Units and Measurements Test

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Units and Measurements Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

Which method is used by astronomers to calculate how far the star is?

A
Parallax

B
Equillax

C
Purplex

D
None

Solution

Parallax is used by astronomers to measure the distance between the stars. Parallax is also known as trigonometric parallax.
Question 2
1 / -0

The dimensions of Planck's constant is the same as the product of

A
force and time

B
force, displacement and time

C
force and distance

D
time and displacement

Solution

$$E=h\nu$$
$$\implies $$ Planck's constant $$h=\dfrac{E}{\nu}$$
Dimension of energy=$$[ML^2T^{-2}]$$
Dimension of $$\nu=[T^{-1}]$$
Thus dimension of $$h=[ML^2T^{-1}]$$
Dimension of Force $$=[MLT^{-2}]$$
Dimension of Displacement=$$[L]$$
Dimension of Time$$=[T]$$
Thus dimension of $$Force\times displacement\times time=[ML^2T^{-1}]$$=Dimension of Planck's constant
Question 3
1 / -0

The physical quantity having the dimensions $$\displaystyle \left[ { M }^{ -1 }{ L }^{ -3 }{ T }^{ 3 }{ A }^{ 2 } \right] $$ is :

A
Resistance

B
Resistivity

C
Electrical conductivity

D
Electromotive force

Solution

Electrical conductivity is given by : $$\rho =R\dfrac { A }{ l }$$ =$$\dfrac { {n e }^{ 2 }\tau }{ m } $$
Dimensions of $$A$$ = $${ L }^{ 2 }$$
Dimensions of $$l$$ = $${ L }^{ 1 }$$
Dimensions of $$R$$ = $${ M }^{ -1 }{ L }^{ -4 }{ T }^{ 3 }{ A }^{ 2 }$$
n $$=$$ number of electrons per unit volume $$={ L }^{ -3 }$$
Dimensions of $$e=C^2={AT}^2$$
where, $$C=$$coulomb; $$A=$$Ampere;
$$T=$$ time
Dimensions of $$\tau$$ $$=T$$
Dimension of mass $$m =M$$
Dimensions of $$\rho =\dfrac { { A }^{ 2 }{ T }^{ 2 }{ T }^{ 1 } }{ M } \times \dfrac { 1 }{ { L }^{ 3 } } $$ $$={ M }^{ -1 }{ L }^{ -3 }{ T }^{ 3 }{ A }^{ 2 }$$

Hence, option C is correct.
Question 4
1 / -0

The dimensional formula for Reynold's number is ________.

A
$$[L^0M^0T^0]$$

B
$$[L^1M^1T^1]$$

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

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

Solution

The formula for Reynold's number is:
$$Re=\dfrac { \rho vd }{ \mu } \\ \Rightarrow [Re]=\dfrac { [\rho ][v][d] }{ [\mu ] } =\dfrac { [M{ L }^{ -3 }][L{ T }^{ -1 }][L] }{ [{ ML }^{ -1 }{ T }^{ -1 }] } =[{ M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 }]$$.
Thus, Reynold's number is a dimensionless quantity. It is defined as the ratio of inertial force to viscous force, so it has to be a dimensionless number.
Question 5
1 / -0

Parallax method is useful

A
for measuring speed of the light.

B
for measuring distances of star.

C
for finding the intensity of the light.

D
None of the above.

Solution

$$Answer:-$$ B
Parallax is a displacement or difference in the apparent position of an object viewed along two different lines of sight, and is measured by the angle or semi-angle of inclination between those two lines. Astronomers use the principle of parallax to measure distances to the closer stars.
Question 6
1 / -0

Which of the following is the parallax angle in the shown diagram?

A
d

B
p

C
q

D
r

Solution

The parallax angle is the angle between the Earth at one time of year, and the Earth six months later, as measured from a nearby star. Astronomers use this angle to find the distance between the Earth and the Star.
Thus $$p$$ is the parallax angle.
Question 7
1 / -0

Parallax method is suitable for the measurement of:

A
ball

B
planet

C
vehicle

D
distance between cities

Solution

Parallax method is used for the measurement of planet distance and also used to measure distance of starts and planets
option B is correct.
Question 8
1 / -0

What is the approximation made in the parallax method?

A
All distances measured between two points on earth is zero.

B
All distances measured between two points on earth is constant.

C
Distance between a point on the earth and the planet is very large as compared to the distance between two points on earth's surface.

D
No approximation is made.

Solution

In parallax method, an approximation is made that distance between a point on the earth and the planet is very large as compared to the distance between two points on the earth's surface.
Question 9
1 / -0

$$1$$ femtometre is equivalent to

A
$$10^{15} m$$

B
$$10^{-12} m$$

C
$$10^{-15} m$$

D
$$10^{-9} m$$

Solution

$$1 $$ femtometre $$=10^{-15} m$$
Question 10
1 / -0

The relation between $$\theta$$ , $$b$$ and $$D$$ is:
(where $$\theta$$ is the parallax angle, $$b$$ is the distance between two points of separation, $$D$$ is the distance of source from any point of observation)

A
$$\theta = \dfrac{b}{D}$$

B
$$\theta=\dfrac{D}{b}$$

C
$$\theta = b\times D$$

D
$$\theta = b^2\times D$$

Solution

From the figure, $$\tan\theta=\dfrac{b}{D}$$
For small angle, we have $$\tan\theta \sim \theta$$
So, we get $$\theta=\dfrac{b}{D}$$

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Units and Measurements Test - 34

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Units and Measurements Test - 34

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SHARING IS CARING

Question 1

Parallax

Equillax

Purplex

None

Solution

Question 2

force and time

force, displacement and time

force and distance

time and displacement

Solution

Question 3

Resistance

Resistivity

Electrical conductivity

Electromotive force

Solution

Question 4

$$[L^0M^0T^0]$$

$$[L^1M^1T^1]$$

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

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

Solution

Question 5

for measuring speed of the light.

for measuring distances of star.

for finding the intensity of the light.

None of the above.

Solution

Question 6

d

p

q

r

Solution

Question 7

ball

planet

vehicle

distance between cities

Solution

Question 8

All distances measured between two points on earth is zero.

All distances measured between two points on earth is constant.

Distance between a point on the earth and the planet is very large as compared to the distance between two points on earth's surface.

No approximation is made.

Solution

Question 9

$$10^{15} m$$

$$10^{-12} m$$

$$10^{-15} m$$

$$10^{-9} m$$

Solution

Question 10

$$\theta = \dfrac{b}{D}$$

$$\theta=\dfrac{D}{b}$$

$$\theta = b\times D$$

$$\theta = b^2\times D$$

Solution

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