Units and Measurements Test

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Units and Measurements Test - 65

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Weekly Quiz Competition

Question 1
1 / -0

The Van der Waal's equation of state for some gases can be expressed as:
$$(P + \dfrac{a}{V^2})(V - b) = RT$$
where P is the pressure, V is the molar volume, and T is the absolute temperature of the given sample of gas and a,b and R are constants.
The dimensions of a are

A
$$ML^5T^{-2}$$

B
$$ML^{-1}T^{-2}$$

C
$$L^3$$

D
none of the above

Solution

$$\left[ P \right] =\left[ \dfrac { a }{ { V }^{ 2 } } \right] $$
$$\left[ a \right] =\left[ P \right] \left[ { V }^{ 2 } \right] $$
$$\left[ a \right] =\left[ \dfrac { Force }{ Area } \times { \left( Volume \right) }^{ 2 } \right] $$
$$\left[ a \right] =\left[ \dfrac { { MLT }^{ -2 }\times { L }^{ 6 } }{ { L }^{ 2 } } \right] $$
$$\left[ a \right] =\left[ { ML }^{ 5 }{ T }^{ -2 } \right] $$
Question 2
1 / -0

Dimensions of $$\dfrac {1}{\mu_{0}\epsilon_{0}}$$, where symbols have their usual meaning, are:

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

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

C
$$[L^{2}T^{-2}]$$

D
$$[LT^{-1}]$$

Solution

We know that the speed of light is represented by:
$$c=\dfrac{1}{\sqrt{\mu_0\epsilon_0}}$$

So, we can write:
$$\dfrac {1}{\mu_{0}\epsilon_{0}} = c^{2}$$
The dimension of speed of light is represented as $$[LT^{-1}]$$

So, for $$\dfrac{1}{\mu_0\epsilon_0}= [L^{2}T^{-2}]$$.
Question 3
1 / -0

The dimension of magnetic field in $$M, L, T$$ and $$C$$ (Coulomb) is given as:

A
$$MLT^{-1}C^{-1}$$

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

C
$$MT^{-1}C^{-1}$$

D
$$MLT^{-2}C^{-1}$$

Solution

$$[B] = \dfrac {F}{il} = \dfrac {Ft}{ql} = \dfrac {MLT^{-2}T}{CL} = [MT^{-1}C^{-1}]$$.
Question 4
1 / -0

In terms of resistance $$R$$ and time $$T$$, the dimensions of ratio $$\dfrac {\mu}{\epsilon}$$ of the permeability $$\mu$$ and permittivity $$\epsilon$$ is:

A
$$[R^{2}T^{2}]$$

B
$$[RT^{-2}]$$

C
$$[R^{2}T^{-1}]$$

D
$$[R^{2}]$$

Solution

The electrostatic force can be given by:
$$F_e=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{r^2}$$

So, permittivity is:
So, $$\varepsilon_0=\dfrac{q^2}{4\pi r^2F_e}$$

So, dimensions : $$\varepsilon_0=\dfrac{[Q^2]}{[L^2][MLT^{-2}]}$$

The dimensions of the permittivity $$\varepsilon$$ is $$=[M^{-1}L^{-3}T^2Q^2]$$

The magnetic force can be given by:
$$F_m=\dfrac{\mu_0}{4\pi}\dfrac{q_{m_1}q_{m_2}}{r^2}$$

So, permeability is:
So, $$\mu_0=\dfrac{q_{m_1}q_{m_2}}{4\pi r^2F_m}$$

The dimensions of the permeability $$\mu$$ is $$=[M^{1}L^{1}Q^{-2}]$$

So, the ratio of the two dimensions is:
$$\dfrac{\mu}{\varepsilon}=\dfrac{[M^{1}L^{1}Q^{-2}]}{[M^{-1}L^{-3}T^2Q^2]}$$

$$\Rightarrow M^2L^4T^{-2}Q^{-4}$$

The dimension of the resistance is $$R=[ML^2T^{-1}Q^{-2}]$$

comparing the dimension with the dimesion of resistance:
$$\therefore \dfrac{\mu}{\varepsilon}=[R^2]$$
Question 5
1 / -0

