If the unit of length be doubled then the numerical value of the universal gravitation constant $G$ will become (with respect to present value)

Question

If the unit of length be doubled then the numerical value of the universal gravitation constant G will become -with respect to present value-8-timesdouble-1-8-timeshalf

Toppr Admin · Accepted Answer

-x2235-n1u1-n2u2-x2234-n2n1-u1u2-M-x2212-11L31T-x2212-21M-x2212-12L32T-x2212-22-M-x2212-1L3T-x2212-2M-x2212-1-2L-3T-x2212-2-18

If the unit of length be doubled then the numerical value of the universal gravitation constant $G$ will become (with respect to present value)
$8 t i m e s$
$d o u b l e$
$1 / 8 t i m e s$
$h a l f$

$∵ n_{1} u_{1} = n_{2} u_{2}$
$∴ \frac{n_{2}}{n_{1}} = \frac{u_{1}}{u_{2}} = \frac{M_{1}^{- 1} L_{1}^{3} T_{1}^{- 2}}{M_{2}^{- 1} L_{2}^{3} T_{2}^{- 2}} = \frac{M^{- 1} L^{3} T^{- 2}}{M^{- 1} (2 L)^{3} T^{- 2}} = \frac{1}{8}$

If the unit of length be doubled then the numerical value of the universal gravitation constant G will become (with respect to present value)8timesdouble1/8timeshalf

∵n1u1=n2u2∴n2n1=u1u2=M−11L31T−21M−12L32T−22=M−1L3T−2M−1(2L)3T−2=18

If the unit of length be doubled then the numerical value of the universal gravitation constant $G$ will become (with respect to present value)
$8 t i m e s$
$d o u b l e$
$1 / 8 t i m e s$
$h a l f$

$∵ n_{1} u_{1} = n_{2} u_{2}$
$∴ \frac{n_{2}}{n_{1}} = \frac{u_{1}}{u_{2}} = \frac{M_{1}^{- 1} L_{1}^{3} T_{1}^{- 2}}{M_{2}^{- 1} L_{2}^{3} T_{2}^{- 2}} = \frac{M^{- 1} L^{3} T^{- 2}}{M^{- 1} (2 L)^{3} T^{- 2}} = \frac{1}{8}$