Units & Measurements — Page 3 | JEE Physics

Q21

Dimensions of

{1 \over {{\mu _0}{\varepsilon _0}}}

, where symbols have their usual meaning, are

A [ L-1T ]

B [ L-2T2 ]

C [ L2T-2 ]

D [ LT-1 ]

Correct Answer

Option C

Solution

The velocity of light in vacuum is c =

{1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}

;

\therefore[{1 \over {{\mu _0}{\varepsilon _0}}}]

= [c2] = [L2T-2] $\therefore$ Dimension of

{1 \over {{\mu _0}{\varepsilon _0}}}

= [L2T-2]

Q22

A body of mass m = 3.513 kg is moving along the x-axis with a speed of 5.00 ms−1. The magnitude of its momentum is recorded as

A 17.6 kg ms-1

B 17.565 kg ms-1

C 17.56 kg ms-1

D 17.57 kg ms-1

Correct Answer

Option A

Solution

Momentum, p = m $\times$ v = 3.513 $\times$ 5.00 = 17.565 kg m/s

\simeq

17.6 kg m/s Here number of significant digits in m is 4 and in v is 3, so, p must have minimum (which is 3) significant digit.

Note : In this case, since we are calculating the magnitude of momentum, which is given by the absolute value of the product of mass and velocity, we should follow the rules for significant figures as if we were multiplying the two values together.

The least precise value in the calculation has three significant figures (the velocity), so we should round the answer to three significant figures as well.

Therefore, the correct answer with three significant figures is 17.6 kg m/s.

Q23

Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale. The total number of divisions on the circular scale is 50. Further, it is found that the screw gauge has a zero error of − 0.03 mm while measuring the diameter of a thin wire, a student notes the main scale reading of 3 mm and the number of circular scale divisions in line with the main scale as 35. The diameter of the wire is

A 3.32 mm

B 3.73 mm

C 3.67 mm

D 3.38 mm

Correct Answer

Option D

Solution

Least count of screw gauge =

{{0.5} \over {50}}mm

= 0.01mm Main scale reading = 3 mm Vernier scale reading = 35 $\therefore$ Reading = [Main scale reading + circular scale reading $\times$ L.C] - (zero error) = [3 + 35 $\times$ 0.01] - (-0.03) = 3.38 mm

Q24

An experiment is performed to find the refractive index of glass using a travelling microscope. In this experiment distances are measured by

A a vernier scale provided on the microscope

B a standard laboratory scale

C a meter scale provided on the microscope

D a screw gauge provided on the microscope

Correct Answer

Option A

Solution

To find the refractive index of glass using a travelling microscope, a vernier scale is provided on the microscope

Q25

In an experiment the angles are required to be measured using an instrument, 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the main scale is half-a degree(=

0.5^\circ

), then the least count of the instrument is:

A one minute

B half minute

C one degree

D half degree

Correct Answer

Option A

Solution

30 vernier scale divisions coincide with 29 main scale divisions. Therefore 1 V.S.D =

{{29} \over {30}}

M.S.D Least count = 1 M.S.D - 1 V.S.D = 1 M.S.D -

{{29} \over {30}}

M.S.D =

{{1} \over {30}}

M.S.D =

{{1} \over {30}}

$\times$ 0.5o =

{{1} \over {30}}

$\times$

{1 \over 2}

o =

{1 \over {60}}

o = 1 min

Q26

The respective number of significant figures for the numbers 23.023, 0.0003 and 2.1

\times

10–3 are

A 5, 1, 2

B 5, 1, 5

C 5, 5, 2

D 4, 4, 2

Correct Answer

Option A

Solution

23.023 has 5 significant figures. 0.0003 has 1 significant figure. 2.1 $\times$ 10–3 has 2 significant figures.

Q27

A screw gauge gives the following reading when used to measure the diameter of a wire. Main scale reading : 0 mm Circular scale reading : 52 divisions Given that 1 mm on main scale corresponds to 100 divisions of the circular scale. The diameter of wire from the above date is:

A 0.052 cm

B 0.026 cm

C 0.005 cm

D 0.52 cm

Correct Answer

Option A

Solution

Least count of screw gauge =

{{Pitch} \over {Number\,of\,division\,on\,circular\,scale}}

=

{1 \over {100}}mm

= 0.01 mm Diameter of the wire = M.S.R + C.S.R $\times$ L.C = 0 + 52 $\times$ 0.01 = 0.52 mm = 0.052 cm

Q28

A spectrometer gives the following reading when used to measure the angle of a prism. Main scale reading: 58.5 degree Vernier scale reading : 09 divisions Given that 1 division on main scale corresponds to 0.5 degree. Total divisions on the vernier scale is 30 and match with 29 divisions of the main scale. The angle of the prism from the above data

A 58.59 degree

B 58.77 degree

C 58.65 degree

D 59 degree

Correct Answer

Option C

Solution

30 vernier scale divisions coincide with 29 main scale divisions. Therefore 1 V.S.D =

{{29} \over {30}}

M.S.D Least count = 1 M.S.D - 1 V.S.D = 1 M.S.D -

{{29} \over {30}}

M.S.D =

{{1} \over {30}}

M.S.D =

{{1} \over {30}}

$\times$ 0.5o Reading of Vernier = Main Scale Reading + Vernier scale reading $\times$ Least count Given that, Main Scale Reading = 58.5 Vernier scale reading = 09 division $\therefore$ Reading of Vernier = 58.5o + 9 $\times$

{{0.5^\circ } \over {30}}

= 58.65o

Q29

Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then error in the value of resistance of the wire is

A 6 %

B zero

C 1 %

D 3 %

Correct Answer

Option A

Solution

We know R =

{V \over I}

$\therefore$

{{\Delta R} \over R} = {{\Delta V} \over V} + {{\Delta I} \over I}

Percentage error in R =

{{\Delta R} \over R} \times 100 = {{\Delta V} \over V} \times 100 + {{\Delta I} \over I} \times 100

$\therefore$

{{\Delta R} \over R} \times 100 =

3% + 3 % = 6%

Q30

Let [

{\varepsilon _0}

] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then:

A

{\varepsilon _0} = \left[ {{M^{ - 1}}{L^{ - 3}}{T^2}A} \right]

B

{\varepsilon _0} =

\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]

C

{\varepsilon _0} = \left[ {{M^1}{L^2}{T^1}{A^2}} \right]

D

{\varepsilon _0} = \left[ {{M^1}{L^2}{T^1}A} \right]

Correct Answer

Option B

Solution

From Coulomb's law we know,

F = {1 \over {4\pi { \in _0}}}{{{q_1}{q_2}} \over {{r^2}}}

$\therefore$

{ \in _0} = {1 \over {4\pi }}{{{q_1}{q_2}} \over {F{r^2}}}

Hence,

\left[ {{ \in _0}} \right] = {{\left[ {AT} \right]\left[ {AT} \right]} \over {\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]}}

=

\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}} \right]