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PHXI02:UNITS AND MEASUREMENTS

367272 The speed $v$ of ripples on the surface of water depends on surface tension $σ$ , density $ρ$ and wavelength $λ$ . The square of speed $v$ is proportional to

1

\frac{σ}{ρ λ}

2

\frac{ρ}{σ λ}

3

\frac{λ}{σ ρ}

4

ρ λ σ

Explanation:

$v \to L T^{- 1}$
$σ \to M T^{- 2}$
$ρ \to M L^{- 3}$
$λ \to L$
$ρ λ \to M L^{- 2}$
$\frac{σ}{ρ λ} \to L^{2} T^{- 2}$
$\Rightarrow [v^{2}] = \frac{σ}{ρ λ}$

AIIMS - 2007

PHXI02:UNITS AND MEASUREMENTS

367273 If force $(F)$ , velocity $(V)$ and time $(T)$ are considered as fundamental physical quantity, then dimensional formula of density will be

1

F^{2} V^{- 2} T^{6}

2

F V^{- 2} T^{2}

3

F V^{4} T^{- 6}

4

F V^{- 4} T^{- 2}

Explanation:

Dimension of the given physical quantities are:
Force $[F] = M L T^{- 2}$
Velocity, $[V] = L T^{- 1}$
Time $[T] = T$
Density $[D] = M L^{- 3}$
We can write
$D = [M L^{- 3}] = [F]^{x} [V]^{y} [T]^{x}$ or
$M^{1} L^{- 3} T^{0} = {[M L T^{- 2}]}^{x} {[L T^{- 1}]}^{y} [T]^{2}$
$M^{1} L^{- 3} T^{0} = [M]^{x} [L]^{x + y} [T]^{- 2 x - 2 y + z}$
Now compare both sides
$x = 1 (1)$
$x + y = - 3 (2)$
$- 2 x - 2 y + z = 0 (3)$
After solving the equations
$x = 1, y = - 4, z = - 2$
$D = F V^{- 4} T^{- 2}$

JEE - 2023

PHXI02:UNITS AND MEASUREMENTS

367274 Which of the following product $e, h, μ, G$ (where $μ$ is the permeability) be taken, so that the dimensions of the product are the same as that of speed of light?

1

h e^{- 2} μ^{- 1} G^{0}

2

h^{2} e G^{0} μ

3

h^{0} e^{2} G^{- 1} μ

4

h G e^{- 2} μ^{0}

Explanation:

Here, $v = e^{a} h^{b} μ^{c} G^{d}$ . Taking the dimensions $[M^{0} L T^{- 1} A^{0}]$ . There will be 4 simultaneous equations by equating the dimensions of $M, L, T$ and $A$ . These are $a - 2 c = 0$ ,
$\begin{aligned} a - b - 2 c - 2 d = - 1 \\ b + c - d = 0 and 2 b + c + 3 d = 1 \end{aligned}$
Solving for ' $a$ ', ' $b$ ', ' $c$ ' and ' $d$ ' we get
$a = - 2, b = 1, c = - 1, d = 0$
Thus $v = e^{- 2} h μ^{- 1} G^{0}$ .

PHXI02:UNITS AND MEASUREMENTS

367275 The velocity $v$ of water may depend on their wavelength $λ$ , density of water $ρ$ and the acceleration due to gravity $g$ . Which of the following relations is correct?

1

v^{2} \propto \frac{λ}{g ρ}

2

v^{2} \propto g λ

3

v^{2} \propto \frac{g ρ}{λ}

4

v^{2} \propto \frac{g}{λ}

Explanation:

$: v = k λ^{a} P^{b} g^{c}$
Here $k$ is a dimensionless quantity.
$[{L T}^{- 1}] = [L]^{a} {[{M L}^{- 3}]}^{b} {[{L T}^{- 2}]}^{c}$
On comparing the two sides we get
$1 = a - 3 b + c, - 1 = - 2 c, b = 0$
$\Rightarrow b = 0, c = \frac{1}{2}, a = \frac{1}{2}$
$v = k λ^{1 / 2} g^{1 / 2} \Rightarrow v^{2} \propto λ g$ .
So correct option is (2)

