Experimental Skills | Physics JEE previous year questions with Solutions

Q 1 :
While measuring diameter of wire using screw gauge the following readings were noted. Main scale reading is 1 mm and circular scale reading is equal to 42 divisions. Pitch of screw gauge is 1 mm and it has 100 divisions on circular scale. The diameter of the wire is $\frac{x}{50}$ mm. The value of $x$ is [2024]

142
21
42
71

1

2

3

4

(4)

MSR = 1 mm, CSR = 42, pitch = 1 mm

$L C = \frac{pitch}{No . of CSD} = (\frac{1}{100}) = 0.01 mm$

Reading by screw gauge = Main scale reading + Least count $\times$ circular scale reading

Diameter = MSR + LC $\times$ CSD = 1 mm + 0.1 mm $\times$ 42 = 1.42 mm

Diameter = 1.42 mm = $\frac{x}{50} ∴ x = 71$

Q 2 :
There are 100 divisions on the circular scale of a screw gauge of pitch 1 mm. With no measuring quantity in between the jaws, the zero of the circular scale lies 5 divisions below the reference line. The diameter of a wire is then measured using this screw gauge. It is found that 4 linear scale divisions are clearly visible while 60 divisions on circular scale coincide with the reference line. The diameter of the wire is [2024]

4.60 mm
3.35 mm
4.65 mm
4.55 mm

1

2

3

4

(4)

Pitch P = 1 mm, N = 100

least count $C = \frac{P}{N} = \frac{1 mm}{100} = 0.01 mm$

The instrument has a positive zero error.

$e = N C = 5 \times 0.01 = 0.05 mm$

Main scale reading is 4 $\times$ (mm) = 4 mm

Circular scale reading is 60 $\times$ 0.01 = 0.6 mm

Observed reading is $R_{0}$ = 4 + 0.6 = 4.6 mm

So True reading $R_{0} - e$ = 4.6 - 0.05 = 4.55 mm

Q 3 :
The least count of a screw gauge is 0.01 mm. If the pitch is increased by 75% and number of divisions on the circular scale is reduced by 50%, the new least count will be ____ $\times 10^{- 3}$ mm. [2025]

0%

(35)

Given least count of Screw Gauge = 0.01 mm

$L . C = \frac{pitch}{No. of circular turns} = \frac{P}{N} = 0.01 mm$

$New pitch = \frac{P (1 + 0.75)}{N (1 - 0.5)} = \frac{P}{N} [\frac{1.75}{0.5}]$

$= (0.01) 3.5 = 0.035 mm = 35 \times 10^{- 3} mm$

Q 4 :
A tiny metallic rectangular sheet has length and breadth of 5 mm and 2.5 mm, respectively. Using a specially designed screw gauge which has pitch of 0.75 mm and 15 divisions in the circular scale, you are asked to find the area of the sheet. In this measurement, the maximum fractional error will be $\frac{x}{100}$ where $x$ is _____ . [2025]

0%

(3)

Least count of screw gauge $= \frac{Pitch}{Number of circular scale divisions}$

$\Rightarrow LC = \frac{0.75 mm}{15} = 0.05 mm$

$Length, l = 5 mm$

$Breadth, b = 2.5 mm$

$Area = l b$

$\Rightarrow % error in area = (% error in length) + (% error in breadth)$

$= \frac{Δ l}{l} \times 100 + \frac{Δ b}{b} \times 100$

$\frac{d A}{A} % = \frac{0.05}{5} \times 100 + \frac{0.05}{2.5} \times 100 = 3 %$

Topic Question Set

Q 3 :
The least count of a screw gauge is 0.01 mm. If the pitch is increased by 75% and number of divisions on the circular scale is reduced by 50%, the new least count will be ____ $\times 10^{- 3}$ mm. [2025]

0%

0%

Want Us to Email you About Special Offers & Updates?

Topic Question Set

Q 3 : The least count of a screw gauge is 0.01 mm. If the pitch is increased by 75% and number of divisions on the circular scale is reduced by 50%, the new least count will be ____ ×10-3 mm. [2025]

0%

0%

Want Us to Email you About Special Offers & Updates?

Q 3 :
The least count of a screw gauge is 0.01 mm. If the pitch is increased by 75% and number of divisions on the circular scale is reduced by 50%, the new least count will be ____ $\times 10^{- 3}$ mm. [2025]