Units and Measurements Test

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Units and Measurements Test - 64

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Weekly Quiz Competition

Question 1
1 / -0

Write the dimensions of $$a \times b$$ in the relation. $$E = \dfrac{b - x^2}{at}$$, where $$E$$ is the energy, $$x$$ is the displacement, and $$t$$ is the time.

A
$$ML^2T$$

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

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

D
$$MLT^{-2}$$

Solution

$$E=\dfrac { b-{ x }^{ 2 } }{ at } $$ $$E$$ is energy $$x$$ is displacement $$t$$ time
$$E=\left[ { ML }^{ 2 }{ T }^{ -2 } \right] $$
$$x=\left[ L \right] $$
$$t=\left[ T \right] $$
$$b$$ will have unit same $${ x }^{ 2 }$$
$$b=\left[ { L }^{ 2 } \right] $$
$$a=\left[ \dfrac { b-{ x }^{ 2 } }{ Et } \right] =\dfrac { \left[ { L }^{ 2 } \right] }{ \left[ { ML }^{ 2 }{ T }^{ -2 }\times T \right] } =\left[ { M }^{ -1 }T{ L }^{ 0 } \right] $$
$$a\times b=\left[ { L }^{ 2 } \right] \left[ { M }^{ -1 }{ L }^{ 0 }T \right] $$
$$=\left[ { M }^{ -1 }{ L }^{ 2 }T \right] $$
Question 2
1 / -0

The percentage errors in the measurement of mass and speed are $$2\%$$ and $$3\%$$, respectively. How much will be the maximum error in the estimation of KE obtained by measuring mass and speed?

A
$$5\%$$

B
$$1\%$$

C
$$8\%$$

D
$$11\%$$

Solution

We know $$K=\dfrac { 1 }{ 2 } { mV }^{ 2 }$$

$$\dfrac { \Delta K }{ K } =\dfrac { \Delta M }{ M } +\dfrac { 2\Delta V }{ V } $$

$$\dfrac { \Delta K }{ K }=2$$%$$+2\times3$$%$$=(2+6)$$%$$=8$$%
Question 3
1 / -0

The mass of the liquid flowing per second per unit area of cross section of the tube if proportional to $$P^x$$ and $$v^y$$, where $$P$$ is the pressure difference and $$v$$ is the velocity then the relation between $$x$$ and $$y$$ is

A
$$x = y$$

B
$$x = - y$$

C
$$y^2 = x$$

D
$$y = -x^2$$

Solution

Given $$m\alpha { P }^{ x }\times { V }^{ y }$$
$$\left[ { ML }^{ -2 }{ T }^{ 0 } \right] ={ \left[ { ML }^{ -1 }{ T }^{ -2 } \right] }^{ x }{ \left[ { M }^{ 0 }{ LT }^{ 1 } \right] }^{ y }$$
$$x=1$$ } Comparing coefficients
$$-2=-x+y$$ and $$-1=-2x-y$$
$$-2=-4x-2y$$
$$0=\left( -4x-2y \right) -\left( -x+y \right) $$
$$0=-3x-3y$$
$$\Rightarrow$$ $$3x=-3y$$ $$\Rightarrow$$ $$\boxed { x=-y } $$.
Question 4
1 / -0

If frequency $$F$$, velocity $$V$$, and density $$D$$ are considered fundamental units, the dimensional formula for momentum will be

A
$$DVF^2$$

B
$$DV^2F^{-1}$$

C
$$D^2V^2F^2$$

D
$$DV^4F^{-3}$$

Solution

Momentum $$=[F]^a [V]^b [D]^c$$

$$\implies [MLT^{-1}]=[T^{-1}]^a [LT^{-1}]^b[ML^{-3}]^c$$

Comparing both sides,

$$c=1$$

$$b-3c=1$$

$$\therefore b=4$$

Also,

$$a+b=1$$

$$\therefore a=-3$$

Therefore,

Momentum $$=[F]^{-3}[V]^4[D]^1=DV^4F^{-3}$$
Question 5
1 / -0

If the velocity of light $$C$$, the universal gravitational constant $$G$$, and Planck's constant $$h$$ are chosen as fundamental units, the dimension of mass in this system are

