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Units and Measurements

139964 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

1

2.121 cm

2

2.124 cm

3

2.125 cm

4

2.123 cm

Explanation:

B Let t be the thickness of the object
So, $t =$ MSR $+ (n \times LC)$
Where, $n =$ No. of circular scale division lie over reference line
MSR $=$ Main scale reading
$LC =$ Least count of screw gauge
Least count of gauge is
$(LC) = \frac{Pitch}{Number of division on circular scale}$
$= \frac{0.1}{50} = 0.002 cm$
When we multiply by division number it must be even because LC is 0.02 so any measurement will be an integral multiple of LC.
Therefore, the digit at the third decimal place will be an even number.
So, the answer is $2.124 cm$ .

JEE Mains- 2020

Units and Measurements

139965 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 $0.5 mm$ is noticed on the pitch scale. The nature of zero error involved and the least count of the screw gauge, are respectively

1 negative,

2 μ m

2 positive,

10 μ m

3 positive,

0.1 mm

4 positive,

0.1 μ m

Explanation:

B Given,
Number of division on circular scale of screw gauge $=$ 50 pitch $= 0.5 mm$
Least count of screw gauge
$(LC) = \frac{Pitch}{No. of division on circular scale}$
$= \frac{0.5}{50} = mm$
$= 1 \times 10^{- 5} m$
$= 10 μ m$
The nature is positive and least count is $10 μ m$

JEE Mains- 2020

Units and Measurements

139966 A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings; 5.50 $mm, 5.55 mm, 5.45 mm$ and $5.65 mm$ . The average of these four readings is $5.5375 mm$ and the standard deviation of the data is $0.07395 mm$ . The average diameter of the pencil should be therefore recorded as

1

(5.5375 \pm 0.0739) mm

2

(5.5375 \pm 0.0740) mm

3

(5.538 \pm 0.074) mm

4

(5.54 \pm 0.07) mm

Explanation:

D Four reading of vernier calipers are
$5.50 mm, 5.55 mm, 5.45 mm, 5.65 mm$ .
The average of these four readings
$= \frac{5.5 + 5.55 + 5.45 + 5.65}{4}$
$= 5.5375 mm$
Standard deviation of the data is 0.07395
The significant rule says that reading should have the same significant figure as that of the reading given.
It we do round off 5.5375 then we will get 5.54 as $7 > 5$
If we do round off 0.07395 then we will get 0.07 as $3 <$ 5
So, Average diameter will be
$(5.54 \pm 0.07) m$ as $0.07 mm$ is standard diviation of the data.

JEE Mains- 2020

Units and Measurements

139967 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

1

0.001 cm

2

0.01 cm

3

0.02 cm

4

0.001 cm

Explanation:

A Given,
On six rotation, readings of main scale changed by 3 $mm$
No. of division on circular scale $= 50$
Pitch $= \frac{Distance moved by screw}{Number of rotation}$
$= \frac{3 mm}{6} = 0.5 mm$
Least count of screw gauge
$(LC) = \frac{Pitch}{No. of circular scale division}$
$= \frac{0.5 mm}{50}$
$= 0.01 mm$
$= 0.001 cm$

JEE Mains- 2020

Units and Measurements

139964 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

1

2.121 cm

2

2.124 cm

3

2.125 cm

4

2.123 cm

Explanation:

B Let t be the thickness of the object
So, $t =$ MSR $+ (n \times LC)$
Where, $n =$ No. of circular scale division lie over reference line
MSR $=$ Main scale reading
$LC =$ Least count of screw gauge
Least count of gauge is
$(LC) = \frac{Pitch}{Number of division on circular scale}$
$= \frac{0.1}{50} = 0.002 cm$
When we multiply by division number it must be even because LC is 0.02 so any measurement will be an integral multiple of LC.
Therefore, the digit at the third decimal place will be an even number.
So, the answer is $2.124 cm$ .

