QBManager

PHXI02:UNITS AND MEASUREMENTS

367263 In terms of resistance $R$ and time $T,$ the dimensions of ratio $\frac{μ}{ε}$ of the permeability $μ$ and permittivity $ε$ is:

1

[R^{2}]

2

[R T^{- 2}]

3

[R^{2} T^{2}]

4

[R^{2} T^{- 1}]

Explanation:

Dimensions $μ = [M L T^{- 2} A^{- 2}]$
Dimensions of $ε = [M^{- 1} L^{- 3} T^{4} A^{2}]$
Dimensions of $R = [M L^{2} T^{- 3} A^{- 2}]$
$\begin{aligned} ∴ \frac{Dimensions of μ}{Dimensions of ε} & = \frac{[M L T^{- 2} A^{- 2}]}{[M^{- 1} L^{- 3} T^{4} A^{2}]} \\ = [M^{2} L^{4} T^{- 6} A^{- 4}] = [R^{2}] \end{aligned}$

JEE - 2014

PHXI02:UNITS AND MEASUREMENTS

367264 If mass is written as $m = k c^{P} G^{- 1 / 2} h^{1 / 2}$ then the value of $P$ will be: (Constants have their usual meaning with $k$ , a dimensionless constant)

1

- \frac{1}{3}

2

\frac{1}{2}

3 2

4

\frac{1}{3}

Explanation:

Given that : $m = k c^{p} G^{- 1 / 2} h^{1 / 2}$
Using principle of homogeneity,
$\begin{aligned} [M] = {[L T^{- 1}]}^{P} {[M^{- 1} L^{3} T^{- 2}]}^{- 1 / 2} {[M L^{2} T^{- 1}]}^{1 / 2} \\ [M] = {[L T^{- 1}]}^{P} [M^{1 / 2} L^{- 3 / 2} T^{1}] [M^{1 / 2} L^{1} T^{- 1 / 2}] \end{aligned}$
$\begin{aligned} [M] = {[L T^{- 1}]}^{P} [M^{1} L^{- 1 / 2} T^{1 / 2}] \\ \Rightarrow & {[L T^{- 1}]}^{P} = {[L T^{- 1}]}^{1 / 2} \end{aligned}$
So, $P = 1 / 2$

JEE - 2024

PHXI02:UNITS AND MEASUREMENTS

367265 Assertion :
Power of a engine depends on mass, angular speed, torque and angular momentum, then the formula of power is not derived with the help of dimensional method.
Reason :
In mechanics, if a particular quantity depends on more than three quantities, then we can not derive the formula of the quantity by the help of dimensional method.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI02:UNITS AND MEASUREMENTS

367266 What is dimensions of energy in terms of linear momentum $[p]$ , area $[A]$ and time $[T]$ ?

1

[p^{1} A^{1} T^{1}]

2

[p^{2} A^{2} T^{- 1}]

3

[p^{1} A^{1 / 2} T^{- 1}]

4

[p^{1 / 2} A^{1 / 2} T^{- 1}]

Explanation:

Dimensions of energy in terms of linear
momentum $(p)$ area ( $A$ ) and time $(T)$ , is
related of $E = p^{a} A^{b} T^{c} (1)$
Writing dimensional formula of both side, we
get
$[M L^{2} T^{- 2}] = {[M L T^{- 1}]}^{a} {[L^{2}]}^{b} [T]^{c}$
$[M L^{2} T^{- 2]}] = [M^{a} L^{a + 2 b} T^{- a + c}]$
Comparing the exponents,
$a = 1, a + 2 b = 2$
$2 b = 1$
$b = \frac{1}{2}$
$- a + c = - 2 - 1 + c = - 2$
$\Rightarrow c = - 1$
$∴ F r o m E q . (1), w e h a v e$
$\Rightarrow E = [p A^{1 / 2} T^{- 1}]$

PHXI02:UNITS AND MEASUREMENTS

367267 If the energy $(E)$ , velocity $(v)$ , and force $(F)$ be taken as the fundamental quantity, then the dimensions of mass will be

1

F v^{- 2}

2

F v^{- 1}

3

E v^{- 2}

4

E v^{2}

Explanation:

Let $m \propto E^{a} v^{b} F^{c}$
Dimensionally, we write it as
$[M] = K {[M L^{2} T^{- 2}]}^{a} {[L T^{- 1}]}^{b} {[M L T^{- 2}]}^{c}$
For dimensional balance,
$a + c = 1; 2 a + b + c = 0; 2 a + b + 2 c = 0$
solving these equations, we get
$a = 1, b = - 2, c = 0$
$∴ m \propto E v^{- 2}$

