Question

Wires A and B are connected with blocks P and Q, as shown. The ratio of lengths, radii and Young's modulus of wires A and B are $r$ , $2r$ and $3r$ respectively ( $r$ is a constant). Find the mass of block P if the ratio of increase in their corresponding lengths is $\dfrac{1}{{6{r^2}}}$ . The mass of block Q is $3M$.

A. $M$
B. $3M$
C. $6M$
D. $9M$

Answer 1

Hint: In the case, when a problem is based on the mechanical properties of a solid, we know that Young’s modulus plays a significant role in establishing a relationship between different parameters hence, identify the formula of Young’s modulus that should be used to calculate the mass of block P in order to provide an accurate solution.

Formula used:
The formula of Young’s modulus is,
$\text{Young's Modulus (Y)} = \dfrac{\text{Stress}(\sigma )}{\text{Strain}( \in )}$
The formula of stress is,
$\text{Stress}(\sigma ) = \dfrac{\text{Force}}{\text{Area}}$
The formula of strain is,
$\text{Strain}( \in ) = \dfrac{{\Delta l}}{l}$

Complete step by step solution:
The ratio of lengths of wire A and B is $\dfrac{{{l_A}}}{{{l_B}}} = r$ (given)
The ratio of radii of wire A and B is $\dfrac{{{r_A}}}{{{r_B}}} = 2r$ (given)
The ratio of Young’s modulus of wire A and B is $\dfrac{{{Y_A}}}{{{Y_B}}} = 3r$ (given)
The mass of block Q or the mass on wire B is ${m_B} = 3M$ (given)

Now, we know that $\text{Young's Modulus (Y)} = \dfrac{\text{Stress}(\sigma )}{\text{Strain}( \in )}$… (1)
But $\text{Stress}(\sigma ) = \dfrac{\text{Force}}{\text{Area}} = \dfrac{\text{Weight}}{\text{Area}} = \dfrac{{mg}}{{\pi {r^2}}}$ and $\text{Strain}( \in ) = \dfrac{{\Delta l}}{l}$
From eq. (1), we get
$\text{Strain}( \in ) = \dfrac{\text{Stress}(\sigma )}{\text{Young's Modulus(Y)}}$
$ \Rightarrow \dfrac{{\Delta l}}{l} = \dfrac{{mg}}{{\pi {r^2}}}.\dfrac{1}{Y}$
Or, it can also be rearranged as: -
$ \Rightarrow \Delta l = \dfrac{{mg}}{{\pi {r^2}}}.\dfrac{l}{Y}$

Now, if we take the ratio of increase in lengths of wire A and wire B, it will be as follows: -
$\dfrac{{\Delta {l_A}}}{{\Delta {l_B}}} = \dfrac{{\dfrac{{{m_A}g}}{{\pi {r_A}^2}}.\dfrac{{{l_A}}}{{{Y_A}}}}}{{\dfrac{{{m_B}g}}{{\pi {r_B}^2}}.\dfrac{{{l_B}}}{{{Y_B}}}}} \\ $
$ \Rightarrow \dfrac{{\Delta {l_A}}}{{\Delta {l_B}}} = \left( {\dfrac{{{m_A}}}{{{m_B}}}} \right).{\left( {\dfrac{{{r_B}}}{{{r_A}}}} \right)^2}.\left( {\dfrac{{{l_A}}}{{{l_B}}}} \right).\left( {\dfrac{{{Y_B}}}{{{Y_A}}}} \right) \\ $ … (2)
But, the ratio of increase in the lengths of wires A and B is $\dfrac{{\Delta {l_A}}}{{\Delta {l_B}}} = \dfrac{1}{{6{r^2}}}$ (given)
Substituting all the required values in eq. (2), we get
$ \Rightarrow \dfrac{1}{{6{r^2}}} = \left( {\dfrac{{{m_A}}}{{3M}}} \right).{\left( {\dfrac{1}{{2r}}} \right)^2}.\left( r \right).\left( {\dfrac{1}{{3r}}} \right) \\ $
On further calculation, we get
$ \Rightarrow {m_A} = \dfrac{{3M \times 4{r^2} \times 3r}}{{6{r^2} \times r}} \\ $
$ \therefore {m_A} = 6M \\ $
Thus, the mass of block P is $6M$.

Hence, the correct option is C.

Note: Since this is a problem related to stress-strain analysis in mechanics hence, given conditions are analyzed very carefully and quantities that are required to calculate the mass of block P, such as the ratio of length and radius; must be identified on a prior basis as it gives a better understanding of the problem. Units must be put after each end result.