Units & Measurements — Page 15 | NEET Physics

Q141

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

Q142

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]

Q143

If the dimensions of a physical quantity are given by M a L b T c , then the physical quantity will be

A velocity if a = 1, b = 0, c =

-

1

B acceleration if a = 1, b = 1, c =

-

2

C force if a = 0, b =

-

1, c =

-

2

D pressure if a = 1, b =

-

1, c =

-

2

Correct Answer

Option D

Solution

Velocity = [LT -1 ] so a = 0, b = 1, c = -1 Acceleration = [LT -2 ] so a = 0, b = 1, c = -2 Force = [MLT -2 ] so a = 1, b = 1, c = -2 Pressure =

{{force} \over {Area}} = {{\left[ {ML{T^{ - 2}}} \right]} \over {\left[ {{L^2}} \right]}} = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]

, so a = 1, b = - 1, c = - 2

Q144

Which two of the following five physical parameters have the same dimensions ? 1. Energy density 2. Refractive index 3. Dielectric constant 4. Young's modulus 5. Magnetic field

A 1 and 4

B 1 and 5

C 2 and 4

D 3 and 5

Correct Answer

Option A

Solution

Energy density =

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

Young's Modulus(Y) =

{{F \over A} \times {l \over {\Delta l}}}

$\Rightarrow$ [Y] =

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

=

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

Q145

If the error in the measurement of radius of a sphere is 2%, then the error in the determination of volume of the sphere will be

A 8%

B 2%

C 4%

D 6%

Correct Answer

Option D

Solution

As we know,

V = {4 \over 3}\pi {R^3}

$\therefore$

{{dV} \over V} = 3{{dR} \over R}

Error in the dimension of the volume is = 3 $\times$ 2% = 6 %

Q146

Dimensions of resistance in an electrical circuit, in terms of dimension of mass M, of length L, of time T and of current I, would be

A [ML 2 T

-

2 ]

B [ML 2 T

-

1 I

-

1 ]

C [ML 2 T

-

3 I

-

2 ]

D [ML 2 T

-

3 I

-

1 ]

Correct Answer

Option C

Solution

We know,

R = {V \over I}

$\therefore$

\left[ R \right] = {{\left[ V \right]} \over {\left[ I \right]}}

=

{{\left[ {M{L^2}{T^{ - 3}}{I^{ - 1}}} \right]} \over {\left[ I \right]}}

=

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

Q147

The velocity v of a particle at time t is given by v = at +

{b \over {t + c}}

, where a, b and c are constants. The dimensions of a, b and c are

A [L], [LT] and [LT

-

2 ]

B [LT

-

2 ], [L] and [T]

C [L 2 ], [T] and [LT

-

2 ]

D [LT

-

2 ], and [LT] and [L]

Correct Answer

Option B

Solution

Given, v = at +

{b \over {t + c}}

Dimension of at = Dimension of velocity v $\therefore$ [at] = [LT -1 ] $\Rightarrow$ [a] =

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

= [LT -2 ] As c is added to t, $\therefore$ [c] = [T]

{{\left[ b \right]} \over {\left[ T \right]}}

= [LT -1 ] $\therefore$ [b] = [L]

Q148

The ratio of the dimensions of Planck's constant and that of moment of inertia is the dimensions of

A time

B frequency

C angular momentum

D velocity

Correct Answer

Option B

Solution

Plank constant (h) =

{{E\lambda } \over c}

=

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

= [ML 2 T -1 ] Moment of inertia (I) = mr 2 = [ML 2 ] $\therefore$

{h \over I} =

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

= [T -1 ] = frequency

Q149

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

Q150

The dimensions of universal gravitational constant are

A [M

-

1 L 3 T

-

2 ]

B [ML 2 T

-

1 ]

C [M

-

2 L 3 T

-

2 ]

D [M

-

2 L 2 T

-

1 ]

Correct Answer

Option A

Solution

F =

{{G{m_1}{m_2}} \over {{r^2}}}

$\Rightarrow$ G =

{{F \times {r^2}} \over {{m_1}{m_2}}}

$\therefore$ Dimension of G =

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

=

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