Units & Measurements — Page 14 | NEET Physics

Q131

Planck's constant (h), speed of light in vacuum (c) and Newton's gravitional constant (G) are three fundamental constants. Which of the following combinations of these has the dimension of length ?

A

{{\sqrt {hG} } \over {{c^{3/2}}}}

B

{{\sqrt {hG} } \over {{c^{5/2}}}}

C

\sqrt {{{hc} \over G}}

D

\sqrt {{{Gc} \over {{h^{3/2}}}}}

Correct Answer

Option A

Solution

According to question, L $\propto$ h p c q G r [M 0 LT 0 ] = [ML 2 T -1 ] p [LT -1 ] q [M -1 L 3 T -2 ] r Equating power both sides, we get p - r = 0 ......(1) 2p + q + 3r = 1 ......(2) - p - q - 2r = 0 ......(

3) Solving equation (1), (2), (3), we get p = r =

{1 \over 2}

, q =

- {3 \over 2}

$\therefore$ L =

{{\sqrt {hG} } \over {{c^{3/2}}}}

Q132

If dimensions of critical velocity

\upsilon

c of a liquid flowing through a tube are expressed as

\left[ {{\eta ^x}{\rho ^y}{r^z}} \right]

where

\eta ,\rho

and r are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of x, y and z are given by

A

-

1,

-

1,

-

1

B 1, 1, 1

C 1,

-

1,

-

1

D

-

1,

-

1, 1

Correct Answer

Option C

Solution

\upsilon

c =

\left[ {{\eta ^x}{\rho ^y}{r^z}} \right]

Put dimensions of various quantities, [M 0 LT -1 ] = [ML -1 T -1 ] x [ML -3 T 0 ] y [M 0 LT 0 ] z = [M x + y L - x - 3y + z T - x ] Equating power both sides, we get x + y = 0; - x - 3y + z = 1; - x = - 1 On solving, we get x = 1, y = -1, z = -1

Q133

If energy (E), velocity (V) and time (T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be

A

\left[ {E{V^{ - 2}}{T^{ - 2}}} \right]

B

\left[ {{E^{ - 2}}{V^{ - 1}}{T^{ - 3}}} \right]

C

\left[ {E{V^{ - 2}}{T^{ - 1}}} \right]

D

\left[ {E{V^{ - 1}}{T^{ - 2}}} \right]

Correct Answer

Option A

Solution

Let surface tension S =kE

a

V b T c where k is a dimensionless constant Writing the dimensions on both sides,

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

=

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

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

{\left[ T \right]^c}

\left[ {M{L^0}{T^{ - 2}}} \right]

=

\left[ {{M^a}{L^{2a + b}}{T^{ - 2a - b + c}}} \right]

Comparing both sides of the equation we get,

a

= 1 ....(1) 2

a

+ b = 0 ....(2) -2

a

- b + c = - 2 ....(3) Solving equation (1), (2) and (3), we get

a

= 1, b = - 2, c = - 2 $\therefore$ Dimension of surface tension =

\left[ {E{V^{ - 2}}{T^{ - 2}}} \right]

Q134

If force (F), velocity (V) and time (T) are taken as fundamental units, then dimensions of mass are

A

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

B

\left[ {FV{T^{ - 2}}} \right]

C

\left[ {F{V^{ - 1}}{T^{ - 1}}} \right]

D

\left[ {F{V^{ - 1}}T} \right]

Correct Answer

Option D

Solution

Let mass m =kF

a

V b T c where k is a dimensionless constant Writing the dimensions on both sides, we get

\left[ {{{M{L^0}{T^{ 0}}}}} \right]

=

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

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

{\left[ T \right]^c}

\left[ {M{L^0}{T^{0}}} \right]

=

\left[ {{M^a}{L^{a + b}}{T^{ - 2a - b + c}}} \right]

Comparing both sides of the equation we get,

a

= 1 ....(1) 2

a

+ b = 0 ....(2) -2

a

- b + c = 0 ....(3) Solving equation (1), (2) and (3), we get

a

= 1, b = - 1, c = 1 $\therefore$ Dimension of mass =

\left[ {F{V^{ - 1}}{T}} \right]

Q135

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 ]

Q136

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%

Q137

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 ]

Q138

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}}

Q139

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

Q140

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 ]