Units and Measurements Test

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Units and Measurements Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

The physical quantity which has the dimensional formula $$\displaystyle \left[ { M }^{ }{ T }^{ -3 } \right] $$ is:

A
Surface tension

B
Density

C
Solar constant

D
Compressibility

Solution

Surface tension, $$S = \dfrac{F}{l}$$
Dimensional formula of force, $$F = [ML T^{-2}]$$
$$\therefore$$ Dimensional formula for surface tension, $$S = \dfrac{MLT^{-2}}{[L]} = [M T^{-2}]$$

Dimensional formula for density, $$\rho = \dfrac{M}{V} = \dfrac{M}{L^3} = [ML^{-3}]$$

Solar constant $$ = \dfrac{\sigma R^2_s T_s^4}{R^2_{se}}$$ where all symbols have their usual meaning

Dimension of Stefan's constant, $$\sigma = [MT^{-3} K^{-4}]$$
$$\therefore$$ Dimensional formula for solar constant $$ = \dfrac{[MT^{-3} K^{-4}] \times L^2 \times K^4}{L^2} = [M T^{-3}]$$

Compressibility, $$k = -\dfrac{\Delta V}{V} \dfrac{1}{\Delta P}$$
$$\therefore$$ Dimensional formula for compressibility, $$ k = \dfrac{[L^3]}{[L^3] \times [ML^{-1} T^{-2}]} = [M^{-1} L T^{2}]$$
Question 2
1 / -0

When a wave traverses a medium the displacement of a particle located at x at a time is given by y = a sin (bt - cx), where a, band are constants of the wave, which of the following is a quantity with dimensions?

A
$$\displaystyle \frac {y}{a}$$

B
$$bt$$

C
$$cx$$

D
$$\displaystyle \frac {b}{c}$$

Solution

Given, y = a sin (bt - cx)
Comparing the given equation with general wave equation
y = a sin ($$\displaystyle \frac {2 \pi t}{T}\, -\, \displaystyle \frac {2 \pi x}{\lambda}$$)
we get $$b\,=\, \displaystyle \frac {2 \pi}{T},c\,= \displaystyle \frac {2 \pi}{\lambda}$$
(a) Dimensions of $$\displaystyle \frac {y}{a}\, =\, \displaystyle \frac {metre}{metre}\, =\, \displaystyle \frac {L}{L}$$
(b) Dimensions of bt $$=\, \displaystyle \frac {2 \pi}{T}\, \cdot\, t\, =\, \displaystyle \frac {T}{T}$$
(c) Dimensions of cx $$=\, \displaystyle \frac {2 \pi}{\lambda}\, \cdot\, x\, =\, \displaystyle \frac {L}{L}$$
(d) Dimensions of $$\displaystyle \frac {b}{c}\, =\, \displaystyle \frac {2 \pi}{T}/\displaystyle \frac {2 \pi}{\lambda}\, =\, \lambda/T\, =\, LT^{-1}$$
Thus, option (d) has dimensions.
Question 3
1 / -0

Ampere-hour is the unit of

A
Quantity of charge

B
Potential

C
Energy

D
Current

Solution

Charge = current $$\times$$ time
Coulomb = ampere $$\times$$ hour
Question 4
1 / -0

If accuracy of weighing machine is better than 1/2 percent, a student record his own mass. Choose the option that explains the recording most accurately.

A
6.43 kg

B
60 kg

C
64.3 kg

D
600kg

E
643 kg

Solution

This question is intended to assess your knowledge of realistic values for physical quantities as well as an understanding of experimental error and significant figures. Only 60 kg and 64.3 kg are likely masses for an adult human. When recording measurements, it is assumed that there is some measurement error in the last digit recorded. For both possible values for the student's weight, 1/2 percent is about 0.3 kg. Therefore, a mass recorded to better than this accuracy will indicate tenths of a kilogram.

Question 5

1 / -0

If energy (E) , force (F) and linear momentum (P) are fundamental quantities, then match the following and give correct answer.

A	B
Physical quantity	Dimensional formula
a) Mass	d) $$\displaystyle { E }^{ 0 }{ F }^{ -1 }{ p }^{ 1 }$$
b) Length	e) $$\displaystyle { E }^{ -1 }{ F }^{ 0 }{ p }^{ 2 }$$
c) Time	f) $$\displaystyle { E }^{ 1 }{ F }^{ -1 }{ p }^{ 0 }$$

a-d, b-e, c-f

a-f, b-e, c-d

a-e, b-f, c-d

a-e, b-d, c-f

Solution

Dimensions of $$E=[ML^2T^{-2}]$$

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

Dimensions of $$P=[MLT^{-1}]$$

Thus $$E^0F^{-1}P^1=[T]$$

$$E^{-1}F^0P^2=[M]$$

$$E^{-1}F^0P^2=[L]$$

Question 6
1 / -0

What is the dimensional formula of Inductance?

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

B
$$[ML^2TA^{-2}]$$

C
$$[ML^2T^{-1}A^{-2}]$$

D
$$[ML^2T^{-2}A^{-1}]$$

Solution

Hint:
Use the formula of inductance in terms of voltage and current and and use there dimensions.

