Units and Measurements Test

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Units and Measurements Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

If $$C$$ and $$R$$ denote capacitance and resistance, then dimensions of $$CR$$ will be

A
$$\displaystyle \left [ M^{0}L^{0}TA^{0} \right ]$$

B
$$\displaystyle \left [ ML^{0}TA^{-2} \right ]$$

C
$$\displaystyle \left [ ML^{0}TA^{2} \right ]$$

D
$$\displaystyle \left [ ML^{0}T^{2}A^{-2} \right ]$$

Solution

We know that value $$RC$$ gives time constant in an $$R-C$$ circuit and that $$V(t)={V}_{0}{e}^{\frac{-t}{RC}}$$
Exponents are always dimensionless, thus $$[\dfrac{-t}{RC}]={M}^{0}{L}^{0}{T}^{0}{A}^{0}$$ or $$[RC]=[t]=[{M}^{0}{L}^{0}{T}^{1}{A}^{0}]$$
Question 2
1 / -0

It takes time $$8$$ min for light to reach from sun to the earth surface. If speed of light is taken to be $$3 \times 10^8$$ m $$s^{-1}$$, find the order of magnitude of distance from the sun to the earth in km.

A
$$10^7 km$$

B
$$15\times10^8 km$$

C
$$10^8 km$$

D
$$10^9 km$$

Solution

Time $$t=8\ min=480\ sec$$
Speed of light $$c=3\times10^8\ ms^{-1}$$
$$Distance=t\times c=1.44\times10^{11}m=1.44\times10^8km$$
So order of magnitude os distance is $$10^8km$$
Question 3
1 / -0

Write the dimensions of $$\cfrac{d{\phi}_{B}}{dt}$$ where $$\phi_B$$ is magnetic flux through a coil.

A
$$[ M{ L }^{ 2 }{ A }^{ -1 }{ T }^{ -3 }] $$

B
$$[ M{ L }^{ 3 }{ A }^{ -1 }{ T }^{ -2 }] $$

C
$$[ M{ L }^{ 2 }{ A }^{ -2 }{ T }^{ -3 }] $$

D
None of these

Solution

$$\cfrac{d{\phi}_{B}}{dt}=$$ [potential or EMF] $$=\left[ M{ L }^{ 2 }{ A }^{ -1 }{ T }^{ -3 } \right] $$
Question 4
1 / -0

The length, breadth and height of a glass slab are respectively $$50 cm$$, $$20 cm$$ and $$25 cm$$. Find the order of magnitude of volume of slab in Sl unit.

A
$$10^{-2} m^3$$

B
$$10^{-4} m^3$$

C
$$10^{-3} m^3$$

D
$$10^{-3} cm^3$$

Solution

$$L=0.5 m , b=0.2 m ,h=0.25 m$$
Volume $$=$$ length $$\times$$ breadth $$\times$$ height
Volume $$=l\times b\times h=0.025 m^{3 }\ = \ 2.5\times 10^{-2}\ m^3 $$
So, the order of magnitude is $$10^{-2} m^3$$.
Question 5
1 / -0

The radius of a hydrogen atom is $$0.5\mathring{A}$$. Find the order of magnitude of volume of $$1$$ mole hydrogen in $$m^3$$. Given that 1 mole of hydrogen has $$6.02 \times 10^{23}$$ hydrogen atoms.

A
$$2\times10^{-4} m^3$$

B
$$10^{-3} m^3$$

C
$$10^{-7} m^3$$

D
$$10^{-2} m^3$$

Solution

Volume of 1 hydrogen atom, $$V=4\pi r^3/3; r=0.5\times10^{-10}m$$
So, $$V=0.5245\times10^{-30}m^3$$
So, volume of 1 mole ($$6.02\times10^{23}$$) is $$V\times6.02\times10^{23}=3.157\times10^{-7}m^3$$
So, order of magnitude is $$10^{-7}m^3$$.
Question 6
1 / -0

Find the order of magnitude of the mass of the star, whose radius is $$384 \times 10^6$$ m and average density is $$4 \times 10^3$$ kg $$m^{-3}$$ :

