Units & Measurements
230 Questions
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th July Morning Shift
Assertion A : If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is 5 mm and there are 50 total divisions on circular scale, then least count is 0.001 cm.
Reason R :
Least Count = ${{Pitch} \over {Total\,divisions\,on\,circular\,scale}}$
In the light of the above statements, choose the most appropriate answer from the options given below :
Reason R :
Least Count = ${{Pitch} \over {Total\,divisions\,on\,circular\,scale}}$
In the light of the above statements, choose the most appropriate answer from the options given below :
A.
A is not correct but R is correct.
B.
Both A and R are correct and R is the correct explanation of A.
C.
A is correct but R is not correct.
D.
Both A and R are correct and R is NOT the correct explanation of A.
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th July Evening Shift
The force is given in terms of time t and displacement x by the equation
F = A cos Bx + C sin Dt
The dimensional formula of ${{AD} \over B}$ is :
F = A cos Bx + C sin Dt
The dimensional formula of ${{AD} \over B}$ is :
A.
$[{M^0}L{T^{ - 1}}]$
B.
$[M{L^2}{T^{ - 3}}]$
C.
$[{M^1}{L^1}{T^{ - 2}}]$
D.
$[{M^2}{L^2}{T^{ - 3}}]$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 20th July Evening Shift
If time (t), velocity (v), and angular momentum (l) are taken as the fundamental units. Then the dimension of mass (m) in terms of t, v and l is :
A.
$[{t^{ - 1}}{v^1}{l^{ - 2}}]$
B.
$[{t^1}{v^2}{l^{ - 1}}]$
C.
$[{t^{ - 2}}{v^{ - 1}}{l^1}]$
D.
$[{t^{ - 1}}{v^{ - 2}}{l^1}]$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 17th March Morning Shift
The vernier scale used for measurement has a positive zero error of 0.2 mm. If while taking a measurement it was noted that '0' on the vernier scale lies between 8.5 cm and 8.6 cm, vernier coincidence is 6, then the correct value of measurement is ___________ cm. (least count = 0.01 cm)
A.
8.58 cm
B.
8.54 cm
C.
8.56 cm
D.
8.36 cm
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 16th March Evening Shift
In order to determine the Young's Modulus of a wire of radius 0.2 cm (measured using a scale of least count = 0.001 cm) and length 1m (measured using a scale of least count = 1 mm), a weight of mass 1 kg (measured using a scale of least count = 1 g) was hanged to get the elongation of 0.5 cm (measured using a scale of least count 0.001 cm). What will be the fractional error in the value of Young's Modulus determined by this experiment?
A.
0.14%
B.
9%
C.
1.4%
D.
0.9%
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 16th March Morning Shift
One main scale division of a vernier callipers is 'a' cm and nth division of the vernier scale coincide with (n $-$ 1)th division of the main scale. The least count of the callipers in mm is :
A.
${{10a} \over n}$
B.
${{10na} \over {(n - 1)}}$
C.
$\left( {{{n - 1} \over {10n}}} \right)a$
D.
${{10a} \over {(n - 1)}}$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Evening Shift
If 'C' and 'V' represent capacity and voltage respectively then what are the dimensions of $\lambda$ where C/V = $\lambda$ ?
A.
$[{M^{ - 3}}{L^{ - 4}}{I^3}{T^7}]$
B.
$[{M^{ - 2}}{L^{ - 3}}{I^2}{T^6}]$
C.
$[{M^{ - 2}}{L^{ - 4}}{I^3}{T^7}]$
D.
$[{M^{ - 1}}{L^{ - 3}}{I^{ - 2}}{T^{ - 7}}]$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Morning Shift
In a typical combustion engine the workdone by a gas molecule is given by $W = {\alpha ^2}\beta {e^{{{ - \beta {x^2}} \over {kT}}}}$, where x is the displacement, k is the Boltzmann constant and T is the temperature. If $\alpha$ and $\beta$ are constants, dimensions of $\alpha$ will be :
A.
$[{M^0}L{T^0}]$
B.
$[ML{T^{ - 1}}]$
C.
$[ML{T^{ - 2}}]$
D.
$[{M^2}L{T^{ - 2}}]$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Evening Shift
If e is the electronic charge, c is the speed of light in free space and h is Planck's constant, the quantity ${1 \over {4\pi {\varepsilon _0}}}{{|e{|^2}} \over {hc}}$ has dimensions of :
A.