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

B
$$h^{-1}C^{-1}G$$

C
$$hCG^{-1}$$

D
$$hCG$$

Solution

Speed of light $$C=\left[ { LT }^{ -1 } \right] $$
gravitational constant $$G=\left[ { M }^{ -1 }{ L }^{ 3 }{ T }^{ -2 } \right] $$
planck's constant $$h={ ML }^{ 2 }{ T }^{ -1 }$$

Let $$M={ C }^{ x }{ G }^{ y }{ h }^{ z }$$
$$\left[ M \right] ={ \left[ { LT }^{ -1 } \right] }^{ x }{ \left[ { M }^{ -1 }{ L }^{ 3 }{ T }^{ -2 } \right] }^{ y }{ \left[ { ML }^{ 2 }{ T }^{ -1 } \right] }^{ z }$$

$$\left[ M \right] =\left[ { M }^{ -y+z }{ L }^{ x+3y+2z }{ T }^{ -x-2y-z } \right] $$

$$-y+z=1,\quad x+3y+2z=0,\quad -x-2y-z=0$$

On solving, we get
$$x=\dfrac { 1 }{ 2 } ;\quad y=\dfrac { -1 }{ 2 } ;\quad z=\dfrac { 1 }{ 2 } $$
So, dimension of $$M=\left[ { C }^{ 1/2 }{ G }^{ -1/2 }{ h }^{ 1/2 } \right] $$
Question 6
1 / -0

The number of significant figures in $$5.69 \times 10^{15}$$kg is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$18$$

Solution

No. of significant figures in $$\underbrace { 5.69 } \times { 10 }^{ 15 }$$ Kg is $$3$$.
Digits that cannot meaning to measurement and measurement resolution.
Question 7
1 / -0

The potential energy of a particle varies with distance $$x$$ as $$U = \dfrac{Ax^{1/2}}{x^2 + B}$$, where A and B are constants. The dimensional formula for A x B is:

A
$$M^1L^{7/2}T^{-2}$$

B
$$M^1L^{11/4}T^{-2}$$

C
$$M^1L^{5/2}T^{-2}$$

D
$$M^1L^{9/2}T^{-2}$$

Solution

Given,  $$U=\dfrac { { Ax }^{ 1/2 } }{ x^{ 2 }+B } $$
$${ x }^{ 2 }+B$$ should have some dimension $$\Rightarrow$$  $${ L }^{ 2 }=dim\left( B \right) $$
$$dim\left( A \right) =dim\left( \dfrac { U\times \left( { x }^{ 2 }+B \right) }{ { x }^{ 1/2 } } \right) =\dfrac { \left[ { ML }^{ 2 }{ T }^{ -2 } \right] \left[ { L }^{ 2 } \right] }{ { L }^{ 1/2 } } $$
  $$={ ML }^{ 7/2 }{ T }^{ -2 }$$
$$dim\left( A\times B \right) =\left[ { ML }^{ 7/2 }{ T }^{ -2 } \right] \left[ { L }^{ 2 } \right] =\left[ { ML }^{ 11/2 }{ T }^{ -2 } \right] $$
Question 8
1 / -0

The time dependence of a physical quantity $$P$$ is given by $$P = P_0e^{-\alpha t^2}$$, where $$\alpha$$ is a constant and t is time. Then constant $$\alpha$$ is / has

A
Dimensionless

B
Dimensions of $$T^{-2}$$

C
Dimensions of P

D
Dimensions of $$T^{2}$$

Solution

$$P={ P }_{ 0 }{ e }^{ -{ \alpha t }^{ 2 } }$$
$${ \alpha t }^{ 2 }$$ must be dimension less in $${ e }^{ -{ \alpha t }^{ 2 } }$$
$$\Rightarrow \quad dim\left( { \alpha t }^{ 2 } \right) ={ M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 }\Rightarrow dim\left( \alpha \right) =\dfrac { { M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 } }{ { T }^{ 2 } } $$
$$\therefore$$ $$dim\left( \alpha \right) ={ M }^{ 0 }{ L }^{ 0 }{ T }^{ -2 }$$
Question 9
1 / -0