A vernier callipers has its main scale of $$10 cm$$ equally divided into $$200$$ equal parts. Its vernier scale of $$25$$ divisions coincides with $$12 mm$$ on the main scale. The least count of the instrument is -

A
$$0.020 cm$$

B
$$0.002 cm$$

C
$$0.010 cm$$

D
$$0.001 cm$$

Solution

$$10\ cm$$ of Main Scale $$=200$$ div of Main scale
$$\Rightarrow IMSD=0.05\ cm$$
given $$25$$ vernier scale division coincide with $$12\ mm$$ on main scale
$$\Rightarrow 25VSD=12\ mm$$ on main scale
$$=\dfrac{12}{0.5}=24\ MSD$$
$$\Rightarrow 25VSD=24MSD$$
$$\Rightarrow 1VSD=\dfrac{24}{25}MSD=0.048\ cm$$
Least Count $$\Rightarrow 1MSD-1VSD$$
$$\Rightarrow (0.050-0.048)cm$$
$$\Rightarrow 0.002\ cm$$
Question 6
1 / -0

The dimensions of angular momentum, latent heat and capacitance are, respectively.

A
$$ML^{2}T^{1}A^{2}, L^{2}T^{-2}M^{-1}L^{-2}T^{2}$$

B
$$MLT^{2}, L^{2}T^{2}, M^{-1}L^{-2}T^{4}A^{2}$$

C
$$ML^{2}T^{-1}, L^{2}T^{-2}, ML^{2}TA^{2}$$

D
$$ML^{2}T^{-1}, L^{2}T^{-2}, M^{-1}L^{-2}T^{4}A^{2}$$

Solution

For angular momentum:
$$[L] = mvr = [ML^{2}T^{-1}]$$

The latent heat:
Latent Heat $$= \dfrac {Q}{m} = [L^{2}T^{-2}]$$

The capacitance is given as:
$$[C] = \dfrac {q}{V} =\dfrac{q^2}{W}$$

$$\dfrac {ATAT}{ML^{2}T^{-2}} = [M^{-1}L^{-2}T^{4}A^{2}]$$.
Question 7
1 / -0

The dimension of electrical conductivity is _________.

A
$$M^{-2}L^{-4}T^3A^2$$.

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

C
$$M^{-1}L^{-3}T^4A^2$$.

D
$$M^{-2}L^{-3}T^6A^2$$.

Solution

Question 8

1 / -0

Match the physical quantities given in Column I with dimensions expressed in terms of mass (M), length (L) time (T), and charge (Q) given in Column II.

Column I	Column II
i. Angular momentum	a. $$ML^2T^{-2}$$
ii. Torque	b. $$ML^2T^{-1}$$
iii. Inductance	c. $$M^{-1}L^{-2}T^2Q^2$$
iv. Latent heat	d. $$ML^2Q^{-2}$$
v. Capacitance	e. $$ML^3T^{-1}Q^{-2}$$
vi. Resistivity	f. $$L^2T^{-2}$$

$$i \rightarrow c., ii \rightarrow a., iii \rightarrow d., iv \rightarrow f., v \rightarrow b., vi \rightarrow e$$

$$i \rightarrow b., ii \rightarrow a., iii \rightarrow d., iv \rightarrow f., v \rightarrow c., vi \rightarrow e$$

$$i \rightarrow b., ii \rightarrow e., iii \rightarrow d., iv \rightarrow f., v \rightarrow c., vi \rightarrow a$$

$$i \rightarrow b., ii \rightarrow a., iii \rightarrow f., iv \rightarrow d., v \rightarrow c., vi \rightarrow e$$

Solution

Question 9
1 / -0

The dimension of permittivity of vacuum

A
$$[M^{ -2} L^ {-3} T^4 A^2]$$

B
$$[M^{ -1} L^ {-3} T^4 A^3]$$

C
$$[M^{ -1} L^ {-3} T^4 A^2]$$

D
$$ [M^{ -1} L^ {-5} T^4 A^2]$$

Solution

The permittivity of the vacuum can be given by:
$$F_e=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{r^2}$$
$$permitivity\ (\varepsilon_0)= \dfrac{(q)^2}{(F_e)\times(r)^2}$$

$$=\dfrac{[TA]^2}{[MLT^ {-2}] [L]^2}$$

$$=\dfrac{[TA]^2}{[ML^3T^ {-2}]}$$

$$= [M^{-1} L^{-3} T^4 A^2]$$

$$= [M^{ -1} L^ {-3} T^4 A^2]$$
This is the required solution.
Question 10
1 / -0

