Units and Measurements Test

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Units and Measurements Test - 70

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Weekly Quiz Competition

Question 1
1 / -0

The dimensions of magnetic flux are?

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

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

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

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

Solution

The force experienced by a moving charge in a magnetic field is given bu $$F=q(\vec{v}\times \vec{B})$$

Here, $$\vec{B}$$ is the magnetic field or magnetic flux density.

Thus, dimensions of magnetic flux, $$\phi=[B]\times [Area]=\dfrac{[F][A]}{[q][v]}$$

We know the dimension of the following as,

$$[F]=[MLT^{-2}]$$

$$[A]=[L^2]$$

$$[q]=[AT]$$

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

Thus,

$$\phi=\dfrac{[MLT^{-2}][L^2]}{[AT][LT^{-1}]}=[ML^2T^{-2}A^{-1}]$$

Option D
Question 2
1 / -0

The dimension formula of torque is

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

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

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

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

Solution
Question 3
1 / -0

A physical quantity $$P$$ is given by the relation, $$P={P}_{0}{e}^{(-\alpha {t}^{2})}$$. If $$t$$ denotes the time, the dimensions of constant $$\alpha$$ are

A
$$[T]$$

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

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

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

Solution

Given,

$$P=P_0e^{(-\alpha t^2)}$$

As we know, both $$P$$ and $$P_0$$ are pressure, it have the same units. Therefore, $$\alpha t^2$$ must be dimensionless for which,

$$\alpha=\dfrac{1}{T^2}=T^{-2}$$

So the dimension of $$\alpha$$ is $$[T^{-2}]$$.
Question 4
1 / -0

if $$L=2.331 cm$$, $$B = 2.1 cm$$, then L+B=

A
$$4.431cm$$

B
$$4.43cm$$

C
$$4.5cm$$

D
$$4.2cm$$

Solution

Step 1
$$L = 2.331 cm = 2.33$$ (corrected to two decimal places) and $$B = 2.1 cm$$
Step 2
$$L + B = 2.33 + 2.1 = 4.43cm. = 4.4cm$$
(by the rule of addition the sum is expressed in minimum decimal places of terms in addition).
Question 5
1 / -0

A balloon of mass $$1$$ g has $$100$$ g of water in it. If it is completely immersed in water, its mass will be

A
$$0.5$$ g

B
$$1.2$$ g

C
$$1$$ g

D
$$1.5$$ g
Question 6
1 / -0

If mass (M), velocity (V) and time (T) are taken as fundamental units, then the dimensions of force (F) are

A
$$[MVT]$$

B
$$[MVT^{-1}]$$

C
$$[M^2VT]$$

D
$$[M^{-1}V^{-1}T]$$

Solution
Question 7
1 / -0

While measuring $$g$$ using simple pendulum method $$\%$$ error in measurement of $$\ell$$ is $$\pm\ 2\%$$ and in measurement time is $$\pm\ 1\%$$ then relative error in $$g$$ is:

A
$$\pm 0.04$$

B
$$0.00$$

C
$$\pm 0.03$$

D
$$\pm 0.01$$

Solution

$$ \begin{aligned} \text { Given }-& * \% \text { error } \operatorname{ in} \ell=2 \% \\ & * \% \text { errole in time } T=1 \% \end{aligned} $$
$$ \begin{array}{l} \text { we Know that - } \\ \qquad T=2 \pi \cdot \sqrt{\frac{\ell}{g}} \\ \Rightarrow \text { on rearanging the formula } \\ \qquad g=4 \pi^{2} \cdot \frac{\ell}{T^{2}} \end{array} $$
$$ \begin{array}{l} \text { Taking logarithm in above eq- } \\ \log (g)=\log (l)-2 \cdot \log (T)+\log \left(4 \pi^{2}\right) \end{array} $$
$$ \begin{array}{l} \text { Ditterentiating both Sides - } \\ \qquad \frac{d g}{g}=\frac{d l}{l}-2 \cdot \frac{d T}{T} \end{array} $$
$$ \frac{d g}{g}=200+2 \cdot(1 \%)=4 \%=\frac{4}{100}=0.04 $$
$$\begin{array} \Rightarrow \text { Relative error in } g=\pm 0.04 \end{array}$$
Question 8
1 / -0

Find the dimension of $$\dfrac {G}{4\pi \epsilon_{0}}$$, where $$G$$ is universal gravitation constant and $$\epsilon_{0}$$ is permittivity
of free space.

