NCERT Class 11 Physics Chapter 1 Units and Measurements Long Answer Type Questions

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Question 36:

A new system of units is proposed in which unit of mass is , unit of length and unit of time . How much will measure in this new system?

Answer:

Since energy has dimensions of in new units becomes . Hence .

Question 37:

The volume of a liquid flowing out per second of a pipe of length l and radius r is written by a student as

where P is the pressure difference between the two ends of the pipe and η is coefficient of viscosity of the liquid having dimensional formula . Check whether the equation is dimensionally correct.

Answer:

The dimensional part in the expression is . Therefore, the dimensional of the right hand side comes out to be

Which is volume upon time. Hence, the formula is dimensionally correct.

Question 38:

A physical quantity X is related to four measurable quantities a, b, c and d as follows:

The percentage error in the measurement of a, b, c and d are and , respectively. What is the percentage error in quantity ? If the value of calculated on the basis of the above relation is , to what value should you round off the result.

Answer:

Percentage error in quantity is given by,

According to the problem, physical quantity is

Percentage error in

Percentage error in

Percentage error in

Percentage error in

Maximum percentage error in is

Mean absolute error in (rounding –off upto two significant digits)

On the basis of these values, the value of X should have two significant digits only.

Question 39:

In the expression and denote energy, mass, angular momentum and gravitational constant, respectively. Show that P is a dimensionless quantity.

Answer:

Since E. l and G have dimensional formulas

Hence, will have dimensions:

Thus, P is Dimensionless.

Question 40:

If velocity of light c, Planck’s constant h and gravitational content G are taken as fundamental quantities then express mass, length and time in terms of dimensions of these quantities.

Answer:

We have to apply principle of homogeneity to solve this problem. Principle of homogeneity states that in a correct equation, the dimensions of each term added or subtracted must be same, i.e., dimensions of LHS and RHS should be equal, We know that, dimensions of

(i)

Let

Where, k is a dimensionless constant of proportionality.

Substituting dimensions of each term in Eq. (i), we get

Comparing powers of same terms on both sides, we get

Adding Eqs. (ii), (iii) and (iv), we get

Substituting value of we get

Form

Substituting value of we get

Putting value of a, b and c in Eq. (i), we get

(ii) Let

Where is a dimensionless constant.

Substituting dimensions of each term in , we get

On comparing powers of same terms, we get

Adding Eqs. (vi), (vii) and (viii), we get

Substituting value of in Eq. (vi), we get

From Eq. (viii),

Substituting value of we get

Putting values of a, b, and c in Eq. (v), we get

(iii) Let

Where is a dimensionless constant.

Substituting dimensions of each term in , we get

On comparing powers of same terms, we get

Adding Eqs. (x), (xi) and (xii), we get

Substituting the value of in Eq. (x), we get

From Eq. (xii),

Substituting value of we get

Putting values of a, b, and c in Eq. (ix), we get

Question 41:

An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit of radius r. From Kepler’s Third law about the period of a satellite around a common central body, square of the period of revolution T is proportional to the cube of the radius of the orbit r. Show using dimensional analysis, that

Where k is a dimensionless constant and g is acceleration due to gravity.

Answer:

Given T is also function of g and

For

For

Therefore.

Thus,

Question 42:

In an experiment to estimate the size of a molecule of oleic acid 1 mL of oleic acid is dissolved in of alcohol. Then of this solution is diluted to by adding alcohol. Now 1 drop of this diluted solution is placed on water in a shallow trough. The solution spreads over the surface of water forming one molecule thick layer. Now, Lycopodium powder is sprinkled evenly over the film and its diameter is measured. Knowing the volume of the drop and area of the film we can calculate the thickness of the film which will give us the size of oleic acid molecule.

Read the passage carefully and answer the following questions:

(a) Why do we dissolve oleic acid in alcohol?

(b) What is the role of Lycopodium powder?

(c) What would be the volume of oleic acid in each mL of solution prepared?

(d) How will you calculate the volume of n drops of this solution of oleic acid?

(e) What will be the volume of oleic acid in one drop of this solution?

Answer:

(a) Because oleic acid dissolves in alcohol but does not dissolve in water.

(b) When Lycopodium powder is spread on water, it spreads on the entire surface. When a drop of the prepared solution is dropped on water, oleic acid does not dissolve in water, it spreads on the water surface pushing the Lycopodium powder away to clear a circular area where the drop falls. This allows measuring the area where oleic acid spreads.

(c)

(d) By means of a burette and measuring cylinder and measuring the number of drops.

(e) If drops of the solution make the volume of oleic acid in one drop will be .

Question 43:

(a) How many astronomical units (A.U.) make 1 parsec?(b) Consider a sun like star at a distance of parsecs. When it is seen through a telescope with magnification, what should be the angular size of the star? Sun appears to be from the earth. Due to atmospheric fluctuations, eye can’t resolve objects smaller than 1 arc minute.(c) Mars has approximately half of the earth’s diameter. When it is closest to the earth it is at about A.U. from the earth. Calculate what size it will appear when seen through the same telescope.(Comment: This is to illustrate why a telescope can magnify planets but not stars.)

Answer:

(a) By definition of parsec

(b) At distance, sun is in diameter.

Therefore, at parsec, star is degree in diameter .

With magnification, it should look . However due to atmospheric fluctuations. It will still look of about arcmin.

It can’t be magnified using telescope.

(c)

At sun is seen as degree in diameter, and mars will be seen as degree in diameter.

At will be seen as degree in diameter. With magnification mars will be seen as arcmin.

This is larger than resolution limit due to atmospheric fluctuations. Hence, it looks magnified.

Question 44:

Einstein’s mass – energy relation emerging out of his famous theory of relativity relates mass (m) to energy ( E ) as , where c is speed of light in vacuum. At the nuclear level, the magnitudes of energy are very small. The energy at nuclear level is usually measured in MeV, where ; the masses are measured in unified atomic mass unit (u) where .(a) Show that the energy equivalent of 1 u is .(b) A student writes the r elation as . The teacher points out that the relation is dimensionally incorrect. Write the correct relation.

Answer

(a) We can apply Einstein’s mass-energy relation in this problem, to Calculate the energy equivalent of the given mass.

Here

Applying

Energy

(b) As

According to this,

Hence the dimensionally correct relation

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