# Dimensional analysis and its applications MCQ Quiz in తెలుగు - Objective Question with Answer for Dimensional analysis and its applications - ముఫ్త్ [PDF] డౌన్లోడ్ కరెన్

Last updated on Aug 4, 2024

## Latest Dimensional analysis and its applications MCQ Objective Questions

## Top Dimensional analysis and its applications MCQ Objective Questions

#### Dimensional analysis and its applications Question 1:

The position x of a body depends on the time t according to the equation x = at + bt^{2}. Here x is in metre and t in second. The units of a and b are respectively

#### Answer (Detailed Solution Below)

^{-1}; ms

^{-2}

#### Dimensional analysis and its applications Question 1 Detailed Solution

__ CONCEPT__:

- Dimensional formula: It is a compound expression showing how and which of the fundamental quantities are involved in making that physical quantity.
- The
**principle of homogeneity**says that an equation is dimensionally correct only**if each term has the same dimension on both sides of the equation.**

__ CALCULATION__:

Given equation is x = at + bt2

The dimension of [x] will be equal to the dimension of [at] and [bt^{2}]

[x] = [at]

[a] = [x/t]

**$x$ has dimension of length and ****$t$ has dimension of time so unit of $a$ will be ms ^{-1}.**

[x] = [bt2]

[b] = [x/t2]

$x$ has dimension of length and $t$ has dimension of time so unit of b will be **ms-2**.

So the correct answer is option 4.

#### Dimensional analysis and its applications Question 2:

If v = u + at , where v is final velocity and u is initial velocity and t is time in seconds, a is acceleration, then the dimensions of a is

#### Answer (Detailed Solution Below)

^{-2}

#### Dimensional analysis and its applications Question 2 Detailed Solution

__ CONCEPT__:

**Principle of homogeneity of dimensions:**

**According to this principle**, a physical equation**will be dimensionally correct**if the**dimensions of all**

** the terms occurring **on **both sides** of the equation are** the same.**

- This principle is based on the fact that only the
**physical quantities of the same kind can be**

** added, subtracted, or compared.**

- Thus,
**velocity can be added to velocity**but**not to force.**

__EXPLANATION__:

Given that, v = u + at

- From the
**principle of dimensional homogeneity**, the**left-hand side of the equation**dimensionally

**equal to the right-hand** side of the equation.

- The
**dimension**formula of**velocity (v) = [LT**^{-1}]

Hence, [LT^{-1}] = u

- Therefore
**dimension of 'u' is [LT**^{-1}].

For the second term

[LT-1] = aT

=> a = [LT^{-1}] / T = [LT^{-2}]

**Therefore the dimension of a is [LT ^{-2}]**

#### Dimensional analysis and its applications Question 3:

The displacement of a particle is given by x = αt^{2} + βt^{3} where x is in meter and t is the time in second. The units of α and β will be

#### Answer (Detailed Solution Below)

^{-2}, ms

^{-3}

#### Dimensional analysis and its applications Question 3 Detailed Solution

The correct answer is option 1) i.e. ms-2, ms-3

__CONCEPT__**:**

- The dimensional formula is used to express any physical quantity in terms of fundamental quantities.
**Dimensional Analysis:**It is a technique adopted to derive a relationship between various physical quantities using their dimensions and units of measurement. The two basic rules followed in the dimensional analysis are:- Only those quantities that have the
**same dimensions**can be added and subtracted from each other. - Two
**physical quantities**are equal if they have the same dimensions.

- Only those quantities that have the

__EXPLANATION__:

- LHS and RHS of the expression will have the same dimensions.
- The terms αt2 and βt3 are added to each other. Therefore, they both will have the same dimensions which are equal to the dimension of x i.e. meter (m).

On equating,

αt^{2} = m ⇒ α(s^{2}) = m ⇒** α = ms ^{-2}**

βt3 = m ⇒ β(s3) = m ⇒ **β = ms-3**

The units of α and β will be ms-2 and ms-3 respectively.

