Units and Measurements Test

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Units and Measurements Test - 68

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Weekly Quiz Competition

Question 1
1 / -0

Dimension of $$\epsilon_0$$ are

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

B
$$ [M^0 L^{-3} T^3 A^3 ] $$

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

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

Solution

The electrostatic force is given as,
$$F = \dfrac{{{Q_1}{Q_2}}}{{4\pi {\varepsilon _0}{R^2}}}$$

$${\varepsilon _0} = \dfrac{{{Q_1}{Q_2}}}{{4\pi F{R^2}}}$$

Dimensions of $${\varepsilon _0}$$ are,

$$ = \dfrac{{{A^2}{T^2}}}{{ML{T^{ - 2}} \times {L^2}}}$$

$$ = \dfrac{{{A^2}{T^4}}}{{M{L^3}}}$$

$$ = {M^{ - 1}}{L^{ - 3}}{T^4}{A^2}$$
Question 2
1 / -0

Which of the following is dimensionally incorrect ?

A
$$u=v-at$$

B
$$s-ut=\dfrac{1}{2}at^2$$

C
$$u^2=2a(gt-1)$$

D
$$v^2-u^2=2as$$

Solution

A> We know acceleration a=rate of change of velocity
$$a=\dfrac{v-u}{t}$$ where v=final velocity and u = initial velocity and t= time
Thus we can rewrite the above equation as,
$$u=v-at$$
$$[LT^{-1}]=[LT^{-1}]-[LT^{-2}][T]$$
$$[LT^{-1}]=[LT^{-1}]$$ (Dimensionally correct)

B> From equation of motion we get displacement
$$s=ut+\dfrac{1}{2}a{t}^{2}$$
$$s-ut=\dfrac{1}{2}a{t}^{2}$$
$$[L]-[LT^{-1}][T]=[LT^{-2}][T^2]$$
Here also, LHS =RHS

C>The equation $$u^2=2a(gt-1)$$
Substituting the dimensions:
$$[LT^{-1}]^2=[LT^{-2}]([LT^{-2}][T])$$
$$[L^2T^{-2}]=[L^2T^{-3}]$$
LHS is not equal to RHS.

D> From equation of motion,
$${v}^{2}={u}^{2}+2as$$
$${v}^{2}-{u}^{2}=2as$$
$$[LT^{-1}]^2-[LT^{-1}]^2=[LT^{-2}][L]$$
So, LHS=RHS

Whereas the dimensions of the equation are not the same on the LHS and RHS.
Thus option C is incorrect
Question 3
1 / -0

The dimensions of energy per unit volumes are the same as those of :

A
Work

B
Stress

C
Pressure

D
Modulus

Solution

The dimension of energy per unit volume is equal to the dimension of pressure.

Firstly, the dimension of energy is : $$[ML^2T^{−2}]$$

The dimension of volume: $$[L^3]$$

The dimension of pressure : $$[L^{-1}M T^{-2}]$$

Now, $$\dfrac{energy}{volume} = \dfrac{[ML^2T^{−2}]}{[L^3]} = [L^{-1}M T^{-2}]$$
Question 4
1 / -0

If temperature of sun is decreased by $$1\%$$ then the value of solar constant will change by

A
$$2\%$$

B
$$-4\%$$

C
$$-2\%$$

D
$$4\%$$

Solution

Solar constant $$=G=\sigma T^4\left[\cfrac{4\pi R^2}{4\pi O^2}\right]$$
$$\therefore$$ change of G w.r.t $$O$$ on both sides
$$\triangle G=4\sigma T^3 \triangle T \left[\cfrac{4\pi R^2}{4\pi O^2}\right]$$
$$\cfrac{\triangle G}{G}=\cfrac{4\triangle T}{T}=-4\%$$
Question 5
1 / -0

Pressure P varies as $$P=\cfrac { \alpha }{ \beta } exp\left( \cfrac { -\alpha }{ { K }_{ B }\theta } Z \right) $$ where Z denotes distance. $${ K }_{ B }$$ Boltzman's constant.$$\theta $$ temperature and $$\alpha ,\beta$$ are constants.Derive the dimensions of $$\beta $$

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

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

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

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

Solution

$$\dfrac { \alpha Z }{ { K }_{ B }\theta } ={ M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 }{ K }^{ 0 }\\ \alpha =\dfrac { { K }_{ B }\theta }{ Z } =\dfrac { \left[ M{ L }^{ 2 }{ T }^{ -2 }{ K }^{ -1 } \right] \left[ K \right] }{ \left[ L \right] } \\ \alpha =ML{ T }^{ -2 }\\ \left[ P \right] =\dfrac { \left[ \alpha \right] }{ \left[ \beta \right] } \\ \beta =\dfrac { \alpha }{ P } =\dfrac { ML{ T }^{ -2 } }{ M{ L }^{ -1 }{ T }^{ -2 } } ={ M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 }\\ $$
Ans: $${ M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 }$$
Question 6
1 / -0

The dimensional formula of angular velocity $$\left(\omega=\cfrac{v}{r}\right)$$ is