Step 1: Formula used,
The formula of the inductance is written as,

$$L=\dfrac { V }{ \dfrac { dI }{ dt } } $$

step 2: Finding the dimensions,
Substituting the dimensions of voltage and the rate of change of current in the above relation,
$$[L]=[\dfrac { M{ L }^{ 2 }{ T }^{ -3 }{ A }^{ -1 } }{ { A }{ T }^{ -1 } }] =[M{ L }^{ 2 }{ T }^{ -2 }{ A }^{ -2 }]$$

$$\textbf{Hence option A correct}$$
Question 7
1 / -0

If $$x=at+b{ t }^{ 2 }$$ where $$x$$ is measured in $$m$$ and $$t$$ in $$s$$, then the dimension of $$\left( { b }/{ a } \right) $$ is:

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

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

C
$$[T]$$

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

Solution

Given equation is: $$x=at+b{ t }^{ 2 }$$
The standard equation is: $$S=ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 }$$

$$\Rightarrow \left[ a \right] =\left[ u \right] =L{ T }^{ -1 }$$
$$\Rightarrow \left[ b \right] =\left[ a \right] =L{ T }^{ -2 }$$

$$\Rightarrow \left[ \dfrac { b }{ a } \right] =\dfrac { L{ T }^{ -2 } }{ L{ T }^{ -1 } } =[{ T }^{ -1 }]$$
Question 8
1 / -0

The given rectangular shaped object pictured is being measured. Its left end is perfectly lined up with "0 m" on the measuring tape. Which of the following is the best and most precise length of the green, rectangular shaped object that can be obtained, using the measuring tape?

A
2.0 m

B
2.4 m

C
2.37 m

D
2.377 m

E
2.5 m

Solution

Least count of the given scale, $$LC =\dfrac{1}{10} =0.1$$ m
From figure, the end of the green rectangular object is close to $$4th$$ line after $$2\ m$$ mark.
Thus most precise length of the object, $$L = 2.0 + 0.4 =2.4\ m$$
Question 9
1 / -0

The dimensions of $$\displaystyle \frac { a }{ b } $$ in the equation $$\displaystyle P=\frac { a-{ t }^{ 2 } }{ bx } $$ where $$P$$ is pressure, $$x$$ is distance and $$t$$ is time, are:

A
$$\displaystyle \left[ { M }^{ 2 }{ LT }^{ -3 } \right] $$

B
$$\displaystyle \left[ { MT }^{ -2 } \right] $$

C
$$\displaystyle \left[ { LT }^{ -3 } \right] $$

D
$$\displaystyle \left[ M{ L }^{ 3 }{ T }^{ -1 } \right] $$

Solution

The given relation is $$P = \dfrac{a- t^2}{bx}$$
Thus dimension of $$a = [T^2]$$
Dimensions of pressure $$P = [ML^{-1} T^{-2}]$$ and that of $$x = [L]$$
Thus dimensions of $$b = \dfrac{[T^2]}{[L] \times [M L^{-1} T^{-2}]} = [M^{-1} T^4]$$

$$\therefore$$ Dimensions of $$\dfrac{a}{b} = \dfrac{[T^2]}{[M^{-1} T^4]} = [M T^{-2}]$$
Question 10
1 / -0

Students A, B and C measure the diameter of a wire using three different screw gauges of least count 0.01 cm, 0.05 cm and 0.001 cm respectively. Each one makes 10 measurements.The measurements will be more precise for

A
A

B
B

C
C

D
all

Solution

For a screw gauge, the precision in measurement is depends on the least count. The smaller least count will give the more precision. Since student C has measured diameter using screw gauge with smallest least count so the measurement will more precise for student C.

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Units and Measurements Test - 55

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Units and Measurements Test - 55

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Question 1

Surface tension

Density

Solar constant

Compressibility

Solution

Question 2

$$\displaystyle \frac {y}{a}$$

$$bt$$

$$cx$$

$$\displaystyle \frac {b}{c}$$

Solution

Question 3

Quantity of charge

Potential

Energy

Current

Solution

Question 4

6.43 kg

60 kg

64.3 kg

600kg

643 kg

Solution

Question 5

a-d, b-e, c-f

a-f, b-e, c-d

a-e, b-f, c-d

a-e, b-d, c-f

Solution

Question 6

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

$$[ML^2TA^{-2}]$$

$$[ML^2T^{-1}A^{-2}]$$

$$[ML^2T^{-2}A^{-1}]$$

Solution

Question 7

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

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

$$[T]$$

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

Solution

Question 8

2.0 m

2.4 m

2.37 m

2.377 m

2.5 m

Solution

Question 9

$$\displaystyle \left[ { M }^{ 2 }{ LT }^{ -3 } \right] $$

$$\displaystyle \left[ { MT }^{ -2 } \right] $$

$$\displaystyle \left[ { LT }^{ -3 } \right] $$

$$\displaystyle \left[ M{ L }^{ 3 }{ T }^{ -1 } \right] $$

Solution

Question 10

A

B

C

all

Solution

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