PHXI02:UNITS AND MEASUREMENTS

367263 In terms of resistance $R$ and time $T,$ the dimensions of ratio $\frac{μ}{ε}$ of the permeability $μ$ and permittivity $ε$ is:

1

[R^{2}]

2

[R T^{- 2}]

3

[R^{2} T^{2}]

4

[R^{2} T^{- 1}]

Explanation:

Dimensions $μ = [M L T^{- 2} A^{- 2}]$
Dimensions of $ε = [M^{- 1} L^{- 3} T^{4} A^{2}]$
Dimensions of $R = [M L^{2} T^{- 3} A^{- 2}]$
$\begin{aligned} ∴ \frac{Dimensions of μ}{Dimensions of ε} & = \frac{[M L T^{- 2} A^{- 2}]}{[M^{- 1} L^{- 3} T^{4} A^{2}]} \\ = [M^{2} L^{4} T^{- 6} A^{- 4}] = [R^{2}] \end{aligned}$

JEE - 2014

PHXI02:UNITS AND MEASUREMENTS

367264 If mass is written as $m = k c^{P} G^{- 1 / 2} h^{1 / 2}$ then the value of $P$ will be: (Constants have their usual meaning with $k$ , a dimensionless constant)

1

- \frac{1}{3}

2

\frac{1}{2}

3 2

4

\frac{1}{3}

Explanation:

Given that : $m = k c^{p} G^{- 1 / 2} h^{1 / 2}$
Using principle of homogeneity,
$\begin{aligned} [M] = {[L T^{- 1}]}^{P} {[M^{- 1} L^{3} T^{- 2}]}^{- 1 / 2} {[M L^{2} T^{- 1}]}^{1 / 2} \\ [M] = {[L T^{- 1}]}^{P} [M^{1 / 2} L^{- 3 / 2} T^{1}] [M^{1 / 2} L^{1} T^{- 1 / 2}] \end{aligned}$
$\begin{aligned} [M] = {[L T^{- 1}]}^{P} [M^{1} L^{- 1 / 2} T^{1 / 2}] \\ \Rightarrow & {[L T^{- 1}]}^{P} = {[L T^{- 1}]}^{1 / 2} \end{aligned}$
So, $P = 1 / 2$

JEE - 2024

PHXI02:UNITS AND MEASUREMENTS

367265 Assertion :
Power of a engine depends on mass, angular speed, torque and angular momentum, then the formula of power is not derived with the help of dimensional method.
Reason :
In mechanics, if a particular quantity depends on more than three quantities, then we can not derive the formula of the quantity by the help of dimensional method.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI02:UNITS AND MEASUREMENTS

367266 What is dimensions of energy in terms of linear momentum $[p]$ , area $[A]$ and time $[T]$ ?

1

[p^{1} A^{1} T^{1}]

2

[p^{2} A^{2} T^{- 1}]

3

[p^{1} A^{1 / 2} T^{- 1}]

4

[p^{1 / 2} A^{1 / 2} T^{- 1}]

Explanation:

Dimensions of energy in terms of linear
momentum $(p)$ area ( $A$ ) and time $(T)$ , is
related of $E = p^{a} A^{b} T^{c} (1)$
Writing dimensional formula of both side, we
get
$[M L^{2} T^{- 2}] = {[M L T^{- 1}]}^{a} {[L^{2}]}^{b} [T]^{c}$
$[M L^{2} T^{- 2]}] = [M^{a} L^{a + 2 b} T^{- a + c}]$
Comparing the exponents,
$a = 1, a + 2 b = 2$
$2 b = 1$
$b = \frac{1}{2}$
$- a + c = - 2 - 1 + c = - 2$
$\Rightarrow c = - 1$
$∴ F r o m E q . (1), w e h a v e$
$\Rightarrow E = [p A^{1 / 2} T^{- 1}]$

PHXI02:UNITS AND MEASUREMENTS

367267 If the energy $(E)$ , velocity $(v)$ , and force $(F)$ be taken as the fundamental quantity, then the dimensions of mass will be

1

F v^{- 2}

2

F v^{- 1}

3

E v^{- 2}

4

E v^{2}

Explanation:

Let $m \propto E^{a} v^{b} F^{c}$
Dimensionally, we write it as
$[M] = K {[M L^{2} T^{- 2}]}^{a} {[L T^{- 1}]}^{b} {[M L T^{- 2}]}^{c}$
For dimensional balance,
$a + c = 1; 2 a + b + c = 0; 2 a + b + 2 c = 0$
solving these equations, we get
$a = 1, b = - 2, c = 0$
$∴ m \propto E v^{- 2}$

