A stream of water flowing with a constant speed v manages to move some pebbles. If the density of water is p and the acceleration due to gravity is g, then the mass of the heaviest pebble that it can move is (here, A is a constant)​

Assuming that the mass of the largest (M) stone that can be moved by flowing river depends on velocity (v), the density (ρ), and acceleration due to gravity (g) show that (M) varies directly as the

velocity (v), the density (ρ), and acceleration due to gravity (g) show that (M) varies directly as thesixth power of velocity flow (v).

• Let M (proportional to) Va db gc …………. (I)

Let M (proportional to) Va db gc …………. (I)M = kVadbgc, where, k is a proportionality constant

adbgc, where, k is a proportionality constant[M] = [V]a[d]b[g]c

Writing the dimensions of each physical quantity.

M1L0T0 = (L1T-1)a(M1L-3)b(L1T-2)c

M1L0T0 = Mb La - 3b + c T-a - 2c

M1L0T0 = Mb La - 3b + c T-a - 2cOn Comparing the powers on both sides of the above dimensional equation:

M1L0T0 = Mb La - 3b + c T-a - 2cOn Comparing the powers on both sides of the above dimensional equation:b = 1, a - 3b + c = 0 Hence, a - 3(1) + c = 0

M1L0T0 = Mb La - 3b + c T-a - 2cOn Comparing the powers on both sides of the above dimensional equation:b = 1, a - 3b + c = 0 Hence, a - 3(1) + c = 0a + c = 3 ------------(i)

M1L0T0 = Mb La - 3b + c T-a - 2cOn Comparing the powers on both sides of the above dimensional equation:b = 1, a - 3b + c = 0 Hence, a - 3(1) + c = 0a + c = 3 ------------(i)-a - 2c = 0 ----------(ii)

M1L0T0 = Mb La - 3b + c T-a - 2cOn Comparing the powers on both sides of the above dimensional equation:b = 1, a - 3b + c = 0 Hence, a - 3(1) + c = 0a + c = 3 ------------(i)-a - 2c = 0 ----------(ii)On solving (i) and (ii)

M1L0T0 = Mb La - 3b + c T-a - 2cOn Comparing the powers on both sides of the above dimensional equation:b = 1, a - 3b + c = 0 Hence, a - 3(1) + c = 0a + c = 3 ------------(i)-a - 2c = 0 ----------(ii)On solving (i) and (ii)a + c = 3 -------------(iii)

M1L0T0 = Mb La - 3b + c T-a - 2cOn Comparing the powers on both sides of the above dimensional equation:b = 1, a - 3b + c = 0 Hence, a - 3(1) + c = 0a + c = 3 ------------(i)-a - 2c = 0 ----------(ii)On solving (i) and (ii)a + c = 3 -------------(iii)-a - 2c = 0------------(iv)

M1L0T0 = Mb La - 3b + c T-a - 2cOn Comparing the powers on both sides of the above dimensional equation:b = 1, a - 3b + c = 0 Hence, a - 3(1) + c = 0a + c = 3 ------------(i)-a - 2c = 0 ----------(ii)On solving (i) and (ii)a + c = 3 -------------(iii)-a - 2c = 0------------(iv)On adding (iii) and (iv)

M1L0T0 = Mb La - 3b + c T-a - 2cOn Comparing the powers on both sides of the above dimensional equation:b = 1, a - 3b + c = 0 Hence, a - 3(1) + c = 0a + c = 3 ------------(i)-a - 2c = 0 ----------(ii)On solving (i) and (ii)a + c = 3 -------------(iii)-a - 2c = 0------------(iv)On adding (iii) and (iv)-c = 3, Hence, c = - 3.

M1L0T0 = Mb La - 3b + c T-a - 2cOn Comparing the powers on both sides of the above dimensional equation:b = 1, a - 3b + c = 0 Hence, a - 3(1) + c = 0a + c = 3 ------------(i)-a - 2c = 0 ----------(ii)On solving (i) and (ii)a + c = 3 -------------(iii)-a - 2c = 0------------(iv)On adding (iii) and (iv)-c = 3, Hence, c = - 3.Substituting c = -3 in equation (I),

M1L0T0 = Mb La - 3b + c T-a - 2cOn Comparing the powers on both sides of the above dimensional equation:b = 1, a - 3b + c = 0 Hence, a - 3(1) + c = 0a + c = 3 ------------(i)-a - 2c = 0 ----------(ii)On solving (i) and (ii)a + c = 3 -------------(iii)-a - 2c = 0------------(iv)On adding (iii) and (iv)-c = 3, Hence, c = - 3.Substituting c = -3 in equation (I),a = 6.

M1L0T0 = Mb La - 3b + c T-a - 2cOn Comparing the powers on both sides of the above dimensional equation:b = 1, a - 3b + c = 0 Hence, a - 3(1) + c = 0a + c = 3 ------------(i)-a - 2c = 0 ----------(ii)On solving (i) and (ii)a + c = 3 -------------(iii)-a - 2c = 0------------(iv)On adding (iii) and (iv)-c = 3, Hence, c = - 3.Substituting c = -3 in equation (I),a = 6.Thus, M = k V6dg

M1L0T0 = Mb La - 3b + c T-a - 2cOn Comparing the powers on both sides of the above dimensional equation:b = 1, a - 3b + c = 0 Hence, a - 3(1) + c = 0a + c = 3 ------------(i)-a - 2c = 0 ----------(ii)On solving (i) and (ii)a + c = 3 -------------(iii)-a - 2c = 0------------(iv)On adding (iii) and (iv)-c = 3, Hence, c = - 3.Substituting c = -3 in equation (I),a = 6.Thus, M = k V6dg∴ M is proportional to the 6th power of V if the mass of the largest (M) stone that can be moved by flowing river depends (i.e., directly proportional to) on velocity (v), the density (ρ), and acceleration due to gravity (g)

Answer:

From Given:-

since Mass ∝V

a

d

b

g

c

⇒M=kV

a

d

b

g

c

Equating dimensions.

[M

′

]=k[L

1

T

−1

]

a

[ML

−3

]

b

[L

1

T

−2

]

c

⇒ on comparing:-

b=1,a−3b+c=0

⇒a+c=3......(1)

and −a−2c=0......(2)

on solving:-

c=−3 and a=6

∴M=k

g

3

V

6

d

∴ Mass ∝ sixth Power of Velocity