$A$ wooden wheel of radius $R$ is made of two semicircular parts (see figure). The two parts are held together by a ring made of a metal strip of cross-sectional area $S$ and length $L$. $L$ is slightly less than $2\pi R$. To fit the ring on the wheel,it is heated so that its temperature rises by $\Delta T$ and it just slips over the wheel. As it cools down to the surrounding temperature,it presses the semicircular parts together. If the coefficient of linear expansion of the metal is $\alpha$,and its Young's modulus is $Y$,the force that one part of the wheel applies on the other part is:
- A$2SY\alpha\Delta T$
- B$2\pi SY\alpha\Delta T$
- C$SY\alpha \Delta T$
- D$\pi SY\alpha \Delta T$
Explore More
Similar Questions
$A$ brass wire of length $2 \ m$ and radius $1 \ mm$ at $27 ^\circ C$ is held taut between two rigid supports. Initially,it was cooled to a temperature of $-43 ^\circ C$,creating a tension $T$ in the wire. The temperature to which the wire has to be cooled in order to increase the tension in it to $1.4 \ T$ is . . . . . . $^\circ C$.
Two rods of different materials having coefficients of linear expansion $\alpha_1, \alpha_2$ and Young's moduli $Y_1$ and $Y_2$ respectively are fixed between two rigid massive walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of rods. If $\alpha_1 : \alpha_2 = 2 : 3$,and the thermal stresses developed in the two rods are equal,then the ratio $Y_1 : Y_2$ is equal to:
MediumIIT 1989
View SolutionAn object of mass $15 \,kg$ is attached to the end of a metal wire of unstretched length $1.0 \,m$. The object is then whirled in a vertical circle with an angular velocity of $4 \,rad/s$ at the bottom of the circle. If the cross-sectional area of the wire is $0.05 \,cm^2$ and Young's modulus of the metal is $2 \times 10^{11} \,N/m^2$, then calculate the elongation of the wire when the mass is at the lowest point of its path. (Take $g = 10 \,m/s^2$) (in $\,mm$)
The mean distance between the atoms of iron is $3 \times 10^{-10} \ m$ and the interatomic force constant for iron is $7 \ N/m$. The Young's modulus of elasticity for iron is:
Medium
View SolutionOne end of the steel rod is clamped to the roof and the other end is attached to a mass of $1000 \,kg$ as shown in the figure. The length of the rod is $50 \,cm$ and its cross-sectional area is $1000 \,mm^2$. The change in the length of the rod due to the weight of the mass is (Young's modulus of steel $= 2 \times 10^{11} \,Nm^{-2}$ and acceleration due to gravity $= 10 \,ms^{-2}$) (in $\,mm$)
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo