Question:
The velocity \( v \) of a particle at time \( t \) is given by \( v = at + \frac{b}{t+c} \), where \( a \), \( b \) and \( c \) are constants. The dimension of \( a \), \( b \) and \( c \) are, respectively
The velocity \( v \) of a particle at time \( t \) is given by \( v = at + \frac{b}{t+c} \), where \( a \), \( b \) and \( c \) are constants. The dimension of \( a \), \( b \) and \( c \) are, respectively
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When performing dimensional analysis, always start with terms that are added or subtracted (like \( t + c \)). This immediately gives the dimension of one of the variables (here, \( [c] = T \)) without needing to analyze the rest of the equation first.
Updated On: May 25, 2026
- \( L T^{-2}, L T, L \)
- \( L, L T, T^2 \)
- \( L T^{-2}, L, T \)
- \( L^2, T, L T^2 \)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
The problem gives a velocity equation containing unknown constants \( a \), \( b \), and \( c \). We need to find the dimensions of these constants using dimensional analysis.
Step 2: Key Formula or Approach:
According to the principle of homogeneity of dimensions, the dimensions of all terms on both sides of a physical equation must be identical.
- Only quantities with the same dimensions can be added or subtracted.
- The dimension of velocity \( v \) is \( [v] = L T^{-1} \).
- The dimension of time \( t \) is \( [t] = T \).
Step 3: Detailed Explanation:
1. Finding the dimension of \( c \):
In the denominator of the second term, \( t \) and \( c \) are added. Therefore, \( c \) must have the same dimension as time \( t \):
\[ [c] = [t] = T \]
2. Finding the dimension of \( a \):
The term \( at \) must have the same dimension as velocity \( v \):
\[ [at] = [v] \implies [a][t] = L T^{-1} \]
\[ [a] T = L T^{-1} \implies [a] = L T^{-2} \]
3. Finding the dimension of \( b \):
The entire term \( \frac{b}{t+c} \) must also have the same dimension as velocity \( v \):
\[ \left[\frac{b}{t+c}\right] = [v] \implies \frac{[b]}{T} = L T^{-1} \]
\[ [b] = L T^{-1} \times T = L \]
Thus, the dimensions of \( a \), \( b \), and \( c \) are \( L T^{-2} \), \( L \), and \( T \) respectively.
Step 4: Final Answer:
The correct dimensions are \( L T^{-2}, L, T \), which corresponds to option (C).
The problem gives a velocity equation containing unknown constants \( a \), \( b \), and \( c \). We need to find the dimensions of these constants using dimensional analysis.
Step 2: Key Formula or Approach:
According to the principle of homogeneity of dimensions, the dimensions of all terms on both sides of a physical equation must be identical.
- Only quantities with the same dimensions can be added or subtracted.
- The dimension of velocity \( v \) is \( [v] = L T^{-1} \).
- The dimension of time \( t \) is \( [t] = T \).
Step 3: Detailed Explanation:
1. Finding the dimension of \( c \):
In the denominator of the second term, \( t \) and \( c \) are added. Therefore, \( c \) must have the same dimension as time \( t \):
\[ [c] = [t] = T \]
2. Finding the dimension of \( a \):
The term \( at \) must have the same dimension as velocity \( v \):
\[ [at] = [v] \implies [a][t] = L T^{-1} \]
\[ [a] T = L T^{-1} \implies [a] = L T^{-2} \]
3. Finding the dimension of \( b \):
The entire term \( \frac{b}{t+c} \) must also have the same dimension as velocity \( v \):
\[ \left[\frac{b}{t+c}\right] = [v] \implies \frac{[b]}{T} = L T^{-1} \]
\[ [b] = L T^{-1} \times T = L \]
Thus, the dimensions of \( a \), \( b \), and \( c \) are \( L T^{-2} \), \( L \), and \( T \) respectively.
Step 4: Final Answer:
The correct dimensions are \( L T^{-2}, L, T \), which corresponds to option (C).
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