$[ML{T^{ - 1}}]$
B.
$[ML{T^0}]$
C.
$[{M^0}{L^0}{T^0}]$
D.
$[L{C^{ - 1}}]$
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Morning Shift
Match List - I with List - II :
Choose the correct answer from the options given below :
| List I | List II | ||
|---|---|---|---|
| (a) | h (Planck's constant) | (i) | $[ML{T^{ - 1}}]$ |
| (b) | E (kinetic energy) | (ii) | $[M{L^2}{T^{ - 1}}]$ |
| (c) | V (electric potential) | (iii) | $[M{L^2}{T^{ - 2}}]$ |
| (d) | P (linear momentum) | (iv) | $[M{L^2}{I^{ - 1}}{T^{ - 3}}]$ |
Choose the correct answer from the options given below :
A.
(a) $ \to $ (ii), (b) $ \to $ (iii), (c) $ \to $ (iv), (d) $ \to $ (i)
B.
(a) $ \to $ (i), (b) $ \to $ (ii), (c) $ \to $ (iv), (d) $ \to $ (iii)
C.
(a) $ \to $ (iii), (b) $ \to $ (ii), (c) $ \to $ (iv), (d) $ \to $ (i)
D.
(a) $ \to $ (iii), (b) $ \to $ (iv), (c) $ \to $ (ii), (d) $ \to $ (i)
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th February Morning Shift
The pitch of the screw gauge is 1 mm and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while 72nd division on circular scale coincides with the reference line. The radius of the wire is :
A.
1.80 mm
B.
0.90 mm
C.
0.82 mm
D.
1.64 mm
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Morning Shift
The work done by a gas molecule in an isolated system is
given by, $W = \alpha {\beta ^2}{e^{ - {{{x^2}} \over {\alpha kT}}}}$, where x is the displacement, k is the Boltzmann constant and T is the temperature. $\alpha$ and $\beta$ are constants. Then the dimensions of $\beta$ will be :
given by, $W = \alpha {\beta ^2}{e^{ - {{{x^2}} \over {\alpha kT}}}}$, where x is the displacement, k is the Boltzmann constant and T is the temperature. $\alpha$ and $\beta$ are constants. Then the dimensions of $\beta$ will be :
A.
$[{M^0}L{T^0}]$
B.
$[M{L^2}{T^{ - 2}}]$
C.
$[ML{T^{ - 2}}]$
D.
$[{M^2}L{T^2}]$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Evening Shift
The diameter of a spherical bob is measured using a vernier callipers. 9 divisions of the main scale, in the vernier callipers, are equal to 10 divisions of vernier scale. One main scale division is 1 mm. The main scale reading is 10 mm and 8th division of vernier scale was found to coincide exactly with one of the main scale division. If the given vernier callipers has positive zero error of 0.04 cm, then the radius of the bob is ___________ $\times$ 10$-$2 cm.
Correct Answer: 52
Explanation:
To understand the measurement of the spherical bob's diameter using Vernier calipers, let's break it down step by step:
Vernier Scale Calculation:
We are given that 9 divisions on the main scale (MSD) are equivalent to 10 divisions on the Vernier scale (VSD).
One main scale division is 1 mm. Therefore, 9 MSD = 9 mm.
Since 9 mm = 10 VSD, one VSD equals $ \frac{9}{10} = 0.9 \, \text{mm} $.
Least Count (LC) of the Vernier Callipers:
The least count is the smallest measurable value and is calculated as:
$ \text{LC} = 1 \, \text{MSD} - 1 \, \text{VSD} = 1 \, \text{mm} - 0.9 \, \text{mm} = 0.1 \, \text{mm} $
Calculate the Initial Reading:
The main scale reading (MSR) is 10 mm.
The 8th division of the Vernier scale coincides with the main scale division, meaning the Vernier scale reading (VSR) is 8.
Total reading before zero error correction = MSR + VSR × LC:
$ 10 \, \text{mm} + 8 \times 0.1 \, \text{mm} = 10.8 \, \text{mm} $
Adjust for the Zero Error:
The Vernier calipers have a positive zero error of 0.04 cm (or 0.4 mm, as 1 cm = 10 mm).
