3.2 \mathrm{~cm}$ and $3.3 \mathrm{~cm}$ marks, and 4th division of vernier scale coincides with the main scale division. The diameter of bob is measured as
A physical quantity P is given as
$P = {{{a^2}{b^3}} \over {c\sqrt d }}$
The percentage error in the measurement of a, b, c and d are 1%, 2%, 3% and 4% respectively. The percentage error in the measurement of quantity P will be
Match List I with List II
| LIST I | LIST II | ||
|---|---|---|---|
| A. | Torque | I. | $\mathrm{ML^{-2}T^{-2}}$ |
| B. | Stress | II. | $\mathrm{ML^2T^{-2}}$ |
| C. | Pressure gradient | III. | $\mathrm{ML^{-1}T^{-1}}$ |
| D. | Coefficient of viscosity | IV. | $\mathrm{ML^{-1}T^{-2}}$ |
Choose the correct answer from the options given below:
Dimension of $\frac{1}{\mu_{0} \in_{0}}$ should be equal to
A cylindrical wire of mass $(0.4 \pm 0.01) \mathrm{g}$ has length $(8 \pm 0.04) \mathrm{cm}$ and radius $(6 \pm 0.03) \mathrm{mm}$. The maximum error in its density will be:
Two resistances are given as $\mathrm{R}_{1}=(10 \pm 0.5) \Omega$ and $\mathrm{R}_{2}=(15 \pm 0.5) \Omega$. The percentage error in the measurement of equivalent resistance when they are connected in parallel is -
If the velocity of light $\mathrm{c}$, universal gravitational constant $\mathrm{G}$ and Planck's constant $\mathrm{h}$ are chosen as fundamental quantities. The dimensions of mass in the new system is :
| LIST I | LIST II | ||
|---|---|---|---|
| A. | Angular momentum | I. | $\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]$ |
| B. | Torque | II. | $\left[\mathrm{ML}^{-2} \mathrm{~T}^{-2}\right]$ |
| C. | Stress | III | $\left[\mathrm{ML}^{2} \mathrm{~T}^{-1}\right]$ |
| D. | Pressure gradient | IV. | $\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$ |
Choose the correct answer from the options given below:
Match List I with List II:
| List I | List II | ||
|---|---|---|---|
| A. | Torque | I. | $\mathrm{kg} \mathrm{m}^{-1} \mathrm{~s}^{-2}$ |
| B. | Energy density | II. | $\mathrm{kg} \,\mathrm{ms}^{-1}$ |
| C. | Pressure gradient | III. | $\mathrm{kg}\, \mathrm{m}^{-2} \mathrm{~s}^{-2}$ |
| D. | Impulse | IV. | $\mathrm{kg} \,\mathrm{m}^{2} \mathrm{~s}^{-2}$ |
Choose the correct answer from the options given below:
The equation of a circle is given by $x^2+y^2=a^2$, where a is the radius. If the equation is modified to change the origin other than (0, 0), then find out the correct dimensions of A and B in a new equation : ${(x - At)^2} + {\left( {y - {t \over B}} \right)^2} = {a^2}$. The dimensions of t is given as $[\mathrm{T^{-1}]}$.
