Question 16: Properties of Matter

Question

A metal string $A$ is suspended from a rigid support and its free end is attached to a block of mass $M$. Second block having mass 2 M is suspended at the bottom of the first block using a string $B$. The area of cross sections of strings $A$ and $B$ are same. The ratio of lengths of strings of $A$ to B is 2 and the ratio of their Young's moduli $\left(Y_A / Y_B\right)$ is 0.5 . The ratio of elongations in $A$ to $B$ is $\_\_\_\_$ .

Accepted Answer

For a string, elongation is given by
$$ \Delta L = \frac{FL}{AY} $$
where
$F$ = tension in the string
$L$ = length of the string
$A$ = area of cross-section
$Y$ = Young's modulus
We need the ratio
$$ \frac{\Delta L_A}{\Delta L_B} $$
So,
$$ \frac{\Delta L_A}{\Delta L_B} = \frac{\frac{F_A L_A}{A Y_A}}{\frac{F_B L_B}{A Y_B}} = \frac{F_A}{F_B}\cdot \frac{L_A}{L_B}\cdot \frac{Y_B}{Y_A} $$
Since area is same for both strings, it cancels.
Now let us find each ratio.
1. Tension in string $A$
String $A$ supports both masses $M$ and $2M$, so
$$ F_A = (M+2M)g = 3Mg $$
2. Tension in string $B$
String $B$ supports only the lower mass $2M$, so
$$ F_B = 2Mg $$
Thus,
$$ \frac{F_A}{F_B} = \frac{3Mg}{2Mg} = \frac{3}{2} $$
3. Ratio of lengths
Given
$$ \frac{L_A}{L_B} = 2 $$
4. Ratio of Young's moduli
Given
$$ \frac{Y_A}{Y_B} = 0.5 = \frac{1}{2} $$
So,
$$ \frac{Y_B}{Y_A} = 2 $$
Now substitute:
$$ \frac{\Delta L_A}{\Delta L_B} = \frac{3}{2}	imes 2 	imes 2 = 6 $$
Hence, the ratio of elongations is
$$ \boxed{6} $$
So, the correct option is
D
.

Answer

1

Answer

4

Answer

8

Answer

6

Properties of Matter

Choose the correct answer: