JEE Main Units and Measurements Previous year Question

6.(JEE Main 2021 (Online) 26th August Morning Shift )

In a Screw Gauge, fifth division of the circular scale coincides with the reference line when the ratchet is closed. There are 50 divisions on the circular scale, and the main scale moves by 0.5 mm on a complete rotation. For a particular observation the reading on the main scale is 5 mm and the 20^th division of the circular scale coincides with reference line. Calculate the true reading.

(A) 5.00 mm

(B) 5.25 mm

(C) 5.15 mm

(D) 5.20 mm

Correct answer is (C)

Least count (L. C.) = $\frac{0.5}{50}$

True reading = $5 + \frac{0.5}{50} \times 20 - \frac{0.5}{50} \times 5$

$= 5 + \frac{0.5}{50} (15) = 5.15$ mm

7.(JEE Main 2021 (Online) 27th July Morning Shift )

Assertion A : If in five complete rotations of the circular scale, the distance travelled on main scale of the screw gauge is 5 mm and there are 50 total divisions on circular scale, then least count is 0.001 cm.

Reason R :

Least Count = $\frac{P i t c h}{T o t a l d i v i s i o n s o n c i r c u l a r s c a l e}$

In the light of the above statements, choose the most appropriate answer from the options given below :

(A) A is not correct but R is correct.

(B) Both A and R are correct and R is the correct explanation of A.

(C) A is correct but R is not correct.

(D) Both A and R are correct and R is NOT the correct explanation of A.

Correct answer is (A)

Least Count = $\frac{P i t c h}{T o t a l d i v i s i o n s o n c i r c u l a r s c a l e}$

In 5 revolution, distance travel, 5 mm

In 1 revolution, it will travel 1 mm.

So least count = $\frac{1}{50}$ = 0.02

8.(JEE Main 2021 (Online) 25th February Morning Shift )

The pitch of the screw gauge is 1 mm and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while 72^nd division on circular scale coincides with the reference line. The radius of the wire is :

(A) 1.80 mm

(B) 0.90 mm

(C) 0.82 mm

(D) 1.64 mm

Correct answer is (C)

Least count = $\frac{1 m m}{100} = 0.01 m m$

zero error = + 8 $\times$ LC = + 0.08 mm

True reading (Diameter)

= (1 mm + 72 $\times$ LC) $-$ (Zero error)

= (1 mm + 72 $\times$ 0.01 mm) $-$ 0.08 mm

= 1.72 mm $-$ 0.08 mm

= 1.64 mm

Therefore, radius = $\frac{1.64}{2}$ = 0.82 mm.

9. (JEE Main 2021 (Online) 25th July Morning Shift )

Student A and student B used two screw gauges of equal pitch and 100 equal circular divisions to measure the radius of a given wire. The actual value of the radius of the wire is 0.322 cm. The absolute value of the difference between the final circular scale readings observed by the students A and B is ______________.

[Figure shows position of reference 'O' when jaws of screw gauge are closed]

Given pitch = 0.1 cm.

JEE Main 2021 (Online) 25th July Morning Shift Physics - Units & Measurements Question 65 English

Correct answer is (13)

For (A)

Reading = MSR + CSR + Error

0.322 = 0.300 + CSR + 5 $\times$ LC

0.322 = 0.300 + CSR + 0.005

CSR = 0.017

For (B)

Reading = MSR + CSR + Error

0.322 = 0.200 + CSR + 0.092

CSR = 0.030

Difference = 0.030 $-$ 0.017 = 0.013 cm

Division on circular scale = $\frac{0.013}{0.001} = 13$

10.(JEE Main 2020 (Online) 6th September Morning Slot )

A screw gauge has 50 divisions on its circular scale. The circular scale is 4 units ahead of the pitch scale marking, prior to use. Upon one complete rotation of the circular scale, a displacement of 0.5mm is noticed on the pitch scale. The nature of zero error involved and the least count of the screw gauge, are respectively :

(A) Positive, 0.1 mm

(B) Positive, 0.1 $μ$ m

(C) Positive, 10 $μ$ m

(D) Negative, 2 $μ$ m

Correct answer is (C)

Least count of screw gauge

= $\frac{0.5}{50}$

= 1 $\times$ 10^-5 m

= 10 $μ$ m

Zero error in positive.