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

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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

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A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?

  • 0.75 mm

  • 0.80 mm

  • 0.70 mm

  • 0.70 mm


A.

0.75 mm

Given that the screw gauge has zero error.
So, least count of screw gauge = 0.5/50 mm

The thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line. we have

 =0 .50 mm + (25) x 0.5/50 mm
= 0.50 mm + 0.25 mm = 0.75 mm

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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

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Let [ε0] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then

  • 0] = [M-1L-3T2A]

  • 0] = [M-1L-3T4A2]

  • 0] =[M-2L2T-1A-2]

  • 0] =[M-2L2T-1A-2]


B.

0] = [M-1L-3T4A2]

From Coulomb's Law, F



On Substituting the units, we get


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