#### Dimensional analysis and its applications Question 4:

Frequency f of oscillations of a mass m suspended from a spring of force constant k is given by f = cm^{x}k^{y}, where c is a dimensionless constant. The values of x and y are:

#### Answer (Detailed Solution Below)

#### Dimensional analysis and its applications Question 4 Detailed Solution

__CONCEPT:__

**Frequency**

- It is defined as the
**number of cycles per unit of time**. - Its unit is Hz or sec
^{-1}.

Force constant of spring

- It is defined as the force required for unit displacement of the spring.
- Its unit is
**N/m**.

__CALCULATION:__

Given f = cm^{x}k^{y} -----(1)

- The dimension of frequency,

\(\Rightarrow \left [ f \right ]=\left [ M^{0}L^{0}T^{-1} \right ]\) -----(2)

- The dimension of mass,

\(\Rightarrow \left [ m \right ]=\left [ M^{1}L^{0}T^{0} \right ]\) -----(3)

- The dimension of force constant of spring,

\(\Rightarrow \left [ k \right ]=\left [ M^{1}L^{0}T^{-2} \right ]\) -----(4)

- The dimension of c is,

\(\Rightarrow \left [ c \right ]=\left [ M^{0}L^{0}T^{0} \right ]\) -----(5)

By equation 1, equation 2, equation 3, equation 4, and equation 5,

\(\Rightarrow \left [ f \right ]=\left [ c \right ]\left [ m \right ]\left [ k \right ]\)

\(\Rightarrow \left [ M^{0}L^{0}T^{-1} \right ]=\left [ M^{0}L^{0}T^{0} \right ]\left [ M^{1}L^{0}T^{0} \right ]^x\left [ M^{1}L^{0}T^{-2} \right ]^y\)

\(\Rightarrow \left [ M^{0}L^{0}T^{-1} \right ]=\left [ M^{x+y}L^{0}T^{-2y} \right ]\) -----(6)

By equating both the sides of equation 6,

\(\Rightarrow x+y=0\) -----(7)

\(\Rightarrow -2y=-1\)

\(\Rightarrow y=\frac{1}{2}\)

\(\Rightarrow x=-\frac{1}{2}\)

- Hence, option 1 is correct.

#### Dimensional analysis and its applications Question 5:

A physical quantity is given by \(A =\dfrac{mv}{qB}\), where m is the mass, v is the speed, q is the charge, and B is the magnetic field. The dimension of A will be-

#### Answer (Detailed Solution Below)

#### Dimensional analysis and its applications Question 5 Detailed Solution

__CONCEPT__:

- Dimensional formula: It is a compound expression showing which of the fundamental quantities are involved and how are they involved in making that physical quantity.

Quantity | Dimensional formula |

Force (F) | [MLT^{-2}] |

velocity (v) | [LT-1] |

Inertia (I) | [M^{1}L^{2}T^{3}] |

Magnetic Field (B) |
[M^{1} L^{0} T^{-}^{2} A^{-}^{1}]. |

Charge (Q) |
[AT] |

Spring constant (K) | [ML0T-2] |

Length | [L] |

Work (W) | [ML2T-2] |

Any constant (C) | No dimension |

__CALCULATION__:

Given quantity \(A =\dfrac{mv}{qB}\)

Take dimensions of both sides

\([A] =\dfrac{[m][v]}{[q][B]}\)

\([X] =\dfrac{[M][LT^{-1}]}{[AT][MT^{-2}A^{-1}]}\)

\([X] =[L]\)

- So the correct answer is option 1.

#### Dimensional analysis and its applications Question 6:

Possible units for the disintegration constant λ are

#### Answer (Detailed Solution Below)

#### Dimensional analysis and its applications Question 6 Detailed Solution

__CONCEPT:__

Radioactivity:

**Radioactivity**is a process by which the**nucleus of an unstable atom loses energy by emitting radiation**.- Two forces, namely the force of repulsion that is electrostatic and the powerful forces of attraction of the nucleus acts in the nucleus.
- These two forces are considered extremely strong in nature.
- The instability of the nucleus increases as the size of the nucleus increases because the mass of the nucleus becomes a lot to concentrate. That’s the reason why atoms of Plutonium, Uranium are extremely unstable and show radioactivity.