The position x of a particle at time t is given by $$x = \dfrac{V_0}{a}(1 - e^{-at})$$, where $$V_0$$ is constant and $$a > 0$$. The dimensions of $$V_0$$ and $$a$$ are

A
$$M^0LT^{-1}$$ and $$T^{-1}$$

B
$$M^0LT^0$$ and $$T^{-1}$$

C
$$M^0LT^{-1}$$ and $$LT^{-2}$$

D
$$M^0LT^{-1}$$ and $$T$$

Solution

$$x=\dfrac { { V }_{ 0 } }{ a } \left( 1-{ e }^{ -at } \right) $$
at should be dimensions $$\Rightarrow Dim\left( a \right) =\dfrac { 1 }{ Dim\left( T \right) } ={ T }^{ -1 }$$
$$x$$ is position $$\Rightarrow dim\left( x \right) ={ M }^{ 0 }L{ T }^{ 0 }=dim\left( \dfrac { { V }_{ 0 } }{ a } \right) $$
$$\therefore$$ $$dim\left( { V }_{ 0 } \right) ={ M }^{ 0 }L{ T }^{ -1 }$$
Question 10
1 / -0

The specific resistance $$\rho$$ of a circular wire of radius r, resistance R, and length l is given by $$\rho = \pi r^2 R/l$$. Given: $$r = 0.24\pm 0.02 cm, R = 30 \pm 1 \Omega,$$ and $$l=4.80\pm 0.01cm.$$ The percentage error in $$\rho$$ is nearly:

A
$$7\%$$

B
$$9\%$$

C
$$13\%$$

D
$$20\%$$

Solution

The specific resistance is given as :
$$\rho =\dfrac{\pi r^{2}R}{l}$$

Differentiating both sides given (in terms of relative error ):-
$$\displaystyle \frac{\Delta P}{\rho }=2\frac{\Delta r}{r}+\frac{\Delta R}{R}+\frac{\Delta l}{l}\rightarrow (2)$$

From given data,
$$\displaystyle r=0.24\pm 0.02 cm \Rightarrow \frac{\Delta r}{r}=\frac{0.02}{0.24}=\frac{1}{12}$$
$$\displaystyle R=30\pm 1\Omega \Rightarrow \frac{\Delta R}{R}=\frac{1}{30}$$
$$\displaystyle l=4.80\pm 0.01cm\Rightarrow \frac{\Delta L}{L}=\frac{0.07}{4.80}=\frac{1}{480}$$

Hence from (1),
$$\displaystyle \frac{\Delta P}{\rho }=2\times \frac{1}{12}+\frac{1}{30}+\frac{1}{480}=20.2\%$$

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Units and Measurements Test - 64

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Units and Measurements Test - 64

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Question 1

$$ML^2T$$

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

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

$$MLT^{-2}$$

Solution

Question 2

$$5\%$$

$$1\%$$

$$8\%$$

$$11\%$$

Solution

Question 3

$$x = y$$

$$x = - y$$

$$y^2 = x$$

$$y = -x^2$$

Solution

Question 4

$$DVF^2$$

$$DV^2F^{-1}$$

$$D^2V^2F^2$$

$$DV^4F^{-3}$$

Solution

Question 5

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

$$h^{-1}C^{-1}G$$

$$hCG^{-1}$$

$$hCG$$

Solution

Question 6

$$1$$

$$2$$

$$3$$

$$18$$

Solution

Question 7

$$M^1L^{7/2}T^{-2}$$

$$M^1L^{11/4}T^{-2}$$

$$M^1L^{5/2}T^{-2}$$

$$M^1L^{9/2}T^{-2}$$

Solution

Question 8

Dimensionless

Dimensions of $$T^{-2}$$

Dimensions of P

Dimensions of $$T^{2}$$

Solution

Question 9

$$M^0LT^{-1}$$ and $$T^{-1}$$

$$M^0LT^0$$ and $$T^{-1}$$

$$M^0LT^{-1}$$ and $$LT^{-2}$$

$$M^0LT^{-1}$$ and $$T$$

Solution

Question 10

$$7\%$$

$$9\%$$

$$13\%$$

$$20\%$$

Solution

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