Units and Measurements Test

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Units and Measurements Test - 80

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Weekly Quiz Competition

Question 1
1 / -0

Taking frequency $$f$$, velocity $$v$$ and density $$\rho$$ to be the fundamental quantities then the dimensional formula for momentum will be?

A
$$\rho v^4f^{-3}$$

B
$$\rho v^3f^{-1}$$

C
$$\rho v f^2$$

D
$$\rho^2v^2f^2$$

Solution
Question 2
1 / -0

In the equation $$y=a\sin(\omega t+kx)$$, the dimensional formula of $$\omega$$ is?

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

B
$$[M^0LT^{-1}]$$

C
$$[ML^0T^0]$$

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

Solution
Question 3
1 / -0

The equation of alternating current $$I=I_0e^{-t/CR}$$, where $$t$$ is time, $$C$$ is capacitance and $$R$$ is resistance of coil, then the dimensions of $$CR$$ is?

A
$$[MLT^{-1}]$$

B
$$[M^0LT]$$

C
$$[M^0L^0T]$$

D
None of these

Solution

$$CR$$ is known as time constant
$$CR=[M^{-1}L^{-2}T^4I^2][ML^2T^{-3}I^{-2}]$$
$$=[M^0L^0T]$$.
Question 4
1 / -0

The dimensions of time constant are:

A
$$[M^0L^0T^0]$$

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

C
$$[MLT]$$

D
None of these

Solution
Question 5
1 / -0

The physical quantities not having same dimensions are

A
torque and work

B
momentum and Planck's constant

C
stress and young's modulus

D
speed and $$ \left ( \mu_{0}\varepsilon _{0} \right )^{-1/2} $$

Solution

Momentum = $$m$$ * $$v$$ = [$$kgM^1S^-1$$]
Planck`s constant =[$$kgM^2S^-2$$]
Question 6
1 / -0

Which of the following statement is correct about conversion of units, for example $$1 m = 100 cm$$

A
Conversion of units have identical dimensions on each side of the equal sign but not the same units.

B
Conversion of units have identical dimensions on each side of the equal sign but the same units.

C
If a larger unit is used then numerical value of physical quantity is large

D
Due to conversion of units physical quantity to be measured will change.

Solution
Question 7
1 / -0

Out of the following the only pair that does not have identical dimensions is

A
angular momentum and Planck's constant

B
moment of inertia and moment of a force

C
work and torque

D
impulse and momentum

Solution

Moment of inertia and moment of force doesn't have same dimension. Moment of inertia ,I= $$Mr^2$$. And moment of force = r x F.

As impulse is change in momentum so both have same dimensions.

Work = F.S and torque = Fxr so both have same dimensions .

Angular momentum and Planck's constant have same dimensions Angular momentum= Planck's constant/(2π)

Hence, option B is correct.
Question 8
1 / -0

Directions For Questions

The van der Waals' equation of state for some gases can be expressed as $$\left(P + \dfrac{a}{V^2} \right) (V - b) = RT$$ where P is the pressure, V is the molar volume, and T is the absolute temperature of the given sample of gas and a, b and R are constants.

...view full instructions

The dimensional representation of $$ab/RT$$ is

A
$$ML^5 T^{-2}$$

B
$$M^0 L^3 T^0$$

C
$$ML^{-1} T^{-2}$$

D
None of these

Solution
Question 9
1 / -0

If the momentum (P), area (A) and time (T) are taken to be fundamental quantities, then energy has dimensional formula.

A
$$\left [P^{1} A^{-1} T^{1} \right ]$$

B
$$\left [P^{2} A^{-1} T^{1} \right ]$$

C
$$\left [P^{1} A^{-1/2} T^{1} \right ]$$

D
$$\left [P^{1} A^{1/2} T^{-1} \right ]$$

Solution

Let, energy $$E = k P^a A^b t^c$$ ...(i)
where k is a dimensionless constant of proportionality
Equating dimensions on both sides of (i), we get
$$[ML^2T^{-2}] = [MLT^{-1}]^a [M^0L^2 T^0]^b [M^0L^0T]^c$$
$$= [M^a L^{a+2b} T^{-a+c}]$$
Applying the principle of homogeneity of dimensions,
we get
$$a = 1$$ ...(ii)
$$a+2b=2$$ ...(iii)
$$-a+c=-2$$ ...(iv)
On solving eqs. (ii), (iii) and (iv), we get
$$\displaystyle a =1, b = \frac{1}{2}, c = -1$$
$$\therefore [E] = [P^1 A^{1/2} t^{-1}]$$
Question 10
1 / -0

(d) The correct relation is:

A
d= $$ M\times V $$

B
M= $$ d\times V $$

C
V= $$ d\times M $$

D
d= $$M+V$$

Solution

$$Density$$ = $$\dfrac{Mass}{Volume}$$