- The
**spontaneous breakdown of an atomic nucleus****of a radioactive substance**causing the**emission of radiation from the nucleus**is known as Radioactive decay.

__EXPLANATION:__

- In radioactive decay, the
**number of remaining nuclei after decay time t**is given as,

⇒ N = N_{o}e^{-λt} -----(1)

Where N_{o} = initial amount of nuclei, and $λ=$ disintegration or decay constant

- We know that the exponential term is always unitless quantity. So the power of the exponential term is also a unitless quantity.

The dimension of t is given as,

⇒ [t] = [M^{0} L^{0} T^{1}] -----(2)

Let the dimension of disintegration constant is,

⇒ [λ] = [Mx Ly Tz] -----(3)

The λt term is a unitless quantity. So,

⇒ [λt] = [M0 L0 T0] -----(4)

by equation 2, equation 3, and equation 4,

⇒ [Mx Ly Tz][M0 L0 T1] = [M0 L0 T0]

⇒ [Mx Ly Tz+1] = [M0 L0 T0] -----(5)

By equation 5,

⇒ x = 0,

⇒ y = 0,

⇒ z + 1 = 0

⇒ z = -1

Therefore the dimension of disintegration constant is,

⇒ [λ] = [M0 L0 T-1] -----(6)

- By equation 6 it is clear that the possible units for the
**disintegration constant λ is 1/day**. - Hence, option 4 is correct.

#### Dimensional analysis and its applications Question 7:

If m is the mass of a particle, and v is the velocity of that particle, find the dimensional formula of mv^{2}.

#### Answer (Detailed Solution Below)

^{2}T

^{-2}]

#### Dimensional analysis and its applications Question 7 Detailed Solution

__CONCEPT__:

- Dimensional formula: It is a compound expression showing which of the fundamental quantities are involved and how are they involved in making that physical quantity.

Quantity | Dimensional formula |

Mass | [M] |

Length | [L] |

Time | [T] |

Force (F) | [MLT-2] |

Electric charge (Q) | [AT] |

Energy (E) | [ML2T-2] |

Any constant (C) | No dimension |

__CALCULATION__:

(1/2)mv2 is the formula of Kinetic Energy.

And the dimensional formula for Energy is [ML2T-2]

So the dimensional formula for mv2 is [ML2T-2].

Hence the correct answer is **option 3.**

__Alternate Method__

Let any quantity X = mv2

[X] = [m][v]^{2} = [M] [LT-1]^{2} = [ML2T-2]

- So the correct answer is option 3.

#### Dimensional analysis and its applications Question 8:

If we validate v = displacement / time using dimensional analysis than number of repeating variables (m) is __________.

#### Answer (Detailed Solution Below)

#### Dimensional analysis and its applications Question 8 Detailed Solution

**Explanation:**

**Buckingham π-Method:**

- The Buckingham's π-theorem states that, if there are n-dimensional variables involved in a phenomenon, which can be completely described by
**m fundamental quantities or dimensions (such as mass, length, time etc.)**and are related by a dimensionally homogenous equation, then the relationship among the n quantities can always be expressed in terms of exactly (n-m) dimensionless and independent π terms.

****Mathematically, If any variable Q1 depends on the independent variables, Q_{2}, Q_{3}, Q_{4}...............Q_{n}; the functional equation may be written as,

Q_{1} = f(Q_{2}, Q_{3}, Q_{4}.................Q_{n})

Which can be transformed into another functional relationship

f_{1}(Q_{1}, Q_{2}, Q_{3},................Q_{n}) = C

where C is a dimensionless constant

In accordance with the π-theorem, a non-dimensional equation can thus be obtained in the following form.

f_{2}(π_{1}, π_{2}, π_{3}.................π_{n-m}) = C_{1}

Where in each dimensionless π-term is formed by combining m variables out of the total n variables with one of the remaining (n-m) variables. **These m variables which appear repeatedly in each of the π terms are consequently called repeating variables and are chosen from among the variables such that they together involve all the m fundamental quantities and they themselves do not form a dimensionless parameter.**

