Question

# If the ratio of lengths, radii and Youngs modulii of steel and brass wires in the figure are a, b and c respectively. Then the corresponding ratio of increase in their lengths would be:

A
2acb2
B
3a2b2c
C
3c2ab2
D
2a2cb

Solution

## The correct option is B $$\displaystyle \dfrac{3a}{2b^{2}c}$$Ref image Given ; $$l_S / l_B = a , \, \dfrac{y_S}{y_B} = c$$$$\dfrac{r_S}{r_B} = b$$By definition, young's Modulus = $$Y = \dfrac{linear \, stress}{linear \, strain}$$$$\Rightarrow Y = \dfrac{T/A}{\Delta l/l}$$ where, T - tension in wire            A - cross sectional area of wire           $$\Delta l$$ - change in length            l - original length$$\Rightarrow \boxed {\Delta l = \dfrac{T.l}{Y.A}}$$ __(i)FBD : Ref. image     $$T_3 = mg + T_B$$ __(ii)    Ref. image $$T_B = 2mg$$ __(iii)Put (iii) in (ii) : $$T_S = mg + 2mg = 3mg$$$$\Rightarrow \dfrac{T_S}{T_B} = \dfrac{3mg}{2mg} = \dfrac{3}{2}$$ __(iv)From (i) :$$\dfrac{\Delta l_S}{\Delta l_B} = \dfrac{\left(\dfrac{T_S l_S}{Y_S A_S} \right)}{\left(\dfrac{T_B l_B}{Y_B A_B} \right)} \begin{matrix} = \left(\dfrac{T_S}{T_B} \right) . \left(\dfrac{l_S}{l_B} \right) \dfrac{Y_B \, A_B}{Y_S \, A_S} & / for \, a \, wire: A = \pi r^2 \\ = \dfrac{3}{2} \times a \times \dfrac{1}{c} \times \dfrac{\pi r_B^2}{\pi r_S^2} &/ from \, given \, ratios \end{matrix}$$$$\boxed {\dfrac{\Delta l_S}{\Delta l_B} = \dfrac{3 a}{2 c.b^2}}$$No correct option givenPhysics

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