PHXI02:UNITS AND MEASUREMENTS

367272 The speed $v$ of ripples on the surface of water depends on surface tension $σ$ , density $ρ$ and wavelength $λ$ . The square of speed $v$ is proportional to

1

\frac{σ}{ρ λ}

2

\frac{ρ}{σ λ}

3

\frac{λ}{σ ρ}

4

ρ λ σ

Explanation:

$v \to L T^{- 1}$
$σ \to M T^{- 2}$
$ρ \to M L^{- 3}$
$λ \to L$
$ρ λ \to M L^{- 2}$
$\frac{σ}{ρ λ} \to L^{2} T^{- 2}$
$\Rightarrow [v^{2}] = \frac{σ}{ρ λ}$

AIIMS - 2007

PHXI02:UNITS AND MEASUREMENTS

367273 If force $(F)$ , velocity $(V)$ and time $(T)$ are considered as fundamental physical quantity, then dimensional formula of density will be

1

F^{2} V^{- 2} T^{6}

2

F V^{- 2} T^{2}

3

F V^{4} T^{- 6}

4

F V^{- 4} T^{- 2}

Explanation:

Dimension of the given physical quantities are:
Force $[F] = M L T^{- 2}$
Velocity, $[V] = L T^{- 1}$
Time $[T] = T$
Density $[D] = M L^{- 3}$
We can write
$D = [M L^{- 3}] = [F]^{x} [V]^{y} [T]^{x}$ or
$M^{1} L^{- 3} T^{0} = {[M L T^{- 2}]}^{x} {[L T^{- 1}]}^{y} [T]^{2}$
$M^{1} L^{- 3} T^{0} = [M]^{x} [L]^{x + y} [T]^{- 2 x - 2 y + z}$
Now compare both sides
$x = 1 (1)$
$x + y = - 3 (2)$
$- 2 x - 2 y + z = 0 (3)$
After solving the equations
$x = 1, y = - 4, z = - 2$
$D = F V^{- 4} T^{- 2}$

JEE - 2023

PHXI02:UNITS AND MEASUREMENTS

367274 Which of the following product $e, h, μ, G$ (where $μ$ is the permeability) be taken, so that the dimensions of the product are the same as that of speed of light?

1

h e^{- 2} μ^{- 1} G^{0}

2

h^{2} e G^{0} μ

3

h^{0} e^{2} G^{- 1} μ

4

h G e^{- 2} μ^{0}

Explanation:

Here, $v = e^{a} h^{b} μ^{c} G^{d}$ . Taking the dimensions $[M^{0} L T^{- 1} A^{0}]$ . There will be 4 simultaneous equations by equating the dimensions of $M, L, T$ and $A$ . These are $a - 2 c = 0$ ,
$\begin{aligned} a - b - 2 c - 2 d = - 1 \\ b + c - d = 0 and 2 b + c + 3 d = 1 \end{aligned}$
Solving for ' $a$ ', ' $b$ ', ' $c$ ' and ' $d$ ' we get
$a = - 2, b = 1, c = - 1, d = 0$
Thus $v = e^{- 2} h μ^{- 1} G^{0}$ .

PHXI02:UNITS AND MEASUREMENTS

367275 The velocity $v$ of water may depend on their wavelength $λ$ , density of water $ρ$ and the acceleration due to gravity $g$ . Which of the following relations is correct?

1

v^{2} \propto \frac{λ}{g ρ}

2

v^{2} \propto g λ

3

v^{2} \propto \frac{g ρ}{λ}

4

v^{2} \propto \frac{g}{λ}

Explanation:

$: v = k λ^{a} P^{b} g^{c}$
Here $k$ is a dimensionless quantity.
$[{L T}^{- 1}] = [L]^{a} {[{M L}^{- 3}]}^{b} {[{L T}^{- 2}]}^{c}$
On comparing the two sides we get
$1 = a - 3 b + c, - 1 = - 2 c, b = 0$
$\Rightarrow b = 0, c = \frac{1}{2}, a = \frac{1}{2}$
$v = k λ^{1 / 2} g^{1 / 2} \Rightarrow v^{2} \propto λ g$ .
So correct option is (2)

