NEET Previous Year Questions (2014-2024): Physical World, Units & Measurement | Physics Class 11 PDF Download

Q1: A Screw Guage gives the following readings when used to measure the diameter of a wire.
Main scale reading = 0.0 mm
Circular scale reading = 52 divisions
Given that: 1 mm on the main scale corresponds to 100 divisions of the circular scale.
The diameter of the wire from the above data is:    [2021]
(a) 0.026 cm
(b) 0.005 cm
(c) 0.52 cm
(d) 0.052 cm
Ans: (d)
Solution: Diameter of wire,


Q2: If force [F], acceleration [A] and time [T] are chosen as the fundamental physical quantities. Find the dimensions of energy.    [2021]
(a) [F][A][T^-1]
(b) [F][A^-1][T]
(c) [F][A][T]
(d) [F][A][T²]
Ans: (d)
Solution:


Q3: If E and G respectively denote energy and gravitational constant, then E/G has the dimensions of :    [2021]
(a) [M] [L⁰] [T⁰]
(b) [M²] [L^–2] [T^–1]
(c) [M²] [L^–1] [T⁰]
(d) [M] [L^–1] [T^–1]
Ans: (c)
Solution:
E = energy = [ML²T^–2]
G = Gravitational constant = [M^–1L³T^–2]
So,
E/G = [E]/[G]
= ML2T–2/M^–1L³T^–2
= [M²L^-1T⁰]

2020

Q1: A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale.   [2020]
The pitch of the screw gauge is :
(a) 0.25 mm
(b) 0.5 mm
(c) 1.0 mm
(d) 0.01 mm
Ans: (b)
Least count of screw gauge =

Q2: Taking into account of the significant figures, what is the value of 9.99 m-0.0099 m?    [2020]
(a) 9.980 m
(b) 9.9 m
(c) 9.9801 m
(d) 9.98 m
Ans: (d)

As per rule
least no. of place in decimal portion of any number 9.98

Q3: Dimensions of stress are :   [2020]
(a) [ML²T^-2]
(b) ML⁰T^-2]
(c) ML^-1T^-2]
(d) MLT^-2]
Ans: (c)

Q4: The energy required to break one bond in DNA is 10^-20 J. This value in eV is nearly :   [2020]
(a) 0.6
(b) 0.06
(c) 0.006
(d) 6
Ans: (b)

2019

Q1: 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  will be     [2019]
(a)
(b) 16 %
(c) -10 %
(d) 10 %
Ans: (b)
Given


 

= 2% + 1%+ 1%+ 12%= 16%

2018

Q1: 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 :-    [2018]
(a) 0.521 cm
(b) 0.525 cm
(c) 0.053 cm
(d) 0.529 cm
Ans: (d)
Solution:
Reading of screw gauge
= MSR + VSR × LC + zero error
= 0.5 cm + 25 × 0.001 cm + 0.004 cm
= 0.529 cm

2017

Q1: A physical quantity of the dimensions of length that can be formed out of  is velocity of light, G is universal constant of gravitation and e is charge] :-    [2017]

(a)

(b)

(c)

(d)
Ans: (d)
Solution:

2016

Q1: 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 ?           [2016]
(a)
(b)
(c)
(d)
Ans: (a)
According to question,
L $\propto$ h^p c^q G^r
[M⁰LT⁰] = [ML²T^-1]^p [LT^-1]^q [M^-1L³T^-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 

2015

Q1: If dimensions of critical velocity v_c of a liquid flowing through a tube are expressed as  where η,ρ 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                  [2015]
Ans: (c)

Put dimensions of various quantities,
[M⁰LT^-1] = [ML^-1T^-1]^x [ML^-3T⁰]^y [M⁰LT⁰]^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 

Question for NEET Previous Year Questions (2014-2024): Physical World, Units & Measurement

Try yourself:Q2: If energy (E), velocity (V) and time (T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be:    [2015]

• A.

  [E^-2 V^-1 T^-3]
• B.

  [E V^-2 T^-1]
• C.

  [E V^-1 T^-2]
• D.

  [E V^-2 T^-2]

Explanation

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2014

Q1: If force (F), velocity (V) and time(T) are taken as fundamental units, then the dimensions of mass are:    [2014]
(a) [F V⁻¹ T⁻¹]
(b) [F V⁻¹ T]
(c) [F V T⁻¹]
(d) [F V T⁻²]
Ans: (b)
Solution: