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2 months ago

Units and Measurement physics ncert exemplar solutions class 12th chapter two

In a Vernier Calipers, 10 divisions of Vernier scale is equal to the 9 divisions of main scale. Then both jaws of Vernier calipers touch each other, the zero of the Vernier scale is shifted to the left of zero of the main scale and 4^th Vernier scale division exactly coincides with the main scale reading. One main scale division is equal to 1mm. While measuring diameter of a spherical body, the body is held between two jaws. It is now observed that zero of the Vernier scale lies between 30 and 31 divisions of main scale reading and 6^th Vernier scale division exactly coincides with the main scale reading. The diameter of the spherical bo

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alok kumar singh

Contributor-Level 10

1 MSD = 1mm

9 MSD = 10 VSD

1VSD = 0.9 MSD = 0.9 mm

Least count = 1 MSD – 1 VSD

= 0.1 mm = 0.01 cm

Zero Error = (10 – 4) * 0.1 = 0.6 mm

Reading = MSR + VSR – Zero error

= 3 cm + 6 * 0.01 – (0.06)

= 3.12 cm $?$ 3.10 cm

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2 months ago

Units and Measurement Physics NCERT Exemplar Solutions Class 12th Chapter Eleven

Consider the efficiency of carnot's engine is given by $η = \frac{α β}{s i n θ} l o g e \frac{β x}{k T},$ where and are constants. If T is temperature, k is Boltzmann constant, is angular displacement and x has the dimensions of length. Then, choose the incorrect option

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Vishal Baghel

Contributor-Level 10

Efficiency $η = \frac{α B}{s i n θ} l o g_{e} \frac{B x}{k T} α, B \to c o n s t a n t s$

As, energy $= \frac{1}{2} k T \Rightarrow$ Dimension of Bx = Dimension of energy

So Dimension as of B = Dimension of 'F'

Dimension of F1

Dimension of $(\frac{1}{α}) x =$ Dimension of energy

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2 months ago

Units and Measurement physics ncert exemplar solutions class 12th chapter two

In an experiment of find out diameter of wire using screw gauge, the following observations were noted:

(A) Screw moves 0.5mm on main scale in one complete rotation

(B) Total divisions on circular scale = 50

(C) Main scale reading is 2.5 mm

(D) 45th division of circular scale is in the pitch line

(E) Instrument has 0.03 mm negative error

Then the diameter of wire is:

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Vishal Baghel

Contributor-Level 10

MS â€“ Reading = 2.5 mm.

50 division on CS = 0.5 mm on MS.

$∴ 4 5^{t h} d i v i s i o n o n C S = \frac{0.5}{50} * 45 m m M . S .$

= .45 mm on M.S.

So, diameter = 2.5 mm

+0.45 mm

+0.03 mm

= 2.98 mm

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New answer posted

2 months ago

Units and Measurement physics ncert exemplar solutions class 12th chapter two

A traveling microscope has 20 divisions per cm on the main scale while its vernier scale has total 50 divisions and 25 vernier scale divisions are equal to 24 main scale divisions, what is the least count of the travelling microscope?

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Vishal Baghel

Contributor-Level 10

MSD = 20 divisions per cm 1 MSD = $\frac{1}{20} c m .$

VSD = 50 divisions

As, 25 VSD = 24 MSD

$∴ 1 V S D = \frac{24}{25} M S D$

L.C. = 1 MSD â€“ 1 VSD

= $= \frac{1}{500} c m = 0.002 c m$

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2 months ago

Units and Measurement physics ncert exemplar solutions class 12th chapter two

Given below are two statements: One is labeled as Assertion (A) and other is labeler as Reason (R).

Assertion (A) : Time period of oscillation of a liquid drop depends on surface tension (S), if density of the liquid is $ρ$ and radius of the drop is r, then

$T = K \sqrt[]{\frac{ρ r^{3}}{S^{\frac{3}{2}}}}$ is dimensionally correct, where K is dimensionless.

Reason (R) : Using dimensional analysis we get R.H.S. having different dimension than that of time period.

In the light of above statements, choose the correct answer from the options given below.