PHXI02:UNITS AND MEASUREMENTS

367272 The speed $v$ of ripples on the surface of water depends on surface tension $σ$ , density $ρ$ and wavelength $λ$ . The square of speed $v$ is proportional to

1

\frac{σ}{ρ λ}

2

\frac{ρ}{σ λ}

3

\frac{λ}{σ ρ}

4

ρ λ σ

Explanation:

$v \to L T^{- 1}$
$σ \to M T^{- 2}$
$ρ \to M L^{- 3}$
$λ \to L$
$ρ λ \to M L^{- 2}$
$\frac{σ}{ρ λ} \to L^{2} T^{- 2}$
$\Rightarrow [v^{2}] = \frac{σ}{ρ λ}$

AIIMS - 2007

PHXI02:UNITS AND MEASUREMENTS

367273 If force $(F)$ , velocity $(V)$ and time $(T)$ are considered as fundamental physical quantity, then dimensional formula of density will be

1

F^{2} V^{- 2} T^{6}

2

F V^{- 2} T^{2}

3

F V^{4} T^{- 6}

4

F V^{- 4} T^{- 2}

Explanation:

Dimension of the given physical quantities are:
Force $[F] = M L T^{- 2}$
Velocity, $[V] = L T^{- 1}$
Time $[T] = T$
Density $[D] = M L^{- 3}$
We can write
$D = [M L^{- 3}] = [F]^{x} [V]^{y} [T]^{x}$ or
$M^{1} L^{- 3} T^{0} = {[M L T^{- 2}]}^{x} {[L T^{- 1}]}^{y} [T]^{2}$
$M^{1} L^{- 3} T^{0} = [M]^{x} [L]^{x + y} [T]^{- 2 x - 2 y + z}$
Now compare both sides
$x = 1 (1)$
$x + y = - 3 (2)$
$- 2 x - 2 y + z = 0 (3)$
After solving the equations
$x = 1, y = - 4, z = - 2$
$D = F V^{- 4} T^{- 2}$

JEE - 2023

PHXI02:UNITS AND MEASUREMENTS

367274 Which of the following product $e, h, μ, G$ (where $μ$ is the permeability) be taken, so that the dimensions of the product are the same as that of speed of light?

1

h e^{- 2} μ^{- 1} G^{0}

2

h^{2} e G^{0} μ

3

h^{0} e^{2} G^{- 1} μ

4

h G e^{- 2} μ^{0}

Explanation:

Here, $v = e^{a} h^{b} μ^{c} G^{d}$ . Taking the dimensions $[M^{0} L T^{- 1} A^{0}]$ . There will be 4 simultaneous equations by equating the dimensions of $M, L, T$ and $A$ . These are $a - 2 c = 0$ ,
$\begin{aligned} a - b - 2 c - 2 d = - 1 \\ b + c - d = 0 and 2 b + c + 3 d = 1 \end{aligned}$
Solving for ' $a$ ', ' $b$ ', ' $c$ ' and ' $d$ ' we get
$a = - 2, b = 1, c = - 1, d = 0$
Thus $v = e^{- 2} h μ^{- 1} G^{0}$ .

PHXI02:UNITS AND MEASUREMENTS

367275 The velocity $v$ of water may depend on their wavelength $λ$ , density of water $ρ$ and the acceleration due to gravity $g$ . Which of the following relations is correct?

1

v^{2} \propto \frac{λ}{g ρ}

2

v^{2} \propto g λ

3

v^{2} \propto \frac{g ρ}{λ}

4

v^{2} \propto \frac{g}{λ}

Explanation:

$: v = k λ^{a} P^{b} g^{c}$
Here $k$ is a dimensionless quantity.
$[{L T}^{- 1}] = [L]^{a} {[{M L}^{- 3}]}^{b} {[{L T}^{- 2}]}^{c}$
On comparing the two sides we get
$1 = a - 3 b + c, - 1 = - 2 c, b = 0$
$\Rightarrow b = 0, c = \frac{1}{2}, a = \frac{1}{2}$
$v = k λ^{1 / 2} g^{1 / 2} \Rightarrow v^{2} \propto λ g$ .
So correct option is (2)

