Units and Measurements Test

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Units and Measurements Test - 63

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Weekly Quiz Competition

Question 1
1 / -0

The dimensional formula of electric flux is:

A
$$\left[ { M }^{ 1 }{ L }^{ 1 }{ T }^{ -2 } \right] $$

B
$$\left[ { M }^{ 1 }{ L }^{ 3 }{ T }^{ -3 }{ A }^{ -1 } \right] $$

C
$$\left[ { M }^{ 2 }{ L }^{ 2 }{ T }^{ -2 }{ A }^{ -2 } \right] $$

D
$$\left[ { M }^{ 1 }{ L }^{ -3 }{ T }^{ 3 }{ A }^{ 1 } \right] $$

Solution

Correct option is (B) $$\left[\mathrm{M}^{1} \mathrm{~L}^{3} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]$$

Hint: Electric flux is the measure of the electric field lines crossing the surface.

Step1: Electric flux $$\phi=\int \vec{E} \cdot \vec{s}$$

The dimension of $$\phi=$$ dimension fo $$\mathrm{E} \times$$ dimension of $$\mathrm{s}$$

$$[\phi]=\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}\right][\mathrm{AT}]^{-1}\left[\mathrm{~L}^{2}\right]=\left[\mathrm{M}^{1} \mathrm{~L}^{3} \mathrm{~T}^{-3} \mathrm{~A}^{-1}\right]$$
Question 2
1 / -0

If E, M, J and G denote energy, mass, angular momentum and gravitational constant respectively, then $$\frac{EJ^2}{M^5 G^2}$$ has the dimensions of

A
length

B
mass

C
time

D
angle

Solution

Dimensional formula of:

$$E : [ML^{2}T^{-2}]\\$$

$$J : [ML^{2}T^{-1}]\\$$

$$G : [M^{-1}L^{3}T^{-2}]\\$$

Thus

$$(E J ^{2})/(M^{5} G^{2}) = \dfrac{[ML^{2}T^{-2}][ML^{2}T^{-1}]^{2}}{M^{5} \times [M^{-1}L^{3}T^{-2}]^{2}} \\$$

$$= \dfrac{[M^{3}L^{6}T^{-4}]}{[M^{3}L^{6}T^{-4}]}\\$$

$$= [M^{0} L^{0} T^{0}]$$

The above quantity is dimensionless. Angle is also a dimensionless quantity while others are having some dimensions.

Thus the correct option is D.
Question 3
1 / -0

Force F is given in terms of time t and distance x by $$F = A \sin Ct + B \cos Dx$$. Then the dimensions of $$\dfrac {A}{B}$$ and $$\dfrac{C}{D}$$ are:

A
$$[M^0L^0T^0], [M^0L^0T^{-1}]$$

B
$$[MLT^{-2}], [M^0L^{-1}T^{0}]$$

C
$$[M^0L^0T^0], [M^0LT^{-1}]$$

D
$$[M^0L^1T^{-1}], [M^0L^0T^0]$$

Solution

$$\textbf{Step 1: Dimensions of C and D}$$
$$F = A \sin \, Ct + B \cos \, Dx$$
Since the argument of trigonometric functions is the angle, which is dimensionless
$$\therefore [Ct] = M^{0} L^{0} T^{0}$$
$$\Rightarrow[CT^1] = M^{0} L^{0} T^{0}$$
$$\Rightarrow [C] = T^{-1}$$ $$....(1)$$

Similarly
$$[Dx] = M^{0} L^{0} T^{0}$$
$$\Rightarrow$$ $$[DL^{1}] = M^{0} L^{0} T^{0}$$
$$\Rightarrow$$ $$[D] = L^{-1}$$ $$....(2)$$

$$\textbf{Step 2: Using principle of homogeneity}$$
Since the value of trigonometric functions is dimensionless,
$$\therefore$$ Using the principle of homogeneity, Dimension of $$A$$ and $$B$$ is same as Force $$F$$

$$[A] = [B] = MLT^{-2}$$ $$....(3)$$

$$\textbf{Step 3: Final calculation}$$
From equation $$(3)$$
$$\dfrac{A}{B} = M^{0} L^{0} T^{0}$$

From equation $$(1)$$ and $$(2)$$
$$\dfrac{C}{D} = M^{0} L T^{-1}$$

Hence Option $$C$$ is correct.
Question 4
1 / -0

The radius of the earth is $$6.37 \times 10^6 m$$ and its mass is $$5.975 \times 10^{24} kg$$. Find the earth's average density to appropriate significant figures.

