Question

A chain of length l is lying in a smooth horizontal tube such that a fraction of its length h hangs freely and the end touches the ground. At a certain moment the other end of chain is set free. The speed of this end of chain when it slips out of the tube is

• A. $\left[ ( 2 g h ) \frac { \mathrm { dl } } { \mathrm { dh } } \right] ^ { 1 / 2 }$
• B. $\sqrt { \mathrm { gh } }$
• C. $\sqrt { 2 \mathrm { gl } }$
• D. $\left( 2 \operatorname { gh } \log _ { \mathrm { e } } \frac { 1 } { \mathrm {~h} } \right) ^ { 1 / 2 }$

Answer 1

Answer: (D)

$\left( 2 \operatorname { gh } \log _ { \mathrm { e } } \frac { 1 } { \mathrm {~h} } \right) ^ { 1 / 2 }$

Answer 2

For hanging part,

mgh – T = mha …….. (i)

and for par in tube,

T = mxa ………….. (ii)

Adding the above equations, we get

mgh = m(h + x)a

or a = $\frac { g h } { h + x }$ or $\frac { d v } { d t } = \frac { g h } { h + x }$

or $\frac { d v } { d x } \frac { d x } { d t } = \frac { g h } { h + x }$ or $\frac { v d v } { d t } = \frac { g h } { h + x }$ or vdv =

$\frac { g h } { h + x }$ dx

∴ $\int _ { 0 } ^ { v } v d v = \int _ { 0 } ^ { 1 - h } \frac { g h } { h + x } d x$

i.e., v = $\left[ 2 \operatorname { gh } \log _ { \mathrm { e } } \frac { 1 } { \mathrm {~h} } \right] ^ { 1 / 2 }$