Units and Dimensions for JEE and NEET | Practice Test

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10 Challenging MCQs on Dimensional Analysis for JEE Mains

Attempt the following questions on finding dimensions. Each correct answer awards +4 marks, and each incorrect answer deducts -1 mark. Select an option and click "Check Answer" to see the result and update your score. Click "Show Explanation" to understand the solution in detail.

1. The dimensions of acceleration are:

a) \([ \text{M L T}^{-2} ]\) b) \([ \text{L T}^{-1} ]\) c) \([ \text{L T}^{-2} ]\) d) \([ \text{M L}^{-1} \text{T}^{-2} ]\)

Correct answer: c) \([ \text{L T}^{-2} ]\)

Explanation:

Acceleration is how quickly velocity changes over time. Let’s find its dimensions step by step.

- **Formula**: Acceleration \( a = \frac{\Delta v}{\Delta t} \) (change in velocity divided by time).

- **Velocity**: Velocity is distance per unit time. So, its dimensions are:

\( v = \frac{\text{length}}{\text{time}} = \frac{\text{L}}{\text{T}} = [ \text{L T}^{-1} ] \).

- **Time**: Time has dimensions \( [ \text{T} ] \).

- **Acceleration**: Now, substitute into the formula:

\( a = \frac{v}{t} = \frac{[ \text{L T}^{-1} ]}{[ \text{T} ]} \).

- **Simplify**: When dividing, subtract the exponents of T:

\( [ \text{L T}^{-1} ] \div [ \text{T} ] = [ \text{L T}^{-1 - 1} ] = [ \text{L T}^{-2} ] \).

So, the dimensions of acceleration are \( [ \text{L T}^{-2} ] \), which matches option c).

2. The dimensions of momentum are:

a) \([ \text{M L T}^{-1} ]\) b) \([ \text{M L T}^{-2} ]\) c) \([ \text{M L}^2 \text{T}^{-2} ]\) d) \([ \text{M L}^{-1} \text{T}^{-1} ]\)

Correct answer: a) \([ \text{M L T}^{-1} ]\)

Explanation:

Momentum is the product of mass and velocity. Let’s calculate its dimensions.

- **Formula**: Momentum \( p = m v \) (mass times velocity).

- **Mass**: Mass has dimensions \( [ \text{M} ] \).

- **Velocity**: Velocity is length per time, so:

\( v = \frac{\text{L}}{\text{T}} = [ \text{L T}^{-1} ] \).

- **Momentum**: Multiply the dimensions:

\( p = m v = [ \text{M} ] \times [ \text{L T}^{-1} ] \).

- **Combine**: There’s no simplification needed here, just multiply:

\( [ \text{M} ] \times [ \text{L T}^{-1} ] = [ \text{M L T}^{-1} ] \).

So, the dimensions of momentum are \( [ \text{M L T}^{-1} ] \), which is option a).

3. Is the equation \( v = u + a t^2 \) dimensionally consistent? (where \( v \) is velocity, \( u \) is initial velocity, \( a \) is acceleration, \( t \) is time)

a) Yes b) No c) Only if \( a \) is in m/s\(^2\) d) Only if \( t \) is in seconds

Correct answer: b) No

Explanation:

For an equation to be dimensionally consistent, all terms must have the same dimensions. Let’s check each term in \( v = u + a t^2 \).

- **Left side (\( v \))**: Velocity has dimensions:

\( v = [ \text{L T}^{-1} ] \).

- **Right side, first term (\( u \))**: Initial velocity also has dimensions:

\( u = [ \text{L T}^{-1} ] \).

- **Right side, second term (\( a t^2 \))**: Let’s calculate this:

- Acceleration \( a = [ \text{L T}^{-2} ] \) (from Question 1).

- Time \( t = [ \text{T} ] \), so \( t^2 = [ \text{T} ]^2 = [ \text{T}^2 ] \).

- Multiply: \( a t^2 = [ \text{L T}^{-2} ] \times [ \text{T}^2 ] \).

- Simplify: \( [ \text{L T}^{-2 + 2} ] = [ \text{L T}^0 ] = [ \text{L} ] \).

- **Compare**: The equation becomes:

\( [ \text{L T}^{-1} ] = [ \text{L T}^{-1} ] + [ \text{L} ] \).

