Units and Measurements Test

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Units and Measurements Test - 71

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Weekly Quiz Competition

Question 1
1 / -0

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

A
resistance

B
resistivity

C
electric cunductivity

D
electromotive force

Solution

$$\textbf{Explanation:-}$$

We know that,

Resistivity, $$\rho = \dfrac{m}{ne^2t}$$

$$\Rightarrow \rho = \dfrac{[M]}{[L^{-3}][AT]^{2}[T]}$$

$$\Rightarrow \rho = [ML^{3}A^{-2}T^{-3}]$$

Now,

Electrical conductivity, $$\sigma = \dfrac{1}{\rho }$$

$$\Rightarrow \sigma = \dfrac{1}{[ML^{3}A^{-2}T^{-3}]}$$

$$\Rightarrow \sigma = [M^{-1}L^{-3}A^{2}T^{3}]$$

$$\textbf{Hence the correct option is C}$$
Question 2
1 / -0

Order of $$\frac { 1 } { 8 \times 10 ^ { 9 } }$$ is:

A
$$1.77\times {{10}^{-10}}$$

B
$$1.25\times {{10}^{-10}}$$

C
$$1.55\times {{10}^{-10}}$$

D
$$1.95\times {{10}^{-10}}$$

Solution

$$\dfrac{1}{8\times 10^{9}}=\left(\dfrac{1}{8}\right)\times 10^{-9}=(0.125)\times 10^{-9}$$
$$=1.25\times 10^{-10}$$
Order of $$\dfrac{1}{8\times 10^{9}}$$ is $$10^{-10}$$
Option $$B$$ is correct
Question 3
1 / -0

A boy wants to buy new shoes. To find out his correct shoe size the ________ of his feet should be measured.

A
Length

B
Thickness

C
Area

D
Volume

Solution

Shoe size basically refers to the length of shoes which go along with the length of feet that have to fit in the shoe. Thus the length of the feet should be measured as only it can help in achieving the right size.
Question 4
1 / -0

The time of 25 oscillations of a simple pendulum is measured to be $$50.0 s$$ by a watch of least count $$0.1 s$$. The percentage error in time is

A
$$0.2 \%$$

B
$$0.02 \%$$

C
$$0.002 \%$$

D
$$2 \%$$

Solution

$$ \begin{array}{l} \text { Given- } \text { Time of } 25 \text { oscillations }=50.0 \mathrm{~s} \\ \text { Least count of the watch }=0.1 \mathrm{~s} \\ \text { Percentage error in Time } \\ \text {%err }=\frac{\Delta T}{T}=\frac{0.1}{50}=\frac{0.1}{50} \times 100 \%=0.2 \% \\ \text { % error in Time is } 0.2 \% \end{array} $$
Question 5
1 / -0

'The parallax angle in radians is: $$\theta = \left( 1 + \frac { 54 } { 60 } \right) \times \frac { \pi } { 180 } = 0.03316 \mathrm { rad }$$
Hence, the distance between moon and earth:

A
$$\dfrac { \text { Diameter of Earth } } { \theta }$$

B
$$\dfrac { 1.276 \times 10 ^ { 7 } m } { 0.03316 }$$

C
$$3.84 \times 10 ^ { 8 } m$$

D
nnone of these

Solution

Given,

$$\theta =0.03316\,\,rad$$

Also $$b = \mathrm { AB } =$$ diameter of earth $$= 1.276 \times 10 ^ { 7 } \mathrm { m }$$

Now d $$=\dfrac{b}{\theta }=\dfrac{1.276\times {{10}^{7}}}{0.03316\,}=3.84\times {{10}^{8}}\text{m}$$

Hence, distance between earth and moon is $$3.84\times {{10}^{8}}\text{m}$$
Question 6
1 / -0

The length of a rod is (11.05 $$\pm$$ 0.2) cm. What is the length of the two rods?

A
(22.1 $$\pm$$ 0.05) cm

B
(22.1 $$\pm$$ 0.1) cm

C
(22.10 $$\pm$$ 0.05) cm

D
(22.10 $$\pm$$ 0.4) cm

Solution

Given,
$$l=(11.05\pm 0.2)cm$$
The length of two rod,
$$l'=l+l$$
$$l'=(11.05\pm0.2)+(11.05\pm0.2)$$
$$l'=(22.10\pm 0.4)cm$$ (error must be added.)
Question 7
1 / -0

If C is the capacity and V is the potential across the condenser, what will be the dimensional representation of $$\dfrac1 2 C V ^ { 2 }$$?

