Three identical spheres of mass m, are placed at the vertices of an equilateral triangle of length a. When released, they interact only through gravitational force and collide after a time T = 4 seconds. If the sides of the triangle are increased to lengt

Question

Three identical spheres of mass m, are placed at the vertices of an equilateral triangle of length $a$ . When released, they interact only through gravitational force and collide after a time T = 4 seconds. If the sides of the triangle are increased to length $2 a$ and also the masses of the spheres are made 2m, then they will collide after ________ seconds. [2025]

Smart Achievers · Accepted Answer

(8)

$a$

$T \propto M^{x} {[M^{- 1} L^{3} T^{- 2}]}^{y} {[L]}^{z}$

$T \propto M^{x - y} L^{3 y + z} T^{- 2 y}$

$x - y = 0 \Rightarrow x = y$

$- 2 y = 1 \Rightarrow y = - \frac{1}{2}, x = - \frac{1}{2}$

$3 y + z = 0 \Rightarrow z = - 3 y = \frac{3}{2}$

Hence, $T \propto m^{- 1 / 2} G^{- 1 / 2} a^{3 / 2}$

$T \propto {(\frac{a^{3}}{m})}^{1 / 2} \Rightarrow T = 4 \times {(\frac{2^{3}}{2})}^{1 / 2} = 8 s$

Solution

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Solution

Ans. (8) T∝mxGyaz T∝Mx[M–1L3T–2]y[L]z T∝Mx–yL3y+zT–2y x–y=0 ⇒ x=y –2y=1 ⇒ y=–12, x=–12 3y+z=0 ⇒ z=–3y=32 Hence, T∝m–1/2G–1/2a3/2 T∝(a3m)1/2 ⇒ T=4×(232)1/2=8 s

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