Velocity (\( [ \text{L T}^{-1} ] \)) and length (\( [ \text{L} ] \)) are not the same, so the equation is not dimensionally consistent. The answer is b) No.

4. The dimensions of power are:

a) \([ \text{M L T}^{-2} ]\) b) \([ \text{M L}^2 \text{T}^{-2} ]\) c) \([ \text{M L}^2 \text{T}^{-3} ]\) d) \([ \text{M L T}^{-3} ]\)

Correct answer: c) \([ \text{M L}^2 \text{T}^{-3} ]\)

Explanation:

Power is the rate of doing work. Let’s find its dimensions.

- **Formula**: Power \( P = \frac{W}{t} \) (work divided by time).

- **Work**: Work is force times distance.

- Force \( F = m a = [ \text{M} ] \times [ \text{L T}^{-2} ] = [ \text{M L T}^{-2} ] \).

- Distance \( d = [ \text{L} ] \).

- Work \( W = F \times d = [ \text{M L T}^{-2} ] \times [ \text{L} ] = [ \text{M L}^2 \text{T}^{-2} ] \).

- **Time**: Time has dimensions \( [ \text{T} ] \).

- **Power**: Now, divide:

\( P = \frac{W}{t} = \frac{[ \text{M L}^2 \text{T}^{-2} ]}{[ \text{T} ]} \).

- **Simplify**: \( [ \text{M L}^2 \text{T}^{-2 - 1} ] = [ \text{M L}^2 \text{T}^{-3} ] \).

So, the dimensions of power are \( [ \text{M L}^2 \text{T}^{-3} ] \), which is option c).

5. The dimensions of the gas constant \( R \) in \( PV = nRT \) are:

a) \([ \text{M L}^2 \text{T}^{-2} \text{K}^{-1} \text{mol}^{-1} ]\) b) \([ \text{M L}^2 \text{T}^{-2} \text{K}^{-1} ]\) c) \([ \text{M L}^2 \text{T}^{-3} \text{K}^{-1} \text{mol}^{-1} ]\) d) \([ \text{L}^2 \text{T}^{-2} \text{K}^{-1} \text{mol}^{-1} ]\)

Correct answer: a) \([ \text{M L}^2 \text{T}^{-2} \text{K}^{-1} \text{mol}^{-1} ]\)

Explanation:

Let’s find the dimensions of \( R \) using the ideal gas law \( PV = nRT \).

- **Rearrange**: \( R = \frac{PV}{nT} \).

- **Pressure (\( P \))**: Pressure is force per unit area.

- Force \( F = [ \text{M L T}^{-2} ] \).

- Area \( A = [ \text{L}^2 ] \).

- \( P = \frac{F}{A} = \frac{[ \text{M L T}^{-2} ]}{[ \text{L}^2 ]} = [ \text{M L}^{-1} \text{T}^{-2} ] \).

- **Volume (\( V \))**: Volume is \( [ \text{L}^3 ] \).

- **\( PV \)**: Multiply:

\( PV = [ \text{M L}^{-1} \text{T}^{-2} ] \times [ \text{L}^3 ] = [ \text{M L}^{2} \text{T}^{-2} ] \).

- **\( n \)**: Number of moles, dimensionless in this context, but we include \( [ \text{mol} ] \) in the final units.

- **\( T \)**: Temperature has dimensions \( [ \text{K} ] \).

- **\( R \)**: Now, divide:

\( R = \frac{PV}{nT} = \frac{[ \text{M L}^2 \text{T}^{-2} ]}{[ \text{mol} ] \times [ \text{K} ]} = [ \text{M L}^2 \text{T}^{-2} \text{K}^{-1} \text{mol}^{-1} ] \).

So, the dimensions are \( [ \text{M L}^2 \text{T}^{-2} \text{K}^{-1} \text{mol}^{-1} ] \), which is option a).

6. From \( F = B I L \sin\theta \), the dimensions of \( B \) (magnetic field) are:

a) \([ \text{M L T}^{-2} \text{A}^{-1} ]\) b) \([ \text{M T}^{-2} \text{A}^{-1} ]\) c) \([ \text{M T}^{-1} \text{A}^{-1} ]\) d) \([ \text{M L}^2 \text{T}^{-2} \text{A}^{-1} ]\)

Correct answer: b) \([ \text{M T}^{-2} \text{A}^{-1} ]\)

Explanation:

Let’s find the dimensions of \( B \) using \( F = B I L \sin\theta \).

