Same force acts on two bodies of different masses 8 kg and 10 kg initially at rest. The ratio of times required to acquire same final velocity is

Question

Same force acts on two bodies of different masses 8 kg and 10 kg initially at rest. The ratio of times required to acquire same final velocity is (A) 1: 1 (B) 8: 11 (C) 4: 5 (D) 2: 3

Tardigrade Admin · Accepted Answer

t=(v/a) ∴ t propto (1/a) [ F = ma, F = same, m propto (1/a)] t propto (1/a) propto m t propto m (t1/t2)=(m1/m2)=(8/10)=(4/5)

Q. Same force acts on two bodies of different masses 8 kg and 10 kg initially at rest. The ratio of times required to acquire same final velocity is

1 : 1

8 : 11

4 : 5

2 : 3

$t = \frac{v}{a}$
$∴ t \propto \frac{1}{a} [F = ma, F =$ same, $m \propto \frac{1}{a}]$
$t \propto \frac{1}{a} \propto m$
$t \propto m$
$\frac{t _{1}}{t _{2}} = \frac{m _{1}}{m _{2}} = \frac{8}{10} = \frac{4}{5}$

Q. Same force acts on two bodies of different masses 8 kg and 10 kg initially at rest. The ratio of times required to acquire same final velocity is

1 : 1

8 : 11

4 : 5

2 : 3

t=av​ ∴t∝a1​[F=ma,F= same, m∝a1​] t∝a1​∝m t∝m t2​t1​​=m2​m1​​=108​=54​

$t = \frac{v}{a}$
$∴ t \propto \frac{1}{a} [F = ma, F =$ same, $m \propto \frac{1}{a}]$
$t \propto \frac{1}{a} \propto m$
$t \propto m$
$\frac{t _{1}}{t _{2}} = \frac{m _{1}}{m _{2}} = \frac{8}{10} = \frac{4}{5}$