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Vishal Baghel

Contributor-Level 10

$T = k \sqrt[]{\frac{ρ r^{3}}{S^{3 / 2}}}$

$[T]$ ${[M L^{- 3}]}^{1 / 2} {[L^{3}]}^{1 / 2} {[M L T^{- 2} L^{- 1}]}^{\frac{- 3}{4}}$

$[T]$ $= [M^{1 / 2} L^{- \frac{3}{2}} L^{\frac{3}{2}} M^{- \frac{3}{4}} T^{\frac{3}{2}}]$

$= [M^{\frac{1}{4}} T^{3 / 2}]$

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New answer posted

2 months ago

Units and Measurement physics ncert exemplar solutions class 12th chapter two

Given below are two statements: One is labeled as Assertion (A) and other is labeler as Reason (R).

Assertion (A) : Time period of oscillation of a liquid drop depends on surface tension (S), if density of the liquid is and radius of the drop is r, then

$T = K \sqrt[]{\frac{ρ r^{3}}{S^{\frac{3}{2}}}}$ is dimensionally correct, where K is dimensionless.

Reason (R) : Using dimensional analysis we get R.H.S. having different dimension than that of time period.

In the light of above statements, choose the correct answer from the options given below.

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

$T = k \sqrt[]{\frac{ρ r^{3}}{S^{3 / 2}}}$

$[T]$ ${[M L^{- 3}]}^{1 / 2} {[L^{3}]}^{1 / 2} {[M L T^{- 2} L^{- 1}]}^{\frac{- 3}{4}}$

$[T]$ $= [M^{1 / 2} L^{- \frac{3}{2}} L^{\frac{3}{2}} M^{- \frac{3}{4}} T^{\frac{3}{2}}]$

$= [M^{\frac{1}{4}} T^{3 / 2}]$

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New answer posted

2 months ago

Units and Measurement physics ncert exemplar solutions class 12th chapter two

A quantity $X$ is given by $({I F v}^{2} M L L^{4})$ in terms of moment of inertia $I$ , force $F$ , velocity $v$ , work $W$ and Length $L$ . The dimensional formula for $x$ is same as that of:

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alok kumar singh

Contributor-Level 10

Sol.

$B_{A} = \frac{μ_{0} I θ}{4 π R}$

$\Rightarrow \frac{B_{A}}{B_{B}} = \frac{I_{A} θ_{A} R_{B}}{I_{B} θ_{B} R_{A}}$

$\Rightarrow \frac{2 (\frac{3 π}{2}) (4)}{3 [\frac{5 π}{3}] [2]}$

$\Rightarrow \frac{6}{5}$ .

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New answer posted

2 months ago

Units and Measurement physics ncert exemplar solutions class 12th chapter two

The distance of the Sun from earth is 1.5 * 10¹¹ m and its angular diameter is (2000) s when observed from the earth. The diameter of the Sun will be

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alok kumar singh

Contributor-Level 10

1° ® 60'

Þ 60' ® 10°

L = 1.5 * 10¹¹ m

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New answer posted

4 months ago

Units and Measurement NEET

Why one should prepare from the NCERT Exemplar for Units and Measurements in Class 11?

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Pallavi Pathak

Contributor-Level 10

The NCERT Exemplar provides a variety of higher-order thinking and application-based questions. It helps students deepen their understanding of Chapter Two Units and Measurements. The chapter helps students build a strong foundation for Physics advanced topics, prepares students for competitive exams, and improves their problem-solving skills.

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New answer posted

5 months ago

Units and Measurement Class 11

Can units and measurements class 11 notes help in understanding the limitations of measurement?

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Pallavi Pathak

Contributor-Level 10

Yes, class 11 physics ch 1 notes clearly explain the limitations of measurement due to human error, instrumental errors and environmental conditions. The solutions show how no measurement is perfectly accurate by providing examples where measured values deviate from actual values. The chapter also shows how to improve precision and minimize errors. This helps students to understand that physics often expresses values with a degree of uncertainty as it involves approximations. By knowing how to handle and report these uncertainties, one can perform better in the experimental and theoretical assessments, and also get a realistic view of

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