Units and Measurements Test

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Units and Measurements Test - 67

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Weekly Quiz Competition

Question 1
1 / -0

Dimensional equation of universal constant of gravitation is:

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

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

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

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

Solution
Question 2
1 / -0

In the context of accuracy to measurement and significant figures in expressing results of experiment, which of the following is/are correct?
(1) Out of the measurements 50.14 cm and 0.00025 ampere, the first one has greater accuracy.
(2) If one travels 478 km by rail and 397 m. by road, the total distance traveled is 478 km.

A
Only (1) is correct

B
Only (2) is correct

C
Both are correct

D
None of these is correct.

Solution

Significant figures are the number of digits in a value, often a measurement, that contribute to the degree of accuracy of the value. We start counting significant figures at the first non-zero digit.
Since for $$50.14cm$$
Now, the significant number$$=4$$
and for $$0.00025$$
Significant number $$=2$$
Both are correct
Question 3
1 / -0

The dimensions of planck's constant equal to that of

A
energy

B
momentum

C
angular momentum

D
power

Solution

Planck's constant, symbolized $$h$$, relates the energy in one quantum (photon) of electromagnetic radiation to the frequency of that radiation.

$$h=\dfrac Ev=\dfrac{[ML^2T^{-2}]}{[T^{-1}]}=[ML^2T^{-1}]$$

Angular momentum $$=$$ Moment of inertia $$\times$$ Angular velocity (Angular momentum)

$$=[ML^2][T^{-1}]=[ML^2T^{-1}]$$

So, here we can see that both the Planck's constant and Angular momentum have the same dimensions.
Question 4
1 / -0

In the formula : y = a sin ( at - b), where y - position, t - time and A, a, b are constants. What will be the dimensional formula of $$\frac{b}{a}$$?

A
T

B
L

C
$$L^{-1}$$

D
$$T^{-1}$$

Solution

Dimensions of at=Dimension of b
$$\left[ a \right] \left[ t \right] =\left[ b \right] \\ \dfrac { b }{ a } =t\\ =T$$
Question 5
1 / -0

The radius of a sphere is $$(2.6 \pm 0.1)$$ cm. The percentage error in its volume is

A
$$\dfrac{0.1}{2.6} \times 100% $$

B
$$ 3 \times \dfrac{0.1}{2.6} \times 100% $$

C
$$ \dfrac{{8.49}}{{73.6}} \times 100$$

D
$$\dfrac{0.1}{2.6} % $$

Solution

Given,
Radius$$ = \left( {2.6 \pm 0.1} \right)cm$$
$$V = \frac{4}{3}\pi {r^3} = \frac{4}{3} \times 3.14 \times {\left( {2.6} \right)^3} = 73.6\,c{m^3}$$
Now, in error
$$\begin{array}{l} \frac { { \Delta V } }{ V } =3\left( { \frac { { \Delta r } }{ r } } \right) \\ \Delta V=v\propto 3\left( { \frac { { \Delta r } }{ r } } \right) \\ \Delta V=73.6\propto 3\left( { \frac { { 0.1 } }{ { 2.6 } } } \right) \\ \Delta V=8.49\, c{ m^{ 3 } } \\ \frac { { \Delta V } }{ V } \times 100=\frac { { 8.49 } }{ { 73.6 } } \times 100 \end{array}$$
$$=11.53$$%
$$\therefore$$ Option $$C$$ is correct.
Question 6
1 / -0

The dimensions of emf in MKS is:

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

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

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

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

Solution

The emf across an inductor is given by:
$$ e = -L\dfrac {di}{dt} $$

$$e = \big[ {ML^{2}}{T^{-2}}{A^{-2}} \big] \big[ \dfrac {A}{T} \big] $$

$$e = \dfrac {{ML^{2}}{T^{-2}}}{A^{-1}{T}} $$

$$e = {ML^{2}}{T^{-2}}{Q^{-1}}$$
Question 7
1 / -0

The dimension of $$\left( \dfrac { 1 }{ 2 } \right) { \varepsilon }_{ 0 }{ E }^{ 2 }$$ is($${ \varepsilon }_{ 0 }$$: permittivity of free space, Electric field)

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

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

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

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

Solution

As we know that
Energy Density$$= \dfrac{energy}{volume}= \left( \dfrac { 1 }{ 2 } \right) { \varepsilon }_{ 0 }{ E }^{ 2 }$$

