Units and Measurements Test

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Units and Measurements Test - 75

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Weekly Quiz Competition

Question 1
1 / -0

The dimension of $$ (\mu _{ 0 }\epsilon _{ 0 })^{-1/2} $$ are:

A
$$ [L^{1/2}T^{1/2}] $$

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

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

D
$$ [LT^{-1}] $$

Solution
Question 2
1 / -0

The electric field in a certain region is given by $$ \overrightarrow E=( \frac {K}{x^3}) \hat i $$. the dimension of K are:-

A
$$ MLT^{-3}A^{-1} $$

B
$$ ML^{-2}T^{-3}A^{-1} $$

C
$$ ML^4T^{-3}A^{-1} $$

D
$$ M^0L^0T^0A^0 $$

Solution

The electric field in the given region is $$\vec E=\dfrac{K}{x^3}\hat{i}$$

The dimension for the electric field is $$[M^1\ L^1\ T^{-3}\ A^{-1}]$$

In order to obtain the dimension of $$K$$, simplify the above expression:
$$K=E\cdot x^3$$

Here $$X$$ denotes the position. So, dimensions of $$x$$ are $$[L]$$.

Simplify the above expression.
$$K=[M^1\ L^1\ T^{-3}\ A^{-1}][L]^3$$

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

Option $$C$$ is correct.
Question 3
1 / -0

A thermometer reads 0$$^{o}$$C as 10 $$^{o}$$C and 100$$^{o}$$C as 90$$^{o}$$C then
Find the correct temperature if the thermometer reads 20$$^{o}$$C

A
$$20^{o}$$C

B
$$12.5^{o}$$C

C
$$25^{o}$$C

D
$$30^{o}$$C

Solution

Temperature on any scale can be easily converted into different scale by the following formula

$$\dfrac{T - T_f}{T_b - T_f} = \text constant$$

where
$$T$$ = Reading of thermometer
$$T_f $$ = Freezing point of water
$$T_b $$ = Boiling Point of water

So here we will equate the reading of two scale namely, thermometer and actual scale.

$$\dfrac{20 - 10}{90 - 10} = \dfrac{x - 0}{100 - 0}$$

$$\dfrac{10}{80} = \dfrac{x}{100}$$

$$x = 12.5 ^oC$$
Question 4
1 / -0

A particle is moving with a velocity $$35\ m/s$$ along positive x-axis.It's acceleration is towards negative X-axis with the magnitude $$4m/s^ {2}$$ then the distance covered by the particle in the $$9th$$ second:-

A
$$1m$$

B
$$\dfrac {9}{8}m$$

C
$$\dfrac {5}{4}m$$

D
$$153m$$

Solution

$$\begin{array}{l}\vec{u}=35 \mathrm{~m} /\mathrm{s}\hat{\imath} \\\vec{a}=4 \mathrm{~m} /\mathrm{s}^{2}\hat{\imath} \\S_{a}=u+\frac{1}{2} a(2 n-1)\end{array}$$

$$\begin{aligned}S_{a} &=u+\frac{1}{2} a(2 \times 9-1) \\&\left.=35+\frac{1}{2} \times(4\right)\times(18-1) \\&=35+\frac{1}{2} \times(-68) \\&=35-34\\&=1\mathrm{~m}\mathrm{~}\mathrm{~}\end{aligned}$$
Question 5
1 / -0

The pair of physical quantities, that have difference dimensions is

A
Planck's constant and angular momentum

B
Energy density and pressure

C
Relative density and plane angle

D
Specific heat capacity and latent heat

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 6
1 / -0

The vector sum of three forces having magnitudes $$ | \overrightarrow F_1 | = 100 N $$, $$ | \overrightarrow F_2 | = 80 N $$ & $$ | \overrightarrow F_3 | = 60 N $$ acting on a particle is zero. the angle between $$ \overrightarrow F_1$$ & $$ \overrightarrow F_2 $$ is nearly:-

A
$$ 53^0 $$

B
$$ 143^0 $$

C
$$ 37^0 $$

D
$$ 127^0 $$

Solution

$$\text { For equilibrium:} \\\begin{array}{c}\left|\vec{F}_{1}+\vec{F}_{2}\right|=\left|\vec{F}_{3}\right| \\\end{array}$$

$$\begin{array}{c}\text { Squaring both side }\\\Rightarrow\left|\vec{F}_{1}+\vec{F}_{2}\right|^{2}=\left|\vec{F}_{3}\right|^{2}\\\Rightarrow\left|\vec{F}_{1}\right|^{2}+\left|\vec{F}_{2}\right|^{2}+2\left|F_{1}\right|\left|F_{2}\right|\cos\theta\\=\left|\vec{F}_{3}\right|^{2} \\\Rightarrow 100^{2}+80^{2}+2\times 100 \times 80\times\cos\theta\\=60^{2}\end{array}$$
$$\begin{aligned}\Rightarrow \cos\theta&=\frac{12800}{16000} \\\Rightarrow \cos \theta&=\frac{4}{5}\end{aligned}$$

$$\theta=37^{\circ}$$
Question 7
1 / -0

Two constant force $$ \overrightarrow F_1 and \overrightarrow F_2 $$ acts on a body.these forces displaces the body from point P(1, -2, 3) to Q (2, 3, 7 ) in 2s starting from rest.force $$\overrightarrow F_1 $$ is of magnitude 9 N and acting along vector $$ ( 2 \hat i - 2 \hat j + \hat k ) $$ . the positions are in meter. find work done by $$ \overrightarrow F_1 $$.