- **Rearrange**: \( B = \frac{F}{I L \sin\theta} \).

- **Force (\( F \))**: \( F = [ \text{M L T}^{-2} ] \).

- **Current (\( I \))**: Current has dimensions \( [ \text{A} ] \).

- **Length (\( L \))**: \( L = [ \text{L} ] \).

- **\( \sin\theta \)**: This is an angle, so it’s dimensionless.

- **Denominator (\( I L \sin\theta \))**: \( [ \text{A} ] \times [ \text{L} ] = [ \text{A L} ] \) (since \( \sin\theta \) has no dimensions).

- **\( B \)**: Divide:

\( B = \frac{[ \text{M L T}^{-2} ]}{[ \text{A L} ]} = [ \text{M L}^{1-1} \text{T}^{-2} \text{A}^{-1} ] = [ \text{M T}^{-2} \text{A}^{-1} ] \).

So, the dimensions of \( B \) are \( [ \text{M T}^{-2} \text{A}^{-1} ] \), which is option b).

7. From Coulomb's law \( F = \frac{1}{4\pi \varepsilon_0} \frac{q_1 q_2}{r^2} \), the dimensions of \( \varepsilon_0 \) (permittivity of free space) are:

a) \([ \text{M}^{-1} \text{L}^{-3} \text{T}^2 \text{A}^2 ]\) b) \([ \text{M}^{-1} \text{L}^{-3} \text{T}^4 \text{A}^2 ]\) c) \([ \text{M L}^3 \text{T}^{-4} \text{A}^{-2} ]\) d) \([ \text{M L}^3 \text{T}^{-2} \text{A}^{-2} ]\)

Correct answer: b) \([ \text{M}^{-1} \text{L}^{-3} \text{T}^4 \text{A}^2 ]\)

Explanation:

Let’s find the dimensions of \( \varepsilon_0 \) using Coulomb’s law.

- **Formula**: \( F = \frac{1}{4\pi \varepsilon_0} \frac{q_1 q_2}{r^2} \).

- **Rearrange**: \( \varepsilon_0 = \frac{q_1 q_2}{4\pi F r^2} \) (4π is dimensionless).

- **Charge (\( q_1, q_2 \))**: Charge is current times time, so \( q = I t = [ \text{A} ] \times [ \text{T} ] = [ \text{A T} ] \).

- **\( q_1 q_2 \)**: \( [ \text{A T} ] \times [ \text{A T} ] = [ \text{A}^2 \text{T}^2 ] \).

- **Force (\( F \))**: \( F = [ \text{M L T}^{-2} ] \).

- **Distance (\( r \))**: \( r = [ \text{L} ] \), so \( r^2 = [ \text{L}^2 ] \).

- **Denominator (\( F r^2 \))**: \( [ \text{M L T}^{-2} ] \times [ \text{L}^2 ] = [ \text{M L}^3 \text{T}^{-2} ] \).

- **\( \varepsilon_0 \)**: Divide:

\( \varepsilon_0 = \frac{[ \text{A}^2 \text{T}^2 ]}{[ \text{M L}^3 \text{T}^{-2} ]} = [ \text{M}^{-1} \text{L}^{-3} \text{T}^{2 - (-2)} \text{A}^2 ] = [ \text{M}^{-1} \text{L}^{-3} \text{T}^4 \text{A}^2 ] \).

So, the dimensions are \( [ \text{M}^{-1} \text{L}^{-3} \text{T}^4 \text{A}^2 ] \), which is option b).

8. From the photoelectric effect equation \( E = h \nu - \phi \), the dimensions of \( h \) (Planck's constant) are:

a) \([ \text{M L}^2 \text{T}^{-2} ]\) b) \([ \text{M L}^2 \text{T}^{-1} ]\) c) \([ \text{M L T}^{-2} ]\) d) \([ \text{M L}^2 \text{T}^{-3} ]\)

Correct answer: b) \([ \text{M L}^2 \text{T}^{-1} ]\)

Explanation:

Let’s find the dimensions of \( h \) using the photoelectric equation.

- **Formula**: \( E = h \nu - \phi \).

- **Energy (\( E \), \( \phi \))**: Energy is work, so:

\( E = [ \text{M L}^2 \text{T}^{-2} ] \) (from Question 4’s work calculation).

