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 steps over the wheel. As it cools down to 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 :

The metal ring is free to expand and contract but its length is fixed by the wooden wheel once it has cooled down.
Initial circumference of the wheel is $$2\pi R$$; the ring is manufactured a little shorter (length $$L$$).
On heating, it expands by $$\Delta L = \alpha L\,\Delta T$$ and just slips over the wheel, so while cooling it contracts by exactly the same amount but is prevented from shortening. Hence the ring is left in uniform tensile strain

Strain in the ring, $$\varepsilon = \alpha\,\Delta T$$

Young’s modulus relates stress and strain: $$\sigma = Y\,\varepsilon = Y\,\alpha\,\Delta T$$

Tension developed in the ring (force along the strip) is therefore
$$T = \sigma S = S\,Y\,\alpha\,\Delta T$$

To find the force with which the two semicircular wooden parts press against each other, draw a free-body diagram of one semicircle. At every point of contact the ring pulls tangentially with magnitude $$T$$. Take radial components and integrate over the half-circumference:

The resultant of tangential tensions around a semicircle equals $$2T$$ and is directed along the diameter joining the two semicircles. (Mathematically, $$\displaystyle \vec F = \int_{-\pi/2}^{\pi/2} T\,\hat t\,d\theta = 2T\,\hat x$$.)

Hence the compressive force exerted by one half of the wheel on the other is
$$F = 2T = 2\,S\,Y\,\alpha\,\Delta T$$

Option D which is: $$2SY\alpha\Delta T$$