Units and Measurement

A screw gauge of pitch 0.5mm is used to measure the diameter of uniform wire of length 6.8cm, the main scale reading is 1.5mm and circular scale reading is 7. The calculated curved surface area wire to appropriate significant figures is:

[Screw gauge has 50 divisions on its circular scale]

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 body will be:

Consider the efficiency of carnot’s engine is given by $\eta =\frac{\alpha \beta }{sin\theta }loge\frac{\beta x}{kT},$ 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

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:

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?

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 $\rho$ and radius of the drop is r, then

$T=K\sqrt[]{\raisebox{1ex}{$\rho r^3$}\!\left/ \!\raisebox{-1ex}{$S^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$}\right.}$ 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.

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 <!-- [if gte mso 9]><xml> </xml><![endif]--> and radius of the drop is r, then

$T=K\sqrt[]{\raisebox{1ex}{$\rho r^3$}\!\left/ \!\raisebox{-1ex}{$S^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$}\right.}$  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.

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

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