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Units & Measurements

NEET Physics · 152 questions · Page 13 of 16 · Click an option or "Show Solution" to reveal answer

Q121
The E and G respectively denote energy and gravitational constant, then EG{E \over G} has the dimensions of :
A [M 2 ] [L −-2 ] [T −-1 ]
B [M 2 ] [L −-1 ] [T 0 ]
C [M] [L −-1 ] [T −-1 ]
D [M] [L 0 ] [T 0 ]
Correct Answer
Option B
Solution

E = energy = [ML 2 T −-2 ] G = Gravitational constant = [M −-1 L 3 T −-2 ] So,

EG=[E][G]=ML2T−2M−1L3T−2=[M2L−1T0]{E \over G} = {{[E]} \over {[G]}} = {{M{L^2}{T^{ - 2}}} \over {{M^{ - 1}}{L^3}{T^{ - 2}}}} = [{M^2}{L^{ - 1}}{T^0}]
Q122
A screw gauge gives the following readings when used to measure the diameter of a wire Main scale reading : 0 mm Circular scale reading : 52 divisions Given that 1 mm on main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is :
A 0.052 cm
B 0.52 cm
C 0.026 cm
D 0.26 cm
Correct Answer
Option A
Solution

L.C. =

PitchCSD{{Pitch} \over {CSD}}
=1mm100= {{1mm} \over {100}}

= 0.01 m = 0.001 cm Radius = M.S. + CSR(L.C) = 0 + 52 (0.001) = 0.052 cm

Q123
If force [F], acceleration [A] and time [T] are chosen as the fundamental physical quantities. Find the dimensions of energy.
A [F] [A −-1 ] [T]
B [F] [A] [T]
C [F] [A] [T 2 ]
D [F] [A] [T −-1 ]
Correct Answer
Option C
Solution

E ∝\propto F a A b T c [M 1 L 2 T −-2 ] ∝\propto [M 1 L 1 T −-2 ] a [LT −-2 ] b [T] c [ML 2 T –2 ] ∝\propto [M a L a + b T –2a – 2b + c ] a = 1 a + b = 2 ⇒\Rightarrow b = 1 −-2a −- 2b + c = −- 2 ⇒\Rightarrow c = 2 a = 1, b = 1, c = 2 E ∝\propto [F] [A] [T 2 ]

Q124
A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale. The pitch of the screw gauge is :
A 0.25 mm
B 0.5 mm
C 1.0 mm
D 0.01 mm
Correct Answer
Option B
Solution

Least count of screw gauge =

PitchNo of divisions on circular scale\frac{Pitch}{No\ of\ divisions\ on\ circular\ scale}

0.01 mm =

Pitch50{{Pitch} \over {50}}

Pitch = 0.5 mm

Q125
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.

Q126
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=ForceAreaStress = {{Force} \over {Area}}
=[MLT−2L2]=[ML−1T−2]= \left[ {{{ML{T^{ - 2}}} \over {{L^2}}}} \right] = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]
Q127
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=1.6×10−19J1eV = 1.6 \times {10^{ - 19}}J
1J=11.6×10−19eV1J = {1 \over {1.6 \times {{10}^{ - 19}}}}eV
10−20J=10−201.6×10−19eV{10^{ - 20}}J = {{{{10}^{ - 20}}} \over {1.6 \times {{10}^{ - 19}}}}eV

= 0.06 eV

Q128
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 = A2B1/2C1/3D3{{{A^2}{B^{1/2}}} \over {{C^{1/3}}{D^3}}}, will be :
A 16%
B -10%
C 10%
D (313)\left( {{3 \over {13}}} \right)%
Correct Answer
Option A
Solution

Given, X =

A2B1/2C1/3D3{{{A^2}{B^{1/2}}} \over {{C^{1/3}}{D^3}}}
ΔXX=2ΔAA+12ΔBB+13ΔCC+3ΔDD{{\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

ΔXX×100=2(1%)+12(2%)+13(3%)+3(4%){{\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 %

Q129
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.

Q130
A physical quantity of the dimensions of length that can be formed out of c, G and e24πε0{{{e^2}} \over {4\pi {\varepsilon _0}}} is [c is velocity of light, G is the universal constant of gravitation and e is charge]
A c2[G−e24πε0]1/2{c^2}{\left[ {G - {{{e^2}} \over {4\pi {\varepsilon _0}}}} \right]^{1/2}}
B 1c2[e2G 4πε0]1/2{1 \over {{c^2}}}{\left[ {{{{e^2}} \over {G\,4\pi {\varepsilon _0}}}} \right]^{1/2}}
C 1cGe2 4πε0{1 \over c}G{{{e^2}} \over {\,4\pi {\varepsilon _0}}}
D 1c2[Ge2 4πε0]1/2{1 \over {{c^2}}}{\left[ {G{{{e^2}} \over {\,4\pi {\varepsilon _0}}}} \right]^{1/2}}
Correct Answer
Option D
Solution

Dimension of

e24πε0{{{e^2}} \over {4\pi {\varepsilon _0}}}

= [ F×\times d 2 ] = [ML 3 T -2 ] Dimension of G = [M -1 L 3 T -2 ], Dimension of c = [LT -1 ] Now assume dimension of length is related as, L = [c] x [G] y [

e24πε0{{{e^2}} \over {4\pi {\varepsilon _0}}}

] z ∴\therefore [L 1 ] = [ML 3 T -2 ] z [M -1 L 3 T -2 ] y [LT -1 ] x Comparing both sides and solving we get, x = -2, y =

12{1 \over 2}

, z =

12{1 \over 2}

∴\therefore L =

1c2[Ge2 4πε0]1/2{1 \over {{c^2}}}{\left[ {G{{{e^2}} \over {\,4\pi {\varepsilon _0}}}} \right]^{1/2}}
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