Practice Start Test

Units & Measurements

NEET Physics · 152 questions · Page 7 of 16 · Click an option or "Show Solution" to reveal answer

Q61
The dimension of 1over2varepsilon0E2{1 \\over 2}{\\varepsilon _0}{E^2}, where varepsilon0{\\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
1over2varepsilon0E2{1 \\over 2}{\\varepsilon _0}{E^2}

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

EnergyoverVolume{{Energy} \\over {Volume}}

=

left[ML2T−2right]overleft[L3right]{{\\left[ {M{L^2}{T^{ - 2}}} \\right]} \\over {\\left[ {{L^3}} \\right]}}

=

left[ML−1T−2right]\\left[ {M{L^{ - 1}}{T^{ - 2}}} \\right]
Q62
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 =

forceoverArea=left[MLT−2right]overleft[L2right]=left[ML−1T−2right]{{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

Q63
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[ML2T−2right]overleft[L3right]=left[ML−1T−2right]{{\\left[ {M{L^2}{T^{ - 2}}} \\right]} \\over {\\left[ {{L^3}} \\right]}} = \\left[ {M{L^{ - 1}}{T^{ - 2}}} \\right]

Young's Modulus(Y) =

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

Rightarrow\\Rightarrow [Y] =

left[MLT−2right]overleft[L3right]left[Lright]overleft[Lright]{{\\left[ {ML{T^{ - 2}}} \\right]} \\over {\\left[ {{L^3}} \\right]}}{{\\left[ L \\right]} \\over {\\left[ L \\right]}}

=

left[ML−1T−2right]\\left[ {M{L^{ - 1}}{T^{ - 2}}} \\right]
Q64
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=4over3piR3V = {4 \\over 3}\\pi {R^3}

therefore\\therefore

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

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

Q65
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=VoverIR = {V \\over I}

therefore\\therefore

left[Rright]=left[Vright]overleft[Iright]\\left[ R \\right] = {{\\left[ V \\right]} \\over {\\left[ I \\right]}}

=

left[ML2T−3I−1right]overleft[Iright]{{\\left[ {M{L^2}{T^{ - 3}}{I^{ - 1}}} \\right]} \\over {\\left[ I \\right]}}

=

left[ML2T−3I−2right]{\\left[ {M{L^2}{T^{ - 3}}{I^{ - 2}}} \\right]}
Q66
The velocity v of a particle at time t is given by v = at + bovert+c{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 +

bovert+c{b \\over {t + c}}

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

left[LT−1right]overleft[Tright]{{\\left[ {L{T^{ - 1}}} \\right]} \\over {\\left[ T \\right]}}

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

left[bright]overleft[Tright]{{\\left[ b \\right]} \\over {\\left[ T \\right]}}

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

Q67
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) =

Elambdaoverc{{E\\lambda } \\over c}

=

left[ML2T−2right]left[Lright]overleft[LT−1right]{{\\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\\therefore

hoverI={h \\over I} =
left[ML2T−1right]overleft[ML2right]{{\\left[ {M{L^2}{T^{ - 1}}} \\right]} \\over {\\left[ {M{L^2}} \\right]}}

= [T -1 ] = frequency

Q68
The unit of permittyvity of free space, varepsilon0{\\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 =

1over4pivarepsilon0q2overr2{1 \\over {4\\pi {\\varepsilon _0}}}{{{q^2}} \\over {{r^2}}}

Rightarrow\\Rightarrow

varepsilon0=q2over4piFr2{\\varepsilon _0} = {{{q^2}} \\over {4\\pi F{r^2}}}

therefore\\therefore Unit of

varepsilon0{\\varepsilon _0}

= coulomb 2 /newton-metre 2

Q69
Taking into account of the significant figures, what is the value of 9.99m−0.0099m9.99 m - 0.0099m?
A 9.98 m
B 9.980 m
C 9.9 m
D 9.9801 m
Correct Answer
Option A
Solution
9.99−0.00999.99 - 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.

Q70
Dimensions of stress are :
A [ML2T−2]\left[ {M{L^{ 2}}{T^{ - 2}}} \right]
B [ML0T−2]\left[ {M{L^{ 0}}{T^{ - 2}}} \right]
C [ML−1T−2]\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]
D [MLT−2]\left[ {M{L^{}}{T^{ - 2}}} \right]
Correct Answer
Option C
Solution
Stress=ForceoverAreaStress = {{Force} \\over {Area}}
=left[MLT−2overL2right]=left[ML−1T−2right]= \\left[ {{{ML{T^{ - 2}}} \\over {{L^2}}}} \\right] = \\left[ {M{L^{ - 1}}{T^{ - 2}}} \\right]
Ready for a full NEET mock test? Timed · full syllabus · instant results
Take a Mock Test →