A
$$5 \times 10^3 kg m^{-3}$$

B
$$5.52 \times 10^3 kg m^{-3}$$

C
$$2 \times 10^3 kg m^{-3}$$

D
$$5.52 \times 10^3 kg m^{-4}$$

Solution

$$\begin{array}{l}\text { radius }=6.37 \times 10^{6}\mathrm{~m} \\\text { Volume }=\frac{4}{3} \pi r^{3}\\V=\frac{4}{3} \times \pi \times\left(6.37 \times 10^{6}\right)^{3} \\\text { Mass }=5.975 \times 10^{24}\mathrm{~kg} \\\text { density }(\rho)=\frac{\operatorname{mass}(\mathrm{m})}{V}\end{array}$$
$$\begin{aligned}\rho &=\frac{5.975 \times 10^{24}}{4\times\pi\times\left(6.37\times10^{6}\right)^{3}}\\&=5.52\times10^{3}\mathrm{kgm}^{-3}\end{aligned}$$
Question 5
1 / -0

In the relation $$y = r sin (\omega t - kx)$$, the dimensions of $$\omega/k$$ are

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

B
$$[M^0L^{1}T^{-1}]$$

C
$$[M^0L^0T^1]$$

D
$$[M^0L^1T^0]$$

Solution

$$y=rsin\left( wt-kx \right) $$
$$\left( wt-kr \right) $$ is angle so it is dimensionless.
$$\left[ wt \right] =\left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 } \right] \quad \quad \left[ kx \right] =\left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 } \right] $$
$$w=\dfrac { 1 }{ t } $$
$$\left[ w \right] =\left[ { T }^{ -1 } \right] $$ $$k=\dfrac { 1 }{ x } $$
$$\dfrac { w }{ k } =\dfrac { \left[ { T }^{ -1 } \right] }{ \left[ { L }^{ -1 } \right] } =\left[ { M }^{ 0 }{ LT }^{ -1 } \right] $$ $$k=\left[ { L }^{ -1 } \right] $$
Question 6
1 / -0

With due regard to significant figures, add the following:
a. 953 and 0.324
b. 953 and 0.625
c. 953.0 and 0.324
d. 953.0 and 0.374

A
a. $$953$$; b. $$954$$ c. $$953.3$$ d. $$953.4$$

B
a. $$953$$; b. $$955$$ c. $$953.3$$ d. $$953.4$$

C
a. $$952$$; b. $$954$$ c. $$953.3$$ d. $$953.4$$

D
a. $$953$$; b. $$954$$ c. $$953.3$$ d. $$952.4$$

Solution

a.  $$953+0.324=953.324\approx \boxed { 953 } $$
b.  $$953+0.625=953.625\approx \boxed { 954 } $$
c.  $$953.0+0.324=953.324\approx \boxed { 953.3 } $$
d.  $$953.0+0.374=953.374\approx \boxed { 953.4 } $$
Question 7
1 / -0

A physical quantity $$x$$ depends on quantities $$y$$ and $$z$$ as follows: $$x = Ay + B \tan (C_z)$$, where $$A, B$$ and $$C$$ are constants. Which of the followings do not have the same dimensions?

A
$$x$$ and $$B$$

B
$$C$$ and $$z^{-}$$

C
$$y$$ and $$B/A$$

D
$$x$$ and $$A$$

Solution

Given $$x=Ay+Btan\left( { C }_{ z } \right) $$
Each term in a equation has same dimensions.
$$\Rightarrow Dim\left( x \right) =Dim\left( Ay \right) =Dim\left( B \right) $$
$$\Rightarrow$$ $$x$$ and $$A$$ cannot have same dimension.
Question 8
1 / -0

If $$L$$ and $$R$$ denote inductance and resistance, respectively, then the dimensions of $$L/R$$ are

