Units & Measurements — Page 2 | NEET Physics

Q11

The pair of quantities having same dimensions is

A Impulse and Surface Tension

B Angular momentum and Work

C Work and Torque

D Young's modulus and Energy

Correct Answer

Option C

Solution

Work = Force $\\times$ distance = [MLT -2 ][L] = [ML 2 T -2 ] Torque = Force $\\times$ Force arm = [MLT -2 ][L] = [ML 2 T -2 ]

Q12

If an experiment four quantities a, b, c and d are measured with percentage error 1%, 2%, 3% and 4% respectively. Quantity P is calculated as follows P

= {{{a^3}{b^2}} \\over {cd}}

% error in P is

A 7%

B 4%

C 14%

D 10%

Correct Answer

Option C

Solution

Given P

= {{{a^3}{b^2}} \\over {cd}}

% error in P :

{{{\\Delta P} \\over P} \\times 100}

=

\\left[ {3{{\\Delta a} \\over a} + 2{{\\Delta b} \\over b} + {{\\Delta c} \\over c} + {{\\Delta d} \\over d}} \\right] \\times 100

= 3 $\\times$ 1% + 2 $\\times$ 2% + 3% + 4% = 14%

Q13

The dimensions of

{{\\left( {{\\mu _0}{\\varepsilon _0}} \\right)}^{ - 1/2}}

are

A

\left[ {{L^{1/2}}{T^{ - 1/2}}} \right]

B

\left[ {{L^{ - 1}}T} \right]

C

\left[ {L{T^{ - 1}}} \right]

D

\left[ {{L^{1/2}}{T^{1/2}}} \right]

Correct Answer

Option C

Solution

{{\\left( {{\\mu _0}{\\varepsilon _0}} \\right)}^{ - 1/2}}

=

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

= Speed of light in vacuum = c So the dimension of

{{\\left( {{\\mu _0}{\\varepsilon _0}} \\right)}^{ - 1/2}}

= [c] =[LT -1 ]

Q14

The damping force on an oscillator is directly proportional to the velocity. The units of the constant of proportionality are

A kg m s

-

1

B kg m s

-

2

C kg s

-

1

D kg s

Correct Answer

Option C

Solution

According to the question, Damping force, F $\\propto$ v $\\Rightarrow$ F = kv where k is constant of proportionality. $\\therefore$ k =

{F \\over v}

=

{N \\over {m\\,{s^{ - 1}}}}

=

{{kg\\,m\\,{s^{ - 2}}} \\over {m\\,{s^{ - 1}}}}

=

kg\\,{s^{ - 1}}

Q15

The density of a material in CGS system of units is 4 g cm

-

3 . In a system of units in which unit of length is 10 cm and unit of mass is 100 g, the value of density of material will be

A 0.04

B 0.4

C 40

D 400

Correct Answer

Option C

Solution

In CGS system, Density = 4

{g \\over {c{m^3}}}

When unit of mass is 100g and unit of length is 10 cm, then Density = x

{{100g} \\over {{{\\left( {10cm} \\right)}^3}}}

$\\therefore$ 4

{g \\over {c{m^3}}}

= x

{{100g} \\over {{{\\left( {10cm} \\right)}^3}}}

$\\Rightarrow$ x = 40 unit

Q16

A student measures the distance traversed in free fall of a body, initially at rest, in a given time. He uses this data to estimate g, the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are e 1 and e 2 respectively, the percentage error in the estimation of g is

A e 2

-

e 1

B e 1 + 2e 2

C e 1 + e 2

D e 1

-

2e 2

Correct Answer

Option B

Solution

Initially body is at rest. $\\therefore$

h = {1 \\over 2}g{t^2}

$\\Rightarrow$

g = {{2h} \\over {{t^2}}}

Maximum percentage error,

{{{\\Delta g} \\over g} \\times 100} = \\left[ {{{\\Delta h} \\over h} + 2{{\\Delta t} \\over t}} \\right] \\times 100

Given that

{{{{\\Delta h} \\over h} \\times 100}}

= e 1 and

{{{{\\Delta t} \\over t} \\times 100}}

= e 2 $\\therefore$

{{{\\Delta g} \\over g} \\times 100}

= e 1 + 2e 2

Q17

The dimension of

{1 \\over 2}{\\varepsilon _0}{E^2}

, where

{\\varepsilon _0}

is permittivity of free space and E is electric field, is

A ML 2 T

-

2

B ML

-

1 T

-

2

C ML 2 T

-

1

D MLT

-

1

Correct Answer

Option B

Solution

{1 \\over 2}{\\varepsilon _0}{E^2}

= Energy density of an electric field E Also Energy density =

{{Energy} \\over {Volume}}

=

{{{\\left[ {M{L^2}{T^{ - 2}}} \\right]}} \\over {\\left[ {{L^3}} \\right]}}

=

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

Q18

The unit of permittyvity of free space,

{\\varepsilon _0}

, is

A coulomb/newton-metre

B newton-metre 2 /coulomb 2

C coulomb 2 /newton-metre 2

D coulomb 2 /(newton-metre) 2

Correct Answer

Option C

Solution

F =

{1 \\over {4\\pi {\\varepsilon _0}}}{{{q^2}} \\over {{r^2}}}

$\\Rightarrow$

{\\varepsilon _0} = {{{q^2}} \\over {4\\pi F{r^2}}}

$\\therefore$ Unit of

{\\varepsilon _0}

= coulomb 2 /newton-metre 2

Q19

The speed of light in vacuum is taken as unity. If light takes 6 min 40 s to reach the Earth from the Sun, the distance between the Sun and the Earth in new unit is:

A

3 \times 10^8

B 500

C

3 \times 10^{10}

D 400

Correct Answer

Option D

Solution

Time, $t=6 \\min 40 \\mathrm{~s}$

\\begin{aligned} & =360+40 \\ & =400 \\mathrm{~s} \\end{aligned}

Distance in new system $d=v t$

=1 \\times 400=400

Q20

Each side of a metallic cube of mass 5.580 kg is measured to the 9.0 cm . Keeping the significant figures in view, the density of the material of the cube can be best expressed as

X \\times 10^3 \\mathrm{~kg} \\mathrm{~m}^{-3}

where the value of

X

is:

A 7.654

B 7.6

C 7.65

D 7.7

Correct Answer

Option B

Solution

Mass of cube: $m = 5.580 \\text{ kg}$ (4 significant figures) Side of cube: $a = 9.0 \\text{ cm} = 9.0 \\times 10^{-2} \\text{ m}$ (2 significant figures)

V = a^3 = (9.0 \\times 10^{-2})^3 \\text{ m}^3

V = 729 \\times 10^{-6} \\text{ m}^3 = 7.29 \\times 10^{-4} \\text{ m}^3

\\rho = \\frac{m}{V} = \\frac{5.580}{7.29 \\times 10^{-4}} \\text{ kg/m}^3

\\rho = 7654.3 \\text{ kg/m}^3 = 7.6543 \\times 10^3 \\text{ kg/m}^3

According to NCERT rules, in multiplication or division, the final result should have the same number of significant figures as the quantity with the least number of significant figures.

Mass has 4 significant figures.

Side has only 2 significant figures.

So, the answer must be rounded off to 2 significant figures .

\\rho = 7.6 \\times 10^3 \\text{ kg/m}^3

Final Answer

\\boxed{X = 7.6}

The correct option is (B) 7.6