Q.1. 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.2. 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.3. 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.4. 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.5. 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.6. 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.7. 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.8. 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.9. 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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