JEE Mains- 2020

Units and Measurements

139965 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 $0.5 mm$ is noticed on the pitch scale. The nature of zero error involved and the least count of the screw gauge, are respectively

1 negative,

2 μ m

2 positive,

10 μ m

3 positive,

0.1 mm

4 positive,

0.1 μ m

Explanation:

B Given,
Number of division on circular scale of screw gauge $=$ 50 pitch $= 0.5 mm$
Least count of screw gauge
$(LC) = \frac{Pitch}{No. of division on circular scale}$
$= \frac{0.5}{50} = mm$
$= 1 \times 10^{- 5} m$
$= 10 μ m$
The nature is positive and least count is $10 μ m$

JEE Mains- 2020

Units and Measurements

139966 A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings; 5.50 $mm, 5.55 mm, 5.45 mm$ and $5.65 mm$ . The average of these four readings is $5.5375 mm$ and the standard deviation of the data is $0.07395 mm$ . The average diameter of the pencil should be therefore recorded as

1

(5.5375 \pm 0.0739) mm

2

(5.5375 \pm 0.0740) mm

3

(5.538 \pm 0.074) mm

4

(5.54 \pm 0.07) mm

Explanation:

D Four reading of vernier calipers are
$5.50 mm, 5.55 mm, 5.45 mm, 5.65 mm$ .
The average of these four readings
$= \frac{5.5 + 5.55 + 5.45 + 5.65}{4}$
$= 5.5375 mm$
Standard deviation of the data is 0.07395
The significant rule says that reading should have the same significant figure as that of the reading given.
It we do round off 5.5375 then we will get 5.54 as $7 > 5$
If we do round off 0.07395 then we will get 0.07 as $3 <$ 5
So, Average diameter will be
$(5.54 \pm 0.07) m$ as $0.07 mm$ is standard diviation of the data.

JEE Mains- 2020

Units and Measurements

139967 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

1

0.001 cm

2

0.01 cm

3

0.02 cm

4

0.001 cm

Explanation:

A Given,
On six rotation, readings of main scale changed by 3 $mm$
No. of division on circular scale $= 50$
Pitch $= \frac{Distance moved by screw}{Number of rotation}$
$= \frac{3 mm}{6} = 0.5 mm$
Least count of screw gauge
$(LC) = \frac{Pitch}{No. of circular scale division}$
$= \frac{0.5 mm}{50}$
$= 0.01 mm$
$= 0.001 cm$

JEE Mains- 2020

Units and Measurements

139964 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

1

2.121 cm

2

2.124 cm

3

2.125 cm

4

2.123 cm

Explanation:

B Let t be the thickness of the object
So, $t =$ MSR $+ (n \times LC)$
Where, $n =$ No. of circular scale division lie over reference line
MSR $=$ Main scale reading
$LC =$ Least count of screw gauge
Least count of gauge is
$(LC) = \frac{Pitch}{Number of division on circular scale}$
$= \frac{0.1}{50} = 0.002 cm$
When we multiply by division number it must be even because LC is 0.02 so any measurement will be an integral multiple of LC.
Therefore, the digit at the third decimal place will be an even number.
So, the answer is $2.124 cm$ .

JEE Mains- 2020

Units and Measurements

139965 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 $0.5 mm$ is noticed on the pitch scale. The nature of zero error involved and the least count of the screw gauge, are respectively

1 negative,

2 μ m

2 positive,

10 μ m

3 positive,

0.1 mm

4 positive,

0.1 μ m

Explanation:

B Given,
Number of division on circular scale of screw gauge $=$ 50 pitch $= 0.5 mm$
Least count of screw gauge
$(LC) = \frac{Pitch}{No. of division on circular scale}$
$= \frac{0.5}{50} = mm$
$= 1 \times 10^{- 5} m$
$= 10 μ m$
The nature is positive and least count is $10 μ m$

JEE Mains- 2020

Units and Measurements

139966 A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings; 5.50 $mm, 5.55 mm, 5.45 mm$ and $5.65 mm$ . The average of these four readings is $5.5375 mm$ and the standard deviation of the data is $0.07395 mm$ . The average diameter of the pencil should be therefore recorded as

1

(5.5375 \pm 0.0739) mm

2

(5.5375 \pm 0.0740) mm

3

(5.538 \pm 0.074) mm

4

(5.54 \pm 0.07) mm

Explanation:

D Four reading of vernier calipers are
$5.50 mm, 5.55 mm, 5.45 mm, 5.65 mm$ .
The average of these four readings
$= \frac{5.5 + 5.55 + 5.45 + 5.65}{4}$
$= 5.5375 mm$
Standard deviation of the data is 0.07395
The significant rule says that reading should have the same significant figure as that of the reading given.
It we do round off 5.5375 then we will get 5.54 as $7 > 5$
If we do round off 0.07395 then we will get 0.07 as $3 <$ 5
So, Average diameter will be
$(5.54 \pm 0.07) m$ as $0.07 mm$ is standard diviation of the data.