Two particles of mass $${m}_{1}$$ and $${m}_{2}$$ are approaching towards each other under their mutual gravitational field. If the speed of the particle is $$v$$, the speed of the center of mass of the system is equal to:

A
$$\cfrac { { m }_{ 1 } }{ { m }_{ 2 } } v$$

B
zero

C
$$\cfrac { \left( { m }_{ 1 }-{ m }_{ 2 } \right) v }{ { m }_{ 1 }+{ m }_{ 2 } } $$

D
$$\cfrac { { m }_{ 2 } }{ { m }_{ 1 } } v$$

Solution

Taking both particles as a system,
$$F_{ext}=0$$
as mutual gravitational force becomes internal force here.
$$a_{cm}=\dfrac{F_{ext}}{m}=0$$

$$v_{cm}=u_{cm}+a_{cm}t$$

$$v_{cm}=0$$

$$u_{cm}$$ of system is zero as system starts from rest.

Hence v_{cm}=0$$

Option $$\textbf B$$ is the correct answer.

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Units and Measurements Test - 65

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Units and Measurements Test - 65

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Question 1

$$ML^5T^{-2}$$

$$ML^{-1}T^{-2}$$

$$L^3$$

none of the above

Solution

Question 2

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

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

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

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

Solution

Question 3

$$MLT^{-1}C^{-1}$$

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

$$MT^{-1}C^{-1}$$

$$MLT^{-2}C^{-1}$$

Solution

Question 4

$$[R^{2}T^{2}]$$

$$[RT^{-2}]$$

$$[R^{2}T^{-1}]$$

$$[R^{2}]$$

Solution

Question 5

$$0.020 cm$$

$$0.002 cm$$

$$0.010 cm$$

$$0.001 cm$$

Solution

Question 6

$$ML^{2}T^{1}A^{2}, L^{2}T^{-2}M^{-1}L^{-2}T^{2}$$

$$MLT^{2}, L^{2}T^{2}, M^{-1}L^{-2}T^{4}A^{2}$$

$$ML^{2}T^{-1}, L^{2}T^{-2}, ML^{2}TA^{2}$$

$$ML^{2}T^{-1}, L^{2}T^{-2}, M^{-1}L^{-2}T^{4}A^{2}$$

Solution

Question 7

$$M^{-2}L^{-4}T^3A^2$$.

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

$$M^{-1}L^{-3}T^4A^2$$.

$$M^{-2}L^{-3}T^6A^2$$.

Solution

Question 8

$$i \rightarrow c., ii \rightarrow a., iii \rightarrow d., iv \rightarrow f., v \rightarrow b., vi \rightarrow e$$

$$i \rightarrow b., ii \rightarrow a., iii \rightarrow d., iv \rightarrow f., v \rightarrow c., vi \rightarrow e$$

$$i \rightarrow b., ii \rightarrow e., iii \rightarrow d., iv \rightarrow f., v \rightarrow c., vi \rightarrow a$$

$$i \rightarrow b., ii \rightarrow a., iii \rightarrow f., iv \rightarrow d., v \rightarrow c., vi \rightarrow e$$

Solution

Question 9

$$[M^{ -2} L^ {-3} T^4 A^2]$$

$$[M^{ -1} L^ {-3} T^4 A^3]$$

$$[M^{ -1} L^ {-3} T^4 A^2]$$

$$ [M^{ -1} L^ {-5} T^4 A^2]$$

Solution

Question 10

$$\cfrac { { m }_{ 1 } }{ { m }_{ 2 } } v$$

zero

$$\cfrac { \left( { m }_{ 1 }-{ m }_{ 2 } \right) v }{ { m }_{ 1 }+{ m }_{ 2 } } $$

$$\cfrac { { m }_{ 2 } }{ { m }_{ 1 } } v$$

Solution

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