Previous Year Questions (2014-22): Physical World, Units & Measurement - Notes | Study Physics Class 11 - NEET

Q.5. 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,


Q.6. 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:


Q.7. 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]

Q.8. 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

Q.9. 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%

Q.10. 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

A:

B:

C:

D:
Ans: D
Solution:

Q.12. 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]
Ans: D
Solution:


Q.13. 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: