If the velocity of light C, the gravitational | vedclass

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

Question

(B) Let the density $\rho$ be expressed as $\rho = k C^a G^b h^c$.Dimensional formulas are:$[\rho] = M L^{-3}$$[C] = L T^{-1}$$[G] = M^{-1} L^3 T^{-2}

Vedclass · Accepted Answer

$C^5 G^{-2} h^{-1}$

Vedclass · Answer

(B) Let the density $\rho$ be expressed as $\rho = k C^a G^b h^c$.Dimensional formulas are:$[\rho] = M L^{-3}$$[C] = L T^{-1}$$[G] = M^{-1} L^3 T^{-2}$$[h] = M L^2 T^{-1}$Substituting these into the equation:$M^1 L^{-3} T^0 = (L T^{-1})^a (M^{-1} L^3 T^{-2})^b (M L^2 T^{-1})^c$$M^1 L^{-3} T^0 = M^{-b+c} L^{a+3b+2c} T^{-a-2b-c}$Comparing the exponents of $M, L,$ and $T$:$1$) $-b + c = 1 \implies c = 1 + b$$2$) $a + 3b + 2c = -3$$3$) $-a - 2b - c = 0 \implies a = -2b - c$Substitute $c = 1 + b$ into $(3)$: $a = -2b - (1 + b) = -3b - 1$Substitute $a$ and $c$ into $(2)$: $(-3b - 1) + 3b + 2(1 + b) = -3$$-3b - 1 + 3b + 2 + 2b = -3$$2b + 1 = -3 \implies 2b = -4 \implies b = -2$Now find $c$: $c = 1 + (-2) = -1$Now find $a$: $a = -3(-2) - 1 = 6 - 1 = 5$Thus, the dimension of density is $[\rho] = C^5 G^{-2} h^{-1}$.

If the velocity of light $C$, the gravitational constant $G$, and Planck's constant $h$ are chosen as the fundamental units, the dimension of density in the new system is

Similar Questions

If velocity of light $c$,Planck's constant $h$,and gravitational constant $G$ are taken as fundamental quantities,then express length in terms of dimensions of these quantities.

Consider the equation $H = \frac{x^p \epsilon^q E^r}{t^s}$, where $H = \text{magnetic field}$, $E = \text{electric field}$, $\epsilon = \text{permittivity}$, $x = \text{distance}$, and $t = \text{time}$. The values of $p, q, r$, and $s$ respectively are:

An experiment measures quantities $a, b,$ and $c$ and then $x$ is calculated from $x = \frac{a^{1/2} b^2}{c^3}.$ If the percentage errors in $a, b,$ and $c$ are $\pm 1\%, \pm 3\%,$ and $\pm 2\%$ respectively,then the percentage error in $x$ is:

In an experiment,four quantities $a, b, c$ and $d$ are measured with percentage errors of $1\%, 2\%, 3\%$ and $4\%$ respectively. Quantity $w$ is calculated as follows: $w = \frac{a^4 b^3}{c^2 \sqrt{d}}$. The percentage error in the measurement of $w$ is .......... $\%$.

If the acceleration due to gravity is $10\,m/s^2$ and the units of length and time are changed to kilometer and hour respectively,the numerical value of the acceleration is

Vedclass Test Series

Exam Paper Generator

Online Exam Module