Question

In the figure shown strings $A B$ and $B C$ have masses $m$ and $2\, m$ respectively. Both are of same length $l$. Mass of each string is uniformly distributed on its length. The string is suspended vertically from the ceiling of a room. A small jerk wave pulse is given at the end ' $C$. It goes up to upper end ' $A$ ' in time $' t'$. If $m=2\, kg\,, l=\frac{9610}{1681} m,\, g=10\, m / s ^{2},\, \sqrt{2}=1.4, \sqrt{3}=1.7$ then $' t$ ' is equal to :

• A. $\frac{620}{697} s $
• B. $\frac{434}{205} s$
• C. 2 s
• D. None of these

Answer 1

Answer: (C)

We use $v_{B C}=\sqrt{g x}$
$\int\limits_{0}^{l} \frac{d x}{\sqrt{x}} =\int\limits_{0}^{t_{B C}} \sqrt{g} d t$
$2 \sqrt{l} =g t_{B C}$
$t_{B C} =2 \sqrt{\frac{l}{g}}$
$v_{A B} =\sqrt{\frac{\left(2 m+\frac{m}{l} x\right) g}{\frac{m}{l}}}$
$=\sqrt{\left(2+\frac{x}{l}\right) g l}$
$\int\limits_{0}^{l} \frac{d x}{\sqrt{2+\frac{x}{l}}} =\int\limits_{0}^{t_{A B}} \sqrt{g l} d t$
$2 l \sqrt{2+\frac{x}{l}} =\sqrt{g l} t_{A B}$
$2l(\sqrt{3}-\sqrt{2}) =\sqrt{g l} t_{A B}$
$t_{A B} =2(\sqrt{3}-\sqrt{2}) \sqrt{\frac{l}{g}}$
$t =t_{B C} +t_{A B}=2 \sqrt{\frac{l}{g}}(1+\sqrt{3}-\sqrt{2})$
$=1.3 \times 2 \sqrt{\frac{961}{1181}}=2.6 \times \frac{31}{41}$
$=1.96\, s \simeq 2\, s$