iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 30th June Morning Shift
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 :
A.
${2 \over {n + 1}}$ mm
B.
${5 \over {n + 1}}$ mm
C.
${1 \over {2n}}$ mm
D.
${1 \over {5n}}$ mm
Correct Answer: A
Explanation:
5 parts of main scale division = 1 cm
$\therefore$ 1 part of main scale division = ${1 \over 5}$ cm
$\therefore$ 1 M.S.D. = ${1 \over 5}$ cm
(n + 1) vernier scale division = n main scale division.
$\therefore$ 1 V.S.D. = ${n \over n+1}$ M.S.D.
= ${n \over n+1}$ $\times$ 1 M.S.D.
= ${n \over n + 1}$ $\times$ ${1 \over 5}$ cm
We know,
L.C. = 1 M.S.D. $-$ 1 V,S.D.
= ${1 \over 5}$ cm $-$ ${n \over {5(n + 1)}}$ cm
= ${{n + 1 - n} \over {5(n + 1)}}$ cm
= ${1 \over {5(n + 1)}}$ cm
2022
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 28th June Evening Shift
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 :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 27th June Morning Shift
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 :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 26th June Morning Shift
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 :
Decay constant $\lambda$ has dimensions of $[{T^{ - 1}}]$ because of the relation ${{dN} \over {dt}} = - \lambda $ N
$\Rightarrow$ Velocity gradient and decay constant have same dimensions.
2022
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2022 (Online) 24th June Morning Shift
Identify the pair of physical quantities which have different dimensions:
A.
Wave number and Rydberg's constant
B.
Stress and Coefficient of elasticity
C.
Coercivity and Magnetisation
D.
Specific heat capacity and Latent heat
Correct Answer: D
Explanation:
$[S] = {{[C]} \over {[m] \times [\Delta T]}}$
and, $[L] = {{[Q]} \over {[m]}}$
$\Rightarrow$ They have different dimensions.
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 1st September Evening Shift
A student determined Young's Modulus of elasticity using the formula $Y = {{Mg{L^3}} \over {4b{d^3}\delta }}$. The value of g is taken to be 9.8 m/s2, without any significant error, his observation are as following.
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 $)
0.01 mm
5 mm
Then the fractional error in the measurement of Y is :
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 31st August Evening Shift
If velocity [V], time [T] and force [F] are chosen as the base quantities, the dimensions of the mass will be :
A.
[FT$-$1 V$-$1]
B.
[FTV$-$1]
C.
[FT2 V]
D.
[FVT$-$1]
Correct Answer: B
Explanation:
[M] = K[F]a [T]b [V]c
[M1] = [M1L1T$-$2]a [T1]b [L1T$-$1]c
a = 1, b = 1, c = $-$1
$\therefore$ [M] = [FTV$-$1]
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 31st August Morning Shift
Match List - I with List - II.
List - I
List - II
(a)
Torque
(i)
MLT$^{ - 1}$
(b)
Impulse
(ii)
MT$^{ - 2}$
(c)
Tension
(iii)
ML$^{ 2}$T$^{ - 2}$
(d)
Surface Tension
(iv)
MLT$^{ - 2}$
Choose the most appropriate answer from the option given below :
A.
(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
B.
(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
C.
(a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
D.
(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
Correct Answer: A
Explanation:
torque $\tau$ $\to$ ML2T$-$2 (iii)
Impulse I $\Rightarrow$ MLT$-$1 (i)
Tension force $\Rightarrow$ MLT$-$2 (iv)
Surface tension $\Rightarrow$ MT$-$2 (ii)
Option (a)
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 31st August Morning Shift
Which of the following equations is dimensionally incorrect?
Where t = time, h = height, s = surface tension, $\theta$ = angle, $\rho$ = density, a, r = radius, g = acceleration due to gravity, v = volume, p = pressure, W = work done, T = torque, $\in$ = permittivity, E = electric field, J = current density, L = length.
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 26th August Morning Shift
In a Screw Gauge, fifth division of the circular scale coincides with the reference line when the ratchet is closed. There are 50 divisions on the circular scale, and the main scale moves by 0.5 mm on a complete rotation. For a particular observation the reading on the main scale is 5 mm and the 20th division of the circular scale coincides with reference line. Calculate the true reading.
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 26th August Morning Shift
If E, L, M and G denote the quantities as energy, angular momentum, mass and constant of gravitation respectively, then the dimensions of P in the formula P = EL2M$-$5G$-$2 are :
iCON Education HYD, 79930 92826, 73309 72826JEE 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 :
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.
Correct Answer: A
Explanation:
Least Count = ${{Pitch} \over {Total\,divisions\,on\,circular\,scale}}$
In 5 revolution, distance travel, 5 mm
In 1 revolution, it will travel 1 mm.