To get the correct measurement, subtract the zero error:
$ 10.8 \, \text{mm} - 0.4 \, \text{mm} = 10.4 \, \text{mm} $
Compute the Radius:
The diameter calculated is 10.4 mm. To find the radius, divide by 2:
$ \text{radius} = \frac{10.4 \, \text{mm}}{2} = 5.2 \, \text{mm} $
Convert the radius from millimeters to centimeters:
$ 5.2 \, \text{mm} = 0.52 \, \text{cm} = 52 \times 10^{-2} \, \text{cm} $
The radius of the spherical bob is $ 52 \times 10^{-2} $ cm.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Evening Shift
The acceleration due to gravity is found upto an accuracy of 4% on a planet. The energy supplied to a simple pendulum to known mass 'm' to undertake oscillations of time period T is being estimated. If time period is measured to an accuracy of 3%, the accuracy to which E is known as ..............%
Correct Answer: 14
Explanation:
$T = 2\pi \sqrt {{l \over g}} \Rightarrow l = {{{T^2}g} \over {4{\pi ^2}}}$
$E = mgl{{{\theta ^2}} \over 2} = m{g^2}{{{T^2}{\theta ^2}} \over {8{\pi ^2}}}$
${{dE} \over E} = 2\left( {{{dg} \over g} + {{dT} \over T}} \right)$
$ = (4 + 3) = 14\% $
$E = mgl{{{\theta ^2}} \over 2} = m{g^2}{{{T^2}{\theta ^2}} \over {8{\pi ^2}}}$
${{dE} \over E} = 2\left( {{{dg} \over g} + {{dT} \over T}} \right)$
$ = (4 + 3) = 14\% $
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Morning Shift
Student A and student B used two screw gauges of equal pitch and 100 equal circular divisions to measure the radius of a given wire. The actual value of the radius of the wire is 0.322 cm. The absolute value of the difference between the final circular scale readings observed by the students A and B is ______________.
[Figure shows position of reference 'O' when jaws of screw gauge are closed]
Given pitch = 0.1 cm.
[Figure shows position of reference 'O' when jaws of screw gauge are closed]
Given pitch = 0.1 cm.
Correct Answer: 13
Explanation:
Zero error occurs when the screw gauge does not read zero when its jaws are closed. This error needs to be accounted for in the final measurement. The least count (LC) of a screw gauge, calculated as the pitch (p) divided by the number of divisions (N) on the circular scale, helps us determine the value of this error.
The least count of both screw gauges is LC = $p / N=0.1 / 100=0.001 \mathrm{~cm}$.
For screw gauge A, the zero error is calculated by adding the main scale reading (MSR) at zero and the circular scale reading (CSR) at zero, multiplied by the least count. Screw gauge A's MSR at zero is 0 and its CSR at zero is 5. Therefore, its zero error is ( 0 + 5 $\times$ 0.001 = 0.005 ) cm.
For screw gauge B, the zero error involves a negative main scale reading and a circular scale reading. The MSR at zero is -1 (since the main scale is 0.1 cm per division, this corresponds to -0.1 cm) and the CSR at zero is 92. Thus, the zero error for B is ( (-1 $\times$ 0.1) + 92 $\times$ 0.001 = -0.008 ) cm.
To find the difference between the final circular scale readings of screw gauges A and B, we look at the difference in their zero errors. The absolute difference is ( |0.005 - (-0.008)| = |0.005 + 0.008| = 0.013 ) cm. This value represents the absolute difference in the final readings of the two screw gauges, accounting for their individual zero errors.
However, to express this difference in terms of the circular scale divisions (since the least count is 0.001 cm per division), we multiply this difference by 1000 (since 1 cm = 1000 mm and each division represents 0.001 cm) :
$ \text{Difference in Divisions} = 0.013 \text{ cm} \times 1000 $
$ = 13 \text{ divisions} $
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 22th July Evening Shift
Three students S1, S2 and S3 perform an experiment for determining the acceleration due to gravity (g) using a simple pendulum. They use different lengths of pendulum and record time for different number of oscillations. The observations are as shown in the table.