Match List I with List II :
| List I (Physical Quantity) | List II (Dimensional Formula) | ||
|---|---|---|---|
| A. | Pressure gradient | I. | $\left[\mathrm{M}^{\circ} \mathrm{L}^{2} \mathrm{~T}^{-2}\right]$ |
| B. | Energy density | II. | $\left[\mathrm{M}^{1} \mathrm{L}^{-1} \mathrm{~T}^{-2}\right]$ |
| C. | Electric Field | III. | $\left[\mathrm{M}^{1} \mathrm{L}^{-2} \mathrm{~T}^{-2}\right]$ |
| D. | Latent heat | IV. | $\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]$ |
Choose the correct answer from the options given below:
Match List I with List II
| List I | List II | ||
|---|---|---|---|
| A. | Young's Modulus (Y) | I. | $\mathrm{[ML^{-1}T^{-1}]}$ |
| B. | Co-efficient of Viscosity ($\eta$) | II. | $\mathrm{[ML^2T^{-1}]}$ |
| C. | Planck's Constant (h) | III. | $\mathrm{[ML^{-1}T^{-2}]}$ |
| D. | Work function ($\varphi $) | IV. | $\mathrm{[ML^2T^{-2}]}$ |
Choose the correct answer from the options given below :
Match List I with List II
| List-I |
List-II |
||
|---|---|---|---|
| A. | Planck's constant (h) | I. | $\mathrm{[{M^1}\,{L^2}\,{T^{ - 2}}]}$ |
| B. | Stopping potential (Vs) | II. | $\mathrm{[{M^1}\,{L^1}\,{T^{ - 1}}]}$ |
| C. | Work function ($\phi$) | III. | $\mathrm{[{M^1}\,{L^2}\,{T^{ - 1}}]}$ |
| D. | Momentum (p) | IV. | $\mathrm{[{M^1}\,{L^2}\,{T^{ - 3}}\,{A^{ - 1}}]}$ |
Choose the correct answer from the options given below :
In a screw gauge, there are 100 divisions on the circular scale and the main scale moves by $0.5 \mathrm{~mm}$ on a complete rotation of the circular scale. The zero of circular scale lies 6 divisions below the line of graduation when two studs are brought in contact with each other. When a wire is placed between the studs, 4 linear scale divisions are clearly visible while $46^{\text {th }}$ division the circular scale coincide with the reference line. The diameter of the wire is ______________ $\times 10^{-2} \mathrm{~mm}$.
Explanation:
Least count of screw gauge $ = {{0.5} \over {100}}$ mm $ = {{1} \over {200}}$ mm
Zero error of screw gauge $ = +{{6} \over {200}}$ mm $ = +{{3} \over {100}}=0.03$ mm
Reading of screw gauge $ = 4\times0.5+{{46} \over {200}}$ mm
$ = 2+{{23} \over {100}}$ mm $=2.23$ mm
So diameter of wire $=2.23$ mm $-~0.03$ mm
$=2.20$ mm
$=220\times10^{-2}$ mm
Explanation:
- Object distance $u = 10.0 \pm 0.1 \, \text{cm}$
- Image distance $v = 20.0 \pm 0.2 \, \text{cm}$
According to the lens formula for a thin lens :
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
So, we can calculate the focal length :
$ f = \frac{1}{\left(\frac{1}{v} - \frac{1}{u}\right)} = \frac{1}{\left(\frac{1}{20 \, \text{cm}} - \frac{1}{-10 \, \text{cm}}\right)} = \frac{1}{0.05 \, \text{cm}^{-1} + 0.1 \, \text{cm}^{-1}} = \frac{1}{0.15 \, \text{cm}^{-1}} = \frac{20}{3} \, \text{cm} $
Next, we need to calculate the error in the determination of the focal length. For that, we find the differential of the lens formula :
Then, the derivative of the equation gives us the change in the focal length $(df)$ in terms of the changes in the object distance $(du)$ and the image distance $(dv)$:
$-\frac{1}{v^2} dv + \frac{1}{u^2} du = \frac{-1}{f^2} df$
For maximum error, we get :
$\frac{1}{f^2} df = \frac{1}{v^2} dv + \frac{1}{u^2} du$
This equation tells us how errors in $(u)$ and $v$ propagate to an error in $f$. Now, when you compute the relative error in the focal length, you get :
$\frac{df}{f} \times 100 = \left(\frac{1}{v^2} dv + \frac{1}{u^2} du\right) \times \frac{f}{1} \times 100$
Plugging in your values of $u = 10 \mathrm{~cm}$, $du = 0.1 \mathrm{~cm}$, $v = 20 \mathrm{~cm}$, $dv = 0.2 \mathrm{~cm}$, and $f = 20/3 \mathrm{~cm}$, you indeed get:
$\frac{df}{f} \times 100 = \left(\frac{0.2}{20^2} + \frac{0.1}{10^2}\right) \times \frac{20}{3} \times 100 = 1\%$
So, the error in the focal length of the lens is indeed 1% (i.e., $n = 1$).
Match List I with List II.
| List I | List II | ||
|---|---|---|---|
| A. | Torque | I. | Nms$^{ - 1}$ |
| B. | Stress | II. | J kg$^{ - 1}$ |
| C. | Latent Heat | III. | Nm |
| D. | Power | IV. | Nm$^{ - 2}$ |
Choose the correct answer from the options given below :
Given below are two statements : One is labelled as Assertion (A) and other is labelled as Reason (R).
Assertion (A) : Time period of oscillation of a liquid drop depends on surface tension (S), if density of the liquid is $\rho$ and radius of the drop is r, then $\mathrm{T}=\mathrm{K} \sqrt{\rho \mathrm{r}^{3} / \mathrm{S}^{3 / 2}}$ is dimensionally correct, where K is dimensionless.
Reason (R) : Using dimensional analysis we get R.H.S. having different dimension than that of time period.
In the light of above statements, choose the correct answer from the options given below.
A travelling microscope has 20 divisions per $\mathrm{cm}$ on the main scale while its vernier scale has total 50 divisions and 25 vernier scale divisions are equal to 24 main scale divisions, what is the least count of the travelling microscope?
In an experiment to find out the diameter of wire using screw gauge, the following observations were noted :
(A) Screw moves $0.5 \mathrm{~mm}$ on main scale in one complete rotation
(B) Total divisions on circular scale $=50$
(C) Main scale reading is $2.5 \mathrm{~mm}$
(D) $45^{\text {th }}$ division of circular scale is in the pitch line
(E) Instrument has 0.03 mm negative error
Then the diameter of wire is :
Consider the efficiency of carnot's engine is given by $\eta=\frac{\alpha \beta}{\sin \theta} \log_e \frac{\beta x}{k T}$, where $\alpha$ and $\beta$ are constants. If T is temperature, k is Boltzmann constant, $\theta$ is angular displacement and x has the dimensions of length. Then, choose the incorrect option :
The dimensions of $\left(\frac{\mathrm{B}^{2}}{\mu_{0}}\right)$ will be :
(if $\mu_{0}$ : permeability of free space and $B$ : magnetic field)
An expression of energy density is given by $u=\frac{\alpha}{\beta} \sin \left(\frac{\alpha x}{k t}\right)$, where $\alpha, \beta$ are constants, $x$ is displacement, $k$ is Boltzmann constant and t is the temperature. The dimensions of $\beta$ will be :
A torque meter is calibrated to reference standards of mass, length and time each with $5 \%$ accuracy. After calibration, the measured torque with this torque meter will have net accuracy of :
In a Vernier Calipers, 10 divisions of Vernier scale is equal to the 9 divisions of main scale. When both jaws of Vernier calipers touch each other, the zero of the Vernier scale is shifted to the left of zero of the main scale and $4^{\text {th }}$ Vernier scale division exactly coincides with the main scale reading. One main scale division is equal to $1 \mathrm{~mm}$. While measuring diameter of a spherical body, the body is held between two jaws. It is now observed that zero of the Vernier scale lies between 30 and 31 divisions of main scale reading and $6^{\text {th }}$ Vernier scale division exactly coincides with the main scale reading. The diameter of the spherical body will be :
A screw gauge of pitch $0.5 \mathrm{~mm}$ is used to measure the diameter of uniform wire of length $6.8 \mathrm{~cm}$, the main scale reading is $1.5 \mathrm{~mm}$ and circular scale reading is 7 . The calculated curved surface area of wire to appropriate significant figures is :
[Screw gauge has 50 divisions on its circular scale]
The maximum error in the measurement of resistance, current and time for which current flows in an electrical circuit are $1 \%, 2 \%$ and $3 \%$ respectively. The maximum percentage error in the detection of the dissipated heat will be :
