A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. what is the area of the surface of water in $m^{2}$ through which light form the bulb can emerge out : $(μ_{w a t e r} = 1.33)$ (rounding off to nearest integer)

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

answer is 3.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Radius of circle if illumination

$\begin{array}{l} (r) = \frac{h}{\sqrt{μ^{2} - 1}} \\ r = \frac{80}{\sqrt{{(1.33)}^{2} - 1}} = 0.9 m \\ Area = {πr}^{2} = 3.14 \times {(0.9)}^{2} = 2.58 \end{array}$

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!