A
30

B
29

C
24

D
21

Solution

Radius of star $$r=384\times10^6m$$
So, Volume of star is $$V=4\pi r^3/3=2.371\times10^{26}m^3$$
Density of star is $$\rho=4\times10^3kgm^{-3}$$
Mass of star is $$M=V\times\rho\approx9.5\times10^{29}kg=0.95\times10^{30}kg$$
So the mass is of order of $$10^{30}kg$$
Question 7
1 / -0

The electric potential existing in space is $$V(x, y, z) = A (xy+ yz + zx)$$. Write the dimensional formula of $$A$$ :

A
$$MT^{-2} I^{-1}$$

B
$$MT^{-3} I^{-1}$$

C
$$MT^{-2} I^{-2}$$

D
$$MT^{-3} I^{-2}$$

Solution

Since $$x, y$$ and $$z$$ have units of length i.e. $$m$$.
Thus Unit of A =unit of potential $$ \times m^{-2} $$ = joules per coulomb $$\times m^{-2} $$
Dimension of joule $$ = ML^2 T^{-2}$$
Dimension of coulomb $$ = I T$$ where $$I$$ is the dimension of current.
Dimension of A = dimension of joule per coulomb $$ \times L^{-2} $$
= $$ M L^2 T^{-2} (I T)^{-1} \times L^{-2} = M T^{-3} I^{-1}$$
Question 8
1 / -0

The value due regard to significant figure $$ 4.0\times 10^{-4}-2.5\times 10^{-6}$$

A
$$\displaystyle 4.0\times 10^{-4}$$

B
$$\displaystyle 3.975\times 10^{-4}$$

C
$$\displaystyle 3.9\times 10^{-4}$$

D
none of the above

Solution

The value of this case is,
$$4.0\times 10^{-4}-2.5\times 10^{-6}$$
$$=3.975\times 10^{-4}$$.
Question 9
1 / -0

Round off 2.0082 to four significant figures.

A
2.008

B
2.0082

C
2.009

D
2

Solution

$$2.0082=2.008$$,
Here the digit $$2$$ is less than $$5$$ so $$2.008$$ is the closest value in four significant figures.
Question 10
1 / -0

Dimension of relative density is -

A
$$ kg m^{-3} $$

B
$$ ML^{-3} $$

C
Dimensionless

D
$$ M^2L^{-6} $$

Solution

Relative density is equal to the ratio of density of substance to density of water. Since both have the same dimensions, thus their ratio is dimensionless.

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Units and Measurements Test - 49

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Units and Measurements Test - 49

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Question 1

$$\displaystyle \left [ M^{0}L^{0}TA^{0} \right ]$$

$$\displaystyle \left [ ML^{0}TA^{-2} \right ]$$

$$\displaystyle \left [ ML^{0}TA^{2} \right ]$$

$$\displaystyle \left [ ML^{0}T^{2}A^{-2} \right ]$$

Solution

Question 2

$$10^7 km$$

$$15\times10^8 km$$

$$10^8 km$$

$$10^9 km$$

Solution

Question 3

$$[ M{ L }^{ 2 }{ A }^{ -1 }{ T }^{ -3 }] $$

$$[ M{ L }^{ 3 }{ A }^{ -1 }{ T }^{ -2 }] $$

$$[ M{ L }^{ 2 }{ A }^{ -2 }{ T }^{ -3 }] $$

None of these

Solution

Question 4

$$10^{-2} m^3$$

$$10^{-4} m^3$$

$$10^{-3} m^3$$

$$10^{-3} cm^3$$

Solution

Question 5

$$2\times10^{-4} m^3$$

$$10^{-3} m^3$$

$$10^{-7} m^3$$

$$10^{-2} m^3$$

Solution

Question 6

30

29

24

21

Solution

Question 7

$$MT^{-2} I^{-1}$$

$$MT^{-3} I^{-1}$$

$$MT^{-2} I^{-2}$$

$$MT^{-3} I^{-2}$$

Solution

Question 8

$$\displaystyle 4.0\times 10^{-4}$$

$$\displaystyle 3.975\times 10^{-4}$$

$$\displaystyle 3.9\times 10^{-4}$$

none of the above

Solution

Question 9

2.008

2.0082

2.009

2

Solution

Question 10

$$ kg m^{-3} $$

$$ ML^{-3} $$

Dimensionless

$$ M^2L^{-6} $$

Solution

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