Q.1. A metal wire has mass (0.4 ± 0.002) g, radius (0.3 ± 0.001) mm and length (5 ± 0.02) cm. The maximum possible percentage error in the measurement of density will nearly be [2023]
A: 1.2%
B: 1.3%
C: 1.6%
D: 1.4%
Ans: C
Solution:
Q.2. The errors in the measurement which arise due to unpredictable fluctuations in temperature and voltage supply are [2023]
A: Instrumental errors
B: Personal errors
C: Least count errors
D: Random errors
Ans: d
Solution: The errors which cannot be associated with any systematic or constant cause are called random errors. These errors can arise due to unpredictable fluctuations in experimental conditions. e.g., random change in pressure, temperature, voltage supply etc.
Q.3. The dimensions [MLT^{–2}A^{–2} ] belong to the [2022]
A: Selfinductance
B: Magnetic permeability
C: Electric permittivity
D: Magnetic flux
Ans: B
Solution:
The dimensional formula of magnetic permeability is [MLT^{–2}A^{–2} ]
Q.4. The area of a rectangular field (in m2 ) of length 55.3 m and breadth 25 m after rounding off the value for correct significant digits is [2022]
A: 1382
B: 1382.5
C: 14 * 10^{2 }
D: 138 * 10^{1}
^{Ans: C}
^{Solution:}
Area = Length × Breadth
= 55.3 × 25 m^{2}
= 1382.5 m^{2}
= 14 × 10^{2} m^{2}
(Rounding off of two significant figures)
Q.5. Match ListI with ListII
Choose the correct answer from the options given below [2022]
A: (a)  (ii), (b)  (iv), (c)  (i), (d)  (iii)
B: (a)  (ii), (b)  (iv), (c)  (iii), (d)  (i)
C: (a)  (iv), (b)  (ii), (c)  (i), (d)  (iii)
D: (a)  (ii), (b)  (i), (c)  (iv), (d)  (iii)
Ans: A
Solution:
Q.6. Plane angle and solid angle have [2022]
A: Dimensions but no units
B: No units and no dimensions
C: Both units and dimensions
D: Units but no dimensions
Ans: D
Solution:
Q.7. 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.8. 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][T^{2}]
Ans: D
Solution:
Q.9. If E and G respectively denote energy and gravitational constant, then E/G has the dimensions of : [2021]
A: [M] [L^{0}] [T^{0}]
B: [M^{2}] [L^{–2}] [T^{–1]}
C: [M^{2}] [L^{–1}] [T^{0}]
D: [M] [L^{–1}] [T^{–1}]
Ans: C
Solution:
E = energy = [ML^{2}T^{–2}]
G = Gravitational constant = [M^{–1}L^{3}T^{–2}]
So,
E/G = [E]/[G]
= ML2T–2/M^{–1}L^{3}T^{–2}
= [M^{2}L^{1}T^{0}]
Q.10. Taking into account of the significant figures, what is the value of 9.99 m0.0099 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.11. 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.12. 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
Q.13. 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]
Q.14. 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.15. 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:
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1. What is the importance of understanding the concept of physical world, units, and measurement in the NEET exam? 
2. How can I improve my understanding of units and measurement for the NEET exam? 
3. Can you provide some examples of commonly used units in physics? 
4. How can I ensure accurate measurements in physics experiments for the NEET exam? 
5. What is the significance of understanding the physical world in the NEET exam? 
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