PHXI02:UNITS AND MEASUREMENTS

367263 In terms of resistance $R$ and time $T,$ the dimensions of ratio $\frac{μ}{ε}$ of the permeability $μ$ and permittivity $ε$ is:

1

[R^{2}]

2

[R T^{- 2}]

3

[R^{2} T^{2}]

4

[R^{2} T^{- 1}]

Explanation:

Dimensions $μ = [M L T^{- 2} A^{- 2}]$
Dimensions of $ε = [M^{- 1} L^{- 3} T^{4} A^{2}]$
Dimensions of $R = [M L^{2} T^{- 3} A^{- 2}]$
$\begin{aligned} ∴ \frac{Dimensions of μ}{Dimensions of ε} & = \frac{[M L T^{- 2} A^{- 2}]}{[M^{- 1} L^{- 3} T^{4} A^{2}]} \\ = [M^{2} L^{4} T^{- 6} A^{- 4}] = [R^{2}] \end{aligned}$

JEE - 2014

PHXI02:UNITS AND MEASUREMENTS

367264 If mass is written as $m = k c^{P} G^{- 1 / 2} h^{1 / 2}$ then the value of $P$ will be: (Constants have their usual meaning with $k$ , a dimensionless constant)

1

- \frac{1}{3}

2

\frac{1}{2}

3 2

4

\frac{1}{3}

Explanation:

Given that : $m = k c^{p} G^{- 1 / 2} h^{1 / 2}$
Using principle of homogeneity,
$\begin{aligned} [M] = {[L T^{- 1}]}^{P} {[M^{- 1} L^{3} T^{- 2}]}^{- 1 / 2} {[M L^{2} T^{- 1}]}^{1 / 2} \\ [M] = {[L T^{- 1}]}^{P} [M^{1 / 2} L^{- 3 / 2} T^{1}] [M^{1 / 2} L^{1} T^{- 1 / 2}] \end{aligned}$
$\begin{aligned} [M] = {[L T^{- 1}]}^{P} [M^{1} L^{- 1 / 2} T^{1 / 2}] \\ \Rightarrow & {[L T^{- 1}]}^{P} = {[L T^{- 1}]}^{1 / 2} \end{aligned}$
So, $P = 1 / 2$

JEE - 2024

PHXI02:UNITS AND MEASUREMENTS

367265 Assertion :
Power of a engine depends on mass, angular speed, torque and angular momentum, then the formula of power is not derived with the help of dimensional method.
Reason :
In mechanics, if a particular quantity depends on more than three quantities, then we can not derive the formula of the quantity by the help of dimensional method.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI02:UNITS AND MEASUREMENTS

367266 What is dimensions of energy in terms of linear momentum $[p]$ , area $[A]$ and time $[T]$ ?

1

[p^{1} A^{1} T^{1}]

2

[p^{2} A^{2} T^{- 1}]

3

[p^{1} A^{1 / 2} T^{- 1}]

4

[p^{1 / 2} A^{1 / 2} T^{- 1}]

Explanation:

Dimensions of energy in terms of linear
momentum $(p)$ area ( $A$ ) and time $(T)$ , is
related of $E = p^{a} A^{b} T^{c} (1)$
Writing dimensional formula of both side, we
get
$[M L^{2} T^{- 2}] = {[M L T^{- 1}]}^{a} {[L^{2}]}^{b} [T]^{c}$
$[M L^{2} T^{- 2]}] = [M^{a} L^{a + 2 b} T^{- a + c}]$
Comparing the exponents,
$a = 1, a + 2 b = 2$
$2 b = 1$
$b = \frac{1}{2}$
$- a + c = - 2 - 1 + c = - 2$
$\Rightarrow c = - 1$
$∴ F r o m E q . (1), w e h a v e$
$\Rightarrow E = [p A^{1 / 2} T^{- 1}]$

PHXI02:UNITS AND MEASUREMENTS

367267 If the energy $(E)$ , velocity $(v)$ , and force $(F)$ be taken as the fundamental quantity, then the dimensions of mass will be

1

F v^{- 2}

2

F v^{- 1}

3

E v^{- 2}

4

E v^{2}

Explanation:

Let $m \propto E^{a} v^{b} F^{c}$
Dimensionally, we write it as
$[M] = K {[M L^{2} T^{- 2}]}^{a} {[L T^{- 1}]}^{b} {[M L T^{- 2}]}^{c}$
For dimensional balance,
$a + c = 1; 2 a + b + c = 0; 2 a + b + 2 c = 0$
solving these equations, we get
$a = 1, b = - 2, c = 0$
$∴ m \propto E v^{- 2}$

PHXI02:UNITS AND MEASUREMENTS

367263 In terms of resistance $R$ and time $T,$ the dimensions of ratio $\frac{μ}{ε}$ of the permeability $μ$ and permittivity $ε$ is:

1

[R^{2}]

2

[R T^{- 2}]

3

[R^{2} T^{2}]

4

[R^{2} T^{- 1}]

Explanation:

Dimensions $μ = [M L T^{- 2} A^{- 2}]$
Dimensions of $ε = [M^{- 1} L^{- 3} T^{4} A^{2}]$
Dimensions of $R = [M L^{2} T^{- 3} A^{- 2}]$
$\begin{aligned} ∴ \frac{Dimensions of μ}{Dimensions of ε} & = \frac{[M L T^{- 2} A^{- 2}]}{[M^{- 1} L^{- 3} T^{4} A^{2}]} \\ = [M^{2} L^{4} T^{- 6} A^{- 4}] = [R^{2}] \end{aligned}$

JEE - 2014

PHXI02:UNITS AND MEASUREMENTS

367264 If mass is written as $m = k c^{P} G^{- 1 / 2} h^{1 / 2}$ then the value of $P$ will be: (Constants have their usual meaning with $k$ , a dimensionless constant)

1

- \frac{1}{3}

2

\frac{1}{2}

3 2

4

\frac{1}{3}

Explanation:

Given that : $m = k c^{p} G^{- 1 / 2} h^{1 / 2}$
Using principle of homogeneity,
$\begin{aligned} [M] = {[L T^{- 1}]}^{P} {[M^{- 1} L^{3} T^{- 2}]}^{- 1 / 2} {[M L^{2} T^{- 1}]}^{1 / 2} \\ [M] = {[L T^{- 1}]}^{P} [M^{1 / 2} L^{- 3 / 2} T^{1}] [M^{1 / 2} L^{1} T^{- 1 / 2}] \end{aligned}$
$\begin{aligned} [M] = {[L T^{- 1}]}^{P} [M^{1} L^{- 1 / 2} T^{1 / 2}] \\ \Rightarrow & {[L T^{- 1}]}^{P} = {[L T^{- 1}]}^{1 / 2} \end{aligned}$
So, $P = 1 / 2$

JEE - 2024

PHXI02:UNITS AND MEASUREMENTS

367265 Assertion :
Power of a engine depends on mass, angular speed, torque and angular momentum, then the formula of power is not derived with the help of dimensional method.
Reason :
In mechanics, if a particular quantity depends on more than three quantities, then we can not derive the formula of the quantity by the help of dimensional method.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI02:UNITS AND MEASUREMENTS

367266 What is dimensions of energy in terms of linear momentum $[p]$ , area $[A]$ and time $[T]$ ?

1

[p^{1} A^{1} T^{1}]

2

[p^{2} A^{2} T^{- 1}]

3

[p^{1} A^{1 / 2} T^{- 1}]

4

[p^{1 / 2} A^{1 / 2} T^{- 1}]

Explanation:

Dimensions of energy in terms of linear
momentum $(p)$ area ( $A$ ) and time $(T)$ , is
related of $E = p^{a} A^{b} T^{c} (1)$
Writing dimensional formula of both side, we
get
$[M L^{2} T^{- 2}] = {[M L T^{- 1}]}^{a} {[L^{2}]}^{b} [T]^{c}$
$[M L^{2} T^{- 2]}] = [M^{a} L^{a + 2 b} T^{- a + c}]$
Comparing the exponents,
$a = 1, a + 2 b = 2$
$2 b = 1$
$b = \frac{1}{2}$
$- a + c = - 2 - 1 + c = - 2$
$\Rightarrow c = - 1$
$∴ F r o m E q . (1), w e h a v e$
$\Rightarrow E = [p A^{1 / 2} T^{- 1}]$

PHXI02:UNITS AND MEASUREMENTS

367267 If the energy $(E)$ , velocity $(v)$ , and force $(F)$ be taken as the fundamental quantity, then the dimensions of mass will be

1

F v^{- 2}

2

F v^{- 1}

3

E v^{- 2}

4

E v^{2}

Explanation:

Let $m \propto E^{a} v^{b} F^{c}$
Dimensionally, we write it as
$[M] = K {[M L^{2} T^{- 2}]}^{a} {[L T^{- 1}]}^{b} {[M L T^{- 2}]}^{c}$
For dimensional balance,
$a + c = 1; 2 a + b + c = 0; 2 a + b + 2 c = 0$
solving these equations, we get
$a = 1, b = - 2, c = 0$
$∴ m \propto E v^{- 2}$

PHXI02:UNITS AND MEASUREMENTS

367263 In terms of resistance $R$ and time $T,$ the dimensions of ratio $\frac{μ}{ε}$ of the permeability $μ$ and permittivity $ε$ is:

1

[R^{2}]

2

[R T^{- 2}]

3

[R^{2} T^{2}]

4

[R^{2} T^{- 1}]

Explanation:

Dimensions $μ = [M L T^{- 2} A^{- 2}]$
Dimensions of $ε = [M^{- 1} L^{- 3} T^{4} A^{2}]$
Dimensions of $R = [M L^{2} T^{- 3} A^{- 2}]$
$\begin{aligned} ∴ \frac{Dimensions of μ}{Dimensions of ε} & = \frac{[M L T^{- 2} A^{- 2}]}{[M^{- 1} L^{- 3} T^{4} A^{2}]} \\ = [M^{2} L^{4} T^{- 6} A^{- 4}] = [R^{2}] \end{aligned}$

JEE - 2014

PHXI02:UNITS AND MEASUREMENTS

367264 If mass is written as $m = k c^{P} G^{- 1 / 2} h^{1 / 2}$ then the value of $P$ will be: (Constants have their usual meaning with $k$ , a dimensionless constant)

1

- \frac{1}{3}

2

\frac{1}{2}

3 2

4

\frac{1}{3}

Explanation:

Given that : $m = k c^{p} G^{- 1 / 2} h^{1 / 2}$
Using principle of homogeneity,
$\begin{aligned} [M] = {[L T^{- 1}]}^{P} {[M^{- 1} L^{3} T^{- 2}]}^{- 1 / 2} {[M L^{2} T^{- 1}]}^{1 / 2} \\ [M] = {[L T^{- 1}]}^{P} [M^{1 / 2} L^{- 3 / 2} T^{1}] [M^{1 / 2} L^{1} T^{- 1 / 2}] \end{aligned}$
$\begin{aligned} [M] = {[L T^{- 1}]}^{P} [M^{1} L^{- 1 / 2} T^{1 / 2}] \\ \Rightarrow & {[L T^{- 1}]}^{P} = {[L T^{- 1}]}^{1 / 2} \end{aligned}$
So, $P = 1 / 2$

JEE - 2024

PHXI02:UNITS AND MEASUREMENTS

367265 Assertion :
Power of a engine depends on mass, angular speed, torque and angular momentum, then the formula of power is not derived with the help of dimensional method.
Reason :
In mechanics, if a particular quantity depends on more than three quantities, then we can not derive the formula of the quantity by the help of dimensional method.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXI02:UNITS AND MEASUREMENTS

367266 What is dimensions of energy in terms of linear momentum $[p]$ , area $[A]$ and time $[T]$ ?

1

[p^{1} A^{1} T^{1}]

2

[p^{2} A^{2} T^{- 1}]

3

[p^{1} A^{1 / 2} T^{- 1}]

4

[p^{1 / 2} A^{1 / 2} T^{- 1}]

Explanation:

Dimensions of energy in terms of linear
momentum $(p)$ area ( $A$ ) and time $(T)$ , is
related of $E = p^{a} A^{b} T^{c} (1)$
Writing dimensional formula of both side, we
get
$[M L^{2} T^{- 2}] = {[M L T^{- 1}]}^{a} {[L^{2}]}^{b} [T]^{c}$
$[M L^{2} T^{- 2]}] = [M^{a} L^{a + 2 b} T^{- a + c}]$
Comparing the exponents,
$a = 1, a + 2 b = 2$
$2 b = 1$
$b = \frac{1}{2}$
$- a + c = - 2 - 1 + c = - 2$
$\Rightarrow c = - 1$
$∴ F r o m E q . (1), w e h a v e$
$\Rightarrow E = [p A^{1 / 2} T^{- 1}]$

PHXI02:UNITS AND MEASUREMENTS

367267 If the energy $(E)$ , velocity $(v)$ , and force $(F)$ be taken as the fundamental quantity, then the dimensions of mass will be

1

F v^{- 2}

2

F v^{- 1}

3

E v^{- 2}

4

E v^{2}

Explanation:

Let $m \propto E^{a} v^{b} F^{c}$
Dimensionally, we write it as
$[M] = K {[M L^{2} T^{- 2}]}^{a} {[L T^{- 1}]}^{b} {[M L T^{- 2}]}^{c}$
For dimensional balance,
$a + c = 1; 2 a + b + c = 0; 2 a + b + 2 c = 0$
solving these equations, we get
$a = 1, b = - 2, c = 0$
$∴ m \propto E v^{- 2}$