Q.

A solid sphere having volume V and density ρ floats at the interface of two immiscible liquids of densities ρ1 and ρ2 respectively. lf ρ1<ρ <ρ2, then the ratio of volume of the parts of the sphere in upper and lower liquid is

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

a

ρ-ρ1ρ2-ρ

b

ρ2-ρρ-ρ1

c

ρ+ρ1ρ+ρ2

d

ρ+ρ2ρ+ρ1

answer is B.

(Unlock A.I Detailed Solution for FREE)

Detailed Solution

According to law of flotation,Vρg = V 1ρ1g+V2ρ2g------(i)and V = V1 + V2----(ii)Hence from Eqs. (i) and (ii),V1ρg + V2ρg = V1ρ1g+V2ρ2gOr V1(ρ-ρ1)g = V2(ρ2-ρ) or V1V2 = ρ2-ρρ-ρ1
Watch 3-min video & get full concept clarity
Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Online Course for IIT JEE
Best Online Course for NEET
Best Online Course for CBSE
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Online Course for IIT JEE
Best Online Course for NEET
Best Online Course for CBSE
whats app icon