(Least count of length = 0.1 cm and Least count for time = 0.1 s)
If E1, E2 and E3 are the percentage errors in 'g' for students 1, 2 and 3 respectively, then the minimum percentage error is obtained by student no. ______________.
| Student No. |
Length of Pendulum (cm) |
No. of oscillations (n) |
Total time for n oscillations |
Time period (s) |
|---|---|---|---|---|
| 1 | 64.0 | 8 | 128.0 | 16.0 |
| 2 | 64.0 | 4 | 64.0 | 16.0 |
| 3 | 20.0 | 4 | 36.0 | 9.0 |
(Least count of length = 0.1 cm and Least count for time = 0.1 s)
If E1, E2 and E3 are the percentage errors in 'g' for students 1, 2 and 3 respectively, then the minimum percentage error is obtained by student no. ______________.
Correct Answer: 1
Explanation:
$T = {t \over n} = 2\pi \sqrt {{l \over g}} $
$ \Rightarrow g = {{4{\pi ^2}l} \over {{T^2}}}$
$ \Rightarrow {{\Delta g} \over g} \times 100 = {{\Delta l} \over l} \times 100 + 2{{\Delta T} \over T} \times 100$
$ = \left( {{{\Delta l} \over l} + {{2\Delta T} \over {T}}} \right)100\% $
${E_1} = {{20} \over {64}}\% $
${E_2} = {{30} \over {64}}\% $
${E_3} = {{19} \over {18}}\% $
$ \Rightarrow g = {{4{\pi ^2}l} \over {{T^2}}}$
$ \Rightarrow {{\Delta g} \over g} \times 100 = {{\Delta l} \over l} \times 100 + 2{{\Delta T} \over T} \times 100$
$ = \left( {{{\Delta l} \over l} + {{2\Delta T} \over {T}}} \right)100\% $
${E_1} = {{20} \over {64}}\% $
${E_2} = {{30} \over {64}}\% $
${E_3} = {{19} \over {18}}\% $
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Evening Shift
The radius of a sphere is measured to be (7.50 $\pm$ 0.85) cm. Suppose the percentage error in its volume is x.
The value of x, to the nearest x, is __________.
The value of x, to the nearest x, is __________.
Correct Answer: 34
Explanation:
Given, radius of sphere = (7.50 $\pm$ 0.85) cm
$ \therefore $ r = 7.50 and dr = 0.85
We know, volume of a sphere $v = {4 \over 3}\pi {r^3}$
taking log both sides, we get
$\ln v = \ln {{4\pi } \over 3} + 3\ln r$
Differentiating both sides,
${{dv} \over v} = 0 + 3{{dr} \over r}$
$ \therefore $ Fractional error in volume ${{dv} \over v} = 3{{dr} \over r}$
$ \therefore $ % error in volume,
${{dv} \over v} \times 100 = 3{{dr} \over r} \times 100$
$ = 3 \times {{0.85} \over {7.50}} \times 100$
= 34%
$ \therefore $ r = 7.50 and dr = 0.85
We know, volume of a sphere $v = {4 \over 3}\pi {r^3}$
taking log both sides, we get
$\ln v = \ln {{4\pi } \over 3} + 3\ln r$
Differentiating both sides,
${{dv} \over v} = 0 + 3{{dr} \over r}$
$ \therefore $ Fractional error in volume ${{dv} \over v} = 3{{dr} \over r}$
$ \therefore $ % error in volume,
${{dv} \over v} \times 100 = 3{{dr} \over r} \times 100$
$ = 3 \times {{0.85} \over {7.50}} \times 100$
= 34%
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Morning Shift
The resistance R = ${V \over I}$, where V = (50 $\pm$ 2)V and I = (20 $\pm$ 0.2)A. The percentage error in R is 'x'%. The value of 'x' to the nearest integer is _________.
Correct Answer: 5
Explanation:
$R = {V \over I}$
${{\Delta R} \over R} \times 100 = {{\Delta V} \over V} \times 100 + {{\Delta I} \over I} \times 100$
% error in $R = {2 \over {50}} \times 100 + {{0.2} \over {20}} \times 100$
% error in R = 4 + 1
$ \therefore $ % error in R = 5%
${{\Delta R} \over R} \times 100 = {{\Delta V} \over V} \times 100 + {{\Delta I} \over I} \times 100$
% error in $R = {2 \over {50}} \times 100 + {{0.2} \over {20}} \times 100$
% error in R = 4 + 1
$ \therefore $ % error in R = 5%
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Evening Slot
A student measuring the diameter of a pencil of circular cross-section with the help of a vernier
scale records the following four readings 5.50 mm, 5.55 mm, 5.45 mm, 5.65 mm. The average of
these four readings is 5.5375 mm and the standard deviation of the data is 0.07395 mm. The
average diameter
of the pencil should therefore be recorded as :
of the pencil should therefore be recorded as :
A.
