JEE Main Units and Measurements Previous year Question

11.(JEE Main 2020 (Online) 3rd September Morning Slot )

Using screw gauge of pitch 0.1 cm and 50 divisions on its circular scale, the thickness of an object is measured. It should correctly be recorded as

(A) 2.123 cm

(B) 2.124 cm

(C) 2.125 cm

(D) 2.121 cm

Correct answer is (B)

Using a screw gauge of pitch 0.1 cm and 50 divisions on its circular scale, the thickness of an object is measured as:

Measurement = (Main scale reading) + (Circular scale reading × Least count)

where the least count is calculated as the pitch of the screw gauge divided by the number of divisions on the circular scale:

Least count = (Pitch of screw gauge) / (Number of circular scale divisions)

Least count = $\frac{0.1}{50}$ = 0.002 cm

Now if we multiply division of circular scale with least count then we get 0^th digit of fraction part even.

Here only option B has 0^th digit of fraction part even.

12.(JEE Main 2020 (Online) 9th January Morning Slot )

If the screw on a screw-gauge is given six rotations, it moves by 3 mm on the main scale. If there are 50 divisions on the circular scale the least count of the screw gauge is :

(A) 0.001 mm

(B) 0.01 cm

(C) 0.02 mm

(D) 0.001 cm

Correct answer is (D)

Pitch = $\frac{3}{6}$ mm = 0.5 mm

Least count = $\frac{0.5}{50}$ mm

= $\frac{1}{100}$ = 0.01 mm = 0.001 cm

13.(JEE Main 2019 (Online) 12th January Morning Slot )

The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5 $μ$ m diameter of a wire is :

(A) 500

(B) 100

(C) 200

(D) 50

Correct answer is (C)

Least count = $\frac{P i t c h}{N u m b e r o f d i v i s i o n o n c i r c u l a r s c a l e}$

5 $\times$ 10^{$-$ 6} = $\frac{10^{- 3}}{N}$

N = 200

14.(JEE Main 2019 (Online) 9th January Evening Slot )

The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line.

The readings of the main scale and the circular scale, for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is :

(A) 5.755 mm

(B) 5.950 mm

(C) 5.725 mm

(D) 5.740 mm

Correct answer is (C)

We know,

Least count (LC) = $\frac{P i t c h}{n o . o f d i v i s i o n s}$

$∴$   LC = $\frac{0.5}{100}$

= 0.5 $\times$ 10^{$-$ 2} mm

Reading = MSR + CSR $-$ positive error

Given, Main scale reading (MSR) = 5.5 mm

Circular scale reading (CSR)

= 48 $\times$ 0.5 $\times$ 10^{$-$ 2} mm

= 0.24

As zero of its circular scale lines 3 division below the mean line, it means error is position error.

$∴$   positive error

= 3 $\times$ 0.5 $\times$ 10^{$-$ 2} mm

= 0.015 mm

$∴$   Reading = 5.5 + 0.24 $-$ 0.015

= 5.725 mm

15.(JEE Main 2018 (Online) 15th April Morning Slot )

In a screw gauge, $5$ complete rotations of the screw cause it to move a linear distance of $0.25$ $c m .$ There are $100$ circular scale divisions. The thickness of a wire measured by this screw gauge gives a reading of $4$ main scale divisions and $30$ circular scale divisions. Assuming negligible zero error, the thickness of the wire is :

(A) $0.4300$ $c m$

(B) $0.2150$ $c m$

(C) $0.3150$ $c m$

(D) $0.0430$ $c m$

Correct answer is (B)

5 complete rotations = 0.25 cms

So, 1 complete rotation of screw = 0.05 cm

$∴$ 1 main scale division = 0.05 cm

1 circular scale = $\frac{0.05}{100}$ = 5 $\times$ 10^{$-$ 4} cm

Thickness of a wire

= 4 main scale and 30 circular scale divisions

= 4 $\times$ 0.05 + 30 $\times$ 5 $\times$ 10 ^{$-$ 4}

= 0.2150 cm.