Units and Measurements

Units and Measurements Physics NCERT Exemplar Solutions Class 11th Chapter Two

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This is a long answer type question as classified in NCERT Exemplar

Dimensions of energy is E= [ML²T^-2]

Mass m= [M]

Dimension of E= [ML²T^-2]

Dimensions of L= [ML^-2T^-1]

Dimensions of G= [M^-1L³T^-2]

By using these values [P]= [ML²T^{-2] $* [M L^{2} T^{- 1}]$ 2 $[{M]}^{- 5} * [M^{- 1} L^{3} T^{- 2}]$ -2}

= [M^1+2-5+2L^2+4-6T^-2-2+4]

= [M⁰L⁰T⁰]

After we know that P is dimensionless quantity

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3 months ago

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

X = a² b³ c^5/2 d^-2

The percentage error in the measurement of a, b, c and d are 1%, 2%, 3% and 4%, respectively. What is the percentage error in quantity X ? If the value of X calculated on the basis of the above relation is 2.763, to what value should you round off the result.

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Units and Measurements Physics NCERT Exemplar Solutions Class 11th Chapter Two

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

As we know that X= a² b³ c^5/2 d^-2

Maximum percentage error in X is $\frac{? X}{X} * 100 = \pm [2 (\frac{? a}{a} * 100) + 3 (\frac{? b}{b} * 100) + \frac{5}{2} (\frac{? c}{C} * 100) + 2 (\frac{? d}{d} * 100)]$

= $\pm [2 (1) + 3 (2) + \frac{5}{2} (3) + 2 (4)] %$

$\pm [2 + 6 + \frac{15}{2} + 8]$

$= \pm 23.5 %$

Mean absolute error in X= $\pm 0.24$ rounding off to significant value.

And calculated value would be 2.8 rounding off upto two digits.

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3 months ago

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

V= $\frac{π}{8} \frac{p r^{4}}{η l}$

where P is the pressure difference between the two ends of the pipe and η is coefficent of viscosity of the liquid having dimensional formula ML^-1 T^-1. Check whether the equation is dimensionally correct.

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Units and Measurements Physics NCERT Exemplar Solutions Class 11th Chapter Two

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

Rate of flow is equal to V= $\frac{π}{8} \frac{p r^{4}}{η l}$

Dimensions of V or LHS= volume/time=L³/T= [L³T^-1]

Dimensions of P= [ML^-1T^-2]

Dimensions of $η$ = [ML^-1T^-1]

Dimensions of L= [L]

Dimensions of r= [L]

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

So they are in equal in dimensions.

So equation is correct dimensionally.

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3 months ago

A new system of units is proposed in which unit of mass is α kg, unit of length β m and unit of time γ s. How much will 5 J measure in this new system?

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Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

Energy E= [ML²T^-2]

Let M₁, L₁and T₁ and M₂, L₂and T₂ are fundamental quantities for two units

M₁=1kg and L₁=1m and T₁=1s

M₂= α kg, L₂= β m and T₂= γ s

And n₁u₁=n₂u₂

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4 months ago

2.32 It is a well known fact that during a total solar eclipse the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.

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2.32

Distance of the moon from Earth = 3.84 * 10⁸ m

Distance of the Sun from Earth = 1.496 * 10¹¹ m

Sun's diameter = 1.39 * 10⁹ m

Sun's angular diameter = 1920° = 1920 * 4.85 * 10^-6 rad = 9.3 * 10^-3 rad

During total Solar eclipse, the moon completely covers the Sun, then the angular diameter of both Sun and the

Moon will be equal

So the angular diameter of moonΘ = 9.3 * 10^-3 rad, distance d of moon from Earth = 3.84 * 10⁸ m and

diameter of the moon = angular diameter * distance

So the approximate diameter of the Moon = 9.3 * 10^-3* 3.84 * 10⁸ m = 35.712 * 10⁵ m

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4 months ago

2.31 The farthest objects in our Universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us?

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2.31

Speed of light = 3 * 10⁸ m/s, Time taken = 3 billion years = 3 * 10⁹years = 3 * 365 * 24 * 60 * 60 * 10⁹ s

The distance = speed x time = 3 * 10⁸* 3 * 365 * 24 * 60 * 60 * 10⁹ m

= 2.838 * 10²⁵ m

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4 months ago

2.30 A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects under water. In a submarine equipped with a SONAR, the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine?

(Speed of sound in water = 1450 m s^–1).

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2.30

Speed of the sound in water = 1450 m s^–1

Time taken from generation + reflection = 2t = 77 s

From the relation s = ut, where s is the distance of the enemy ship, we get

s = 1450 * 77/2 m = 55825 m = 55.8 km

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4 months ago

2.29 A LASER is a source of very intense, monochromatic, and unidirectional beam of light. These properties of a laser light can be exploited to measure long distances. The distance of the Moon from the Earth has been already determined very precisely using a laser as a source of light. A laser light beamed at the Moon takes 2.56 s to return after reflection at the Moon's surface. How much is the radius of the lunar orbit around the Earth?

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2.29 Time taken by the laser beam after reflection = 2.56 s

The speed of laser = 3 * 10⁸ m/s

If the distance between Earth and the Moon is d, the distance covered by laser beam is 2d

From the relation distance = speed * time, we get

d = (2.56 * 3 * 10⁸)/2 = 3.84 * 10⁸m

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4 months ago

2.28 The unit of length convenient on the nuclear scale is a fermi : 1 f = 10^–15 m. Nuclear sizes obey roughly the following empirical relation :

r = r₀ A^1/3

where r is the radius of the nucleus, A its mass number, and r₀ is a constant equal to about, 1.2 f. Show that the rule implies that nuclear mass density is nearly constant for different nuclei. Estimate the mass density of sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.

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Kindly go through the solution

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2.27 Estimate the average mass density of a sodium atom assuming its size to be about 2.5 Å. (Use the known values of Avogadro's number and the atomic mass of sodium). Compare it with the mass density of sodium in its crystalline phase: 970 kg/m³. Are the two densities of the same order of magnitude? If so, why?

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