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The velocity of a particle (v) at an instant t is given by v = at + bt2. The dimension of b is the
- a)[L]
- b)[LT-1]
- c)[LT-2]
- d)[LT-3]
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
The velocity of a particle (v) at an instant t is given by v = at + bt...
v = at + bt2
[v] = [bt2] or [LT−1] = [bT2] or [b] = [LT−3]
[v] = [bt2] or [LT−1] = [bT2] or [b] = [LT−3]
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The velocity of a particle (v) at an instant t is given by v = at + bt...
Explanation:
To determine the dimension of the variable "b" in the given equation v = at + bt^2, we need to analyze the dimensions of both sides of the equation.
The dimension of velocity (v) is given by [L][T]^-1, where [L] represents length and [T] represents time.
Dimension of the left-hand side (LHS):
The dimension of velocity (v) is [L][T]^-1.
Dimension of the right-hand side (RHS):
The equation v = at + bt^2 consists of two terms: at and bt^2.
1. Dimension of the first term (at):
The dimension of acceleration (a) is [L][T]^-2.
The dimension of time (t) is [T].
Therefore, the dimension of the first term (at) is [L].
2. Dimension of the second term (bt^2):
The dimension of time (t) is [T].
Therefore, the dimension of the second term (bt^2) is [L][T]^2.
Combining the dimensions:
Since both terms on the RHS have different dimensions, we cannot directly add them. However, for addition to be possible, the dimensions of both terms must be the same.
Comparing the dimensions of the first term (at) and the second term (bt^2), we can see that they have the same dimension of [L].
Therefore, the dimension of b must be such that the dimension of bt^2 is [L]. To cancel out the [T]^2 term, the dimension of b must be [L][T]^-2.
Hence, the correct dimension of b is [LT]^-2, which can be rearranged to [LT]^-3.
Answer:
The correct answer is option D, [LT]^-3.
To determine the dimension of the variable "b" in the given equation v = at + bt^2, we need to analyze the dimensions of both sides of the equation.
The dimension of velocity (v) is given by [L][T]^-1, where [L] represents length and [T] represents time.
Dimension of the left-hand side (LHS):
The dimension of velocity (v) is [L][T]^-1.
Dimension of the right-hand side (RHS):
The equation v = at + bt^2 consists of two terms: at and bt^2.
1. Dimension of the first term (at):
The dimension of acceleration (a) is [L][T]^-2.
The dimension of time (t) is [T].
Therefore, the dimension of the first term (at) is [L].
2. Dimension of the second term (bt^2):
The dimension of time (t) is [T].
Therefore, the dimension of the second term (bt^2) is [L][T]^2.
Combining the dimensions:
Since both terms on the RHS have different dimensions, we cannot directly add them. However, for addition to be possible, the dimensions of both terms must be the same.
Comparing the dimensions of the first term (at) and the second term (bt^2), we can see that they have the same dimension of [L].
Therefore, the dimension of b must be such that the dimension of bt^2 is [L]. To cancel out the [T]^2 term, the dimension of b must be [L][T]^-2.
Hence, the correct dimension of b is [LT]^-2, which can be rearranged to [LT]^-3.
Answer:
The correct answer is option D, [LT]^-3.
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