PHXI02:UNITS AND MEASUREMENTS

367272 The speed $v$ of ripples on the surface of water depends on surface tension $σ$ , density $ρ$ and wavelength $λ$ . The square of speed $v$ is proportional to

1

\frac{σ}{ρ λ}

2

\frac{ρ}{σ λ}

3

\frac{λ}{σ ρ}

4

ρ λ σ

Explanation:

$v \to L T^{- 1}$
$σ \to M T^{- 2}$
$ρ \to M L^{- 3}$
$λ \to L$
$ρ λ \to M L^{- 2}$
$\frac{σ}{ρ λ} \to L^{2} T^{- 2}$
$\Rightarrow [v^{2}] = \frac{σ}{ρ λ}$

AIIMS - 2007

PHXI02:UNITS AND MEASUREMENTS

367273 If force $(F)$ , velocity $(V)$ and time $(T)$ are considered as fundamental physical quantity, then dimensional formula of density will be

1

F^{2} V^{- 2} T^{6}

2

F V^{- 2} T^{2}

3

F V^{4} T^{- 6}

4

F V^{- 4} T^{- 2}

Explanation:

Dimension of the given physical quantities are:
Force $[F] = M L T^{- 2}$
Velocity, $[V] = L T^{- 1}$
Time $[T] = T$
Density $[D] = M L^{- 3}$
We can write
$D = [M L^{- 3}] = [F]^{x} [V]^{y} [T]^{x}$ or
$M^{1} L^{- 3} T^{0} = {[M L T^{- 2}]}^{x} {[L T^{- 1}]}^{y} [T]^{2}$
$M^{1} L^{- 3} T^{0} = [M]^{x} [L]^{x + y} [T]^{- 2 x - 2 y + z}$
Now compare both sides
$x = 1 (1)$
$x + y = - 3 (2)$
$- 2 x - 2 y + z = 0 (3)$
After solving the equations
$x = 1, y = - 4, z = - 2$
$D = F V^{- 4} T^{- 2}$

JEE - 2023

PHXI02:UNITS AND MEASUREMENTS

367274 Which of the following product $e, h, μ, G$ (where $μ$ is the permeability) be taken, so that the dimensions of the product are the same as that of speed of light?

1

h e^{- 2} μ^{- 1} G^{0}

2

h^{2} e G^{0} μ

3

h^{0} e^{2} G^{- 1} μ

4

h G e^{- 2} μ^{0}

Explanation:

Here, $v = e^{a} h^{b} μ^{c} G^{d}$ . Taking the dimensions $[M^{0} L T^{- 1} A^{0}]$ . There will be 4 simultaneous equations by equating the dimensions of $M, L, T$ and $A$ . These are $a - 2 c = 0$ ,
$\begin{aligned} a - b - 2 c - 2 d = - 1 \\ b + c - d = 0 and 2 b + c + 3 d = 1 \end{aligned}$
Solving for ' $a$ ', ' $b$ ', ' $c$ ' and ' $d$ ' we get
$a = - 2, b = 1, c = - 1, d = 0$
Thus $v = e^{- 2} h μ^{- 1} G^{0}$ .

PHXI02:UNITS AND MEASUREMENTS

367275 The velocity $v$ of water may depend on their wavelength $λ$ , density of water $ρ$ and the acceleration due to gravity $g$ . Which of the following relations is correct?

1

v^{2} \propto \frac{λ}{g ρ}

2

v^{2} \propto g λ

3

v^{2} \propto \frac{g ρ}{λ}

4

v^{2} \propto \frac{g}{λ}

Explanation:

$: v = k λ^{a} P^{b} g^{c}$
Here $k$ is a dimensionless quantity.
$[{L T}^{- 1}] = [L]^{a} {[{M L}^{- 3}]}^{b} {[{L T}^{- 2}]}^{c}$
On comparing the two sides we get
$1 = a - 3 b + c, - 1 = - 2 c, b = 0$
$\Rightarrow b = 0, c = \frac{1}{2}, a = \frac{1}{2}$
$v = k λ^{1 / 2} g^{1 / 2} \Rightarrow v^{2} \propto λ g$ .
So correct option is (2)

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