Question

The volume of metallic hallow sphere is
constant. If the outer radius is increasing at
the rate of V cm/sec. Then the rate at which
the inner radius increasing when the radii are
a +d, a is
1a
V(a+d), V(a+d) 3) V(a+d) 4) a+d​

Answer 1

$\huge\mathfrak\red{Answer}$

THIS QUESTION ACTUALLY CHECK YOUR DIFFERENTIABLITY TALENT.

$volume \: of \: hollow \: sphere \\ \frac{4}{3} \pi ({m}^{3} - {n}^{3} ) \\ where \: m \: is \: outer \: radii \: and \: n \: is \\ \: inner \: radii \\ \\ v = \frac{4}{3} \pi( {m}^{3} - {n}^{3} ) \\ differentiate \: the \: equation \: with \\ respect \: to \: time \\ \\ \frac{d(v)}{dt} = \frac{4}{3} \pi( \frac{d( {m}^{3}) }{dt} - \frac{d( {n}^{3}) }{dt} ) \\ as \: we \: know \: \frac{dv}{dt} = 0 \: because \: volume \: \\ is \: constant \\ \\ 0 = \frac{4}{3} \pi(3 {m}^{2} \frac{dm}{dt} - 3 {n}^{2} \frac{dn}{dt} ) \\ \\ \frac{dm}{dt} = v \\ m = a + d \\ n = a \\ \\ 0 = 3{(a + d)}^{2} \times v - 3{(a)}^{2} \frac{dn}{dt} \\ 3 {a}^{2} \frac{dn}{dt} = 3 {(a + d)}^{2} </strong><strong>v</strong><strong> \\ \frac{dn}{dt} = \frac{ {(a + d)}^{2}</strong><strong>v</strong><strong> }{ {a}^{2} } \\ \frac{dn}{dt} = ({ 1 + \frac{d}{a} })^{2} </strong><strong>v</strong><strong>$