\(Velocity=\frac {Displacement}{Time}\)

\(V=\frac st= \frac LT=LT^{-1}\)

As **repeating variables are always fundamental variables and Fundamental variable in the dimensional formula of velocity is 2. Hence no. of repeated variable here is 2.**

#### Dimensional analysis and its applications Question 9:

If momentum [P], area [A] and time [T] are taken as fundamental quantities, then the dimensional formula for the coefficient of viscosity is

#### Answer (Detailed Solution Below)

^{–1}T

^{0}]

#### Dimensional analysis and its applications Question 9 Detailed Solution

**Concept:**

The coefficient of viscosity (η) in terms of momentum [P], area [A], and time [T].

The coefficient of viscosity (η) is given by,

η = Force / Area ×Velocity gradient

**Momentum [P]: **Momentum is given by

P=M×V,

Where V is velocity.

The dimensional formula of momentum is

[P] = MLT^{−1}

**Area [A]:** Area has the dimensional formula

[A]=L^{2}

**Time [T]: **Time has the dimensional formula

[T]=T

**Calculation:**

The dimensional formula for viscosity is:

[η] = [ML-1T-1]

[η] = [P]a [A]b [T]c

⇒ [ML-1T-1] = [ML1T-1]a [L2]b [T]c

⇒ a = 1, a + 2b = – 1, – a + c = – 1

⇒ a = 1, b = –1, c = 0

⇒ [η] = [P] [A]–1 [T]0

= [PA–1T0]

**∴ The correct option is 1**

#### Dimensional analysis and its applications Question 10:

The equation of the velocity is given as v = ax sin(bωt), where x, ω and t are displacement, angular velocity and time respectively. Find the dimension of a and b.

#### Answer (Detailed Solution Below)

^{0}L

^{0}T

^{-1}] and [M

^{0}L

^{0}T

^{0}]

#### Dimensional analysis and its applications Question 10 Detailed Solution

__CONCEPT__:

**Dimensions:**

**Dimensions**of a physical quantity are the powers to which the fundamental units are raised to obtain one unit of that quantity.

**CALCULATION**__:__

Given v = ax sin(bωt)

- Since the
**trigonometric function is a dimensionless quantity**. So,**bωt**is also a**dimensionless quantity.**

Here, t = time and ω = angular velocity

⇒ [bωt] = [M^{0} L^{0} T^{0}] -----(1)

- The dimension of t is given as,

⇒ [t] = [M^{0} L^{0} T^{1}] -----(2)

- The dimension of ω is given as,

⇒ [ω] = [M^{0} L^{0} T^{-1}] ----(3)

Let the dimension of b is,

⇒ [b] = [M^{x} L^{y} T^{z}] ----(4)

By equation 1, equation 2, equation 3, and equation 4,

⇒ [M^{x} L^{y} T^{z}].[M^{0} L^{0} T^{-1}].[M^{0} L^{0} T^{1}] = [M^{0} L^{0} T^{0}]

⇒ [M^{x} L^{y} T^{z}] = [M^{0} L^{0} T^{0}] ----(5)

By equation 5,

⇒ x = 0

⇒ y = 0

⇒ z = 0

So the dimension of b is,

⇒ [b] = [M^{0} L^{0} T^{0}]

- The dimension of v is given as,

⇒ [v] = [M^{0} L^{1} T^{-1}] ----(6)

- The dimension of x is given as,

⇒ [x] = [M^{0} L^{1} T^{0}] ----(7)

Let the dimension of a is,

⇒ [a] = [M^{p} L^{q} T^{r}] ----(8)

By equation 6, equation 7, and equation 8,

⇒ [M^{0} L^{1} T^{-1}] = [M^{p} L^{q} T^{r}].[M^{0} L^{1} T^{0}]

⇒ [M^{p} L^{q} T^{r}] = [M^{0} L^{0} T^{-1}] ----(9)

By equation 9,

⇒ p = 0

⇒ q = 0

⇒ r = -1

So the dimension of a is,

⇒ [a] = [M^{0} L^{0} T^{-1}]

- Hence, option 2 is correct.