Question 15: Properties of Matter

Question

Two wires as shown in the figure below, made of steel and have breaking stress of $12 	imes 10^8 \mathrm{~N} / \mathrm{m}^2$. Area of cross-section of upper wire is $0.008 \mathrm{~cm}^2$ and of lower wire is $0.004 \mathrm{~cm}^2$. The maximum mass that can be added to pan without breaking any wire is $\_\_\_\_$ kg.
$$ \left(	ext { take } g=10 \mathrm{~m} / \mathrm{s}^2ight) $$

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

56

Answer

38

Answer

96

Answer

5.6

Properties of Matter

Choose the correct answer: