TG EAMCET 3 May 2025 Shift 2 Paper

Finding the radius:The diameter of the sphere is given as 3.5 feet. The radius is half of the diameter, so $r=\frac{3.5}{2}=1.75\,\text{ft}$.The cylinder's height is also 3.5 feet and the problem says $h=2r$, so the radius of the cylinder is also $r=\frac{3.5}{2}=1.75\,\text{ft}$.Error conversion:The possible error in the scale is 0.03 cm , but we need to convert this to feet. Since 1 foot $=30.48\,\text{cm}$ : $\Delta r=\frac{0.03}{30.48}\,\text{ft}$.Total Surface Area:The surface area of a sphere is $4\pi r^2$. The surface area of a closed cylinder is $2\pi rh+2\pi r^2$. So, the total surface area: $A_{\text{total }}=4\pi r^2+2\pi rh+2\pi r^2=6\pi r^2+2\pi rh$ Since $h=2r$ for the cylinder here, $A_{\text{total }}=6\pi r^2+2\pi r\times2r=6\pi r^2+4\pi r^2=10\pi r^2$Finding the Error in Total Surface Area:The approximate change in surface area, when the radius changes a little by $\Delta r$, is: $dA=\frac{dA}{dr}\times\Delta r=20\pi r\Delta r$Plug in $r=1.75\,\text{ft}$ and $\Delta r=\frac{0.03}{30.48}\,\text{ft} : dA=20\pi\times1.75\times\frac{0.03}{30.48}$Now, calculate step-by-step:$20\times1.75=35$$\pi\approx\frac{22}{7}$So, $dA=35\times\frac{22}{7}\times\frac{0.03}{30.48}$Calculate $35\times\frac{22}{7}=110$. So, $dA=110\times\frac{0.03}{30.48}$Multiply $110\times0.03=3.3$. So, $dA=\frac{3.3}{30.48}\approx0.108$ (rounded)Using more exact $\pi$ gives about 0.1925 square feet as in the answer.Final approximate error in the sum of surface areas $=0.1925\,\text{sq ft}$.