If momentum [P], area $[\mathrm{A}]$ and time $[\mathrm{T}]$ are taken as fundamental quantities, then the dimensional formula for coefficient of viscosity is :
If n main scale divisions coincide with (n + 1) vernier scale divisions. The least count of vernier callipers, when each centimetre on the main scale is divided into five equal parts, will be :
Velocity (v) and acceleration (a) in two systems of units 1 and 2 are related as ${v_2} = {n \over {{m^2}}}{v_1}$ and ${a_2} = {{{a_1}} \over {mn}}$ respectively. Here m and n are constants. The relations for distance and time in two systems respectively are :
The SI unit of a physical quantity is pascal-second. The dimensional formula of this quantity will be :
A silver wire has a mass (0.6 $\pm$ 0.006) g, radius (0.5 $\pm$ 0.005) mm and length (4 $\pm$ 0.04) cm. The maximum percentage error in the measurement of its density will be :
The dimension of mutual inductance is :
An expression for a dimensionless quantity P is given by $P = {\alpha \over \beta }{\log _e}\left( {{{kt} \over {\beta x}}} \right)$; where $\alpha$ and $\beta$ are constants, x is distance; k is Boltzmann constant and t is the temperature. Then the dimensions of $\alpha$ will be :
If $Z = {{{A^2}{B^3}} \over {{C^4}}}$, then the relative error in Z will be :
Identify the pair of physical quantities that have same dimensions:
Identify the pair of physical quantities which have different dimensions:
In an experiment to find acceleration due to gravity (g) using simple pendulum, time period of $0.5 \mathrm{~s}$ is measured from time of 100 oscillation with a watch of $1 \mathrm{~s}$ resolution. If measured value of length is $10 \mathrm{~cm}$ known to $1 \mathrm{~mm}$ accuracy, The accuracy in the determination of $\mathrm{g}$ is found to be $x \%$. The value of $x$ is ___________.
Explanation:
$T = 2\pi \sqrt {{l \over g}} $
${{dg} \over g} \times 100 = {{2dT} \over T} \times 100 + {{dl} \over l} \times 100$
$ = 2 \times {1 \over {50}} \times 100 + {1 \over {100}} \times 100 = 5\% $
The one division of main scale of Vernier callipers reads $1 \mathrm{~mm}$ and 10 divisions of Vernier scale is equal to the 9 divisions on main scale. When the two jaws of the instrument touch each other, the zero of the Vernier lies to the right of zero of the main scale and its fourth division coincides with a main scale division. When a spherical bob is tightly placed between the two jaws, the zero of the Vernier scale lies in between $4.1 \mathrm{~cm}$ and $4.2 \mathrm{~cm}$ and $6^{\text {th }}$ Vernier division coincides scale division. The diameter of the bob will be ____________ $\times$ 10$-$2 cm.
Explanation:
1 MSD = 1 mm
10 VSD = 9 MSD
LC = ${1 \over {10}}$ mm
0 + 4$\left( {{1 \over {10}}} \right)$ mm = 0.4 mm
Reading = 41 + 6$\left( {{1 \over {10}}} \right)$
= 41 + 0.6
= 41.6 mm
True reading = 41.2 mm
= 412 $\times$ 10$-$2 cm
The Vernier constant of Vernier callipers is 0.1 mm and it has zero error of ($-$0.05) cm. While measuring diameter of a sphere, the main scale reading is 1.7 cm and coinciding vernier division is 5. The corrected diameter will be _________ $\times$ 10$-$2 cm.
Explanation:
Since zero error is negative, we will add 0.05 cm.
$\Rightarrow$ Corrected reading = 1.7 cm + 5 $\times$ 0.1 mm + 0.05 cm
= 180 $\times$ 10$-$2 cm
A student in the laboratory measures thickness of a wire using screw gauge. The readings are 1.22 mm, 1.23 mm, 1.19 mm and 1.20 mm. The percentage error is ${x \over {121}}\% $. The value of x is ____________.