A
$$-12J$$

B
$$+12J$$

C
$$36J$$

D
$$-36J$$

Solution

$$\begin{aligned}\vec{F}_{1}&=k(2\hat{i}-2\hat{\jmath}+\hat{k}) \\\vec{d}=&(2-1) \hat{1}+(3-(-2))\hat{j}\\&+(7-3) \hat{k} \\=&(1) \hat{1}+(5)\hat{j}+(4)\hat{k}\\\vec{W}_{1}=&\vec{F}_{1}\cdot\vec{d}\\&=k(2\hat{i}-2\hat{j}+\hat{k})\cdot(i+5\hat{j}+4\hat{k})......(i)\end{aligned}$$

$$|\vec{F}_1|=9$$
$$g=\sqrt{4 k^{2}+4 k^{2}+k^{2}}$$
$$9=3 k$$
$$k=3...(ii)$$
$$\text{putting
(ii) in (i)
we get}$$
$$\overrightarrow{W_{1}}=-12 \mathrm{~J}$$
Question 8
1 / -0

A force is given by $$F = at + b{t^2}$$, where $$t$$ is time, the dimensions of $$a$$ and $$b$$ are respectively :

A
$$\left[ {ML{T^{ - 4}}} \right]$$ and $$\left[ {ML{T^{ - 1}}} \right]$$

B
$$\left[ {ML{T^{ - 1}}} \right]$$ and $$\left[ {ML{T^0}} \right]$$

C
$$\left[ {ML{T^{ - 3}}} \right]$$ and $$\left[ {ML{T^{ - 4}}} \right]$$

D
$$\left[ {ML{T^{ - 3}}} \right]$$ and $$\left[ {ML{T^0}} \right]$$

Solution

$$\Rightarrow \ $$ Since $$F=at+bt^2---(1)$$
$$\therefore $$ Dimension of $$ at$$ and $$bt^2$$ must be equal to force only.

Hence $$[F]=[M^{1} L^{1} T^{-2}]---(2)$$

$$F$$ from $$(1)$$ and $$(2)$$

$$\Rightarrow \ [at]=a[T] =[F]$$

$$\therefore \ a[T] =[M^{1} L^{1} T^{-2}]$$

$$\therefore \ [a]=[M^{1} L^{1} T^{-3}]$$

and per $$bt^2=b[T^{2}]=[M^{1} L^{1} T^{-2}]$$

$$\therefore \ \boxed {[b]= [M^{1} L^{1} T^{-4}]}$$
Question 9
1 / -0

The radius of a sphere is measured to be $$(4.0 +2.0)$$ cm the maximum percentage error in the measurement of the volume of the sphere is

A
$$10$$ %

B
$$15$$ %

C
$$20 $$%

D
$$25$$ %

Solution

Given,
$$r=(4\pm0.2)cm$$
$$\dfrac{\Delta r}{r}=\dfrac{0.20}{4}=0.05$$
Volume, $$V=\dfrac{4}{3}\pi r^3$$
$$\dfrac{\Delta V}{V}\times 100=3\dfrac{\Delta r}{r}\times 100=3\times 0.05\times 100=15$$%
The correct option is B.
Question 10
1 / -0

If the constant of gravitation $$(G)$$, Planck's constant $$(h)$$ and the velocity of light $$(c)$$ be chosen as fundamental units. The dimension of the radius of gyration is

A
$$h^{-1/2} c^{1/2} G^{1/2}$$

B
$$h^{1/2} c^{-3/2} G^{1/2}$$

C
$$h^{1/2} c^{-3/2} G^{-1/2}$$

D
$$h^{-1/2} c^{-3/2} G^{1/2}$$

Solution

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Re-Attempt Weekly Quiz Competition

Units and Measurements Test - 75

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Units and Measurements Test - 75

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Question 1

$$ [L^{1/2}T^{1/2}] $$

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

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

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

Solution

Question 2

$$ MLT^{-3}A^{-1} $$

$$ ML^{-2}T^{-3}A^{-1} $$

$$ ML^4T^{-3}A^{-1} $$

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

Solution

Question 3

$$20^{o}$$C

$$12.5^{o}$$C

$$25^{o}$$C

$$30^{o}$$C

Solution

Question 4

$$1m$$

$$\dfrac {9}{8}m$$

$$\dfrac {5}{4}m$$

$$153m$$

Solution

Question 5

Planck's constant and angular momentum

Energy density and pressure

Relative density and plane angle

Specific heat capacity and latent heat

Solution

Question 6

$$ 53^0 $$

$$ 143^0 $$

$$ 37^0 $$

$$ 127^0 $$

Solution

Question 7

$$-12J$$

$$+12J$$

$$36J$$

$$-36J$$

Solution

Question 8

$$\left[ {ML{T^{ - 4}}} \right]$$ and $$\left[ {ML{T^{ - 1}}} \right]$$

$$\left[ {ML{T^{ - 1}}} \right]$$ and $$\left[ {ML{T^0}} \right]$$

$$\left[ {ML{T^{ - 3}}} \right]$$ and $$\left[ {ML{T^{ - 4}}} \right]$$

$$\left[ {ML{T^{ - 3}}} \right]$$ and $$\left[ {ML{T^0}} \right]$$

Solution

Question 9

$$10$$ %

$$15$$ %

$$20 $$%

$$25$$ %

Solution

Question 10

$$h^{-1/2} c^{1/2} G^{1/2}$$

$$h^{1/2} c^{-3/2} G^{1/2}$$

$$h^{1/2} c^{-3/2} G^{-1/2}$$

$$h^{-1/2} c^{-3/2} G^{1/2}$$

Solution

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