Question

The bar ABC fixed at both ends consists of a solid circular portion AB and hollow circular portion BC as shown in figure. The ratio $\left(\frac{a}{L}\right)$ for which reactive torque at A and C are of equal magnitude is

• A. $\frac{d_1}{d_2}$
• B. ${\left(\frac{d_1}{d_2}\right)}^4$
• C. ${\left(\frac{d_1}{d_2}\right)}^2$
• D. ${\left(\frac{d_1}{d_2}\right)}^3$

Answer 1

The correct option is B ${\left(\frac{d_1}{d_2}\right)}^4$



$T_{AB}=T_A$
$T_{BC}=T_C$

$T_A+T_C=T$

$T_A=T_C=\frac{T}{2}$ (given)

and ${\theta }_{A}B={\theta }_{B}C$

$T=\frac{G\theta J}{L}$

$T_{AB}=T_{CB}$

${\left(\frac{G\theta J}{L}\right)}_{AB}={\left(\frac{G\theta J}{L}\right)}_{BC}$

$\Rightarrow \frac{J_{AB}}{L_{AB}}=\frac{J_{BC}}{L_{BC}}$

$\Rightarrow \frac{\frac{\pi }{32}{d}_1^4}{a}=\frac{\frac{\pi }{32}(d_2^4-d_1^4)}{(L-a)}$

$\Rightarrow d_1^4(t-a)=a(d_2^4-d_1^4)$

$\Rightarrow d_1^{4}L=(d_2^2-d_1^2+d_1^2)a$

$\Rightarrow \frac{a}{L}=\frac{d_1^4}{d_2^4}={\left(\frac{d_1}{d_2}\right)}^4$