(5.54 $ \pm $ 0.07) mm
B.
(5.5375 $ \pm $ 0.0740) mm
C.
(5.5375 $ \pm $ 0.0739) mm
D.
(5.538 $ \pm $ 0.074) mm
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Morning Slot
A screw gauge has 50 divisions on its circular scale. The circular scale is 4 units ahead of the
pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of 0.5mm is noticed on the pitch scale. The nature of zero error involved and the least
count of the screw gauge, are respectively :
A.
Positive, 0.1 mm
B.
Positive, 0.1 $\mu $m
C.
Positive, 10 $\mu $m
D.
Negative, 2 $\mu $m
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Evening Slot
The quantities x = ${1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}$, y = ${E \over B}$ and z = ${l \over {CR}}$ are
defined where C-capacitance, R-Resistance, l-length, E-Electric field, B-magnetic field and ${{\varepsilon _0}}$, ${{\mu _0}}$, - free space permittivity and permeability respectively. Then :
defined where C-capacitance, R-Resistance, l-length, E-Electric field, B-magnetic field and ${{\varepsilon _0}}$, ${{\mu _0}}$, - free space permittivity and permeability respectively. Then :
A.
Only y and z have the same dimension
B.
x, y and z have the same dimension
C.
Only x and y have the same dimension
D.
Only x and z have the same dimension
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Morning Slot
A physical quantity z depends on four
observables
a, b, c and d, as z = ${{{a^2}{b^{{2 \over 3}}}} \over {\sqrt c {d^3}}}$. The percentages of error in the measurement of a, b, c and d are 2%, 1.5%, 4% and 2.5% respectively. The percentage of error in z is :
a, b, c and d, as z = ${{{a^2}{b^{{2 \over 3}}}} \over {\sqrt c {d^3}}}$. The percentages of error in the measurement of a, b, c and d are 2%, 1.5%, 4% and 2.5% respectively. The percentage of error in z is :
A.
13.5 %
B.
14.5%
C.
16.5%
D.
12.25%
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Evening Slot
A quantity x is given by $\left( {{{IF{v^2}} \over {W{L^4}}}} \right)$ in terms of moment of inertia I, force F, velocity v, work W and
Length L. The dimensional formula for x is same as that of :
A.
Coefficient of viscosity
B.
Force constant
C.
Energy density
D.
Planck's constant
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Morning Slot
Dimensional formula for thermal conductivity is (here K denotes the temperature):
A.
MLT–3K–1
B.
MLT–2K–2
C.
MLT–2K
D.
MLT–3K
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Evening Slot
Amount of solar energy received on the earth’s surface per unit area per unit time is defined a solar
constant. Dimension of solar constant is :
A.
MLT–2
B.
ML0T–3
C.
M2L0T–1
D.
ML2T–2
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Morning Slot
Using screw gauge of pitch 0.1 cm and
50 divisions on its circular scale, the thickness
of an object is measured. It should correctly be
recorded as
A.
2.123 cm
B.
2.124 cm
C.
2.125 cm
D.
2.121 cm
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Evening Slot
If momentum (P), area (A) and time (T) are
taken to be the fundamental quantities then the
dimensional formula for energy is
A.
[P2AT–2]
B.
$\left[ {{P^{{1 \over 2}}}A{T^{ - 1}}} \right]$
C.
$\left[ {P{A^{{1 \over 2}}}{T^{ - 1}}} \right]$
D.
[PA–1T–2]
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Morning Slot
If speed V, area A and force F are chosen as
fundamental units, then the dimension of
Young’s modulus will be
A.
FA–1V0
B.
FA2V–1
C.
FA2V–2
D.
FA2V–3
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Morning Slot
The least count of the main scale of a vernier
callipers is 1 mm. Its vernier scale is divided
into 10 divisions and coincide with 9 divisions
of the main scale. When jaws are touching
each other, the 7th division of vernier scale
coincides with a division of main scale and the
zero of vernier scale is lying right side of the
zero of main scale. When this vernier is used to
measure length of a cylinder the zero of the
vernier scale between 3.1 cm and 3.2 cm and
4th VSD coincides with a main scale division.
