Question 19: Properties of Matter

Question

A string $A$ of length 0.314 m and Young's modulus $2 	imes 10^{10} \mathrm{~N} / \mathrm{m}^2$ is connected to another string $B$ of length and Young's modulus both twice of those of $A$. This series combination of strings is then suspended from a rigid support and its free end is fixed to a load of mass 0.8 kg . The net change in length of the combination is $\_\_\_\_$ mm.
(radius of both the strings is 0.2 mm and acceleration due to gravity $=10 \mathrm{~m} / \mathrm{s}^2$ ) (Mass of both strings is to be neglected as compared to the mass of load)

Accepted Answer

For a string under tension, extension is given by
$$ \Delta L=\frac{FL}{AY} $$
where
$F$ = tension,
$L$ = length,
$A$ = area of cross-section,
$Y$ = Young's modulus.
Since the two strings are connected in series and mass of strings is neglected, the same force acts in both strings.
The load is $0.8\ 	ext{kg}$, so tension is
$$ F=mg=0.8 	imes 10=8\ 	ext{N} $$
Radius of each string:
$$ r=0.2\ 	ext{mm}=0.2	imes 10^{-3}\ 	ext{m}=2	imes 10^{-4}\ 	ext{m} $$
So cross-sectional area is
$$ A=\pi r^2=\pi (2	imes 10^{-4})^2=4\pi 	imes 10^{-8}\ 	ext{m}^2 $$
Now for string $A$:
$$ L_A=0.314\ 	ext{m}, \qquad Y_A=2	imes 10^{10}\ 	ext{N/m}^2 $$
Its extension:
$$ \Delta L_A=\frac{8	imes 0.314}{(4\pi 	imes 10^{-8})(2	imes 10^{10})} $$
$$ \Delta L_A=\frac{2.512}{8\pi 	imes 10^2} =\frac{2.512}{800\pi} $$
Using $\pi \approx 3.14$,
$$ \Delta L_A=\frac{2.512}{2512}=10^{-3}\ 	ext{m}=1\ 	ext{mm} $$
Now for string $B$:
Its length and Young's modulus are both twice those of $A$.
So,
$$ L_B=2L_A,\qquad Y_B=2Y_A $$
Hence
$$ \Delta L_B=\frac{F L_B}{A Y_B} =\frac{F(2L_A)}{A(2Y_A)} =\frac{F L_A}{A Y_A} =\Delta L_A $$
Therefore,
$$ \Delta L_B=1\ 	ext{mm} $$
Net extension of the combination:
$$ \Delta L=\Delta L_A+\Delta L_B=1+1=2\ 	ext{mm} $$
Therefore, the correct answer is
$$ \boxed{2\ 	ext{mm}} $$
So, Option B is correct.

Answer

3

Answer

2

Answer

1.9

Answer

1

Properties of Matter

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