Question

time period of a liquid drop is given by T=Kd^m r^n sigma^l . where d is density , r is radius and sigma is surface tension of liquid then the value of m, n and l respectively are​

Answer 1

Answer:

Let Tρrσ  

Dimersim of T=[T]    ρ is density  

Dimersim of r=[L]    r is radius  

Dimersim of σ=[mT  

−2

]  σ is surface Tension  

Dimersim of ρ=[mL  

−3

]    Then T  

2

=?

T=r  

a

σ  

b

ρ  

c

 

[T]=[L]  

a

[mT  

−2

]  

b

[mL  

−3

]  

c

 

comparative on both sides

a−3c=0  12−2b

b+c=0

Solving =b=−1/2,c=1/2a=3/2

∴T=r  

3/2

σ  

−1/2

ρ  

1/2

 

T=  

σ

r  

3

ρ

​  

 

​  

 

T  

2

=  

σ

r  

3

ρ

​  

 

​  

     

AnswerisA

​