Question

The linear momentum p of a particle is given as a
Function of time t as p = At2 + Bt + C. The
dimensions of constant B are
1) [M L-1 T-11
2) [M L-11-21
3) [ML T-2]
4)[M LT-1]​

Answer 1

Answer:

we know that p=At=BT=C

p=BT

B=p/t

MLT^-1/T(we know the dimensional formula for linear momentum is MLT^-1)

MLT^-2is the ans

Answer 2

Answer:

The dimensions of constant B are option 3

Explanation:

Given: The linear momentum p of a particle is given as a Function of time t as $p = At^{2} + Bt + C$

To find: The dimensions of constant

Solution:

Given Momentum $P=a+bt^{2}.$

We know force F= $\frac{dt}{dp}$​ =rate of change of momentum.

$== > F=2bt$

or $Force\ ∝ \ time.$

$P=At^{2} + Bt +C$

Dimension of$P = MLT^{2}$ And now dimension of Bt and C must be same as P So dimension of C is $MLT^{2}$ Dimension of B will be $MLT^{-2}$

Given,

$P=At^{2}+Bt+C$, where A,B and C are constants.

Also, linear momentum(P)=MLT^-1

So, $MLT^{-1}$=$At^{2}+Bt+C$,$C=MLT^{-2}$ [since it is a constant]

therefore,$MLT^{-1}=Bt$

                          $=\frac{MLT^{-1}}{T^{1}}=MLT^{-2},$

Dimensions of constant B is $MLT^{-2}$