Question

JEE ADVANCED PHYSICS QUESTION SEPTEMBER 2020

Answer 1

AnswEr :

Dimensions of the following physical quantities are represented as :

• $\sf Speed \longrightarrow X^{\beta}$
• $\sf Position \longrightarrow X^{\alpha}$
• $\sf Acceleration \longrightarrow X^{p}$
• $\sf Momentum \longrightarrow X^{q}$
• $\sf Force \longrightarrow X^{r}$

Dimensional Formula of :

• $\sf Position \longrightarrow L = X^{\alpha}$
• $\sf Speed \longrightarrow LT^{-1}$
• $\sf Acceleration \longrightarrow LT^{-2}$
• $\sf Momentum \longrightarrow MLT^{-1}$
• $\sf Force \longrightarrow MLT^{-2}$

Consider the dimensions of acceleration :

$\longrightarrow \sf \: X {}^{p} = X {}^{ \alpha }T {}^{ - 2} - - - - - - (1)$

Also,

Speed = Position/Time

$\longrightarrow \sf \: T {}^{ - 1} = \dfrac{X {}^{ \beta }}{ X {}^{ \alpha }}\\ \longrightarrow \sf \: T = X {}^{ \alpha - \beta }- - - - - - (2)$

Putting equation (2) in (1),

$\longrightarrow \sf \: X {}^{p} = { X}^{ \alpha }. {X}^{ - 2( \alpha - \beta ) } \\ \\ \longrightarrow \sf \: X {}^{p} = {X}^{ 2\beta - 2\alpha + \alpha } \\ \\ \longrightarrow \sf \: X {}^{p} = {X}^{ 2\beta - \alpha }$

Since, the bases are equal.

$\longrightarrow \sf \: p = 2 \beta - \alpha \\ \\ \longrightarrow \boxed{ \boxed{ \sf \alpha + p = 2 \beta }}$

Option (A) is correct

Answer 2

Sometimes it is convenient to construct a system of units is that all quantities can be expressed in tern of only one physical quantity. in one such system, dimensions of different quantities are given in terms of quantity X follows : [ position] = [x^α] ; [speed] = [x^β] ; [ acceleration] = [x^p] ; [linear momentum] = [x^q] ; [force ] = [x^r]

Then...

solution : dimension of position = [L] = [x^α]....(1)

dimension of speed = [LT¯¹] = [x^β] ....(2)

dimension of acceleration = [LT¯²] = [x^p] .....(3)

dimension of linear momentum = [MLT¯¹] = [x^q] ....(4)

dimension of force = [MLT¯²] = [x^r] ......(5)

now from equation (1) and (2),

[L]/[LT¯¹] = x^(α-β)

T = x^(α -β)......(6)

from equation (3),

[x^α][x^(α - β)]¯² = x^p

⇒x^(α - 2α + 2β) = x^p

⇒-α + 2β = p

⇒α + p = 2β..........(7)

so option (A) is correct choice..

now from equation (4) and (2),

M × x^β = x^q

⇒M = x^(q - β)

from equation (5),

x^(q - β) × x^α × x^-2(α - β) = x^r

⇒x^{q - β + α - 2α + 2β} = x^r

⇒q - α + β = r

⇒q - 2β + p + β = r [ from equation (7) ]

⇒q + p - β = r

⇒p + q - r = β.....(8) , so option (B) so also correct choice.

Therefore options (A) and (B) are correct choices.