Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
57 views
in Physics by (67.7k points)
closed by
Test if the following equations are dimensionally correct:
a` h=(2Scostheta)/(rhorg)` b. `nu= sqrt(P/rho),
c. `V=(pi P r^4t)/(8etal),`
d. `v=(1)/(2pi) sqrt(mgl)/(I)`
where h height, S= surface tension, `rho`= density, P= pressure, V=volume, `eta` = coefficient of viscosity, v= frequency and I = moment of inertia.

1 Answer

0 votes
by (75.2k points)
selected by
 
Best answer
Correct Answer - A::B::C::D
a. `h=(2Scostheta)/(rhorg)
LHS=h=[L]`
Surface Tension
`S=F/L=(MT^-2)/L=[MT^-2]
Density
rho= M/V=[ML^-3T^0], Radius
r=[L], g=[LT^-2]`
Now
`RHS= (2Scostheta)/(rhorg)
= ([ML^-2])/([ML^-3T^0][L][LT^-2])
=[M^0L^1M^0]=[L]` Hence the relation is correct., b. Velocity, `v=sqrt((P/rho))
LHS= Demension of v, =[LT^-1], Dimension of P=F/A=[ML^-1T^-2]
Dimension of rho= m/V=[ML^3], RHS=sqrt(P/rho),][([ML^-1T^-2])/([ML^-3])]^(1/2), =[L^2T^-2]^(1/2), =[LT^-10=LHS` Hence relation is correct.
c. `V=((pi Pr^4 t))/((8etal))`, LHS= Dimension of `V=[L^3]`
Dimension of `P=[ML^-1T^-2],r^4=[L^4],t=[T]`, Coefficient of viscosity
`=[ML^-1T^-1]
Dimensions of RHS= ((pi Pr^4t))/(8 etal))
=([ML^-1T^-2][L^4][T])/([ML^-1T^-1][L])
= L^3=LHS`
Hence the relation is correct. d. `v= 1/(2pi) sqrt((mgl)/I),` RHS= dimension fo `v=[T^-1], RHS= sqrt((mgl)/I)
=sqrt(]M][LT^-2][L])/([ML^2]), [([ML^2T^-2])/([ML^2])]^(1/2)
=[T^-1]=LHS`
Hence the relation is correct.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...