If mass is written as m=kc^{p} G^{-1 / 2} ,h^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the formula for mass: $m = k c^{P} G^{-1/2} h^{1/2}$.We use the dimensional formulas for each constant:$c$ (speed of light) $= [L T^{-1}]$$G

Vedclass · Accepted Answer

$1 / 2$

Vedclass · Answer

(A) Given the formula for mass: $m = k c^{P} G^{-1/2} h^{1/2}$.We use the dimensional formulas for each constant:$c$ (speed of light) $= [L T^{-1}]$$G$ (gravitational constant) $= [M^{-1} L^3 T^{-2}]$$h$ (Planck's constant) $= [M L^2 T^{-1}]$Substituting these dimensions into the equation:$[M^1 L^0 T^0] = [L T^{-1}]^{P} [M^{-1} L^3 T^{-2}]^{-1/2} [M L^2 T^{-1}]^{1/2}$$[M^1 L^0 T^0] = [L^P T^{-P}] [M^{1/2} L^{-3/2} T^1] [M^{1/2} L^1 T^{-1/2}]$Combining the powers of $M$,$L$,and $T$ on the right side:$M: 1/2 + 1/2 = 1$$L: P - 3/2 + 1 = P - 1/2$$T: -P + 1 - 1/2 = -P + 1/2$Comparing the powers of $L$ on both sides:$P - 1/2 = 0 \implies P = 1/2$.Thus,the value of $P$ is $1/2$.

If mass is written as $m=kc^{p} G^{-1 / 2} \,h^{1 / 2}$ then the value of $P$ will be : (Constants have their usual meaning with $k$ a dimensionless constant)

Similar Questions

If energy is given by $U = \frac{A\sqrt{x}}{x^2 + B}$,then the dimensions of $AB$ are:

The physical quantity which has the dimensional formula ${M^1}{T^{ - 3}}$ is

If the unit of force is $1 \,kN$,the unit of length is $1 \,km$,and the unit of time is $100 \,s$,what will be the unit of mass in $kg$?

If $\mu_0$ and $\varepsilon_0$ are the permeability and permittivity of free space,respectively,then the dimension of $\left(\frac{1}{\mu_0 \varepsilon_0}\right)$ is

If $x = at + bt^2$,where $x$ is the distance travelled by the body in kilometres while $t$ is the time in seconds,then the units of $b$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module