TSEAMCET
Hello,
Based on your education history, having studied from 2nd to 9th grade in Telangana (TS) and completing your 10th class and intermediate education in Andhra Pradesh (AP), you would be considered a local candidate for the TS EAMCET (Telangana State Engineering, Agriculture, and Medical Common Entrance Test). The local status is typically determined by your place of study for a significant period, and since you spent the majority of your schooling years (8 out of 12) in Telangana, you qualify as a local candidate for TS EAMCET.
hope this helps,
Thank you
Question : The curved surface area of the sphere is 154 cm2. Find the volume of the sphere (rounded off to one digit after decimal).
Option 1: 156.9 cm3
Option 2: 179.7 cm3
Option 3: 161.1 cm3
Option 4: 147.8 cm3
Correct Answer: 179.7 cm3
Solution : Given, the curved surface area of the sphere is 154 cm2 Let the radius of the sphere be $r$. $4πr^2 = 154$ ⇒ $4 × \frac{22}{7} × r^2 = 154$ ⇒ $r^2 = \frac{49}{4}$ ⇒ $r = \frac{7}{2}$ So, the volume of the sphere $= \frac{4}{3} × \frac{22}{7} × (\frac{7}{2})^3$ $=\frac{4}{3} × 22 × \frac{49}{8}$ $=11 × \frac{49}{3}$ $=\frac{539}{3}$ $=179.7$ cm$^3$ Hence, the correct answer is 179.7 cm$^3$.
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