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

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

C
$$[M^{0}L^{6}T^{-6}A^{-2}]$$

D
$$[M^{2}A^{2}T^{2}]$$

Solution

We know that ,
Gravitational force, $$F_g= \dfrac{Gm_1m_2}{r^2}$$

Dimension of $$G= \dfrac{F_g r^2}{m_1 m_2} = \dfrac{MLT^{-2}\times L^2}{M^2}=M^{-1}L^3T^{-2}$$

Also
Elecrostatic force , $$F_e= \dfrac{q_1 q_2}{4\pi \epsilon_0 r^2}$$

Dimension of $$\dfrac1{4\pi \epsilon_0 }= \dfrac{F_er^2}{q_1 q_2}=\dfrac{MLT^{-2}\times L^2}{AT\times AT}= ML^3T^{-4}A^2$$

So, Dimension of $$\dfrac G{4\pi \epsilon_0 }= ML^3T^{-4}A^2 \times M^{-1}L^3T^{-2}= M^0 L^6T^{-6}A^{-2}$$
Question 9
1 / -0

The period of osciflation of a simple pendulum in the recorded as $$2.63 \,sec, 2.42 \,sec, 2.42 \,sec, 2.71 \,sec$$ and $$2.80 \,sec$$ respectively. The average absolute error is

A
$$0.1 \,sec$$

B
$$0.14 \,sec$$

C
$$0.01 \,sec$$

D
$$1.0 \,sec$$

Solution

Average value $$=\dfrac{2.63+2.42+2.42+2.71+2.80}{5}$$
$$=2.6$$ sec
Now for
Average absolute error
$$|\Delta T_1|=|2.63-2.42|=0.21$$
$$|\Delta T_2|=2.42-2.42|=0.00$$
$$|\Delta T_3|=|2.42-2.71|=0.29$$
$$|\Delta T_4|=|2.71-2.80|=0.09$$
Mean absolute error
$$\Delta T=\dfrac{|\Delta T_1|+|\Delta T_2|+|\Delta T_3|+|\Delta T_4|}{4}$$
$$=\dfrac{0.59}{4}$$
$$=0.1475$$ sec.
Question 10
1 / -0

$$X = 3YZ^{2}$$ find dimension of $$Y$$ in $$(MKSA)$$ system, if $$X$$ and $$Z$$ are the dimension of capacity and magnetic field respectively.

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

B
$$ML^{-2}$$

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

D
$$M^{-3}L^{-2}T^{8}A^{4}$$

Solution

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Units and Measurements Test - 70

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Units and Measurements Test - 70

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Question 1

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

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

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

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

Solution

Question 2

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

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

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

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

Solution

Question 3

$$[T]$$

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

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

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

Solution

Question 4

$$4.431cm$$

$$4.43cm$$

$$4.5cm$$

$$4.2cm$$

Solution

Question 5

$$0.5$$ g

$$1.2$$ g

$$1$$ g

$$1.5$$ g

Question 6

$$[MVT]$$

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

$$[M^2VT]$$

$$[M^{-1}V^{-1}T]$$

Solution

Question 7

$$\pm 0.04$$

$$0.00$$

$$\pm 0.03$$

$$\pm 0.01$$

Solution

Question 8

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

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

$$[M^{0}L^{6}T^{-6}A^{-2}]$$

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

Solution

Question 9

$$0.1 \,sec$$

$$0.14 \,sec$$

$$0.01 \,sec$$

$$1.0 \,sec$$

Solution

Question 10

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

$$ML^{-2}$$

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

$$M^{-3}L^{-2}T^{8}A^{4}$$

Solution

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