The length of the cylinder is : (VSD is vernier
scale division)
A.
3.21 cm
B.
2.99 cm
C.
3.07 cm
D.
3.2 cm
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Evening Slot
For the four sets of three measured physical
quantities as given below. Which of the
following options is correct ?
(i) A1 = 24.36, B1 = 0.0724, C1 = 256.2
(ii) A2 = 24.44, B2 = 16.082, C2 = 240.2
(iii) A3 = 25.2, B3 = 19.2812, C3 = 236.183
(iv) A4 = 25, B4 = 236.191, C4 = 19.5
(i) A1 = 24.36, B1 = 0.0724, C1 = 256.2
(ii) A2 = 24.44, B2 = 16.082, C2 = 240.2
(iii) A3 = 25.2, B3 = 19.2812, C3 = 236.183
(iv) A4 = 25, B4 = 236.191, C4 = 19.5
A.
A1 + B1 + C1 = A2 + B2 + C2 = A3 + B3 + C3
= A4 + B4 + C4
B.
A4 + B4 + C4 $ < $ A1 + B1 + C1 $ < $ A3 + B3 + C3
$ < $ A2 + B2 + C2
C.
A4 + B4 + C4 $ < $ A1 + B1 + C1 = A2 + B2 + C2
= A3 + B3 + C3
D.
A4 + B4 + C4 $ > $ A3 + B3 + C3 = A2 + B2 + C2 $ > $
A1 + B1 + C1
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Morning Slot
A quantity f is given by $f = \sqrt {{{h{c^5}} \over G}} $ where c is
speed of light, G universal gravitational
constant and h is the Planck's constant.
Dimension of f is that of :
A.
Energy
B.
Momentum
C.
Area
D.
Volume
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Morning Slot
If the screw on a screw-gauge is given six
rotations, it moves by 3 mm on the main scale.
If there are 50 divisions on the circular scale
the least count of the screw gauge is :
A.
0.001 mm
B.
0.01 cm
C.
0.02 mm
D.
0.001 cm
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Morning Slot
The dimension of stopping potential V0 in photoelectric effect in units of Planck's constant 'h', speed of light 'c' and Gravitational constant 'G' and ampere A is :
A.
h1/3 G2/3 c1/3 A–1
B.
h0 c5 G-1 A-1
C.
h2/3 c5/3 G1/3 A–1
D.
h2 G3/2 c1/3 A–1
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Evening Slot
The dimension of ${{{B^2}} \over {2{\mu _0}}}$, where B is magnetic field and ${{\mu _0}}$
is the magnetic permeability of vacuum,
is :
A.
ML2T–2
B.
MLT–2
C.
ML-1T–2
D.
ML2T–1
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 6th September Morning Slot
The density of a solid metal sphere is determined by measuring its mass and its diameter. The
maximum error in the density of the sphere is $\left( {{x \over {100}}} \right)$ %. If the relative errors in measuring the mass
and the diameter are 6.0% and 1.5% respectively, the value of x is_______.
Correct Answer: 1050
Explanation:
$\rho $ = ${M \over V}$ = ${M \over {{4 \over 3}\pi {{\left( {{D \over 2}} \right)}^3}}}$
$ \Rightarrow $ $\rho $ = ${6 \over \pi }M{D^{ - 3}}$
For maximum error
${{d\rho } \over \rho } \times 100 = {{dM} \over M} \times 100 + 3{{dD} \over D} \times 100$
= 6 + 3 $ \times $ 1.5
= 10.5 %
= $\left( {{{1050} \over {100}}} \right)$ %
$ \therefore $ x = 1050
$ \Rightarrow $ $\rho $ = ${6 \over \pi }M{D^{ - 3}}$
For maximum error
${{d\rho } \over \rho } \times 100 = {{dM} \over M} \times 100 + 3{{dD} \over D} \times 100$
= 6 + 3 $ \times $ 1.5
= 10.5 %
= $\left( {{{1050} \over {100}}} \right)$ %
$ \therefore $ x = 1050
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
Which of the following combinations has the dimension of electrical resistance ($ \in $0 is the permittivity of
vacuum and $\mu $0 is the permeability of vacuum)?
A.
$\sqrt {{{{ \in _0}} \over {{\mu _0}}}} $
B.
${{{{ \in _0}} \over {{\mu _0}}}}$
C.
