Practice Start Test

Units & Measurements

NEET Physics · 152 questions · Page 6 of 16 · Click an option or "Show Solution" to reveal answer

Q51
If dimensions of critical velocity upsilon\\upsilon c of a liquid flowing through a tube are expressed as left[etaxrhoyrzright]\\left[ {{\\eta ^x}{\\rho ^y}{r^z}} \\right] where eta,rho\\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\\upsilon

c =

left[etaxrhoyrzright]\\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

Q52
If energy (E), velocity (V) and time (T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be
A [EV−2T−2]\left[ {E{V^{ - 2}}{T^{ - 2}}} \right]
B [E−2V−1T−3]\left[ {{E^{ - 2}}{V^{ - 1}}{T^{ - 3}}} \right]
C [EV−2T−1]\left[ {E{V^{ - 2}}{T^{ - 1}}} \right]
D [EV−1T−2]\left[ {E{V^{ - 1}}{T^{ - 2}}} \right]
Correct Answer
Option A
Solution

Let surface tension S =kE

aa

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

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

=

left[ML2T−2right]a{\\left[ {M{L^2}{T^{ - 2}}} \\right]^a}
left[LT−1right]b{\\left[ {L{T^{ - 1}}} \\right]^b}
left[Tright]c{\\left[ T \\right]^c}
left[ML0T−2right]\\left[ {M{L^0}{T^{ - 2}}} \\right]

=

left[MaL2a+bT−2a−b+cright]\\left[ {{M^a}{L^{2a + b}}{T^{ - 2a - b + c}}} \\right]

Comparing both sides of the equation we get,

aa

= 1 ....(1) 2

aa

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

aa

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

aa

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

left[EV−2T−2right]\\left[ {E{V^{ - 2}}{T^{ - 2}}} \\right]
Q53
If force (F), velocity (V) and time (T) are taken as fundamental units, then dimensions of mass are
A [FVT−1]\left[ {FV{T^{ - 1}}} \right]
B [FVT−2]\left[ {FV{T^{ - 2}}} \right]
C [FV−1T−1]\left[ {F{V^{ - 1}}{T^{ - 1}}} \right]
D [FV−1T]\left[ {F{V^{ - 1}}T} \right]
Correct Answer
Option D
Solution

Let mass m =kF

aa

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

left[ML0T0right]\\left[ {{{M{L^0}{T^{ 0}}}}} \\right]

=

left[MLT−2right]a{\\left[ {M{L}{T^{ - 2}}} \\right]^a}
left[LT−1right]b{\\left[ {L{T^{ - 1}}} \\right]^b}
left[Tright]c{\\left[ T \\right]^c}
left[ML0T0right]\\left[ {M{L^0}{T^{0}}} \\right]

=

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

Comparing both sides of the equation we get,

aa

= 1 ....(1) 2

aa

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

aa

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

aa

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

left[FV−1Tright]\\left[ {F{V^{ - 1}}{T}} \\right]
Q54
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\\times distance = [MLT -2 ][L] = [ML 2 T -2 ] Torque = Force times\\times Force arm = [MLT -2 ][L] = [ML 2 T -2 ]

Q55
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 =a3b2overcd = {{{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

=a3b2overcd= {{{a^3}{b^2}} \\over {cd}}

% error in P :

DeltaPoverPtimes100{{\\Delta P} \\over P} \\times 100

=

left[3Deltaaovera+2Deltaboverb+Deltacoverc+Deltadoverdright]times100\\left[ {3{{\\Delta a} \\over a} + 2{{\\Delta b} \\over b} + {{\\Delta c} \\over c} + {{\\Delta d} \\over d}} \\right] \\times 100

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

Q56
The dimensions of left(mu0varepsilon0right)−1/2{\\left( {{\\mu _0}{\\varepsilon _0}} \\right)^{ - 1/2}} are
A [L1/2T−1/2]\left[ {{L^{1/2}}{T^{ - 1/2}}} \right]
B [L−1T]\left[ {{L^{ - 1}}T} \right]
C [LT−1]\left[ {L{T^{ - 1}}} \right]
D [L1/2T1/2]\left[ {{L^{1/2}}{T^{1/2}}} \right]
Correct Answer
Option C
Solution
left(mu0varepsilon0right)−1/2{\\left( {{\\mu _0}{\\varepsilon _0}} \\right)^{ - 1/2}}

=

1oversqrtmu0varepsilon0{1 \\over {\\sqrt {{\\mu _0}{\\varepsilon _0}} }}

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

left(mu0varepsilon0right)−1/2{\\left( {{\\mu _0}{\\varepsilon _0}} \\right)^{ - 1/2}}

= [c] =[LT -1 ]

Q57
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\\propto v Rightarrow\\Rightarrow F = kv where k is constant of proportionality. therefore\\therefore k =

Foverv{F \\over v}

=

Noverm,s−1{N \\over {m\\,{s^{ - 1}}}}

=

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

=

kg,s−1kg\\,{s^{ - 1}}
Q58
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

govercm3{g \\over {c{m^3}}}

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

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

therefore\\therefore 4

govercm3{g \\over {c{m^3}}}

= x

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

Rightarrow\\Rightarrow x = 40 unit

Q59
The dimensions of left(mu0varepsilon0right)−1/2{\\left( {{\\mu _0}{\\varepsilon _0}} \\right)^{ - 1/2}} are
A [L1/2T−1/2]\left[ {{L^{1/2}}{T^{ - 1/2}}} \right]
B [L−1T]\left[ {{L^{ - 1}}T} \right]
C [LT−1]\left[ {L{T^{ - 1}}} \right]
D [L1/2T1/2]\left[ {{L^{1/2}}{T^{1/2}}} \right]
Correct Answer
Option C
Solution
left(mu0varepsilon0right)−1/2{\\left( {{\\mu _0}{\\varepsilon _0}} \\right)^{ - 1/2}}

=

1oversqrtmu0varepsilon0{1 \\over {\\sqrt {{\\mu _0}{\\varepsilon _0}} }}

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

left(mu0varepsilon0right)−1/2{\\left( {{\\mu _0}{\\varepsilon _0}} \\right)^{ - 1/2}}

= [c] =[LT -1 ]

Q60
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\\therefore

h=1over2gt2h = {1 \\over 2}g{t^2}

Rightarrow\\Rightarrow

g=2hovert2g = {{2h} \\over {{t^2}}}

Maximum percentage error,

Deltagovergtimes100=left[Deltahoverh+2Deltatovertright]times100{{\\Delta g} \\over g} \\times 100 = \\left[ {{{\\Delta h} \\over h} + 2{{\\Delta t} \\over t}} \\right] \\times 100

Given that

Deltahoverhtimes100{{{\\Delta h} \\over h} \\times 100}

= e 1 and

Deltatoverttimes100{{{\\Delta t} \\over t} \\times 100}

= e 2 therefore\\therefore

Deltagovergtimes100{{\\Delta g} \\over g} \\times 100

= e 1 + 2e 2

Ready for a full NEET mock test? Timed · full syllabus · instant results
Take a Mock Test →