So least count = ${1 \over {50}}$ = 0.02
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE Main 2021 (Online) 25th July Evening Shift
The force is given in terms of time t and displacement x by the equation
${M^1}{L^0}{T^0} = {M^c}{L^{b + 2c}}{T^{a - b - c}}$
comparing powers
c = 1, b = $-$2, a = $-$1
$m \propto {t^{ - 1}}{v^{ - 2}}{l^1}$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE 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)
iCON Education HYD, 79930 92826, 73309 72826JEE 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?
iCON Education HYD, 79930 92826, 73309 72826JEE 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 :
iCON Education HYD, 79930 92826, 73309 72826JEE 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.
$[{M^2}L{T^{ - 2}}]$
Correct Answer: A
Explanation:
kT has dimension of energy
${{\beta {x^2}} \over {kT}}$ is dimensionless
$[\beta ][{L^2}] = [M{L^2}{T^{ - 2}}]$
$[\beta ] = [M{T^{ - 2}}]$
${\alpha ^2}\beta $ has dimensions of work
$[{\alpha ^2}][M{T^{ - 2}}] = [M{L^2}{T^{ - 2}}]$
$[\alpha ] = [{M^0}L{T^0}]$
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE 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 :
iCON Education HYD, 79930 92826, 73309 72826JEE 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
Correct Answer: C
Explanation:
Least count = ${{1mm} \over {100}} = 0.01mm$
zero error = + 8 $\times$ LC = + 0.08 mm
True reading (Diameter)
= (1 mm + 72 $\times$ LC) $-$ (Zero error)
= (1 mm + 72 $\times$ 0.01 mm) $-$ 0.08 mm
= 1.72 mm $-$ 0.08 mm
= 1.64 mm
Therefore, radius = ${{1.64} \over 2}$ = 0.82 mm.
2021
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE 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 :
A.
$[{M^0}L{T^0}]$
B.
$[M{L^2}{T^{ - 2}}]$
C.
$[ML{T^{ - 2}}]$
D.
$[{M^2}L{T^2}]$
Correct Answer: C
Explanation:
where, k is Boltzmann constant,
T is temperature and x is displacement.
We know that, ${{{x^2}} \over {\alpha kT}}$ is a dimensionless quantity.
iCON Education HYD, 79930 92826, 73309 72826JEE 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 :
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
Correct Answer: A
Explanation:
Given, dav = 5.5375 mm
$\Delta $d = 0.07395 mm
Significant rule says that reading should has same significant figure as that of reading given.
$ \because $ Measured data are up to two digits after
decimal.
$ \therefore $ 5.5375 rounded to $ \to $ 5.54
2020
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE 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
Correct Answer: C
Explanation:
Least count of screw gauge
= ${{0.5} \over {50}}$
= 1 $ \times $ 10-5 m
= 10 $\mu $m
Zero error in positive.
2020
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE 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 :
iCON Education HYD, 79930 92826, 73309 72826JEE 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.
13.5 %
B.
14.5%
C.
16.5%
D.
12.25%
Correct Answer: B
Explanation:
z = ${{{a^2}{b^{{2 \over 3}}}} \over {\sqrt c {d^3}}}$
iCON Education HYD, 79930 92826, 73309 72826JEE 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 :
iCON Education HYD, 79930 92826, 73309 72826JEE 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
Correct Answer: A
Explanation:
Y = k [F]x
[A]y
[V]z
[M1L1T
–2] = [MLT–2]x [L2]y [LT–1]z
[M1L1T
–2] = [M]x [L]x+2y+z[T]–2x–z
Comparing power of M, L and T
x = 1 ……(1)
x + 2y + z = –1 ……(2)
–2x – z = –2 ……(3)
After solving
x = 1
y = –1
z = 0
$ \therefore $ Y = FA–1V0
2020
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE 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
Correct Answer: C
Explanation:
Least count = 1 mm or 0.01 cm
Zero error = 0 + 0.01 × 7 = 0.07 cm
Reading = 3.1 + (0.01 × 4) – 0.07
= 3.1 + 0.04 – 0.07
= 3.1 – 0.03
= 3.07 cm
2020
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE 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
iCON Education HYD, 79930 92826, 73309 72826JEE 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 :
iCON Education HYD, 79930 92826, 73309 72826JEE 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
Correct Answer: D
Explanation:
Pitch = ${3 \over 6}$ mm = 0.5 mm
Least count = ${{0.5} \over {50}}$ mm
= ${1 \over {100}}$ = 0.01 mm = 0.001 cm
2020
JEE Mains
MCQ
iCON Education HYD, 79930 92826, 73309 72826JEE 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 :