$\sqrt {{{{\mu _0}} \over {{ \in _0}}}} $
D.
${{{{\mu _0}} \over {{ \in _0}}}}$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
In the formula X = 5YZ2
, X and Z have dimensions of capacitance and magnetic field, respectively. What are
the dimensions of Y in SI units?
A.
[M–3L–2T8A4]
B.
[M–2L–2T6A3]
C.
[M–1L–2T4A2]
D.
[M–2L0 T–4A–2]
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Evening Slot
The area of a square is 5.29 cm2. The area of
7 such squares taking into account the
significant figures is :-
A.
37.0 cm2
B.
37 cm2
C.
37.030 cm2
D.
37.03 cm2
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Morning Slot
In the density measurement of a cube, the mass and
edge length are measured as (10.00 ± 0.10) kg and
(0.10 ± 0.01) m, respectively. The error in the
measurement of density is :
A.
0.01 kg/m3
B.
0.10 kg/m3
C.
0.31 kg/m3
D.
0.07 kg/m3
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Evening Slot
In a simple pendulum experiment for
determination of acceleration due to gravity (g),
time taken for 20 oscillations is measured by
using a watch of 1 second least count. The
mean value of time taken comes out to be
30 s. The length of pendulum is measured by
using a meter scale of least count 1 mm and the
value obtained is 55.0 cm. The percentage
error in the determination of g is close to :-
A.
0.2%
B.
3.5%
C.
0.7%
D.
6.8%
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Evening Slot
If surface tension (S), Moment of inertia (I) and
Planck's constant (h), were to be taken as the
fundamental units, the dimensional formula for
linear momentum would be :-
A.
S1/2I1/2h0
B.
S3/2I1/2h0
C.
S1/2I1/2h-1
D.
S1/2I3/2h-1
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Morning Slot
In SI units, the dimensions of $\sqrt {{{{ \in _0}} \over {{\mu _0}}}} $ is :
A.
A–1 TML3
B.
A2T3M–1L–2
C.
AT–3ML3/2
D.
AT2M–1L–1
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Evening Slot
Let $\ell $, r, C and V represent inductance, resistance, capacitance and voltage, respectively. The dimension of ${\ell \over {rCV}}$ in SI units will be :
A.
[A–1]
B.
[LTA]
C.
[LA–2]
D.
[LT2]
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Morning Slot
The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5 $\mu $m diameter of a wire is :
A.
500
B.
100
C.
200
D.
50
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Evening Slot
If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young,s modulus will be:
A.
V$-$2A2F2
B.
V$-$4A$-$2F
C.
V$-$4A2F
D.
V$-$2A2F$-$2
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Morning Slot
The force of interaction between two atoms is given by F = $\alpha $$\beta $exp $\left( { - {{{x^2}} \over {\alpha kt}}} \right)$; where x is the distance, k is the Boltzmann constant and T is temperature and $\alpha $ and $\beta $ are two constants. The dimension of $\beta $ is :
A.
M2L2T$-$2
B.
M2LT$-$4
C.
MLT$-$4
D.
M0L2LT$-$4
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th January Evening Slot
The diameter and height of a cylinder are measured by a meter scale to be 12.6 $ \pm $ 0.1 cm and 34.2 $ \pm $ 0.1 cm, respectively. What will be the value of its volume in appropriate significant figures ?
A.
4264.4 $ \pm $ 81.0 cm3
B.
4264 $ \pm $ 81 cm3
C.
4300 $ \pm $ 80 cm3
D.
4260 $ \pm $ 80 cm3
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th January Morning Slot
The density of a material in SI units is 128 kg m–3
. In certain units in which the unit of length is 25 cm and the unit of mass is 50 g, the numerical value of density of the material is -
A.
40
B.
640
C.
16
D.
410
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th January Evening Slot
The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line.
The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is :
The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is :
A.
5.755 mm
B.
5.950 mm
C.
5.725 mm
D.
5.740 mm
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th January Evening Slot
Expression for time in terms of G(universal gravitional constant), h (Planck constant) and c (speed of light) is proportional to :
A.
$\sqrt {{{h{c^5}} \over G}} $
B.
$\sqrt {{{{c^3}} \over {Gh}}} $
C.
$\sqrt {{{Gh} \over {{c^5}}}} $
D.
$\sqrt {{{Gh} \over {{c^3}}}} $