A
$$M^1L^0T^0Q^{-1}$$

B
$$M^0L^0TQ^{0}$$

C
$$M^0L^1T^{-1}Q^{0}$$

D
$$M^{-1}LT^0Q^{-1}$$

Solution

We know $$E=\dfrac { 1 }{ 2 } L{ i }^{ 2 }\quad \Rightarrow \quad L=\dfrac { 2E }{ { i }^{ 2 } } $$
$$\Rightarrow$$ dimension of $$L=\dfrac { { MLT }^{ -2 } }{ { Q }^{ 2 } } .{ T }^{ 2 }={ MLT }^{ 0 }{ Q }^{ -2 }$$
Also we know $$E={ i }^{ 2 }RT\Rightarrow R=\dfrac { E }{ { i }^{ 2 }t } $$
dimension of $$R=\dfrac { { MLT }^{ -2 } }{ { Q }^{ 2 } } .{ T }^{ -1 }={ MLT }^{ -1 }{ Q }^{ -2 }$$
$$\therefore$$ $$dim\left( \dfrac { L }{ R } \right) =\dfrac { { MLT }^{ 0 }{ Q }^{ -2 } }{ { MLT }^{ -1 }{ Q }^{ -2 } } =\left[ T \right] $$
Question 9
1 / -0

In the relation $$\dfrac{dy}{dt} = 2 \omega \sin (\omega t + \phi_0)$$, the dimensional formula for $$\omega t + \phi_0$$ is

A
$$MLT$$

B
$$MLT^0$$

C
$$ML^0T^0$$

D
$$M^0L^0T^0$$

Solution

The given ralation is:
$$\dfrac { dy }{ dt } =2wsin\left( wt+{ \phi }_{ 0 } \right) $$

It is the equation for the velocity of the wave.

Arguments of trignometric functions are dimension less
$$\Rightarrow \left( wt+\phi \right) $$ has dimension $$\left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 } \right] $$.
Question 10
1 / -0

The following observations were taken for determining the surface tension of water by capillary tube method : diameter of capillary , $$D = 1.25 \times 10^{-2} m$$. Taking $$ g = 9.80 m s^{-2}$$ and using the relation $$T = (rgh/2) \times 10^3 Nm^{-1}$$. what is the possible error in measurement of surface tension T?

A
$$2.4\%$$

B
$$15\%$$

C
$$1.6\%$$

D
$$0.15\%$$

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Units and Measurements Test - 63

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Units and Measurements Test - 63

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Question 1

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

$$\left[ { M }^{ 1 }{ L }^{ 3 }{ T }^{ -3 }{ A }^{ -1 } \right] $$

$$\left[ { M }^{ 2 }{ L }^{ 2 }{ T }^{ -2 }{ A }^{ -2 } \right] $$

$$\left[ { M }^{ 1 }{ L }^{ -3 }{ T }^{ 3 }{ A }^{ 1 } \right] $$

Solution

Question 2

length

mass

time

angle

Solution

Question 3

$$[M^0L^0T^0], [M^0L^0T^{-1}]$$

$$[MLT^{-2}], [M^0L^{-1}T^{0}]$$

$$[M^0L^0T^0], [M^0LT^{-1}]$$

$$[M^0L^1T^{-1}], [M^0L^0T^0]$$

Solution

Question 4

$$5 \times 10^3 kg m^{-3}$$

$$5.52 \times 10^3 kg m^{-3}$$

$$2 \times 10^3 kg m^{-3}$$

$$5.52 \times 10^3 kg m^{-4}$$

Solution

Question 5

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

$$[M^0L^{1}T^{-1}]$$

$$[M^0L^0T^1]$$

$$[M^0L^1T^0]$$

Solution

Question 6

a. $$953$$; b. $$954$$ c. $$953.3$$ d. $$953.4$$

a. $$953$$; b. $$955$$ c. $$953.3$$ d. $$953.4$$

a. $$952$$; b. $$954$$ c. $$953.3$$ d. $$953.4$$

a. $$953$$; b. $$954$$ c. $$953.3$$ d. $$952.4$$

Solution

Question 7

$$x$$ and $$B$$

$$C$$ and $$z^{-}$$

$$y$$ and $$B/A$$

$$x$$ and $$A$$

Solution

Question 8

$$M^1L^0T^0Q^{-1}$$

$$M^0L^0TQ^{0}$$

$$M^0L^1T^{-1}Q^{0}$$

$$M^{-1}LT^0Q^{-1}$$

Solution

Question 9

$$MLT$$

$$MLT^0$$

$$ML^0T^0$$

$$M^0L^0T^0$$

Solution

Question 10

$$2.4\%$$

$$15\%$$

$$1.6\%$$

$$0.15\%$$

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