Question

1-A student claims that the time t of free fall of a body depends upon the height h from which its falls, the mass m of the body and the acceleration due to gravity g. Use the method of dimensions to check his claim.

2-Derive the relation E=mc² using the method of dimensions.

3-Derive the method of dimensions , an expression for the energy off a body executing SHM, assuming that this energy depends upon:
(i)the mass m, (ii)the frequency v, (iii)the amplitude of vibration r

4-Assuming that the mass m of the largest stone that can be moved by a flowing river depends upon v (the velocity), d (the density of water), and on g (the acceleration due to gravity), show that m varies sixth power of the velocity of flow.

Answer 1

1 . according to student time depends upon (1) height ( h) dimension of h =[L]
(2) mass of body ( m) dimension = [M]
(3) acceleration due to gravity (g) dimension of g = [LT-²]

use, dimension rule ,

[T ] = K[M ]^a [L]^b [LT-²]^c
[T] = K[M^aL^(b +c)T^-2c]
compare both sides,
a = 0
b + c = 0
-2c = 1 => c = -1/2 so, b = 1/2

hence, time = K√h/g
hence, student is wrong , time is i independent of mass .

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2. dimension of E = [ML² T-²]
dimension of C = [LT-¹]
dimension of m = [M ]

now ,use dimension rule ,
[ML²T-²] =K [M]^a [LT-¹]^b
[ML²T-²] =K[M^a L^b T^-b]
compare both sides,
a = 1 , b = 2
hence,
E = Kmc²
where K is proportionality constant after experiment , got K = 1 by Einstein
so, E = mc² ///

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3. Energy { [ML²T-²] } of SHM depends upon ,
(1) mass of body ( m) => [ M ]
(2) frequency (v) => [ T-¹]
(3) amplitude ( r) => [L ]

use dimension rule ,
[ML²T-² ] = K [M]^a [T-¹]^b [ L]^c
[ML²T-²] =K [M^a L^c T^-b ]
compare both sides,
a = 1 , b = 2 and c = 2

hence,
energy of SHM = Kmv²r²
where K is proportionality constant.
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4. A/c to question
mass depends upon
(1) velocity ( v) => [LT-¹ ]
(2) density (d ) => [ ML-³]
(3) accⁿ due to gravity (g) => [LT-²]

use dimension rule ,
[M ] = K[LT-¹]^a [ML-³]^b [LT-²]^c
[M ] = K[M^b L^(a -3b +c ) T^(-a-2c) ]
compare both sides,
b = 1
a -3b + c = 0 and
a -3 + c = 0
a + c = 3 -----(1)

(-a -2c) = 0
a = -2c -------(2) put in eqn (1)

-2c + c = 3
c = -3 put it in eqn (2)
a = 6
hence, a = 6, b = 1 and c = -3
so,

m =K v^6.d.g^-3
= K d(v²/g)³
= K(v^6/g³)

hence , proved that m is directly proportional to 6th power of velocity of flow .

Answer 2

Explanation:

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