Units and Measurements Test

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Units and Measurements Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

The ratio of the dimensions of Planck constant and that of moment of inertia has the dimensions of

A
Time

B
Frequency

C
Angular momentum

D
Velocity

Solution

Step 1: Dimension formula of Planck constant:

Energy of photon is given by:
$$E=h \nu $$
Where $$h$$ is Planck's constant and $$ \nu $$ is frequency of photon

Now, Planck's constant will be $$h = \dfrac{E}{\nu}$$

Dimension of energy $$[E] = [ML^2T^{-2}]$$

Dimension of frequency $$[ \nu ]=[T^{-1}]$$ {frequency is reciprocal of time period)

Dimensions of Planck's constant $$[h]= [ML^2T^{-2}]/[T^{-1}] = [ML^2T^{-1}]$$

Step 1: Dimension formula of moment of inertia:

The moment of inertia $$(I)$$ is given by:
$$I=MR^2$$
Where $$M$$ is mass of the obect and $$R$$ is distance.

Dimensions of moment of inertia $$[I] = [ML^2]$$

Step 3: Dimension formula of their ratio:
$$\therefore$$ Dimensions of $$\dfrac{h}{I}$$ $$ =\dfrac{[ML^2T^{-1}]}{[ML^2]}=[T^{-1}]$$
Thus ratio of Planck's constant and moment of inertia has dimensions of frequency.

Option B is correct.
Question 2
1 / -0

If the angular distance between the stars turns out to be approximately 1100 arcseconds, or 0.30 degrees. The moon appears to shift 0.3 degrees when we observe it from two vantage points 2360 km apart, then find the distance of the moon from the surface of the earth.Given angular diameter of moon is 0.5 degrees.

A
450642 km

B
450392 km

C
325684 km

D
480264 km

Solution

The parallax angle in radians is given by:
$$\theta_{rad}=\theta_{deg}\times \dfrac{\pi}{180}$$$$=0.3\times \dfrac{\pi}{180}rad$$

Thus the distance between moon and surface of earth is:
$$D=\dfrac{d}{\theta}$$$$=\dfrac{2360km}{0.3\times \dfrac{\pi}{180}}$$$$=450642km$$
Question 3
1 / -0

Parallax angles _______ $$0.01/ arcsec$$ are very difficult to measure from Earth.

A
more than

B
less than

C
equal to

D
greater than or equal to

Solution

Parallax effect depends upon the path of light that travels from object to the observer. For distant objects observed from Earth, the light that reaches Earth has to go through a number of layers in Earth's atmosphere, during which it refracts and disperses and hence decreases the accuracy of the method. Resultantly parallax angles less than $$0.01/arcsec$$ are very difficult to measure from Earth.
Question 4
1 / -0

If a star is $$5.2\times 10^{16}\ m$$ away. What is the parallax angle in degrees?

A
$$1.67 \times 10^{-4}$$ degrees

B
$$1.67 \times 10^{-5}$$ degrees

C
$$0.67 \times 10^{-4}$$ degrees

D
$$2.3 \times 10^{-4}$$ degrees

Solution

Given : $$1$$ AU $$ = 1.5\times 10^11$$ m $$d = 5.2\times 10^{15}$$ m
Parallax angle: $$\alpha = \dfrac{1 AU}{d} =\dfrac{1.5\times 10^{11}}{5.2\times 10^{16}} = 0.288\times 10^{-5}$$ radians
$$\implies$$ $$\alpha = \dfrac{180}{\pi} \times 0.288\times 10^{-5} = 1.67\times 10^{-4}$$ degrees
Question 5
1 / -0

A star is $$1.45\ parsec$$ light years away. How much parallax would this star show when viewed from two locations of the earth six months apart in its orbit around the sun?

A
$$2\ Parsec$$

B
$$0.725\ Parsec$$

C
$$1.45\ Parsec$$

D
$$2.9\ Parsec$$

Solution

One light year $$=$$ speed of light $$\times$$ one year
or $$1 ly=3\times 10^8\times (24\times 3600)=94608\times 10^{11} m$$
So, $$4.29 ly=4.29\times (94608\times 10^{11})=4.058\times 10^{16} m$$
As $$1 $$ parsec $$=3.08\times 10^{16} m$$
(Parsec is a unit of length used to measure large distances to objects outside our Solar System)
Thus, $$4.29 ly=\dfrac{4.058\times 10^{16}}{3.08\times 10^{16}}=1.32$$ parsec
Now angular displacement , $$\theta=\dfrac{d}{D}$$
where $$d=$$ diameter of earth's orbit $$= 3\times 10^{11} m$$ and
$$D=$$ distance of star from the earth $$=4.058\times 10^{16} m$$
So, $$\theta=\dfrac{3\times 10^{11}}{4.058\times 10^{16}}=7.39\times 10^{-6}$$ rad
As $$1 sec=4.85\times 10^{-6} rad$$ so, $$7.39\times 10^{-6} rad=\dfrac{7.39\times 10^{-6}}{4.85\times 10^{-6}}=1.52 sec$$
Question 6
1 / -0

Which of the following is a possible dimensionless quantity?

