As shown in the figure,an equilateral triangl | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the length of each rod be $l$. In the right-angled triangle $ADC$,by Pythagoras theorem:$DC^2 = AC^2 - AD^2$Since $D$ is the midpoint of $AB$,

Vedclass · Accepted Answer

$\alpha_1 = 4\alpha_2$

Vedclass · Answer

(B) Let the length of each rod be $l$. In the right-angled triangle $ADC$,by Pythagoras theorem:$DC^2 = AC^2 - AD^2$Since $D$ is the midpoint of $AB$,$AD = \frac{l}{2}$.Thus,$DC^2 = l^2 - (\frac{l}{2})^2 = l^2 - \frac{l^2}{4} = \frac{3l^2}{4}$.When the temperature increases by $\Delta T$,the new lengths are $l' = l(1 + \alpha \Delta T)$.$AC' = l(1 + \alpha_2 \Delta T)$ and $AD' = \frac{l}{2}(1 + \alpha_1 \Delta T)$.The new length $DC'$ is given by:$DC'^2 = AC'^2 - AD'^2 = [l(1 + \alpha_2 \Delta T)]^2 - [\frac{l}{2}(1 + \alpha_1 \Delta T)]^2$$DC'^2 = l^2(1 + 2\alpha_2 \Delta T + \alpha_2^2 \Delta T^2) - \frac{l^2}{4}(1 + 2\alpha_1 \Delta T + \alpha_1^2 \Delta T^2)$Neglecting higher-order terms of $\Delta T$ (i.e.,$\alpha^2 \Delta T^2 \approx 0$):$DC'^2 \approx l^2(1 + 2\alpha_2 \Delta T) - \frac{l^2}{4}(1 + 2\alpha_1 \Delta T)$$DC'^2 \approx (l^2 - \frac{l^2}{4}) + (2l^2\alpha_2 \Delta T - \frac{l^2}{2}\alpha_1 \Delta T)$For $DC$ to remain constant,the change in $DC^2$ must be zero:$2l^2\alpha_2 \Delta T - \frac{l^2}{2}\alpha_1 \Delta T = 0$$2\alpha_2 = \frac{\alpha_1}{2} \Rightarrow \alpha_1 = 4\alpha_2$.

Similar Questions

The volume of a metal sphere increases by $0.33 \%$ when its temperature is raised by $50^{\circ} C$. The coefficient of linear expansion of the metal is

The density of a substance at $0^{\circ}C$ is $10 \, g/cm^3$ and at $100^{\circ}C$,its density is $9.7 \, g/cm^3$. The coefficient of linear expansion of the substance is:

The moment of inertia $I$ of a uniform rod about a perpendicular bisector increases to $I+\Delta I$, if the temperature is increased slightly by $\Delta T$. If the coefficient of linear expansion is $\alpha$, then $\frac{\Delta I}{I}$ is (Assume $\frac{\Delta T}{T} \ll 1$)

Three rods of length $\ell$ each are joined to form an equilateral triangle $PQR$. $O$ is the midpoint of $PQ$. For a small change in temperature,the distance $OR$ remains constant. If the coefficient of linear expansion for $PR$ and $RQ$ is $\alpha_2$ and for $PQ$ is $\alpha_1$,then:

$A$ solid metallic cube having a total surface area of $24 \; m^{2}$ is uniformly heated. If its temperature is increased by $10 \; ^{\circ}C$,calculate the increase in volume of the cube. (Given: $\alpha = 5.0 \times 10^{-4} \; ^{\circ}C^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module