A
$$ML^oT^{-2}$$

B
$$MLT^{-1}$$

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

D
$$ML^oT^{1}$$

Solution

The angular velocity of a body is given by:
$$\omega=\cfrac{V}{r}$$

The dimension of $$V=[LT^{-1}]$$
The dimension of $$r=[L]$$

$$\omega=\cfrac{LT^{-1}}{L}=T^{-1}$$

$$\omega=M^0L^0T^{-1}$$
Question 7
1 / -0

Pressure P varies as $$P=\dfrac{\alpha}{\beta}$$ exp$$(\dfrac{-\alpha}{K_B \theta}Z)$$ where Z denotes distance,$$K_B$$ Boltzman's constant,$$\theta$$ absolute temperature and $$\alpha,\beta$$ are constants.Derive the dimensions of $$\beta$$

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

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

C
$$[M^0 L^2 T^0]$$

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

Solution

$$P=\dfrac{\alpha}{\beta}$$ exp$$(\dfrac{-\alpha}{k_B \theta}Z)$$
Since the terms inside the exponential do not have any dimension,we can find out the dimension of $$\alpha$$
$$\dfrac{\alpha \times L' \times k}{J \times k}$$
$$\alpha$$ has dimension of $$\dfrac{J}{L'}=\dfrac{ML^2 T^{-2}}{L'}=ML T^{-2}$$
For $$\dfrac{\alpha}{\beta}$$ to have dimension of pressure ,i.e $$ML^{-1}T^{-2}$$
$$\beta$$ should have the dimension $$L^2$$ i.e,$$M^0L^2T^0$$
Question 8
1 / -0

If there is a positive error of $$50\%$$ in the measurement of velocity of a body, then the error in the measurement of kinetic energy is :

A
$$25\%$$

B
$$50\%$$

C
$$100\%$$

D
$$125\%$$

Solution

Given that,

$$\dfrac{\Delta v}{v}\times 100$$ is the % error in velocity = 50%

Kinetic energy $$K.E=\dfrac{1}{2}m{{v}^{2}}$$

Error in the kinetic energy

$$\dfrac{\Delta K.E}{K.E}\times 100=m\times 2\dfrac{\Delta v}{v}\times 100$$

$$m$$ is as a constant

Now, percentage error

$$ \dfrac{\Delta K.E}{K.E}\times 100=2\dfrac{\Delta v}{v}\times 100 $$

$$ \dfrac{\Delta K.E}{K.E}\times 100=2\times 50$$%

$$ \dfrac{\Delta K.E}{K.E}\times 100=100$$%

Hence, the error in the measurement of kinetic energy is $$100$$%
Question 9
1 / -0

A student uses a metre scale, measuring upto $$1\ mm$$, to measure the length an breadth of a retangular plate. He finds that length$$=5.7\ cm$$ and breadth $$=3.4\ cm$$. What is the percentage error in the area of the plate?

A
$$3.58\%$$

B
$$4.7\%$$

C
$$5.3\%$$

D
$$6.2\%$$

Solution

$$ \begin{array} \text { Given } \rightarrow \text { Measuring scale with least count } 1 \mathrm{~mm} \\ \text { length }=5.7 \mathrm{~cm}=57 \mathrm{~mm} \\ \text { Breadth }=3.4 \mathrm{~cm}=34 \mathrm{~mm} \end{array} $$
$$ \begin{aligned} \text { Now } \rightarrow \text { Area } &=\text { Length } \times \text { Bredth. } \\ A &=l \times b=57 \times 34=1938 \mathrm{~mm}^{2} \end{aligned} $$
$$ \begin{array}{l} \text { Using error analysis } \rightarrow \\ \text { Take logarithm both sides } \\ \qquad \log A=\log (l)+\log (b) \end{array} $$
$$ \begin{array}{l} \text { difterentiate both sides - } \\ \frac{d A}{A}=\frac{d l}{l}+\frac{d b}{b} \quad \Rightarrow d A=A\left[\frac{d l}{l}+\frac{d b}{b}\right] \end{array} $$
$$ \frac{d A}{A}=1938\left[\frac{1}{57}+\frac{1}{34}\right]=0.047=4.7 \% $$
Question 10
1 / -0

A dimensionless quantity :

A
may have a unit

B
never has a unit

C
always has a unit

D
does not exist

Solution

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Units and Measurements Test - 68

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Units and Measurements Test - 68

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Question 1

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

$$ [M^0 L^{-3} T^3 A^3 ] $$

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

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

Solution

Question 2

$$u=v-at$$

$$s-ut=\dfrac{1}{2}at^2$$

$$u^2=2a(gt-1)$$

$$v^2-u^2=2as$$

Solution

Question 3

Work

Stress

Pressure

Modulus

Solution

Question 4

$$2\%$$

$$-4\%$$

$$-2\%$$

$$4\%$$

Solution

Question 5

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

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

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

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

Solution

Question 6

$$ML^oT^{-2}$$

$$MLT^{-1}$$

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

$$ML^oT^{1}$$

Solution

Question 7

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

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

$$[M^0 L^2 T^0]$$

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

Solution

Question 8

$$25\%$$

$$50\%$$

$$100\%$$

$$125\%$$

Solution

Question 9

$$3.58\%$$

$$4.7\%$$

$$5.3\%$$

$$6.2\%$$

Solution

Question 10

may have a unit

never has a unit

always has a unit

does not exist

Solution

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