A
Velocity gradient

B
Pressure gradient

C
Displacement gradient

D
Force gradient

Solution

Displacement gradient in x-direction is given by $$\dfrac{dr}{dx}$$
Since $$r$$ and $$x$$ have same dimension, this results the displacement gradient to be dimensionless quantity.
Question 7
1 / -0

$$ 5\times 10^7 \mu s$$ is equivalent to ____

A
0.5 s

B
5 s

C
50 s

D
500 s

Solution

We know that $$1\mu s = 10^{-6} \ s$$
Thus, $$5\times 10^{7}\mu s = 5\times 10^{7}\mu s\times \dfrac{10^{-6} \ s}{\mu s} = 50 \ s$$
Question 8
1 / -0

The dimension of $$\dfrac{h}{mv}$$ is: ($$h$$$$=$$ planks constant, $$m$$= mass, $$v$$=velocity)

A
$$M$$

B
$$L$$

C
$$T$$

D
none of the above

Solution

Debroglie wavelength is given by: $$\lambda=\dfrac{h}{mv}$$
Wavelength has dimension of length $$[L]$$.
Question 9
1 / -0

The dimensional formula of coefficient of viscosity is

A
$$[MLT^{-1}]$$

B
$$[M^{-1}L^{2}T^{-2}]$$

C
$$[ML^{-1}T^{-1}]$$

D
none

Solution

Coefficient of viscosity is given by $$\eta =\dfrac{F}{A}\dfrac{dz}{dv}$$
Dimension of force $$F = [MLT^{-2}]$$
Dimension of Area $$A = [L^2]$$
Dimension of velocity gradient $$\dfrac{dv}{dz} = [T^{-1}]$$
Thus dimension of coefficient of viscosity $$\eta = \dfrac{[MLT^{-2}]}{[L^2][T^{-1}]} = [ML^{-1}T^{-1}]$$
Question 10
1 / -0

A physical quantity $$Q$$ is found to depend on observables $$x, y$$ and $$z$$, obeying relation $$Q=\dfrac{x^{2/5}z^{3}}{y}$$. The percentage error in the measurements of $$x, y$$ and $$z$$ are $$1\%$$, $$2\%$$ and $$4\%$$ respectively. What is percentage error in the quantity $$Q$$ will be :

A
$$5\%$$

B
$$4.5\%$$

C
$$11\%$$

D
$$7.75\%$$

Solution

The given quantity $$Q = \dfrac{x^3y^2}{z}$$

Percentage error in Q $$\dfrac{\Delta Q}{Q} \times 100 = 3\dfrac{\Delta x}{x}\times 100$$ $$+2\dfrac{\Delta y}{y} \times 100 + \dfrac{\Delta z}{z}\times 100$$

Percentage error in $$x,y$$ and $$z$$ are $$1\%$$, $$2\%$$ and $$4\%$$ respectively.

$$\implies$$ $$\dfrac{\Delta Q}{Q} \times 100 = 3(1) + 2(2)+1(4) = 11$$ %

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Units and Measurements Test - 57

Result

Units and Measurements Test - 57

Score

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SHARING IS CARING

Question 1

Time

Frequency

Angular momentum

Velocity

Solution

Question 2

450642 km

450392 km

325684 km

480264 km

Solution

Question 3

more than

less than

equal to

greater than or equal to

Solution

Question 4

$$1.67 \times 10^{-4}$$ degrees

$$1.67 \times 10^{-5}$$ degrees

$$0.67 \times 10^{-4}$$ degrees

$$2.3 \times 10^{-4}$$ degrees

Solution

Question 5

$$2\ Parsec$$

$$0.725\ Parsec$$

$$1.45\ Parsec$$

$$2.9\ Parsec$$

Solution

Question 6

Velocity gradient

Pressure gradient

Displacement gradient

Force gradient

Solution

Question 7

0.5 s

5 s

50 s

500 s

Solution

Question 8

$$M$$

$$L$$

$$T$$

none of the above

Solution

Question 9

$$[MLT^{-1}]$$

$$[M^{-1}L^{2}T^{-2}]$$

$$[ML^{-1}T^{-1}]$$

none

Solution

Question 10

$$5\%$$

$$4.5\%$$

$$11\%$$

$$7.75\%$$

Solution

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