Units & Measurements Questions — NEET Physics

Q1

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 imes 10^8

B 500

C

3 imes 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

Q2

Taking into account of the significant figures, what is the value of

9.99 m - 0.0099m

?

A 9.98 m

B 9.980 m

C 9.9 m

D 9.9801 m

Correct Answer

Option A

Solution

9.99 - 0.0099

= 9.9801 m Here, 9.98 is having the least number of decimal places of two, so answer should also have only two decimal places.

So answer is 9.98 m.

Q3

Dimensions of stress are :

A

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

B

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

C

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

D

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

Correct Answer

Option C

Solution

Stress = {{Force} \\over {Area}}

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

Q4

The energy required to break one bond in DNA is 10 -20 J. This value in eV is nearly :

A 0.6

B 0.06

C 0.006

D 6

Correct Answer

Option B

Solution

1eV = 1.6 \\times {10^{ - 19}}J

1J = {1 \\over {1.6 \\times {{10}^{ - 19}}}}eV

{10^{ - 20}}J = {{{{10}^{ - 20}}} \\over {1.6 \\times {{10}^{ - 19}}}}eV

= 0.06 eV

Q5

In an experiment, the percentage of error occurred in the measurement of physical quantities A, B, C and D are 1%, 2 %, 3 %and 4 % respectively. Then the maximum percentage of error in the measurement X, where X =

{{{A^2}{B^{1/2}}} \\over {{C^{1/3}}{D^3}}}

, will be :

A 16%

B -10%

C 10%

D

\left( {{3 \over {13}}} \right)

%

Correct Answer

Option A

Solution

Given, X =

{{{A^2}{B^{1/2}}} \\over {{C^{1/3}}{D^3}}}

{{{\\Delta X} \\over X}} = {{2{\\Delta A} \\over A}} + {1 \\over 2}{{{\\Delta B} \\over B}} + {1 \\over 3}{{{\\Delta C} \\over C}} + 3{{{\\Delta D} \\over D}}

$\\Rightarrow$

{{{\\Delta X} \\over X}} \\times 100 = 2\\left( {1\\% } \\right) + {1 \\over 2}\\left( {2\\% } \\right) + {1 \\over 3}\\left( {3\\% } \\right) + 3\\left( {4\\% } \\right)

= 16 %

Q6

A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of –0.004 cm, the correct diameter of the ball is

A 0.521 cm

B 0.525 cm

C 0.053 cm

D 0.529 cm

Correct Answer

Option D

Solution

Diameter of the ball = MSR + CSR × (Least count) – Zero error = 5 mm + 25 × 0.001 cm – (–0.004) cm = 0.5 cm + 25 × 0.001 cm – (–0.004) cm = 0.529 cm.

Q7

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}}}}

Q8

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

Q9

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]

Q10

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]