Hence ,
Dimesion of $$\left( \dfrac { 1 }{ 2 } \right) { \varepsilon }_{ 0 }{ E }^{ 2 }$$ is, =$$\dfrac{[ML^2T^{-2}]}{L^3}= ML^{-1}T^{-2}$$
Question 8
1 / -0

Write down the significant figures in the following
1. $$5238$$ N
2. $$4200 $$kg

A
$$4 , 4$$

B
$$4 , 2$$

C
$$4, 3$$

D
$$5, 3$$

Solution

All non-zero digits are significant and last zeros are not considered as significant.
(a) : The given number $$5238$$ has four significant figures.
(b) : The given number $$4200$$ has two significant figures.
Question 9
1 / -0

The dimension of the magnetic field intensity B is

A
$$(a)\,ML{T^{ - 2\,}}\,{A^{ - 1}}$$

B
$$(b)\,M{T^{ - 2\,}}\,{A^{ - 1}}$$

C
$$(c)\,M{L^{2\,}}\,T{A^{ - 2}}$$

D
$$(d)\,{M^{2\,\,}}L{T^{ - 2}}\,{A^{ - 1}}$$

Solution

$$F=qVB\sin { \theta } $$

$$\left[ F \right] =\left[ ML{ T }^{ -2 } \right] $$

$$\left[ V \right] =\left[ L{ T }^{ -1 } \right] $$

$$\left[ q \right] =\left[ AT \right] $$

$$\sin \theta =$$ dimensionless

$$\left[ B \right] =\cfrac { \left[ F \right] }{ \left[ q \right] \left[ V \right] } $$

$$=\cfrac { \left[ ML{ T }^{ -2 } \right] }{ \left[ AT \right] \left[ L{ T }^{ -1 } \right] } $$

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

$$\left[ B \right] =\left[ M{ T }^{ -2 }{ A }^{ -1 } \right] $$
Question 10
1 / -0

The dimension of combination $$\dfrac{L}{CVR}$$ are same as dimensions of

A
Change

B
Current

C
Charge$$^{-1}$$

D
Current $$^{-1}$$

Solution

Dimension of $$Resistance = \dfrac{Potential \,Difference}{Current} = \dfrac{ML^2T^{-3}A^{-1}}{ A} = [ML^2T^{-3}A^{-2}[$$

Dimension of $$Capacitance = \dfrac{Charge}{ Potential\, Difference} = \dfrac{AT}{ML^2T^{-3}A^{-1}} = [M^{-1}L^{-2}T^4A^2[$$

Dimension of $$Self Inductance [L]=\dfrac {[\phi]}{[I]}= \dfrac{MT^{−2}L^2A^{−1}}{ A} = [ML^2T^{-2}A^{-2}[$$

Dimension of $$Voltage=\dfrac{Energy}{Charge}=\dfrac{[ML^2T^{-2}]}{AT}=[ML^2T^{-3}A^{-1}]$$

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

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Units and Measurements Test - 67

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Units and Measurements Test - 67

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Question 1

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

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

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

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

Solution

Question 2

Only (1) is correct

Only (2) is correct

Both are correct

None of these is correct.

Solution

Question 3

energy

momentum

angular momentum

power

Solution

Question 4

T

L

$$L^{-1}$$

$$T^{-1}$$

Solution

Question 5

$$\dfrac{0.1}{2.6} \times 100% $$

$$ 3 \times \dfrac{0.1}{2.6} \times 100% $$

$$ \dfrac{{8.49}}{{73.6}} \times 100$$

$$\dfrac{0.1}{2.6} % $$

Solution

Question 6

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

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

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

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

Solution

Question 7

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

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

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

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

Solution

Question 8

$$4 , 4$$

$$4 , 2$$

$$4, 3$$

$$5, 3$$

Solution

Question 9

$$(a)\,ML{T^{ - 2\,}}\,{A^{ - 1}}$$

$$(b)\,M{T^{ - 2\,}}\,{A^{ - 1}}$$

$$(c)\,M{L^{2\,}}\,T{A^{ - 2}}$$

$$(d)\,{M^{2\,\,}}L{T^{ - 2}}\,{A^{ - 1}}$$

Solution

Question 10

Change

Current

Charge$$^{-1}$$

Current $$^{-1}$$

Solution

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

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

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

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