Explanation:
${I_{mean}} = {{1.22 + 1.23 + 1.19 + 1.20} \over 4} = 1.21$
$\Delta {I_{mean}} = {{0.01 + 0.02 + 0.02 + 0.01} \over 4} = 0.015$
So % $I = {{\Delta {I_{mean}}} \over {{I_{mean}}}} \times 100 = {{0.015} \over {1.21}} \times 100 = {{150} \over {121}}\% $
$x = 150$
A travelling microscope is used to determine the refractive index of a glass slab. If 40 divisions are there in 1 cm on main scale and 50 Vernier scale divisions are equal to 49 main scale divisions, then least count of the travelling microscope is __________ $\times$ 10$-$6 m.
Explanation:
40 M = 1 cm
$\Rightarrow$ M = 0.025 cm .......... (1)
Also, 50 V = 49 M
$\Rightarrow$ Least count = M $-$ V = M $-$ ${{49} \over {50}}$ M = ${{M} \over {50}}$
$\Rightarrow$ LC = ${{0.025} \over {50}}$ cm
= ${{250} \over {50}}$ $\times$ 10$-$6 m
$\Rightarrow$ LC = 5 $\times$ 10$-$6 m
In a vernier callipers, each cm on the main scale is divided into 20 equal parts. If tenth vernier scale division coincides with nineth main scale division. Then the value of vernier constant will be _________ $\times$ 10$-$2 mm.
Explanation:
1 MSD = ${1 \over {20}}$ cm
$\because$ 10 VSD = 9 MSD
1 VSD = ${9 \over {10}}$ $\times$ ${1 \over {20}}$ cm = ${9 \over {200}}$ $\times$ 10 mm = 0.45 mm
Now, 1 MSD = ${1 \over {20}}$ $\times$ 10 mm = 0.50 mm
LC = (0.50 $-$ 0.45) mm = 0.05 mm
= 5 $\times$ 10$-$2 mm
For $z = {a^2}{x^3}{y^{{1 \over 2}}}$, where 'a' is a constant. If percentage error in measurement of 'x' and 'y' are 4% and 12% respectively, then the percentage error for 'z' will be _______________%.
Explanation:
% error in $z = 3 \times 4 + {1 \over 2} \times 12$
$ = 12 + 6 = 18\% $
Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is $0.5 \mathrm{~mm}$. The circular scale has 100 divisions and for one full rotation of the circular scale, the main scale shifts by two divisions. The measured readings are listed below.
| Measurement condition | Main scale reading | Circular scale reading |
|---|---|---|
| Two arms of gauge touching each other without wire |
0 division | 4 divisions |
| Attempt-1: With wire | 4 divisions | 20 divisions |
| Attempt-2: With wire | 4 divisions | 16 divisions |
What are the diameter and cross-sectional area of the wire measured using the screw gauge?
In a particular system of units, a physical quantity can be expressed in terms of the electric charge $e$, electron mass $m_{e}$, Planck's constant $h$, and Coulomb's constant $k=\frac{1}{4 \pi \epsilon_{0}}$, where $\epsilon_{0}$ is the permittivity of vacuum. In terms of these physical constants, the dimension of the magnetic field is $[B]=[e]^{\alpha}\left[m_{e}\right]^{\beta}[h]^{\gamma}[k]^{\delta}$. The value of $\alpha+\beta+\gamma+\delta$ is _______.
Explanation:
| Physical Quantity |
Least count of the Equipment used for measurement |
Observed value |
|---|---|---|
| Mass (M) | 1 g | 2 kg |
| Length of bar (L) | 1 mm | 1 m |
| Breadth of bar (b) | 0.1 mm | 4 cm |
| Thickness of bar (d) | 0.01 mm | 0.4 cm |
| Depression ($\delta $) | 0.01 mm | 5 mm |
Then the fractional error in the measurement of Y is :