A
$$M L ^ { 2 } T ^ { -2 } A ^ { 2 }$$

B
$$\mathrm { ML } ^ { - 2 } \mathrm { T } ^ { - 2 }$$

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

D
$$\mathrm { ML } ^ { 2 } \mathrm { T } ^ { - 2 }$$

Solution

Let $$X$$ be the quantity represented.
$$X = \dfrac 12 CV^2$$
$$X = \dfrac 12 \dfrac QV \times V^2$$
$$X = \dfrac12 Q.V$$
$$[X] = [Q][V]=[IT][ML^2T^{-3}I^{-1}]=[ML^2T^{-2}]$$
Question 8
1 / -0

The dimensions of electric potential are:

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

B
$$\left[ {ML{T^{ - 2}}{Q^{ - 1}}} \right]$$

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

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

Solution

The electric potential is expressed as:
$$V=\dfrac {Work}{charge}$$

So, the dimensional formula will be $$[ML^{-2}T^{-2}Q^{-1}]$$
Question 9
1 / -0

The dimensions of elelctrical permittivity are

A
$$\left[ {{A^{ - 2}}{M^0}{L^{ - 3}}{T^4}} \right]$$

B
$$\left[ {{A^2}{M^{ - 1}}{L^{ - 3}}{T^0}} \right]$$

C
$$\left[ {A{M^{ - 1}}{L^{ - 3}}{T^4}} \right]$$

D
$$\left[ {{M^{ - 1}}{L^{ - 3}}{T^4}}{A^2} \right]$$

Solution
Question 10
1 / -0

The quantity $$X$$ is given by $${\varepsilon _o}L\frac{{\Delta V}}{{\Delta T}}$$, where $${\varepsilon _o}$$ is the permittivity of free space, $$L$$ is a length, $${\Delta V}$$ is a potential difference and $${\Delta T}$$ is a time interval. The dimensional formula for $$X$$ is the same as that of

A
resistance

B
charge

C
voltage

D
current

Solution

We know than, from coulomb's law
$$E_0 =\dfrac{1}{4\pi}\,\,\,\,\,\, \dfrac {Q_1 Q_2}{F R^2}$$
$$ \Rightarrow [E_0]= \dfrac {[Q_1][Q_2]}{[F][R^2]}$$
$$\Rightarrow [E_0]=\dfrac {A^{2}T^{2}}{mLT^{-2}L^2}=\left [ \dfrac {A^2T^2}{mLT^{3}L^{-2}} \right ]$$
$$ V = - \int \vec {E}. \vec{d\pi}$$
$$[V]= [E][L]$$
$$[V]= mLT^{-2}A^{-1}T^{-1}L= [mL^2A^{-1}T^{-3}]$$
$$ [T]= T$$
$$ [L]= L$$
$$X = \epsilon _0L\dfrac {\Delta V}{\Delta T}$$
$$[X]= \left [ \dfrac {A^2T^2}{mLT^{3}L^{-2}} \right ] [L]\left [ \dfrac {mL^2A^{-1}T^{-3}}{T} \right ]$$
$$[X]= [A]$$
$$[A]$$ is the dimension of current.

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Units and Measurements Test - 71

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Units and Measurements Test - 71

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SHARING IS CARING

Question 1

resistance

resistivity

electric cunductivity

electromotive force

Solution

Question 2

$$1.77\times {{10}^{-10}}$$

$$1.25\times {{10}^{-10}}$$

$$1.55\times {{10}^{-10}}$$

$$1.95\times {{10}^{-10}}$$

Solution

Question 3

Length

Thickness

Area

Volume

Solution

Question 4

$$0.2 \%$$

$$0.02 \%$$

$$0.002 \%$$

$$2 \%$$

Solution

Question 5

$$\dfrac { \text { Diameter of Earth } } { \theta }$$

$$\dfrac { 1.276 \times 10 ^ { 7 } m } { 0.03316 }$$

$$3.84 \times 10 ^ { 8 } m$$

nnone of these

Solution

Question 6

(22.1 $$\pm$$ 0.05) cm

(22.1 $$\pm$$ 0.1) cm

(22.10 $$\pm$$ 0.05) cm

(22.10 $$\pm$$ 0.4) cm

Solution

Question 7

$$M L ^ { 2 } T ^ { -2 } A ^ { 2 }$$

$$\mathrm { ML } ^ { - 2 } \mathrm { T } ^ { - 2 }$$

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

$$\mathrm { ML } ^ { 2 } \mathrm { T } ^ { - 2 }$$

Solution

Question 8

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

$$\left[ {ML{T^{ - 2}}{Q^{ - 1}}} \right]$$

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

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

Solution

Question 9

$$\left[ {{A^{ - 2}}{M^0}{L^{ - 3}}{T^4}} \right]$$

$$\left[ {{A^2}{M^{ - 1}}{L^{ - 3}}{T^0}} \right]$$

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

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

Solution

Question 10

resistance

charge

voltage

current

Solution

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