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If a screw gauge has a pitch of 1.5 mm and 300 divisions on circular scale, which of the following reading can be made from this screw gauge?

1.05 mm

3.01 cm

0.030000 m

2.1185 dm

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Explanation

Let us first find the least count.

L.C. = $\frac{P i t c h}{N o . o f C i r c u l a r s c a l e d i v i s i o n s}$ = $\frac{1.5 m m}{300} = 0.005 m m$

In (C) option, 0.030000m = 30.000 mm

The instrument from which reading has been taken has a least count of 0.001 mm or more.

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A Screw Guage gives the following readings when used to measure the diameter of a wire.
Main scale reading = 0mm
Circular scale reading = 52 divisions
Given that: 1 mm on main scale corresponds to 100 divisions of the circular scale.
The diameter of wire from the above data is:

0.026 cm

0.005 cm

0.52 cm

0.052 cm

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Explanation

$L e a s t C o u n t = \frac{P i t c h}{N o . o f c i r c u l a r s c a l e d i v i s i o n s} = \frac{1 m m}{100} = 0.01 m m$

Reading = Main scale reading (MSR) + Circular scale reading (CSR)

= 0 + 0.01 mm X 52

= 0.52 mm or 0.052 cm

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Two full turns of the circular scale of gauge cover a distance of 1 mm on scale. The total number of divisions on circular scale is 50. Further, it is found that screw gauge has a zero error of -0.03 mm. While measuring the diameter of a thin wire a student notes the main scale reading of 3 mm and the number of circular scale division in line, with the main scale as 35. The diameter of the wire is

3.32 mm

3.73 mm

3.67 mm

3.38 mm

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Explanation

Pitch = 0.5 mm

$L e a s t c o u n t = \frac{P i t c h}{N o . o f C S D}$

$= \frac{0.5 m m}{50} = 0.01 m m$

Observed Reading = MSR + CSR

= 3 mm + 35 X 0.01mm

= 3.35 mm

Correct Reading = Observed reading- Error

= 3.35 mm - (-0.03 mm)

= 3.38 mm

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The pitch of a screw gauge is 1mm and there are 100 divisions on the circular scale. While measuring the diameter of a wire, the linear scale reads 1 mm and 47^th division on the circular scale coincides with the reference line. The length of the wire is 5.6 cm. Find the curved surface area (in cm²) of the wire in appropriate number of significant figures.

2.4 cm²

2.56 cm²

2.6 cm²

2.8 cm²

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Explanation

$L . C . = \frac{1 m m}{100} = 0.01 m m$

Diameter (d) = 1mm + 0.01 mm X 47

= 1.47 mm

Length (l) = 5.6 cm

Curved surface area A = 2 $π r l$ = 2.586 $c m^{2}$

The final answer should have the same no. of significant figures (S.F.) as quantity having least no. of significant figures.

d = 1.47 mm ( 3 S.F.)

l = 5.6 cm (2 S.F.)

Ans : 2.6 $c m^{2}$ (rounded to 2 decimal places)

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One cm on the main scale of vernier callipers is divided into ten equal parts. If 20 divisions of vernier scale coincide with 8 small divisions of the main scale. What will be the least count of callipers?

0.06 cm

0.6 cm

0.5 cm

0.7 cm

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Explanation

(A)

20 VSD = 8 MSD

1 VSD = 0.4 MSD

L.C. = 1 MSD - 0.4 MSD

= 0.6 MSD = 0.6 X 1 mm

= 0.6 mm or 0.06 cm

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Component of $3 \hat{i} + 4 \hat{j}$ perpendicular to $\hat{i} + \hat{j}$ and in the same plane as that of $3 \hat{i} + 4 \hat{j}$ is:

$\frac{1}{2} (\hat{j} - \hat{i})$

$\frac{3}{2} (\hat{j} - \hat{i})$

$\frac{5}{2} (\hat{j} - \hat{i})$

$\frac{7}{2} (\hat{j} - \hat{i})$

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Given that $\overset{⇀}{C} = \overset{⇀}{A} + \overset{⇀}{B} and \overset{⇀}{C}$ makes an angle $α with \overset{⇀}{A} and β with \overset{⇀}{B}$ . Which of the following options is correct?

$α cannot be less then β$

$α < β, if A < B$

$α < β, if A > B$

$α < β, if A = B$

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If the angle between the vector $\vec{A} a n d \vec{B}$ is $θ$ , the value of the product $(\vec{B} \times \vec{A}) . \vec{A}$ is equal to:

$B A^{2} \cos θ$

$B A^{2} \sin θ$

$B A^{2} \sin θ \cos θ$

zero

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Explanation

Using vector product:

$Let \vec{R} = (\vec{B} \times \vec{A}) \Rightarrow \vec{R} ⊥ \vec{A} \Rightarrow (\vec{B} \times \vec{A}) . \vec{A} \Rightarrow \vec{R} . \vec{A} = 0$

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If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is

90°

45°

180°

0°

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Explanation

(a)

$Suppose two vectors are P and Q . It is given that - |P + Q| = |P - Q| Let the angle between P and Q is θ . \Rightarrow P^{2} + Q^{2} + 2 PQcosθ = P^{2} + Q^{2} - 2 PQcosθ \Rightarrow cosθ = 0 = \cos 90^{0} \Rightarrow θ = 90^{0}$

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A force of 6 N and another of 8 N can be applied together to produce the effect of a single force of -

1 N

11 N

15 N

20 N

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Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.