A screw gauge gives the following reading when used to measure the diameter of a wire. Main scale reading : 0 mm
Circular scale reading: 52 division
Given that 1 mm on the main scale corresponds to 100 divisions of the circular scale. The diameter of wire from the above data is
0.052 cm
0.026 cm
0.005
0.005
A.
0.052 cm
Least count of screw gauge =1/100 mm = 0.01 mn
Diameter - Divisions on cirular scale × least count + main scale reading = 52 × 1/100 + 0
100 = 0.52 mm
diameter = 0.052 cm
Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale. The total number of divisions on the circular scale is 50. Further, it is found that the screw gauge has a zero error of − 0.03 mm while measuring the diameter of a thin wire, a student notes the main scale reading of 3 mm and the number of circular scale divisions in line with the main scale as 35. The diameter of the wire is
3.32 mm
3.73 mm
3.67 mm
3.67 mm
D.
3.67 mm
Diameter = M.S.R. + C.S.R × L.C. + Z.E.
= 3 + 35 × (0.5/50) + 0.03
= 3.38 mm
In an experiment, the angles are required to be measured using an instrument. 29 divisions of the main scale exactly coincide with the 30 divisions of the vernier scale. If the smallest division of the main scale is half–a–degree (= 0.5o° ), then the least count of the instrument is
one minute
half minute
One degree
One degree
A.
one minute
1VSD = 29/30 MSD
L.C. = 1 MSD – 1 VSD
= 1/30 MSD
Let [ε_{0}] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then
[ε_{0}] = [M^{-1}L^{-3}T^{2}A]
[ε_{0}] = [M^{-1}L^{-3}T^{4}A^{2}]
[ε_{0}] =[M^{-2}L^{2}T^{-1}A^{-2}]
[ε_{0}] =[M^{-2}L^{2}T^{-1}A^{-2}]
B.
[ε_{0}] = [M^{-1}L^{-3}T^{4}A^{2}]
From Coulomb's Law, F
On Substituting the units, we get
A spectrometer gives the following reading when used to measure the angle of a prism.
Main scale reading: 58.5 degree
Vernier scale reading: 09 divisions
Given that 1 division on the main scale corresponds to 0.5 degree. Total divisions on the vernier scale is 30 and match with 29 divisions of the main scale. The angle of the prism from the above data
58.59 degree
58.77 degree
58.65 degree
58.65 degree
C.
58.65 degree
The dimension of magnetic field in M, L, T and C (Coulomb) is given as
MLT^{−1}C^{−1}
MT^{2}C^{−2}
MT^{−1}C^{−1}
MT^{−1}C^{−1}
C.
MT^{−1}C^{−1}
F = qvB
B = F/qv
= MC^{−1} T^{−1}
The respective number of significant figures for the numbers 23.023, 0.0003 and 2.1 × 10^{–3} are
5,1,2
5,1,5
5,5,2
5,5,2
A.
5,1,2
Which one of the following represents the correct dimensions of the coefficient of viscosity?
ML^{−1} T^{−2}
MLT^{−1}
ML^{−1} T^{−1}
ML^{−1} T^{−1}
C.
ML^{−1} T^{−1}
Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale. The total number of divisions on the circular scale is 50. Further, it is found that the screw gauge has a zero error of − 0.03 mm while measuring the diameter of a thin wire, a student notes the main scale reading of 3 mm and the number of circular scale divisions in line with the main scale as 35. The diameter of the wire is
3.32 mm
3.73 mm
3.67 mm
3.67 mm
D.
3.67 mm
Diameter = M.S.R. + C.S.R × L.C. + Z.E.
= 3 + 35 × (0.5/50) + 0.03 = 3.38 mm
A student measured the length of a rod and wrote it as 3.50 cm. Which instrument did he use to measure it?
A meter scale.
A vernier calliper where the 10 divisions in vernier scale match with 9 division in the main scale and main scale have 10 divisions in 1 cm.
A screw gauge having 100 divisions in the circular scale and pitch as 1 mm.
A screw gauge having 100 divisions in the circular scale and pitch as 1 mm.
B.
A vernier calliper where the 10 divisions in vernier scale match with 9 division in the main scale and main scale have 10 divisions in 1 cm.
If student measures 3.50 cm, it means that there is an uncertainty of order 0.01 cm
For vernier scale with 1 MSD = 1mm
and 9 MSD = 10 VSD
LC of vernier scale with 1 MSD - 1 VSD