Units and Measurements Test

Result

Units and Measurements Test - 87

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Weekly Quiz Competition

Question 1
1 / -0

Dimensions of magnetic flux density is-

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

B
$$ M^1L^0T^{-2}A^{-1} $$

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

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

Solution
Question 2
1 / -0

The figure shows the position-time (x-t) graph of one-dimenstional motion of a body of mass $$0.4kg$$. The magnitude of each impulse is

A
$$0.2Ns$$

B
$$0.4Ns$$

C
$$0.8Ns$$

D
$$1.6Ns$$

Solution

$$\text { impulse }=\Delta p=m \Delta v$$

$$u=\dfrac{2-0}{2-0}=1$$

$$v=\dfrac{2-4}{2-0}=-1$$

$$\Delta v=1-(-1)$$$$=2$$

$$\begin{aligned}\Delta p &=0.4 \times 2\\&=0.8\mathrm{NS}\end{aligned}$$
Question 3
1 / -0

The dimensions of the quantity $$ \dfrac {L}{RCV} $$ are-

A
$$ M^0L^0T^1A^1 $$

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

C
$$ M^0L^0T^0A^1 $$

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

Solution
Question 4
1 / -0

The dimensions of emissive power are:

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

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

C
$$ [M^1L^0T^{-3} ] $$

D
$$ [M^1L^0T6{-2} ] $$

Solution
Question 5
1 / -0

The dimensional formula
for the physical quantity $$\cfrac {B \mu_0 \epsilon_0}{E}$$ is : (E=intensity
of electric field, B-magnetic induction and symbols have their usual meaning)

A
$$L^oMT $$

B
$$L^1MT^{-1}$$

C
$$L^{-3}MT^3$$

D
$$L^{\cfrac {1}{2}}MT^{-\cfrac {1}{2}}$$
Question 6
1 / -0

The diameter and height of a cylinder are measured by a meter scale to be $$12.6\pm 0.1$$ cm and $$34.2\pm 0.1$$ cm ,respectively. What will be the value of its volume in appropriate significant figures?

A
$$4264\pm 81{ cm }^{ 3 }$$

B
$$4260\pm 80{ cm }^{ 3 }$$

C
$$4264.4\pm 81.0{ cm }^{ 3 }$$

D
$$4300\pm 80{ cm }^{ 3 }$$

Solution

$$\begin{aligned}\text { Given } \rightarrow & \text{ diameter }(d)=12.6 \pm 0.1 \mathrm{~cm} \\& \text { height }(h)=34.2 \pm 0.1 \mathrm{~cm}\end{aligned}$$

$$\text { Volume }=\pi \dfrac{d^{2}}{4} h=\dfrac{22}{7}\times \dfrac{12.6 \times 12.6 \times 34.2}{4}=4260\mathrm{~cm}^{3}$$

$$\begin{array}{l}\text { Taking log both side }\\\log V=\log \left(\pi \frac{d^{2}}{4}h\right)\\\log V=\log \pi+2 \log d-\log4+\operatorname{logh} \\\text { Differentiate both side } \\\frac{\delta V}{V}=0+2 \frac{\delta d}{d}-0+\frac{\delta h}{h}\end{array}$$
$$\begin{aligned}\delta V &=\left(\frac{2 \times 0.1}{12.6}+\frac{0.1}{34.2}\right) V \\&=\left(\frac{0.2}{12.6}+\frac{0.1}{34.2}\right){4260} \\&=8\mathrm{~cm}^{3} \\\therefore \text { Volume } &=4260 \pm 80 \mathrm{~cm}^{3}\end{aligned}$$
Question 7
1 / -0

Let $$[{\varepsilon _o}]$$ denotes the dimensional formula of the permittivity of the vacuum. If $$M$$ = mass, $$L$$ = Length, $$T$$ = time and $$A$$ = electric current, then:

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

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

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

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

Solution
Question 8
1 / -0

The moon's distance from the earth is 360000 km and its diameter subtends an angle of 42' at the eye of the observer. The diameter of the moon in kilometers is :

A
4400

B
1000

C
3600

D
8800

Solution
Question 9
1 / -0

The periodic time $$(T)$$ of a simple pendulum of length $$(L)$$ is given by $$T = 2\pi \sqrt {\dfrac {L}{g}}$$. What is the dimensional formula of $$T = \sqrt {\dfrac {g}{L}}$$?

A
$$[M^{0}L^{0}T^{1}]$$

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

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

D
$$[M^{0}L^{-1}T^{1}]$$
Question 10
1 / -0

The dimensional formula for $$ \frac{1}{2} \varepsilon_{0} \mathrm{E}^{2} $$ is identical to that of

A
$$ \frac{\mathrm{B}^{2}}{2 \mu_{0}} $$

B
$$ \frac{1}{2} B^{2} \mu_{0} $$

C
$$ \frac{\mu_{0}^{2}}{2 B} $$

D
$$ \frac{1}{2} B \mu_{0}^{2} $$

Solution