JEE Mains- 2020

Units and Measurements

139967 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

1

0.001 cm

2

0.01 cm

3

0.02 cm

4

0.001 cm

Explanation:

A Given,
On six rotation, readings of main scale changed by 3 $mm$
No. of division on circular scale $= 50$
Pitch $= \frac{Distance moved by screw}{Number of rotation}$
$= \frac{3 mm}{6} = 0.5 mm$
Least count of screw gauge
$(LC) = \frac{Pitch}{No. of circular scale division}$
$= \frac{0.5 mm}{50}$
$= 0.01 mm$
$= 0.001 cm$

JEE Mains- 2020

Units and Measurements

139964 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

1

2.121 cm

2

2.124 cm

3

2.125 cm

4

2.123 cm

Explanation:

B Let t be the thickness of the object
So, $t =$ MSR $+ (n \times LC)$
Where, $n =$ No. of circular scale division lie over reference line
MSR $=$ Main scale reading
$LC =$ Least count of screw gauge
Least count of gauge is
$(LC) = \frac{Pitch}{Number of division on circular scale}$
$= \frac{0.1}{50} = 0.002 cm$
When we multiply by division number it must be even because LC is 0.02 so any measurement will be an integral multiple of LC.
Therefore, the digit at the third decimal place will be an even number.
So, the answer is $2.124 cm$ .

JEE Mains- 2020

Units and Measurements

139965 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 $0.5 mm$ is noticed on the pitch scale. The nature of zero error involved and the least count of the screw gauge, are respectively

1 negative,

2 μ m

2 positive,

10 μ m

3 positive,

0.1 mm

4 positive,

0.1 μ m

Explanation:

B Given,
Number of division on circular scale of screw gauge $=$ 50 pitch $= 0.5 mm$
Least count of screw gauge
$(LC) = \frac{Pitch}{No. of division on circular scale}$
$= \frac{0.5}{50} = mm$
$= 1 \times 10^{- 5} m$
$= 10 μ m$
The nature is positive and least count is $10 μ m$

JEE Mains- 2020

Units and Measurements

139966 A student measuring the diameter of a pencil of circular cross-section with the help of a vernier scale records the following four readings; 5.50 $mm, 5.55 mm, 5.45 mm$ and $5.65 mm$ . The average of these four readings is $5.5375 mm$ and the standard deviation of the data is $0.07395 mm$ . The average diameter of the pencil should be therefore recorded as

1

(5.5375 \pm 0.0739) mm

2

(5.5375 \pm 0.0740) mm

3

(5.538 \pm 0.074) mm

4

(5.54 \pm 0.07) mm

Explanation:

D Four reading of vernier calipers are
$5.50 mm, 5.55 mm, 5.45 mm, 5.65 mm$ .
The average of these four readings
$= \frac{5.5 + 5.55 + 5.45 + 5.65}{4}$
$= 5.5375 mm$
Standard deviation of the data is 0.07395
The significant rule says that reading should have the same significant figure as that of the reading given.
It we do round off 5.5375 then we will get 5.54 as $7 > 5$
If we do round off 0.07395 then we will get 0.07 as $3 <$ 5
So, Average diameter will be
$(5.54 \pm 0.07) m$ as $0.07 mm$ is standard diviation of the data.

JEE Mains- 2020

Units and Measurements

139967 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

1

0.001 cm

2

0.01 cm

3

0.02 cm

4

0.001 cm

Explanation:

A Given,
On six rotation, readings of main scale changed by 3 $mm$
No. of division on circular scale $= 50$
Pitch $= \frac{Distance moved by screw}{Number of rotation}$
$= \frac{3 mm}{6} = 0.5 mm$
Least count of screw gauge
$(LC) = \frac{Pitch}{No. of circular scale division}$
$= \frac{0.5 mm}{50}$
$= 0.01 mm$
$= 0.001 cm$

JEE Mains- 2020

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