- **Frequency (\( \nu \))**: Frequency is 1/time, so \( \nu = [ \text{T}^{-1} ] \).

- **\( h \nu \)**: This term must match energy’s dimensions:

\( h \nu = [ \text{M L}^2 \text{T}^{-2} ] \).

- **Solve for \( h \)**: \( h = \frac{[ \text{M L}^2 \text{T}^{-2} ]}{\nu} = \frac{[ \text{M L}^2 \text{T}^{-2} ]}{[ \text{T}^{-1} ]} \).

- **Simplify**: \( [ \text{M L}^2 \text{T}^{-2 - (-1)} ] = [ \text{M L}^2 \text{T}^{-1} ] \).

So, the dimensions of \( h \) are \( [ \text{M L}^2 \text{T}^{-1} ] \), which is option b).

9. In the relation \( v = k a^m t^n \), where \( v \) is velocity, \( a \) is acceleration, \( t \) is time, and \( k \) is dimensionless, the values of \( m \) and \( n \) are:

a) \( m=1, n=2 \) b) \( m=2, n=1 \) c) \( m=1, n=1 \) d) \( m=0, n=1 \)

Correct answer: c) \( m=1, n=1 \)

Explanation:

The equation must be dimensionally consistent. Let’s match both sides.

- **Left side (\( v \))**: Velocity \( v = [ \text{L T}^{-1} ] \).

- **Right side (\( k a^m t^n \))**: \( k \) is dimensionless, so:

- Acceleration \( a = [ \text{L T}^{-2} ] \).

- Time \( t = [ \text{T} ] \).

- \( a^m = [ \text{L T}^{-2} ]^m = [ \text{L}^m \text{T}^{-2m} ] \).

- \( t^n = [ \text{T} ]^n = [ \text{T}^n ] \).

- Total: \( a^m t^n = [ \text{L}^m \text{T}^{-2m} ] \times [ \text{T}^n ] = [ \text{L}^m \text{T}^{-2m + n} ] \).

- **Equate**: \( [ \text{L T}^{-1} ] = [ \text{L}^m \text{T}^{-2m + n} ] \).

- **Exponents**:

- Length: \( m = 1 \).

- Time: \( -2m + n = -1 \).

- **Solve**: Substitute \( m = 1 \):

\( -2(1) + n = -1 \Rightarrow -2 + n = -1 \Rightarrow n = 1 \).

So, \( m = 1, n = 1 \), which is option c).

10. From the Stefan-Boltzmann law \( P = \sigma A T^4 \), where \( P \) is power, \( A \) is area, \( T \) is temperature, the dimensions of \( \sigma \) (Stefan-Boltzmann constant) are:

a) \([ \text{M L}^2 \text{T}^{-3} \text{K}^{-4} ]\) b) \([ \text{M T}^{-3} \text{K}^{-4} ]\) c) \([ \text{M L T}^{-3} \text{K}^{-4} ]\) d) \([ \text{L}^2 \text{T}^{-3} \text{K}^{-4} ]\)

Correct answer: b) \([ \text{M T}^{-3} \text{K}^{-4} ]\)

Explanation:

Let’s find the dimensions of \( \sigma \) using the Stefan-Boltzmann law.

- **Formula**: \( P = \sigma A T^4 \).

- **Rearrange**: \( \sigma = \frac{P}{A T^4} \).

- **Power (\( P \))**: \( P = [ \text{M L}^2 \text{T}^{-3} ] \) (from Question 4).

- **Area (\( A \))**: \( A = [ \text{L}^2 ] \).

- **Temperature (\( T \))**: \( T = [ \text{K} ] \), so \( T^4 = [ \text{K}^4 ] \).

- **Denominator (\( A T^4 \))**: \( [ \text{L}^2 ] \times [ \text{K}^4 ] = [ \text{L}^2 \text{K}^4 ] \).

- **\( \sigma \)**: Divide:

\( \sigma = \frac{[ \text{M L}^2 \text{T}^{-3} ]}{[ \text{L}^2 \text{K}^4 ]} = [ \text{M L}^{2-2} \text{T}^{-3} \text{K}^{-4} ] = [ \text{M T}^{-3} \text{K}^{-4} ] \).

So, the dimensions are \( [ \text{M T}^{-3} \text{K}^{-4} ] \), which is option b).

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