A hollow disc of aluminum whose external and internal radii are R and r respectively, is floating on the surface of a liquid whose surface tension is T. The maximum weight of disc can be

Correct Option: 2π (R+r)T

Step-by-step Explanation:

When a hollow disc (an annular ring) of external radius R and internal radius r is placed on the surface of a liquid, it makes contact with the liquid along two boundaries:
1. The outer circular boundary of radius R.
2. The inner circular boundary of radius r.

Surface tension (T) is defined as the force per unit length acting perpendicular to a line drawn on the liquid surface. Therefore, the force due to surface tension acts along both the inner and outer circumferences of the hollow disc, pulling upwards to support its weight.

The circumference of the outer boundary is:
$L_{\mathrm{outer}}=2\pi R$

The circumference of the inner boundary is:
$L_{\mathrm{inner}}=2\pi r$

The total length of the boundary in contact with the liquid is:
$L=L_{\mathrm{outer}}+L_{\mathrm{inner}}=2\pi R+2\pi r=2\pi (R+r)$

The maximum upward force that surface tension can exert on the disc to keep it floating is given by the product of the surface tension and the total contact length:
$F_{\max }=T\times L$

Substituting the expression for the total length:
$F_{\max }=2\pi (R+r)T$

For the disc to float, the maximum weight of the disc ($W$) must be balanced by this maximum force of surface tension. Therefore, the maximum weight of the disc is:
$W=2\pi (R+r)T$